Solution

Solutions to Problemsin

Fundamentals ofStructural Mechanics

An instructor’s manual to accompany the bookFundamentals of Structural Mechanics

Keith D. HjelmstadUniversity of Illinois at Urbana-Champaign

Solutions to Problemsin

Fundamentals ofStructural Mechanics

An instructor’s manual to accompany the bookFundamentals of Structural Mechanics

by Keith D. Hjelmstad

Keith D. HjelmstadUniversity of Illinois at Urbana-Champaign

With assistance fromGhadir Haikal, Dan Turner, and

Cara (Liverant) Phillips

Solutions to Problems in Fundamentals ofStructural MechanicsAn instructor’s manual to accompany the bookFundamentals of Structural Mechanicsby Keith D. HjelmstadbyKeith D. HjelmstadUniversity of Illinois at Urbana-ChampaignWith assistance fromGhadir Haikal, Dan Turner, andCara (Liverant) Phillips

E 2007 by Keith D. Hjelmstad

All rights reserved. No part of this document may be translated or reproduced in any formwithout the written permission of the author.

Contents

Preface vii

1 Vectors and Tensors 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 The Geometry of Deformation 39. . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 The Transmission of Force 77. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 Linear Elastic Constitutive Theory 95. . . . . . . . . . . . . . . . . . . . . . . . .

5 Boundary Value Problems in Elasticity 129. . . . . . . . . . . . . . . . . . . . .

6 The Ritz Method of Approximation 141. . . . . . . . . . . . . . . . . . . . . . . .

7 The Linear Theory of Beams 173. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 The Linear Theory of Plates 239. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 Energy Principles and Static Stability 263. . . . . . . . . . . . . . . . . . . . . . .

10 Fundamental Concepts in Static Stability 285. . . . . . . . . . . . . . . . . . . .

11 The Planar Buckling of Beams 309. . . . . . . . . . . . . . . . . . . . . . . . . . . .

12 Numerical Computation for Nonlinear Problems 359. . . . . . . . . . . . . .

Appendix A. Fortran Codes 399. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Preface

Asignificant part of the book Fundamentals of StructuralMechanics lives in the problemsat the back of each chapter. No student can hope tomaster thematerial covered in the bookwithout solving at least some of these problems. An instructor who uses the book as a textfor a course on structural mechanics may want to use these problems as a source of exam-ples to illuminate points in the text or as a source of homework problems for the studentto work out. This manual should relieve some of the burden of preparation of coursemate-rials.

This manual provides detailed solutions to almost all of the 302 problems in Funda-mentals of Structural Mechanics 2e. The style inwhich the solutions are presented is simi-lar to the examples presented in the book, with narrative to help get from one step to thenext as well as some discussion of the results. The notation agrees with that used in thebook.

In solving the problems it became quite apparent that a program likeMATHEMATICAT

is indispensable in carrying out some of the more tedious computations. For those prob-lems that we used MATHEMATICAT we have provided a listing of the script. Most of theseare fairly short, but might be useful to someone who is not very familiar withMATHEMATI-CAT.

The problems in Chapter 12, in almost every case, involve modifying the programspresented in the book. When I revised the book for the second edition I converted all fothe programs from FORTRAN to MATLAB. However, I have not found time to changethe problem solutions. Therefore, those are still in FORTRAN. I have included the FOR-TRAN versions from the first edition of the book at the back so that you can follow thesolutions presented. You can get the FORTRAN source, for the programs in the text aswellas the programs in the solutions manual, from me. Make your request by e-mail to:[email protected]. I will send you the files.

Prefaceviii

We have tried very hard to get the answers right and to get them into this book withthe highest possible fidelity. Alas, I am sure some errors remain. Please forgive those er-rors. I would appreciate learning of the errors that you discover so that I can eliminate themin future editions.

Putting this manual together was hard work, the burden of which I gladly shared withGhadir Haikal, Cara (Liverant) Phillips, and Dan Turner. Ghadir managed to locate andorganize the solutions to quiz problems that I had collected over the years. She then com-pleted Chapters 1 through 5 while I did 6 through 12. The problems from the first editionwere available to us as were the problems that Cara had typset from old hand-written ex-amination solutions a few years ago. Dan proofread a large portion of the document. Iwould also like to (re)acknowledge the contributors to the solutions manual for the firstedition: Ertugrul Taciroglu, Jiwon Kim, Eric Williamson, and Quihai (Ken) Zuo. Theproblems they worked out are still correct! I am pleased to acknowledge the contributionsof Ghadir, Dan, Cara, Ertugrul, Jiwon, Eric and Ken to the preparation of this manual.

Keith D. Hjelmstad

Chapter 1Vectors and Tensors

1. Compute the values of the following expressions(a) δii(b) δijδij(c) Cijδikδjk(d) δabδbcδcd ... δxyδyz (enough terms to exhaust the whole alphabet)

(a) δii. Compute the value by summing on the repeated index i as

δii= δ11+δ22+δ33 = 3.

(b) δijδij. The two indices are repeated. Contract first on index j, then on index i.

δijδij= δii= 3.

(c) Cijδikδjk. The three indices are repeated. Contract first on index i, then on index j,and finally on index k.

Cijδikδjk= Ckj δjk= Ckk= C11+C22+C33.

(d) δabδbcδcd ... δxyδyz (enough terms to exhaust the whole alphabet). All of theindices are repeated except for a and z. Contract first on index b, then on index c, etc.until all indices have been summed

δabδbcδcd ... δxyδyz= δacδcd ... δxyδyz= δad ... δxyδyz= ... = δaz

Fundamentals of Structural Mechanics2

2. Let two vectors, u and v, have components relative to some basis as u= (5, −2, 1)and v= (1, 1, 1). Compute the lengths of the vectors and the angle between them. Findthe area of the parallelogram defined by u and v.

The length of a vector can be computed as

‖ u ‖= u ⋅ u = ui ui .

Thus, the lengths of the vectors can be compute as

‖ u ‖ = 25+4+1 = 30 ‖ v ‖ = 1+1+1 = 3

The dot product of the vectors is

u ⋅ v = 5−2+1= 4

cos θ(u, v) = u ⋅ v∕ ‖ u ‖ ‖ v ‖ = 4∕ 30 3 , θ= 1.136 rad

The area of the parallelogram can be computed either of two ways:

A = ‖ (u× v) ‖ = ‖ (−3,−4, 7) ‖ = 9+16+49 = 8.6023

A = ‖ (u× v) ‖ = ‖ u ‖ ‖ v ‖ sin θ = 30 3 sin(1.136 rad) = 8.6023

3. The vertices of a triangle are given by the position vectors a, b, and c. The componentsof these vectors in a particular basis are a= (0, 0, 0), b= (1, 4, 3), and c= (2, 3, 1). Usinga vector approach, compute the area of the triangle. Find the area of the triangle projectedonto the plane with normal n= (0, 0, 1). Find the unit normal vector to the triangle.

(a) Using a vector approach, compute the area of the triangle. The area of the originaltriangle can be computed as

A1 =12 ‖ (c× b) ‖ = 1

2 ‖ (5,−5, 5) ‖ =12 25+25+25 = 4.330

(b) Find the area of the triangle projected onto the plane with normal n= (0, 0, 1).The projected vectors can be found as follows

c′ = c− (c ⋅ n)n = (2, 3, 1)− 1(0, 0, 1)= (2, 3, 0)

b′ = b− (b ⋅ n)n = (1, 4, 3)− 3(0, 0, 1)= (1, 4, 0)

The area of the projected triangle can be computed as

A2 =12 ‖ (c′ × b′) ‖ = 1

2 ‖ (0, 0, 5) ‖ =12 0+0+25 = 2.5

c

ax1

x2

x3

b

d

3Chapter 1 Vectors and Tensors

x1

x2

x3

a

b

c

b′

c′

nA1

A2

(c) Find the unit normal vector to the triangle.

n = ac× ab‖ ac× ab ‖

where ab is the vector pointing from a to b. Noting that a is located at the origin of thecoordinate system, we find

ab = b = (1, 4, 3) ac = c = (2, 3, 1)

ac× ab= 2, 3, 1 × 1, 4, 3 = 5,−5, 5 ; ‖ ac× ab ‖= 5 3

⇒ n = 13(1,− 1, 1)

4. Let the coordinates of four points a, b, c and d be given bythe following position vectors a=(1, 1, 1), b=(2, 1, 1), c=(1, 2,2), and d=(1, 1, 3) in the coordinate system shown. Find vectorsnormal to planes abc and bcd. Find the angle between those vec-tors. Find the area of the triangle abc. Find the volume of the tet-rahedron abcd.

(a) Find vectors normal to planes abc and bcd.

nabc = ab× ac= b− a × c− a = (1, 0, 0)× (0, 1, 1)= (0,−1, 1)

nbcd = bc× bd= c− b × d− b = (−1, 1, 1)× (− 1, 0, 2)= (2, 1, 1)

(b) Find the angle between those vectors.

cos θ=nabc ⋅ nbcd‖ nabc ‖‖ nbcd ‖

nabc ⋅ nbcd = 0⇒ θ= π2


(c) Find the area of the triangle abc.

Aabc = ‖ nabc ‖ = 2

(d) Find the volume of the tetrahedron abcd.

Vabcd =16ab× ac ⋅ ad = 1

6(0,−1, 1) ⋅ (0, 0, 2)= 1

3

5. Demonstrate that (u× v) ⋅ w= uivjwkε ijk from basic operations on the base vectors.

Equality can be demonstrated by simply substituting the component forms of thevectors into the triple scalar product, collecting the scalars to the front and recognizingthe definition of the components of the permutation tensor. To wit,

(u× v) ⋅ w = (uiei × vjej) ⋅ wkek = uivjwk(ei× e j) ⋅ ek = uivjwkÁ ijk

6. Show that the triple scalar product is skew-symmetric with respect to changing the or-der in which the vectors appear in the product. For example, show that

(u× v) ⋅ w = − (v× u) ⋅ w

To generalize this notion, any cyclic permutation (e.g., u, v, w →w, u, v) of the order ofthe vectors leaves the algebraic sign of the product unchanged, while any acyclic permuta-tion (e.g., u, v, w → v, u, w) of the order of the vectors changes the sign. How does thisobservation relate to swapping rows of a matrix in the computation of the determinant ofthat matrix?

For the case given the demonstration is straightforward as, by definition of thecross product, the vector u× v=−(v× u), i.e., a vector of the same magnitude,pointing in the opposite direction. The negative sign persists in the triple product. Tosee the more general case, consider the component form of the triple scalar product

(u× v) ⋅ w = Á ijk uivjwk

To swap an entry, for example v and w we would have

(u×w) ⋅ v = Á ijk uiwjvk = Áikj uivjwk

Clearly, then, swapping entries in the triple scalar product is the same as swappingindices on the permutation tensor. A cyclic permutation of indices leaves the sign of


the product unchanged while an acyclic permutation changes the sign. Two successiveacyclic permutations change the sign and then change it back and hence must be equiv-alent to a single cyclic permutation. The determinant of a matrix and the triple scalarproduct are equivalent because

u× v ) ⋅ w = detuvw

where the components of u, v, and w comprise the rows of the matrix. The effects ofpermutation of rows of the matrix on the determinant are clearly the same as swappingelements of the triple scalar product.

7. Use the observation that ‖ u−v ‖2= u−v ⋅ u−v along with the distributive lawfor the dot product to show that

u ⋅ v ≡ 12 ‖ u ‖2 + ‖ v ‖2− ‖ v−u ‖2

The solution depends upon the computation of the length of u−v

‖ u− v ‖2= u−v ⋅ u−v

= u ⋅ u− 2u ⋅ v+ v ⋅ v

= ‖ u ‖2− 2u ⋅ v+ ‖ v ‖2

Rearrange to give desired result. Note that ‖ u−v ‖ = ‖ v−u ‖.

8. Prove the Schwarz inequality, |u ⋅ v |≤ ‖ u ‖ ‖ v ‖. Try to prove this inequality with-out using the formula u ⋅ v=‖ u ‖ ‖ v ‖ cos θ(u, v).

Consider any two vectors u≠ 0 and v≠ 0 (equality holds if either one of the vectorsis zero). The length of any vector is greater than or equal to zero.

0 ≤ ‖ αu+βv ‖2= αu+βv ⋅ αu+βv

= α2 ‖ u ‖2 + 2αβu ⋅ v+ β2 ‖ v ‖2

Let α=‖ v ‖2 and β=−u ⋅ v. Making this specialization gives

0 ≤ ‖ αu+βv ‖2= ‖ v ‖2 ‖ v ‖2 ‖ u ‖2 − 2 ‖ v ‖2 u ⋅ v 2+ u ⋅ v 2 ‖ v ‖2

= ‖ v ‖2 ‖ v ‖2 ‖ u ‖2 − u ⋅ v 2


Since ‖ v ‖2≥ 0 we can conclude that

0 ≤ ‖ v ‖2 ‖ u ‖2 − u ⋅ v 2 ⇒ ‖ v ‖2 ‖ u ‖2≥ u ⋅ v 2

9. Show that [u v] T= v u using the definition of the transpose of a tensor and bydemonstrating that the two tensors give the same result when acting on arbitrary vectorsa and b.

A tensor A has transpose AT defined by a ⋅ Ab= b ⋅ ATa for any vectors a and b.Let A≡ u v and proceed as follows:

a ⋅ u vb = a ⋅ b ⋅ v u

= a ⋅ u b ⋅ v

= b ⋅ v a ⋅ u

= b ⋅ a ⋅ u v

= b ⋅ v u a

By definition tensor product of vectors

Dot product is a scalar

Scalars are commutative

Dot product is a scalar

By definition tensor product of vectors

Therefore AT= v u and hence [u v]T = v u.

10. Show that the components of a tensor T and its transpose TT satisfy [TT] ij= [T] ji.

a ⋅ Tb = a ⋅ Tjk e j ek b Component form of T

= Tjka ⋅ ej ek b= Tjka ⋅ e j ek ⋅ b

= Tjk ek ⋅ b a ⋅ e j

= Tjk b ⋅ ek ej ⋅ a

Gather scalars to front

By definition of tensor product

Scalars are commutative

Dot product is commutative

= Tjkb ⋅ ek eja= b ⋅ Tjk ek eja= b ⋅ Tkj e j eka= b ⋅ TTa

Definition of tensor product

Scatter scalars

Rename dummy indices

Definition of tensor transpose


By identification with the penultimate line of the derivation we can observe that thecomponents satisfy [TT]kj = [T] jk.

11. Show that the tensor TTT is symmetric.

A tensor A is symmetric if u ⋅ Av= v ⋅ Au for any vectors u and v. Let A≡ TTTand proceed as follows:

u ⋅ TTTv = u ⋅ TT Tv

= Tu ⋅ Tv

By definition of composition of tensors

By definition of transpose of tensor

= TT Tu) ⋅ v

= TTTu ⋅ v

= v ⋅ TTTu

By definition of transpose of tensor

By definition composition of tensors

By commutativity of dot product of vectors

12. Consider any two tensors S and T. Prove the following:

(a) det TT = det T

(b) det ST = det S det T

(c) [ST] T= TT ST

(d) [ST]−1= T−1 S−1

(a) det TT = det T . The determinant of a tensor T can be computed from its com-ponents as

det T = 16Tim Tjn Tkp Áijk Ámnp

Noting that the components of TT are [TT ]ij= Tji

det TT = 16Tmi Tnj Tpk Á ijk Ámnp

= 16Tmi Tnj Tpk Ámnp Áijk

Changing the dummy index i to m, j to n, k to p, and vice--versa

⇒ det TT = 16Tim Tjn Tkp Áijk Ámnp= det T


(b) det ST = det S det T . Recall that

u× v ) ⋅ w = det

uvw

Therefore, we can compute the determinant of a tensor S as

det (S )= det

S1S2S2

= S1× S2 ) ⋅ S3

where the components of a vector Si are Si ]j= Sij. Thus, we can compute the deter-minant of S as

det S = S1 i S2 j S3 k Áijk= S1i S2j S3k Á ijk

det T = T1iT2jT3kÁijk= T1i T2j T3k Á ijk

det S det T = S1i S2j S3k Á ijk det T

Note that

Áijk det T = Áijk T1p T2m T3n Ápmn = Tip Tjm Tkn Ápmn

⇒ det S det T = S1i S2j S3k Tip Tjm Tkn Ápmn

= (S1i Tip ) (S2j Tjm )(S3k Tkn )Ápmn

= [ST]1p [ST]2m [ST ]3nÁpmn= det ST

(c) [ST] T= TT ST. From the definition of the product of two tensors

ST = Sik Tkj [e i ej ]

[ST] T = Sik Tkj [e j ei ]= Sjk Tki [ei e j ]

Similarly,

TTST = [TT ]ik [ST ]kj [ei e j ]

= Tki Sjk [ei e j ]

= Sjk Tki [ei e j ]= [ST] T

(d) [ST]−1= T−1 S−1. From the definition of the inverse of a tensor, for this rela-tionship to hold, the following has to be true


[ST]−1[ST]= [T−1 S−1 ][ST]= I

But,

[T−1 S−1 ][ST ]= T−1 [S−1 S ] T= T−1 [I ] T= T−1 T= I

⇒ [ST]−1 = T−1 S−1

13. Consider two Cartesian coordinate systems, one with basis e1, e2, e3 and the otherwith basis g1, g2, g3. Let Qij≡ gi ⋅ e j be the cosine of the angle between gi and ej.(a) Show that gi= Qije j and ej= Qijgi relate the two sets of base vectors.(b) We can define a rotation tensor Q such that ei= Qgi. Show that this tensor can be

expressed as Q ≡ Qij [gi gj], that is, Qij are the components of Q with respect tothe basis [gi gj]. Show that the tensor can also be expressed in the formQ = [e i gi].

(c) We can define a rotation tensor QT, such that gi= QTe i (the reverse rotation frompart (b). Show that this tensor can be expressed as QT≡ Qij [e j ei], that is, Qij arethe components of QTwith respect to the basis [ej e i]. Show that the tensor canalsobe expressed in the form QT= [gi e i].

(d) Show that QTQ = I, which implies that the tensor Q is orthogonal.

(a) Show that gi= Qije j and ej= Qijgi relate the two sets of base vectors.

Qik = gi ⋅ ek = Qijej ⋅ ek = Qijδjk = Qik

Qkj = gk ⋅ e j = gk ⋅ Qijgi = Qijδki = Qkj

(b) We can define a rotation tensor Q such that ei= Qgi. Show that this tensor can beexpressed as Q ≡ Qij [gi gj], that is, Qij are the components of Q with respect to thebasis [gi gj]. Show that the tensor can also be expressed in the form Q = [e i gi].

Qgk = Qijgi gjgk= Qij

gj ⋅ gkgi = Qijδjkgi = Qikgi = ek

Qgk = ei gigk= gi ⋅ gk ei = δike i = ek

(c) We can define a rotation tensor QT, such that gi= QTe i (the reverse rotation frompart (b). Show that this tensor can be expressed as QT≡ Qij [e j ei], that is, Qij arethe components of QT with respect to the basis [ej e i]. Show that the tensor can alsobe expressed in the form QT= [gi e i].

QTek = Qijej e iek= Qij

e i ⋅ ek e j = Qijδike j = Qkjej = gk

QTek = gi eiek= e i ⋅ ek gi = δikei = ek

(d) Show that QTQ = I, which implies that the tensor Q is orthogonal.


QTQ = gi e i [ej gj]= e i ⋅ ej [gi gj]= δij[gi gj]= [gi gi]= I

14. The components of tensors T and S and the components of vectors u and v are

u~112

v~111

S~0 --2 12 0 --1--1 1 0

T~1 2 02 0 10 1 2

Compute the components of the vector Su. Find the cosine of the angle between u and Su.Compute the determinants of T, S, and TS. Compute Tij Tij and uiTikSkj vj.

(a) Compute the components of the vector Su.

=112

Su~0 --2 12 0 --1--1 1 0

000

(b) Find the cosine of the angle between u and Su.

Since Su has zero length, the angle between u and Su cannot be determined.

(c) Compute the determinants of T, S, and TS.

det (T) = --9; det (S) = 0; det(TS) = det(T)det(S) = 0

(d) Compute Tij Tij and uiTikSkj vj. Note that Tij Tij= trT TT = trT2 since T issymmetric

1 2 02 0 10 1 2

T2 ~1 2 02 0 10 1 2

=5 2 22 5 22 2 5

⇒ TijTij= trT2 = 15

Similarly uiTikSkj vj= u ⋅ TSv

0 --2 12 0 --1--1 1 0

TSv~1 2 02 0 10 1 2

111=

1--21

uiTikSkj vj= u ⋅ TSv= 1

15. Verify that, for the particular case given here, the components of the tensor T and thecomponents of its inverse tensor T−1 are


2 --1 0--1 2 --10 --1 2

T ~ T−1 ~ 14

3 2 12 4 21 2 3

That this is actually the inverse can be verified by multiplying the components of thetensor and its inverse together to get the identity

14

3 2 1

2 4 2

1 2 3

2 --1 0

--1 2 --1

0 --1 2

1 0 0

0 1 0

0 0 1

=

16. Consider two bases: e1, e2, e3 and g1, g2, g3 . The basis g1, g2, g3 is given interms of the base vectors e1, e2, e3 as

g1 =1

3e1+e2+e3 , g2=

1

62e1−e2−e3 , g3 =

1

2e2−e3

The components of the tensor T and vector v, relative to the basis e1, e2, e3 are

T~0 --1 11 0 --1--1 1 0

v~123

Compute the components of the vector Tv in both bases. Compute the nine values ofTij TjkTkl (i.e., for i, l = 1, 2 ,3). Find the components of the tensor [T+TT]. Compute Tii.

(a) Compute the components of the vector Tv in both bases. In the basis e1, e2, e3

Tv= =1--21

123

0 --1 11 0 --1--1 1 0

Let bi= Tvi be the i--th component of Tv relative to the basis g1, g2, g3

bi = Tv ⋅ gi = (e1--2e2+e3) ⋅ gi

⇒

b1 =13-- 23+ 2

3= 0

b2 =26+ 2

6-- 16= 3

6

b3 = -- 22-- 12= -- 3

2


(b) Compute the nine values of Tij TjkTkl (i.e., for i, l = 1, 2 ,3).

--2 1 11 --2 11 1 --2

Tij Tjk =0 --1 11 0 --1--1 1 0

0 --1 11 0 --1--1 1 0

=

Tij Tjk Tkl=0 --1 11 0 --1--1 1 0

0 3 --3--3 0 33 --3 0

=--2 1 11 --2 11 1 --2

(c) Find the components of the tensor [T+TT].

T+ TT =0 --1 11 0 --1--1 1 0

0 0 00 0 00 0 0

=0 1 --1--1 0 11 --1 0

+

(d) Compute Tii.

Tii = T11+ T22+ T33 = 0

17. Consider two bases: e1, e2, e3 and g1, g2, g3 , where

g1 = e1+e2+e3, g2 = e2+e3, g3 = e2−e3Compute Qij for the given bases. Compute the value of QikQjk. Explain why the identityQikQjk= δij does not hold in this case.

Now consider a vector v= e1+2e2+3e3 and a tensor T given as

T= e2 e1−e1 e2 + e3 e1−e1 e3 + e3 e2−e2 e3

Compute the components of the vector Tv in both bases, i.e., find vi and v^i so that the fol-

lowing relationship holds Tv= viei= v^igi. Find the cosine of the angle between the vec-tor v and the vector Tv. Find the length of the vector Tv.

(a) Compute Qij for the given bases. Recall that Qij= gi ⋅ e j or the ij--th componentof Q is the j--th component of gi in the basis e1, e2, e3

⇒ Qij=1 1 10 1 10 1 --1

(b) Compute the value of QikQjk. Explain why the identity QikQjk= δij does not holdin this case.

3 2 02 2 00 0 2

Qij Qkj =1 1 10 1 10 1 --1

=1 0 01 1 11 1 --1


For the idendity QikQjk= δij to hold, the base vectors g1, g2, g3 have to be orthogo-nal and of unit length. It can be observed that g1 and g2 are not orthogonal, which leadsto the non--zero off diagonal terms in Q. The diagonal entries in Q are not equal to 1because the base vectors are not of unit length.

(c) Now consider a vector v= e1+2e2+3e3 and a tensor T given as

T= e2 e1−e1 e2 + e3 e1−e1 e3 + e3 e2−e2 e3

Compute the components of the vector Tv in both bases, i.e., find vi and vî so that

the following relationship holds Tv= viei= vîgi.

In the basis e1, e2, e3

Tv= =--5--23

123

0 --1 --11 0 --11 1 0

Let vî= Tvi be the i--th component of Tv relative to the basis g1, g2, g3

vî = Tv ⋅ gi = (--5e1--2e2+3e3) ⋅ gi

⇒v^1 = − 5− 2+ 3=− 4

v^2 =− 2+ 3= 1

v^3 =− 2− 3=− 5

(d) Find the cosine of the angle between the vector v and the vector Tv.

cos θ(Tv, v) = Tv ⋅ v‖ Tv ‖ ‖ v ‖

= − 5(1)− 2(2)+ 3(3)‖ Tv ‖ ‖ v ‖

= 0

(e) Find the length of the vector Tv.

‖ Tv ‖= Tv ⋅ Tv = 38

18. A general nth-order tensor invariant can be defined as follows

fn(T) ≡ Ti1 i2Ti2 i3⋅⋅⋅ Tin i1

where i1, i2, . . ., in are the n indices. For example, when n= 2 we can use i, j to givef2(T)= TijTji; when n = 3 we can use i, j, k to give f3(T)= TijTjkTki. Prove thatfn(T) is invariant with respect to coordinate transformation.

Let the sets of indices j1, j2, . . ., jn and k1, k2, . . ., kn be identical to the set i1, i2, . . ., in. Use the formula for change of basis to transform the components of


the tensor T from one coordinate system to another as T^ij= QimQjnTmn. Observe that

the Qij terms can be rearranged and the identity QikQij= δij can be used pairwise toeliminate the Qij. To wit,

fn(T) ≡ Ti1 i2Ti2 i3⋅⋅⋅ Tin i1

= Qi1 j1Qi2 k2

T^

j1 k2Qi2 j2

Qi3 k3T^

j2 k3⋅⋅⋅ Qin jn

Qi1 k1T^

jn k1

= Qi1 j1Qi1 k1

T^

j1 k2Qi2 k2

Qi2 j2T^

j2 k3⋅⋅⋅ Qin kn

Qin jnT^

jn k1

= δj1 k1T^

j1 k2δj2 k2T

^

j2 k3⋅⋅⋅ δjn knT

^

jn k1

= T^

k1 k2T^

k2 k3⋅⋅⋅ T^kn k1

Since the final expression does not depend upon the basis it is invariant.

19. Use the Cayley-Hamilton theorem to prove that for n≥ 4 all of the invariants fn(T),defined in Problem 18, can be computed from f1(T), f2(T), and f3(T).

First let n = 4, f4(T)= tr(T4). According to the Cayley--Hamilton theorem

T3− ITT2+ IITT− IIITT = 0

Operating on both sides with T

T4− ITT3+ IITT

2− IIITT = 0

⇒ T4 = ITT3− IITT

2+ IIITT

Taking the trace of both sides leads to

f4T)= trT4)= 2 IT IIIT− (IIT)2

Following a similar procedure we can find that

T5 = ITT4− IITT

3+ IIITT2

f5T)= II f4(T)= 2 I2T IIIT− II2T IT

By extension, it can be seen that for any n≥ 4 all of the invariants fn(T) can be com-puted from f1(T), f2(T), and f3(T)


20. From any tensor T one can compute an associated deviator tensor Tdevwhich has theproperty that the deviator tensor has no trace, i.e., tr Tdev

= 0. Such a tensor can be ob-tained from the original tensor T simply by subtracting α≡ 1

3tr T times the identity

from the original tensor, i.e., Tdev= T−αI. Show that tr Tdev = 0. Show that the prin-

cipal directions of TdevandT are identical, but that the principal values of Tdev are reducedby an amount α from those of the tensor T.

(a) Show that tr Tdev = 0.

trT′ = tr T−pI

= trT− 3p

= trT− trT = 0

(b) Show that the principal directions of Tdev and T are identical, but that the princi-pal values of Tdev are reduced by an amount α from those of the tensor T. Let m and nbe a principal value and a principal direction of the stress tensor S. By definition, thenSn= mn. Now S= S′+pI. Therefore, the eigenvalue problem can be recast as

S′+pI n = mn, ⇒ S′n = m−p n

By definition of the eigenvalue problem we can observe that n is, indeed, a principaldirection of the deviatoric stress tensor S′ corresponding to principal value m−p .

21. Consider a tensorT that has all repeated eigenvalues m1 = m2 = m3≡ m. Show thatthe tensor T must have the form T= mI.

Let n1, n2, and n3 be eigenvectors of T. Further assume that these vectors areorthogonal (remember, if they are not already orthogonal they can always be ortho-gonalized by the Gram-Schmidt process). By the spectral decomposition theorem

T = 3i=1

mi ni ni = m3i=1

ni ni = mI

Since the sum of outer products of any orthogonal set of vectors is equal to the identitytensor.

22. Prove that the product of a tensor with itself n times can be represented as

Tn = 3i=1

mi nni ni


Hint: Observe that [ni ni][nj nj]= δij [ni nj] (no summation implied).

Let us assume that we have proven the result for Tn−1. Compute Tn as the prod-uct of Tn−1 and T:

Tn = Tn−1T = 3i=1

mn−1i ni ni3

j=1

mj nj nj

= 3i=1

3j=1

mn−1i m j ni ni nj nj

= 3i=1

3j=1

mn−1i m j ni nj δij

= 3i=1

mni ni ni

Since the spectral decomposition clearly holds for n=1, the proof is complete by in-duction.

23. Show that the determinant of the tensor T can be expressed as follows

det T = 13tr T3 − 1

2IT tr T2 + 1

6 IT 3

where IT= tr T = Tii is the first invariant of T. Use the Cayley-Hamilton theorem.

The Cayley-Hamilton theorem says that

T3− ITT2+ IITT− IIITI = 0

where IT= trT, IIT, and IIIT= detT are invariants of T. Taking the trace of theabove expression gives

tr(T3)− IT tr(T2)+ IIT tr(T)− 3 detT = 0

Note that ( trT) 2−tr(T2)= 2IIT (one can show this identity with a component com-putation). Substituting for IIT in the above expression gives

tr(T3)− IT tr(T2)+ 1

2 I2T−tr(T2) IT− 3 detT = 0

Finally, rearranging the expression we get

detT = 13tr(T3)− 1

2IT tr(T

2)+ 16I3T


As an alternate approach we can observe that if the identity is proven in one coor-dinate system then it is proven in all coordinate systems. Let us select the principalcoordinates. Then tr(T3)= m31+m32+m33 and tr(T2)= m21+m22+m23. Now we can com-pute

= 13m31+m32+m33 − 1

2m1+m2+m3 m21+m22+m23 + 1

6m1+m2+m3 3

= m1m2m3 = detT

detT = 13tr(T3)− 1

2IT tr(T

2)+ 16I3T

24. Acertain state of deformation at a point in a body is described by the tensor T, havingthe components relative to a certain basis of

3 --1 0--1 5 10 1 2

T~

Find the eigenvalues and eigenvectors of T. Show that the invariants of the tensor T arethe same in the given basis and in the basis defined by the eigenvectors for the present case.

(a) Find the eigenvalues and eigenvectors of T.The coefficients of the characteristicequation are the invariants of the tensor: IT= 10, IIT= 29 and IIIT= 25 . Thus, thecharacteristic equation is

−m3+ 10m2− 29m+ 25 = 0

The three roots of the characteristic equation are m1 =1.6228, m2 =2.7261 and m3 =5.6511. The corresponding principal directions are

n1 = 0.2482, 0.3419,−0.9064

n2 = 0.9064, 0.2482, 0.3419

n3 = −0.3419, 0.9064, 0.2482

(b) Show that the invariants of the tensor T are the same in the given basis and in thebasis defined by the eigenvectors for the present case. The tensor T in the basis definedby eigenvectors is a diagonal matrix and its components are the eigenvalues them-selves.

1.6228 0

2.7261

0 5.6511

T~ 0

0

0

0

Invariants of this tensor can be obtained by Eqn. (63) in the text.


IT = m1+ m2+ m3 = 10

IIT = m1m2+ m1m3+ m2m3 = 29

IIIT = m1m2m3 = 25

They are the same as the invariants computed above.

25. Find the tensorT that has eigenvalues m1=1, m2=2, and m3=3with two of the asso-ciated eigenvectors given by

n1 =1

2e1+e2, n2 =

13−2e1+2e2+e3

Is the tensor unique (i.e., is there another one with these same eigenproperties)?

Since the eigenvalues are distinct we can find n3 by taking the cross product ofthe other two eigenvectors

n3 = n1× n2 =1

18e1−e2+4e3

Based on the spectral decomposition theorem in the text, T can be computed as

T = 3i=1

mi ni ni

In components we have

T ~ 12

1 1 01 1 00 0 0

+ 29

4 --4 --2--4 4 2--2 2 1

+ 318

1 --1 4--1 1 --44 --4 16

= 118

28 --10 4--10 28 --44 --4 52

The tensor T is unique by the spectral decomposition theorem.

26. Find the tensorT that has eigenvalues m1=1, m2=3, and m3=3,with twoof the asso-ciated eigenvectors given by

n1 =1

3e1+e2+e3, n2=

1

2−e2+e3

Are the eigenvectors unique?


The solution is similar to Problem 25.

n3 = n1× n2 =1

62e1−e2−e3

T = 3i=1

mi ni ni

T ~ 13

1 1 11 1 11 1 1

+ 32

0 0 00 1 --10 --1 1

+ 36

4 --2 --2--2 1 1--2 1 1

= 16

14 --4 --4--4 14 --4--4 --4 14

The eigenvectors n2 and n3 are not unique because the eigenvalues are repeated. Anylinear combination of n2 and n3 is an eigenvector. The tensor T, however, is unique.

27. Acertain state of deformation at a point in a body is described by the tensor T, havingthe components relative to a certain basis of

T~ 10−214 2 142 --1 --1614 --16 5

Let the principal values and principal directions be designated as m and n. Show that n1= (--1, 2, 2) is a principal direction and find m1. The second principal value is m2 =9× 10−2, find n2. Find m3 and n3 with as little computation as possible.

(a) Let the principal values and principal directions be designated as m and n. Showthat n1= (--1, 2, 2) is a principal direction and find m1. Assume that n1 is a principaldirection of T, with eigenvalue m1× 10−2. Then

=5−m1

14−m1 2 142 −1−m1 − 16

14 − 16

10−2

2

−12

0

00

From the first equation we have−14− m1 + 4+ 28 = 0, or m1 =−18. Toverify, substitute into the second and third equations

− 2+ 2 −1+ 18 − 32 = 0

−14− 2(16)+ 25+18 = 0


Therefore n1 is a principal direction.

(b) The second principal value is m2= 9× 10−2, find n2.

=(5−9)

(14−9) 2 142 (−1−9) − 16

14 − 16

10−2

b

1a

0

00

From the first and second rows

2a+ 14b = −510a+ 16b = 2

a = 1

b = −0.5⇒

The corresponding eigenvector is n2 =132, 2−1 .

(c) Find m3 and n3 with as little computation as possible. The eigenvalue m3 can befound from the first invariant of Lagrangian strain tensor and n3 can be found from thecross product of the two other principal directions.

IT = T11+ T22+ T33 = 18 = m1+ m2+ m3 = −18+ 9+ m3

The third eigenvalue is m3 = 27× 10−2

n3 = n1× n2 =13−2, 1,−2

28. The equation for balance of angular momentum can be expressed in terms of a tensorT and the base vectors eias ei× Tei = 0 (sum on repeated index implied). What specif-ic conditions must the components of the tensor T satisfy in order for this equation to besatisfied?

Carry out the computation for the tensor S as follows

ek× Sij [e i ej]ek = Sijek× (ej ⋅ ek)e i = Sijδjkek× ei

∴ ek× Sek = Sikek × e i= 0

If k = i then ek× e i= 0 ⇒ S11, S22, S33 can be anything. For the rest of the doublesummation we have

S12(e2× e1)+ S13(e3× e1)+ S21(e1× e2)+ S23(e3× e2)+

S31(e1× e3)+ S32(e2× e3) = 0

S12(−e3)+ S13(e2)+ S21(e3)+ S23(−e1)+ S31(−e2)+ S32(e1) = 0

[S32--S23]e1+ [S13--S31]e2+ [S21--S12]e3 = 0

n v


Since e1, e2, e3 ≠ 0, we must have

S32 = S23 S13= S31 S21= S12

29. The tensor R that operates on vectors and reflects them(as in a mirror) with unit normal n is given by

R≡ I−2 n n

Compute the vector that results from [RR]v. Compute thelength of the vector Rv in terms of the length of v. What is the inverse of the tensor R?Compute the eigenvalues and eigenvectors of R.

(a) Compute the vector that results from [RR]v.

RRv = I− 2n n I− 2n n v = I− 2n n v− 2(n ⋅ v)n

= v− 2(n ⋅ v)n− 2(n ⋅ v)n+ 4[n n]n(n ⋅ v)

= v− 4(n ⋅ v)n+ 4(n ⋅ n)n(n ⋅ v) = v

(b) Compute the length of the vector Rv in terms of the length of v.

Rv 2 = Rv ⋅ Rv = (v− 2(n ⋅ v) n) ⋅ (v− 2(n ⋅ v) n)

v ⋅ v− 4(n ⋅ v) 2+ 4(n ⋅ v) = v ⋅ v = v 2

∴ Rv = v

(c) What is the inverse of the tensor R?

RRv = I v = v ⇒ R --1 = R

(d) Compute the eigenvalues and eigenvectors of R.

R= I− 2 n n= I− n n − n n

From the spectral decomposition theorem, we have one distinct eigenvalue and a pairof repeated eigenvalues

m1 = 1; n1= n

m2 = m3 =− 1; n2, n3 are any two vectors in the plane ⊥ n


30. Let v(x) and u(x) be two vector fields, andT(x) be a tensor field. Compute the follow-ing expressions in terms of the components (vi, ui, and Tij) of these fields relative to thebasis e1, e2, e3 : div Tv , ∇u ⋅ Tv , ∇Tv , u Tv

(a) div Tv

Tv = Tij [ei e j ] vk ek= Tij vk [e i ej ] ek= Tij vk δjk ei= Tij vj e i

div(Tv) = ∂∂xi

(Tijvj ) =∂Tij∂xi

vj + Tij∂vj∂xi

(b) ∇u ⋅ Tv

∇(u ⋅ Tv) = ∂∂xk

(ui Tij vj ) ek = ∂ui∂xk Tij vj+ ui∂Tij∂xk

vj+ ui Tij∂vj∂xk ek

(c) ∇Tv

∇(Tv) = ∂∂xkTij vje i ek = ∂Tij∂xk vj+ Tij

∂vj∂xk e i ek

(d) u Tv.

u Tv = ukek Tij vj e i = uk Tij vj ek e i

31. Evaluate the following expressions:

(a) div div x x (e) ∇xdivx (f) ∇x ⋅ ∇x ⋅ x (d) div x div x x

(c) ∇‖ ∇ ‖ x ‖ 2 ‖2 (b) div x div xdivx

where x= x1e1+x2e2+x3e3 is the position vector in space and all derivatives are withrespect to the coordinates xi.

(a) div div x x

div(x x) = (xi xj),j ei = [xi,j xj+ xi xj,j ] ei = [δij xj+ 3xj ] e i = 4x

div div[x x] = div(4x) = 4(xiei),j ⋅ ej = 4 xi,j ei ⋅ e j = 4 δij δij = 12

(b) div x div xdivx

div x = (xi ei ),j ⋅ ej = 3


div(x div x ) = div(3x ) = 3div x = 9

div x div xdivx = div(9x ) = 9 div x = 27

(c) ∇‖ ∇ ‖ x ‖ 2 ‖2

x 2 = xi xi

∇ x 2 = (xi xi),j e j = [xi,j xi+ xi xi,j] e j = 2 δij xi ej = 2x

2x 2 = 4 x 2

∇ 2x 2 = 4 ∇ x 2 = 8x

(d) div x div x x

[x x]ij = xi xj

div[x x] = (xi xj ),j ei = (xi,j xj+ xi xj,j ) ei = (δij xj+ xi δjj) ei = 4xi ei = 4x

div x div x x = div[x 4x] = 4div[x x]= 16x

(e) ∇xdivx

div x = xi,i = δii = 3

∇(x div x ) = ∇(3x) = 3∇x = 3I

(f) ∇x ⋅ ∇x ⋅ x

∇(x ⋅ x) = (xi xi),j ej = (xi,j xi+ xi xi,j ) ej = 2xje j = 2x

∇(x ⋅ 2x ) = 2∇(x ⋅ x ) = 4x

32. Let v(x)= x2−x3 e1+ x3−x1 e2+ x1−x2 e3. Evaluate the following expres-sions: ∇v, ∇x ⋅ v , div x v , and ∇x× v , where x= xiei is the position vector.Evaluate the expressions at the point x= e1+2e2+e3.

(a) ∇v

∇v=0 1 --1--1 0 11 --1 0

x1

x2

x3

n

B


(b) ∇x ⋅ v

x ⋅ v = (x1x2−x1x3 )+ (x2x3−x2x1 )+ (x3x1−x3x2 ) = 0

∴∇(x ⋅ v) = 0

(c) div x v

x3x1--x2x3x2--x3

x1x3--x1x v~

x1x2--x3 x1x1--x2x2x3--x1x2x2--x3 x2x1--x2x3x3--x1

x1--x2div x v ~

x2--x3x3--x1

(d) ∇x× v

x1x3--x1 − x2x2--x3x× v~

x2x1--x2 − x3x3--x1x3x2--x3 − x1x1--x2

x1+x2--2 x1

--2 x2∇ x× v ~

x2+x3 --2 x3x1+x3--2 x1 --2 x3--2 x2

(e) Evaluate the expressions at the point x= e1+2e2+e3.

3--2

--4

∇ x× v ~3 --2

2--2 --2

--4− 1div x v ~

1

0

33. Let v(x) be given by the following explicit function

v(x) = x21+x2x3 e1+ x22+x1x3 e2+ x23+x1x2 e3where x is the position vector of any point and has componentsx1, x2, x3 relative to the Cartesian coordinate system shown.The vector field is defined on the spherical region B of unit ra-dius as shown in the sketch. Give an explicit expression for theunit normal vector field n(x) to the surface of the sphere. Compute the gradient of the vec-tor field v(x). Compute the product [∇v]n, i.e., the gradient of the vector field acting onthe normal vector. Compute the divergence of the vector field v(x). Compute the integralof div v over the volume of the sphere. Compute the integral of v ⋅ n over the surface ofthe sphere.


(a) Give an explicit expression for the unit normal vector field n(x) to the surface ofthe sphere.

n= x= xie i

(b) Compute the gradient of the vector field v(x).

∇v=∂vi∂xj

[e i ej ]~x1

2x1

x2

x2x3x3 2x2 x1

2x3

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on thenormal vector.

[∇v] n~x1

2x1

x2

x2x3x3 2x2 x1

2x3

x1x2x3

=2 x 21 + 2 x2x32 x 22 + 2 x1x32 x 23 + 2 x1x2

(d) Compute the divergence of the vector field v(x).

div v= tr (∇v)= 2(x1+x2+x3)

(e) Compute the integral of div v over the volume of the sphere. Given the geometryof the problem, it would be best to compute the integral in spherical coordinates usingthe following transformations (see figure below)

x1 = r cos φ cos θ x2 = r cos φ sin θ x3 = r sin φ

dV= (r dφ ) (r cos φ dθ )dr= r2 cos φ dφ dθ dr

r* dθ

dφ

θr* = r cos φ

dθ

d r r dφ

r

zφ

x1

x2 x3

B

div v dV= 2 π∕2

−π∕2

2π

0

1

0

r cos φ cos θ+ r cosφ sin θ+ r sinφ r2 cos φ dr dθ dφ= 0

(f) Compute the integral of v ⋅ n over the surface of the sphere.

v ⋅ n = x21+x2 x3 x1+ x22+x1x3 x2+ x23+x1x2 x3 = x31+ x32+ x33+ 3 x1x2x3

x1

x2

x3

x1

x2

h

RB


At the surface we have

x1 = cos φ cos θ x2 = cos φ sin θ x3 = sin φ

dA= (1 dφ ) (1 cos φ dθ )= cos φ dφ dθ

Ω

v ⋅ n dA= π∕2

−π∕2

2π

0

[cos3φ cos3 θ+ cos3φ sin3 θ+ sin3φ

+ 3 cos2 φ sin φ cos θ sin θ ] cos φ dθ dφ= 0

34. Let v(x) be a vector field given by the follow-ing explicit function

v(x)= x1e1+x2e2 lnx21+x22

where ln(⋅) indicates the natural logarithm of (⋅).The vector field is defined on the cylindrical re-gion B of height h and radius R as shown in the sketch. Give an expression for the unitnormal vector field n(x) to the for the cylinder (including the ends). Compute the diver-gence of the vector field v(x) and the integral of div v over the volume of the cylinder.

(a) Give an expression for the unit normal vector field n(x) to the for the cylinder(including the ends).

nt= e3 for the top surface

nb =− e3 for the bottom surface

ns= 1R(x1e1+ x2e2 ) for the side surface

(b) Compute the divergence of the vector field v(x) and the integral of div v over thevolume of the cylinder. Let r2 ≡ x21+ x22. Since v has no components in the directionof, we will assume that the index range is 1 to 2 instead of 1 to 3 for this problem.Thus,

∂r2∂xi= 2xi

Note that

vi= xi ln r2

⇒ div v=∂vi∂xi= δii ln r2+ xi

1r2∂r2∂xi= δii ln r2+ 2

xi xir2= 2 ln r2+ 2

x1

x2

x3

n

B

g(x)= 0


where a repeated index implies summation from 1 to 2 .

Computing the integral of div v over the volume would be most conveniently donein cylindrical coordinates. However, we can apply the divergence theorem to find theresult more easily. Note that v ⋅ n= 0 on the top and bottom surfaces. On the sidesurface

r2 = R2 = const.

v ⋅ n= ln R2 (x1e1+ x2e2 ) ⋅1R(x1e1+ x2e2 )= R lnR2

⇒ B

div v dV= Ω

v ⋅ n dA=Ωs

v ⋅ ns dA= R lnR2 As= 2 π R2 h ln R2

35. Consider the scalar field g(x)= x ⋅ x 2. Compute div∇div ∇g(x ) .

g(x)= (xi xi)2

∇g= 2(xixi)∂∂xj

(xkxk)ej= 2(xixi)[δkj xk+ xkδkj]e j= 4xi xixjej

div(∇g)= ∂∂xj

(4xi xi xj)= 4[δij xi xj+ xi δij xj+ xi xi δjj]= 20 xi xi

∇(div(∇g))= ∂∂xj

(20 xi xi)e j= 20[δij xi+ xi δij] ej= 40 xj e j

div(∇(div(∇g)))= ∂∂xj

(40 xj)= 40δjj= 120

36. Let v(x) be given by the following explicit function

v(x) = x2+x3 e1+ x1+x3 e2+ x1+x2 e3where x is the position vector of any point and has componentsx1, x2, x3 relative to the Cartesian coordinate system asshown. The vector field is defined on the ellipsoidal regionBwhose surface is described by the equation g(x)= 2x21+x22+2x23−4= 0. Give an ex-pression for the unit normal vector field n(x) to the ellipsoid. Compute the gradient of thevector field v(x). Compute the product [∇v]n, i.e., the gradient of the vector field actingon the normal vector. Compute the divergence of the vector field v(x).


(a) Give an expression for the unit normal vector field n(x) to the ellipsoid.

n= ∇g ∇g ∇g~

4 x12 x24 x3

∇g = (4 x1)2+(2 x2)

2+(4 x3)2

(b) Compute the gradient of the vector field v(x).

∇v~0 1 11 0 11 1 0

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on thenormal vector.

[∇v]n= 1 ∇g

4x12x24x3

2 x2+ 4 x34 x1+ 4 x34 x1+ 2 x2

0 1 11 0 11 1 0

= 1 ∇g

(d) Compute the divergence of the vector field v(x).

div v= tr(∇v)= 0

37. Evaluate the expression div∇x ⋅ Ax , whereA is a constant tensor (i.e., it does notdepend upon x), and the vector x has components x= xiei. The derivatives are to be takenwith respect to the independent variables xi. Express the results in terms of the componentsof A and x.

x ⋅ Ax= xi Aij xj ∇g= ∂g∂xkek div v=

∂vk∂xk

∇(x ⋅ Ax)= ∂∂xk

(xi Aij xj)ek= (δij Aij xj+ xi Aij δjk) ek= (Akj xj+ xi Aik)ek

div(∇(x ⋅ Ax))= ∂∂xk

(Akj xj+ xi Aik )= Akj δjk+ δik Aik= Akk+ Akk= 2Akk

38. Let g(x)= e−‖x ‖2be a scalar field in three-dimensional space, where ‖ x ‖ is the dis-

tance from the origin to the point x. Qualitatively describe the behavior of the function (aone- or two-dimensional analogymight be helpful). Compute the gradient ∇g of the field.Where does the gradient of the function go to zero?

1x1

2

3

x2

x3


Let r2 ≡‖ x ‖2= xixi be the radial distance from the origin. Note that

r2, j = 2rr, j = xi, j xi+ xixi, j = 2δij xi = 2xj ⇒ r,i=xir

∇g(x) = ∂g∂r∂r∂xi

ei = e−r2(−2r)

xir e i = −2e−r2xie i = −2g(x)x

Therefore, the gradient of the given function is ∇g(x)=−2xe−‖x ‖2. The function hasvalue g(0)= 1 at the origin and decays exponentially to zero with distance from theorigin. The function has a zero gradient at the origin.

39. Consider a tensor field T defined on a tetrahedral re-gion bounded by the coordinate planes x1 = 0, x2 = 0,x3 = 0, and the oblique plane 6x1+3x2+2x3 = 6, asshown in the sketch. The tensor field has the particular ex-pression T= b x, where b is a constant vector and xis the position vector x= xi ei. Compute the integral ofdiv T over the volume and the integral of Tn over thesurface of the tetrahedron (and thereby show that theygive the same result, as promised by the divergence theo-rem). Note that the volume of the tetrahedron of the givendimensions is one.

(a) Compute the integral of div T over the volume.

Tij = bi xj⇒ divT = bi∂xi∂xj

ei= biδjj ei= 3 b

B

div T dV=B

3b dV= 3bV= 3 b

(b) Compute the integral of Tn over the surface of the tetrahedron.

Face x1 = 0 ; n=−e1

Tn = b x (−e1)=−(x ⋅ e1 )b=−x1 b= 0

Face x2 = 0 ; n=−e2

Tn = b x (−e2)=−(x ⋅ e2 )b=−x2 b= 0

Face x3 = 0 ; n=−e3

Tn = b x (−e3)=−(x ⋅ e3 )b=−x3 b= 0


Face 6x1+3x2+2x3 = 6 ;

The normal to the oblique face can be found as

n= ∇6x1+3x2+2x3− 6 = 6e1+ 3e2+ 2e3

Normalizing to unit length

n= 176e1+ 3e2+ 2e3

Tn = (x ⋅ n )b= 176x1+3x2+2x3 b=

67b

Ω

TndA= Ω

67 bdA=

67 Ab= 6

7 (72) b= 3 b

40. Let v(x)= x on a spherical region of radius R, centered at the origin. Compute theintegral of div v over the volume of the sphere and compute the integral of the flux v ⋅ n,where n is the unit normal to the sphere, over the surface of the sphere. Give the result interms of the radius R.

div v =∂vi∂xi=∂xi∂xi= δii= 3

⇒ B

div v dV= 3B

dV= 3 V= 3 4 π R3

3= 4 π R3

The normal to the sphere surface is n= x∕R

⇒ Ω

v ⋅ n dA= 1RΩ

x ⋅ x dA= 1RΩ

R2 dA= R A= 4 π R3

What does this calculation tell you about the ratio of surface area to volume of asphere? From the divergence theorem

B

div v dV=Ω

v ⋅ n dA

Substituting with the results obtained above

⇒ 3 V= R A VA= R

3


41. TheLaplacian of a scalar field is a scalarmeasure of the second derivative of the field,defined as ∇2g(x)≡ div ∇g(x) . Write the component (index) form of the Laplacian ofg in Cartesian coordinates. Compute the Laplacian of the scalar field of Problem 38.

In components, ∇2g(x) ≡ div(∇g(x)) = g, ii. From Problem 38 note that

g,i=−2g(r)xi where r2 ≡‖ x ‖2= xixi.

Thus,

g,ii=−2 g,i xi+ gxi, i = −2 −2gxixi+ 3g = 4r2−6 g(x)

Therefore, the Laplacian is ∇2g(x) = 4r2−6 e−‖x ‖2

42. Computediv T , where T(x)= x ⋅ x I−2x x is a tensor field.

Tij = (xkxk ) δij− 2 xi xj

divT =∂Tij∂xj

e i = 2xkδjkδij− 2 δij xj− 2 xi δjj e i

= 2xjδij− 2 xj δij− 6 xi ei=− 6xi e i=− 6 x

43. Let u(x), v(x), and w(x) be vector fields and let T(x) be a tensor field. Compute thecomponent forms of the following derivatives of products of vectors

(a) ∇(u ⋅ v)= (uivi), j e j= ui,j vi+ uivi, je j(b) div(u× v)= uivjÁijk,k= ui,k vj+ uivj,kÁ ijk(c) ∇(u× v)= uivjÁijm,n em en = ui,n vj+ uivj,n Áijm em en

(d) div(Tv)= Tijvj ,i= Tij,i vj+ Tijvj,i

(e) ∇(u ⋅ Tv)= uiTij vj ,k ek= ui,k Tij vj+ uiTij,k vj+ uiTij vj,kek(f) ∇(Tv)= Tijvj ,k ei ek = Tij,k vj+ Tijvj,k ei ek

(g) div(u v)= uivj ,j ei= ui, j vj+ uivj,j ei(h) div [u v]w = ui vjwj

,i= ui, i vjwj+ uivj,i wj+ uivjwj,i


(i) ∇ (u× v) ⋅ w = uivjwkÁ ijk,m em= ui,m vjwk+ uivj,m wk+ uivjwk,mÁ ijkem

44. Use the same reasoning that was used to derive the three-dimensional version of thedivergence theorem to develop

(a) a one-dimensional version

f (x)

xa bxi−1 xi

n(xi)n(xi−1)

∆xi

Noting that in one dimension n= 1 for the normal on the right side and n=−1 forthe normal on the left side of the segment

div1 f (x) ≡ lim∆xi→0

1∆xi f (xi)− f (xi−1) ≡

dfdx

Divide the region [a,b] into segments with endpoints defined by the x values

xo, x1, x2, . . ., xn−1, xn

where xo ≡ a and xn ≡ b. Let us do the following simple computation

ni=1

f (xi)− f (xi−1) = f (x1)− f (a)+ f (x2)− f (x1)+ . . .

+ f (xn−1)− f (xn−2)+ f (b)− f(xn−1) = f (b)− f (a)

Each term appears once with a plus sign and once with a minus sign except for the twoend points. Define the “almost divergence” as follows

Di ≡1∆xi f (xi)− f (xi−1)

Now we can write

ni=1

D i∆xi = n

i=1

f (xi)− f (xi−1) = f (b)− f (a) (*)


In the limit as ∆xi→ 0, Di→ div1 f (x), n→∞, and the above sum becomes an inte-gral (this is the definition of the Riemann integral, the limit of a sum!) as follows

lim∆xi→0ni=1

Di∆xi = b

a

dfdxdx

From (*) we can see that the result is independent of the limiting process and we getthe one-dimensional version of the divergence theorem

ba

dfdxdx = f (b)− f (a)

a very familiar result from the integral calculus.

(b) The two-dimensional version version is identical to the three-dimensional versionif the domain B is interpreted as a planar region, the boundary Ω a closed curve in theplane. Also, let v(B) be the area of the domain, the integral over B an area integral,and the integral over Ω a line integral. With these reinterpretations of notation, theargument proceeds exactly as the three-dimensional case.

45. Consider a vector field v(x) on a region B with surface Ω having unit normal fieldn. The “curl” of the vector field can be defined as

curl v ≡ limV(B)→0

1V(B)Ω

v× n dA

Show (using the cuboid for B, as in the text) that the expression for curl v is

curl v = ∂vxi× ei = ∂v2∂x3−∂v3∂x2e1+∂v3∂x1−∂v1∂x3e2+∂v1∂x2−∂v2∂x1e3

Note thatmany authors define the curl tobe thenegative of the definitiongiven here,whichis easily achieved by using the flux n× v instead. The form presented here seems to bemore consistent with our other definitions of derivatives of vector fields.

Let Ωi be the rectangular region with area Ai over which xi is constant. The integralover the surface is then the sum of the integrals over the six faces.

Ω

v× n dA=3i=1

Ωi

v(x+ ∆xie i )× ei+ v(x ) × (−e i ) dAi

=3i=1

Ωi

v(x+ ∆xie i )− v(x ) × e i dAi

e2− e2

e3

e1− e3

− e1

x1

x2

x3


The ratio of flux to volume is

1V(B)Ω

v× n dA=3i=1

1AiΩi

v(x+ ∆xie i )− v(x ) ∆xi

× ei dAi

The curl can be computed as

curl v = 3i=1

limAi→0

1AiΩi

lim∆xi→0

v(x+∆xiei )− v(x ) ∆xi

× e i dAi =3

i=1

∂v∂xi× ei

= ∂v1∂x1 e1+∂v2∂x1 e2+∂v3∂x1 e3× e1+∂v1∂x2 e1+∂v2∂x2 e2+∂v3∂x2 e3× e2

+∂v1∂x3 e1+∂v2∂x3 e2+∂v3∂x3 e3× e3

Therefore,

curl v = ∂v2∂x3−∂v3∂x2e1+∂v3∂x1−∂v1∂x3e2+∂v1∂x2−∂v2∂x1e3

46. Consider variously a scalar field g(x), a vector field v(x), and a tensor field T(x) ona region B with surface Ωwith unit normal vector field n. Prove the following theorems

B

∇g dV= Ω

g n dA, B

∇v dV= Ω

v n dA, B

divT dV= Ω

Tn dA

(a) Prove that B

∇g dV= Ω

g n dA

Chop the region B into N small subvolumes Bi with surfaces Ωi. Let each subvo-lume be centered at location xi and have volume v(Bi). The surface of each subvolumecan be characterized by its outward normal field ni. Now the surface of each subvo-lume either will be adjacent to another subvolume, or will lie on the original boundaryΩ. Since the normal vector points outward from each subvolume, the normals of twoadjacent subvolumes point in exactly the opposite direction. Consequently, if we sumthe fluxes over all of the subvolumes, we get simply

Ni=1

Ωi

gni dAi = Ω

gn dA (*)

because all of the interior surfaces cancel each other out. Let us define the “almostgradient” of the scalar field to be


Gi[g] ≡1

v(Bi)Ωi

gni dAi

The only difference between G[g] and ∇g is that ∇g is the limit of G[g] as the regionbecomes vanishingly small. From (*) we can see that

Ω

gn dA = Ni=1

Gi[g]v(Bi)

Upon taking the limit as the number of subvolumes becomes large, the discrete ele-ments pass to their infinitesimal limits, that is, Gi[g]→ ∇g and v(Bi)→ dV, the infini-tesimal volume element. The limit of the sum is the integral over the volume

limN→∞Ni=1

Gi[g]v(B i) = B

∇g dV

thereby completing the proof.

(b) Prove that B

∇v dV= Ω

v n dA

Chop the region B into N small subvolumes Bi with surfaces Ωi. Let each subvo-lume be centered at location xi and have volume v(Bi). The surface of each subvolumecan be characterized by its outward normal field ni. Now the surface of each subvo-lume either will be adjacent to another subvolume, or will lie on the original boundaryΩ. Since the normal vector points outward from each subvolume, the normals of twoadjacent subvolumes point in exactly the opposite direction. Consequently, if we sumthe fluxes over all of the subvolumes, we get simply

Ni=1

Ωi

v ni dAi = Ω

v n dA (*)

because all of the interior surfaces cancel each other out. Let us define the “almostgradient” of the vector field to be

Gi[v] ≡1

v(Bi)Ωi

v ni dAi

The only difference between G[v] and ∇v is that ∇v is the limit of G[v] as the regionbecomes vanishingly small. From (*) we can see that

Ω

v n dA = Ni=1

Gi[v] v(Bi)

Upon taking the limit as the number of subvolumes becomes large, the discrete ele-ments pass to their infinitesimal limits, that is, Gi[v]→ ∇v and v(Bi)→ dV, the infini-tesimal volume element. The limit of the sum is the integral over the volume


limN→∞Ni=1

Gi[v]v(B i) = B

∇v dV

thereby completing the proof.

(c) Prove that B

divT dV= Ω

Tn dA

First observe that div (TTh) = h ⋅ divT and that (TTh) ⋅ n = h ⋅ Tn. Sincethe vector h is constant it can be taken outside the integral. Therefore, from the origi-nal statement of the divergence theorem we have

h ⋅ B

divT dV = h ⋅ Ω

Tn dA

Since this result must hold for any h the theorem is proved.

47. Use the divergence theorem for a vector field to show the following identities

(a) Green’s first identity for scalar functions u(x) and v(x), (Hint: Let v(x)= u∇v)

B

u∇2v+∇u ⋅ ∇v dV = Ω

n ⋅ u∇v dA

(b) Green’s second identity for scalar functions u(x) and v(x),(Hint: Let v(x)= u∇v−v∇u)

B

u∇2v−v∇2u dV = Ω

n ⋅ u∇v−v∇u dA

(a) Green’s first identity for scalar functions u(x) and v(x)

B

u∇2v+∇u ⋅ ∇v dV = Ω

n ⋅ u∇v dA

Let v= u∇v. In components div v= (uv,i ),i= u, i v, i+ uv,ii. Observe that, by defini-tion ∇2v= v,ii. Thus, div v= ∇u ⋅ ∇v+ u∇2v.(b) Green’s second identity for scalar functions u(x) and v(x)

B

u∇2v−v∇2u dV = Ω

n ⋅ u∇v−v∇u dA

Let v= u∇v− v∇u. In components div v= (uv,i−vu,i ),i= uv, ii− vu, ii.

x1

x2

x3

er (θ)

ez

eθ (θ)

r

z

θ


48. Many problems are more conveniently formulated andsolved in cylindrical coordinates (r, θ, z). In cylindrical coor-dinates, the components of a vector v can be expressed as

v(r, θ, z) = vrer+ vθeθ+ vzezwhere the components vr, vθ, and vz are each functions of thecoordinates (r, θ, z). However, now the base vectors er (θ)and eθ (θ) depend upon the coordinate θ. We must accountfor this dependence of the base vectors on the coordinateswhen computing derivatives of the vector field.

Using the coordinate-free definition of the divergence of a vector field, Eqn.(78), showthat the divergence of v in cylindrical coordinates is given by

divv(r, θ, z) = 1r∂∂rr vr + 1

r∂vθ∂θ +

∂vz∂z

(Hint: Observe from the figure that n1 = eθ(θ+∆θ) and n2 =−eθ(θ) and are constantover the faces 1 and 2, respectively. The normal vectors n3 = er (ξ) and n4 =−er (ξ),with ξ∈ [θ, θ+∆θ], vary over faces 3 and 4. Finally, note that n5 = ez and n6 =−ezare constant over faces 5 and 6.)

θr

n3

n2

n1

n4∆θ

∆r ∆zr∆θ

∆r

n4 n2

n5

n6z

r

z

Note that the volume of the wedge is V(B)= r∆θ∆r∆z plus terms of higher orderthat vanish more quickly in the limit as V(B)→ 0.

The contribution of the two surfaces with normal vectors n1 and n2 give

lim∆θ, ∆ r, ∆ z→0

1r∆θ∆ r∆ z

F1

v(x+∆θn1) ⋅ n1 drdz + F2

v(x) ⋅ n2 drdz

Noting that the region of integration is the same for both integrals, F1 = F2 with area∆ r∆ z, and substituting the expressions for the normal vector fields we get

lim∆θ, ∆ r, ∆ z→0

1r∆ r∆ zF1

v(x+∆θeθ) ⋅ eθ(θ+∆θ)− v(x) ⋅ eθ(θ)∆θ

drdz

Taking the limit, and noting that v ⋅ eθ= vθ we find that the above expression reducesto

1r∂(v ⋅ eθ)∂θ = 1

r∂vθ∂θ


Similarly, the contribution of the faces with normal vectors n3 and n4 give

lim∆θ, ∆ r, ∆ z→0

1r∆θ∆ r∆ z

F3

v(x+∆ rn3) ⋅ n3 (r+∆r)dθdz + F4

v(x) ⋅ n4 rdθdz

In this case, the regions of integration are not the same, but F3 = F4 since the vari-ables of integration are θ and z, and at every point in the region, n3 =−n4 = er(θ).Thus, we have

lim∆θ, ∆ r, ∆ z→0

1r∆θ∆ zF4

(r+∆r)v(x+∆ r er) ⋅ er(θ)− rv(x) ⋅ er(θ)∆ r

dθdz

Taking the limit, and noting that v ⋅ er= vr we find that the above expression reducesto

1r∂(rv ⋅ er)∂r = 1

r∂(rvr)∂r = vr

r +∂vr∂r

Finally, the contribution of the faces with normal vectors n5 and n6 give

lim∆θ, ∆ r, ∆ z→0

1r∆θ∆ r∆ z

F5

v(x+∆ zn5) ⋅ n5 rdθdr + F6

v(x) ⋅ n6 rdθdr

In this case, the regions of integration are the same, but F5 = F6, and at every point inthe region, n5 =−n6 = ez. Thus, we have

lim∆θ, ∆ r, ∆ z→0

1∆θ∆ rF5

v(x+∆ z ez) ⋅ ez− v(x) ⋅ ez∆ z

dθdr

Taking the limit, and noting that v ⋅ ez= vz we find that the above expression reducesto

∂(v ⋅ ez)∂z = ∂vz

∂z

Adding the contributions together gives the desired result.

Chapter 2The Geometry ofDeformation

Note: Unless otherwise indicated we shall assume that the base vectors in the deformedand undeformed configurations coincide, i.e., that e i = gi .

49. Consider a unit cube in the positive octant with a vertex positioned at the origin ofcoordinates subjected to the following deformation map

φ(z)= z1+Á z2 z3 e1+ z2+Á z1 z3 e2+ z3+Á z1 z2 e3

where Á is a constant. Compute the deformation gradient F, the Green deformation tensorC, and the Lagrangian strain tensor E for the given deformation. Using graph paper, plotthe deformed position of a square in the x1−x2 plane by locating the positions of a gridof points. (Select a value of Á to execute the plot.)

(a) Compute the deformation gradient F, the Green deformation tensor C, and theLagrangian strain tensor E for the given deformation. The components of the deforma-tion gradient are given by F

ij= ∂φi(z)∕∂z j. The components of F can, therefore, be

computed as

F ~Áz3 Áz2

Áz3 Áz1Áz2 Áz1

11

1

The Green deformation tensor has components Cij= FkiFkj, which can be computed inmatrix format as follows

P

z1

z2

z3n


C= FTF ~Áz3 Áz2

Áz3 Áz1Áz2 Áz1

11

1

Áz3 Áz2Áz3 Áz1Áz2 Áz1

11

1

1+Á2z12+Á2z321+Á2z22+Á2z32

= 2Áz3+Á2z1z2

2Áz2+Á2z1z32Áz3+Á2z1z22Áz1+Á2z2z3

2Áz2+Á2z1z3 2Áz1+Á2z2z3 1+Á2z12+Á2z22

The components of the Lagrangian strain tensor can be computed in matrix format asfollows:

Á2z12+Á2z32

Á2z22+Á2z32

E = 12C−I ~ 1

22Áz3+Á2z1z2

2Áz2+Á2z1z32Áz3+Á2z1z22Áz1+Á2z2z3

2Áz2+Á2z1z3 2Áz1+Á2z2z3 Á2z12+Á2z22

(b) Using graph paper, plot the deformed position of a square in the x1−x2 plane bylocating the positions of a grid of points. (Select a value of Á to execute the plot.)

1

1

0.25

0.25

z1, x1

z2, x2

0

z3 = 1

Á= 0.25

x1 = z1+0.25z2

x2 = z2+0.25z1

50. The deformation gradient that results from deforming thebody shown through a deformationmap φ(z) has the followingcomponents relative to the standard basis at the point P

F(P) ~1.1 0.3 0.10.1 1.2 0.20.2 0.3 1.3

Find the stretch of a line oriented in the direction of the vector n= (1,1,0) at the point P.What is the value of the Lagrangian strain of that same line at that same point? Calculatethe tensors C and E.

(a) Find the stretch of a line oriented in the direction of the vector n= (1,1,0) at thepoint P. First, the vector n must be normalized to unit length. The length of n is 2 .

z1

z2

0

4

4 x1

x2φ(z)

Chapter 2 The Geometry of Deformation 41

Define the unit vector m≡ n∕ ‖ n ‖. The tangent vector m in the deformed configura-tion is

F(P)m = 1

2(1.4, 1.3, 0.5)

λ2(m) = Fm ⋅ Fm = 1.95

λ(m) = 1.396

(b) What is the value of the Lagrangian strain of that same line at that same point?Calculate the tensors C and E.

E = 12λ2(m)−1 = 0.475

The components of the Green deformation tensor are

C= FTF ~1.26 0.51 0.390.51 1.62 0.660.39 0.66 1.74

The components of the Lagrangian strain tensor are

E = 12C−I ~

0.13 0.255 0.1950.255 0.31 0.330.195 0.33 0.37

51. Consider a square piece ofmateri-al of unit thickness with a round holein it of radius 1. The material is sub-jected to a deformation described bythe map shown in the diagram. The de-formation map shown has the follow-ing explicit expression

φ(z) = z1 1+βz2 e1+ z2 1+3βz1 e2+ z3e3

Compute the the volume of the hole in the undeformed and deformed configurations.Compute the perimeter area of the square in the undeformed and deformed configurations.Compute the perimeter area of the circle in the undeformed and deformed configurations.

(a) Compute the the volume of the hole in the undeformed and deformed configura-tions. The volume of the undeformed hole can be obtained easily as V= π. The vol-ume of the deformed hole can be obtained as


v = φ(Ω)

dv = Ω

detF dV

The components of the deformation gradient are

F ~βz1

1

03βz2 1+3βz1 00 0

1+βz2

detF = 1+βz2 1+3β z1 − 3β2 z1 z2 = β 3 z1+z2 + 1

v = β 3 z1+z2 + 1 dz1dz2dz3

Let us carry out the volume integral in cylindrical coordinates r and θ, centered atz1 = 2, z2 = 2. The relationship can be described as

z1 = 2+r cos θ z2 = 2+r sin θ

Substituting into the equation for deformed volume we have

v = 10

2π0

10

β 8+3r cos θ+r sin θ + 1rdrdθdz3 = π (1+8β)

(b) Compute the perimeter area of the square in the undeformed and deformed config-urations. The perimeter area of the undeformed square is A= 16. To obtain the de-formed perimeter area we must integrate over each of the four faces. Based on Eqn.(156) in the text, the deformed area can be obtained from

a = φ(Ω)

da = Ω

J F–TN dA

J F−T ~ −βz1J

0−3βz21+3βz10

0 0

1+βz2

where J= det F is the Jacobian of F. The faces with normals −e1 and −e2 are posi-tioned at z1 = 0 and z2 = 0, respectively. Because we have JF−Te1 = e1 alongz1 = 0 and JF−Te2 = e2 along z2 = 0 these faces are not deformed. Therefore, weonly need to integrate the two faces with normals e1 and e2. At the face with normal e1with z1 = 4

J F−Te1 = 1+12β 2+ 16β 2

and the area of the face can be computed as


ae1 = 1

0

10

1+12β 2+ 16β 2 dz2dz3 = 1+24β+160β2

Similarly, at the face with normal e2 with z2 = 4

ae2 = 1

0

10

12β 2+ 1+4β 2 dz2dz3 = 1+8β+160β2

Thus, total deformed perimeter area of the square is

a = 8+ 1+24β+160β2 + 1+8β+160β2

(c) Compute the perimeter area of the circle in the undeformed and deformed config-urations. The undeformed perimeter area of the circle is simply A= 2π. To obtain thedeformed perimeter area, we follow the same procedure as part (b) of this problem.The normal to the undeformed circle can be parameterized in terms of the angle θ andrepresented with the vector N= cos θe1+ sin θe2. Thus

J F−TN = 1+3βz1 cos θ−3βz2 sin θ 2+−βz1 cos θ+1+βz2 sin θ 2We can change to cylindrical coordinates to carry out the integration. Noting thatr= 1 is constant along the undeformed circle we have

J F–TN = c21+ c22

where

c1 ≡ 1+6β cos θ−6β sin θ+3β cos 2θ

c2 ≡ 1+2β sin θ−2β cos θ−β cos 2θ

The deformed area of the cylinder can be computed numerically for specific values ofthe deformation parameter β as

a=10

2π0

J F−TN dθdz3

β a

0.0 6.28310.5 22.59611.0 40.14832.0 75.6558

52. Prove that (unit) eigenvectors n1 and n2, of the tensor C, associated with distinct ei-genvalues m1 and m2, respectively, point in the direction of extreme stretch by computingthe stretch for a unit vector m= sin θn1+ cos θn2, where θ is a parameter. Plot thestretch in the direction m as a function of θ.

z1

z2

0

1

1 x1

x2

0

1

1

φ(z)


Let us compute the stretch in the direction of the vector m

m ⋅ Cm = sin θn1+ cos θn2 ⋅ C sin θn1+ cos θn2

= sin θn1+ cos θn2 ⋅ sin θCn1+ cos θCn2

= m1 sin2 θ n1 ⋅ n1 + m2 cos

2 θ n2 ⋅ n2 + m1+ m2 sin θ cos θ n1 ⋅ n2

= sin θn1+ cos θn2 ⋅ m1 sin θn1+m2 cos θn2

Since the eigenvectors corresponding to distinct eigenvalues are orthogonal, and, bydefinition, of unit length, we have

n1 ⋅ n2 = 0, n1 ⋅ n1= n2 ⋅ n2= 1

⇒ λ2(m)= m ⋅ Cm = m1 sin2 θ+ m2 cos

2 θ

It can be seen from this equation that,

for θ= 0, λ2(m)= m2

for θ= π2,

λ2(m)= m1

for any 0≤ θ≤ 2π, minm1,m2 ≤ λ2(m)≤ maxm1,m2

Therefore m1 and m2 are the extreme values of stretch (squared). This result can befurther verified by plotting the stretch in the direction m as a function of θ as shown inthe sketch below.

0 θπ2

3π2

π 2π

λ2(m)

m1 < m2

m1

m2

0 θπ2

3π2

π 2π

λ2(m)

m1 > m2

m2

m1

53. Consider a square piece of material ofunit thickness. The material is subjected to adeformation described by the map shown inthe diagram. The deformation map shownhas the following explicit expression

φ(z) = z1 1+z2 e1+ z2 1+3z1 e2+ z3e3


Compute the components (with respect to the standard basis e1, e2, and e3) of the Greendeformation tensor C and the Lagrangian strain tensorE at the point z= (1,1,0). Find theprincipal stretches and principal directions of C at z= (1,1,0). Find the eigenvalues andeigenvectors of E at z= (1,1,0).

(a) Compute the components (with respect to the standard basis e1, e2, and e3) of theGreen deformation tensor C and the Lagrangian strain tensor E at the point z=(1,1,0). Obtain the components of the deformation gradient and evaluate it at the posi-tion z= (1, 1, 0)

F ~z1

1

1+z2 0

3z2 1+3z1 00 0

=2 1 03 4 00 0 1


C= FTF ~13 14 014 17 00 0 1


E = 12C−I ~

6 7 07 8 00 0 0

(b) Find the principal stretches and principal directions of C at z= (1,1,0). The prin-cipal stretches are the solutions to the characteristic Eqn. (49), with coefficientsIC= 31, IIC= 55 and IIIC= 25. The three solutions of the characteristic equationare m1 = 0.8579, m2 = 1 and m3 = 29.1421. The corresponding principal directionsare

n1 = 0.7555,− 0.6552, 0

n2 = 0, 0, 1

n3 = 0.6552, 0.7555, 0

Note that, from Gershgorin’s theorem one can deduce, by inspection, that one of thestretches is 1. One can use this information to reduce the cubic to a quadratic by syn-thetic division.

(c) Find the eigenvalues and eigenvectors of E at z= (1,1,0). The eigenvalues andeigenvectors of E can be found easily by Eqn. (136) in the text. The eigenvalues areγ1 =− 0.07105, γ2 = 0 and γ3 = 14.0711 and eigenvectors are the same as the prin-cipal directions in (b).

φ(z)

a a

z1z2

z3

B


54. Prove that it is impossible to deform the vertexof a solid cube into a flat face (e.g., the deformationmap shown in the sketch deforms the cube into atetrahedron with the vertex at a deformed onto theflat plane). Hint: You do not need to find an explicitexpression for the map to do this problem. Considera neighborhood of the point a.

Let N1, N2 and N3 be unit vectors normal to the three faces at point a in the unde-formed configuration. These vectors are orthogonal by virtue of the initial geometry.These vectors deform to vectors n1, n2 and n3 according to

ni = FNi

If the corner is deformed to a flat face at a then n1 = n2 = n3≡ n. Observe that(n× n) ⋅ n= 0. Therefore

n1× n2 ⋅ n3 = FN1× FN2 ⋅ FN3 = 0

Since N1× N2 ⋅ N3 = 1 we have

FN1× FN2 ⋅ FN = detF N1× N2

⋅ N3 = det F

Thus, detF= 0, implying that a finite volume of material in the vicinity of the point amust be deformed to zero volume, which is impossible.

55. A semi-infinite half-space (i.e., the body occupies everypoint in space that satisfies z3 > 0) has a deformation mapgiven by the following explicit expression

φ(z) = 1+βe−R z+ γe−Re3where β and γare constants andR is the distance from the ori-gin to any point with position vector z, that is, R2 ≡ z ⋅ z. Plot the variation of displace-ment along the coordinate axes. Compute the displacement of the point that was originallylocated at z= (0, 0, ln2). Compute the deformation gradient F(z) of the motion in generaland evaluate it at z= (0, 0, ln2). Compute the Green deformation tensor C(z). Find thevalue of the stretch of a line in the neighborhood of z=(0, 0, ln2) and initially orientedin the direction e1.

(a) Plot the variation of displacement along the coordinate axes. Compute the dis-placement of the point that was originally located at z= (0, 0, ln2). The displacementmap can be obtained from deformation map as

u(z) = φ(z)− z = βe−R z+ γe−Re3


The displacement at the point z= (0, 0, ln2) is

u(0, 0, ln2) = β ln2+ γ e−ln 2e3 =12β ln2+ γe3

(b) Compute the deformation gradient F(z) of the motion in general and evaluate it atz= (0, 0, ln2). Compute the Green deformation tensor C(z). The deformation gradientcan be computed as follows

F = ∇φ = 1+β e–R I+ β∇e–R z+ γ∇e–R e3

Noting that ∇R= z∕Rwe have ∇e–R=−e–Rz∕R. Thus,

F = 1+βe–R I− e–RRβz z+ γ z e3

Evaluating F at the position z= (0, 0, ln2) we get

F(0, 0, ln2) = 1+ 12 β I− 1

2β ln2+γ e3 e3

To obtain Green’s deformation tensor, first compute the transpose of the deformationgradient

FT = 1+βe–R I− e–RRβ z z+ γ e3 z

Noting that z z z z = R2 z z and e3 z z e3 = R2 e3 e3, we canrepresent Green’s deformation tensor as

C = FTF = aI+ bz z+ c z e3+ e3 z + de3 e3

where

a = 1+β e−R 2 b =βe−2R

RβR−eR−β

c = γe−2R

RβR−eR−β d = γ2e−2R

(c) Find the value of the stretch of a line in the neighborhood of z=(0, 0, ln2) andinitially oriented in the direction e1.

λ2(e1) = Fe1 ⋅ Fe1 = 1+ 12 β 2

λ(e1) = 1+ 12β

z1, x1

z2, x2

0

ℓ

w(z1)

θ(z1)φ(z)

u(z1)


56. Abeam theory is characterized by a spe-cific deformation map that is parameterizedby a set of deformation variables that dependonly on the axial coordinate z1. The depen-dence of the map on z2 and z3 is explicit. Letu(z1) represent the displacement of the cen-troid of the beam in the z1 direction, w(z1)the displacement of the centroid of the beamin the z2 direction, and θ(z1) the rotation ofa vector normal to the deformed cross section relative to the horizontal. The deformationmap for finite planar motion of the beam then takes the form shown in the diagram. Thedeformation map has the following mathematical expression

φ(z) = z1+u(z1)−z2 sin θ(z1) e1+ w(z1)+z2 cos θ(z1) e2+ z3e3

Compute the deformation gradient F of the given deformation map. Compute the Greendeformation tensor C, and the Lagrangian strain E. Linearize the deformation map by as-suming that cos θ≈ 1 and sin θ≈ θ, and computeF,C, andE for the linearized kinemat-ic description. Is the strain linear in the displacement variables u(z1), w(z1), and θ(z1)?Linearize E by neglecting all squares and products of the generalized variables u, w, andθ. What are the consequences of neglecting the higher-order terms?

(a) Compute the deformation gradient F of the given deformation map. Compute theGreen deformation tensor C, and the Lagrangian strain E. The components of the de-formation gradient are given by F

ij= ∂φi(z)∕∂z j. In matrix format we have

F ~1+u′−z2 θ′ cos θ − sin θ 0w′+z2 θ′ sin θ cos θ 0

0 0 1


C= FTF ~0

0

0 0 1

1

C11 C12

C21

where

C11 = 1+u′−z2θ′ cos θ 2 + w′+z2θ′ sin θ 2

C12 = C21 = − sin θ (1+u′)+ w′ cos θ + 2z2θ′ sin θ cos θ


E = 12C−I ~ 1

2

0

0

0 0

C11−1 C12

C21 0

0


(b) Linearize the deformation map by assuming that cos θ≈ 1 and sin θ≈ θ, andcompute F, C, and E for the linearized kinematic description. Is the strain linear in thedisplacement variables u(z1), w(z1), and θ(z1)? The linearized deformation map can bewritten as

φ(z) = z1+ u(z1)− z2 θ(z1)e1+ w(z1)+ z2e2+ z3e3

The components of the deformation gradient of the linearized deformation map are, inmatrix form

F ~1+u′−z2 θ′ −θ 0

w′ 00 0 1

1


C= FTF ~ θ2+1

C11 C12

C21

0

0

0 0 1


E = 12C−I ~ 1

2

C11−1 C12

C21 θ2

0

0

0 0 0

where

C11 = 1+u′−z2θ′ 2 + w′ 2

C12 = C21 = −θ (1+u′)+ w′ + z2θθ′

The strain is not linear in the displacement variables because strain tensor containssquares and products of the displacements and derivatives.

(c) Linearize E by neglecting all squares and products of the generalized variables u,w, and θ. What are the consequences of neglecting the higher-order terms?

2u′−2z2 θ′ w′−θ 0w′−θ 0

0 0

EL =12CL−I ~

12

00

The consequences of neglecting higher-order terms is that the accuracy of the strainsdecrease as the deformations increase.

57. Does the linearized strain tensor ever have the same eigenvectors and eigenvalues asthe Lagrangian strain tensor? If so, provide an explicit example.

x1

x2

0

2π

z1

z2

0

2π

φ(z)


The Lagrangian strain tensor cannot have the same eigenvalues as the linearized straintensor. However, for a rigid body motion they can have the same eigenvectors. SupposeQ is an orthogonal tensor, then a rigid body motion can be expressed as

φ(z) = z+ αQz

The deformation gradient is F= I+αQ and the Lagrangian strain tensor is

E = 12FTF− I

= 12αQ+αQT+α2QTQ

= 12αQ+ αQT+ α2 I

= Elinear+12α2 I

If the difference of two tensors is a multiple of the identity tensor, then the two tensorshave identical eigenvectors. Suppose that the identical eigenvector is ni, then

Eni = Elinear+12 α

2IniThe relationship between eigenvalues of Lagrangian and linearized strain tensors are

γi = (γlinear) i+12α2

58. Find the mathematical expres-sion for the map that takes a strip oflength 2π and deforms it into a semi-circular arc without changing thedepth of the strip. The deformationmap is illustrated in the sketch. Com-pute the deformation gradient F, theGreen deformation tensor C, and the Lagrangian strain tensor E for the map.

This problem is similar to the example in the text. The difference is that this beam isbent to a half-circle instead of a complete circle. Since the fiber along the initial z1 axisdoes not change length we can determine the radius of the arc. The arc length isπR= 2π, implying that the radius is R= 2. The position of a point, say P in thesketch, is easily described in polar coordinates (r, θ). The position can be described as

x1 = r cosθ− π2, x2= R+ r sinθ− π

2

Since the arc length from the origin to the point P is z1 and the radius of the arc isR= 2 we can compute the angle θ= z1∕2. Furthermore, the radial distance to the

z2

1 z1

1

032

32

4

0

a

bc


x1

x2

0

2π

θ

r

z1R

P

z2

point P is r= 2−z2. Substituting these into the above equations, noting thatcos(θ−π∕2)= sin θ and sin(θ−π∕2)=− cos θ, gives the resulting deformationmap as

φ(z) = (2−z2) sin(z1∕2)e1+ 2−(2−z2) cos(z1∕2)e2+ z3e3

To verify the solution, check out the location of the point originally located at positionz=(2π, 0, 0) which gives φ(2π, 0, 0)= 4e2.

59. Consider the rectangular piece of material with thetriangular cutout. The body is subjected to the deformationmap

φ(z) = z1+βz2 e1+ z2e2+ z3e3

Find the angle of the triangle at the vertex at a before and afterdeformation. Find the equation describing the inclined line a-bbefore and after deformation. Find the area of the triangle abcbefore and after deformation.

(a) Find the angle of the triangle at the vertex at a before and after deformation. LetN1 = (0,−1, 0) and N2 =

1

5(1,−2, 0) be the unit vectors before deformation

cos θ = N1 ⋅ N2 =2

5θ = 26.57⇒

Let n1 = FN1 and n2 = FN2 be the deformed vectors. The angle between them can becomputed as

cosψ =n1 ⋅ n2n1 n2

=FN1 ⋅ FN2FN1

FN2

F ~β

1

0

0

0 0

1

10 FN1 =− β

0

− 1 FN2 =1

5

1−2β

0

− 2


cosψ =2–β (1−2β )

1+β2 4+ (1−2β )2

(b) Find the equation describing the inclined line a-b before and after deformation.The coordinates of the points a and b are (1, 3, 0) and (2, 1, 0), respectively. The equa-tion of the line connecting them before deformation is z2 = 5−2z1. From the de-formation map we have

x1 = z1+βz2x2 = z2

z1 = x1−βx2z2 = x2

⇒

The equation after deformation can be obtained by substitution

x2 = −2(x1−βx2)+ 5 = 11−2β (5−2x1)

(c) Find the area of the triangle abc before and after deformation. The area beforedeformation is simply 1. The area after deformation is

a = φ(Ω)

da = Ω

detF F–TN dA

The normal vector of plane is e3

F−Te3 = − β1

0

0

0 0

1

1

0

=100

100

Therefore, detF ‖ F–TN ‖= 1 and the deformation preserves the area at 1.

60. Consider a square piece of material of unit thickness. The material is subjected to adeformation described by the following explicit expression

φ(z) = αz1+βz2 e1+ γz1+δz2 e2+ z3e3

where α, β, γ, and δ are constants. For what values of the constants is the given deforma-tionmap physically impossible to realize? Assume that we have scribed a line on the bodybefore deforming it according to the above map. The equation of that line in the unde-formed configuration was z2 = 1−3z1. What is the equation of the line after deforma-tion? Will the given map ever deform straight lines into curved lines? Why or why not?

(a) For what values of the constants is the given deformation map physically impossi-ble to realize? The determinant of the deformation gradient must always be positive ifthe deformation is physically realizable. The components of the deformation map giv-en above are

r

z3

z1

z2

ℓ

x3

x1

x2

φ(z)


0F ~βα

γ δ

0

0 0 1

The condition for a physically reasonable deformation is detF= αδ−βγ> 0. Anoth-er way to look at the deformation is that the stretch in any direction should not be lessthan zero. Let n have components n1, n2, and n3. The square of the length of the vectorFn is

αn1+βn2 2+ γn1+δn2 2+ n23 > 0, ∀ n1, n2, n3

subject to the constraint that n21+n22+n23 = 1.

(b) Assume that we have scribed a line on the body before deforming it according tothe above map. The equation of that line in the undeformed configuration wasz2 = 1−3z1. What is the equation of the line after deformation? The relationship canbe obtained by inverting the deformation map.

x1 = αz1+ β z2

x2 = γz1+ δ z2z1 =

δx1−βx2αδ−βγ z2 =

αx2−γx1αδ−βγ⇒

Substituting these into the equation of the original line gives

x2 =αδ−βγα−3β +

γ−3δα−3β x1

(c) Will the given map ever deform straight lines into curved lines? Why or why not?Straight lines always deform to straight lines because the deformation map is linear inthe undeformed coordinates z1, z2 and z3.

61. The deformation map for the pure twist of a circularshaft of length ℓ and radius r can be expressed in terms ofthe rate of twist β (a constant) as follows

+ z1 sin(βz3 )+z2 cos(βz3 ) e2+ z3 e3

φ(z)= z1 cos(βz3 )−z2 sin(βz3 ) e1

Compute the deformation gradient F(z). Find the displace-ment of the point initially located at the position z=(r, 0,ℓ) in the undeformed configuration. Find the volume of thedeformed shaft in terms of the angle of twist β. Ahorizontalline is etched on the surface of the undeformed shaft, paral-lel to the z3 axis as shown. Find the length of the line in thedeformed configuration.

e1z2

z3

1

1

1z1


(a) Compute the deformation gradient F(z). Find the displacement of the point initial-

ly located at the position z=(r, 0, ℓ) in the undeformed configuration.

− sin βz3F ~

1

cosβz3

0 0

−βz1 sin βz3+z2 cosβz3

sin βz3 cosβz3 βz1 cosβz3−z2 sin βz3

u(z) = φ(z)− z

u(r, 0, ℓ) = rcos(βℓ)− 1e1+ r sin(βℓ) e2

(b) Find the volume of the deformed shaft in terms of the angle of twist β.

v = φ(B)

dv = B

detF dV

detF = cos2 βz3+ sin2 βz3 = 1

The map is, therefore, volume preserving, and the deformed volume is πr2ℓ.

(c) A horizontal line is etched on the surface of the undeformed shaft, parallel to thez3 axis as shown. Find the length of the line in the deformed configuration.

Fe3 =1

−β z1 sin βz3+z2 cos βz3

β z1 cosβz3−z2 sin βz3

To obtain the length of the deformed line, we must integrate the stretch along the origi-nal line. The normal vector of the original line is n= e3. The line is etched on thesurface so that z21+ z22 = R2.

λ2(e3) = Fe3 ⋅ Fe3 = β 2 z21+ z22 + 1 = β 2R2+1

The deformed length can be obtained by integration

L=ℓ0

λ(e3)dz3 = ℓ β 2R2+1

62. Consider the unit cube shown. Let the cube besubjected to the deformation map given by

φ(z) = z1+z1 z2 e1+ z2+z1 z2 e2+ z3e3

x3

T

T


Compute the volume of the cube in the deformed configura-tion. Find the area in the deformed configuration of the facewith normal e1 in the undeformed configuration.

(a) Compute the volume of the cube in the deformed configuration.

F ~z1

1

1+z2 0

z2 1+z1 00 0

detF = 1+z1+z2

v = 10

10

10

1+z1+z2 dz1dz2 dz3 = 2

(b) Find the area in the deformed configuration of the face with normal e1 in the un-deformed configuration. The unit normal of shaded face is N= e1 with z1 = 1 on thatface

F−TN = 12+z2

−z22 0

−1 1+z2 0

0 0 2+z2

0

0

1= 1

2+z2 --1

0

2

Therefore, detF F–TN = 5 on this face. The deformed area is then

a = 10

10

det F F–TN dz2dz3 = 5

63. A thin flexiblewire of initial length ℓ, originally ori-ented along the z3 axis, is wrapped around ahub (withnegli-gible friction between the wire and the hub). The deforma-tion map that accomplishes the motion is given by

φ(z)= sinαz3e1+ cosαz3e2+βz3e3where α and β are known constants. What is the radius of the hub? Howmany times doesthe wire wrap around it? What is the spacing between adjacent passes of the wire? Whatis the length of the wire after it is wrapped?

(a) What is the radius of the hub? How many times does the wire wrap around it?What is the spacing between adjacent passes of the wire? The magnitude of


sinαz3e1+ cosαz3 e2 is 1, therefore the radius of hub is 1. The wrapping goes ac-cording to a periodic function. The original distance required to make one cycle isz3 = 2π∕α. The total number of wraps is the initial length ℓ divided by the distanceper wrap. Therefore, the number of wraps N is given byN= ℓα∕2π. When∆ z3= 2π∕α the wire makes a complete wrap. In this distance, the x3 coordinate ad-vances by β∆ z3= 2πβ∕α, which is the spacing between adjacent wires.

(b) What is the length of the wire after it is wrapped? The deformed length of the wirecan be obtained in a manner similar to Problem 61

F ~α cosαz3

0 0

−α sin αz30 0

0 0C ~

0 00 0

0 0

0

0

α2+β 2β

λ2(e3) = e3 ⋅ Ce3 = α2+β 2

L=ℓ0

λ(e3)dz3 = ℓ α2+β 2

64. The displacement map of a certain solid body can be expressed as follows:

u(z) = αz2 z3e1+ αz1 z3e2+ αz1 z2e3

where α is a constant. Compute the deformation gradient of the motion. Find the principalstretches at the point originally located at z= 0, 0, 1, in terms of α. Is n~ (1, 1, 0)a principal direction for the specified motion? Find the principal (Lagrangian) strains atz= 0, 0, 1 in terms of α.

(a) Compute the deformation gradient of the motion.

1 αz3 αz2αz3 1 αz1αz2 αz1 1

∇u~ α F= I+∇u~0 αz3 αz2αz3 0 αz1αz2 αz1 0

(b) Find the principal stretches at the point originally located at z= 0, 0, 1, interms of α.

2α 1+α2 0C= FTF= =1+α2 2α 0

0 0 1

1 α 0α 1 00 0 1

1 α 0α 1 00 0 1

Solving the characteristic equation of C

x1

x2

z1

z2

z x

φ(z)


detC− mI = 1+ α2− m2− 4α21− m = 0

⇒ 1− m = 0 ⇒ m1= 1

1+ α2− m= 2α ⇒ m2,m3= α2 2α+ 1= (α 1)2

(c) Is n~ (1, 1, 0) a principal direction for the specified motion?

1+α2 2α 02α 1+α2 0 = α2+ 2α+ 1=0 0 1

110

1+α2+2α2α+1+α2

0

110

Therefore n is an eigenvector and the corresponding eigenvalue is α2+ 2α+ 1.

(d) Find the principal (Lagrangian) strains at z= 0, 0, 1 in terms of α. The princi-pal values of strain are the eigenvalues of E and can be computed from the eigenvaluesof C as follows

γi=12(m i− 1)

Thus,

γ1 = 0

γ2, γ3 =12α2 2α+ 1− 1 = 1

2(α2 2α)

65. The expansion of a hollow sphere can bedescribed by the deformation map

φ(z) = λ(r) z

where z is the position vector of a point in theundeformed configuration and λ(r) is a givenfunction of the radial distance r(z)≡ z ⋅ z .Compute the deformation gradient F for themap. Compute the stretch through the thickness of the sphere in terms of λ, r, and dλ∕dr.

(a) Compute the deformation gradient F for the map.

F= ∇φ= λ′( r) ∇r x+ λ(r) ∇x

∇xij=∂xi∂xj= δij⇒ ∇x= I

P

z1

z2

z3m1

m2

UndeformedConfiguration


∴ F= λ′(r)r x x+ λ(r) I

(b) Compute the stretch through the thickness of the sphere in terms of λ, r, anddλ∕dr.

Fn= λ′(r)r x x xr+ λIxr =

λ′(r)r2x ⋅ x x+ λr x= λ′ + λr x

where λ′(r)≡ dλ∕dr . Let us denote the stretch by m (since λ is already used).

m2 = Fn 2 = Fn ⋅ Fn= λ′ + λr2

x ⋅ x= λ′ + λr2

r2

⇒ m= λ′r+ λ

66. The Green deformation tensor that results fromdeforming the body shown through a deformationmap φ(z) has the following components relative tothe standard basis at the point P:

C(P)~1.0 0.2 0.50.2 3.0 0.20.5 0.2 2.0

Find the stretch of a line oriented in the direction of the vector m1 = (1,1,1) at the pointP. Find the angle, after deformation, between two lines with tangent vectors m1 = (1,1,1)and m2 = (0,1,1) in the undeformed configuration at the pointP. Is the vector m1 = (1,1,1)an eigenvector of the tensor C at the point P?

(a) Find the stretch of a line oriented in the direction of the vector m1 = (1,1,1) at thepoint P. First we need to normalize m1 to unit length

m1 =13(1, 1, 1)

λ2m1 = m1 ⋅ Cm1 =

13(1, 1, 1) ⋅ (1.7, 3.4, 2.7)= 7.8

3

⇒ λm1 = 1.612

(b) Find the angle, after deformation, between two lines with tangent vectors m1 =

(1,1,1) and m2 = (0,1,1) in the undeformed configuration at the point P.

cos θ(Fm1,Fm2)=m2 ⋅ Cm1

λ(m1) λ(m2)

z1

z24 in.

3 in.

1 in.

line


m2 ⋅ Cm1 =1213(0, 1, 1) ⋅ (1.7, 3.4, 2.7)= 2.4903

λ2m2 = m2 ⋅ Cm2 =

12(0, 1, 1) ⋅ (0.7, 3.2, 2.2)= 2.7⇒ λm2

= 1.6432

cos θ(Fm1,Fm2)=2.4903

(1.612) (1.6432)= 0.9399 , θ= 0.348 rad

(c) Is the vector m1 = (1,1,1) an eigenvector of the tensor C at the point P? SinceCm1 = (1.7, 3.4, 2.7)∕ 3 ≠ m (1, 1, 1) where m is a scalar, then m1 is not an eigen-vector of C.

67. A 4 by 3 by 1 in. block of material is scribed with a straightline from corner to corner on one of its broad faces as shown. Theblock is then subjected to a deformation described by the follow-ing map:

φ(z)= z1+0.2z2 e1+ z2+0.3z3 e2+ z3+0.1z1 e3Compute the length of the line in the deformed configuration.Compute the Lagrangian strain of the line due to the motion. Compute the Lagrangianstrain tensor E associated with the motion. Compute the volume of the block in the de-formed configuration.

(a) Compute the length of the line in the deformed configuration.

F= ∇φ~1.0 0.2 0.00.0 1.0 0.30.1 0.0 1.0

2.2--40.8

n~ 15

3--40

Fn~ 15

λ2(n)= Fn ⋅ Fn= 125(2.2)2+ (4)2+ (0.8)2 = 0.8375⇒ λ(n)= 0.914

Since the deformation is homogeneous l = λlo= 4.575 in.

(b) Compute the Lagrangian strain of the line due to the motion.

E(n)= 12λ2(n)− 1 = 1

20.8372− 1 = − 0.0816

(c) Compute the Lagrangian strain tensor E associated with the motion.

E= 12FTF− I ~ 1

2

0.01 0.20 0.100.20 0.04 0.300.10 0.30 0.09


(d) Compute the volume of the block in the deformed configuration.

v=B

det FdV= detFB

dV= det F * V= 1.006(12)= 12.072 in3

68. The components of the deformation tensorC at a certain point in a solid body, relativeto the basis e1, e2, e3 , are given as

C~ 110

11 --1 0--1 11 00 0 10

Compute the eigenvalues and eigenvectors ofC. What is the direction in which the stretchof the body is greatest at the given point? What is the magnitude of that stretch? What isthe ratio of deformed volume to undeformed volume in the neighborhood of the point?

(a) Compute the eigenvalues and eigenvectors of C. What is the direction in whichthe stretch of the body is greatest at the given point? What is the magnitude of thatstretch?

detC− mI = det11∕10--m --1∕10 0--1∕10 11∕10--m 00 0 1− m

= 0

⇒ (1− m)1110− m2− 1

100 = 1− m 1

100121− 220m+ 100m2− 1 = 0

⇒ 110(1− m)10m2− 22m+ 12 = 0

∴ m1 = 1, m2= 1, m3=65

Stretch is greatest in direction n3 where λ2(n3)= 6∕5⇒ λmax= λ(n3)= 1.095.

(b) What is the ratio of deformed volume to undeformed volume in the neighborhoodof the point?

dv= det F dV

The determinant of F can be found from the determinant of C by noting that

C= FTF⇒ detC= detFTF = detFT detF = detF2

z1

z2

1

z3

2

3


∴ det F= detC

From the spectral decomposition of C we can find detC= m1 m2 m3 = 6∕5

⇒ dvdV= detF= 6

5 = 1.095= const ⇒ v

V= 1.095

69. A right tetrahedral block of material, with edges of length 1,2, and 3 along the coordinate axes, is subjected to a deformationdescribed by the following map:

φ(z) = 6z1e1+ 3z2e2+ 2z3e3

Find the volumes of undeformed and deformed bodies. Find theareas of the four faces in the deformed and undeformed configura-tions. Compute the principal stretches and principal directions.Compute the volume of the block in the deformed configuration.

(a) Find the volumes of undeformed and deformed bodies. The volume of the unde-formed body is

V= 16(1)(2)(3)= 1

The deformed volume can be computed using dv= det F dV

F~6 0 00 3 00 0 2

detF= (6)(3)(2)= 36 v= (detF) V= 36

(b) Find the areas of the four faces in the deformed and undeformed configurations.First compute undeformed areas. Let Ai be the undeformed area of the surface withnormal −ei.

A1 =12(2)(3)= 3 A2=

12(1)(3)= 3

2A3 =

12(1)(2)= 1

The undeformed area of the oblique face can be computed from the cross product ofany two of the three vectors defining the edges to that face

AN= (0,− 2, 3)× (1,− 2, 0)= (6, 3, 2)

⇒ AN=12 AN = 7

2

And the normal to that face is N~ 176e1+3e2+2e3 .

z1

z2z3

π

π

0


Now compute the deformed areas.

aN= Ω

JF−TN dAN

JF−T~ 36

1/6 0 00 1/3 00 0 1/2

6 0 00 12 00 0 18

=

0120

⇒ a2 = 12 A2= 18For N=−e2, JF−TN~

--600

⇒ a1 = 6 A1= 18For N=−e1, JF−TN~

36/736/736/7

⇒ aN=367

3 AN= 10 3JF−TN~

--600

⇒ a3 = 18 A3= 18For N=−e1, JF−TN~

For N=6/73/72/7

(c) Compute the principal stretches and principal directions.

C= FTF~36 0 00 9 00 0 4

By Gershgorin’s theorem

m1 = 36⇒ λ1 = 6; n1 = e1

m2 = 9⇒ λ2 = 3; n2 = e2

m3 = 4⇒ λ3 = 2; n3 = e3

(d) Compute the volume of the block in the deformed configuration. Note, this wascomputed in the first part of the problem.

70. A thin square plate of dimension π (the number3.14...) and thickness t is subjected to the deformation

+ z2e2+ z3+β sin z1 e3

φ(z) = z1−β z3 cos z1 e1

n

a b


where ei is the ith base vector in the deformed configuration and β≪1 (very smallcompared to 1) is a constant that describes the motion. Compute the strain tensor associat-ed with the map (you can neglect all terms of order β2 and higher). Where is the strain thegreatest? Sketch the deformed shape of the plate.

(a) Compute the strain tensor associated with the map (you can neglect all terms oforder β2 and higher). Where is the strain the greatest? Since β << 1 the linearizedstrain is sufficient E= 1

2∇u+∇uT

u= φ(z)− z=− β z3 cos z1 e1+ β sin z1e3

--β cos z10∇u~

00

000β z3 sin z1

β cos z10⇒ E~

00

000β z3 sin z1 0

0

The strain is greatest at z3 = t and sin z1= 1⇒ z1= π∕2 giving E11 = β t.(b) Sketch the deformed shape of the plate.

x1

x3

0π

β

71. The unit cube shown is subjected to a homogeneous de-formation (i.e., the deformation gradient is constant). The de-formation tensor C is given by

C ≡ γ I− n n

where γ is a constant that characterizes the deformation and nis a unit vector normal to one of the faces of the cube, as shownon the sketch. Find the principal stretches associated with this state of deformation. Findthe stretch λ and the (scalar) Lagrangian strain E of the line ab. What is the smallest pos-sible value of the constant γ for which the deformation is physically reasonable? Explainwhy smaller values are not possible. If detC= 1 then the volume of the deformed cubeis the same as the volume of the undeformed cube. For what value of γ is the volume un-changed?

(a) Find the principal stretches associated with this state of deformation.

z2

z3

z12

2

2

a

b


C= γI− n n= γ I− n n + γ− 1 n n

From the spectral decomposition theorem, we have one distinct eigenvalue and a pairof repeated eigenvalues m1 = γ− 1⇒ λ1 = γ− 1 ; n1= n. The other two aregiven by m2 = m3 = γ⇒ λ2 = λ3= γ ; n2, n3 are any two vectors in the plane⊥ n. Also,

Cn= γ n− n ⋅ n n= γ−1 n

Cm= γ m− m ⋅ n n= γ m if m⊥ n

(b) Find the stretch λ and the (scalar) Lagrangian strain E of the line ab. First we de-fine a unit vector along the line ab in the undeformed configuration

nab =12m+n , where m⊥ n and m is a unit vector

Note that nab ⋅ n= nab ⋅m= 1∕ 2 due to the orthonormality of m and n. Thus,

Cnab = γ nab− nab ⋅ n n= γ nab− n

λ2(nab)= nab ⋅ Cnab = γ nab ⋅ nab− nab ⋅ n2 = γ− 1∕2

λ(nab)= γ−1∕2 , E(nab)=12λ2−1 = 1

42γ−3

(c) What is the smallest possible value of the constant γ for which the deformation isphysically reasonable? Explain why smaller values are not possible. γ> 1 because ifγ≤ 1 lines along n deform from finite length to zero length.

(d) If detC= 1 then the volume of the deformed cube is the same as the volume ofthe undeformed cube. For what value of γ is the volume unchanged?

detC= γ2 γ−1 = 1⇒ γ3− γ2− 1= 0

Solving this equation by Newton’s method leads to γ= 1.4656

72. A 2 by 2 by 2 unit solid cube is subjected to the deformationdescribed by the map (the center of the block is at the origin ofcoordinates):

φ(z) = z1 1+αz1 e1+ z2 1+αz1 e2+ βz3e3Compute the values of the constants α and β that are consistentwith the observation that the total volume of the block is un-changed by the deformation. Compute the length of the line ab in the deformed configura-tion. Compute the Lagrangian strain tensorE associated with the motion. Compute the de-formed area of the side with original normal e1.


(a) Compute the values of the constants α and β that are consistent with the observa-tion that the total volume of the block is unchanged by the deformation.

1+z1 0 0

z2 1+z1 0

0 0 β

F= ∇φ~

detF= 1+z1 2β

The volume of the undeformed block is V = 8. After deformation, the volume of theblock is

v=V

det F dV= 1−1

1−1

1−1

β 1+z1 2 dz1dz2dz3

= 4β1−1

1+2z1+z21 dz1= 4βz1+z21+ 13 z

311--1= 32

3β

If v= V⇒ β= 4∕3.(b) Compute the length of the line ab in the deformed configuration. Letn= −e1+e2∕ 2 be a unit vector pointing in the direction of ab in the undeformedconfiguration

−(1+z1)−z2+(1+z1)

0

Fn~ 12

−(1+z1)1+2 z1

0

= 12

given that z2 =−z1 along ab.

λ2(n)= Fn ⋅ Fn= 121+z12+1+2 z12 = 1

22+6 z1+5 z21

Therefore the length after deformation is

l =1--1

λ(z1)dz1 =1

21--1

2+6 z1+5 z21 dz1 = 2.3144

(c) Compute the Lagrangian strain tensor E associated with the motion.

1+z1 2+z22 z2 1+z1 0

0 0 β2z2 1+z1 1+z1 2 0C= FTF~

⇒ E~ 12

z1 2+z1 +z22 z2 1+z1 0

0 0 β2−1z2 1+z1 z1 2+z1 0

z1

z2

z3

z1

z2

hB

1

2


(d) Compute the deformed area of the side with original normal e1.

1+z1 −z2 0

0 1+z1 0

0 01+z1 2β

F−T~ 11+z1 2

J= det F= β 1+z1 2

Given that z1 = 1 for the side with original normal e1

⇒ JF−Te1 = β1+z1 = 2β

Therefore, the area after the deformation is a= 2β A= 8.

73. Acircular cylinderwith initial inside radius of 1 and outside ra-dius of 2 is subjected to a deformation with displacement map

u(z) = z1e1+z2e2 ln z21+z22

where ln(⋅) indicates the natural logarithm of (⋅). Find the de-formation gradient F for the given motion. Compute the stretch ofthe cylinder in the radial direction. Compute the Lagrangian strainof a line in the radial direction. What are the height, inside radius,and outside radius of the cylinder after the deformation?

(a) Find the deformation gradient F for the given motion. Let r2 ≡ z21+z22 so that

d ln r2dzi

= 1r2dr2dzi= 2

zir2

∇u~ ln r2+2z2

1

r22z1z2r2

0

0 0 0

2z1z2r2

ln r2+2z2

2

r20

We can define the unit vector n= 1rz1e1+z2e2 so that

∇u= ln r2 I− e3 e3 + n n

F= I+∇u= I+ ln r2 I− e3 e3 + n n

(b) Compute the stretch of the cylinder in the radial direction.


Fn= n+ ln r2 n− n ⋅ e3 e3 + n ⋅ n n= 3+ ln r2 n

λ2(n)= Fn ⋅ Fn= 3+ ln r2 2 ⇒ λ(n)= 3+ ln r2

(c) Compute the Lagrangian strain of a line in the radial direction.

E(n)= 12λ2−1 = 1

23+ ln r2 2− 1

(d) What are the height, inside radius, and outside radius of the cylinder after the de-

formation? Fe3 = e3 λ2(e3)= e3 ⋅ e3 = 1 the height does not change

rinside= 1

0

λ(n) dr= 10

3+ ln r2 dr= 1

Thus, the inside radius does not change

routside= 2

0

λ(n) dr= 20

3+ ln r2 dr= 4.7726

74. Consider a deformation map φ(z) given by the explicit expression

φ(z) = 1+ε z ⋅ z zCompute the deformation gradientF of the givenmotion. Compute the stretch in the radialdirection (i.e., in the direction z). Compute the Lagrangian strain tensor E for the givenmotion. Is the direction z an eigenvector of E or not?

(a) Compute the deformation gradient F of the given motion. Let r2 ≡ z ⋅ z= zizi sothat

∇r2 = ∂r2

∂zjej= 2 zi

∂zi∂zj

e j= 2zir2e i= 2

r2z.

Also, note that ∇z= I so that

F= ∇φ= 1+εr2 I+ 2ε z z

(b) Compute the stretch in the radial direction (i.e., in the direction z). The unit vectorin the direction z is n= z∕r. Thus,

Fn= 1r 1+εr2 z+

2εr r2 z= 1r+3ε r z

λ(n)= Fn = 1r+3εr z = 1+3εr2


(c) Compute the Lagrangian strain tensor E for the given motion.

C= FTF= 1+εr2 I+ 2ε z z 1+ε r2 I+ 2ε z z

= 1+εr2 2 I+ 4ε 1+εr2 z z+4ε2z zz z

= 1+εr2 2 I+ 4ε+8ε2 r2 z z

E= 12C− I = 1

2 1+εr2 2 I+ 4ε+8ε2 r2 z z − I

= εr2+ 12ε2 r4 I+ 2ε+4ε2 r2 z z

(d) Is the direction z an eigenvector of E or not?

Ez= ε r2+ ε22r4 z+ 2ε+4ε2 r2 z z z = 3εr2+ 9

2ε2r4 z

Therefore z is an eigenvector of E and the corresponding eigenvalue is 3εr2+ 92ε2r4

75. Aspherical shell in the undeformed configuration has an inside radius of R and an out-side radius of 2R. The shell is subjected to a deformation described by the followingmap:

φ(z)= 1+α4R2−z ⋅ z z

where α is a given constant of themotion and z is the position vector of a point in the unde-formed configuration. Find the displacement of the point originally located at z=(0, 0,R)? Compute the deformation gradient F of the motion. What is the change in thicknessof the shell? How much does the inside surface of the shell stretch? (Note: the stretch isthe same in all directions because of the spherical symmetry).

(a) Find the displacement of the point originally located at z=(0, 0, R)?

u= φ− z= α4R2−z ⋅ z z

At z=(0, 0, R) we have z ⋅ z= R2

⇒ u(0, 0,R)= α4R2−R2 R e3 = 3αR3 e3

(b) Compute the deformation gradient F of the motion. Let r2 ≡ z ⋅ z= zizi so that

∇r2 = ∂r2

∂zjej= 2 zi

∂zi∂zj

e j= 2 zi e i= 2 z.

⇒ F= ∇φ= 4R2−z ⋅ z I− 2α z z

z1

z2

z1

z2

B

12

z3


(c) What is the change in thickness of the shell?

u(0, 0, 2R)= α(4R2− 4R2) 2R e3 = 0

⇒ ∆t= u(0, 0, 2R)− u(0, 0, 2) ⋅ e3 =− 3αR3

(d) How much does the inside surface of the shell stretch? (Note: the stretch is thesame in all directions because of the spherical symmetry). On the inside surfacez ⋅ z= R2 and z points in the radial direction. Therefore any vector m⊥n is tangentto that surface. Let m = 1

Fm= 1+ 3α R2 m

λ(m)= Fm = 1+ 3α R2

76. A sphere (exploded view shown in sketch) with initial insideradius of 1 and outside radius of 2 is subjected to a deformationwitha radially symmetric displacement map given by

u(z)= βz lnz ⋅ z

where z is the position vector and ln(⋅) indicates the natural loga-rithm of (⋅). Find the deformation gradient F for the givenmotion.Compute the stretch of the sphere in the radial direction. What isthe inside radius and the outside radius of the sphere after the de-formation? Compute the stretch of the sphere in any direction per-pendicular to the radial direction and evaluate that stretch at thesurface.

(a) Find the deformation gradient F for the given motion. Let r2 ≡ z ⋅ z so that

d ln r2dzi

= 1r2dr2dzi= 2

zir2

⇒ F= I+∇u= 1+ β ln r2 I+ 2βr2

z z

= 1+ β ln r2 I+ 2β n n

where n= z∕r is a unit vector pointing in the radial direction

(b) Compute the stretch of the sphere in the radial direction.

Fn= 1+ β ln r2 n+ 2βn ⋅ n n= 1+ β ln r2+ 2 n

z1

z2

θ


λ(n)= Fn = 1+ β ln r2+ 2

(c) What is the inside radius and the outside radius of the sphere after the deforma-tion?

rinside= 1

0

λ(n)dr= 10

1+ β ln r2+ 2dr= 1

Therefore, the inside radius does not change. The outside radius is

routside= 2

0

λ(n)dr= 20

1+ β ln r2+ 2dr= 2+ 2.7726 β

(d) Compute the stretch of the sphere in any direction perpendicular to the radial di-rection and evaluate that stretch at the surface. For any m⊥n

Fm= 1+ β ln r2 n+ 2βm ⋅ n n= 1+ β ln r2 n

λ(m)= Fm = 1+ β ln r2

At the surface r = 2, λ(m)= Fm = 1+ β ln 4

77. A circle of unit radius is etched on a plate. The plate is thensubjected to a homogeneous deformation that stretches accordingto the following map:

φ(z)= 2z1e1+ z2e2+ z3e3

Find the expression for the stretch of the line under the deforma-tion map (as a function of θ). Find the length of the etched line inthe deformed configuration.

(a) Find the expression for the stretch of the line under the deformation map (as afunction of θ).

2 0 00 1 00 0 1

F=

Let n=− sin θ e1+ cos θ e2 be a unit vector tangent to the line

4 0 00 1 00 0 1

cos θ− sin θ

0

λ2(θ)= n(θ) ⋅ Cn(θ)= − sin θ cos θ 0

R

ℓφ(z)

z1

z2z3


λ2(θ)= 4 sin2 θ+ cos2 θ= 1+ 3 sin2 θ ⇒ λ(θ)= 1+ 3 sin2 θ

(b) Find the length of the etched line in the deformed configuration.

l =2π0

λ(θ) Rdθ= 2π0

1+ 3 sin2 θ dθ= 9.688

78. A circular cylinder of length ℓ and radius R experiencesthe deformation characterized by the following map:

φ(z) = αz1e1+ βz2e2+ γz3 e3where α, β, and γ are constants of the motion. Find the vol-ume of the deformed cylinder. Find the total surface area ofthe deformed cylinder. Find theprincipal stretchesof themo-tion. What are the limits on the constants α, β, and γ?

(a) Find the volume of the deformed cylinder.

α 0 0

0 β 0

0 0 γ

F=

So that J= det F= α β γ. Since J is constant, v= J V= αβγπR2ℓ.(b) Find the total surface area of the deformed cylinder. The total area of the de-formed cylinder is the sum of the areas of the top, bottom and side surfaces of the cyl-inder after deformation. For each of these surfaces, da= JF−TN dA, where

1∕α 0 0

0 1∕β 0

0 0 1∕γ

β γ 0 0

0 α γ 0

0 0 α β

J F−T=F−T=

For the top and bottom surfaces

N= e3⇒ JF−TN = αβ; ae3 = αβ Ae3 = αβ πR2

For the side surface

Ns= cos θ e1+ sin θ e2⇒ JF−TN = βγ cos θ e1+αγ sin θ e2;

atotal= 2π

0

l0

(βα)2 cos2 θ+ (αγ)2 sin2 θ R dθ dz3+ 2αβ πR2

x2

x3

x1

z1

z2

R

R

z2

z3

z1

R

r


(c) Find the principal stretches of the motion.

α2 0 0

0 β2 0

0 0 γ2C= FTF=

By the spectral decomposition theorem, the principal stretches are α, β, and γ.

(d) What are the limits on the constants α, β, and γ?

α> 0 β> 0 γ> 0

79. Consider a thin (i.e., it has essentially no thickness in thez3 direction) circular membrane of radius R initially lying inthe z1--z2 plane as shown in the sketch. Under pressure themembrane deforms into a bubble according to the followingdeformation map

φ(z) = z1e1+z2e2+ β cosπ z21+z22 ∕2R e3where β is a known constant and R is the radius of the circle.Compute the deformation gradient of the given map. Com-pute the stretch in the initial radial direction (i.e., the direc-tion of the vector r= z1e1+z2e2). Also compute the stretchin the direction that is in the initial plane of the membranebut is orthogonal to r (i.e., tangent to a circle centered at theorigin). Are these two directions principal directions? Whyor why not? What is the deformed length of the line that wasthe radial line from the origin to the edge of the circle alongthe z1 direction in the undeformed configuration? What is the slope of the membrane atthe edge after deformation?Note: The stretch through the thickness of the membrane is zero, but that is acceptable be-cause we are assuming that the thickness is very small compared to the diameter of themembrane.

(a) Compute the deformation gradient of the given map.

F = ∇φ~ 10

αz1∕r

01

αz2∕r

000

where α= α(r) is defined as follows. Let r= z21+ z22 and compute the partial de-

rivatives


∂r∂z1=

z1r ,

∂r∂z2=

z2r

∂∂ziβ cos πr

2R = −

πβ2R

sin πr2R

zir = α(r)

zir for i= 1, 2

where

α(r) = −πβ2R

sin πr2R

(b) Compute the stretch in the initial radial direction (i.e., the direction of the vectorr= z1e1+z2e2). Also compute the stretch in the direction that is in the initial plane ofthe membrane but is orthogonal to r (i.e., tangent to a circle centered at the origin).Are these two directions principal directions? Why or why not?

C = FTF = I+ α2n n

where n= 1rz1e1+z2 e2. Note that n is the unit vector in the direction r.

Cn = I+α2n n n= n+α2n ⋅ n n= 1+α2n

⇒ n ⋅ Cn= 1+ α2, λ(n)= 1+ α2

Let m= 1r−z2e1+z1 e2 ⇒ m ⋅ n= 0

Cm = I+α2n n m= m+α2n ⋅m n= m

⇒ m ⋅ Cm= 1, λ(m)= 1

Therefore m and n are the eigenvectors of C.

(c) What is the deformed length of the line that was the radial line from the origin tothe edge of the circle along the z1 direction in the undeformed configuration?

l = R0

λ(n)ds = R0

1+ α2(r) dr

(d) What is the slope of the membrane at the edge after deformation? The slope is theangle between n and Fn

cos θn, Fn = n ⋅ Fnλ(n)

= 1λ(n)= 1

1+α2

θn

Fn

z1

z2

z3

B

Ω

n


80. Consider the deformationmap defined on a sphere of unit radius

φ(z) = (1+ε) z− β (n ⋅ z) nwhere ε and β are known (small) constants of the motion and n is aknown constant direction. Compute the ratio of the volume of thesphere after deformation to the volume of the sphere before de-formation. Compute the surface area of the sphere after deformation. Whatmakes this cal-culation complicated? Is the deformed area larger or smaller than the original area?Whatis the stretch of the sphere in the radial direction? What is the radius of the sphere afterthe deformation?

(a) Compute the ratio of the volume of the sphere after deformation to the volume ofthe sphere before deformation. To compute the ratio of the volumes we need the dete-minant of the deformation gradient F

φi= (1+ ε )zi− εnk zkni

∂φi∂zj= (1+ ε )δij− εnkniδkj= (1+ ε )δij− εnknjδkj

⇒ F= (1+ ε )I− εn n

The eigenvalues of F can be easily determined. Note that Fn= γn. First consider m =n as a possible eigenvector

Fn= (1+ ε )n− ε n ⋅ n n= 1 n⇒ γ1 = 1

Now consider m⊥n as a possible eigenvector

Fm= (1+ ε )m− ε m ⋅ n n= (1+ ε ) n⇒ γ2= γ3 = ( 1+ ε )

Therefore

det (F)= (1+ ε )2 v=B

det (F)dV= det (F)dV vV= (1+ ε )2

(b) Compute the surface area of the sphere after deformation. What makes this cal-culation complicated? Is the deformed area larger or smaller than the original area? Tocompute the deformed area we must compute

a=Ω

det (F) ‖ F−Tm ‖ dA

The normal vector in the undeformed configuration is simply m= z. Computing theinverse transpose of F is not particularly difficult. Executing the integral over the sur-face would be done in spherical coordinates, which may be difficult. The deformedarea would be larger than the undeformed area because all of the principal stretches arelarger than or equal to 1.


(c) What is the stretch of the sphere in the radial direction? What is the radius of thesphere after the deformation? Let m= z∕‖ z ‖= z∕r be a unit vector in the radialdirection

Fm= (1+ ε )m− ε m ⋅ n n= (1+ ε ) n⇒ γ2= γ3 = ( 1+ ε )

λ2 = Fm ⋅ Fm= (1+ ε )m− ε m ⋅ n n ⋅ (1+ ε )m− ε m ⋅ n n

(1+ ε )2m ⋅m− 2ε 1+ ε n ⋅m+ ε2n ⋅ n

(1+ ε )2+ ε2− 2ε 1+ ε n ⋅m

λ(z )= (1+ ε )2+ ε2− 2ε 1+ ε n ⋅m (z )

The radius of the sphere after deformation is

Rdef= 1

0

λ( z )dr=10

(1+ ε )2+ ε2−2εr 1+ ε n ⋅ z

dr

where

r=‖ z ‖= z ⋅ z

Chapter 3The Transmissionof Force

81. The stress tensor S at a certain point in a body has components with respect to a setof coordinate axes x1, x2, x3 of

S~5 3 --83 0 --3--8 --3 11

On a plane whose normal n makes equal acute angles with the coordinate axes, find thetraction vector tn, the component of the traction vector that is normal to the plane, and theshearing component of the traction vector.

The cosine of the angles between the normal vector and the base vectors must all bethe same. Hence, cos θ= n ⋅ ei= ni, where ni is the i--th component of the vector n.Hence unit normal vector has components n= 1

3(1, 1, 1). The traction can be com-

puted as

tn = Sn = 1

3

5

3

–8

3

0

–3

–8

–3

11

1

1

1

=0

0

0

indicating that the surface is traction free.

σ = tn ⋅ n = 0

τ = tn 2− σ 2 = 0


82. Resolve Problem 81 with S22 changed to 10 3 .

tn = Sn = 1

3

5

3

–8

3

10 3

–3

–8

–3

11

1

1

1

=0

0

10

σ = tn ⋅ n = 10

3

τ = tn 2− σ 2 = 10 23

83. Find the principal values and principal directions of the two stress tensors havingcomponents with respect to the standard basis of

S~3 1 21 --6 02 0 15

S~20 --5 0--5 --10 00 0 0

(a) The first stress tensor. The principal values can be obtained by solving the charac-teristic equation with coefficients IS = 12, IIS =−68, and IIIS =−261. Thus,

−m3+ 12m2+ 68m− 261 = 0

The roots are m1 =−6.1121, m2 = 2.7863, and m3 = 15.3258. The correspondingprincipal directions are

n1 = 0.1114,−0.9937,−0.0105

n2 = 0.9807, 0.1116,−0.1606

n3 = 0.1608, 0.0075, 0.9870

(b) The second stress tensor. The principal values can be obtained by solving thecharacteristic equation with coefficients IS = 10, IIS =−225, and IIIS = 0 as above.Note that, by Gershgorin’s theorem one of the eigenvalues must be zero because theoff-diagonals corresponding to the zero diagonal are all zeros. Thus, the characteristicequation can be factored and reduced to a quadratic

−m m2−10m−225 = 0

The roots are m1 =−10.8114, m2 = 0, and m3 = 20.8114. The corresponding prin-cipal directions are

Chapter 3 The Transmission of Force 79

n1 = 0, 0, 1

n2 = 0.1602, 0.9871, 0

n3 = −0.9871, 0.1602, 0

84. The condition called plane stress is characterized by the stress state S33 = S23 =S13 = 0. Show that if the remaining stress components are given by

S11 =∂2ψ(x1, x2)∂x2

2

, S22=∂2ψ(x1, x2)∂x2

1

, S12 =−∂2ψ(x1, x2)∂x1∂x2

and the body forceb= 0, then the equations of equilibrium are satisfied for any sufficient-ly smooth function ψ(x1, x2). How smooth must the function be?

The components of the stress tensor should satisfy the equilibrium equations

S11,1+ S12,2 = 0

S21,1+ S22,2 = 0

∂3ψ(x1, x2)∂x1∂x22

− ∂3ψ(x1, x2)∂x1∂x22

= 0

∂3ψ(x1, x2)∂x2

1∂x2

− ∂3ψ(x1, x2)∂x2

1∂x2

= 0

⇒

Thus, the equations are automatically satisfied when the stress components are definedin this way. The function ψ(x1, x2) must be sufficiently differentiable to satisfy equilib-rium.

85. The state of stress at a point is characterized by the stress tensor S, given below

S~4 --4 0--4 4 00 0 8

Consider the vectors n and m given by

n = 1

3e1−e2−e3 , m = 1

2 e1+e2

Are the two given vectors n and m eigenvectors of S? Find the principal stresses for thegiven stress tensor S.

(a) Are the two given vectors n and m eigenvectors of S? If n and m are eigenvectorsof S, then Sn and Sm should be proportional to n and m, respectively, i.e., Sn= σ1nand Sm= σ2m where σ1 and σ2 are eigenvalues.

1

3

4 -- 4 1

4 --1-- 4

--100

0

0

8

= 8

3

1

--1

--1

x1

x2

x3

n

42

1


Proportionality holds, so n is an eigenvector. The eigenvalue is σ1 = 8.

1

3

4 -- 4 1

4 1-- 4

000

0

0

8

= 0

1

1

0

Proportionality holds, so m is an eigenvector. The eigenvalue is σ2 = 0.

(b) Find the principal stresses for the given stress tensor S. From part (a), two of theprincipal stresses are already found. They are 0 and 8. One last principal stress can beobtained from the first invariant.

IS = S11+ S22+ S33 = σ1+ σ2+ σ3 = 16

Therefore last one is σ3 = 8.

86. Consider the tetrahedron shown in the figure, with edgesalong the coordinate axes of length 4, 2, and 1, respectively.The state of stress in the tetrahedron is given by the expression

S(x) = So x x

where So is a constant and x is the position vector. The equa-tion of the oblique plane is x1+2x2+4x3= 4. Compute thebody force b required for the tetrahedron to be in equilibrium.Compute the tractions on the four faces of the tetrahedron re-quired for equilibrium.

(a) Compute the body force b required for the tetrahedron to be in equilibrium.

b = − div S = −Sij, j e i

= −Soxixj ,j ei= −Soδij xj+xiδjj ei= −4Soxiei = −4Sox

(b) Compute the tractions on the four faces of the tetrahedron required for equilibri-um. First note that the traction on any face with normal ei is given by

tei = Sei = So x x ei = So x ⋅ e i x = Soxix

On the face with normal −ei the coordinate xi= 0. Therefore, by the above formula,the traction on those faces is identically zero. The vector normal to the oblique face isn = 1

21(1, 2, 4). The traction on that face is


tn = Sn = So x x n = So x ⋅ n x

= So21x1+2x2+4x3x =

4So21

x

where the position vector x is restricted to lie in the plane.

87. Find an expression for the following derivatives of the principal invariants with re-spect to tensor components

∂IS∂Smn

,∂IIS∂Smn

, and∂IIIS∂Smn

The derivatives of the first and second invariant can be easily computed in compo-nents.

∂IS∂Smn

=∂Sii∂Smn

= δimδin = δmn

The second invariant has derivative

∂IIS∂Smn

= 12Sii ∂Sjj∂Smn

+ ∂Sii∂SmnSjj− 2Sij

∂Sij∂Smn

= 12Siiδmn+ Sjjδmn− 2Sijδimδjn

= Siiδmn− Smn

We can compute the derivative of detS= IIIS formally as follows

∂IIIS∂Smn

= 16Á ijkÁlpq

∂SilSjpSkq∂Smn

= 16ÁijkÁlpq ∂Sil∂Smn

SjpSkq+ Sil∂Sjp∂Smn

Skq+ SilSjp∂Skq∂Smn

= 16 ÁijkÁlpq

SjpSkqδimδln+ Sil Skqδjmδpn+ SilSjpδkmδqn

= 16ÁmjkÁnpqSjpSkq+

16ÁimkÁlnqSilSkq+

16ÁijmÁ lpnSilSjp

Alternatively, from Problem 44 (and the Cayley-Hamilton theorem) we have the rela-tionship

det S = 13tr(S3)− 1

2IStr(S

2)+ 16I3S

Noting that ∂(tr(S3))∕∂S= 3S2 and ∂(tr(S2))∕∂S= 2S we can compute the derivativeof detS as follows:

x1x2

x3

m

m

a

bc

d


∂∂Sdet S = S2− ISS−

12tr(S2)I+ 1

2I2S I

= S2− ISS+ IIS I

where we have noted that 2 IIS= I2S−tr(S2). One can see that the component form isthe same as the direct form by multiplying the direct form by S as follows

S ∂∂Sdet S = S3− ISS

2+ IISS = IIIS I

where we have used the Cayley-Hamilton theorem. Multiplying through by the inverseof S we get

∂∂Sdet S = IIISS

−1

Note the similarity with the component form of the derivation.

88. Consider the sphere of radius R shown in thefigure. The state of stress in the sphere is given bythe stress field

S(x) = So m x+ xm

where Sois a constant, x is the position vector ofthe point in question, andm is a constant unit vec-tor field. What is the body force b(x) required for equilibrium? Compute the tractions act-ing on the surface of sphere. Sketch the traction vectors at points a, b, c, and d shown onthe figure (line segment ca points in the direction of m).

(a) What is the body force b(x) required for equilibrium?

b = − div S = −Sij, j e i= −Somixj+ ximj

, j e i= −Somiδjj+ δijmj

ei= −4Somi ei = −4So m

(b) Compute the tractions acting on the surface of sphere. A vector normal to the sur-face of a sphere is a radial vector parallel to the position vector x. Hence,n = x∕ x = x∕R≡ r. The tractions on the sphere are

tn = Sn = So m x+ xm r

= So x ⋅ r m+ m ⋅ r x

= SoR m+ m ⋅ r r

n3

n2

n1

m1

m1 =1

3n1+n2+n3


(c) Sketch the traction vectors at points a, b, c, and d shown on the figure (line seg-ment ca points in the direction of m). At point A, r= m and r ⋅m= 1. At points Band D, r⊥m and r ⋅m= 0. At point C, r=−m and r ⋅m=−1. The tractionvectors are shown on the sketch.

x1

x2

x3

m

m

A

BC

D

SoRm

SoRm

2SoRm

2SoRm

89. Consider a state of stress S that has principal valuesσ1, σ2, σ3with corresponding (orthogonal) principal di-rections n1, n2, n3. Let us consider one of the eight (oc-tahedral) planes whose normal vector mi makes equalangles with the principal directions (one of the eight vec-tors is shown in the sketch). Show that the normal compo-nent of the traction on any of the eight octahedral planes isgiven by σ= σ1+σ2+σ3 ∕3. Show that the shearingcomponent of the traction on any of the eight octahedralplanes is

τ2 = 19(σ1−σ2)2+ (σ2−σ3)2+ (σ3−σ1)2 ≡ τ2oct

Express τoct in terms of the principal invariants of the stress tensor IS and IIS.

(a) Show that the normal component of the traction on any of the eight octahedralplanes is given by σ= σ1+σ2+σ3 ∕3. The traction on the face with normal vectormi can be computed as

σ(i) = mi ⋅ tmi = mi ⋅3

j=1

σj nj nj ⋅mi = 3

j=1

σj nj ⋅m i 2

The only difference among the eight normal vectors is sign in front of each vectorcomponent

mi=1

3γik nk

(sum implied on k), i.e., γik= 1. There are eight choices of the three pairings ofplus and minus values. Therefore, noting the identity nj ⋅m i=

1

3γij

σ(i) = 3j=1

σj nj ⋅m i 2 = 3

j=1

13σj γ

2ij


Thus, for all i (i.e., for any of the eight faces) we have

σ(i) = 13σ1+σ2+σ3

(b) Show that the shearing component of the traction on any of the eight octahedralplanes is

τ2 = 19(σ1−σ2)2+ (σ2−σ3)2+ (σ3−σ1)2 ≡ τ2oct

We can compute the normal traction as

tmi = 3

j=1

σj nj ⋅m i nj =

3

j=1

1

3σjγij nj

We can compute the length of this vector as follows

‖ tmi ‖2= 1

33j=1

3k=1

σjγij nj σkγik nk = 133j=1

σ2j γ2ij =

133j=1

σ2j

Finally, we can compute the square of the shearing component on the face as

τ2 = tmi 2− σ 2

= 13σ21+ σ22+ σ23 − 1

9σ1+σ2+σ3 2

= 192σ21+ 2σ22+ 2σ23− 2σ1σ2− 2σ2σ3− 2σ1σ3

= 19(σ1−σ2)2+ (σ2−σ3)2+ (σ3−σ1)2 ≡ τ2oct

(c) Express τoct in terms of the principal invariants of the stress tensor IS and IIS.First, compute the values of the invariants

I 2S= σ1+σ2+σ3 2 = σ21+ σ22+ σ23+ 2σ1σ2+ 2σ1σ3+ 2σ2σ3

IIS = σ1σ2+ σ1σ3+ σ2σ3

From part (b) we found that

τ2oct = 29 σ21+ σ22+ σ23− σ1σ2− σ2σ3− σ1σ3

= 29 I2S− 3 IIS

Therefore,

τoct = 23

I 2S− 3 IIS

x1

x2

r

σ3

σ1 or σ2


90. A thick-walled sphere of inside radius 1 and outsideradius 2 is subjected to an internal pressure of magnitudep. The principal directions of stress are the radial and tan-gential directions. The principal values of stress are givenby the expressions

σ1 = σ2 =p71+ 4

r3, σ3=

p71− 8

r3

where r is the radial distance to an arbitrary point (withposition vector x) from the center of the sphere, i.e., r2 = x21+x22+x23. Find the expressionfor the stress tensorS in the cartesian coordinate system x1, x2, x3 . Prove that the outsidesurface of the sphere is traction free. Find the body forceb that must be present tomaintainequilibrium.

(a) Find the expression for the stress tensor S in the cartesian coordinate systemx1, x2, x3 . Using the spectral decomposition theorem in the case of a repeated eigen-value

S = σ1 I− n3 n3 + σ3 n3 n3

where n3 = x∕r is the unit radial vector associated with the distinct eigenvalue σ3.Simplifying

S = σ1 I+ (σ3− σ1 )n3 n3

= σ1 I+1r2(σ3− σ1 )x x

(b) Prove that the outside surface of the sphere is traction free. The normal to theoutside surface is n3 with r= 2

Sn3 = σ1n3+ (σ3− σ1 )n3 ⋅ n3 n3

= σ3n3=p71− 8

23n3 = 0

(c) Find the body force b that must be present to maintain equilibrium.

b=− divS=−∂Sij∂xj

ei

Sij = σ1δij+1r2(σ3− σ1 )xixj

∂Sij∂xj= σ′1

∂r∂xjδij+− 2

r3(σ3− σ1 )+ 1

r2(σ′3− σ′1 ) ∂r∂xj xixj

+ 1r2(σ3− σ1 )δij xj+

1r2(σ3− σ1 )xi δjj

n1

n2

n3

mθ

θ

n2

n1

m

n3


where

σ′1 =dσ1dr=− 12 p

7 r4, σ′3 =

dσ3dr= 24 p

7 r4,

∂r∂xj= ∂∂xj

(xk xk )1∕2 = 1

2(xk xk )

−1∕2 (2xk δkj )=1r xj

∂Sij∂xj= 1

r σ′1xi+− 2r3(σ3−σ1 )+ 1

r2(σ′3−σ′1 ) (xjxj )xi+ 4

r2(σ3−σ1 )xi= 0

b=−div S= 0

91. Consider a state of stress S that has prin-cipal values σ1, σ2, σ3 with corresponding(orthogonal) principal directions n1, n2,n3. Let us consider a plane parallel to n3 de-scribed by the normal vector

m≡ cos θn1+ sin θn2parameterized by the angle θ as shown in thefigure. Show, using the spectral decomposition theorem, that the traction vector and itsnormal component, acting on this plane, are given by

tm(θ) = σ1 cos θn1+ σ2 sin θn2, σ(θ) = σ1 cos2 θ+ σ2 sin2 θ

Show that the shear and normal components of the traction vector satisfy the relationship

τ2+σ2 = σ21 cos2 θ+σ22 sin2 θ

Now let

σ≡ 12σ1+σ2 , and ρ≡ 1

2σ1−σ2

Show that the shear and normal components of the traction vector satisfy the relationship

τ2+ σ−σ 2 = ρ2

Note that this problem proves that the shear and normal components of the traction vectoron a plane with any value of θ lies on a circle of radius ρ, centered at (σ,0) in the τ−σplane. This result, discovered by Otto Mohr, is usually called Mohr’s circle and is shownin the figure above. Clearly, the same results hold for all three pairings of the principaldirections.

(a) Show, using the spectral decomposition theorem, that the traction vector and itsnormal component, acting on this plane, are given by

tm(θ) = σ1 cos θn1+ σ2 sin θn2, σ(θ) = σ1 cos2 θ+ σ2 sin2 θ


From the spectral decomposition theorem, the stress tensor can be constructed as

S = 3i=1

σi ni ni

The traction on a face with normal m can be computed as

tm(θ) = Sm = 3i=1

σi ni ni cos θn1+ sin θn2

= σ1 cos θn1+ σ2 sin θn2

= 3i=1

σi cos θδi1ni+ sin θδi2ni

The normal component of the traction vector is

σ(θ) = tm(θ) ⋅m

= σ1 cos θn1+ σ2 sin θn2 ⋅ cos θn1+ sin θn2

= σ1 cos2 θ+ σ2 sin2 θ

(b) Show that the shear and normal components of the traction vector satisfy the rela-tionship τ2+σ2 = σ21 cos2 θ+σ22 sin2 θ. The shear and normal components satisfyτ2+σ2 = tn 2. Thus, from above,

τ2+ σ 2 = tn 2 = σ21 cos2 θ + σ22 sin

2 θ

(c) Now let σ≡ σ1+σ2 ∕2 and ρ≡ σ1−σ2 ∕2. Show that the shear and normalcomponents of the traction vector satisfy the relationship τ2+ σ−σ 2 = ρ2. Theidentity can be shown by the following computation

τ2+ (σ− σ)2 = τ2+ σ2− 2σσ+ σ2

= σ21 cos2 θ+ σ22 sin2 θ− 2 σ1 cos2 θ+ σ2 sin2 θ 12 σ1+σ2 + σ

2

= −σ1σ2 cos2 θ+ σ1σ2 sin2 θ + 14σ1+σ2

2

= 14σ1−σ2 2 = ρ2

92. Prove the identity S ⋅ ∇(h× x)= h ⋅ [e j× Sej ], where h is constant.

First compute the derivative ∇(h× x) as

∇(h× x)= ∂∂xk

(h× x) ek= (h× ek ) ek= [h×] (ek ek)= [h×]

3ℓ4ℓ

x1

x2


where we have noted that (ek ek)= I and that h× is the skew-symmetric tensorassociated with the vector h. Since A ⋅ B= trATB = e i ⋅ ATB e i we can write

S ⋅ ∇(h× x) = S ⋅ [h×] = ei ⋅ ST h×ei= Sei ⋅ h× e i

= h ⋅ ei× Se i

93. A block of material is subjected to a homogeneousstate of stress described by the constant stress tensor withS= 10[ e1 e1 ]−2[e1 e2+e2 e1 ]+5[e2 e2 ].The triangular wedge shown is cut out of the block as afreebody. Compute the tractions thatmust act on each sideof the freebody diagram. Demonstrate that the freebody is in overall equilibrium. Assumethat the block has unit width.

(a) Compute the tractions that must act on each side of the freebody diagram.

n= 45e1+

35e2

−e1

−e2

e3

The normals to the wedge surfaces are shown in the figure. The corresponding tractionvectors are,

t−e1 =−Se1 =−10--20

t−e2 =−Se2 =−--250

tn= Sn=8 -- 6/5--8/5 + 3

0=

34/57/50

There are no tractions on the faces with normal e3.

(b) Demonstrate that the freebody is in overall equilibrium. Assume that the block hasunit width. Since S is constant and in the absence of body forces (b = 0), the equilibri-um equation div S+ b= 0 is satisfied. Also, S= ST by inspection, therefore equilib-rium of angular momentum is satisfied. We further need verify that the total tractionsacting on the wedge surfaces are in equilibrium.

4 t−e1+ 3 t−e2+ 5 tn=--4080+

6--150

+3470

= 0

x1

x2

x3

h

b1

x1

x2

x3

Face 1

Face 3

Face 5

Face 4

Face 2


Therefore the freebody is in overall equilibrium.

94. A triangular prism of material (with base b, height h, andunit thickness) has an internal stress given by the stress field

S(x)= ρbhx1−h bx2+hx1−bh e2 e2

where ρ is the (constant) unit weight of the material and ei is theunit base vector in the direction of the coordinate axis xi . Findthe body forceb required for equilibrium. Find the tractions of all of the faces of the prism.Sketch the normal (σ) and tangential (τ) components of traction on the three faces whosenormals are orthogonal to the e3 direction.

(a) Find the body force b required for equilibrium. For the body to be in equilibrium,the body force vector has to satisfy the equation

div S+ b= 0

or in component form Sij,j+ bi= 0.

S11,1+ S12,2+ S13,3+ b1 = 0 ⇒ b1= 0

S21,1+ S22,2+ S23,3+ b2 = 0 ⇒ b2=−ρhx1−h

S31,1+ S32,2+ S33,3+ b3 = 0 ⇒ b3= 0

Therefore the body force vector required for equilibrium is

b=− ρhx1−h e2

(b) Find the tractions of all of the faces of the prism.

Face 1 (x3 = 1): t1 = Se3 = 0

Face 2 (x3 = 0): t2 =−Se3 = 0

Face 3 (x1 = 0): t3 =−Se1 = 0

Face 4 (bx2+hx1−bh= 0): t4 = Sn4 = 0because S = 0 everywhere on that face.

Face 5 ( x2 = 0): The traction is

t5 =−Se2 =−ρb x1−h x1−b e2

(c) Sketch the normal (σ) and tangential (τ) components of traction on the three faceswhose normals are orthogonal to the e3 direction.

x1x2

x3

B

x1x2

x3

t (traction)


x1

x2

x1

x2

σ τ

0

0

000

−ρh

95. A spherical shell has an inside radius ofR and an outside radiusof 2R. In the center of the sphere there is a magnetic core that setsup a stress field in the shell. The state of stress in the shell is

S(x)=ρ3r1− 8R3

r3x x

where r is a magnetic constant of the material, x is the position vec-tor, and r is the radial distance to the point x defined as

r≡ x ⋅ x (Note that ∂r∕∂xj= xj∕r)

Find the body force vector field b in the shell. What is the pressureat the inside surface of the shell? Take a freebody of the shell byslicing it along the plane x3 = 0. What are the tractions t on theshell that must act at the slice?

(a) Find the body force vector field b in the shell. Let

f (r)=ρ3r1− 8R3

r3

The components of the stress tensors are given by Sij = f (r) xi xj. Then we can com-pute

div S i= Sij,j = f′(r) ∂r∂xj

xi xj+ f (r)xi,j xj+ xi xj,j

= f′(r)xjr xi xj+ f (r)δij xj+ xi δjj

= f′(r)x ⋅ xxir + 4f (r) xi

Thus, divS= r f′(r)+4f (r)x, in which

r f′(r)+ 4f (r)= ρ3r-- 1

r2+ 32R3

r4 + 41r− 8R3

r4 = fr

Therefore, the body force vector required for equilibrium is


b=− div S=− fr x

(b) What is the pressure at the inside surface of the shell? The normal to the insidesurface of the shell is n=−x∕r , with r= R. The traction acting on that surface is

Sn= ρ3R

(−7)[x x]− xR = 7

3ρx

which acts along the normal to the surface, therefore the pressure is

Sn = 73ρR

(c) Take a freebody of the shell by slicing it along the plane x3 = 0. What are thetractions t on the shell that must act at the slice? The normal to the surface along theplane x3 = 0 is n=− e3. The traction acting on that surface is

Se3 = f (r)x xe3= f (r)x ⋅ e3 x

Also, on the slice x3 = 0, x= x1e1+x2e2

⇒ x ⋅ e3 = 0 and t= Se3= 0

96. The stress tensor S can be expressed in cylindrical coordinates (r, θ, z) as

S(r, θ, z)= Srr er er + Srθ er eθ + Srz er ez

+ Sθr eθ er + Sθθ eθ eθ + Sθz eθ ez

+ Szr ez er + Szθ ez eθ + Szz ez ez

where the components (e.g., Srz) are each functions of the coordinates (r, θ, z). However,now the base vectors er (θ) and eθ (θ) depend upon the coordinate θ.

θr

n3

n2

n1

n4∆θ

∆r∆zr∆θ∆r

n4 n2

n5

n6

z

r

zx1

x2

x3 ≡ z

er (θ)

ezeθ (θ)

rz

θ

Using the coordinate-free definition of the divergence of a tensor field, Eqn. (87), showthat the divergence of S in cylindrical coordinates is given by

divS(r, θ, z) = 1r∂∂rrSer + 1

r∂∂θSeθ + ∂

∂zSez


Observe from the figure that n1 = eθ(θ+∆θ) and n2 =−eθ(θ), andare constant overthe faces 1 and 2, respectively. The normal vectors n3 = er (ξ) and n4 =−er (ξ) withξ∈ [θ, θ+∆θ] varying over faces 3 and 4. Finally, note that n5 = ez and n6 =−ez areconstant over faces 5 and 6. The volume of the wedge is ∆V= r∆θ∆r∆z plus terms ofhigher order that vanish more quickly in the limit as ∆V→ 0.

To compute the component expression for the divergence of the stress tensor, we mustexpand the vectors Ser, Seθ, and Sez. Show that

Ser = Srrer+ Sθreθ+ Szr ezSeθ = Srθer+ Sθθeθ+ SzθezSez = Srzer+ Sθzeθ+ Szz ez

Beforewe take derivatives, wemust observe that in terms of the standard constant basise1, e2, e3, the radial and angular base vectors have the form

er (θ) = cos θe1+ sin θe2eθ (θ) =− sin θe1+ cos θe2

and, therefore, ∂er∕∂θ= eθ and ∂eθ∕∂θ=−er. Show that the component expression ofthe divergence of S is

divS = ∂Srr∂r + 1r∂Srθ∂θ +

∂Srz∂z +

1r Srr−Sθθ er

+ ∂Sθr∂r + 1r∂Sθθ∂θ +

∂Sθz∂z +

1r Srθ+Sθr eθ

+ ∂Szr∂r + 1r∂Szθ∂θ +

∂Szz∂z +

1r Szr ez

Begin with Eqn. (87) in text.

+Ω1

S(r, θ+∆θ, z )eθ(θ+∆θ) drdz−Ω2

S(r, θ, z )eθ(θ) drdz

divS = limv(B)→0

1v(B)

Ω3

S(r+∆ r, θ, z )er(θ) (r+∆r)dθdz−Ω4

S(r, θ, z ) er (θ) r dθdz

+Ω5

S(r, θ, z+∆ z )ez rdθdr−Ω6

S(r, θ, z )ez rdθdr

Substituting v(B) = r∆ r∆θ∆ z we get


+ lim∆r∆z→0

1r∆r∆z∆r∆z

S(r, θ+∆θ, z)eθ(θ+∆θ)− S(r, θ, z )eθ(θ)∆θ

drdz

divS(x) = lim∆θ∆z→0

1r∆θ∆zr∆θ∆z

(r+∆ r)S(r+∆ r, θ, z )− rS(r, θ, z )∆ r

er(θ) dθdz

+ lim∆θ∆r→0

1∆θ∆rr ∆θ∆r

S(r, θ, z+∆ z )− S(r, θ, z )∆ z

ez dθdr

Taking the limits we find that

divS = 1r∂∂r rSer + 1

r∂∂θSeθ + ∂

∂zSez

Substituting the components expression for the tractions we get

+ ∂∂zSrzer+Sθzeθ+Szzez

divS = 1r∂∂r r Srrer+Sθreθ+Szrez + 1

r∂∂θSrθer+Sθθeθ+Szθez

Carrying out the explicit differentiations we get

divS = 1r Srr er+ Sθreθ+ Szrez +

∂Srr∂r er+

∂Sθr∂r eθ+

∂Szr∂r ez

+ ∂Srz∂z er+∂Sθz∂z eθ+

∂Szz∂z ez

+ 1r Srθeθ− Sθθer + 1

r∂Srθ∂θ er+ 1

r∂Sθθ∂θ eθ+

1r∂Szθ∂θ ez

Finally, rearranging the terms we arrive at the desired result.

Chapter 4Linear ElasticConstitutive Theory

97. The constitutive equations for a three-dimensional isotropic, linearly elastic materialcan be expressed in the formSij= λEkkδij+2mEij where the subscripts i, j, and k rangeover the values 1, 2, and 3. Find equivalent expressions for the constitutive equations thatalready reflect the plane stress condition S33 = S23 = S13= 0, that is, find newmaterialconstants λ* and m* such that the two-dimensional relationship can be written as

Sαβ = λ*Eγγδαβ+ 2m*Eαβ

where theGreek indices range only over the values 1 and 2. Express the new constants ( λ*,m*) in terms of the constants ( λ, m) of the three-dimensional theory.

To find equivalent constitutive equations for plane stress condition, use S33 = 0 to getE33 which is not zero.

S33 = λ E11+E22 + λ+2m E33 = 0

E33 =−λ(E11+E22)λ+2m

Substitute E33 into S11 to get

S11 = λ(E22+E33)+ (λ+2m)E11

= λE22−λ2 (E11+E22)λ+2m + (λ+2m)E11

=2λmλ+2m

E11+E22 + 2mE11

Similarly, for S22


S22 = λ(E11+E33)+ (λ+2m)E22

= λE11−λ2 (E11+E22)λ+2m + (λ+2m)E22

=2λmλ+2m

E11+E22 + 2mE22

Finally for S12 we have S12 = 2mE12. Combine these three equations and make asingle constitutive equation with indices α, β and γ, which have the range (1, 2).

Sαβ =2λmλ+2m Eγγδαβ+ 2mEαβ

which are in the proposed form if we let

λ =2λmλ+2m , m = m

98. Demonstrate that the following relationships between the elastic constants λ, m, C,K, and ν hold for an isotropic, linearly elastic material

λ =2mν1−2ν =

m(C−2m)3m−C = Cν

(1+ν)(1−2ν) =3Kν1+ν

C = 2m(1+ν) =m(3λ+2m)λ+m = λ(1+ν)(1−2ν)

ν =9Km3K+m

K = λ+ 23m =

mC3(3m−C) =

λ(1+ν)3ν

= C3(1−2ν)

m = C2(1+ν) =

32(K− λ) = 3K(1−2ν)

2(1+ν) = λ(1−2ν)2ν

ν = λ2(λ+m )

= C2m−1 =

3K−2m2(3K−m )

= 3K−C6K

We can observe that, in each case, one of the elastic constants is expressed in terms of twoof the others from the set of five constants. There are some natural limits to the values thatthe constitutive parameters can take. Assume that under compressive hydrostatic pressureit is impossible for the volume to increase, and that in uniaxial tension it is impossible fora bar to get shorter. What do these hypotheses imply about the other moduli?

Start from Eqn. (240), (241), (242) and what we have already proved in Chapter 4:

Chapter 4 Linear Elastic Constitutive Theory 97

λ = Cν(1+ν)(1−2ν)

m = C2(1+ν)

ν = λ2(λ+m)

C =m(3λ+2m)λ+m

K = λ+ 23m

(a)

(b)

(c)

(d)

(e)

Noting that C= 2m(1+ν) we can eliminate C from (a) to give

λ =2m(1+ν)ν

(1+ν)(1−2ν) =2mν1−2ν

Noting that ν= (C−2m)∕2m we can eliminate ν from (a) to give

λ =C

C−2 m2 m

1+ C−2 m2 m1−2 C−2 m

2 m =

Cm (C−2m)(2m−C−2m) (m−C+2m)

=m(C−2m)3m−C

Noting that C= m(3λ+2m)∕(λ+m) we can eliminate C from (a) to give

λ = ν(1+ν)

C(1−2ν) =

ν(1+ν)

m(3λ+2m)λ+m

1− λλ+m =

ν1+ν (3λ+2m) =

3Kν1+ν

Noting that λ= m(C−2m)∕(3m−C) we can eliminate λ from (e) to give

K = λ+ 23m =

m(C−2m)3m−C + 2

3m =

3m(C−2m)+ 2m(3m−C)3(3m−C) =

mC3(3m−C)

Noting that 2m= λ(1−2ν)∕ν from (b), we can eliminate m from (e) to give

K = λ+ 23m = λ+ λ(1−2ν)

3ν= λ(1+ν)

3ν

Substitute (a) and (b) into (e) to give

K = λ+ 23m = Cν

(1+ν)(1−2ν)+C

3(1+ν) =3Cν+ C(1−2ν)3(1+ν)(1−2ν) =

C3(1−2ν)

From (b), we can directly manipulate C

C = 2m(1+ν)

From (a), we can directly manipulate C

C = λ(1+ν)(1−2ν)ν


Noting that λ= K− 23m we can eliminate λ from (d) to give

C =m(3λ+2m)λ+m =

m (3K− 2m+2m)K− 2

3 m+m=

3mK

K+ 13m=

9Km3K+m

From (e), we can directly manipulate m

m = 32(K− λ)

Noting that C= 3K(1−2ν) we can eliminate C from (b) to give

m = C2(1+ν) =

3K(1−2ν)2(1+ν)

Noting that C = λ(1+ν)(1−2ν)∕ν we can eliminate C from (b) to give

m = C2(1+ν) =

λ(1+ν)(1−2ν)2ν(1+ν) = λ(1−2ν)

2ν

99. Show that the isotropic elasticity tensor with components

Cijkl = λδijδkl+ m δikδjl+δilδjkis invariant with respect to coordinate transformation since the components of the tensorin the two coordinate systems are related by cabcd = CijklQaiQbjQckQdl,, where, as usual,Qij= gi ⋅ e j are the components of the orthogonal change-of-basis tensor. (Hint: anotherway to view change of basis is gi= Qije j.)

Substitute (1) into (2)

cabcd = λδijδkl+ m δikδjl+δilδjkQaiQbjQckQdl

= λδijδklQaiQbjQckQdl+ mδikδjlQaiQbjQckQdl+ mδilδjkQaiQbjQckQdl

= λQaiQbiQckQdk+ mQaiQbjQciQdj+ mQaiQbjQcjQdi

= λδabδcd+ m δacδbd+δadδbc

Therefore the isotropic elasticity tensor is invariant with respect to coordinate trans-formation.

100. Consider a linearly elastic, isotropic material with Lamé parameters λ and m, sub-jected to the following displacement map u(x)= β x21e1+x22e2 . Assume that the linea-rized strain tensor is adequate to characterize the strain field, and compute the body forcesrequired to satisfy equilibrium.


The linearized strain tensor can be given by Eqn. (129).

E = 12∇u(x)+∇uT(x) = 2βx1e1 e1+ 2βx2 e2 e2

The stress tensor can be obtained from the strain tensor by Eqn. (236) as

S = λ( trE)I + 2mE

Noting that

trE= 2β (x1+x2 )

we have

S = 2β λ(x1+x2 )+ 2mx1e1 e1+ 2β λ(x1+x2 )+ 2mx2e2 e2

The body force required for equilibrium is

b = −divS = −2β λ+2m e1+e2

101. Arrange the six independent stress and strain components in columnmatrices as fol-lows: S= (S11, S22, S33, S12, S23, S13 )

T and E= (E11,E22,E33, 2E12, 2E23, 2E13 )T. As-

sume that the constitutive equations of linear elasticity hold. Show that the constitutiveequations can be expressed in matrix form as S= DE, where D is a six by six matrix.

Expand constitutive relationship equation as a component form.

S11 = (λ+2m )E11 + λE22 + λE33

S22 = λE11 + ( λ+2m )E22 + λE33

S33 = λE11 + λE22 + (λ+2m )E33

S12 = 2mE12S13 = 2mE13S23 = 2mE23

The normal stress terms and shear stress terms are independent of each other, thereforethe upper right and lower left blocks of D are zero

D =

λ+2mλ+2m

λ+2mλ

λ λ

λ

λ λ 0 0 0

0 0 0

0 00

0

0

0

0

0

0

0

0

0

m

m

m

0 0

0 0

0 0

15 psi

15 psi

25 psi25 psi

35 psi 35 psi

15 psi

15 psi

35 psi35 psi


102. Consider the thin rectangular sheet with Young’smodulus C=1000 ksi and Poisson’s ratio ν=0.45. Thesheet is subjected to a uniform state of stress through thetractions given in the sketch. The thickness of the sheetbefore the tractions were applied was 0.1 in. What is thethickness of the sheet after the tractions are applied?

The stress tensor can be determined from the given tractions. We can note that the trac-tions Se1 = 25e1, Se2 = 15e2, and Se3 = 0. Thus, the components of the stress ten-sor must be

25 0 00 15 00 0 0

S =

The change in thickness is tE33, where t is undeformed thickness. From Eqn. (244) inthe text

E33 = − νC( trS)+ 1+ν

CS33 = − 0.45

1000(25+15) = −0.018

Thus, the thickness change is -- 0.0018.

103. Consider the thin rectangular sheet with Young’smodulus C=1000 ksi and Poisson’s ratio ν=0.2. Thesheet is fixed between two immovable frictionless platesand is subjected to a uniform state of stress through thetractions around the edges as shown in the sketch. Thethickness of the sheet before the tractions were appliedwas 0.3 in. What is the state of stress in the sheet afterthe tractions are applied?What are the reacting tractionsprovided by the plates? Find the ratio of the change in volume to the original volume ofthe sheet.

(a) What is the state of stress in the sheet after the tractions are applied?

S~--35 0 00 --15 00 0 S33

The unknown component in S can be computed as follows.

E= 1+ νC

S− νCtr(S)

p

p p

t

t

Side Top

t

t

p p


where

tr(S)=− 50+ S33

Since the plates cannot move in the vertical direction

E33 = 0 ⇒ 1+ νC

S33−νC(− 50+ S33 )= 0

S33 =− ν(50)=− 10 ksi

(b) What are the reacting tractions provided by the plates?

t= Sn

On the top face n= e3⇒ t=−10e3On the bottom face n=−e3⇒ t= 10e3

10 psi

(c) Find the ratio of the change in volume to the original volume of the sheet. Thechange in volume is

∇VV≈ tr(E)

The strain components are

E11 =1+ νC

(−35)− νC(−60)= --42+12

1000=− 0.030

E22 =1+ νC

(−15)− νC(−60)= --18+12

1000=− 0.006

E33 = 0

Therefore,

∇VV≈ tr(E)=− 0.030− 0.006=− 0.036

104. A disk made of isotrop-ic, linearly elastic material issubjected to a known uniformpressure p around its perime-ter. The faces of the disk areclamped between immovable, frictionless plates so that the strain through the thicknessis zero. Assume that the stress state is homogeneous throughout the disk and that the Laméconstants are known. Find the tractions t acting on the faces.

σ1

σ1

σ2 σ2

σ1 =− 50 psiσ2 =− 20 psi

ε1 =− 0.00367ε2 =+ 0.00133


We know that E11 = 0 because the top and bottom of the disk are immovable. There-fore, the traction t = S11 is an unknown reaction force. The applied tractions aroundthe cylinder suggest that S22 = S33 = p . To find p, use Eqn. (237) in the text

E11 = − λ2m (3λ+2m)

trS + 12m

S11

Substituting trS= t+2p gives

E11 = − λ2m (3λ+2m)

( t+2p) + 12m

t = 0

Solving for t we find

t = λλ+m

p

105. In a triaxial test, a cylindrical specimenis subjected to a uniform pressure σ1 on theends of the cylinder and a uniform pressure σ2on the sides. The change in height ∆h and thechange in diameter ∆d are measured. Letε1 ≡ ∆h∕h and ε2 ≡ ∆d∕d, where h is theoriginal height and d the original diameter. Thevalues measured in a test are given in the diagram. Assume that the material is linear, iso-tropic, and elastic. What is the volume of the deformed cylinder? Compute the value ofthe bulk and shear moduli (K and m) for this sample.

(a) What is the volume of the deformed cylinder? The deformed volume can be ob-tained (approximately) as

V= Vo (1+trE) = Vo (1+Á1+2Á2 ) = 0.99899Vo

(b) Compute the value of the bulk and shear moduli (K and m) for this sample. Thebulk and shear moduli can be determined from the parameters λ and m.

S ~σ1 0

σ2 0

0 0

00

σ2

Á1 0Á2 0

0 0

00

Á2

E ~

From the constitutive relationship

Sij = λEkkδij + 2mEij

The trace of the stress tensor is Sii= (3λ+2m)Eii. The bulk modulus can be foundfrom the pressure and volume change as follows:


K = λ+ 23m = 1

3SiiEii= 1

3σ1+2σ2Á1+2Á2

= 29703psi

To determine the shear modulus, use the constitutive relationship

S11 = λEkk+ 2mE11 = (λ+2m )E11+ λ(E22+E33)

Substitute λ= K− 2m∕3, then

σ1 = K+ 43 mÁ1 + 2K− 2

3 mÁ2

Solving for m and observing that σ1+2σ2 = K(Á1+2Á2) we find that

m = 34σ1−K(Á1+2Á2 )Á1−Á2 = 1

4 3σ1−(σ1+2σ2 )Á1−Á2 = 3000psi

106. Prove the identity det[I+2E]= 1+2 IE+4 IIE+8 IIIE. (Hint: Use the componentexpression for the determinant of a tensor. The Á−δ identity from Chapter 1 may also beuseful.)

det[I+2E] = 16Á ijkÁlmn (δil+Eil ) (δjm+Ejm ) (δkn+Ekn )

= 16 Á ijkÁlmn

δilδjmδkn + 2 (Eilδjmδkn+Ejmδilδkn+Eknδilδjm )

+ 4(EilEjmδkn+EilEknδjm+EjmEknδil ) + 8EilEjmEkn

Noting that IIIE= det(E)= 16ÁijkÁ lmnEil EjmEkn and det(I)=

16ÁijkÁlmnδilδjmδkn = 1

we have

det[I+2E] = 1 + 8 IIIE +162 EilÁijkÁ ljk+EjmÁijkÁ imk+EknÁ ijkÁ ijn

+ 4 EilEjmÁ ijkÁlmk+EilEkn ÁijkÁ ljn+EjmEknÁ ijkÁimn

From the ε—δ identity proved in Chapter 1

ÁijkÁimn = δjmδkn− δjnδkm

Substitute j = m into the ε—δ identity, then

ÁijkÁijn = δjjδkn−δjnδkj = 3δkn−δkn = 2δkn

Substitute them into the original expression to get

det[I+2E] = 1 + 8 IIIE + 164(Eilδil+Ejmδjm+Eknδkn )

+ 4 EilEjm (δilδjm−δimδjl )+ EilEkn (δilδkn−δinδkl )

+ EjmEkn (δjmδkn−δjnδkm )


Contracting the terms involving the Kronecker deltas we get

+ 4 EiiEjj−EmlElm+EiiEkk−EnlEln+EjjEkk−EnmEmn det[I+2E] = 1 + 8 IIIE +

164 Eii+Ejj+Ekk

We know that IE = Ekk and IIE=12EiiEjj−EijEij

. Substituting these into the aboveexpression we obtain

det[I+2E] = 1+ 2 IE+ 4 IIE+ 8 IIIE

107. Consider a beam of length ℓ with its axis oriented along the z3 direction. The crosssection of the beam lies in the z1−z2 plane, and its second moment of the area is equalto I. The beam is subjected to equal and opposite end moments of magnitudeM, bendingit about the axis with second moment of the area I. The beam is made of elastic materialwith moduli C and ν. The displacement field in the beam is given by the expression

u(z) = MCI12z23+νz21−ν z22 e1+ νz1 z2e2− z1 z3e3

Assume that the applied moment is small enough relative toCI that the displacements arequite small. Compute the components of the strain tensor. Compute the components of thestress tensor from the strain tensor and the linear elastic constitutive equations. Verify thatthe stress field satisfies the equations of equilibrium.

(a) Compute the components of the strain tensor. Use linearized theory for the straintensor.

∇u ~ MCI

--νz2νz1νz2 νz1 0

0

z3

--z3 --z1

E = 12[∇u+∇uT] ~ M

CI

νz1νz1 0

0 −z10

00 0

(b) Compute the components of the stress tensor from the strain tensor and the linearelastic constitutive equations. From the linear elastic constitutive relationship in Eqn.(243) in text

S = Cν(1+ν) (1−2ν)

( trE) I + C(1+ν) E

The trace of the strain tensor given above is

trE = MCI

(2ν−1)z1

z1

z3

B

Ω

z2


Substituting into the constitutive relationship we get

S = MCI−Cνz1

1+ν I +C

1+ν E = −M

Iz1 e3 e3

(c) Verify that the stress field satisfies the equations of equilibrium. Suppose b=0,and from stress equation of equilibrium since S33 is the only nonzero component of thestress tensor

b = −divS = −MIz1,3 e3 = 0

108. Consider the displacement map u(z) for a sphere of unit radius,given by the explicit expression

u(z) = ε z ⋅ z z

where ε is a (very small) constant of the motion. Assume that the ma-terial is isotropic and linearly elastic with material constants λ and m(i.e., the Lamé parameters). Compute the body force b required to maintain equilibrium.Compute the traction forces t that must be acting on the surface of the sphere. Determinethe principal stress field associated with the given motion.

(a) Compute the body force b required to maintain equilibrium.

∇u= ∂∂zjε zk zk zi e i ej= ε2zkδkj zi+ zkzkδij ei e j

= ε2zj zi+ zkzkδij e i ej= ε2 z z+ z ⋅ z I

The linearized strain tensor can be given by Eqn. (129).

E = 12∇u+∇uT = ε2 z z+ z ⋅ z I

tr (E) = ε2zi zi+ zkzkδii = 5ε z ⋅ z

The stress tensor can be obtained from the strain tensor as

S = λ( trE)I + 2mE= 5 λεz ⋅ z I+ 2mε2 z z+ z ⋅ z I

= ε 5 λ+ 2m z ⋅ z I+ 4mεz z

The body force required for equilibrium is

b = −divS = − ∂∂zjε 5 λ+ 2m zk zk δij+ 4mεzi zj e i


= −2ε 5 λ+ 2m zkδkj δij+ 4mεδij zj+ 4m εδjj zi ei= −ε 10 λ+ 20m z

(b) Compute the traction forces t that must be acting on the surface of the sphere.Note that the radial vector z is normal to the surface of the sphere and is of unit length(since the radius of the sphere is 1).

Sz= ε 5 λ+ 2m z ⋅ z z+ 4mεz z z

= ε 5 λ+ 6m z ⋅ z z

(c) Determine the principal stress field associated with the given motion. First we canrewrite the stress tensor as

S= αI+ β z z = α I− z z + α+β z z

where α= ε 5 λ+ 2m z ⋅ z , β= 4mε. From the spectral decomposition theorem,we have one distinct eigenvalue and a pair of repeated eigenvalues m1 = α; n1= z(radial direction) and m2 = m3 = α+ β; n2, n3 are any two vectors in the plane⊥ z (tangent to the sphere surface).

109. Let the elasticity tensor be given by Cijkl= λδijδkl+ m δikδjl+δilδjk. Show thatthe expression Sij= Cijkl Ekl reduces to Eqn. (236).

Substitute the elasticity tensor into the constitutive equation.

Sij = λδijδkl+ m δikδjl+δilδjk Ekl= λδijδklEkl+ mδikδjlEkl+ mδilδjkEkl

= λδijEkk+ mEij+ mEji

The strain tensor is symmetric, therefore

Sij = λEkkδij+ 2mEij

110. A cube of isotropic elastic material, having Lamé constants λ= 1000 psi andm= 1000 psi is in a homogeneous (i.e., does not vary with position) state of stress givenby a stress tensor with components

10 2 12 5 11 1

S~S33

σσ

Frictionless

ho h

Ao


Find the stress component S33 that is consistent with the observation that the cube de-creases in volume (from the stress-free state) by 5%. Now compute the components of thedeviatoric stress tensor and the strain components E13 and E33.

(a) Find the stress component S33 that is consistent with the observation that the cubedecreases in volume (from the stress-free state) by 5%.

S= λ tr(E)I+ 2mE ⇒ tr(S)= (3λ+ 2m) tr(E)

tr(E)= --0.05 ⇒ 15+ S33= (5000)(--0.05)=− 250

∴ S33 =− 265

(b) Now compute the components of the deviatoric stress tensor

∴ S33 =− 265

S′ = S− 13( trS ) I~

1210+ 2503

5+ 2503

−265+ 2503

1

11

2

1

22803

2653

5433

1

11

2S′ ~

(c) and the strain components E13 and E33.

S13 = 2m E13 ⇒ E13 =1

2000= 0.0005

S33 = λ( trE)+ 2mE33 ⇒ E33=12m[S33− λ tr(E) ]

E33 =1

2000(− 365− 1000(−0.05) )=− 0.1075

111. A block of elastic material, havingLamé constants λ= 1000 psi andm= 1000 psi is subjected to a lateralcompressive pressure of σ = 80 psi andclamped between two frictionless rigidplates that reduce the height of the block to 99% of its original height. Compute the totalforce required on the plates to accomplish the motion. Compute the volume of the blockafter deformation. Compute the change in the area of the block on the faces in contact withthe plates.


(a) Compute the total force required on the plates to accomplish the motion. To com-pute the total force acting on the plates, we need to find the stress component normal tothe plates S11. Given that

S22 = S33 =− 80 psi

Let E22 = E33 = Á. From the change of height we can deduce that

E11 =− 0.01, tr(E)= 2Á− 0.01

Substituting in the constitutive equation

⇒ S22 = λ(2Á− 0.01)+ 2mÁ =− 80

2 λ+m Á=− 80+ 0.001λ=− 80+ 10=− 70

⇒ Á=− 704000

=− 7400

S11 = λ− 7200− 2

200 − 0.01 2m = − 45− 20=− 65 psi

And the total force =− 65Ao lbs.

(b) Compute the volume of the block after deformation.

∆VV≈ tr(E )= E11+ E22+ E33 =−

7200− 2

200=− 9

200

v= V− 9200

V= 191200

V

(c) Compute the change in the area of the block on the faces in contact with theplates.

V= hoA; v= ha= 0.99hoa

Using the deformed volume of the block calculated in the previous part

v= 191200

V= 191200

Aho

Substituting,

0.99aho= 191200

Aho ⇒ a= 191198

A

a− A=− 7198

A

112. Consider a body B subjected to the following displacement map:


u(z) = β z21−2z2 z3 e1+ β z22+2z1 z3 e2+ β z23−2z1 z2 e3where β is a (very small) constant. Find the stress tensor associated with this motion, as-suming that the material is linear, isotropic, and elastic with moduli λ and m and that thestress is zero when the displacement is zero. Find the body force field required tomaintainequilibrium for the givenmotion. Find the principal values of the (linearized) strain tensor.

(a) Find the stress tensor associated with this motion, assuming that the material islinear, isotropic, and elastic with moduli λ and m and that the stress is zero when thedisplacement is zero. The gradient of displacement is

z1 −z3 −z2z3 z2 z1

−z2 −z1 z3

∇u~ 2β

Therefore, the strain is

z1 0 −z20 z2 0

−z2 0 z3

E= 12(∇u+∇uT)~ 2β

with a trace of

tr(E)= 2β(z1+ z2+ z3)

The stress can be computed from teh constitutive equation

S= λ tr(E) I+ 2m E

λ(z1+z2+z3)+2m z1

0

λ(z1+z2+z3)+2mz2λ(z1+z2+z3)+2mz3

0

0

0

−2mz2

−2mz2

S~ 2β

(b) Find the body force field required to maintain equilibrium for the given motion.

div S+ b= 0 b=− div S

div S=∂Se i∂zi~ 2β λ+2m

111

b=−2β λ+2m 111

(c) Find the principal values of the (linearized) strain tensor.

det E−γI = 0


Since E is proportional to 2β it is sufficient to find the eigenvalues of E∕2β then mul-tiply the result by 2β

z1− γ 0 −z20 z2− γ 0

−z2 0 z3− γ= (z2− γ)( z1− γ)(z3− γ)− z22 = 0det

One root of this equation is γ1 = z2. The other two are the roots of the quadratic

γ2− ( z1+ z3 )γ+ (z1 z3− z22)= 0

which has solutions

⇒ γ2,3 =(z1+ z3) (z1+ z3 )2− 4( z1 z3− z2

2)

2

The principal values of strain are 2β γi i = 1, 2, 3

113. The state of the deformation at a certain point in a solid body is such that it has thefollowing principal strains Á1 = ε and Á2 = Á3 = 3ε, where ε is a known value. The prin-cipal directions associated with these principal strains are known to be n1, n2, and n3. As-suming linear, isotropic response, find the principal values and principal directions of thestress tensor S. The material constants are λ= 1000 psi and m= 1000 psi. What is theaverage pressure p at the point in question?What is the change in volume in the neighbor-hood of the point in question?

(a) Assuming linear, isotropic response, find the principal values and principal direc-tions of the stress tensor S. The material constants are λ= 1000 psi and m= 1000psi.

S= λ tr (E)I+ 2mE

Sn= σn= λ tr (E)n+ 2mEn= λ tr (E)+ 2mÁ n

∴ σi= λ Á1+Á2+Á3 + 2mÁi

The principal directions of S and E are the same.

tr (E)= Á+ 3Á+ 3Á= 7Á

⇒ σ1 = 1000( 7Á )+ 2000(Á )= 9000 Á

σ2 = σ3 = 1000 (7Á )+ 2000(3Á )= 13000Á


(b) What is the average pressure p at the point in question?

p= 13tr (S )= 1

3(σ1+ σ2+ σ3 )=

35000 Á3

(c) What is the change in volume in the neighborhood of the point in question?

∆VV≈ tr (E)= 7Á ⇒ ∆V= 7ÁV

114. The strain energy function of a nonlinear hyperelastic material is given by

W(E) = 12aEiiEjj+

12bEijEij+

13cEijEjkEki

where a, b, and c are material constants and E= Eij [e i ej ] is the strain tensor. Findthe stress tensor S as a function of the strain E implied by the strain energy function.

Smn =∂W (E)∂Emn

= 12a (δimδin Ejj+ Eiiδjmδjn )+

12b (δimδjnEij+ Eijδimδjn )+

13c (δimδjnEjkEki+ EijδjmδknEki+ EijEjkδkmδin )

Smn = 12a (δmn Ejj+ δmn Eii )+

12b (Emn+ Emn )+ 1

3c (Enk Ekm+ Eim Eni+ Emj Ejm )


W(E) = aEii ln1+Ejj + 3

2 bEijEij

where ln(.) indicates the natural logarithm of (.), a and b are known material constants,and E= Eij [e i ej ] is the strain tensor. Find the stress tensorS as a function of the strainE implied by the strain energy function. Consider a hydrostatic state of stresswith pressurep in which the stress tensor is given by S= pI. Set up a relationship between the changein volume and the pressure p. What is the pressure required to decrease the volume to 95%of the original volume (assume that the linearized strain tensor is adequate)?

(a) Find the stress tensor S as a function of the strain E implied by the strain energyfunction.

Skℓ =∂W (E)∂Ekℓ

= a ∂∂EkℓEii ln(1+Ejj ) + 3

2b ∂∂Ekℓ

(EijEiℓ )


= a ln(1+ Ejj )+Eii

1+ Ejjδkℓ + 3bEkℓ

Let e≡ tr (E)≈ ∆V∕V. Then,

S= a ( ln(1+ e)+ e1+ e

) I+ 3bE

(b) Consider a hydrostatic state of stress with pressure p in which the stress tensor isgiven by S= pI. Set up a relationship between the change in volume and the pressurep. What is the pressure required to decrease the volume to 95% of the original volume(assume that the linearized strain tensor is adequate)?

S= pI ⇒ tr (S )= 3p

Thus, taking the trace of S from above

tr (S )= 3p= 3a ( ln(1+ e)+ e1+ e

)+ 3be

p= a ( ln(1+ e)+ e1+ e

)+ 3be

For a 5% volume increase

e=− 0.05⇒ p=− 0.1039a− 0.05b

116. The strain energy function of a nonlinear hyperelastic material is given by the (com-ponent) expressionW(E)= a0EiiEjjEkk+a1EijEjkEki , where a0 and a1 are knownmate-rial constants, and Eij is the ijth component of the strain tensor E. Find the stress tensorS as a function of the strainE implied by the strain energy function. Is the material isotrop-ic? Explain your answer. Is thematerial linear? Explain. Consider a uniform state of shear-ing in which the strain tensor has components E12 = E21 = γ and all other componentsequal zero, where γ is a given constant. Find the principal values of the stress tensor S forthe given constitutive model under the given state of strain.


Smn =∂W (E)∂Emn

= a0∂∂EmnEiiEjj Ekk + a1

∂∂EmnEijEjkEki

= a0δimδin EjjEkk+ EiiδjmδjnEkk+ EiiEjjδkmδkn

+ a1δimδjnEjkEki+ Eijδjmδkn Eki+ EijEjkδkmδin


Smn = a0 EjjEkk+ EiiEkk+ EiiEjjδmn+ a1 EnkEkm+ EniEim+ EnjEjm

(b) Is the material isotropic? Explain your answer. Is the material linear? Explain. Thestrain energy function is written in terms of the invariants of the strain tensor, thereforethe material is isotropic. The stress components are quadratic in strain, therefore thematerial is not linear.

(c) Consider a uniform state of shearing in which the strain tensor has componentsE12 = E21 = γ and all other components equal zero, where γ is a given constant. Findthe principal values of the stress tensor S for the given constitutive model under thegiven state of strain.

Eii= 0⇒ Smn = a1 EnkEkm+ EniEim+ EnjEjm

which can be alternatively written as (given the symmetry of E)

S= 3a1(EE)T= 3a1E

TET= 3a1EE

⇒ S~ 3a1

0 γ 0

γ 0 0

0 0 0

0 γ 0

γ 0 0

0 0 0

= 3a1

γ2 0 0

0 γ2 0

0 0 0

Therefore the principal values of stress are m1 = m2 = γ2, m3 = 0


W(E) = ln1+αEij Eij + β eEii− Eii

where ln(.) indicates the natural logarithm of (.) and e(.) indicates the exponential of (.),α and β are known material constants, and Eij is the ijth component of the strain tensorE. Find the stress tensor S as a function of the strain E implied by the strain energy func-tion. Howdo the constantsα and β relate to the Lamé parameters of linear isotropic elastic-ity? Consider a uniform state of dilatation in which the strain tensor is given by E= ε I,where ε is a constant. Find the principal values of the stress tensor as a function of ε.


Smn =∂W (E)∂Emn

= ∂∂Emn

ln1+αEij Eij + β ∂∂Emn eEii− Eii

=2α

1+αEkℓEkℓEijδimδjn+ β eEkk− 1 δimδin


=2α

1+αEkℓEkℓEmn+ β eEkk− 1 δmn

In direct notation the stress has the form

S=2α

1+α trE2E+ β e tr(E)− 1 I

(b) How do the constants α and β relate to the Lamé parameters of linear isotropicelasticity? When strains are small, the quadratic term EkℓEkℓ becomes negligible. Fur-thermore, expanding eEkk in a Taylor series around Ekk= 0 gives

etr(E)≈ 1+ trE + 12trE

2+

Dropping the terms of order 2 and higher, and substituting in the stress tensor obtainedabove

⇒ S= 2αE+ β tr E I

from which it can be observed that a→ m, b→ λ for small strains.

(c) Consider a uniform state of dilatation in which the strain tensor is given byE= ε I, where ε is a constant. Find the principal values of the stress tensor as a func-tion of ε. Let tr(E)= 3ε so that

S=2α

1+9αε2 ε I+ β e3 ε− 1 I= 2αε

1+9αε2+ β e3 ε− 1 I

The principal values of the stress tensor are

m1 = m2 = m3=2αε

1+9αε2+ β e3 ε− 1

118. The strain energy function of a nonlinear hyperelasticmaterial is givenby the expres-sion W(e, γ)= a0e

2+a1γ+a2eγ, where a0, a1, and a2 are known material constants,and the scalar invariant strain measures e and γ, which are functions of the strain tensorE, are defined as e≡ tr E and γ≡ tr E′E′ , where E′ ≡ E−eI∕3 is the deviatorstrain. Observe that ∂e∕∂E = I and ∂γ∕∂E = 2E′. Find the stress tensor S as a functionof the strain E implied by the strain energy function. Consider a state of hydrostatic pres-sure S= pI, where p is a given pressure. Find the relationship among p, e and γ. Nextconsider a sample of the material subjected to a state of pure shear strain described byE= g nm+m n , where g is a given constant describing the motion andm andn are given orthogonal unit vectors. Will there be a change in volume of the sample? Doyou expect that you would need a confining pressure to execute this motion? Why?



The stress can be computed as

S= ∂W∂E =∂W∂e∂e∂E+

∂W∂γ∂γ∂E

= 2a0eI+ 2a1E+ a2 γ I+ 2eE

= 2a0e+a2γ I+ 2a1+2a2e E

(b) Consider a state of hydrostatic pressure S= pI, where p is a given pressure. Findthe relationship among p, e and γ. Compute the trace of the stress tensor

trS = p tr(I) = 2a0e+a2γ tr(I)+ 2a1+2a2e tr(E)

3p = 3 2a0e+a2γ

p = 2a0e+a2γ

(c) Next consider a sample of the material subjected to a state of pure shear straindescribed by E= g nm+m n , where g is a given constant describing themotion and m and n are given orthogonal unit vectors. Will there be a change in vol-ume of the sample? Do you expect that you would need a confining pressure to executethis motion? Why? The two strain measures can be computed for the present case asfollows:

e = tr(E) = g tr[nm]+tr[m n] = g n ⋅m+m ⋅ n = 0

e = ∆V∕V = 0 (no volume change) also E= E

γ = tr(EE) = g2 tr [nm+m n] [nm+m n]

= g2 tr [nm][nm]+ [nm][m n]

+ [m n][nm]+ [m n][m n]

= g2 tr [n ⋅m][nm]+ [m ⋅m][n n]

+ [n ⋅ n][mm]+ [n ⋅m][m n]

= g2 tr [n n]+ [mm] = 2g2

Therefore, the stress is

S = a2γI+ 2a1E, tr(S)= 6a2g2

The confining pressure is proportional to tr S. Therefore, this deformation will requireconfining pressure to execute.

−σ−σ

2σ

3σ 2σ

3σ

P

P

A B A


119. Acube of elasticmaterial, havingLamé constants λ= 1000psi and m= 1000 psi is subjected to purely normal tractions onits faces as shown in the sketch. Compute the value of σ requiredto change the volume of the block by 2% of its original volume.

S~--σ 0 0

0 2σ 0

0 0 3σ

tr (S )= 5σ

S= λ tr (E)I+ 2mE⇒ tr (S )= (3λ+2m ) tr (E)

tr (E)≈∆VV= 3

100

∴ 5σ= 5000 ( 3100

)⇒ σ= 300 psi

120. Three unit cubes (1×1×1) are uniformly com-pressed between two rigid plates with an aggregateforce of P. The change in height is the same for all threecubes. The two outer cubes are made of material A,while the inner cube is made of material B. Both of thematerials are linearly elastic with Lamé constants

λA= 1000 psi, mA= 1000 psi

λB= 500 psi, mB= 2000 psi

Compute the force P required to change the volume of the middle block by 3% of its origi-nal volume. What is the final area of the compressed face of the outer cubes?

(a) Compute the force P required to change the volume of the middle block by 3% ofits original volume. The plates are compressed, therefore the volume of block B mustbe decreased by 3 percent ⇒ tr (EB )=−0.03. The stress tensors have the form,

SA~α 0 0

0 0 0

0 0 0

SB~

β 0 0

0 0 0

0 0 0

⇒ β≡ tr (SB)= 3(λB+23mB ) tr (E

B )= 3[500+ 23(2000)](−0.03)=−165 psi

P

P

AB


Also,

SB11 = λB tr (EB)δ11+ 2mB EB11

− 165= 500(−0.03)+ 2(1000)EB11 ⇒ EB11 =− 0.0375

Since the vertical displacements (and strains) are the same for all the blocks.

EA11 = EB11 =−0.0375

EB22 = EB33 =− EA11 νA= 0.0375× 0.25= 0.009375

where

νA=λA

2(λA+ mA)= 1000

2(1000+ 1000)= 0.25

⇒ tr (EA )=− 0.0375+ 2× 0.009375=− 0.01875

Thus,

α= tr (SA)= 3λA+ 23mA tr (EA)= 3(1000+ 2

31000)(− 0.01875)=− 93.75 psi

P=− 165 psi× 1 in2− 2× 93.75 psi× 1 in2= 352.5 lbs

assuming that the change in area is relatively small.

(b) What is the final area of the compressed face of the outer cubes?

A= (1+ EA22)(1+ EA33)= (1.009375)2= 1.0188 in2

121. Two cubes with dimensions 2×2×2 are uniformlycompressed between two rigid plateswith an aggregate forceof P. Assume that there is no friction between any of the con-tacting surfaces. The top cube is made of material A, whilethe bottom cube is made of material B. Both of the materialsare linearly elastic with Lamé constants

λA= 1000 psi, mA= 1000 psi

λB= 500 psi, mB= 2000 psi

Compute the force P required to change the total volume of the two cubes by 5% of theoriginal volume. What are the final dimensions of the two cubes?

(a) Compute the force P required to change the total volume of the two cubes by 5%of the original volume. Let σA , σB be stresses in blocks a, b along force axis andÁA , ÁB be the corresponding strains.


Si~σi 0 00 0 00 0 0

Ei~Ái 0 00 Ei22 00 0 Ei22

i = A, B

where Ei22 =− νi Ái is the lateral strain. The material constants are

C=m(3λ+ 2m)λ+ m

ν= λ2(λ+ m)

⇒ CA= 2500 CB= 4400 νA= 0.25 νB= 0.10

From equilibrium

σA= CA ÁA= σB= CB ÁB ⇒ ÁA=CB

CAÁB

Since the total change in volume to be 5% of the original volume

∆VA+ ∆VB

VA+ VB=− 0.05

Noting that ∆Vi= ei Vi where ei≡ tr (Ei )

⇒eA VA+ eB VB

VA+ VB=− 0.05

⇒ eA+ eB=− 0.1

Substituting ei≡ (1− 2νi )Ái leads to

(1− 2νA)CB

CA+ (1− 2νB)ÁB=− 0.10 ⇒ ÁB=− 0.0595

ÁA=− 0.1048

P= 4σaA= 4CAÁA=− 1048 lb

assuming that the change in area is relatively small.

(b) What are the final dimensions of the two cubes? Each block decreases in height byÁi and increases in width by −νi Ái times the original dimensions of 2.

Block A is V=(1.7905)(2.0524)(2.0525), thus Vol. = 7.542

Block B is V=(1.8810)(2.0120)(2.0120), thus Vol. = 7.61

Note: ∆V∕V|total=− 0.0527≠− 0.05 because of the small strain assumption

122. The strain at a point in a body is given by

E ~ 10−32 3 43 5 14 1 1

z2

z3

z12

22


Find the components of the stress tensor assuming linear, isotropic, elastic material behav-ior, with λ = 16,000 ksi and m = 11,000 ksi.

Sij= λ Ekk δij+ 2m Eij

Ekk= 10−3(2+ 5+ 1)= 8× 10−3

⇒ Sij= (16× 103)(8× 10--3) δij+ (22× 103) Eij

⇒ S~ 1281 0 00 1 00 0 1

+ 222 3 43 5 14 1 1

=172 66 8866 238 2288 22 150

123. A 2 by 2 by 2 unit solid cube, centered at the origin of coordi-nates, is subjected to the deformation described by the map:

φ(z)= z1+ 13 az

31 e1+ z2+ 1

3 az32 e2+ bz3e3

Compute the values of the constants a and b that are consistent withthe observations that the total volume of the block is unchanged bythe deformation and the total area of the side with original normal e1 decreases by 5% dueto the deformation. Assuming that the cube is made of a linear, elastic, isotropic materialwith Lamé parameters λ andm, find the body forces and surface tractions required for equi-librium. (You may assume that the linearized strain tensor is adequate to describe thestrains for this problem).

(a) Compute the values of the constants a and b that are consistent with the observa-tions that the total volume of the block is unchanged by the deformation and the totalarea of the side with original normal e1 decreases by 5% due to the deformation. Theoriginal volume and area are V = 8 and a = 4. The values after deformation are

v= B

det FdV= 8 a= B

det F ‖ F−T e1 ‖ dA= 0.95(4)

The following quantities can be computed from the deformation

F~

1+az21 0 0

0 1+az22 0

0 0 b

F−T~

11+az2

1

0 0

0 11+a z2

2

0

0 0 1b

F−T e1 ~100


det F= b (1+ az21 ) (1+ az22 ) det F ‖ F−Te1 ‖= b (1+ az22 )

Now the volume can be computed as

v=1−1

1−1

1−1

b 1+az21 1+az22 dz1dz2dz3

= 2b (z1+a3z31 ) |

1

−1(z2+

a3z32 ) |

1

−1= 8b (1+ a

3)2

The area can be computed as

area=1−1

1−1

b 1+az22 dz2dz3 = 2bz2+ a3z32

1

−1= 4b1+ a

3

Substituting in the given constraints on the deformation we get

∴ b (1+ a3)2 = 1 b ( 1+ a

3)= 0.95

⇒ a= 0.158 b= 0.903

(b) Assuming that the cube is made of a linear, elastic, isotropic material with Laméparameters λ and m, find the body forces and surface tractions required for equilibrium.(You may assume that the linearized strain tensor is adequate to describe the strains forthis problem).

u(z)= φ(z)− z= 13az31 e1+

13az32 e2+ bz3e3

∇u~az21 0 0

0 az22 0

0 0 b

E≈ 12[∇u+∇uT ]= ∇u

The stress is given by S = λ tr (E)I+ 2mE. In components we have

⇒

S11 = λ(az21+ az22+ b )+ 2maz21S22 = λ(az21+ az22+ b )+ 2maz22S33 = λ(az21+ az22+ b )+ 2mb

all other Sij= 0. Therefore, the body force required to maintain equilibrium is

b=− divS~2a (λ+ 2m ) z12a (λ+ 2m ) z2

0

The surface tractions are as follows


Face z1 = 1, n= e1,

te1 = S11(z1= 1)e1= [λ(a+ az22+ b )+ 2ma ]e1

Face z1 =−1, n=−e1,

t−e1 =−S11(z1=−1)e1=−[λ(a+ az22+ b )+ 2ma ]e1

Face z2 = 1, n= e2,

te2 = S22(z2= 1)e2= [λ(az21+ a+ b )+ 2ma ]e2

Face z2 =−1, n=−e2,

t−e2 =−S22(z2=−1)e2=−[λ(az21+ a+ b )+ 2ma ]e2

Face z3 = 1, n= e3,

te3 = S3(z3= 1)e3= [λ(az21+ az22+ b )+ 2mb ] e3

Face z3 =−1, n=−e3,

t−e3 =−S33(z3=−1)e3=−[λ(az21+ az22+ b )+ 2mb ] e3

124. Consider a displacement map u(z) given by the explicit expression u(z)= εAz,where A is a given constant tensor and ε is a given scalar (which is very small comparedto 1). The vector z is the position vector of a point in the undeformed configuration. Com-pute the strain tensor E of the given motion. Compute the stress tensor S assuming thatthematerial is linear and elastic and has Lamé parameters λ andm. Compute the body forceb required to maintain equilibrium with the stress.

(a) Compute the strain tensor E of the given motion.

∇u= ∂∂zjεAikzk ei e j= εAikδkj e i ej= εA

E = 12∇u+∇uT = εA

(b) Compute the stress tensor S assuming that the material is linear and elastic andhas Lamé parameters λ and m.

S = λ( trE)I + 2mE = λε ( trA)I + 2mεA constant

(c) Compute the body force b required to maintain equilibrium with the stress.

b = −divS = 0

z2

z3

z11

1

1


125. Consider the unit cube with vertex at the origin of coordinatesas shown in the sketch. The cube is subjected to the following de-formation map:

φ(z)= z1+z2 sin γ e1+ z2 cos γ e2+ z3e3

Note that γ is a constant. Compute the tractions and body forces re-quired to achieve the given deformation for the specific shearingangle of γ=0.2 rad assuming that the material is linear, elastic, and isotropic with Young’smodulus of 1000 psi and Poisson’s ratio of 0.499. Does it make any difference if you usethe linearized strain tensor as opposed to the Lagrangian strain tensor in the constitutiveequation for this problem? Explain.

(a) Compute the tractions and body forces required to achieve the given deformationfor the specific shearing angle of γ=0.2 rad assuming that the material is linear, elas-tic, and isotropic with Young’s modulus of 1000 psi and Poisson’s ratio of 0.499.

F= ∇φ~1 sin γ 0

0 cos γ 0

0 0 1

FTF~1 sin γ 0

sin γ 1 0

0 0 1

;

E= 12[FTF− I ]~ 1

2

0 sin γ 0

sin γ 0 0

0 0 0

tr (E)= 0

The stress can be computed from S = λ( trE)I + 2mE (constant), where

m= C2( 1+ ν)

= 10003= 333 λ=

2mν2( 1− 2ν)

=m

0.002= 166, 6667

⇒ S12 = S21= m sin γ; all other Sij= 0

The body force vector required for equilibrium is,

b = −divS = 0

The surface tractions are as follows

Face z1 = 1, n= e1, te1 = S12 e2= m sin γ e2

Face z2 = 1, n= e2, te2 = S21 e1= m sin γ e1

(b) Does it make any difference if you use the linearized strain tensor as opposed tothe Lagrangian strain tensor in the constitutive equation for this problem? Explain.

z1

z2

ab

z1z2

z3


∇u= F− I~0 sin γ 0

0 cos γ−1 0

0 0 0

Elinear =12∇u+∇uT ~ 1

2

0 sin γ 0

sin γ 2 cos γ−2 0

0 0 0

tr (Elinear )= cos γ− 1

⇒ S12 = S21= m sin γ

S22 = (λ+ 2m ) ( cos γ− 1); S11 = S33 = λ (cos γ− 1)

Comparing these results with those obtained using Lagrangian strain we find that:

⇒ S12 = S21= 67 psi in both cases

whereas

S11 = S33 ≈− 166, 667(0.02)=− 3322 psi

S22 ≈− (166, 667+ 667)(0.02)=− 3333 psi

as opposed to S11 = S22 = S33= 0 when the Lagrangian strain tensor was used.Therefore, even though the term cos γ− 1≈−γ2∕2 is a second--order term, it givesrise to large stresses because ν= 0.499 is a nearly incompressible material withλ≫ m.

126. Abar of length ℓ has an elliptical cross section. Theequation of the ellipse is b2 z21+a2 z22 = a2b2, where aand b are themajor andminor semi-axis dimensions. Thebar experiences the following displacement map:

u(z)=−βz2 z3 e1+ βz1 z3 e2− βcz1 z2e3where β and c are constants. Find the stress tensor associ-ated with this motion, assuming that the material is lin-ear, isotropic, and elastic with moduli λ and m. Find thebody force required for equilibrium.What value must the constant c have in order that thelateral surface of the bar be traction-free?


(a) Find the stress tensor associated with this motion, assuming that the material islinear, isotropic, and elastic with moduli λ and m.

∇u~ β0 −z3 −z2z3 0 z1

−cz2 −cz1 0

E= 12∇u+∇u ~ β

0 0 −x2( 1+c)

0 0 x1( 1−c)

−x2( 1+c) x1(1−c ) 0

We thus have tr (E)= 0. The stress is given by S= λ tr (E)I+ 2mE= 2mE. Thus,

S~ 2mβ

0 0 −x2( 1+c)

0 0 x1( 1−c)

−x2( 1+c) x1(1−c ) 0

(b) Find the body force required for equilibrium.

b=−divS= 0

(c) What value must the constant c have in order that the lateral surface of the bar betraction-free? The normal to the lateral surface is

nΓ= ∇b2 z21+a2 z22− a2b2 ~ γ2b2 z1

2a2 z2

0

tΓ= SnΓ~ 2γ

0

0

− b2 z1 z21+c)+a2 z1 z2 (1−c )

For the lateral surface to be traction free

− b2 z1 z21+c )+a2 z1 z2 (1−c )= 0

⇒ c= a2− b2a2+ b2


127. Consider a displacement map u(z)= z z a, where a is a given constant vector(which has a magnitude very small compared to 1). The vector z is the position vector ofa point in the undeformed configuration. Compute the linearized strain tensorEof the giv-enmotion. Compute the stress tensor S assuming that the material is linear and elastic andhas Lamé parameters λ and m. Compute the body force b required to maintain equilibriumfor the given motion.

(a) Compute the linearized strain tensor E of the given motion.

E= 12∇u+∇uT , ∇u=

∂ui∂zj

[e i ej ]

ui= akzkzi,∂ui∂zj= akzk,j zi+zkzi,j = akδkj zi+akzkδij = ziaj+akzkδij

= akzk,j zi+zkzi,j = akδkj zi+akzkδij = ziaj+akzkδij

In direct notation,

∇u = z a + (a ⋅ z)I

∇uT = a z + (a ⋅ z)I

E = (a ⋅ z) I+ 12z a+ a z

(b) Compute the stress tensor S assuming that the material is linear and elastic andhas Lamé parameters λ and m. The stress is given by S= λ tr(E) I+ 2mE. Thus,

tr(E) = 3(z ⋅ a)+ (z ⋅ a)= 4(z ⋅ a)

S= 4λ(a ⋅ z) I+ 2m (a ⋅ z)I+ m z a+ a z

= 4λ+2m (a ⋅ z)I+ m z a+ a z

(c) Compute the body force b required to maintain equilibrium for the given motion.

divS+ b = 0, b=−divS

divS =∂Sij∂zj

e i = Sij,j ei

Sij= 4λ+2m akzkδij+ m ziaj+ aizj

The divergence of the stress tensor is

Sij, j= 4λ+2m akzk,j δij+ m zi,j aj+ aizj,j = 4λ+2m akδkjδij+ m δij aj+ aiδjj

= 4λ+2m ai+ m ai+ 3ai

= 4λ + 6m ai

z3z2

z1


Therefore, the body force required for equilibrium is

b = − 4λ+6m a

128. A linearly elastic solid body is subjected to forcesthat give rise to the following displacement map:

u(z)= γ 12 z23+βz21−βz22 e1+ γβz1 z2e2− γz1 z3e3

where γ≪ 1 (i.e., very small) and β are constants describing the motion. Assume that theelastic response is adequately characterized by Hooke’s law with known materialconstants λ and m (the Láme parameters). Find β in terms of the constants λ andm such thatS11 = 0 (S is the stress tensor). Find S22 and S33 for the conditions given previously. Findthe traction on the surface with normal−e3 at z3 = 0.

(a) Find β in terms of the constants λ and m such that S11 = 0 (S is the stress tensor).Since the displacements are very small, the linearized strain tensor is appropriate.

E= 12∇u+∇uT , ∇u=

∂ui∂zj

[e i ej ]

∇u~ γβz1 −βz2 z3βz2 βz1 0−z3 0 −z1

E~ γβz1 0 z30 βz1 00 0 −z1

The stress can be determined from Hooke’s Law

S= λ tr(E) I+ 2mE, tr(E)=−z1 1−2β γ

S11 = −1−2β λ+ 2mβ z1 γ= 0 ⇒ β= λ2(λ+m)

(b) Find S22 and S33 for the conditions given previously.

S22 = −1−2β λ+ 2mβ z1 γ= 0 ⇒ S22 = 0

S33 = −1−2β λ−2m z1γ=−2mβ+2m z1γ ⇒ S33=−2m1+β z1γ

(c) Find the traction on the surface with normal−e3 at z3 = 0. The traction on thesurface can be computed with Cauchy’s relationship

tn = Snwith n = −e3

S= S33 e3 e3

S(−e3)= S33 e3 e3 (−e3)=−S33 (e3 ⋅ e3)e3=−S33e3


tn= 2m1+β z1γe3

129. The state of stress S as a function of position x in a certain solid body is given by theexpression S(x)= x Bx, where B is a given constant tensor. Find the body force (asa function of position x) required to maintain equilibrium of the body. Express the resultin both index and direct (vector) notation. What are the restrictions, if any, on the constanttensorB in order for the stress field S to be an admissible stress state? (Please describe anyrestrictions explicitly in terms of the components of B, not in terms of x and S.)

(a) Find the body force (as a function of position x) required to maintain equilibriumof the body. Express the result in both index and direct (vector) notation. Thecomponents of the stress tensor are

S= Sij ei ej = xiei Bjk e j ek xle l

= xiBjkxlei e j ek el

= xiBjkxlδkl e i ej = xiBjkxk ei e j

⇒ Sij= xiBjk xk

The body force vector is, therefore

b=−divS=−∂Sijxj

ei

∂Sij∂xj=∂xi∂xj

Bjkxk+ xiBjk∂xk∂xj= δijBjk xk+ xiBjkδkj

∂Sij∂xj= Bikxk+ xiBkk

bi=− Bik xk+ xiBkk

b = −Bx+ tr(B)x = −B+tr(B)I x

(b) What are the restrictions, if any, on the constant tensor B in order for the stressfield S to be an admissible stress state? (Please describe any restrictions explicitly interms of the components of B, not in terms of x and S.) The stress state must alsosatisfy balance of angular momentum. Hence, the stress tensor must be symmetric.

Sij= xiBjkxk


S12 = x1 B21x1+ B22x2+ B23x3

S21 = x2 B11x1+ B12x2+ B13x3

S13 = x1 B31x1+ B32x2+ B33x3

S31 = x3 B11x1+ B12x2+ B13x3

S23 = x2 B31x1+ B32x2+ B33x3

S32 = x3 B21x1+ B22x2+ B23x3

B21 = B23 = B21= B13= 0

B11 = B22

B31 = B32 = B12= B13= 0

B11 = B33

B31 = B32 = B21= B23= 0

B22 = B33

Therefore, the tensor B must be proportional to the identity tensor, i.e.,B= BI.

The solution above was given in component form. Here is a component-freederivation of the conditions on B implied by symmetry of the stress tensor. Thanks toXiang Ding for pointing out this elegant solution. The stress state must satisfy balanceof angular momentum. Hence, the stress tensor must be symmetric.

S is symmetric if u ⋅ Sv= v ⋅ Su for any vectors u, v

u ⋅ x Bx v = v ⋅ x Bx u

u ⋅ v ⋅ Bx x = v ⋅ u ⋅ Bx x

u ⋅ x v ⋅ Bx = v ⋅ x

This equality can be rewritten as

v ⋅ Bx v ⋅ x

=u ⋅ Bx u ⋅ x

= b

where b is a constant (because that is the only way a function of v and a function of ucan be equal to each other). Therefore, (for either u or v) we must then have

u ⋅ Bx = bu ⋅ x , [or v ⋅ Bx = bv ⋅ x ]

u ⋅ Bx− bx = 0

u ⋅ B−bI x = 0

Since the above equation must hold for an arbitrary vector u and must also be true forany position vector x then we must have

B= bI

In other words, the tensor B must be proportional to the identity.

ℓ

x

b(x)= bo1−x∕ℓ

Chapter 5Boundary ValueProblems in Elasticity

130. Consider the uniaxial rod shown below, fixed at x= 0,free at x= ℓ, and subjected to the linearly varying body forceindicated. The rod is made from a composite material with avariable elastic modulus C(x)= Co 2−x∕ℓ, making ittwice as stiff at x= 0 as it is at x= ℓ. The governing differ-ential equation for a rod with variable modulus is

C(x)u′ ′ + b(x) = 0

where a prime indicates differentiationwith respect to x. Find the exact (classical) solutionto the problem by directly integrating the governing equations.

Let ξ≡ x∕ℓ. The original differential equation is given by

Co2−ξ u′ ′ = −bo 1−ξ

Integrating once we get

Co 2−ξ u′ = −boℓξ− 12 ξ

2 + a1 = σ(ξ)

The constant a1 can be determined from the boundary condition σ(1)= 0 to have thevalue a1 = boℓ∕2. Dividing through by Co 2−ξ we have

u′ = boℓ2Co 12−ξ− ξ

Integrating this equation gives

u(ξ) = boℓ22Co− ln(2−ξ)− 1

2 ξ2 + a2

1

x

b(x)= 15x

C(x)= 2−x2


The constant a2 can be determined from the boundary condition u(0)= 0 to have thevalue a2 = boℓ2 ln 2∕2Co. Thus, the displacement map has the explicit form

u(ξ) = boℓ22Coln 2− ln(2−ξ)− 1

2 ξ2

131. Consider the rod of unit length and modulus C(x) thatvaries as shown in the sketch. The rod is fixed at the left end,is free at the right end, and is subjected to a linearly varyingbody force b(x) as shown. Consider the following displace-mentmap: u(x)= a x3+2x2−3x where a is some constant.Is the displacementmap a solution to the given problem?Whyor why not?

For the given map to be a solution to the problem, it has to satisfy the governing differ-ential equation and boundary condition.

(Cu′ )′ + b= 0 ∀ x, u (0)= 0, Cu′ (1)= 0

Substituting,

u′(x)= a 3x2+4x−3 , u′′(x)= a 6x+4

(Cu′ )′ + b= Cu′′ + C′u′ + b

= a(2− x2 ) 6x+ 4 + a(− 2x ) 3x2+ 4x− 3 + 15x

= a(− 12x3− 12x2+ 18x+ 8)+ 15x≠ 0

u (0)= 0, Cu′ (1)= 4a≠ 0

Therefore the given map is not a solution to the problem.

132. Prove that S ⋅ E= S ⋅ ∇u when the virtual strain is defined to be the strain thatwould occur if the virtual displacement actually took place, i.e., E= 1

2∇u+∇uT. Upon

what property of the stress tensor S does this identity rely?

Chapter 5 Boundary Value Problems in Elasticity 131

S ⋅ ∇u= Sij ui,j

= Sij 12 ui, j+ uj, i + 12ui, j− uj, i

= SijEij+ Sij 12 ui, j− uj, i = SijEij+

12Sijui, j− Sij uj, i

= SijEij+12Sijui, j− Sji uj, i

= SijEij= S ⋅ E

The penultimate line of the proof relies on the symmetry of the stress tensor.

133. Show that ∇u ⋅C∇u= λ divu divu +m ∇u+∇uT ⋅ ∇u for an isotropic, lin-ear, elastic material. Express this equation in component form.

Since the components of C are Cijkl = λδij δkl+ m δikδjl+δilδjk we have

ui, j Cijkluk,l = ui,j λδij δkl+ m δikδjl+δilδjk uk, l= λui, i uk,k+ m ui, j uk,l δikδjl+ui, j uk,l δilδjk

= λui, i uk,k+ m uk, j uk,j+ ui,j uj, i

= λui, i uk,k+ m uk, j uk,j+ uk,j uj,k

= λui, i uk,k+ m uk, j+ uj,k uk, j= λdiv u div u+ m ∇u+∇uT ⋅ ∇u

134. Carry out the derivation of the principle of virtual work for the case in which the realdisplacements are known and a system ofvirtual forces are applied to thebody, and therebydeduce the principle of virtual forces. Specifically, apply virtual body forces b and virtualsurface tractions t, and define the complementary external virtual work as

W^E ≡

B

u ⋅ b dV + Ω

u ⋅ t dA

where u is the real displacement of the body. Perform a derivation similar to the one forthe principle of virtual displacements to demonstrate that an appropriate definition ofcomplementary internal virtual work is

W^I ≡

B

S ⋅ E dV


where S is the virtual stress associated with the applied virtual force system and E is thestrain tensor associated with the real displacements. Prove the principle of virtual forces,which states that if W

Ê= W

Î for all virtual stresses S in equilibriumwith the appliedvirtu-

al forces b and t, then E= 12∇u+∇uT . State precisely the conditions that must hold

in order for the principle to be valid.

The external complementary virtual work has the following equivalent expressions

WÊ =

Ω

t ⋅ u dA + B

b ⋅ u dV

= Ω

t−Sn ⋅ u dA + Ω

Sn ⋅ u dA + B

b ⋅ u dV

= Ω

t−Sn ⋅ u dA + B

div Su + b ⋅ u dV

= Ω

t−Sn ⋅ u dA + B

divS+b ⋅ u dV + B

S ⋅ E dV

− B

S ⋅ E− 12∇u+∇uT dV

= Ω

t−Sn ⋅ u dA + B

divS+b ⋅ u dV + B

S ⋅ ∇u dV

The third term is the complementary internal virtual work. The difference betweencomplementary external and internal virtual work will be defined as the complementa-ry virtual work functional

G^(S, u)≡ W

Ê−W

Î

G^(S, u) =

Ω

t−Sn ⋅ u dA + B

divS+b ⋅ u dV

− B

S ⋅ E− 12∇u+∇uT dV

From the fundamental theorem of the calculus of variations, we obtain the principle ofvirtual work for a three-dimensional solid: If G

^(S, u)= 0 for all virtual stress fields S

that are in equilibrium with the applied virtual forces, i.e., divS+ b= 0 in B andt= Sn on Ω, then the strains are compatible with the displacements, that is, they sat-isfy E= 1

2∇u+∇uT .

ℓ

x

b(x)

to tℓ


135. The virtual-work functional for the littleboundary value problem is given by

G(σ, u)= ℓ0

σu′−bu dx− tou(0)− tℓu(ℓ)

The body force b(x) and the traction at the free end tℓ are known while the stress σ(x) andthe reaction to are unknown. Using the principle of virtual work, select a virtual displace-ment field that allows you to compute the reaction force in terms of only the known appliedforces. Give the expression for to in terms of b(x) and tℓ.

Note that G(σ, u)= 0 represents a statement of equilibrium for any choice of u. If wechose u(x)= uo (a constant) we get an interesting result because u′ = 0. Recognizingthat, for this choice, u(0)= u(ℓ)= uo we get

−ℓ0

buo dx− touo− tℓuo = 0

Since uo factors out of every term we get the result

ℓ0

b dx + to + tℓ = 0

which represents overall horizontal equilibrium of the applied forces and the reactions.

136. In the little boundary value problem, we saw that the virtual-work functional couldbe stated as a weighted residual functional. A weighted residual functional for a three-di-mensional solid body B with boundary Ω (with normal vector field n) can be defined as

G(S,w)≡−B

div S+b ⋅ w dV−Ω

t−Sn ⋅ w dA

where divS+b is the equilibrium residual in the domain B, t−Sn is the equilibrium re-sidual on the boundary Ω, and w is an arbitrary weighting function. Show that theweighted residual functional is identical to the virtual-work functional given in Eqn. (295),and, therefore, that the arbitrary weighting function is identical to the virtual displace-ment, i.e., w= u.

The correspondence can be shown by applying the divergence theorem

x

ℓk

P

C (pile)

F= 2kℓu(ℓ)

(elasticmedium)


G(S,w) ≡ −Ω

t−Sn ⋅ w dA − B

divS+b ⋅ w dV

= −Ω

t−Sn ⋅ w dA − B

div Sw + b ⋅ w dV + B

S ⋅ ∇w dV

= −Ω

t ⋅ w dA − B

b ⋅ w dV + B

S ⋅ ∇w dV

If we interpret w as a virtual displacement then we have G(S,w) = WI−WE.

137. Consider the pile of length ℓ, constant modulus C(w/ unit area), embedded in an elastic medium with mo-dulus k (force per unit displacement per unit length), andsubjected to a load P at x= 0. The pile is elastically re-strained at the end x= ℓ giving an end force of theamount F= 2kℓu(ℓ) as shown. The governing differen-tial equation for the system is Cu′′(x)−ku(x)= 0.Whatmust be the value of the constantα for the solution to havethe form u(x)= Aeαx+ Be−αx?

What are the values of the constants A and B that satis-fy the problem shown in the figure? Does this functionu(x) represent a classical solution to the given problem? Why or why not? Are there anyother solutions to this specific problem?

(a) What must be the value of the constant α for the solution to have the formu(x)= Aeαx+ Be−αx? The given displacement has to satisfy the differential equation

u′ = α(Aeαx− Be--αx ) ∴ Cα2u− ku= 0

u′′ = α2(Aeαx+ Be--αx )= α2x [Cα2− k ] u(x)= 0 ⇒ α= kC

(b) What are the values of the constants A and B that satisfy the problem shown in thefigure? Does this function u(x) represent a classical solution to the given problem?Why or why not? Are there any other solutions to this specific problem?

u′ = α(Aeαx− Be--αx ) ∴ Cα2u− ku= 0

The constants A and B have to satisfy the boundary conditions

σ(0)P

at x= 0

σ(ℓ) F= 2kℓ u(ℓ)

at x= ℓ

1

x

b(x)=−(1−x2)


σ(0)=− P σ(ℓ)=− 2kℓ u(ℓ)

Cu′(0)=− P (1) Cu′(ℓ)=− 2kℓ u(ℓ) (2)

Let m≡ 2αℓ and P≡ P∕αC

(1)→ Cα (A− B)=− P ⇒ A= B− P

(2)→ αAeαℓ − Be--αℓ + 2α2ℓAeαℓ + Be--αℓ = 0

e2αℓ 1+ 2αℓ A− 1− 2αℓ B= 0

em(1+ m) A= (1− m) B

From (1) A= P− B. Substituting in (2)

em(1+ m) (P− B)= (1− m) B

⇒ B= P em

em− 1−m1+m A= P

1−m1+m

em− 1−m1+m

Since u satisfies the classical differential equation and boundary conditions, it is anexact solution to the problem, and this solution is unique.

138. Consider the rod of length ℓ=1 and constantmodu-lusC=1. The rod is restrained by an elastic spring ofmo-dulus k=1 at each end and rests on an elastic foundation,alsowithmodulus k=1. The rod is subjected to a quadrat-ically varying body force as shown. The displacementu(x), positive in the x direction, is governed by the follow-ing differential equation u′′−u= 1−x2. What are the boundary conditions for this prob-lem? Is the following displacement function a classical solution to this problem?

u(x) = 1+ x2− 2ex− 12e−x

Why or why not? If it is not then modify it so that it is.

(a) What are the boundary conditions for this problem?

σ(0)ku(0)

at x= 0

σ(ℓ) ku(ℓ)

at x= 1

Cu′(0)− ku(0)= 0 Cu′(1)+ ku(1)= 0

u′(0)− u(0)= 0 Cu′(1)+ ku(1)= 0


(b) Is the given displacement function a classical solution to this problem? Why orwhy not? If it is not then modify it so that it is. The solution has to satisfy the differen-tial equation and boundary conditions

u′(x) = 2x− 2ex+ 12e−x u′′(x) = 2− 2ex− 1

2e−x

Substituting in the differential equation

u′′(x)− u(x) = (2+ 2ex− 12e−x )− (1+ x2− 2ex− 1

2e−x )

= 1− x2

Therefore the solution satisfies the differential equation for all values of x. Checkingthe boundary conditions:

u′(0)− u(0)= (− 2+ 12)− (1− 2− 1

2)= 0 OK

u′(1)+ u(1)= (2− 2e+ 12e)− (2− 2e− 1

2e)= 4(1− e)≠ 0

The boundary condition at x = 1 is not satisfied. Thus the given function is not a solu-tion to the problem. We can modify the function to

u(x) = 1+ x2− 2e e

x− 12e−x

u′(x) = 2x− 2e e

x+ 12e−x u′′(x) = 2− 2

e ex− 1

2e−x

u′′(x)− u(x) = (2+ 2e e

x− 12e−x )− ( 1+ x2− 2

e ex− 1

2e−x )= 1− x2

u′(0)− u(0)= (− 2e+

12)− (1− 2

e−12)= 0

u′(1)+ u(1)= (2− 2+ 12e)− (2− 2− 1

2e)= 0

The modified function is the exact solution to the problem.

139. The principle of virtual work for a certain boundary value problem can be stated as

G(u, u) ≡ ℓ0

A(x) u′′u′′ + B(x) uu− b(x) u dx= 0 for all u(x)∈ F(0, ℓ)

where A, B, and b are known functions of x, u(x) is the unknown field, and a prime denotesderivative with respect to x. What is the classical differential equation that is equivalentto this variational statement?


To apply the fundamental theorem of the calculus of variations, we need to have afunctional in the form

G(u, u)= ℓ0

⋅ u dx+ Boundary terms

Hence, need to work on Au′′u′′ term using integration by parts. Note that,

Au′′′u′ = Au′′′′u+ Au′′′u′

Au′′u′′ = Au′′′u′ + Au′′u′′

Hence,

Au′′ ′u′ − Au′′ u′′ = Au′′ ′′u− Au′′u′′

or

Au′′u′′ = Au′′′′u+ Au′′u′′ − Au′′′u′

⇒ l0

Au′′u′′dx= l0

Au′′′′u dx+ Au′′u′ℓ0− Au′′ ′u

ℓ

0

Thus,

G(u, u)= l0

Au′′′′ + Bu− b u dx+ Boundary terms

By the fundamental theorem of the calculus of variations:

Au′′ ′′ + Bu− b= 0 for x ∈ (0, l)

is the classical differential equation

140. The classical (4th order) differential equation and boundary conditions for a certainboundary value problem are

Au′′′′ + Bu′′ + Cu = b for all x∈ [0, ℓ]

u(0)= 0, u(ℓ)= 0, Au′′(0)= 0, Au′′(ℓ)= 0

where A, B, C, and b are known constants, u(x) is the unknown field, and a prime denotesderivative with respect to x. Find an expression for the virtual-work functional associatedwith the classical differential equation. In other words, find the functional G that has theproperty that the statement “G(u, u)= 0 for all u∈ Fe” is equivalent to the classical dif-ferential equation and the highest derivative that appears in G is second order. Describeany restrictions that must be placed on Fe.

x1

x3

B

Ω

x2

n


Because Au′′′′ + Bu′′ + Cu− b= 0 we can define

G(u, u)≡ l0

Au′′′′ + Bu′′ + Cu− bu dx= 0

By the fundamental theorem of the calculus of variations, G(u, u)= 0 for all u impliesAu′′′′ + Bu′′ + Cu− b= 0. To balance derivatives integrate by parts

ℓ0

A u′′′′ u dx=−ℓ0

A u′′′ u′ dx+ A u′′′ u ℓ0

= ℓ0

A u′′ u′′ dx− A u′′ u′′ ℓ0+ A u′′′u

ℓ0

= ℓ0

A u′′ u′′ dx+ A u′′′u ℓ0

ℓ0

B u′′ u dx=−ℓ0

B u′ u′ dx+ B u′ u ℓ0

Therefore,

G(u, u)= ℓ0

A u′′ u′′−B u′ u′+C u u−bu dx+ A u′′′+B u′u ℓ

0

We can eliminate the boundary term by restricting u to satisfy essential B.C.’su(0)= 0 and u(ℓ)= 0.

141. Consider the solid spherical regionBwith surfaceΩ havinga unit normal vector field n, as shown in the sketch. Assume thatthere exists a scalar field w(x), of the position vector x, for whichwe can define the functional

G(w, v) ≡ B

∇v ⋅ ∇w− v dV−Ω

t v dA

that has the property that if G(u, v)= 0 for all (virtual) scalar functions v(x) then the clas-sical differential equations governing the real field w(x) are satisfied (i.e., G(w,v) is a“virtual-work” functional). Note that the scalar field t(x) is defined on the surface of thesolid region. Find the classical governing differential equation for w(x) that is implied bythe variational statement “G(u, v)= 0 for all v”.Determinewhat must be the relationshipbetween t(x) and w(x) on the surface of the sphere.


First note that

div(∇w v )= ∂∂xi∂w∂xi v = ∂

2w∂x2i

v+ ∂w∂xi∂v∂xi= ∆wv+∇v ⋅ ∇w

B

∇v ⋅ ∇w dV=B

div(∇w v ) dV−B

∆wv dV

Applying the divergence theorem

B

∇v ⋅ ∇w dV=Ω

∇w ⋅ n v dV−B

∆wv dV

where n is the normal to the surface. Substituting back into the virtual work functional

G(w, v) = −B

∆w+ 1 v dV+Ω

(∇w ⋅ n− t ) v dA= 0 ∀ v

Applying the fundamental theorem of the calculus of variations

∆w+ 1 = 0 in B (governing differential equation)

t = ∇w ⋅ n on Ω (natural boundary condition)

ℓ

bo

x

Chapter 6The Ritz Methodof Approximation

142. Consider the uniaxial rod shown in the sketch, fixed atx= 0, free at x= ℓ, and subjected to the linearly varying bodyforce indicated. The rod has a variable elastic modulusC(x)= Co 2−x∕ℓ, making it twice as stiff at x= 0 as it isat x= ℓ. Using the principle of virtual work, find the expres-sion for the displacement u(x) and stress σ(x) for the given body force and variable modu-lus, approximating the real and virtual displacements with polynomials.

Consider first a two-term polynomial approximation

u(x) = a1xℓ + a2

x2ℓ2 , u(x) = a1

xℓ + a2

x2ℓ2

Let ξ≡ x∕ℓ. The derivative of u is u′ = (a1+2a2ξ)∕ℓ, similarly for u. Substitutinginto the virtual-work functional, noting that b(x)= bo(1−ξ), we get

G(u, u) ≡ 10

Co

ℓ (2−ξ)a1+2a2ξ a1+2a2ξ − boℓ (1−ξ)a1 ξ+a2ξ2 dξ

In matrix form, then the coefficients a1, and a2 must satisfy the equation

= boℓa1

a2

32

43

43

53

16112

= boℓ2Co

a1

a2

313

− 752

⇒Co

ℓ

Thus, the displacement field is given by

u(ξ)= boℓ252Co

12ξ−7ξ2


The stress can be computed from σ= Cu′. Thus, we have

σ(ξ)= boℓ522−ξ 12−14ξ

Note that the approximation of the stress at the free end is σ(1)=−boℓ∕26, about a4% error.

We can examine the accuracy of the solution by using a cubic approximation. Letξ≡ x∕ℓ and let u(ξ)= a1ξ+a2ξ2+a3 ξ3 and u(ξ)= a1ξ+a2ξ2+a3 ξ3 be theapproximation for the real and virtual displacement fields. The derivative of u can becomputed as u′ = (a1+2a2ξ+3a3ξ2)∕ℓ, similarly for u. Substituting into the virtual-work functional, we get

G(u, u) ≡ 10

Co

ℓ (2−ξ)a1+2a2ξ+3a3ξ2 a1+2a2ξ+3a3ξ2 dξ

− 10

boℓ (1−ξ)a1 ξ+a2ξ2+a3ξ3 dξ

In matrix form, then, the coefficients a1, a2, and a3 must satisfy the equations

= boℓ2Co

= boℓ

a1

a2

a354

95

95

2110

120

a1

a2

a310189

54

⇒

32

43

43

53

16112

Co

ℓ

1663

− 53252

Thus,

u(ξ)= boℓ2756Co

192ξ−159ξ2+40ξ3

The stress can be computed from σ= Cu′. Thus,

σ(ξ)= boℓ7562−ξ 192−318ξ+120ξ2

The approximation of the stress at the free end is σ(1)=−boℓ∕126 which is less than1% error.

143. Reconsider the nonprismatic rod of problem 142 subjected to the linearly varyingbody force. However, consider the condition in which the rod is fixed at both ends withprescribed end displacements of u(0)= uo and u(ℓ)= u1. Solve the problem with theRitz method using polynomial base functions.

Let ξ≡ x∕ℓ and let u(ξ)= a0+a1ξ+a2ξ2 and u(ξ)= a0+a1ξ+a2ξ2 be the approx-imation for the real and virtual displacement fields. We must first satisfy the essentialboundary conditions

Chapter 6 The Ritz Method of Approximation 143

u(0)= a0= uo

u(1)= a0+ a1+ a2= u1

Thus, we have a1 = u1−uo−a2. Substituting these expressions into the originalapproximations gives

u(ξ)= uo+ u1−uo ξ + a2 ξ2− ξ

The virtual displacement field must satisfy homogeneous essential boundary condi-tions, u(0)= 0 and u(1)= 0. Therefore, u(ξ)= a2 ξ2−ξ . From these results wecan identify uo(ξ)= uo+ u1−uo ξ and h2(ξ)= ξ2−ξ. The virtual-work functionalis

G(u, u) ≡ 10

Co

ℓ (2−ξ)u1−uo+a2 (2ξ−1) a2(2ξ−1) dξ

− 10

boℓ (1−ξ)a2 (ξ2−ξ) dξ = 0

Carrying out the integrals we find that the coefficient a2 must satisfy the equation

−Co

6ℓu1−uo +

Co

2ℓ a2+boℓ12= 0 ⇒ a2 = 1

3u1−uo −

boℓ26Co

Thus, the displacement field is given by

u(ξ)= uo+ 13u1−uo 2ξ+ξ2 − boℓ2

6Co

ξ2−ξ

144. Consider using a basis for the virtual displacement different from the basis used forthe real displacement. What would be the ramifications of using a different number ofterms in the expansions for real and virtual displacements? That is

u(x) = Nn=1

anhn(x), u(x) = Mn=1

anhn(x)

where N≠ M. What happens if N> M? What happens if N< M? Perform some com-putations on the little boundary value problem to investigate this issue.

The number of terms in the approximation of u determines the degree of refine-ment of the approximation while the number of terms in the expansion of u determinesthe degree to which we satisfy the weak form of equilibrium. If N> M then the sys-tem of equations resulting from the Ritz method is underdetermined. The equations areall valid equilibrium statements, but are insufficient to determine all of the coefficientsof the displacement approximation. As such we get a family of possible displacementapproximations that satisfy equilibrium in a limited sense. If N< M then the system


of equations is overdetermined. Since all of the equations are valid equilibrium equa-tions they should be consistent and therefore solvable. For example, the little boundaryvalue problem with a uniform load with N= 3 and M= 4 gives rise to the followingself-consistent overdetermined equations.

= boℓ

a0

a1

a2

1

1

1

43

64

64

94

121314

= boℓ

a0

a1

a2

1

− 12

0

1 1

⇒

1 85

126

15

Co

ℓ

These equations can be solved by a least-squares projection. For an overdeterminedsystem of equations Ax= b, the least squares solution can be obtained as the solutionto ATAx= ATb (least--squares projection).

145. Consider using a basis for the virtual displacement different from the basis used forthe real displacement. What would be the ramifications of using different base functionsfor the real and virtual displacements? That is

u(x) = Nn=1

anhn(x), u(x) = Nn=1

an gn(x)

where gn(x)≠ hn(x). Perform some computations on the little boundary value problemto investigate this issue using, for example, polynomials for the real displacements andtrigonometric functions for the virtual displacements.

Let ξ≡ x∕ℓ and let u(ξ)= a1ξ+a2ξ2 and u(ξ)= a1 sin πξ+a2 sin(πξ∕2) be theapproximation for the real and virtual displacement fields for the little boundary valueproblem, fixed at the left end free at the right, subjected to a uniform load. The virtual-work functional is

G(u, u) ≡ 10

Co

ℓa1+2a2 ξ πa1 cosπξ+ π2 a2 cosπξ2 dξ− 1

0

boℓa1 sinπξ+a2 sinπξ2 dξ = 0

Integrating we get the following system of equations

= boℓa1

a2

0 − 4π

1 2− 4π

2π

2π

= boℓ2Co

a1

a2

1

− 12

⇒Co

ℓ

Which is the exact solution. Clearly the coefficient matrix and right side are different.The coefficient matrix is not symmetric. If the displacement approximation is not ex-

ℓ∕2 ℓ∕2

bo

x

ℓ∕2 ℓ∕2x

bo


act, then the choice of functions for the virtual displacements will bias the satisfactionof weak equilibrium.

146. The uniaxial rod shown has unit area, length ℓ, andelastic modulus C. The body force is characterized by

b(x) =0 0< x≤ ℓ∕2bo ℓ∕2< x≤ ℓ

Assume that the real and virtual displacements can be approximated by the expressions

u(ξ) = a0+a1ξ+a2ξ2 u(x) = a0+a1ξ+a2ξ2

where ξ≡ x∕ℓ. Using the principle of virtual work, compute the displacement field u(x)and the tractions at the two ends σ(ℓ) and σ(0).

First, the real and virtual approximations must satisfy the essential boundary condition.Setting u(0)= 0 and u(0)= 0 gives a0 = 0 and a0 = 0. The virtual-work functionalis

G(u, u) ≡ 10

Cℓa1+2a2ξ a1+2a2ξ dξ−

1

12

boℓa1ξ+a2ξ2 dξ = 0

resulting in the equations

52

= boℓa1

a2

1 1

1 43

38724

= boℓ2C

a1

a2 − 14

⇒Cℓ

Therefore, the approximate displacement is

u(ξ)= boℓ28C5ξ− 2ξ2

The approximate end tractions are σ(0)= 0.625boℓ and σ(ℓ)= 0.125boℓ.

147. The uniaxial rod shown has unit area and length ℓ.It is fixed at the left end, is free at the right end, and is sub-jected to a constant body force field b(x)= bo along itslength. The elastic modulus C(x) is characterized by

C(x) =2C 0< x≤ ℓ∕2

C ℓ∕2< x≤ ℓ


Find the classical solution to the governing differential equation. Using the principle ofvirtual work, compute a stress field σ(x) assuming a Ritz approximation as follows

u(ξ) = a0+a1ξ+a2ξ2 u(x) = a0+a1ξ+a2ξ2

where ξ≡ x∕ℓ. Because the modulus changes abruptly at x= ℓ∕2, the stresses andstrains are discontinuous at that point. Why is this discontinuity a problem for the polyno-mial base functions suggested? What happens if you increase the order of the approxima-tion?

(a) Within each segment the modulus is constant. Therefore, the solution in each seg-ment satisfies

2Cu1′′ + bo = 0 Cu2′′ + bo = 0

Where u1 is the solution in the region 0< x< ℓ∕2 and u2 is the solution in the regionℓ∕2< x< ℓ. Both equations can be integrated by direct quadrature. Let ξ≡ x∕ℓ, thetwo general solutions are

u1 =boℓ22C− 1

2 ξ2+ a1ξ+ a2 u2 =

boℓ2C− 1

2 ξ2+ b1ξ+ b2

The four constants can be found by the two boundary conditions (1) u1(0)= 0 and (2)Cu2′(1)= 0 and the two continuity conditions (3) u1(1∕2)= u2(1∕2) (continuity ofdisplacement) and equilibrium of force (4) 2Cu1′(1∕2)= Cu2′(1∕2), as shown in thesketch.

σ2(ℓ∕2)σ1(ℓ∕2)

The boundary conditions can be implemented as

u1(0)= 0

Cu2′(1)= 0

u1(1∕2)= u2(1∕2)

a2 = 0

boℓ2−1+b1 = 0

boℓ22C− 1

8+12 a1+a2 =

boℓ2C− 1

8+12 b1+b2

(1)

(2)

(3)

2Cu1′(1∕2)= Cu2′(1∕2) boℓ2− 12+a1 = boℓ2− 1

2+b1 (4)

These four equations can be solved to give a1 = 1, a2 = 0, b1 = 1, b2 =−3∕16.Thus, the displacement map is given by

− 14ξ2+ 1

2ξ

− 12ξ2+ξ− 3

16

0≤ ξ≤ 12

12≤ ξ≤ 1

u(ξ) = boℓ2C

And the stresses are given by σ(ξ)= C(ξ)u′(ξ) to be


1−ξ

1−ξ

0≤ ξ≤ 12

12≤ ξ≤ 1

σ(ξ) = boℓ

(b) The approximations must satisfy the essential boundary condition. Settingu(0)= 0 and u(0)= 0 gives a0 = 0 and a0 = 0. The virtual-work functional is

G(u, u) ≡ 12

0

2Cℓa1+2a2ξ a1+2a2ξ dξ+

1

12

Cℓa1+2a2ξ a1+2a2ξ dξ

− 10

boℓa1 ξ+a2ξ2 dξ = 0

Carrying out the integrations results and invoking the Fundamental Theorem of theCalculus of Variations results in the equations

1633

= boℓa1

a2

32

54

54

32

1213

= boℓ2C

a1

a2 − 211

⇒Cℓ


u(ξ)= boℓ233C16ξ− 6ξ2

The approximate stress is given by

32− 24ξ

16− 12ξ

0≤ ξ≤ 12

12≤ ξ≤ 1

σ(ξ) = boℓ33

3220

104

33σboℓ

Exact

Although the approximate displacement is reasonably accurate, the stress field is not.The base functions are smooth and thus the approximate strain is smooth, forcing thestress to jump at the midpoint. In the exact solution, the stress is smooth and the strainjumps at the midpoint.

148. Using a piecewise linear finite element basis, resolve Problem 147. Does the finiteelement basis suffer from the same problem as the polynomial basis?Why?What generalconclusions can we make about the smoothness of the approximation?

The piecewise linear finite element base functions have the explicit expressions

2ξ

2− 2ξ

0≤ ξ≤ 12

12≤ ξ≤ 1

h1(ξ) = 1

2π

x

b(x)= sin x


0

2ξ− 1

0≤ ξ≤ 12

12≤ ξ≤ 1

h2(ξ) = 1

The virtual-work functional can be expressed as

G(u, u) ≡ 12

0

2Cℓ2a1 2a1 dξ+

1

12

Cℓ−2a1+2a2 −2a1+2a2 dξ

− 12

0

boℓ2a1 ξ dξ − 1

12

boℓa1 (2−2ξ)+a2 (2ξ−1) dξ = 0

Resulting in the equations

316= boℓ

a1

a2

6 −2

−2 2

1214

= boℓ2C

a1

a2 516

⇒Cℓ


u(ξ)= boℓ2C 316 h1(ξ)+

516 h2(ξ)

which is exact at the free end. The approximate stress is given by

12h1′ + 20h2′

6h1′ + 10h2′

0≤ ξ≤ 12

12≤ ξ≤ 1

σ(ξ) = boℓ32

24 24

8 832σboℓ

Exact32

The stress again takes a jump at the midpoint, but this time it jumps because the linearapproximation of displacement within each element implies a constant variation ofstress in each element. The approximation is trying to capture the linear variation ofstress of the exact solution within these constraints.

149. Consider the rod of length 2π and constant modulusC=1, free at both ends and subjected to the sinusoidal bodyforce, as shown. The general classical solution for the givenloading is u(x)= a0+a1x+ sin x. Show that the given solu-tion satisfies the governing differential equation for the bar,and state the essential and natural boundary conditions. Usethe boundary conditions to find the integration constants a0 and a1. Explain any peculiarfeatures of the solution to this problem. Use a polynomial Ritz basis to find a two-termapproximate solution for the displacement field, using the principle of virtual work. Ex-plain any peculiar features of the Ritz approximate solution.

1

x

1 1

h1

h2

b(x)

h3


By direct differentiation of the solution: u′ = a1+ cos x and u′′ = − sin x,which is clearly equivalent to the governing equation for C=1.

The boundary conditions are both natural: u′(0)= 0 and u′(2π)= 0. Substitut-ing we get a1+1= 0 and a1+1= 0, respectively. These two equation give only theconstant a1 =−1, a0 is arbitrary, implying u(x)= a0−x+ sin x. It is not possible toestablish the position of the bar, only the strains. The loading is called a self-equilibrat-ing load.

Let u= a0+a1x be the approximate displacement (u′ = a1) and u= a0+a1xbe the virtual displacement (u′ = a1). The virtual-work functional then gives

G(u, u) ≡ 2π0

a1a1− (a0+a1x) sin x dx = 2πa1+1 a1

The approximate configuration is found by setting G= 0 for all values of the virtualdisplacement. Hence we get a1 =−1, as in the classical solution. Note that, again, theconstant a0 is undetermined.

150. Consider the rod of length 3, constant unit modulus C= 1(and unit area), fixed at x= 0. The rod is subjected to a certain(unspecified) distribution of body force b, as shown. Three piece-wise linear finite element basis functions are shown in the sketch.The functional expressions for the basis function hi (x) is

hi (x)=x− i+ 1, i --1≤ x≤ i

i+ 1− x, i ≤ x≤ i+1

0, elsewhere

An approximate displacement field can be constructed from thebase functions as u(x)= a1h1 (x)+a2h2 (x)+a3h3 (x). Find the stiffnessmatrixK consis-tent with this approximation and the principle of virtual work. Assume that the values ofthe coefficients are a1 = 1, a2 = 2, and a3 = 4. Plot the stress field associated with theapproximation. Find the equivalent force vector f.

(a) Find the stiffness matrix K consistent with this approximation and the principle ofvirtual work. The first derivatives of the base functions are (graphically)

h1′ h2′1

--1

h3′1 1

--1

ℓ

x F

ℓ ℓh1

h2

b(x)= bo

h3

k


Kij= 3

0

hi′ hj′ dx (note: C= 1)

= 10

hi′ hj′ dx+2

1

hi′ hj′ dx+3

2

hi′ hj′ dx

Carrying out the integrations we get

K=2 --1 0--1 2 --10 --1 1

(b) Assume that the values of the coefficients are a1=1, a2=2, and a3=4. Plot the stressfield associated with the approximation and find the equivalent force vector f.

The stress can be computed from the base functions as

a1 0 ≤ x≤ 1

a2−a1 1≤ x≤ 2

a3−a2 2≤ x≤ 3

σ= Cu′ =3i=1

ai hi′(x)=

Substituting the values of the coefficients given the stress is

σ

0

1

321

2

(c) The force vector can be computed as

2 --1 0--1 2 --10 --1 1

0--12

124=f= Ka=

151. Consider the rod of length 3ℓ, constant modulusC= 2kℓ, and unit area, fixed at x = 0 and spring supportedat x= 3ℓ, with spring constant k. The rod is subjected to apoint load F= 3boℓat midspan and a uniform body force bo,as shown. Three piecewise linear finite element basis func-tions are shown in the sketch. The functional expressions forthe basis function hi (x) is

hi (x)=x∕ℓ − i+ 1, i --1≤ x∕ℓ ≤ i

i+ 1− x∕ℓ, i ≤ x∕ℓ ≤ i+1


Set up the equilibrium equations implied by the principle of virtual work using the Ritzmethod (i.e., find K and f). Express your answer in terms of k, bo, and ℓ (not F and C).

The displacement interpolation is

u(x)= a1h1(x)+a2h2(x)+a3h3(x)

and the virtual work functional has the form

G(u, u)= 3ℓ0

Cu′u′ − bu dx− Fu3ℓ∕2 + k u(3ℓ)u(3ℓ)

Therefore, the stiffness matrix and force vector

Kij= 3ℓ

0

C hi′hj′ dx+ k hi (3ℓ) hj(3ℓ) fi= 3ℓ

0

b hi dx+ F hi 3ℓ∕2

The components of K can be computed as

K11= ℓ

0

C1ℓ1ℓ dx+2ℓ

0

C− 1ℓ− 1

ℓ dx= 2C

ℓ = 4k= K22

K12= 2ℓ

0

C− 1ℓ1ℓ dx=−C

ℓ =−2k= K21= K23= K32

K13= K31= 0

K33= 3ℓ

2ℓ

C1ℓ1ℓ dx+ k(1)(1)= Cℓ + k= 3k

The components of f can be computed as

f1= 2ℓ

0

b0 h1(x) dx+ F h13ℓ∕2 = b0ℓ+ 12F= 5

2b0ℓ = f2

f3= 3ℓ

2ℓ

b0 h3(x) dx= 12b0ℓ

Summarizing, we have

4 --2 0--2 4 --20 --2 3

K= k f= b0ℓ2

551

ℓ

bo

x


152. Solve the problem of the rod subjected to a triangularload shown in the sketch using the quadratic Lagrangian finiteelement base functions. Use at least two quadratic elements(i.e., five nodes with four segments of length ℓ∕4) to carry outthe solution.

The virtual-work functional can be put into the form

G(u, u) ≡ 4i=1

4j=1

aiKij aj−4

i=1

ai fi

Where the coefficients are computed as

Kij=Cℓ

1

0

hi′(ξ)hj′(ξ) dξ fi= boℓ10

(1−ξ)hi(ξ) dξ

For example,

K24 =Cℓ

1

12

16ξ−14 (16ξ−10) dξ = 2C3ℓ

f2 = boℓ12

0

(1−ξ)(8ξ2−2ξ) dξ + boℓ112

(1−ξ)(8ξ2−14ξ+6) dξ = 112 boℓ

The remainder of the values are summarized in the following equations

= boℓ

a1

a2

32 −16

−16 28

14

112

⇒C3ℓ

0 0

−16 2

a3

a4

0 −16

0 2

32 −16

−16 14

112

0

= boℓ2384C

a1

a2

37

56

a3

a4

63

64

The approximate displacement is then computed as

u(ξ)= boℓ384C

37h1(ξ)+ 56h2(ξ)+ 63h3(ξ)+ 64h4(ξ)

Note that the displacement at the free end is boℓ∕6C, which is exact.

ℓ

x

F 2F

ℓ ℓ


153. Consider the rod of length 3ℓ and constantmodulusC, fixedat both ends, and subjected to point loads of magnitude F and 2Fat the third points, as shown. Usea piecewise linear finite elementapproximation with nodes at the ends and at the third points.Write the expressions for the base functions hi (x). Compute Kand f associatedwith the discrete virtual-work function.Computethe coefficients a fromKa= f. Sketch the approximatedisplace-ment field. Compute the approximate stress field.

(a) Let ξ≡ x∕ℓ. The base functions are

ξ

2−ξ

0≤ ξ≤ 1

1≤ ξ≤ 2h1(ξ) = h11

0 2≤ ξ≤ 3

h2(ξ) = h21

0

ξ−1

0≤ ξ≤ 1

1≤ ξ≤ 2

3−ξ 2≤ ξ≤ 3

(b) The coefficients are computed as

Kij=Cℓ

3

0

hi′(ξ)hj′(ξ) dξ fi= Fhi(1)+ 2Fhi(2)

The resulting equations are

43

= F

a1

a2

2 −1

−1 2

1

2= Fℓ

C

a1

a253

⇒Cℓ


u(ξ)= FℓC43h1(ξ)+

53h2(ξ) 54

3CuFℓ

(c) The approximate stress is σ= Cu′. Therefore, from the interpolation,

σ(ξ)= F 43 h1′(ξ)+ 53 h2′(ξ)

−5

43σF

1

x

bo

C k

ℓ


154. Consider the rod of length ℓ and constant modulusC. The rod is fixed at the left end and restrained by an elas-tic spring of modulus k at the right end. The spring accruesforce equal to the product of spring constant and stretch ofthe spring, i.e., fs= ku(ℓ). The rod is subjected to aconstant body force b(x)= bo, as shown.What are the essential andnatural boundarycon-ditions for this problem? (Hint: Take a freebody diagram of the right end of the rod to getthe mixed boundary condition at that end). Find the classical solution to the boundary val-ue problem. At what point is the strain in the rod the greatest? Consider the two limitingcases (1) k→∞, and (2) k→ 0. What are the boundary conditions in these two limitingcases?What is the solution in these two cases? Find an approximate solution with the Ritzmethod and a polynomial approximation.

(a) What are the boundary conditions for this problem? The boundary condition onthe left end of the rod is u(0)= 0. To get the condition at the right end consider afreebody of the right end of the rod.

ku(ℓ)σ(ℓ)u(ℓ)

From equilibirum we have

σ(ℓ)+ ku(ℓ)= 0 ⇒ Cu′(ℓ)+ ku(ℓ)= 0

(b) Find the classical solution to the boundary value problem. The classical equationgoverning the response is Cu′′ = −bo. The general solution (homogeous plusparticular) of this equation is

Cu=− 12bo x2+ a1x+ a2

Substituting the condition u(0)= 0 gives a2= 0. Substituting the boundary condi-tion Cu′(ℓ)+ ku(ℓ)= 0 gives

−boℓ+ a1+ kC− 1

2boℓ2+ a1ℓ = 0

Upon simplification we find that

a11+ kℓC = boℓ1+ kℓ

C ⇒ a1= 1

2boℓ2+β1+β

where β≡ kℓ∕C is a nondimensional measure of the spring stiffness relative to themodulus of the rod. Substituting this constant back into the general solution we findthe expression for the displacement as

u(x)= boℓ22C2xℓ 2+β1+β− x2

ℓ2

ℓ

x

b(x)= bo

3uo uo


(c) At what point is the strain in the rod the greatest? Compute the strain as

Á= u′ = boℓC2+β

1+β− x

ℓwhich is a linear function. Therefore, the maxima are at the ends of the rod.

Á(0)= boℓC2+β1+β Á(ℓ)= boℓ

C 11+β

Thus, the strain is greatest at x= 0.

(d) Consider the two limiting cases (1) k→∞, and (2) k→0. What are the boundaryconditions in these two limiting cases? What is the solution in these two cases?

k→∞ u(ℓ)=− 1kσ(ℓ)→ 0 u(x)= boℓ2

2Cxℓ −

x2ℓ2

k→ 0 σ(ℓ)=−ku(ℓ)→ 0 u(x)= boℓ22C2xℓ −

x2ℓ2

The first represents a fixed end, the second is a free end.

155. Consider the rod of length ℓ and constant modulus C,fixed at both ends, and subjected to a uniform body forcebo as shown. The left end moves to the right by an amount3uo and the right end moves to the left by an amount uo.What are the essential and natural boundary conditions.Compute the displacement field u(x) using the principle ofvirtual work and the Ritz method with a quadratic approximation. Sketch the approximatedisplacement field. Compute the approximate stress field.

(a) What are the essential and natural boundary conditions.

Essential B.C.’s: u(0)= 3uo and u(ℓ)=−uoNatural B.C.’s: none.

(b) Compute the displacement field u(x) using the principle of virtual work and theRitz method with a quadratic approximation. Take a solution of the formu(x)= a0+a1x+a2x2. Substituting the essential boundary conditions we get

u(0)= a0= 3uo

u(ℓ)= a0+a1ℓ+a2ℓ2=−uo ⇒ a1=−4uoℓ − a2ℓ

Therefore, the displacment approximation can be expressed in the form

x

ℓk

P

C


u(x)= uo(x)+ a2h2(x)= uo3−4 xℓ+ a2 x2−xℓ

Now, the virtual displacement can be taken to satisfy the homogeous boundary condi-tions u(0)= 0 and u(ℓ)= 0. Hence, the virtual displacement is

u(x)= a2 x2−xℓ

The virtual work functional for this problem is

G(u, u)= ℓ0

Cu′u′ − bou dx= 0 ∀u

Substituting the interpolation we get

a2ℓ

0

C−4u0ℓ 2x−ℓ+ 2x−ℓ 2a2− bo x2−xℓ dx= 0

Carrying out the integrals we get

a20+ 13Ca2ℓ3+ 1

6boℓ3 = 0 ⇒ a2=−

bo2C

Thus, the approximation solution is

u(x)= uo3−4 xℓ−bo2Cx2−xℓ

(c) Compute the approximate stress field.

σ(x)= Cu′(x)= C− 4u0ℓ −

bo2C2x−ℓ

= − 4Cuoℓ −

bo22x−ℓ

156. Consider the rod of unit length ℓ = 1, constant unit modulusC= 1 (w/ unit area), embedded in an elasticmedium that gets stiff-er with depth. The elastic medium has variable modulus given byk(x)= 12(1+x), and the resistance to motion is linearly propor-tional to the displacement. The rod is subjected to a unit load atx= 0, i.e., P= 1, and is traction free at the end x= 1. The classi-cal governing differential equation for the displacement field u(x)of the rod is u′′−12(1+x)u= 0. Calculate an approximate valueof the displacement at the point of load using a linear approximationof the displacement field. Use the Ritzmethod to carry out the calculations. Is the approxi-mate solution a good one? Why or why not? Does the accuracy of the approximation de-pend upon the relative flexibilities of the rod and the elastic medium? How?


(a) Calculate an approximate value of the displacement at the point of load using alinear approximation of the displacement field. The virtual work functional for thisproblem is

G(u, u)= 10

u′u′ + 121+x uu dx− u(0)

Take the displacement approximation in the form

u(x)= a0+ a1x u(x)= a0+ a1x

u′ = a1 u′ = a1

Note that there are no essential boundary conditions. Substituting the approximationinto the virtual work functional we get

10

a1a1+ 12(1+x)a0+a1x a0+a1x dx− a0= 0 ∀ a0, a1


a0 18a0+10a1− 1 + a1 10a0+8a1 = 0

or, invoking the Fundamental Theorem of the Calculus of Variations,

18a0+10a1= 1

10a0+ 8a1= 0

which can be solve to give a0= 8∕44 and a1=−10∕44. Thus, the approximationdisplacement is

u(x)= 1224−5x u(0)= 2

11

(b) Is the approximate solution a good one? Why or why not? Does the accuracy ofthe approximation depend upon the relative flexibilies of the rod and the elasticmedium? How? Compute the error as the residual of the classical differential equation

e≡ u′′ − 12(1+x)u=− 12221+x 4−5x

~ 1

1

~ 2

0

e

x

Also, u′(0)=−5∕22≠− 1 (the value of the applied load at the end) andu′(1)=−5∕22≠ 0 (traction free at the other end). Therefore, this is not a veryaccurate approximation.


157. Consider the rod of Problem 156, now with length ℓ, constant modulus C (w/ unitarea), subjected to a load P at x= 0. The rod is embedded in an elastic medium such thatthe force developed is linearly proportional to the displacement at eachpoint. Themodulusk is constant. The elastic constants are related by kℓ2∕C= 1. The rod is pointed so thatthe traction at the end x= ℓ is zero. Find the virtual-work functional for the given prob-lem. What are the essential and natural boundary conditions? Find an approximation tothe displacement field using the Ritz method with a two-term approximation with the fol-lowing base functions h1(x)= eαx, h2(x)= e−αx, where α≡ k∕C .

The virtual work functional for the given problem is

G(u, u)= ℓ0

Cu′u′ + kuu dx− Pu(0)

There are no essential boundary conditions. The natural boundary conditions areσ(0)=−P and σ(ℓ)= 0.

Approximate the displacement field using the Ritz method with a two-termapproximation with the following base functions h1(x)= eαx, h2(x)= e−αx , withα≡ k∕C . The components of K and f are

Kmn= ℓ0

Chm′hn′+khmhn dx fm= Phm(0)

Substituting the base functions we get

K11= ℓ

0

α2C+k e2α x dx= α2C+k2α

e2α ℓ−1 = kα e

2α ℓ−1

K12= ℓ

0

−α2C−k dx= k−α2C ℓ = 0= K21

K22= ℓ

0

α2C+k e−2α x dx=α2C+k2α

1−e−2αℓ = kα 1−e

−2αℓ

The force vector has components f1= f2= P. Solving Ka= f gives

a1= αPk e2αℓ−1

a2= αPk 1−e−2αℓ

Note that, from the problem description αℓ = kℓ2∕C = 1. Hence,

u(x)= Pkℓe2−1

ex∕ℓ + e(2−x∕ℓ)

x

ℓ f (x)

P

koℓ


158. Consider the rod of length ℓ, constant modulus C (w/ unitarea), subjected to a load P at x = 0. The rod is embedded in anelastic medium that provides a resisting force proportional to thedisplacement at each point with amodulus k(x) that increases lin-early with depth, so that the force is f (x)= koxu(x)∕ℓ. The endresistance can be modeled as a spring of modulus koℓ. The elasticconstants are related by koℓ2∕C= 1. Find the virtual-work func-tional for the given problem. What are the essential and naturalboundary conditions? Find the displacement field using the Ritzmethod with a two-term polynomial approximation.

The virtual work functional for the given problem.

G(u, u)= ℓ0

Cu′u′ + koxℓ uu− Pu(0)+ koℓu(ℓ)u(ℓ)= 0

There are no essential boundary conditions. The natural boundary conditions can befound by taking a freebody diagram of each end as enforcing the equilibirumequations. They are σ(0)=−P and σ(ℓ)=−koℓu(ℓ).

Approximate the displacement field using a two-term polynomial approximation.Let ξ≡ x∕ℓ and

u′ = a1∕ℓ u′ = a1∕ℓ

u= a0+ a1ξ u= a0+ a1ξ

Substituting the interpolation into the virtual work functional gives

− Pa0+ koℓa0+a1 a0+a1

G(u, u)= 10

Cℓ2 a1a1+koξ

a0+a1ξ a0+a1ξ dξ

Carrying out the integrals, noting C= koℓ2, gives

a012 koℓa0+ 13koℓa1− P+ koℓa0+ koℓa1

+ a1koℓa1+ 13koℓa0+ 1

4koℓa1+ koℓa0+ koℓa1 = 0

Invoking the Fundamental Theorem of the Calculus of Variations we obtain

32koℓa0+ 4

3koℓa1= P

43koℓa0+ 9

4koℓa1= 0

These equations can be solved to give

x

bo

C k

ℓ

1

1

1

h1

h2

h3

uouo(x)

uo


a0= 1.409 Pkoℓ

, a1=−0.835 Pkoℓ

Therefore, the approximate displacement field is

u(x)= Pkoℓ1.409−0.835 xℓ

159. Consider the rod of length ℓ and constant modulus C.The rod is fixed at the left end and restrained by a linearelastic spring of modulus k at the right end. The elasticconstants are related by kℓ/C=2. The rod is subjected to aconstant body force b(x) =bo and a prescribed displacementat the left end of uo, as shown. Set up, the discrete equationsKa=f that result from applying theRitz method to the prin-ciple of virtual work using the base functions shown in thesketch. Express the answer in terms of k, ℓ, bo, and uo. Doyou expect the solution using the Ritz method to be the ex-act classical solution to the boundary value problem?Whatbase functions would you need to add to make the Ritzapproximation exact?

(a) Set up the discrete equations Ka=f that result from applying the Ritz method tothe principle of virtual work using the base functions shown in the sketch. The virtualwork functional has the form

G(u, u)= ℓ0

Cu′u′ − bou dx+ ku(ℓ)u(ℓ)

Take a displacement approximation in the form

u(x)= u0(x)+ h(x) ⋅ a, u(x)= h(x) ⋅ a

where h(x)= h1, h2, h3, a= a1, a2, a3, and a= a1, a2, a3. We cantake uo(x)= uo (constant), so that uo′(x)= 0. With these definitions the virtualwork functional takes the form G(a, a)= a T Ka−f , where

K=ℓ0

Ch′ h′ dx+ kh(ℓ) h(ℓ)

f=ℓ0

boh dx− kuo(ℓ)h(ℓ)

1

x

b(x)= 1

C kk


Note that

ℓ0

Ch′ h′ dx= 3Cℓ I h(ℓ)= 1, 1, 1

Carrying out the remaining integrals and making the substitutions we get

1 0 00 1 00 0 1

K= 3Cℓ + k

1 1 11 1 11 1 1

= k2

5 2 22 5 22 2 5

5ℓ∕6

ℓ∕2

ℓ∕6

f= b0 − k u0

1

1

1

= boℓ6

5

3

1

− k u0

1

1

1

(b) Do you expect the solution using the Ritz method to be the exact classical solutionto the boundary value problem? No. The exact solution is quadratic for this problem.Thus, we need quadratic base functions to represent it exactly.

160. Consider the rod of length ℓ = 1 and a constant modu-lus C= 10. The rod is restrained by an elastic spring of mo-dulus k= 2 at each end and is subjected to a constant bodyforce b(x)= 1, as shown. The virtual-work functional is

G(u, u)= ℓ0

Cu′u′−bu dx+ ku(0)u(0)+ ku(ℓ)u(ℓ)

What are the boundary conditions for this problem? Are they essential or natural boundaryconditions? Explain. Set up, the system of equations that result from using a quadraticapproximation of the displacement field with the Ritzmethod. Consider solving this prob-lem using the trigonometric approximation

u(x)= a1 sin πx+ a2 sin 2πx+ a3 sin 3πx

Is this approximation likely to give a good solution to the problem or not?

(a) The boundary conditions for this problem are both natural because they representequations of equilibrium. They can be obtained by taking a freebody diagram of eachend of the beam.

σ(0)− ku(0)= 0 ⇒ Cu′(0)− ku(0)= 0

σ(ℓ)− ku(ℓ)= 0 ⇒ Cu′(ℓ)− ku(ℓ)= 0

x1

x3

B

x2

1

1

1


(b) Set up the system of equations that result from using a quadratic approximation ofthe displacement field with the Ritz method. Take an approximation of the form

u= a0+ a1x+ a2x2 u= a0+ a1x+ a2x2

u′ = a1+ 2a2x u′ = a1+ 2a2x

Substituting the approximations into the virtual work functional gives

0 0 0

0 2x 4x20 1 2xK= C1

0

dx+ k

1 0 0

0 0 0

0 0 0

+ k1 1 1

1 1 1

1 1 1

4 2 2

2 12 12

2 12 463

K=

1

x2xf= b1

0

dx= 1

112

13

The governing equations are then Ka= f.(c) Consider solving this problem using the trigonometric approximation

u(x)= a1 sinπx+ a2 sin 2πx+ a3 sin 3πx

Is this approximation likely to give a good solution to the problem or not? Why orWhy not? This basis is not good for this problem because each base function satisfiesh(0)= 0h(1)= 0. Hence, the springs cannot be mobilized by the approximation.

161. Consider the solid cubical regionB shown in the sketch hav-ing unit dimensions. Let the scalar field w(x) characterize the re-sponse of the system. The field w is a function of the position vec-tor x. If we define the functional

G(w, v) ≡ B

∇w ⋅ ∇v− 3 v dV

then G= 0 (for all v) is a “virtual-work” statement of the equations governingw. The es-sential boundary conditions are such that w= 0 on the coordinate faces. Use the Ritzmethod with a single term approximation of the form w(x)= a0 x1 x2 x3 to determine theunknown field w. Describe how you would improve the approximation.


Let the virtual displacement field be v= a0 x1 x2 x3. Computing the gradients

x2 x3

x2 x1x1 x3∇w= a0

x2 x3

x2 x1x1 x3∇v= a0

∇w ⋅ ∇v= a0 a0 x22 x23+ x21 x23+ x22 x

21

Substituting into the virtual work functional, invoking the Fundamental Theorem of theCalculus of Variations, we get

10

10

10

x22x23+x21x23+x22x21 a0− 3x1x2x3 dx1dx2dx3= 0

13a0− 3

8= 0 ⇒ a0= 9

8


w(x)= 98x1 x2 x3

To improve the approximation add more terms, w(x)= a ⋅ h(x), where each hi(x)satisfies the essential B.C. (i.e., equal zero on the coordinate faces).

162. Reconsider the 2ℓ × 2ℓ × h block shown in Fig. 84. The block is fixed at the base,(i.e., u(x1, x2, 0)= 0) and subjected to a body force (self-weight) of b=−ρoe3 through-out the volume. Let the material have independent elastic constants λ and m. Solve theproblem by the Ritz method using the following assumption about the displacement field

u(x)= a1x3ℓe3+ a2 x1 x3e1+ x2x3e2

The base functions are h1(x)= x3ℓe3 and h2(x)= x1x3e1+ x2x3e2. These base func-tions are identical to the example problem in the book. The necessary derivatives of thefunctions are

divh1(x) = ℓdivh2(x) = 2x3

∇h1(x) = ℓe3 e3∇h2(x) = x3e1 e1+ x3 e2 e2+ x1e1 e3+ x2e2 e3

∇h1+∇hT1 ⋅ ∇h1 = 2ℓ2 ∇h1+∇hT1 ⋅ ∇h2 = 0

∇h2+∇hT2 ⋅ ∇h2 = x21+ x22+ 4x23∇h2+∇hT2 ⋅ ∇h1 = 0

The coefficient matrix Kij and right side fi can be computed from the equations

Kij = h

0

ℓ−ℓ

ℓ−ℓ

λdivhi divhj+ m ∇hi+∇hTi ⋅ ∇hj dx1dx2dx3


fi = h

0

ℓ−ℓ

ℓ−ℓ

−ρoe3 ⋅ hi(x1, x2, x3) dx1dx2

Carrying out the indicated integrations gives the following system of equations for theunknown coefficients a1 and a2

0

3λ+2m 3βλ

3βλ 2+4β2 λ+m 43hℓ4

1= − 2ρohℓ4β

a1

a2

where β≡ h∕ℓ. These equations can be solved to give the constants

a1 =−12ρoβ

2m+4β2(λ+m)2m(λ+2m)+β2(λ2+12λm+8m2)

a2 =32ρoβ

λβ2m(λ+2m)+β2(λ2+12λm+8m2)

It is interesting to note that when λ= 0, 2m= C, and there is no Poisson’s effect. Inthis case we get a1 =−ρoβ∕2C and a2 = 0 so that u3 =−ρo x3h∕2C.

163. Resolve Problem 162 with the following assumed displacement field

u(x) = a1x3ℓe3+ a2x23e3+ a3 x1x3e1+ x2x3e2

What is the contribution of the term x1x3e1+ x2 x3e2 (i.e., the lateral displacement) tothe response in parts (a) and (b)? Compute the stress tensor S implied by the displacementfields of parts (a) and (b). What body forces and surface tractions are implied by thesemaps? What terms would you add to the solution to improve the Ritz approximation?

The base functions are h1(x)= x3ℓe3, h2(x)= x23e3, h3(x)= x1x3e1+ x2x3e2. Thedivergence and gradient for each base function can be computed as follows

divh1(x) = ℓ

divh3(x) = 2x3

∇h1(x) = ℓe3 e3

∇h3(x) = x3e1 e1+ x3 e2 e2+ x1e1 e3+ x2e2 e3

divh2(x) = 2x3 ∇h2(x) = 2x3e3 e3

∇Sh1 ⋅ ∇h1 = 2ℓ2 ∇Sh1 ⋅ ∇h3 = 0

∇Sh3 ⋅ ∇h3 = x21+x22+4x

23∇Sh3 ⋅ ∇h1 = 0

∇Sh2 ⋅ ∇h1 = 4x3ℓ ∇Sh2 ⋅ ∇h3 = 0

∇Sh1 ⋅ ∇h2 = 4x3ℓ

∇Sh3 ⋅ ∇h2 = 0

∇Sh2 ⋅ ∇h2 = 8x23

where ∇Shi≡ ∇hi+∇hTi is the symmetric gradient. Carrying out the indicated integra-tions as in part (a) gives the following system of equations for the unknown coeffi-cients a1, a2 and a3


0

3λ+2m 3 λ+2m β

4λ+2m β243hℓ4

32 β

=− 43ρohℓ4

a1

a2

3λβ

4λβ2

3βλ 2m+4β2 λ+m a3

3 λ+2m β

4λβ2

β2

where β≡ h∕ℓ. These equations can be solved to give the constants

a1 =−ρoβ

λ+2m

a2 = ρom(λ+2m)+ β2(4m2+6λm−λ2)2m(λ+2m)λ+2m+2β2(3λ+2m)

a3 = ρoλβ2

2mλ+2m+ 2β2(3λ+2m)

It is interesting to note that when λ= 0, 2m= C, and there is no Poisson’s effect. Inthis case we get a1 =−ρoβ∕C, a2 = ρo∕2C, and a3 = 0 which is the classical resultfor a truss bar, with u3 = ρox23−2x3h ∕2C.

The contribution of the term x1x3e1+ x2 x3e2 is driven by the Poisson effect.The stress can be computed as

S= λdivu+ m∇Su = i

ai λdivhi+ mai∇Shi

For example, in part (a) the stress tensor has the explicit form (with divergence andtractions as indicated):

λa1ℓ+2ma2x3 0

S=

a2mx1

a2mx2

a1ℓ(λ+2m)+λa2x3

0

a2mx1 a2mx2

λa1ℓ+2ma2x3

0

divS= 0

a2 λ+2m

λa1ℓ+2ma2x3 0

Se1 = 0

a2mx1 a2mx2

λa1ℓ+2ma2x3Se2 =

a2mx1

a2mx2

a1ℓ(λ+2m)+λa2x3

Se3 =

Observe that the given displacement field does not do a very good job of representingthe stresses for this problem (except when Poisson’s effect goes to zero). It is clearly adifficult problem trying to select base functions that improve the representation ofstresses. Perhaps a good choice for the next term in the expansion would have the spe-cific form x1x23e1+ x2 x

23e2 .


164. Reconsider the 2ℓ × 2ℓ × h block shown in Fig. 84. As in Problem 162, the blockis fixed at the base, (i.e., u(x1, x2, 0)= 0) and subjected to a body force b=−ρoe3throughout the volume. Let the material have independent elastic constants λ and m. Letξ1 =

12x1∕ℓ+1 , ξ2 =

12x2∕ℓ+1 and ξ3 = x3∕h be a change of variables that maps

the block onto the unit cube with one vertex at the origin of the coordinate system(ξ1,ξ2,ξ3). Define the following functions

φ2(ξ)= (1−ξ1)ξ2ξ3

φ1(ξ)= ξ1ξ2ξ3

φ3(ξ)= ξ1 (1−ξ2)ξ3

φ5(ξ)= ξ1ξ2 (1−ξ3)

φ7(ξ)= ξ1(1−ξ2) (1−ξ3)

φ4(ξ)= (1−ξ1) (1−ξ2)ξ3

φ6(ξ)= (1−ξ1)ξ2 (1−ξ3)

φ8(ξ)= (1−ξ1) (1−ξ2) (1−ξ3)These functions have the property that, at each of the eight vertices, one of the functionshas unit value while the others are zero. They are, in fact, the finite element base functionsfor a hexahedron element. Let the displacement be approximated as

u(ξ) = 8i=1

aiφi (ξ)

where ai is a vector constant with component expression ai= aije j (no sum on j). Whatare the base functions hi(ξ) associated with this expansion? What is the physical signifi-cance of the coefficient vector ai? What does the essential boundary conditionu(ξ1, ξ2, 0)= 0 imply about the values of the coefficients in the expansion? Solve theblock problem using the base functions identified, as restricted by the essential boundarycondition.

It is convenient to write the expansion of the displacement in terms of scalar coef-ficients as

u(ξ) = 8i=1

3j=1

aij φi (ξ) ej

Then the base functions can be represented as hij= φi (ξ)e j.Let the nodes be numbered as shown in the figure. Then the function φi (ξ) has

unit value at node i and is zero at all of the other nodes. The coefficient aij is the nodaldisplacement at node i in the direction j.

1

2

3

4

85

6

7

ξ1

ξ2

ξ3


Since the base functions φ5, φ6, φ7, and φ8 are not zero at ξ3 = 0 the coefficientsmust be zero to enforce the essential boundary conditions. Therefore, aij= 0 for thevalues i=5, 6, 7, and 8 and j=1, 2, and 3. Since the base functions φ1, φ2, φ3, and φ4are zero at ξ3 = 0 the remaining coefficients are not restricted by the essential bound-ary conditions.

where differentiation of φi is with respect to x which can be done by the chain rule.

∇hij= (φie j ),k ek= φi,k e j ek

∇Shij ⋅ ∇hkl= φi,α φk,α δjl+ φi,l φk, j

The derivatives of the shape functions are

φ2,1=−aξ2ξ3

φ1,1= aξ2ξ3

φ3,1= a (1−ξ2)ξ3φ4,1=−a (1−ξ2)ξ3

φ2,2= a (1−ξ1)ξ3

φ1,2= aξ1ξ3

φ3,2=−aξ1ξ3φ4,2=−a (1−ξ1)ξ3

φ2,3= b (1−ξ1)ξ2

φ1,3= bξ1ξ2

φ3,3= bξ1 (1−ξ2)φ4,3= b (1−ξ1)(1−ξ2)

where a= 1∕2ℓ and b= 1∕h. The components of the coefficient and right side ma-trices can be computed as

K(ij)(kl) = 1

0

10

10

λφi, j φk,l+ m φi,α φk,α δjl+φi,l φk,j Jdξ1dξ2dξ3

f(kl) = 1

0

10

10

−ρoφkδ3l J dξ1dξ2dξ3

where J= 4ℓ2h is the determinant of the Jacobian. The double indices can be put intostandard matrix form by ordering the components in a particular way. For example,[11, 21, 31, 41, 12, 22, 32, 42, 13, 23, 33, 43], which simply runs through the first index(associated with node number) for each value of the second index (the coordinate di-rection). With this ordering we can put the results into matrix form by defining thevectors φ1, φ2, and φ3 to have components [φj] i= φi,j. In other words φj containsthe derivatives of the base functions with respect to xj. In the present case each of thesevectors has length 4 since there are four base functions. Let us further define the ma-trices

A≡φ11φ21φ31

C≡Φ

0

0

0

Φ

Φ0

0

0

φ12φ22φ32

φ13φ23φ33

B≡φ11φ12φ13

φ21φ22φ23

φ31φ32φ33

where

φij= φi φj and Φ≡3i=1

φi φi


Note that A, B, and C are 12×12. The coefficient matrices of the system of equationscan now be written

K = 10

10

10

λA+ m B+C Jdξ1dξ2dξ3

f = 10

10

10

−ρo00φ Jdξ1dξ2dξ3

where φ≡φ1, φ2, φ3, φ4. The problem can be solved as Ka= f to determine theconstants however there is a great deal of symmetry in this problem that one can use to

reduce the system of equations to a 2×2. Let us select a^ ≡a11, a13 as our vector ofprimary unknowns. By symmetry the remaining unknowns can be related to the presentones as a= Γa^ where Γ is given by

ΓT=01 −1 1 −1 1 1 −1 −1 0 0 0

1 1 1 10 0 0 0 0 0 0 0

The problem can now be solved as ΓTKΓ a^ = ΓTf. The computation can be simpli-fied by defining ΓT≡ ΓT1 ΓT2 ΓT3 along the partition lines described above. Further,define cij≡ ΓTi φj. Now we can observe that

A^ ≡ ΓTAΓ = c11+c22+c33 c11+c22+c33

and that

B^ ≡ ΓTBΓ = 3

i=1

3j=1

c ij cji C^ ≡ ΓTCΓ = 3

i=1

ΓTi ΦΓ i

Finally, the system of equations can be put into the form K^a= f

^, where

K^ = 1

0

10

10

λA^ + mB^+C^ Jdξ1dξ2dξ3

f^ = 1

0

10

10

−ρo ΓT3φ Jdξ1dξ2dξ3

The MATHEMATICAt commands needed to form and solve the problem are given be-low:

p = x y z , (1-x) y z , x (1-y) z ,(1-x) (1-y) zp1 = D[p,x]/(2 a)p2 = D[p,y]/(2 a)p3 = D[p,z]/bgam1 = 1,0,-1,0, 1,0,-1,0gam2 = 1,0, 1,0,-1,0,-1,0


gam3 = 0,1, 0,1, 0,1, 0,1c11 = Transpose[gam1].p1c12 = Transpose[gam1].p2c13 = Transpose[gam1].p3c21 = Transpose[gam2].p1c22 = Transpose[gam2].p2c23 = Transpose[gam2].p3c31 = Transpose[gam3].p1c32 = Transpose[gam3].p2c33 = Transpose[gam3].p3AA = c11 + c22 + c33AAA = Outer[Times,AA,AA]BB = Outer[Times,c11,c11] + Outer[Times,c12,c21] + Outer[Times,c13,c31] +

Outer[Times,c21,c12] + Outer[Times,c22,c22] + Outer[Times,c23,c32] +Outer[Times,c31,c13] + Outer[Times,c32,c23] + Outer[Times,c33,c33]

PHI = Outer[Times,p1,p1] + Outer[Times,p2,p2] + Outer[Times,p3,p3]CC = Transpose[gam1].PHI.gam1 +

Transpose[gam2].PHI.gam2 +Transpose[gam3].PHI.gam3

KK = ( lam AAA + mu ( BB + CC )) (4 a^2 b)K = Integrate[KK, x,0,1,y,0,1,z,0,1]FF = -(4 a^2 b)(Transpose[gam3].p)F = Integrate[FF, x,0,1,y,0,1,z,0,1]DD = Simplify[Inverse[K].F]

165. Consider a cube of dimension 2× 2× 2 fixed at the base and subjected to a bodyforce b=−ρoe3. Describe a method for refining the finite element approximation byestablishing a local coordinate system for each element that allows the creation of the fi-nite element base functions from the eight basic element functions φi (ξ) described inProblem 164. Notice that each element is associated with eight nodes while the entireblock is associatedwith 27 nodes. Continuity of displacements can be assured by associat-ing the element base functionswith element nodal displacements (i.e., finite element func-tions) and by associating elements nodal displacements with a common global displace-ment parameter where elements share a common node.

x1

x3

B

x2

2

2

2

x1

x3

x2

1

1

1

x1

x3

x2

1

1ξ1

ξ2

ξ3

1

1

1

11

1

1

Increase the order of interpolation. To refine the approximation a node can beplaced at the center of each edge and face and one at the center of the cube. Then, thequadratic Lagrangian functions can be used to create the functions. There are 27 nodesand hence 27 functions φi that take unit value at node i and are zero at the other nodes.

At each node i there exist two other nodes that have the same (ξ2,ξ3) coordinates.Call the ξ1 coordinates of those nodes ξ

a1 and ξ

b1. There are two other nodes that have


the same (ξ1,ξ3) coordinates. Call the ξ2 coordinates of those nodes ξc2 and ξ

d2. There

are two other nodes that have the same (ξ1,ξ2) coordinates. Call the ξ3 coordinates ofthose nodes ξe3 and ξ

f3. Define

Lj1(ξ1) =

(ξ1−ξa1)(ξ1−ξb1)

(ξj1−ξa

1)(ξj

1−ξb

1)

Lj2(ξ2)=

(ξ2−ξc2)(ξ2−ξd2)

(ξj2−ξc

2)(ξj

2−ξd

2)

Lj3(ξ3)=

(ξ3−ξe3)(ξ3−ξf3)

(ξj3−ξe

3)(ξj

3−ξf

3)

The base function for node j can be constructed by multiplying these three functionstogether

φj(ξ1, ξ2, ξ3) = L j1(ξ1)Lj

2(ξ2)L j

3(ξ3)

For example, consider the case shown in the figure with the node j taken at the cornerwith ξ1= ξ2= ξ3= 1. The functions are

L j1(ξ1) = ξ1 (2ξ1−1), Lj

2(ξ2) = ξ2 (2ξ2−1), L j

3(ξ3) = ξ3 (2ξ3−1)

Therefore, the base function for this node is

φj(ξ1, ξ2, ξ3) = ξ1 (2ξ1−1)ξ2 (2ξ2−1)ξ3 (2ξ3−1)

Observe that this function has unit value at node j and is zero at all other nodes.

ξ1

ξ2

ξ3

ja

b

cd

e

f

There are many other possible higher-order finite element base functions with differentdefinitions of the nodal positions.

Refine the mesh. An alternative to increasing the order of interpolation is to usethe same base functions, but to have them describe the solution over a smaller portionof the domain. For example, we can subdivide the original problem into eight regionsof half the dimension on each side. For each region we can define a change of coordi-nates to map that region onto the unit cube. Now, for each region (for example, theshaded region in the sketch), the base functions given for the eight-node cube provide asatisfactory building block for the finite element base functions. Since the original basefunctions have the property that they have unit value at one node and are zero at theother nodes, they can be pieced together to get base functions that are continuous be-tween adjacent regions.

Let us assume that all of the nodes have the local numbering show in the sketchfor the single isolated element. The base functions can be constructed by piecing to-


gether the local base functions φi(ξ) defined in the problem statement. For example,the global base function that has unit value at node k and is zero at all other nodes canbe defined as follows:

φ3(ξA)e jφ4(ξB)e j

x∈ A

x∈ Bhkj(x) =

0 x∉ A or B

The change of variable is different for each region. For example

ξA =x1∕ℓx2∕ℓ+1

2x3∕h−1

ξB =x1∕ℓ+1

x2∕ℓ+1

2x3∕h−1

These functions can be constructed systematically by establishing a global node num-bering (each node in the global mesh has a distinct number) and establishing the corre-spondence between the local and global node numbers. The details of these transforma-tions are covered in books on finite elements.

ξ1

ξ2

ξ3

x1

x2

x3

1

2

34

578

ABk

ℓ ℓℓℓ

h∕2

h∕2

ξA

166. What is the appropriate definition of f in Eqn. (366)when there is a nonzeroboundarydisplacement term u^(x) in the Ritz approximation?

Substituting Eqn. (352) into the virtual-work functional

G(u, u)= B

λ divu divu + m ∇Su ⋅ ∇u dV−B

b ⋅ u dV−Ωt

t^ ⋅ u dA

we get, for the jth component of f

fj = Ωt

t^ ⋅ hj dA+

B

b ⋅ hj− λ divuo divhj + m ∇Suo ⋅ ∇hj dV


where ∇Su≡ ∇u+∇uT is the symmetric gradient of u.

x

50

x1x2

2

15

4

x2

x1

Chapter 7The Linear Theoryof Beams

167. The beam shown below has a rectangular crosssection of depth 4 and width 2, and has a length of 50length units. It has a uniform mass density that givesrise to a constant body force of b(x)=−2e2 (forceunits per length units cubed), and is subjected to asurface traction on its top surface that is bilinear withrespect to x2 and x3 ≡ x reaching a maximum valueof 15 (force units per length units squared), as shown.

Find an expression for the applied tractions tΓ(x).Find the resultant applied loads q(x) andm(x) equiv-alent to the surface tractions and body force. Find thedistribution of resultant force Q(x) and resultant momentM(x) along the beam. Find thedisplacements w(x) and the rotations θ(x) along the beam.

(a) Find an expression for the applied tractions tΓ(x). The distribution of the tractionson the top surface is bilinear, i.e., at any fixed value of x1 the variation with x3 is linearand at any fixed value of x3 the variation with x1 is linear, as shown in the sketch.

x1

x2c2(x3) 1+x1

x3

x2c1(x1)x3

Since for any x1 the value of the function is zero at x3 = 0 it must have the formc1(x1)x3. Similarly, since for any x3 the value of the function is zero at x1 =−1 itmust also have the form c2(x3) 1+x1 . Combining these two observations we conclude


that the distribution must have the form cx3 1+x1 . At the point x1 = 1, x3 = 50 thefunction takes the value 15. Therefore, we must have c= 3∕20. The tractions act inthe direction −e2. Therefore, the applied tractions are given by

tΓ(x) =− 3

201+x1 x3e2

0

on Ω1

elsewhere

where Ω1 is the top surface of the beam (x2 = 2).

(b) Find the resultant applied loads q(x) and m(x) equivalent to the surface tractionsand body force. The applied loading per unit length is

q x = Ω

b dA + Γ

tΓ ds

= 2−2

1−1

−2e2dx1dx2+1

−1

− 320 x1+x1 e2dx1

=−16+ 310 xe2

The applied moment per unit length is

m x = Ω

p× b dA + Γ

p× tΓ ds

The position vector px1 , x2 on the top surface is given by p= x1e1+2e2. Thus,

m x = Ω

x1 e1+x2e2 × −2e2 dA + 1

−1

x1 e1+2e2 × tΓ dx1

=− 110x e3

= 2−2

1−1

−2x1e3 dx1dx2 −1

−1

320 xx21+x1 e3 dx1

(c) Find the distribution of resultant force Q(x) and resultant momentM(x) along thebeam. The resultant force Q(x) is related to the applied loading q(x) by the equationQ′x+q x = 0. Integrating with respect to x, we get

Qx = 16x+ 320 x

2 e2 + c

Since Q50 = 0 , then c=−1175e2. Thus,

Qx = 16x+ 320 x

2−1175e2

Chapter 7 The Linear Theory of Beams 175

The resultant momentM(x) is related to the applied moment m(x) by

M′x + e3 ×Qx +m x = 0

By plugging in the value of the resultant force Q(x) into the equation above and inte-grating it with respect to x, we get,

Mx = −1175x−8x2− 120 x

3 e1 + 120 x

2 e3 + d

Since M 50 = 0 , then d= 32500e1−125e3. Thus,

Mx = −1175x−8x2− 120 x

3−32500 e1 + 120 x2−125 e3≡ M1

xe1 +M3xe3

(d) Find the displacements w(x) and the rotations θ(x) along the beam. The tensor Sis a zero tensor since the origin is located at the centroid of the cross section. There-fore, the equations for the shear stress and moment resultants are uncoupled. To wit,

Q = AÁo ⇒ Áo= A−1Q

M= Iκo ⇒ κo= I−1M

The cross-sectional constants can be evaluated as

A=Ω

dA= 8 I11 = Ω

x21dA=83

I22 = Ω

x22dA=323

I12 = Ω

x1 x2dA= 0

So,

= 18m−1175+16x+ 3

20x2 e21

8mÁo =

0

0

18m 0

0

0 018E

Q2(x)

0

0

= 332E

M1xe1 +

340m

M3xe33

8Eκo =

0

0

332E 0

0

0 0340m

M3(x)

0

M1(x)

It follows that, since θ′x = κo x, the rotation vector can be computed as

θ(x)= 332E32500x− 1175

2x2+ 8

3x3+ 1

80x4 e1 + 3

40m 160x3− 125xe3

τo

x3

τo

x3

x1

ℓ


where the vector constant of integration vanishes because θ 0 = 0. By using the resultfor the rotation above, we can integrate the following equation

w′x = Áo x − e3× θx = 18m Q2x−θ1 xe2

to get the displacement vector

− 332E16250x2− 1175

6x3+ 2

3x4+ 1

200x5 e2

w(x) = 18m−1175x+ 8x2+ 1

20x3e2

where the vector constant of integration vanishes because w 0 = 0.

168. Consider the beam with square cross section, of di-mension h by h and length ℓ. The beam has Young’s modu-lus C and shear modulus m. The beam is subjected to hori-zontal tractions on its top face, as shown. The body forcesacting on the beam are negligible. The coordinate axesshown are principal and centroidal.

Find expressions for the applied force and moment perunit length of beam, q(x) andm(x), where x= x3 is the axialcoordinate. Find the displacement and rotation field caused by the loading by integratingthe governing beam equations (that is, find the classical solution).

(a) Find expressions for the applied force and moment per unit length of beam, q(x)and m(x), where x= x3 is the axial coordinate. The traction vector can be written as,tΓ = τoe3 which act only at the top surface. Therefore, the applied load is

q=h2

−h2

τoe3 dx2 = τo he3

and the applied moment is

m=h2

−h2

12 he1+x2 e2 × τoe3 dx2 =− 1

2 τo h2e2

(b) Find the displacement and rotation field caused by the loading by integrating thegoverning beam equations (that is, find the classical solution).

1

x1

x2

x3

11

1 tℓ


x2

x3

h

h

I11 = I22 =112h4

A= h2

Sα = 0

The resultant force Q(x) is related to the applied loading q(x) by the differential equa-tion Q′x = −q x. Integrating with respect to x, we get

Qx = −τo h xe3 + c1

Since Qℓ = 0 , we get c1 = τo hℓe3. Thus,

Qx = τo h ℓ−xe3

Since, e3 × Qx = 0, the resultant momentM(x) is related to the applied momentm(x) by M′x+m x = 0. Integrating this with respect to x, we get

Mx = 12τo h2x e2 + c2

Since Mℓ = 0 , we get c2 =−12τo h2ℓe2. Thus,

Mx = − 12τo h2 ℓ−x e2

It follows that, since θ′(x) = I–1M(x), we can integrate to obtain the rotation field

θx = −3 τoEh22ℓx−x2 e2 + c3

Since θ0 = 0 , we get c3 = 0. Also since Áo = A−1Q the displacement can be ob-tained by integrating the equation w′x = A−1Q− e3× θx. Thus,

wx = − τoEh23ℓx2−x3e1 +

τoEhℓx− 1

2x2e3 + c4

Since w0 = 0, we have c4 = 0.

169. Consider a beam of length ℓ withsquare 2×2 cross section. The beam is sub-jected to the applied traction field over thecross section at the end of the beam

tℓ = τ x1 1+x22 e3where τ is the knownmagnitude of the loading. Find the resultant force andmoment actingon the end of the beam. Assume that u(x1, x2, 0), i.e., the beam is fixed at x3 = 0. Find the

x1

x2

x3 ℓ

x2

b

t


resultant force fieldQ(x) and the resultant moment fieldM(x) that equilibrate the appliedforces. Compute the displacement and rotation fields that result from the applied loads.

The resultant force at the right end can be computed as

Q(ℓ)=Ω

tℓ dA=1

−1

1−1

τ x1 1+x22 e3 dx1 dx2

= 1−1

τ 1+x22 e3 dx2 x212 1

−1

= 0

The resultant moment at the right end, noting that p= x1e1+x2e2

M(ℓ)=Ω

p× tℓ dA

= 1−1

1−1

−τx21 1+x22 e2+ τx1x2 1+x22 e1 dx1 dx2

= 1−1

1−1

−τx21 1+x22 e2 dx1 dx2

= τ x3131−1

x2+ x32

31−1

e2 =−169τ e2

170. The hollow box beam shown has a square crosssection of dimension b, thickness t ≪ b, and unitweight ρb. It is submerged in a fluid of unit weight ρo.Recall that the pressure at any point in a static fluid isproportional to the depth h according to the relationshipp = hρo. The unit weight of the air inside the beam canbe taken as zero. The end is capped so that fluid cannotget inside. Plot the typical traction field tΓ acting on thelateral surface of the beam. Compute the resultant ap-plied load q(x) and the resultant applied momentm(x)that would be appropriate in order to treat the problem using beam theory.

(a) Plot the typical traction field tΓ acting on the lateral surface of the beam.


Sketch traction field here.

ρo b ρo b

(b) Compute the resultant applied load q(x) and the resultant applied moment m(x)that would be appropriate in order to treat the problem using beam theory. The traction,pressure, and body force fields are

(1) (3)

(4)

(2)

tΓ=

−ρo x2e1 (1) dx2

ds

−ρo be2 (2) dx1ρo x2e1 (3) dx2

0 (4) dx1

p=

be1+ x2e2 (1)x1 e1+ be2 (2)

x2e2 (3)x1 e1 (4)

b= ρb te2 all segments.

The loading q is defined as

q(x)=Γ

tΓ ds+Ω

ρbe2 t ds

Using the quantities defined previously we have

q=b0

−ρo x2e1 dx2+b

0

− ρobe2 dx1+b

0

ρo x2e1 dx2+ 4bt ρbe2

= 4btρb−ρob2 e2The loading m is defined as

m(x)=Γ

p× tΓ ds+Ω

p× ρbe2 ) t ds

Using the quantities defined previously we have

m=b0

be1+x2e2)× −ρo x2e1+tρb e2) dx2

+b0

x2e2 × −ρo x2e1+tρbbe2 dx2+b

0

x1e1 × tρbe2 dx1

+b0

x1e1+be2 × −ρobe2+tρb e2 dx1



m= ρo b33 e3+ ρb tb2e3+ −ρo b33 e3+ ρb tb22e3

= 2ρb tb2− 12ρo b3e3

+ −ρo b33 e3+ 0 + ρb t b22 e3

171. The Saint-Venant torsion problem is restricted to problemswith constant rate of twistand traction-free lateral surfaces. One feature of this solution is that, at the fixed end, therotation is restrained but the out-of-plane warping is not. Physically, such a boundary con-dition would be very difficult to realize. One solution to this problem is to create a modelin which the amplitude of warping is independent from the rate of twist of the beam. Con-sider the deformation map

u(x) = θ(x3) e3× p + φ(x3)ψ(x1, x2)e3where θ(x3) is the angle of twist, φ(x3) is the amplitude ofwarping, ψ(x1, x2) is thewarpingfunction, and p= x1e1+x2e2. Further, assume that the warping function ψ(x1, x2) is theSaint-Venant warping function derived in the text. Compute the strain tensor and use thelinear elastic constitutive equations to show that the tractions on a cross section are

Se3 = mφ∇ψ+ θ′e3× p + Eφ′ψe3where E≡ λ+2m, and a prime denotes differentiation with respect to x3.In addition to the polar moment of inertia J, define the cross-sectional properties

J1 ≡ Ω

p×∇ψ ⋅ e3 dA, J2≡ Ω

∇ψ ⋅ ∇ψ dA, J3≡ Ω

ψ2 dA

which can be computed once ψ(x1, x2) is known. Define the stress resultants

T≡ e3 ⋅ Ω

p× Se3dA, W≡ e3 ⋅ Ω

ψSe3dA, B≡Ω

∇ψ ⋅ Se3dA

where T is the usual torque. The stress resultantsW and B are often called the bi-momentand the bi-shear. Substitute the expression for Se3 to show that

T= mJ1φ+ mJθ′, W= EJ3φ′, B= mJ2φ+ mJ1θ′Show that if φ(x3)= θ′(x3)= α, then the above results are consistent with the Saint-Ve-nant problem discussed in the text.

Using the definitions of the stress resultants, compute T′ and W′ and show that

T′ + t = 0, W′ − B+ w= 0

What are the appropriate definitions of the applied loads t(x) and w(x)?Substitute the resultant constitutive equations into the resultant equilibrium equations,

and show that the equations


mJθ′′ + mJ1φ′ + t= 0, EJ3φ′′ − mJ2φ− mJ1θ′ + w= 0

govern the spatial variation of rotation θ and warping φ. These equations constitute a pairof second-order ordinary differential equations, and, therefore, we can expect fourconstants of integration thatmust be found from boundary conditions.What are thebound-ary conditions for a free end and a fixed end?

(a) Compute the strain tensor and use the linear elastic constitutive equations to showthat the tractions on a cross section are given by

Se3 = mφ∇ψ+ θ′e3× p + Eφ′ψe3

where E≡ λ+2m, and a prime denotes differentiation with respect to x3.The gradient of the displacement field is given by ∇u= u, i ei where

u,1 = θ x3 e3× e2 + φψ,1 e3 = θ x3 e2 + φψ,1 e3

u,2 =−θ x3 e1 + φψ,2 e3

u,3 = θ′ x3 e3× p + φ′ψ e3

Thus, the strain tensor can be expressed as

E= 12∇u+∇uT

= 12φψ,1 e1 e3+e3 e1+φψ,2 e2 e3+e3 e2

+ θ′ e3 × p e3+e3 e3 × p+2φ′ψ e3 e3

Noting that, from the equation above, trE = φ′ψ, we have (see text)

Se3 = λ φ′ψe3+ 2mEe3

where the term Ee3 can be computed as follows

Ee3 =12φ ψ,1 e1+ψ,2 e2 + θ′ e3× p + θ′ e3× p ⋅ e3e3+ 2φ′ψe3

= 12φ∇ψ+ θ′ e3× p + 2φ′ψe3

since e3× p ⋅ e3 = 0. By plugging this result into the relationship above, we get

Se3 = λ+2m φ′ψ e3+ m φ∇ψ+ θ′ e3× p

= Eφ′ψe3+ m φ∇ψ+ θ′ e3× p

(b) In addition to the polar moment of inertia J, define the cross-sectional properties

J1 ≡ Ω

p×∇ψ ⋅ e3 dA J2 ≡ Ω

∇ψ ⋅ ∇ψ dA J3≡ Ω

ψ2 dA


which can be computed once ψ(x1, x2) is known. Define the stress resultants

T≡ e3 ⋅ Ω

p× Se3dA W≡ e3 ⋅ Ω

ψSe3dA B≡ Ω

∇ψ ⋅ Se3dA

where T is the usual torque. The stress resultants W and B are often called the bi-mo-ment and the bi-shear. Substitute the expression for Se3 to show that

T= mJ1φ+ mJθ′ W= EJ3φ′ B= mJ2φ+ mJ1θ′

Show that if φ(x3)= θ′(x3)= α, then the above results are consistent with the Saint-Venant problem discussed in the text. Using the result obtained in (a), we get

T≡ e3 ⋅ Ω

p× Se3dA

= e3 ⋅ Ω

Eφ′ψ p× e3 + mφ p×∇ψ + mθ′ p× e3× p dA

= m φJ1+ m θ′J

where we have made use of the following

e3 ⋅ p× e3 = 0 J= e3 ⋅ Ω

p× e3× p dA

Similarly, since e3 ⋅ p× e3 = 0 and e3 ⋅ e3 = 1,

W≡ e3 ⋅ Ω

ψSe3dA

= e3 ⋅ Ω

Eφ′ψ2e3+ mφψ∇ψ+ mψθ′ e3× p dA

= EJ3φ′

And finally, since e3 ⋅ ∇ψ= 0, we have

B≡Ω

∇ψ ⋅ Se3dA

= Ω

Eφ′ψ∇ψ ⋅ e3+ mφ∇ψ ⋅ ∇ψ+ mψθ′∇ψ ⋅ e3× p dA

= mJ2φ+ m J1θ′


If φ(x3)= θ′(x3)= α, then θ= αx3+ c= αx3, since θ(0)= 0. If we plug thisresult into the given expression for the displacement field we get

u(x) = αx3 e3× p + αψ(x1, x2) e3

which gives the same results as the Saint-Venant problem discussed in the text.

(c) Using the definition of the stress resultants, compute T′ and W′ and show thatT′ + t = 0 and W′ − B+ w= 0. What are the appropriate definitions of the appliedloads t(x) and w(x)? Using the definition of the stress resultant T, we can obtain itsderivative with respect to the x3 coordinate as follows,

T′ = e3 ⋅ Ω

p× ∂S∂x3e3 dA = e3 ⋅

Ω

p×divS− ∂S∂xα eα dAwhere α = 1, 2. Then,

T′ = e3 ⋅ Ω

p× divS dA− e3 ⋅ Γ

p× tΓ ds− e3 ⋅ e3×Q

= e3 ⋅ Ω

p× divS dA− e3 ⋅ Γ

p× tΓ ds

since e3 ⋅ e3× Q = 0. Let us define t (applied torque per unit length) as

t ≡ e3 ⋅m= e3 ⋅ Ω

p× b dA+ e3 ⋅ Γ

p× tΓ ds

Adding this term to T′, we get the equation

T′ + t = 0

Similarly, by using the definition of the stress resultant W, we can obtain its derivativewith respect to the x3 coordinate as follows

W′ = e3 ⋅ Ω

ψ ∂S∂x3e3 dA = e3 ⋅

Ω

ψ divS− ∂S∂xα eα dAwhere α = 1, 2. Then, from equilibrium (i.e., divS+b= 0), we have

W′ = −e3 ⋅ Ω

ψ ∂S∂xα eα+ b dA

=−e3 ⋅ Ω

∂∂xα ψSeα+ ψb− ∂ψ∂xα SeαdAObserve that

x

ℓ


e3 ⋅ Ω

∂ψ∂xα

Seα dA= e3 ⋅ Ω

∇ψ ⋅ S dA= B

Thus, we get

W′ = −e3 ⋅ Ω

ψb dA− e3 ⋅ Ω

∂∂xαψSeα dA+ B

=−e3 ⋅ Ω

ψb dA− e3 ⋅ Γ

ψSnΓ ds+ B

=−e3 ⋅ Ω

ψb dA− e3 ⋅ Γ

ψ tΓ ds+ B

At this point, let us define w (warping force per unit length) as

w≡ e3 ⋅ Ω

ψb dA+ e3 ⋅ Γ

ψ tΓ ds

Adding this term to W′, we get the equation

W′ − B+ w= 0

(d) Substitute the resultant constitutive equations into the resultant equilibrium equa-tions, and show that the equations

mJθ′′ + mJ1φ′ + t = 0

EJ3φ′′ − mJ2φ− mJ1θ′ + w = 0

govern the spatial variation of rotation θ and warping φ. These equations constitute apair of second-order ordinary differential equations, and, therefore, we can expect fourconstants of integration that must be found from boundary conditions. What are theboundary conditions for a free end and a fixed end?

Using the results we have obtained in (b), we have the constitutive equations

T′ = mφ ′ J1+ mθ′′ J= mφ′J1+ mθ′′J

W′ = EJ3φ′′

Plugging these results into the equations T′+t = 0 and W′−B+w= 0, obtained inpart (c), we get the desired result.

At a free end the tractions are prescribed. At a fixedend displacements are prescribed. Let us illustrate theboundary conditions for a free end and a fixed end on thebeam in the figure. The essential boundary conditions(i.e., no twist and no warping at the fixed end) are givenby,


θ(0)= 0 φ(0)= 0

The boundary conditions at the free end require some care. To be able to tell what hasto be prescribed at a free end, we shall resort to the principle of virtual work. The inter-nal virtual work is

WI= B

S ⋅ ∇udV

where, in the present case

∇u= θ′ e3× p e3+ φ′ψe3 e3+ θ e3× p,α eα+ φψ,α e3 eα

Then, since S ⋅ ∇u= tr S∇u = ei ⋅ ST∇ue i= ei ⋅ S∇uei (since S is symmetric), itfollows that

WI= B

θ′e3 ⋅ Se3× p + φ′ψ e3 ⋅ Se3

+ θeα ⋅ S e3× p,α + φψ,α eα ⋅ Se3 dV

Note that ψ,α eα = ∇ψ and e3 ⋅ Se3× p = e3 ⋅ p× Se3. From balance of angularmomentum we know that ei× Se i= 0. Also, a ⋅ a× b = 0 for any pair of vectorsa and b. Therefore, since p,α= eα we have

eα ⋅ Se3× p,α = eα ⋅ Se3× eα

= e3 ⋅ eα× Seα

= e3 ⋅ −e3× Se3 = 0

Using the definitions of stress resultants T, W, and B, and integrating by parts we get

WI = ℓ

0

θ′T+ φ′W+ φB dx

= ℓ0

−θT′ + φ W′−B dx + Tθ ℓ0+ Wφ ℓ

0

= ℓ0

θt+ φwdx+ Tθ ℓ0+ Wφ ℓ

0

where we have recognized that T′ = −t and W′−B=−w. The end resultant forceresultants can be defined as

to ≡ e3 ⋅ Ωo

to dA tℓ ≡ e3 ⋅ Ωℓ

tℓ dA

wo ≡ e3 ⋅ Ωo

ψ to dA wℓ ≡ e3 ⋅ Ωℓ

ψ tℓ dA


Thus, the external work is given by

WE= ℓ

0

θt+ φwdx+ tℓθ (ℓ)+ to θ (0)+ wℓφ (ℓ)+ woφ (0)

Finally, the virtual work functional G(θ, φ, θ , φ )= WI−WE can be stated as

G(θ, φ, θ , φ )=ℓ0

θ T′ + t + φ W′−B+ w dx

+ T(ℓ)−tℓ θ (ℓ)− T(0)+to θ (0)

+ W(ℓ)−wℓ φ (ℓ)− W(0)+wo φ (0)

We can observe from the above variational statement that, in the sense of virtual work,T is conjugate to θ and W is conjugate to φ. Thus, the appropriate boundary conditionsat a free end are

W(ℓ)= EJ3φ′(ℓ)= 0

T(ℓ)= mφ(ℓ) J1+ mθ′(ℓ)J= 0

172. The method of initial parameters integrates the governing equations and substitutesthe values at x= 0 to give the general form of the displacement function. For theBernoul-li-Euler beam, the transverse deflection can be computed as

wBE(x)= wo+ θo x+Mox2

2EI−

Qox3

6EI+ 1

EIx0

16 (x−ξ)

3q(ξ)− 12 (x−ξ)

2m(ξ) dξ

where wo = w(0), θo = w′(0), Mo = M(0), Qo = Q(0)are the initial parameters.Verifythat the expression satisfies the governing differential equations of Bernoulli-Euler beamtheory. This equation is particularly useful for those cases where Mo and Qo can be deter-mined from overall equilibrium. Use the method of initial parameters to solve the problemof the cantilever beam under uniform load given as an example in the text.

(a) Verify that the expression satisfies the governing differential equations of Ber-noulli-Euler beam theory.

The Leibnitz Rule of the calculus affords us a method for differentiating an inte-gral when the limits of integration are functions of the variable of differentiation. Ingeneral, we have

ddxb(x)a(x)

f ξ, x dξ= b(x)a(x)

dfdx

dξ+ f b(x), x dbdx− f a(x), x da

dx


If we specialize this to the case where a(x)= 0 and b(x)= xwe get,

ddxx0

f ξ, x dξ=x0

dfdx

dξ+ f x, x

Let f (ξ, x)= 16(x−ξ)3q(ξ)− 1

2(x−ξ)2m(ξ). Applying the Leibnitz rule we get

dwBE

dx= θo +

MoxEI−

Qox2

2EI+ 1

EIx0

12 (x−ξ)

2q(ξ)−(x−ξ)m(ξ) dξ

because f (x, x)= 0. Now let f (ξ, x)= 12(x−ξ)2q(ξ)− (x−ξ)m(ξ). Again, applying

the Leibnitz rule we get

EId2wBE

dx2= Mo−Qox+x

0

(x−ξ)q(ξ)− m(ξ) dξ

where, again f (x, x)= 0. Let f (ξ, x)= (x−ξ)q(ξ)− m(ξ). By the Leibnitz rule weget

ddxEI d2wBE

dx2 = −Qo + x

0

q(ξ) dξ− m(x)

And finally, letting f (ξ, x)= q(ξ) we get

d2dx2EI d2wBE

dx2 = q(x)−m′(x)

which is the differential equation for Bernoulli-Euler beam theory.

(b) This equation is particularly useful for those cases where Mo and Qo can be deter-mined from overall equilibrium. Use the method of initial parameters to solve theproblem of the cantilever beam under uniform load given as an example in the text.

For the cantilever beam with length ℓ, loaded with a uniform downward actingload q, we have, from static equilibrium of the entire beam: Mo =− 1

2qℓ2 and

Qo =−qℓ . The boundary conditions give wo = 0 and θo = 0. Plugging these valuesinto the expression for the displacement, we get,

EIq wBE(x) =−

14 ℓ

2x2+ 16 ℓx

3− x0

16 (x−ξ)

3 dξ

= − 1

4ℓ2x2+ 1

6ℓx3− 1

24x4

Letting ξ≡ x∕ℓ, we get

wBE(x) =qℓ424EI

−ξ4+4ξ3−6ξ2


173. The method of initial parameters can be applied to the Timoshenko beam to give

wT(x)= wBE(x)+QoxGA− 1

GAx0

(x−ξ)q(ξ) dξ

where wT(x) is the deflection according to Timoshenko beam theory and wBE(x) is thedeflection according toBernoulli-Euler beam theory (as given in Problem172). Verify thatthe expression satisfies the governing differential equations of Timoshenko beam theory.Use the method of initial parameters to solve the problem of the cantilever beam underuniform load given as an example in the text.

(a) Verify that the expression satisfies the governing differential equations of Timo-shenko beam theory. Differentiating the second of Eqns. (407) twice allows us to solvefor θ′′′ in terms of w as

θ′′′ = wiv+ q′′GA

Substituting into the first of Eqns. (408) gives the alternative form of the Timoshenkobeam equations

EIwiv+ EIGA

q′′ − q+m′ = 0

The displacement wT(x) must satisfy this fourth-order differential equation in order tobe consistent with the Timoshenko beam theory.

Using Leibnitz rule with f (ξ, x)= (x−ξ)q(ξ) we get

dwT

dx=

dwBE

dx+

Qo

GA– 1GAx0

q(ξ) dξ

Similarly, with f (ξ, x)= q(ξ), we can differentiate again to give

d2wT

dx2=

d2wBE

dx2–q(x)GA

Differentiating this result twice and multiplying the result by EI we get

EIwivT = EIwiv

BE−EIGA

q′′

Substituting the expression for the wBE we get the desired result

EIwivT +

EIGA

q′′ − q+m′ = 0

(b) Use the method of initial parameters to solve the problem of the cantilever beamunder uniform load given as an example in the text. For the cantilever beam withlength ℓ, loaded with a uniform downward acting load q, we have, from static equilib-rium of the entire beam: Mo =− 1

2qℓ2 and Qo =−qℓ . The boundary conditions give


wo = 0 and θo = 0. Plugging these values into the expression for the displacement,we get

EIq wT(x) = − 1

4 ℓ2x2+ 1

6 ℓx3− 1

24 x4− 1

12 βℓ3x + 1

12 βℓ2 x

0

(x−ξ) dξ

where β≡ 12EI∕GAℓ2. Let ξ≡ x∕ℓ and multiply through by q∕EI to get

wT(x)=qℓ424EI

−ξ4+ 4ξ3− 6−βξ2− 2βξ

174. The three-dimensional rotation tensor Λ can be expressed in terms of three parame-ters e1, e2, and e3 as

Λ(e1, e2, e3)=

e20+ e21− e22− e23

2(e1 e2− e0 e3)

2(e1 e3+ e0 e2)

2(e1 e2+ e0 e3)

e20− e21+ e22− e23

2(e2 e3− e0 e1)

2(e1 e3− e0 e2)

2(e2 e3+ e0 e1)

e20− e21− e22+ e23

where the parameter e0 has been introduced for convenience. This fourth parameter doesnot represent an independent parameter, but rather satisfies the constraint equatione20+e21+e22+e23 = 1. These parameters are called theEuler parameters. Demonstrate thatthe tensor Λ is orthogonal by showing that Λ−1 = ΛT. Show that for small values of theparameters e1, e2, and e3 the tensor can be expressed in the form Λ≈ I+W, where I isthe identity andW is a skew-symmetric tensor. Show, therefore, that when the three pa-rameters are small, they can be viewed as the components of the rotation vector θ, withθi= 2ei, and that W= θ×.

(a) Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT.

If Λ−1 = ΛT, then ΛTΛ= Imust hold. The tensor ΛT is given by

Λ(e1, e2, e3)~

e20+ e21− e22− e23

2(e1 e2+ e0e3)

2(e1 e3− e0e2)

2(e1 e2− e0e3)

e20− e21+ e22− e23

2(e2 e3+ e0e1)

2(e1 e3+ e0e2)

2(e2 e3− e0e1)

e20− e21− e22+ e23

Since ΛTΛT= ΛTΛ, then ΛTΛ is a symmetric tensor. Therefore, we need only com-pute the upper (or lower) triangle and diagonal entries of this tensor. It follows that,

ΛTΛ11= (e20+e21−e22−e23 )2+ 4(e1e2−e0e3)2+ 4(e1 e3+e0e2)2

= e40+ e41+ e42+ e43 + 2(e20e21− e20e

22− e20e

23− e21e

22− e21 e

23+ e22e

23 )

+ 4(e21e22+e20e23 )− 8e0e1 e2e3+ 4(e21 e

22+e20e22 )+ 8e0e1e2 e3

= (e20+e21+e22+e23 )2= 1


Likewise, ΛTΛ22= ΛTΛ

33= 1. The (1,2) term of ΛTΛ can be computed as

ΛTΛ12= 2 (e20+e21−e22−e23 ) (e1 e2+e0e3 )

+ 2(e20−e21+e22−e23 ) (e1e2−e0e3 )+ 4(e1 e3+e0e2 ) (e2e3−e0e1 )

= 4e20e1e2+4e21e0e3−4e22e0e3−4e23 e1e2

+ 4e23e1e2− 4e20e1e2− 4e21e0e3+ 4e22e0e3 = 0

Likewise, ΛTΛ13= ΛTΛ

23= 0. Therefore, ΛTΛ= I holds.

(b) Show that for small values of the parameters e1, e2, and e3 the tensor can be ex-pressed in the form Λ≈ I+W, where I is the identity andW is a skew-symmetrictensor. Show, therefore, that when the three parameters are small, they can be viewedas the components of the rotation vector θ, with θi= 2ei, and that W= θ×.

Let us define W^ ≡ Λ− I. Then

W^ ~

−2(e22+ e23 )

2(e1 e2− e0e3)

2(e1 e3+ e0e2)

2(e1 e2+ e0e3)

2(e2 e3− e0e1)

2(e1 e3− e0e2)

2(e2 e3+ e0e1)−2(e21+ e23 )

2(e2 e3− e0e1) −2(e21+ e22 )

For small values of the parameters e1 , e2 and e3 , the squared terms above are verysmall. Since e20+ e21+ e22+ e23 = 1 then e20 ≈ 1 or e0 ≈ 1. Furthermore theterms e1 e2 , e1 e3 and e2 e3 are very small. If we neglect the higher order terms andtake e0 < 0, we get

W^ ≈W~

0 −2e30

0

2e22e3 −2e1

2e1−2e2

Then, with θi≡ 2ei we get

W^ ≈W~

0 −θ30

0

θ2θ3 −θ1

θ1−θ2

Thus, W= θ×. Note that, if we had assumed e0 > 0 then, W=−θ×.

175. Reconsider the cantilever beam of length ℓ, fixed at x= 0, with bending modulusEI and shear modulus GA solved as an example in the text (Fig. 96). Examine the resultsof the Ritz method as you increase the number of basis functions taken from the sets

hi (x)∈ x, x2ℓ , x3

ℓ2 , . . .,xNℓN−1, gi (x)∈ xℓ , x

2

ℓ2 ,x3ℓ3 , . . .,

xMℓM


In particular, find general expressions for the ijth components of Kaa, Kab, Kba, Kbb, andthe ith components of f a and f b when hi (x)= xi∕ℓi−1 and gi (x)= xi∕ℓi. Solve the prob-lem for (N,M)= (2, 2), (3, 2), (3, 3), (4, 3), (4, 4). What do you expect to happen for high-er-order approximations? Comment on the differences in the solutions obtained whenM= N−1 versus those obtained with M= N.

(a) Find general expressions for the ijth components of Kaa, Kab, Kba, Kbb, and theith components of f a and f b when hi(x)= xi∕ℓi−1 and gi(x)= xi∕ℓi. The essentialboundary conditions w(0)= 0 and θ(0)= 0 are satisfied if the displacement and rota-tion are approximated as

w(x)=Ni=1

aihi, θ(x)=Mi=1

bi gi

Let ξ≡ x∕ℓ. The base functions and their derivatives can be expressed as

hi(ξ)= ξiℓ

hi′(ξ)= iξi−1

gi(x)= ξi

gi′(ξ)= iξi−1∕ℓ

It follows that

Kaa ij=

1

0

GAℓhi′hj′ dξ = GAℓi j

i+j−1

Kab ij= 1

0

−GAℓhi′gj dξ = −GAℓii+j =

Kba ji

Kbb ij= 1

0

EIℓgi′gj′+GAℓ gi gj dx = GAℓ12 12i+j+1+

i jβi+j−1

where β≡ 12EI∕GAℓ2. The right side vectors have components

f a i= 1

0

q(ξ)hi ℓdξ =−qℓ21

i+1

and f b= 0.

(b) Solve the problem for (N, M)= (2, 2), (3, 2), (3, 3), (4, 3), (4, 4). Solving thesystem of equations

Kaaa+ Kabb= f a

Kbaa+ Kbbb= f b

where the coefficient matrices are computed from part (a) yields, the results presentedin the following table,


151+ 5β

3β

9+ 210β

3β+ 35β26β

b2

−7+ 10 β2+ 10 β

β− 22β

h

h

−2+ 25 β

β+ 5β2

h

−1

β− 52β

1β

h

− 5β

h

−13+ 70 β6+ 70β

−15+ 165 β+ 35β2

6+ 70β2

1β

h

−15+ 210β

3β+ 35β2

− 703+ 35 β

−1

β− 62β

2β

− 12β

– 6β

− 2β

a1

a2

a3

a4

b1

b3

(2, 2 ) (3, 2 ) (3, 3 ) (4, 3 )(N,M)→

where all the coefficients in the table are normalized by qℓ∕GA. The results for thecase (4, 4) are the same as the case (4, 3) with b4 = 0. These results were obtainedusing MATHEMATICAT and the commands needed to solve the problem are given below

n = 4m = 4hi = xî / L^(i - 1)hj = x^j / L^(j - 1)gi = xî / Lîgj = x^j / L^jkaa = G*A*D[hi, x]*D[hj, x]kab = -G*A*gj*D[hi, x]kbb = G*A*(Beta*L^2/12*D[gi, x]*D[gj, x] + gi*gj)fa = (-q)*hifb = 0*iKaa = Table[Integrate[kaa, x, 0, L], i, 1, n, j, 1, n]Kab = Table[Integrate[kab, x, 0, L], i, 1, n, j, 1, m]Kbb = Table[Integrate[kbb, x, 0, L], i, 1, m, j, 1, m]Kba = Transpose[Kab]fa = Table[Integrate[fa, x, 0, L], i, 1, n]fb = Table[Integrate[fb, x, 0, L], i, 1, m]a = Inverse[Kaa - Kab . Inverse[Kbb] . Kba]a = a . (fa - Kab . Inverse[Kbb] . fb)b = Inverse[Kbb] . fb - Kba . aFullSimplify[a]//MatrixFormFullSimplify[b]//MatrixFormhvec = Table[xî/L^(i-1), i,1,n]gvec = Table[x^j/L^j , j,1,m]

xℓ

qo


w = Simplify[a . hvec]theta = Simplify[b . gvec]

(c) What do you expect to happen for higher-order approximations? Since the solu-tion for the case (4, 3) corresponds to the exact solution, higher-order approximationswill yield the same results as the case (4, 3).

(d) Comment on the differences in the solutions obtained when M= N−1 versusthose obtained with M= N. The cases with the rotation interpolated one polynomialorder less than the displacement give results with the rotation independent of the shearstiffness of the beam, in accord with the exact solution, even for low-order approxima-tions.

176. Prove that S ⋅W= 0, or SijWij= 0, whenS is a symmetric tensor andW is a skew-symmetric tensor. Note that S ⋅W= e i ⋅ SWei.

Since S is a symmetric tensor, ST = S, or Sij = Sji . Also, sinceW is a skew--symmetric tensor, WT =−W, or Wij =−Wji . It follows that

SijWij = SjiWij=−SjiWji=−SijWij

where the first step uses the symmetry of S, the second step uses the skew-symmetry ofW, and the third step is just a change of labels. As a consequence we have

SijWij+SijWij = 2SijWij = 0 ⇒ SijWij= 0

An alternative proof is given below. For any tensor A, the difference between the ten-sor and its transpose A−AT is always skew-symmetric. The trace of a skew-symmetrictensor is always zero. Therefore, noting that S ⋅W= trSW , we have

trSW−SW T = 0

Observing that AB T= BTAT, that trA+B = trA+trB, and noting thatW isskew-symmetric, we find that

trSW + trWS = 0

Since tr AB = tr BA for and tensors A and B, we have, finally, 2 trSW = 0, there-by proving the result.

177. The prismatic beam shownhas a cross section thatis symmetric with respect to the plane of the page. Thecross section has axial modulus EA, shear modulusGA,and flexural modulusEI.The beam is subjected to auni-form transverse load q(x)=−qo. Find the displace-ments and rotations for the beam by directly integrating the governing equations.


For the Timoshenko beam, EIθ′′′ = q(x)− m′ (x). Since q(x)=−qo and thereare no applied moments (m(x)= 0) we can integrate the governing equation by directquadrature

θ′′′ = − qoEI

θ′′ = − qoEI

x+ a0

θ′ = − qo2EI

x2+ a0x+ a1

Since M(x)= EIθ′ we can enforce the boundary conditions on the simple ends todetermine the two constants of integration obtained thus far. To wit,

M(0)= 0 ⇒ a1= 0 M(ℓ)= 0 ⇒ a0=qoℓ2EI

Substituting these values and integrating again we get

θ = qoEI− 1

6x3+ 1

4ℓx2 + a2

Also, for the Timoshenko beam, GAw′ = GAθ− EIθ′′ − m . Substituting the resultswe have obtained above for the rotation θ and integrating again we get

w = qoEI− 1

24x4+ 1

12ℓx3 + a2x+

qo2GAx2−ℓx + a3

The displacement boundary conditions at the two ends give

w(0)= 0 ⇒ a3= 0 w(ℓ)= 0 ⇒ a2=−qoℓ324EI

Letting ξ≡ x∕ℓ we get, after some simplification,

w =qoℓ424EI

−ξ4+2ξ3−1 + qo2GAξ2−ξ

θ =qoℓ324EI

−4ξ3+6ξ2−1

178. Resolve Problem 177 after making the Bernoulli-Euler assumption that w′ = θ(i.e., there is no shear deformation). What is the difference between the two solutions?

For the Bernoulli--Euler beam,

d2

dx2EI d2w

dx2 = q(x)−m′(x)


With q(x)=−qo and m(x)= 0, we can integrate the differential twice to get

M(x)= EIw′′ = − 12qox2+ a0x+ a1

The force boundary conditions at the simple ends give

M(0)= 0 ⇒ a1= 0

M(ℓ)= 0 ⇒ a0=12qoℓ

Substituting these values and integrating twice more gives

w = qoEI− 1

24x4+ 1

12ℓx3 + a2x+ a3

The displacement boundary conditions at the ends give

w(0)= 0 ⇒ a3= 0

w(ℓ)= 0 ⇒ a2=−qoℓ324EI

Letting ξ≡ x∕ℓ we get, with some rearrangement

w =qoℓ424EI

−ξ4+2ξ3−1

w′ = θ =qoℓ324EI

−4ξ3+6ξ2−1

Note that, the results we have here are identical to the ones we got for Problem 177 forthe extreme case when GA→∞. In other words, for a beam with a very large shearstiffness, the response of a Timoshenko beam and the Bernoulli-Euler beam are nearlyidentical.

179. Use a polynomial basis to find an approximate solution to Problem 177 using theprinciple of virtual work for the beam including shear deformation. Which terms shouldyou include? What order approximation is adequate?

The virtual work functional is

G(w, θ,w, θ) = ℓ0

EIθ′θ′ +GA(w′−θ)(w′−θ)− qw dx = 0

The essential boundary conditions w(0)= w(ℓ)= 0 are satisfied by the approximatedisplacement field


w(x)=Ni=1

aihi, θ(x)=Mi=1

bigi

when the trial functions are given by

hi(ξ)= ξℓ1−ξi

hi′(ξ)= 1−ξi−1 1−1+iξ

gi(ξ)= ξi−1

gi′(ξ)= ( i−1)ξi−2∕ℓ

where ξ≡ x∕ℓ. The coefficient matrices have the components

Kaa ij=

1

0

GAℓhi′hj′ dξ

Kbb ij= 1

0

EIℓgi′gj′+GAℓ gi gj dξ

Kab ij= −1

0

GAℓhi′gj dξ

f a i= 1

0

q(ξ)hi ℓdξ

Carrying out the indicated integrations yields

13

215

16

110

110

325

symm.

Kaa = GAℓ

0

112

16

115

16

130

Kab = −GAℓ 120

320

3140

0

0120

1

β+412

12

β+312

13

5 β+945

Kbb = GAℓ

5 β+1260

14

3 β+424

21 β+20140

symm.

f a = −qℓ2112

16

120

130

f b =0

0

0

0

where the portion of the matrices for N = 1, M = 2 are shaded. Note that, by symmetry,we have Kba = Kab T and the parameter β≡ 12EI∕GAℓ2. The results for variouscases are presented in the table below.


−1+ β2 β

0

− 12β

1β

0

h

−1+ β2 β

h

− 12β

1β

0

h

−1+ β2 β

h

h

− 12β

1β

h

h

a1

a2

a3

b1

b2

b3

b4

(1, 2 ) (2, 3 ) (3, 3 ) (3, 4 )(N,M)→

−1+ β2 β

− 12β

12β

− 1β

0

3β

− 2β

00

where all the coefficients in the table are normalized by qoℓ∕GA. The result for thecase (3, 4) corresponds to the exact solution which is given by

w(x)= qoℓ424EI

−ξ4+ 2ξ3− βξ2+ 1+βξ

θ(x)= qoℓ324EI

−4ξ3+ 6ξ2− 1

where ξ≡ x∕ℓ. These results were obtained using MATHEMATICAT with the followingcommands

n = 3m = 4hi = x*(1 - x/L^i)hj = x*(1 - x/L^j)gi = x/L^(i - 1)gj = x/L^(j - 1)kaa = G*A*D[hi, x]*D[hj, x]kab = -G*A*gj*D[hi, x]kbb = G*A*(Beta*L^2/12*D[gi, x]*D[gj, x] + gi*gj)fa = -q*hifb = 0*iKaa = Table[Integrate[kaa, x, 0, L], i, 1, n, j, 1, n])Kab = Table[Integrate[kab, x, 0, L], i, 1, n, j, 1, m])Kbb = Table[Integrate[kbb, x, 0, L], i, 1, m, j, 1, m])Kba = Transpose[Kab]fa = Table[Integrate[fa, x, 0, L], i, 1, n]fb = Table[Integrate[fb, x, 0, L], i, 1, m]a = Inverse[Kaa - Kab . Inverse[Kbb] . Kba]a = a . (fa - Kab . Inverse[Kbb] . fbb = Inverse[Kbb] . (fb - Kba . a)FullSimplify[a]//MatrixFormFullSimplify[b]//MatrixForm


hvec = Table[x*(1-x/L)^i, i,1,n]gvec = Table[(x/L)^(j-1), j,1,m]w = FullSimplify[a . hvec]theta = FullSimplify[b . gvec]

180. Repeat the virtual-work computation in Problem 179 for the Bernoulli-Euler beam.


G(w,w) = ℓ0

EIw′′w′′ − qw dx = 0

The essential boundary conditions, w(0)= w(ℓ)= 0, are satisfied for the approximatedisplacement field

w(x)=Ni=1

aihi

when the base functions are given by

hi(ξ)= ξℓ1−ξi

hi′(ξ)= 1−ξi−1 1−1+iξ

hi′′(ξ)=1ℓi−1 1−ξ i−2 1−1+iξ − (1+i )1−ξ i−1

where ξ≡ x∕ℓ. The stiffness matrix and the force vector are given by

K ij= 1

0

EIℓhi′′hj′′ dξ , f i= −10

qℓhi dξ

Thus, we get the system of equations (the shading is to indicate the appropriate partsfor N=1 (darkest), N=2, and N=3 (lightest))

4

4

2

4

2

245

symm.

K = EIℓ f=−qℓ2 1

12

16

120

with which we solve for the vector of unknown coefficients a from the system of equa-tions Ka=f. For the case N=1, we get


a1 =−qoℓ324EI

⇒ w=qoℓ424EI

(ξ2−ξ)

For the case N=2, we get the same result as the case N=1 (i.e., a2 = 0). For the caseN=3, the coefficients are

a1 =−qoℓ324EI

, a2=−qoℓ324EI

, a3=qoℓ324EI

and the displacement is given by

w=qoℓ424EI−ξ4+ 2ξ3− ξ

which corresponds to the exact solution. These results were obtained using MATHEMAT-ICAT and the commands needed to solve the problem are given below

n=3hi = x*(1 - x/L)^ihj = x*(1 - x/L)^jkx = EI*D[hi, x,2]*D[hj, x,2]fx = -q*hiK = Table[Integrate[kx, x, 0,L], i,1,n, j,1,n]f = Table[Integrate[fx, x,0,L], i,1,n]a = Inverse[K].fFullSimplify[a]//MatrixFormhvec = Table[x*(1-x/L)^i, i,1,n]w = FullSimplify[a.hvec]

181. Carry out the integrations in Eqn. (385) to show that the tensors given in Eqn. (387)result.

Throughout the computations we shall assume that the material constants λ and mare constant throughout the cross section. We shall also use the following definitions

A≡Ω

dA Sα≡ Ω

xαdA Iαβ ≡ Ω

xαxβdA

The tensor A is defined as

A≡Ω

(λ+m)e3 e3 + mI dA

Carrying out the integration we have A= (λ+m) e3 e3 + mI A. In componentform this tensor can be expressed as


A~

mA 0

mA

EA

0

0 0

00

The tensor S is defined as

S≡Ω

(λ+m)e3 (p× e3 )− m (p× ) dA

The components of the tensors e3 (p× e3 ) and p× are given as follows

e3 (p× e3 )~0 0

0

0

x2

0 0

−x10 −x2

(p× )~0 0

0

0

x2

0 − x1

x1

Therefore, we get

S ~−S2

0 0

0

0

S2

0 − S1

S1

0 0

0

0

S2

0 0

−S10

(λ+m) − m

ES2

0 0

0

0

−mS20 mS1

ES1

=

The tensor I is defined as

I≡Ω

(λ+m) (p× e3 ) (p× e3 )− m [p× p× ] dA

The components of the tensors (p× e3 ) (p× e3 ) are given as

(p× e3 ) (p× e3 ) ~

x22 −x1 x2x21

0

0

0

00

−x1 x2

and the components of p× p× are given as

−x2

0 0

0

0

x2

0 − x1

x1

(p× p× )~

−x22 x1 x2

−x21−x21−x

22

0

0

00

x1 x2

−x2

0 0

0

0

x2

0 − x1

x1

=

x1ℓ

q

ℓx2


Therefore, we get

I12I ~

I22 −I12I11

0

0

−I12 0

00

(λ+m) − m

−I22 I12

−I11

−J

0

0

00

EI22 −EI12EI11

mJ

0

−EI12 0

00

=

182. Consider the beam in Problem 177. Find an expression for the transverse displace-ment w(x) using the principle of virtual work, using a quartic polynomial basis. Note thatthe problem has two essential boundary conditions and two natural boundary conditions.

See problem 179. The quartic terms add no new information and therefore the solutiondoes not improve.

183. The principle of virtual work does not require that the assumed displacement func-tions satisfy the natural boundary conditions a priori. Is there an advantage to satisfyingthe natural boundary conditions, too? What happens in Problem 182 if we do enforce thenatural boundary conditions?

Enforcing the natural boundary conditions improves the approximation’s ability tosatisfy the exact solution at the end points at the expense of using the additional termsto satisfy the equations in a weak sense throughout the domain. In general, it is diffi-cult to justify enforcing equilibrium exactly at certain points in the domain.

184. A continuous beam is one that has one ormore intermediate supports. The extra boundaryconditions are in excess of the four endconditions.Describe an approach to solving the followingproblem that exactly satisfies the differentialequations everywhere in the domain, as well as the boundary and intermediate conditions.Find the classical solution to the given problem by integrating the governing differentialequations. (Hint: It is useful to describe the solution independently in each segment andto enforce continuity by equating state variables at the placewhere the two segments join.)


Let us divide the beam at the middle support into two segments and refer to thevariables associated with each beam with indices 1 and 2.

x1ℓ

q

ℓx2

M1 (ℓ)M2 (0)

w1 (0)= 0

M1 (0)= 0

w1 (ℓ)= 0

θ1 (ℓ)= θ^

w2 (0)= 0

M2 (ℓ)= 0

w2 (ℓ)= 0

θ2 (0)= θ^

θ^

For the first beam,

EIwiv1 =−q

EIw1′′′ = −qx1+a1

EIw1′′ = −12qx21+ a1 x1+ b1

EIw1′ = −16qx31+

12a1x

21+ b1 x1+ c1

EIw1=−124qx41+

16a1 x

31+

12b1 x

21+ c1 x1+ d1

Since

w1(0)= 0 ⇒ d1 = 0

M1(0)= EIw1 ′′(0) ⇒ b1 = 0

w1(ℓ)= 0 ⇒ c1 =qℓ324− a1

ℓ26

Similarly, for the second beam, since EIwiv2 = 0 we have

EIw2′ =12a2x

22+ b2 x2+ c2

EIw2′′′ = a2

EIw2′′ = a2 x2+ b2

EIw2=16a2x

32+

12b2x

22+ c2 x2+ d2

Evaluating the boundary conditions

w2(0)= 0 ⇒ d2 = 0

M2(ℓ)= EIw2 ′′(ℓ)= 0 ⇒ b2 =−a2ℓ

w2(ℓ)= 0 ⇒ c2 =13a2ℓ2

x

ℓ

qkℓ2

k


Then, since M1 (ℓ)= M2 (0)

a1+ a2 =12qℓ (1)

and since θ1 (ℓ)= θ2 (0) or w1′ (ℓ)= w2′ (0)

a1− a2 =38qℓ (2)

we can use equations (1) and (2) to get

a1 =716qℓ a2 =

116qℓ

Letting ξα≡ xα∕ℓ we get

w1=qℓ424EI−ξ41+ 7

4ξ31−

34ξ1

w2=qℓ424EI14ξ2

3− 34ξ22+

12ξ2

185. The following prismatic beam has a cross sec-tion that is symmetric with respect to the plane of thepage. The cross section has flexural modulus EI.Axialand shear deformations can be neglected (i.e., useBer-noulli-Euler beam theory). The beam is subjected to auniform transverse load q actingdownward. Thebeamhas deformable spring supports at the ends. At the left end, the support prevents translationin the vertical and horizontal directions and the spring elastically restrains rotations. Themoment developed by the spring is related to the rotation at that point by Ms= kℓ2θs,where θs= θ(0) is the rotation experienced by the spring. The right end of the beam isfree to translate horizontally and to rotate, but the spring elastically restrains vertical mo-tion. The force developed by the spring is related to the deflection at that point byFs= kws, where ws= w(ℓ) is the deflection experienced by the spring. What are the ap-propriate boundary conditions for this problem? Solve the problem by integrating the dif-ferential equations and using the boundary conditions to find the constants of integration.Revise the principle of virtual work to account for the work done by the springs. Estimatethe deflection of the beam using a two-term polynomial expansion for the transversedeflection. That is, assume the real and virtual transverse deflections to be of the form

w(x)= a1x+ a2x2ℓ , w(x)= a1x+ a2

x2ℓ

What constraints do the assumed displacement field add to the problem?

(a) What are the appropriate boundary conditions for this problem?


Q(ℓ)kℓ2w′(0)

k

M(0)

kw(ℓ)

There is one essential boundary condition:

w(0)= 0

The natural boundary conditions can be established by taking freebody diagrams ofeach end (see sketch). At the left end we have M(0)= kℓ2w′(0). At the right end wehave M(ℓ)= 0 and Q(ℓ)+kw(ℓ)= 0. Moment and shear in the beam are related tothe transverse displacement of the beam as M= EIw′′ and Q=−EIw′′′. Therefore,the natural boundary conditions can be expressed as

EIw′′(ℓ)= 0

EIw′′(0)− kℓ2w′(0)= 0

EIw′′′(ℓ)− kw(ℓ)= 0

(b) Solve the problem by integrating the differential equations and using the boundaryconditions to find the constants of integration.

EIw=− 124qx4+ dx3ℓ + cx2ℓ2+ bxℓ3+ aℓ4

EIw′ = − 16qx3+ 3dx2ℓ + 2cxℓ2+ bℓ3

EIw′′ = − 12qx2+ 6dxℓ+ 2cℓ2

EIw′′′ = −qx+ 6dℓ

Let us define the parameterη≡ kℓ3∕EI to be the ratio of spring stiffness to bendingstiffness. Then,

w(0)= 0

EIw′′(ℓ)= 0

EIw′′(0)−ηℓ EIw′(ℓ)= 0

EIw′′′(ℓ)−ηℓ3EIw(ℓ)= 0

⇒ a= 0

⇒ − 12qℓ2+ 6dℓ2+ 2cℓ2 = 0

⇒ 2cℓ2−ηbℓ2= 0

⇒ −qℓ+ 6dℓ−η− qℓ24+bℓ+cℓ+dℓ = 0

Solving these equations for b, c, and d we get

b=− q8η12+η6+η , c=− q

1612+η6+η , d= q

4836+5η6+η


Letting ξ≡ x∕ℓ, we get

w= qℓ424EI−ξ4+36+5η

12+2ηξ3−36+3η

12+2ηξ2−36+3η

6η+η2ξ

Note that as η→∞, we get the solution for a propped cantilever beam

xℓ

q

w= qℓ424EI−ξ4+ 5

2 ξ3− 3

2 ξ2

(c) Revise the principle of virtual work to account for the work done by the springs.Recall that the virtual work expression without the springs is

G(w,w) = ℓ0

EIw′′w′′ − qw dx = 0

The work associated with the springs is internal work (see text for a discussion).Thus,

G(w,w) = ℓ0

EIw′′w′′ − qw dx + kw(ℓ)w(ℓ)+ kℓ2w′(0)w′(0)

(d) Estimate the deflection of the beam using a two-term polynomial expansion forthe transverse deflection. That is, assume the real and virtual transverse deflections tobe of the form

w(x)= a1x+ a2x2ℓ , w(x)= a1x+ a2

x2ℓ

What constraints does the assumed displacement field add to the problem?

Let us define the vector of base functions and unknown coefficients as

h=ξℓξ2ℓ

a=a1

a2a=

a1

a2

where ξ≡ x∕ℓ. It follows that, w= aTh and w′′ = aTh′′. By using this notation, thevirtual work functional becomes

G(w,w) = aT10

EIℓh′′ h′′ T a + qℓh dξ

+ k a T h(1)h(1) T+ℓ2 h′(0)h′(0) T a

xℓ

PP P


Thus, the stiffness matrix K and the force vector f are given by

K ≡10

EIℓh′′ h′′ T dξ+ k h(1)h(1) T+ℓ2 h′(0)h′(0) T

f≡10

−qℓh dξ

Letting η≡ kℓ3∕EI, we find that

K ~ EIℓ

2ηf~−qℓ2

1213η

η

4+η

Solving the system of equations Ka= f we get

a1 =−qℓ36EI 12+η8η+η2 , a2 =−

qℓ36EI

18+η

Thus,

w= a1h1+ a2h2 =−qℓ4

48+6ηηEIηξ2+ 12+ηξ

The approximate displacement is quadratic. The second derivative is constant. There-fore, the approximation imposes a constant bending moment field, which is clearly atodds with the exact solution.

186. In the derivations of beam theory, both in a clas-sical sense and in a variational sense, no mention wasmade of concentrated forces. Describe a way to ac-count for concentrated forces in solving the classicaldifferential equations (for example, for a Bernoulli-Euler beam). Describe how the concentrated forces should be implemented into the princi-ple of virtual work.

(a) Describe a way to account for concentrated forces in solving the classical differen-tial equations (for example, for a Bernoulli-Euler beam).

Similar to the previous problem, we can divide the beam into four segments (seethe figure below) and refer to the variables associated with each beam with indicesi = 1,. . ., 4.


P

x1 x2 x3 x4

Q1 (ℓ∕4)

M2 (0)M1 (ℓ∕4)

Q2 (0) Q2 (ℓ∕4)

M2 (ℓ∕4)

Q3 (ℓ∕4)

M3 (ℓ∕4)M3 (0)

Q3 (0)

M4 (0)

Q4 (0)

P P

For each beam beam

EIwivi = 0

EIwi′′′ = −Qi = ai

EIwi′′ = Mi = ai xi+ bi

EIwi′ =12aix

2i + bi xi+ ci

EIwi=16aix

3i +

12bix

2i + ci xi+ di

where, i = 1, , 4 and xi denotes the local coordinate for the ith beam(0< xi< ℓ∕4). So we have 16 (4 per each beam) unknown coefficients to be deter-mined. We get the 16 equations required as follows

Boundary conditions at the ends

w1(0)= 0

M1(0)= 0

w4(ℓ∕4)= 0

M4(ℓ∕4)= EIw4′′(ℓ∕4)= 0

continuity of deflection and slope at the nodes

wi(ℓ∕4)= wi+1(0)

wi′(ℓ∕4)= wi+1′(0)

and finally equilibrium at the nodes

Mi(ℓ∕4)= Mi+1(0)

Qi(ℓ∕4)−Qi+1(0)=−P

where ( i = 1, 2, 3).

(b) Describe how the concentrated forces should be implemented into the principle ofvirtual work. Concentrated forces must be added to the external work. So, for thisbeam the virtual work functional is given by


G(w,w) = ℓ0

EIw′′w′′ dx− −P w(ℓ∕4)+w(2ℓ∕4)+w(3ℓ∕4) = 0

(c) Solve the problem pictured above, assuming that the forces are applied at thequarter points of the span, using the principle of virtual work for the Bernoulli-Eulerbeam theory and the approximation basis functions

h1 = 1−3ξ2+2ξ3, h2= ξ2(3−2ξ)

h3 = ξ(1−2ξ+ξ2)ℓ, h4 = ξ2(ξ−1)ℓ

where ξ≡ x∕ℓ, so that the transverse displacement can be approximated by

w(x)=4i=1

aihi, w(x)=4i=1

aihi

Can you ascribe a physical interpretation to the coefficients ai? The first derivatives ofthe base functions are

h1′ = 6ξ2−6ξ

h3′ = (1–4ξ+3ξ2)ℓ

h2′ = −6ξ2+6ξ

h4′ = (3ξ2−2ξ)ℓ

We can observe that the coefficients ai are related to the end displacement and endrotations by

w(0)= a1

w′(0)= θ(0)= a3ℓ

w(ℓ)= a2

w′(ℓ)= θ(ℓ)= a4ℓ

For this problem, since the essential boundary conditions are w(0)= 0 and w(ℓ)= 0so that a1 = 0 and a2 = 0. Let us define a matrix of shape functions hT≡ [h3, h4 ]

and a matrix of unknown coefficients aT≡ [a3, a4 ]. Thus, w= aTh and

w′′ = aTh′′. By using this notation, the virtual work functional becomes

G(w,w) = aT10

EIℓh′′ h′′ T a dξ + P a T h(1∕4)+h(1∕2)+h(3∕4)


K ≡10

EIℓh′′ h′′ T dξ f≡−P h(1∕4)+h(1∕2)+h(3∕4)

Carrying out the indicated integrations we get

K = EIℓ24

2f= 5

16P

1

−1

2

4


Then solving Ka= f yields

a3 =−532

PEIℓ , a4 =

532

PEIℓ

Thus,

w= a3h3+a4h4 =532

Pℓ3EI−ξ(1−2ξ+ξ2)+ ξ2(ξ−1) = 5

32Pℓ3EIξ2−ξ

For a better solution, we can divide the beam into segments, approximate the displace-ment in each beam segment and enforce continuity of state variables among neighbor-ing segments.

187. Abeamon aWinkler elastic foundation accrues force in the foundation in proportionto the deflection of the beam according to f (x)= kw(x), where k is the modulus of thefoundation.

k

q(x)

ℓ

x kw(x)

Show that a simply supported Bernoulli-Euler beam on an elastic foundation is governedby the following differential equation and boundary conditions

EIwiv+ kw = q(x)

w(0)= 0, w′′(0)= 0, w(ℓ)= 0, w′′(ℓ)= 0

Verify that w(x)= eβx a1 cos βx+a2 sin βx + e−βx a3 cos βx+a4 sin βx is the dis-placement field that satisfies the homogeneous differential equation, if 4β4 ≡ k∕EI.

(a) Show that a simply supported Bernoulli-Euler beam on an elastic foundation isgoverned by the following differential equation and boundary conditions

EIwiv+ kw = q(x)

w(0)= 0, w′′(0)= 0, w(ℓ)= 0, w′′(ℓ)= 0

The reaction force due to the elastic foundation can be treated as an applied dis-tributed load which acts always in the direction opposite the displacement. The loadingshown on the sketch above is downward and hence negative according to our sign con-vention. Let us take q(x) in the positive sense as shown below


q(x)

kw(x)

Then let us replace the load term in Eqn. (417) in the text with the effective loading

q^(x) ≡ q(x)− kw(x)

So, for the Bernoulli-Euler beam we have EIwiv= q^(x)= q(x)− kw(x). Rearrangingterms gives the governing equation for the beam on an elastic foundation (with no ap-plied moment)

EIwiv+ kw = q

The boundary conditions are same as those of a simply supported Bernoulli-Eulerbeam.

(b) Verify that the displacement

w(x)= eβxa1 cos βx+ a2 sin βx + e−βxa3 cos βx+ a4 sin βx

where 4β4 ≡ k∕EI, satisfies the homogeneous differential equation.

w′ = βeβx a1 cos βx− sin βx + a2 cos βx+ sin βx

−βe−βx a3 cos βx + sin βx − a4 cos βx − sin βx

w′′ = −2β2eβx a1 sin βx − a2 cos βx + 2β2e−βx a3 sin βx − a4 cos βx

w′′′ = −2β3eβx a1 cos βx+ sin βx + a2 cos βx− sin βx

+2β3 e−βx a3 cos βx − sin βx + a4 cos βx + sin βx

wiv=−4β4eβx a1 cos βx + a2 sin βx − 4β2 e−βx a3 cos βx − a4 sin βx

= −4β4w

Substituting these results into the governing differential equation we get

EIwiv+ kw = EI wiv+ 4β4w = EI −4β4w+ 4β4w = 0

Thus, the given displacement satisfies the homogeneous differential equation.

188. Consider the beam on a Winkler elastic foundation of Problem 187, subjected to auniform load q. Show that the principle of virtual work, accounting for the work done bythe elastic foundation, is


G(w,w) = ℓ0

EIw′′w′′ + kww− qw dx = 0

Find the displacementmap of the system using theRitz method, assuming that the real andvirtual displacements are approximated as

w(x)=Nn=1

an sinnπxℓ , w(x)=N

n=1

an sinnπxℓ

(a) Show that the principle of virtual work, accounting for the work done by the elas-tic foundation, is

G(w,w) = ℓ0

EIw′′w′′ + kww− qw dx = 0

Similar to the discussion in Problem 187, the work done by the Winkler founda-tion can be treated as work done by an external (applied) load. So, the contribution ofthe Winkler foundation to the external work is,

WWINKLERE =−ℓ

0

kww dx

Then G(w,w)≡ WI−WE yields

G(w,w) = ℓ0

EIw′′w′′ + kww− qw dx

(b) Find the displacement map of the system using the Ritz method, assuming that thereal and virtual displacements are approximated as

w(x)=Nn=1

an sinnπxℓ , w(x)= N

m=1

am sinmπxℓ

The derivatives of the approximate displacement are

w′ =Nn=1

annπℓ cos nπxℓ , w′′ = −Nn=1

an nπℓ 2

sin nπxℓ

The derivatives of the virtual displacement are

w′ = Nm=1

ammπℓ cosmπxℓ , w′′ = −Nm=1

ammπℓ 2

sinmπxℓ

Then, the virtual work statement yields

k

P

x

EI


G(w,w) = Nm=1

amN

n=1

an ℓ0

n2m2π4ℓ4 + 4β4 sin nπxℓ sinmπxℓ dx

− Nm=1

am ℓ0

−q sinmπxℓ dxG(w,w)= 0 for all choices of am is satisfied only if

Ni=n

ann2m2π4ℓ4 + 4β4ℓ

0

sin nπxℓ sinmπxℓ dx= −qℓ0

sinmπxℓ dx

The integrals in the equation above are easily evaluated to be

ℓ0

sin nπxℓ sinmπxℓ dx=

ℓ0

sinmπxℓ dx= ℓmπ1− cosmπ = ℓ

mπ1−−1 m

12ℓ

0

n= m

n≠ m

Substituting these results into the virtual work equation we get

anℓ2n4π4ℓ4 + 4β4 = −q ℓnπ 1− −1n

which can be immediately solved for an to give

an =2qnπ −1n−1n4π4∕ℓ4+ 4β4

= 2qℓ4nπ−1n−1π4n4+γ

where γ≡ kℓ4∕EI is a dimensionless constant. The displacement is given by

w(x)=Nn=1

an sinnπxℓ =Nn=1

2qℓ4nπ−1n−1π4n4+γ sin nπxℓ

189. A semi-infinite beam on a Winkler elasticfoundation extends to infinity in one direction. Find theclassical solution to the problem of a beam of modulusEI on a foundation with modulus k subjected to a con-centrated force P at x= 0. There is no boundary at the right end of the beam, but you canargue that a1 = a2 = 0 to have finite displacements. Plot the deflected shape of the beam.

k

P

ℓ∕3

x

EI

k

ℓ∕3 ℓ∕3


Using the displacement given in Problem 187 for the homogeneous solution andnoting that a1 = a2 = 0 for finite displacements, we have

w(x)= e–βxa3 cos βx+ a4 sin βx

w(x)= eβxa1 cos βx+ a2 sin βx

The boundary conditions are

M(0)= EIw′′(0)= 0

Q(0)− P=−EIw′′′(0)− P= 0

These conditions yield

a3 =−PEI

, a4= 0

So the displacement is

w(x)=− PEI

e–βx cos βx

w(x)

2 4

− PEI

0βx

190. A beam of length ℓ and modulus EI rests on twolinearly elastic springs, each of modulus k. The springsaccrue force in proportion to the amount by which theystretch. The beam is pinned at the left end and is sub-jected to a point load P at the right end. Axial and sheardeformations of the beam can be neglected. What is thevirtual-work form of the equilibrium equations? Whatare the essential and natural boundary conditions? Use the Ritz method to find an approxi-mation of the displacement field using the two-term polynomial w(x)= a1x+a2x2∕ℓ.

(a) What is the virtual work form of the equilibrium equations?

G(w,w) = ℓ0

EIw′′w′′ dx

+ kw(ℓ∕3)w(ℓ∕3)+ kw(2ℓ∕3)w(2ℓ∕3)+ Pw(ℓ)


(b) What are the essential and natural boundary conditions? The essential boundarycondition is

w(0)= 0

Natural boundary conditions are M(0)= M(ℓ)= 0 and Q(ℓ)+P= 0. These bound-ary conditions can be expressed in terms of w as

w′′(0)= 0

w′′(ℓ)= 0

EIw′′′(ℓ)− P= 0

(c) Use the Ritz method to find an approximation of the displacement field using thetwo-term polynomial w(x) = a1x+ a2 x

2∕ℓ.Let ξ≡ x∕ℓ and let us define the following matrices

h=ξℓ

ξ2ℓa=

a1

a2a=

a1

a2

It follows that w= aTh and w′′ = aTh′′. By using this notation, the virtual workfunctional becomes

G(w,w) = aT10

EIℓh′′ h′′ T a + qℓh dξ + P aTh(1)

+ k a T h(1∕3)h(1∕3) T+h(2∕3)h(2∕3) T a


K ≡10

EIℓh′′ h′′ T dξ+ k h(1∕3)h(1∕3) T+h(2∕3)h(2∕3) T

f≡−P h(1)


K = kℓ259

f=−Pℓ1

113

13

4η+ 1781

where, η≡ EI∕kℓ3. Solving the system of equations Ka=f yields

a1 =−9Pkℓ162η−52+810η , a2=−

81Pkℓ 12+810η

Therefore, the approximate displacement map is

w(x) = − 9Pk(2+810η)

162η−5ξ+ 9ξ2

ℓ

k PEI


191. Consider the beam of modulus EI, pinned at oneend, free at the other, and restrained by a rotational springas shown. The beam is subjected to a tip load P at the freeend. Shear and axial deformations can be neglected. Esti-mate the deflection of the beam at the point where load isapplied. Discuss the accuracy of your estimate and discuss one other possible method formaking the estimate.

The simplest method to solve this problem is to use the classical solution to thefixed cantilever beam and add the effect of the discrete rotation at the spring (whichadds a linear motion). The classical solution for the fixed cantilever gives a deflectionsof Pℓ3∕3EI. The moment felt by the spring is Pℓ so the rotation is Pℓ∕k. The deflec-tion at the end caused by that rotation can be computed to give Pℓ2∕k. Thus, the enddeflection is

w(ℓ)= Pℓ33EI+ Pℓ2

k

One can more formally solve the classical equation by noting the homogeneous solu-tion (and its derivatives) is

w= a+bξ+cξ2+dξ3

w′ = 1ℓb+2cξ+3dξ2 w′′′ = 1

ℓ36d

w′′ = 1ℓ22c+6dξ


w(0)= 0, M(0)= kw′(0), M(ℓ)= 0, Q(ℓ)=−P

Substituting the solution into these conditions give

EIw′′(0)= kw′(0), EIℓ2 2c=

kℓ b, b= 2c EI

kℓ

w(0)= 0, a= 0

EIw′′(ℓ)= 0, EIℓ22c+6d = 0, c=−3d

EIw′′′(ℓ)= P, EIℓ3 6d= P, d= Pℓ3

6EI

Thus,

a= 0, b=− Pkℓ , c=− Pℓ3

2EI, d= Pℓ3

6EI

so that the displacement field is

w= Pℓ36EIξ3−3ξ2 − Pℓ2

kξ

k

q

ℓ

x

EIRigid

ℓ


which yields the results mentioned earlier. This solution is exact.

An alternate approach to this problem is to solve it using the Ritz method and thePrinciple of Virtual Work.

192. A flexible beam of length ℓ and modu-lus EI is welded to a rigid beam of length ℓ,and rests on an elastic foundation ofmodulusk=60EI/ℓ4. The beam is simply supportedand is subjected to a transverse force q overthe rigid part of the span. The elastic founda-tion accrues a transverse force in proportion to the transverse displacement w. Shear andaxial deformations in the beam are negligible. Write the virtual-work functionalG for thesystem.What are the essential and natural boundary conditions for the flexible beam?Findan approximate solution for the displacementw(x) usinga two-termpolynomial Ritz basis.

(a) Write the virtual work functional G for the system.

ℓ

x

ℓ

∆

θ

w(x)

q

Q(ℓ)

M(ℓ)

ξ x

k∆+θξ

ξ

Q(ℓ)RA

Let us define ∆ ≡ w(ℓ), θ≡ w′(ℓ), and a coordinate axis ξ, which starts fromthe left end of the rigid part of the beam. From the geometric constraint that the beammust reach the right support we must have

∆+ℓθ= 0

Thus, the effective force on the rigid part of the beam is q^ ≡−q+ kθ ℓ−ξ , takenpositive upward. The “external work done by these forces in a virtual displacementgiven by w(ξ)= ∆+θξ=−θ ℓ−ξ is

WE = ℓ

0

w(ξ)q^(ξ) dξ = ℓ0

qθℓ−ξ − kθθ ℓ−ξ 2 dξ

= 12qℓ2θ− 1

3kℓ3θθ = 1

2qℓ2w′(ℓ)− 1

3kℓ3w′(ℓ)w′(ℓ)

The virtual work functional G(w,w)= WI−WE is, therefore,


G(w,w) = ℓ0

EIw′′w′′ + kww dx + 13 kℓ

3w′(ℓ)w′(ℓ)− 12 qℓ

2 w′(ℓ)

or, noting the constraint ∆+ℓθ= 0, equivalently

G(w,w) = ℓ0

EIw′′w′′ + kww dx + 13 kℓw(ℓ)w(ℓ)+

12 qℓw(ℓ)

(b) What are the essential and natural boundary conditions for the flexible beam?Since the right portion of the beam is rigid (see figure above), it follows that the essen-tial boundary conditions on the beam are

w(0)= 0, w(ℓ)+ ℓw′(ℓ)= 0

Also, taking the rigid part of the beam as a freebody diagram, the sum of the momentsabout the right support is zero. To wit,

M(ℓ)−ℓQ(ℓ)+ℓ0

q^ ℓ−ξ dξ = 0

Substituting the expression for q^ from above, and carrying out the integrations, we getthe expression

M(ℓ)−ℓQ(ℓ)− 12qℓ2+ 1

3kℓ3θ = 0

Finally, substituting the constitutive relationships M= EIw′′ and Q=−EIw′′′ wefind that the natural boundary conditions are

w′′(0)= 0

EI w′′(ℓ)+ℓw′′′(ℓ) + 13kℓ3w′(ℓ)− 1

2qℓ2 = 0

(c) Find an approximate solution for the displacement w(x) using a two-term polyno-mial Ritz basis. Let the approximate displacement be given by the following polyno-mial expansion

w(x)= a0ℓ3+ a1ℓ2 x+ a2ℓx2+ a3 x3

w′(x)= a1ℓ2+ 2a2ℓx+ 3a3x2

The essential boundary conditions must be satisfied a priori. Therefore,

w(0)= 0

w(ℓ)+ℓw′(ℓ)= 0

⇒ a0 = 0

⇒ a1 =−32a2− 2a3

Substituting these relationships into the original expansion we get

w(x)= a2 ℓx2− 32 ℓ

2x + a3 x3−2ℓ2 x

w′(x)= a2 (2ℓx−32ℓ2 )+ a3 ( 3x

2− 2ℓ2 )

w′′(x)= a2 (2ℓ )+ a3 (6x )

ℓ

EI

wℓ


Let ξ≡ x∕ℓ and define the matrices

h=ℓ3ξ2− 3

2 ξ

ℓ3ξ3−2ξ a=

a2

a3a=

a2

a3

It follows that w= aTh and w′′ = aTh′′. Using this notation, the virtual work func-tional becomes

G(w,w) = ℓ0

EIw′′w′′ + kww dx + 13 kℓ

3w′(ℓ)w′(ℓ)− 12 qℓ

2 w′(ℓ)

= a T10

EIℓ h′′ h′′ T+kℓh h T adξ

+ a T13 kℓ3 h′(1)h′(1) T a− a T 12 qℓ2 h′(1) Thus, the stiffness matrix K and the force vector f are given by

K = 10

EIℓh′′ h′′ T+kℓh h T dξ + 13kℓ3 h′(1)h′(1) T

f = 12qℓ2 h′(1)

Carrying out the indicated integrations, noting that k=60EI/ℓ4, we get

K = EIℓ321

f=−qℓ414

12

38

508738

Solving the system of equations Ka=f yields

a2 =3qℓ280EI

, a2 =−qℓ80EI

The approximate displacement is given by

w(ξ) = qℓ4560EI

−7ξ3+6ξ2+5ξ

193. Consider the beam of modulus EI, fixed at oneend, pinned at the other. The beam is subjected to aprescribed displacement of wℓ at the right end. Shearand axial deformations can be neglected. Find theexpression for the displacement w(x) that satisfiesthe governing equations exactly. Approximately solve the problem using the principle ofvirtual work, assuming a cubic polynomial deflection field.


Using the results from Problem 191 without the spring we have

w= Pℓ36EIξ3−3ξ2

where P is an unknown reaction force required to obtain the prescribed motion. Use theboundary condition w(ℓ)= wℓ to get

w(1)=− Pℓ33EI= wℓ, P=−

3wℓEIℓ3

Thus,

w=wℓ23ξ2−ξ3

To solve the problem using the Ritz method we have a virtual work functional

G(w,w)= ℓ0

EIw′′w ′′ dx= 0

Take a Ritz approximation of

w= wℓξ2+ a ξ3−ξ2 , w= a ξ3−ξ2

w′′ = 2wℓ + a (6ξ−2) ∕ℓ2 , w′′ = a 6ξ−2 ∕ℓ2

Note that the fixed function is quadratic so that the displacement function satisfies bothessential boundary conditions at the left end. The base functions are chosen to satisfythe homogeneous essential boundary conditions. Since there are three conditions acubic function only has one degree of freedom left. Therefore, the Ritz approximationis a one--term approximation.

Substituting the approximation into the virtual work functional gives

G= EIℓ410

2wℓ + a 6ξ−2 a 6ξ−2 ℓdξ= 0

Invoking the fundamental theorem of the calculus of variations we get

2wℓ1

0

6ξ−2 dξ+ a10

36ξ2−24ξ+4 dξ= 0

2wℓ 3−2 + a 12−12+4 = 0, ⇒ a=−wℓ∕2

Thus, the Ritz approximation gives

w= wℓξ2−

wℓ2ξ3−ξ2 = wℓ

23ξ2−ξ3

which is the exact solution.

1

x

M


194. Abeam of unit length and variablemodulus is fixed at the rightend and is subjected to a momentM at the left end. The bending andshear stiffnesses of the beam are variable with EI(x)= EI0 1+x and GA(x)= GA0

1+x , where EI0 and GA0 are known constants.Recall that the governing equations for the Timoshenko beam aregiven by Eqns. (406). Find the deflection and rotation at the left end of the beam (i.e., atx= 0) by finding the classical solution to the governing differential equations.

The governing equations for the Timoshenko beam are

Q′ + q= 0, Q= GA(w′−θ)

M′ +Q+m= 0, M= EIθ′

Note the following differentiation relationships

dxx = ln x, ddxln x= 1

x

The boundary conditions are (letting the applied moment at the left be called M0 todistinguish it from the moment field)

Q(0)= 0, M(0)= M0, w(1)= 0, θ(1)= 0

Note that, for this problem, we have q= 0 and m= 0. Therefore, we can integrate theequilibrium equations and use the boundary conditions at the left end to find two of theconstants of integration. To wit,

Q′ = 0 ⇒ Q= a0

Q(0)= a0= 0 ⇒ Q(x)= 0

Similarly for moment we have

M′ = 0 ⇒ M= a1

M(0)= a1= M0 ⇒ M(x)= M0

Now, substituting this result into the constitutive equation for moment we can find therotation by integration

EI0 1+x θ′ = M0

θ′ =M0

EI01

1+x ⇒ θ(x)=M0

EI0ln1+x + a2

Substituting the boundary condition on rotation at the right end we get

θ(1)= 0=M0

EI0ln(2)+ a2 = 0 ⇒ a2=−

M0

EI0ln 2

k

M

ℓ∕3

EI

k

ℓ∕3 ℓ∕3


Therefore, the rotation field is

θ(x)=M0

EI0ln1+x

2

Now we can do the same for the constitutive equation for shear. To wit,

Q= GA(w′ − θ)= 0 ⇒ w′ = θ

Integrating this equation we get

w′(x)=M0

EI0ln1+x

2

w(x)=M0

EI01+x ln1+x

2− 1+x + a3

Substituting the boundary condition for displacement at the right end we get

w(1)= 0=−2M0

EI0+ a3 ⇒ a3=

2M0

EI0

Thus, the displacement field is

w(x)=M0

EI01+x ln1+x

2 + 1−x

Finally, we can evaluate the rotation and displacement at the left end to get

θ(0)=−M0

EI0ln 2, w(0)=

M0

EI01− ln 2

195. A beam of length ℓ and modulus EI rests on twolinearly elastic springs, each of modulus k. The springsaccrue force in proportion to the amount by which theystretch. The beam is pinned at both ends and is subjectedto a concentrated momentM at the right end. Axial andshear deformations of the beam can be neglected. What is the virtual-work form of theequilibrium equations? What are the essential and natural boundary conditions? Use theRitz method to find a polynomial approximation of the displacement field.


G(w,w)= ℓ0

EIw′′w ′′ dx+ k wℓ∕3wℓ∕3+w2ℓ∕3 w2ℓ∕3 −Mw′(ℓ)

x1

x2

2b

2a


(b) What are the essential and natural boundary conditions?

Essential B.C.’s: w(0)= 0, w(ℓ)= 0

Natural B.C.’s: w′′(0)= 0, EI w′′(ℓ)= M

(c) Use the Ritz method to find an approximation of the displacement field using aquadratic approximation.

w(x)= a0+ a1x+ a2x2

The approximation must satisfy the essential boundary conditions. Thus,

w(0)= 0 ⇒ a0= 0

w(ℓ)= 0 ⇒ a1=−a2ℓ

Letting a2 ≡ a, the displacement field, and its derivatives are

w(x)= a x2−xℓ w′ = a 2x−ℓ ) w′′ = 2a

Thus, the discrete form of the virtual work functional is

G(a, a)= ℓ0

EI2a2a dx+ kaa 292ℓ2+ 292ℓ2−M aℓ = 0

Carrying out the integration we get

a4EIℓ+ 881 kℓ

4 a−Mℓ = 0

This equation must hold for all a therefore we get

a=81M

324EI+8kℓ3

Thus, the approximate displacement field is

w(x)= 81M324EI+8kℓ3

x2−xℓ

196. Consider a beam of length ℓ, elastic moduli E and G and rec-tangular cross section of width 2a and depth 2b. Let c be a (verysmall) constant and let ξ≡ x3∕ℓ be the normalized axial coordi-nate. The beam has the following displacement and rotation fields

θ1 = c 6ξ−6ξ2 , θ2= c 3ξ2−2ξ , θ3 = 0

w1 = cℓξ3−ξ2 , w2 = cℓ2ξ3−3ξ2 , w3= 0

P

ℓ

xEI,GA


Find the resultant moment and resultant force at ξ= 1, i.e.M(ℓ) andQ(ℓ). Find the totaldisplacement u of the point located at the position x = (a, b, ℓ).

(a) Find the resultant moment and resultant force at x = ℓ, i.e.M(ℓ) and Q(ℓ).

w′ + e3× θ=w1′ − θ2

w3′w2′ + θ1

= c3ξ2− 2ξ− 3ξ2+ 2ξ

0

6ξ2− 6ξ+ 6ξ− 6ξ2= 0

Therefore, Q = 0. The curvature can be computed from the given field as

θ′(ℓ)=6∕ℓ − 12∕ℓ

0

6∕ℓ − 2∕ℓ c=−6∕ℓ

0

4∕ℓ c

Noting that I11 = 16a3b∕12 and I22 = 16b3a∕12, the moment is

M=43 ab

3E 0 00 4

3 a3bE 0

0 0 GJ

−6∕ℓ

04∕ℓ c=

0

− 243 ab

3c

163 a

3bc

(b) Find the total displacement u of the point located at the position x = (a, b, ℓ).

u= w− p× θ=

w3− x1 θ2+ x2 θ1

w1− x2 θ3w2+ x1 θ3

u(a, b, ℓ)=0

−cℓ

−ca

197. A beam of length ℓ is fixed at the left end, free at the rightend, and is subjected to a concentrated transverse load P at theright end. The bending and shear stiffnesses of the beam are EIandGA, respectively. What is the virtual-work functional for thesystem?What are the essential and natural boundary conditions?Let ξ≡ x∕ℓ. Find the deflection and rotation fields for the givenloading by the Ritz method using the following approximation

w(ξ)= a2ℓξ+12a0ℓξ2+

13a1ℓξ3, θ(ξ)= a0 ξ+a1ξ2

k

P

ℓ∕3

x

EI

k

ℓ∕3 ℓ∕3

P


(a) What is the virtual work functional for the system?

G(w, θ,w, θ)=ℓ0

EIθ′θ′ +GA w′−θ w′−θ ′ dx+ Pw(ℓ)


Essential B.C.’s: w(0)= 0, θ(0)= 0

Natural B.C.’s: M(ℓ)= 0, Q(ℓ)+P= 0

(c) Find the deflection and rotation fields for the given loading by the Ritz methodusing the following approximation

w(ξ)= a2ℓx+12a0ℓξ2+

13a1ℓξ3 θ(ξ)= a0ξ+ a1 ξ

2

w′ = a2ξ+ a0 ξ+ a1ξ2 θ′ = 1

ℓa0+2a1ξ

Note that w′−θ= a2ξ. Take w, θ using the same interpolation as w, θ. Then

G=10

EIℓa0+2a1ξ a0+2a1ξ + 1

γ a2a2 dξ+Pℓ a02 + a13+a2 = 0

where γ≡ EI∕GAℓ2. Carrying out the integrals, and invoking the fundamental theo-rem of the calculus of variations, we get

EIℓ + Pℓ

a0a1

a2

1∕2

1∕31

= 0

1 1 0

1 4∕3 0

0 0 1∕γ

Solving these equations gives

a2 =−Pℓ2EIγ=− P

GAa0 =−Pℓ2

EI a1 = Pℓ22EI

The displacement and rotation fields are

w(ξ)= Pℓ36EI−3ξ2+ ξ3 − 6γξ , θ(ξ)= Pℓ2

2EI−2ξ+ξ2

198. A beam of length ℓ and modulus EI rests on two lin-early elastic springs, each of modulus k. The beam is sub-jected to point loadsP at the ends. Axial and shear deforma-tions of the beam can be neglected.What is the virtual-workform of the equilibrium equations? What are the essentialand natural boundary conditions? Solve the discrete virtual-work equations Ka = f for this system using a three-term polynomial Ritz approximation.

k

P

x

EI



G(w,w)= ℓ0

EIw′′w′′ dx+ kwℓ3wℓ

3+w2ℓ

3w2ℓ

3+Pw(0)+Pw(ℓ)


Essential B.C.’s: None

Natural B.C.’s: M(0)= 0 ⇒ w′′(0)= 0

M(ℓ)= 0 ⇒ w′′(ℓ)= 0

P−Q(0)= 0 ⇒ P+EIw′′′(0)= 0

P+Q(ℓ)= 0 ⇒ P− EIw′′′(ℓ)= 0

(c) Set up, but do not solve, the discrete virtual work equations Ka = f for this systemusing a three-term polynomial Ritz approximation.

Let ξ= x∕ℓ and take an approximation of the form

w(ξ)= a0+ a1ξ+ a2ξ2 w(ξ)= a0+ a1 ξ+ a2ξ

2

The base functions are then h= 1, ξ, ξ2 and h′′ = 0, 0, 2 ∕ℓ2. The stiffness ma-trix is then

K=10

EIh′′ h′′ ℓdξ+ k h1∕3 h1∕3 + h2∕3 h2∕3

and the load vector is

f= P h0+h1


K= EIℓ3

0 0 00 0 0

0 0 4

+ k 1 5∕9 1∕35∕9 1∕3 17∕81

2 1 5∕9, f=−P

2

1

1

199. A semi-infinite (i.e., x extends to infinity) beamof modulus EI rests on an elastic foundation ofmodulusk and is subjected to a concentrated forcePat x= 0.Theclassical differential equation for the beam is given byEIwiv+kw= 0, where wiv means fourth derivative.The shear in the beam is given by Q(x)=−EIw′′′. Use the Ritz method with the princi-ple of virtual work to find an approximation of the displacement field using the single termapproximation w(x)= ae–βx, where β is a given constant and a is the unknown Ritz dis-placement parameter. Find an expression for the error in equilibrium at each point x in thebeam. For what value of β is the shear boundary condition at x=0 satisfied exactly?


(a) Use the Ritz method with the principle of virtual work to find an approximation ofthe displacement field using the single term approximation

w(x)= ae–βx

where β is a given constant and a is the unknown Ritz displacement parameter. TheRitz function and its derivatives are given as

w= a e--βx w′ = − β a e--βx w′′ = β2 a e--βx

w= a e--βx w′ = − β a e--βx w′′ = β2 a e--βx

The virtual work functional is thus

G(w,w)= ∞0

EIw′′w′′+kww dx+ Pw(0)

= a EIβ4+k a∞0

e−2 β x dx+ P= 0

If G= 0 for all a, then the value of a can be determined as (and therefore the approxi-mate displacement as)

a=−2βPEIβ4+k w(x)= −2βP

EIβ4+k e−β x

(b) Find an expression for the error in equilibrium at each point x in the beam. Theerror is the amount by which the approximation fails to satisfy the classical differentialequation. Substituting into the equations we get

error= EIβ4ae−β x+ kae−β x

= EIβ4+k − 2βPEIβ4+k e−β x=−2βPe−β x

(c) For what value of β is the shear boundary condition at x=0 satisfied exactly?The boundary conditions at the left end is P−Q(0)= 0. Substituting the approxi-

mation we get

P−Q(0) = P+ EIw′′′(0) = P+2EIβ4PEIβ4+k

= 0

Thus,

PEIβ4+k

3EIβ4 + k = 0 ⇒ β4 =− 3EIk

Since β is a positive constant, there is no value that will satisfy the boundary conditionexactly.

k

x

EI

ℓ

M

ℓ

k


200. A flexible beam of length ℓ and modulus EI is con-nected to a rigid beam of length ℓ at a point that rests on aroller support. The left end of the rigid part of the beam isrestrained by a linear elastic spring of modulus k as is theright end of the flexible part of the beam. The beam is sub-jected to an end momentM. Axial and shear deformationsof the flexible beam can be neglected. What is the virtual-work form of the equilibriumequations?What are the essential and natural boundary conditions for the flexible segmentof the beam? Solve the discrete virtual-work equations Kaf for this system using athree-term polynomial Ritz approximation.


w′(0)

w(x) w(ℓ)

M

R

kℓw′(0)

ℓw′(0)

Q(0)M(0) M(ℓ)

Q(ℓ)kw(ℓ)

The kinematics and freebody diagrams are given in the sketch. The virtual work formof the differential equations

G(w,w)= ℓ0

EIw′′w′′ dx+ kℓ2w′(0)w′(0)+ kw(ℓ)w(ℓ)−Mw′(0)

The principle of virtual work suggests that if G(w,w)= 0 for all w then equilibriumholds.

(b) What are the essential and natural boundary conditions for the flexible segment ofthe beam?

Essential B.C.’s : w(0)= 0

Natural B.C.’s: From the freebody diagram above.

M(0)+M−kℓ2w′(0)= 0; M(0)= EIw′′(0)

M(ℓ)= EIw′′(ℓ)= 0

Q(ℓ)+kw(ℓ)= 0; Q=−EIw′′′(c) Set up, but do not solve, the discrete virtual work equations Ka = f for this systemusing a three-term polynomial Ritz approximation. Let the Ritz approximation bew= aTh(ξ) and w= a Th(ξ) where the base functions (and derivatives) are

h(ξ)= ξ, ξ2, ξ3 , h′(ξ)= 1ℓ1, 2ξ, 3ξ2 , h′′(ξ)= 1

ℓ20, 2, 6ξ

The stiffness and force vectors then have the form

x

q(x)= 2x

k

1∕21∕2


K=10

EIh′′ h′′ ℓdξ+ kℓ2h′(0) h′(0)+ kh(1) h(1)

f= Mh′(0)

Carrying out the indicated integrations

0 4 60 6 12

0 0 0

K= EIℓ3

1 1 1

2 1 1

1 1 1+ k f= M

ℓ 0

0

1

201. Consider the simply supported beam of length 1 andconstantmodulusEI1, subjected to a linearly varying forceq(x)2x, as shown. The beam is supported by a spring atmidspan that has modulus k64. Shear and axial deforma-tions can be neglected. What are the natural boundary condi-tions? What are the essential boundary conditions? Find anapproximate solution with the Ritz method. Use a cubic polynomial.

(a) What are the natural boundary conditions? What are the essential boundaryconditions?

Natural B.C.’s : w′′(0)= 0 w′′(1)= 0

Essential B.C.’s: w(0)= 0 w(1)= 0

Note: The spring is not a boundary!

(b) Find an approximate solution with the Ritz method. Use a polynomialapproximation of cubic order. The essential boundary conditions can be used to estab-lish a suitable set of base functions. Take a cubic approximation as follows

w(x)= a0+a1x+a2x2+a3x3

w(0)= 0 ⇒ a0= 0

w(1)= 0 ⇒ a1+ a2+ a3= 0

w(x)= a2 x2−x + a3 x3−x

Let the Ritz approximation be w= aTh(ξ) and w= a Th(ξ) where the base functions(and derivatives) are

h(x)= x2−x, x3−x , h′(x)= 2x−1, 3x2−1 , h′′(x)= 2, 6x


G(w,w)= 10

w′′w′′ dx+ 64 w(1∕2)w(1∕2)−10

2x w dx

qo

x

ℓ

EI, GA


Therefore, the stiffness and force vectors then have the form

K=10

h′′ h′′ dx+ 64h(1∕2) h(1∕2), f= 10

2xh dx


12 21

8 12K= f=− 0.10

− 0.1667

Thus, the coefficients satisfy the equations

a3

a2

12 21

8 12 =− 0.10

− 0.1667

−12 8

21 −12

a3

a2 = 124 − 0.10

− 0.1667=

0.050

− 0.0958

Therefore, the approximation solution for the displacement field is

w(x)=−0.0958 x2−x + 0.05 x3−x

202. A beam of length ℓ is fixed at the left end, pinnedat the right end, and is subjected to a uniform load, asshown. The shear and bending moduli are related as EI/GAℓ2=1. Find the displacement and rotation fields forthe beam by solving the classical governing equations.Find the reaction forces at the supports.

(a) Find the displacement and rotation fields for the beam by solving the classicalgoverning equations.

Start by integrating the equation for rotation

EIθ′′′ = −q0 ⇒ θ(x)=−q0ℓ36EIx3ℓ3+ a1

x2

ℓ2+ a2xℓ + a3

θ′(x)=−q0ℓ26EI3 x2ℓ2+ 2a1

xℓ + a2

Substitute the boundary conditions that involve the rotation field:

k

M

ℓ

EI

k


(1) θ(0)= 0 ⇒ a3= 0

(2) θ′(ℓ)= 0 ⇒ 2a1+ a2+ 3= 0

Now integrate the remaining governing equation

w′ = θ−ℓ2θ′′ = −q0ℓ36EIx3ℓ3+ a1

x2

ℓ2+ a2xℓ− 6 xℓ − 2a1

to give

w(x)=−q0ℓ46EI14x4

ℓ4+13a1x3

ℓ3+12a2x2

ℓ2− 2 x2

ℓ2− 2a1xℓ + a4

Now substitute the boundary conditions that involve the displacement field:

(3) w(0)= 0 ⇒ a4= 0

(4) w(l)= 0 ⇒ − 20a1+ 6a2− 33= 0

From (2) and (4): a1 =− 51∕32, a2 = 6∕32. Therefore, The displacement and rota-tion fields have the final form

w(x)=−q0ℓ4192EI8 x4ℓ4− 17 x

3

l3− 93 x

2

ℓ2+ 102 xℓ

θ(x)=−q0ℓ3192EI32 x3

l3− 51 x

2

ℓ2+ 6 xℓ

(b) Find the reaction forces at the supports.

The reactions can be found from: M(0)= EIθ′(0)=− q0ℓ2∕32. To wit,

Q(0)=−EIθ′′(0)=−51q0ℓ96

Q(ℓ)=−EIθ′′(ℓ)=45q0ℓ96

qo

ℓ−

q0ℓ232

5196q0ℓ

4596q0ℓ

203. A beam of length ℓ andmodulus EI rests on two lin-early elastic springs, each of modulus k. The springs ac-crue force in proportion to the amount by which theystretch. The beam is subjected to a concentrated momentM at the right end. Axial and shear deformations of thebeam can be neglected. Assume that motion along the axis of the beam is restrained.Whatis the virtual-work form of the equilibrium equations? What are the essential and naturalboundary conditions? Solve the discrete equations of equilibrium Ka=f using the Ritzmethod with a quadratic approximation of the displacement field. Describe the error inapproximation.



G(w,w)= ℓ0

EIw′′w′′ dx+ k w(0)w(0)+ w(ℓ)w(ℓ) −Mw′(ℓ)

(b) What are the essential and natural boundary conditions? There are no essentialB.C.’s (remember mixed B.C.’s are natural). The natural boundary conditions can beobtained from freebody diagrams of the ends of the beam.

(1) EI w′′′(0)+ k w(0)= 0

(2) EI w′′(0)= 0

M(ℓ)

Q(ℓ) k w(ℓ)

M

M(0)

k w(0)

Q(0)

(3) −EI w′′′(ℓ)+ k w(ℓ)= 0

(4) M− EI w′′(ℓ)= 0

(c) Solve the discrete equations of equilibrium Ka=f using the Ritz method with aquadratic approximation of the displacement field. Take a quadratic approximation tothe displacement field as

w(x)= a0+ a1xℓ+ a2

x2ℓ2 w(x)= a0+ a1

xℓ+ a2

x2ℓ2

Let the Ritz approximation be w= aTh(ξ) and w= a Th(ξ) where the base functions(and derivatives) are

h(x)= 1, xℓ , x2

ℓ2 h′(x)= 0, 1ℓ , 2xℓ h′′(x)= 0, 0, 2ℓ2

Substituting these into the virtual work functional we get the stiffness and force vectors

K=ℓ0

EIh′′ h′′ dx+ k h(0) h(0)+ h(ℓ) h(ℓ) , f= M h′(ℓ)

Carrying out the integrals we find

0 0 00 0 00 0 4

K= EIℓ3 + k

2 1 11 1 11 1 1

f= Mℓ

012

(d) Describe the error in approximation. Shear is constant, moment in linear for theexact solution. The quadratic approximation gives constant moment, no shear.

kx

EI

ℓ

M

ℓ

k

rigid


204. A flexible beam of length ℓ and modulus EI is con-nected to a rigid beam of length ℓ at a point that rests ona roller support. The left end of the rigid beam is restrainedby a linear elastic spring of modulus k. The beam is sub-jected to an end momentM. Axial and shear deformationsof the flexible beam can be neglected. What is the virtual-work form of the equilibrium equations?What are the essential and natural boundary con-ditions for the flexible segment of the beam? Solve the discrete equations of equilibriumKa=f using the Ritz method with a quadratic approximation of the displacement field.Describe the error in approximation.


w′(0)

w(x) w(ℓ)

R

kℓw′(0)

ℓw′(0)Q(0)

M(0) M(ℓ)

Q(ℓ)kw(ℓ)

M

The kinematics and freebody diagrams are given in the sketch. The virtual workform of the differential equations

G(w,w)= ℓ0

EIw′′w′′ dx+ kℓ2w′(0)w′(0)+ kw(ℓ)w(ℓ)−Mw′(ℓ)

The principle of virtual work suggests that if G(w,w)= 0 for all w then equilibriumholds.

(b) What are the essential and natural boundary conditions for the flexible segment ofthe beam?

Essential B.C.: w(0)= 0

Natural B.C.’s : From the freebody diagram above.

M(0)−kℓ2w′(0)= 0; M(0)= EIw′′(0)

M(ℓ)= EIw′′(ℓ)= M

Q(ℓ)+kw(ℓ)= 0; Q=−EIw′′′

(c) Solve, the discrete virtual work equations Ka = f for this system using athree-term polynomial Ritz approximation. Let the Ritz approximation be w= aTh(ξ)and w= a Th(ξ) where the base functions (and derivatives) are

h(ξ)= ξ, ξ2, ξ3 , h′(ξ)= 1ℓ1, 2ξ, 3ξ2 , h′′(ξ)= 1

ℓ20, 2, 6ξ

The stiffness and force vectors then have the form

k

P

x

EI

ℓ


K =10

EIh′′ h′′ ℓdξ+ kℓ2h′(0) h′(0)+ kh(1) h(1)

f= Mh′(1)

Carrying out the indicated integrations

0 4 60 6 12

0 0 0K = EI

ℓ3 + k 1 1 1

2 1 1

1 1 1f= M

ℓ2

3

1

205. A semi-infinite beam (i.e., the beam extends to in-finity in the positive x direction) of modulus EI rests onan elastic foundation of modulus k. The beam is sup-ported at x=0 and is subjected to a concentrated force Pat a distance ℓ from the support. Discuss how youwouldsolve this problem. Include in your discussion com-ments on both classical and variational approaches. Assume that Bernoulli-Euler beamtheory is adequate to describe the response of this system. The classical equations of aBer-noulli-Euler beam on an elastic foundation are

EIwiv+ kw= 0 M(x)= EIw′′ Q(x)=− EIw′′′

Classical Solution: The homogenous solution is given in the form of an exponentialw(x)= eλx. Substituting this expression into the governing equation gives the charac-teristic equation to establish the value of λ

λ4EI+k eλx= 0 ⇒ λ4 = kEI

Let β≡ 4 k∕EI be the positive fourth root of k∕EI. Then the characteristic equationhas four roots

λ1 = β, λ2=−β, λ3= i β, λ4=−i β

where i is the imaginary unit. Then the homogeneous solution has the form

w(x)= a0eβ x+ a1e

−β x+ a2 ei β x+ a3e

−i β x

= b0 sin βx+ b1 cos βx+ b2 sinh β x+ b3 cosh βx

Break the domain into two segments:

P M1(0)

R

Q1(0)

(1) (2) ∞Q1 Q2

Q2(0)

M2(0)


There are 4 constants per segment, which gives 8 constants of integration to establishfrom boundary and continuity conditions.:

2 B.C.’s at x=−ℓ

(1) M1(−ℓ)= 0

(2) Q1(−ℓ)+ P= 0

4 B.C.’s at x= 0

(1) w1 = 0

(2) w2 = 0

(3) w1′(0)= w2′(0)

(4) M1(0)= M2(0)

Note that as x→∞, w2 → 0, w2′ → 0. These two conditions provide the remainingtwo boundary conditions.

These eight conditions would be used to establish the eight constants. The twodisplacement fields are

w1(x)= b10 sin βx+ b11 cos βx+ b12 sinh β x+ b13 cosh βx

w2(x)= b20 sin βx+ b21 cos βx+ b22 sinh β x+ b23 cosh βx

It is possible that a different parameterization of distance in the two domains wouldsimplify the algebra of solving the equations that result from the boundary and conti-nuity conditions (i.e., measure x from the left end of segment 1 and use ξ in segment 2with origin at the left of segment 2).

Variational Approach:

The virtual work functional can be written as

G(w,w)= ∞−ℓ

EIw′′w′′ + k ww dx+ Pw(−ℓ)

We can use a Ritz approximation in the following form

w(x)=Ni=1

ai hi(x) w(x)=Ni=1

ai hi(x)

where hi(x) must satisfy hi(0)= 0. The moment will have a kink and the shear willhave a jump at x= 0, therefore, the FE functions should be used. Also, the solutionmust approach zero asx→∞.

Use the ordinary cubic Hermitian functions in segment (1) then use decayingfunctions, e.g., e−βx in the second segment matching value and derivative at x= 0.

xℓ = 1


h11

h4

h3

h2

1

1

1

206. Consider a beam of unit length and circular cross section, fixed atx= 0 and free at x= 1. The axis of the beam (x3 = x) points along thee3 direction. The origin of the cross sectional coordinates x1 and x2 are atthe center of the centroid of the section. The internal resultant force is giv-en by the explicit expression Q(x)= x2−1 e1+x3−1 e2. Find the ap-plied force q(x) that must be present. Assume that the applied momentm(x)= 0. Find the internal moment fieldM(x). Find the rotation field θ(x)

(a) Find the applied force q(x) that must be present.

The equations of equilibrium of a beam assert that

Q′ + q= 0

Therefore, the applied load that is implied by the net internal force is

q=−Q′ = −2xe1− 3x2e2

(b) Assume that the applied moment m(x)=0. Find the internal moment fieldM(x).The equilibrium equations governing moment equilibrium for the beam are given by(for m= 0) is M′ + e3× Q= 0. First we can compute the second term because Qis given

e3×Q = x2−1 e3× e1+ x3−1 e3× e2= x2−1 e2− x3−1 e1

That establishes the rate of change of the moment.

M′ = −e3× Q= x3−1 e1− x2−1 e2

The moment can be computed from integration

M= 14 x4−xe1− 13 x3−xe2+ c

k x

EI

ℓ

P

ℓ


The boundary condition at x= 1 can be used to establish the constant of integration

M(1)= 14−1e1− 13−1e2+ c= 0

c= 34 e1− 23 e2Thus, the moment field is

M= 14x4−4x+3 e1− 1

3x3−3x+2 e2

(c) Find the rotation field θ(x). The rotation field can be found by integrating theconstitutive equation for moment--rotation. Note that I11 = I22 ≡ I and J= 2 I, whereI is the second moment of the cross--sectional area. The constitutive equation for mo-ment is M= Iθ′ where

0 0 2GI0 EI 0EI 0 0

I=

Consider the equations by components. Integrating we get

EIθ1′ = 14x4−4x+ 3 ⇒ EIθ1= x5

20− x2

2+ 3x

4+ a1

EIθ2′ = 13−x3+ 3x−2 ⇒ EIθ2=− x4

12+ x2

2− 2x

3+ a2

2 GI θ3′ = 0 ⇒ 2 GI θ3 = a3

Using the boundary condition θ(0)= 0 gives a1 = a2 = a3= 0. Therefore, the rota-tion field is

θ(x)= 1EI x5

20− x2

2+ 3x

4 e1+ − x4

12+ x2

2− 2x

3 e2

207. A flexible beam of length 2ℓ and modulus EIrests on an elastic foundation of modulus k. The prop-erties have values such that the dimensionless ratiokℓ4∕EI= 15. The beam is subjected to a loadP at itsmidpoint. Axial and shear deformations of the flex-ible beam can be neglected. Find the deflection at themiddle and ends of the beam using virtual work and the Ritz method with a polynomialapproximation. (Note: due to symmetry odd functions—i.e., linear, cubic, etc.—need notbe included.)


The virtual work form of the equilibrium equations?

G(w,w)= ℓ−ℓ

EIw′′w′′+kww dx− Pw(0)

There are no essential B.C.’s. Let the Ritz approximation be w= aTh(ξ) andw= a Th(ξ) where the base functions (and derivatives) are

h(ξ)= 1, ξ2, ξ4 h′(ξ)= 1ℓ 0, 2ξ, 4ξ3 h′′(ξ)= 1

ℓ2 0, 2, 12ξ2

Substituting these into the virtual work functional we get the stiffness and force vectors

K = 210

EIh′′ h′′ + kh h ℓdξ, f=−P h(0)

Carrying out the integrals we find

1

0

0

K = 2EIℓ3 + 2kℓ f= P

0 0 0

0

0

4 8

8 1445

17

19

17

1 13

15

15

13

15

Noting that kℓ4∕EI= 15 we have

K = 2EIℓ3 + 30EI

ℓ3

0 0 0

0

0

4 8

8 1445

17

19

17

1 13

15

15

13

15

K = EIℓ3 + EI

ℓ3

0 0 0

0

0

8 16

16 2885

307

309

307

30 10 6

610

6

K = EIℓ3

1427

274245

1427

30 10 6

1410

6

Note that for a two--term solutions we can use the upper two by two corner of K andthe upper two terms of f. The two--term solution is

a1 =− 7160

Pℓ3EI

, a2 = 5160

Pℓ3EI


So the approximate displacement is

w(x)=− Pℓ3160EI7−5 x2ℓ2

We can evaluate the displacement at the middle and ends to be

w(0)=− 7Pℓ3160EI

, w(ℓ)=− 2Pℓ3160EI

Chapter 8The Linear Theoryof Plates

208. The three-dimensional rotation tensor Λ without drilling rotation can be obtainedfrom two successive rotations, first ψ about the x1 axis, and then φ about the new x2 axis

Λ(ψ,φ)=cosφ

0

− sin φ

0

1

0

sinφ

0

cosφ

1

0

− sinψ

0

cosψ

0

sinψ

0

cosψ

Compute the product of the two tensors to find Λ.Demonstrate that the tensor Λ is orthogo-nal by showing that Λ−1 = ΛT. Show that for small values of the parameters ψ and φ, thetensor can be expressed in the form Λ≈ I+W, where I is the identity andW is a skew-symmetric tensor. Show, therefore, that when the parameters are small, they can be viewedas the components of the rotation vector θwith ψ = θ1, φ=− θ2, and θ3 = 0 such thatW= θ×.

(a) Demonstrate that the tensor Λ is orthogonal by showing that Λ−1 = ΛT.

If Λ−1 = ΛT, then ΛTΛ= Imust hold. Let us denote Λ= Q1 (φ)Q2 (ψ), where

Q1 (φ)≡cosφ

0

-- sinφ

0

1

0

sinφ

0

cosφ

1

0

-- sinψ

0

cosψ

0

sinψ

0

cosψ

Q2 (ψ)≡

Note that, Q1 (φ) and Q2 (ψ) are orthogonal tensors. To wit,

cosφ

0

sinφ

0

1

0

− sin φ

0

cosφ

1

0

0

0

1

0

0

0

1

QT1 (φ)Q1 (φ)=

cosφ

0

− sin φ

0

1

0

sinφ

0

cosφ

=


1

0

− sin ψ

0

cosψ

0

sinψ

0

cosψ

1

0

sinψ

0

cosψ

0

− sin ψ

0

cosψ

1

0

0

0

1

0

0

0

1

QT2 (ψ)Q2 (ψ)= =

where we have used the trigonometric identity sin2 α+ cos2α= 1. It follows that

ΛTΛ= (Q2Q1 )T (Q2Q1 )

= QT1Q

T2Q2Q1 = QT

1 IQ1

= QT1 Q1= I

(b) Show that for small values of the parameters ψ and φ, the tensor can be expressedin the form Λ≈ I+W, where I is the identity andW is a skew-symmetric tensor.Show, therefore, that when the parameters are small, they can be viewed as the compo-nents of the rotation vector θ with ψ = θ1, φ= -- θ2, and θ3 = 0 such thatW= θ×.

Since sinα≈ α and cosα≈ 1, for small values of the angle α

1

0

0

=

Λ(ψ, θ)≈1

0

-- φ

0

1

0

φ

0

1 -- ψ

0

1 ψ

0

1

1

0

-- φ

-- φψ

1

-- ψ

φ

ψ

1

≈1

0

-- φ

0

1

-- ψ

φ

ψ

1

where we have noted that −φψ ≈ 0, since ψ and φ are very small. It follows that

Λ≈ I+0

0

-- φ

0

0

-- ψ

φ

ψ

0

= I+W

Thus, by letting ψ = θ1, φ= -- θ2, we have

W=0

0

θ2

0

0

-- θ1

-- θ2θ10

So, W= θ× with θ3 = 0.

Chapter 8 The Linear Theory of Plates 241

209. Write out the explicit constitutive expressions for the stress resultants

Nαβ= hλwγ,γ δαβ+ mwβ,α+wα,β

Qα= hm w3,α+Áαβθβ

Mαβ=h312λÁαβ Áγηθη,γ+ m ÁγβÁαηθη,γ+θβ,α

For the normal force resultants we get

N11 = h λwγ,γ+ m (w1,1+ w1,1 )

= h (λ+2m)w1,1+ λw2,2

= h Ew1,1+ λw2,2

N12 = hm w1,2+ w2,1

where E= λ+2m. Similarly

N22 = h Ew2,2+ λw1,1

N21 = hm w1,2+ w2,1 = N12

For the shear force resultants we get

Q1 = hm w3,1+ Á11θ1+ Á12θ2

= hm w3,1+ θ2

Q2 = hm w3,2− θ1

For the moment resultants we get

M12 = h312λÁγηθη,γ+ m Áγ2Á1ηθη,γ+ θ2,1

= h312λθ2,1− θ1,2 + m θ2,1+ θ2,1

= h312Eθ2,1− λθ1,2

M11 = h312 mÁγ1Á1ηθη,γ+ θ1,1

= h312 mθ1,1− θ2,2

And similarly

M22 = h312 mθ2,2− θ1,1

M21 = h312Eθ1,2− λθ2,1


210. Write out the explicit equilibrium expressions for the Mindlin plate equations

h312λÁαβÁγνθν,γα+ mÁγβÁανθγ,να+θβ,αα + hmÁαβ w3,α+Áαγθγ + mβ= 0

hmw3,αα+Áαβθβ,α + q3= 0

Note that β is a free index in this equation. The following simplifications can be made

λÁαβÁγνθν,γα= λÁαβ (θ2,1α− θ1,2α )

mÁγβÁανθγ,να+ θβ,αα = m Áγβ (θγ,12− θγ,21 )+ θβ,11+ θβ,22 = m θβ,11+ θβ,22

Thus, for β= 1

h312λÁα1 (θ2,1α− θ1,2α )+ m (θ1,11+ θ1,22 ) + hmÁα1 w3,α+ Áαγθγ + m1 = 0

h312−λ(θ2,12− θ1,22 )+ m (θ1,11+ θ1,22 ) − hm w3,2−θ1 + m1 = 0

And for β= 2

h312λÁα2 (θ2,1α− θ1,2α )+ m (θ2,11+ θ2,22 ) + hmÁα2 w3,α+ Áαγθγ + m2 = 0

h312−λ(θ2,11− θ1,21 )+ m (θ1,11+ θ2,22 ) − hm w3,1−θ2 + m2 = 0

The equation

hm w3,αα+Áαβθβ,α + q3 = 0

yields

hm w3,11+ w3,22+ θ2,1− θ1,2 + q3 = 0

211. For a smooth boundary, the expression for the external virtual work for the Kirch-hoff-Love plate is

WE= Ω

qw dA+Γ

Q^ sw−M^ sw,γ nγ ds

Modify the equation to account for point loads in the domain and corners on the boundary.What terms need to be added for a plate that has the shapeof a regular polygonwith n sides?


x2

x1

si

Γi+1

Γi

Γi−1

niti

ym

Pm ℓi

The external virtual work is given by (Eqn. 484 in the text)

WE= Ω

qwdA+Γ

mΓ ⋅ θ + nαQαw ds

Let us divide the boundary Γ into subdomains Γi which represent each edge of theplate where i = 1,. . ., n. Thus, the external virtual work becomes

WE= Ω

qwdA+ni=1

Γi

mΓi⋅ θ + niαQαwdsi

For the Kirchhoff-Love plate, we have from Eqn. (488) in the text

Γi

mΓi⋅ θ dsi=

Γi

mΓi⋅ ni ∇w ⋅ t i dsi−

Γi

mΓi⋅ t i ∇w ⋅ ni dsi

and by the Lemma on page 260 of the text we have

Γi

vi ∇w ⋅ t i dsi= Γi

∇viw ⋅ ti dsi−Γi

w ∇vi ⋅ t i dsi

= viwℓi0 −

Γi

w ∇vi ⋅ ti dsi

where we have denoted vi≡ mΓi⋅ ni . Thus, by defining

M^s≡ mΓ ⋅ t

Q^

s≡ nαQα −∇(mΓ ⋅ n ) ⋅ t

and carrying out the summation, we get

WE = Ω

qw dA + Γ

Q^ sw−M^ sw,γ nγ ds+n

i=1

w mΓi⋅ ni ℓi

0

In addition to this, for the concentrated loads Pm, applied at the locations denoted bythe position vectors ym in the domain, external virtual work functional can be aug-mented to


WE = Ω

qw dA + Γ

Q^ sw−M^ sw,γ nγ ds+n

i=1

w mΓi⋅ ni ℓi

0−M

j=1

Pmw(ym)

where M denotes the number of concentrated loads.

212. Consider the simply supported square plate of depth h, sides of length π, moduli λ*

and m, subjected to a downward uniform load of intensity −qo (shown in Fig. 110). As-sume an approximate transverse displacement of the form

w(x1, x2)= ∞

m=1

∞n=1

amn sinmx1 sin nx2

with a similar approximation for the virtual displacement. Assume that shear deformationsare negligible and compute the coefficients amn using the principle of virtual work. Is itpossible to consider an infinite number of terms in the displacement function? Howshouldthe solution be modified to solve the problem of a rectangular plate of dimensionsℓ1×ℓ2?

The virtual displacement field can be written as

w(x1, x2)=∞

k=1

∞l=1

akl sin kx1 sin lx2

The derivatives of the approximate displacement are given by

w,11 =−∞

m=1

∞n=1

amn m2 sinmx1 sin nx2

w,22 =−∞

m=1

∞n=1

amn n2 sinmx1 sin nx2

w,12 =−∞

m=1

∞n=1

amn mn cosmx1 cos nx2

with w,21 = w,12. The derivatives of the virtual displacements can be computed in thesame way. For convenience, let sij≡ sin ix1 sin jx2 and cij≡ cos ix1 cos jx2. Thus,since

w,ααw,ββ=∞

k=1

∞l=1

akl∞

m=1

∞n=1

amn (m2+n2 ) (k2+l2 )smn skl

w,αβw,αβ=∞

k=1

∞l=1

akl∞

m=1

∞n=1

amn (k2m2+ l2n2 )smn skl + 2 klmncmnckl


The virtual work equation G(w,w)= 0 for all w, where

G(w,w) ≡ Ω

h312λw,ααw,ββ+ 2mw,αβw,αβ − qwdA

yields the following system of equations for the coefficients amn

Ω

∞m=1

∞n=1

amn λ(m2+n2 ) (k2+l2 )+2m (k2m2+l2n2 ) smn skl+4mklmncmnckl dA

=−Ω

12qoh3

skl dA

The integrals can be evaluated with the help of the following identities

cosα+β + cosα−β = −2 cos α cos βcosα+β − cosα−β = 2 sin α sin β

To wit,

Ω

smn skl dA=14π0

π0

cos(m+n)x1− cos(m−n)x1 cos(k+l)x2− cos(k−l)x2 dx1dx2

= 0π2∕4

if m≠ k OR n≠ l

if m= k AND n= l

Ω

cmnckl dA=14π0

π0

cos(m+n)x1+ cos(m−n)x1 cos(k+l)x2+ cos(k−l)x2 dx1dx2

= 0π2∕4

if m≠ k OR n≠ l

if m= k AND n= l

Ω

skl dA=π

0

π0

sin kx1 sin lx2 dx1dx2 =1kl (−1)k−1 (−1)l−1

By substituting these results into the virtual work equation above, we get

amn λ(m2+n2 ) 2+2m (m4+n4 )+ 4mm2n2 = –12qomnh32π

2

(−1)m−1 (−1)n−1

Thus, we can find the coefficients explicitly as


amn = –qomnD 2π m2+ n2

2 (−1)m−1 (−1)n−1

amn = –qomnD 4π m2+ n2

2 for all m, n that are odd integers.

where D≡ λ+2mh3∕12.(a) Is it possible to consider an infinite number of terms in the displacement function?

We can, indeed, consider an infinite number of terms because of the orthogonalityproperties of the base functions. The equations for the displacement coefficients amnare uncoupled.

(b) How should the solution be modified to solve the problem of a rectangular plateof dimensions ℓ1×ℓ2?

The displacement approximation can be written as

w(x1, x2)= ∞

m=1

∞n=1

amn sinmπ x1ℓ1 sinnπx2ℓ2 with a similar expression for the virtual displacement. Then we follow the same proce-dure as in part (a) to obtain the coefficients amn .

213. Reconsider the plate in Problem 212.We can compute an approximate solution usingthe Ritz method with a polynomial basis. Note that, in order to satisfy the boundary condi-tions, the polynomial must have the form

w(x1, x2)= x1x2 (x1−π)(x2−π)a00+a10x1+a01x2+a11x1x2+⋅⋅⋅

(a) Compute the approximate displacement considering only the first term a00.(b) Unlike the beam,wherewe can add one term at a time with good results, the next term

wemight want to add to improve the solution is more complicated for the plate. Sincethe displacement is a function of two variables, it is possible to introduce an asymme-try if we are not careful in the introduction of new terms. One strategy is to select allterms with the same exponents, as shown below by the dashed lines.

x1 x2

1

x21 x1 x2 x22

x31 x21 x2 x1 x22 x32

x21 x22x31 x2 x1 x

32 x42x41

Resolve the problem using a four-term and a nine-term approximation.(c) A mixed strategy for selecting the basis functions might also be fruitfully employed.

Solve the problem with the four-term approximation


w(x1, x2) = sin x1 sin x2 a00+ a10x1+ a01x2+ a11 x1x2

Let g(x1 , x2 )≡ x1x2(x1−π)(x2−π), and let us define a matrix containing the basefunctions as hT= g(x1, x2) 1, x1, x2, x1x2, and a matrix containing the unknowncoefficients as aT= a00, a10, a01, a11, so that we can express the displacement as

w(x1, x2)= aTh(x1, x2)

Similarly, we can write w= aTh. The derivatives of the displacement field can becomputed as w,11 = aTh,11 and w,22 = aTh,22 and w,12 = aTh,12. It follows that

w,ααw,ββ= aT (h,11 +h,22 )(h,11 +h,22 )Ta

w,αβw,αβ= aT h,11 hT,11+h,22 hT,22+2h,12 h

T,12 a

We can substitute these results into the virtual work expression to get the standard lin-ear equations Ka= f. After some manipulation, the coefficient matrix K is given by

K = DΩ

h,11+h,22 h,11+h,22 T− (1−ν) h,11 hT,22+h,22 hT,11−2h,12 h

T,12 dA

where D≡ 112λ+2m h3, and the right-side matrix f is given by

f≡−Ω

qohdA

Carrying out the indicated integrals we find the explicit matrix components as

367π11

3150K = D

22π6

45

82π8

315

11π7

45

11π8

90

11π7

45

82π8

315

11π8

90

47π9

210

6π8

35

3π9

35

6π8

35

47π9

210

3π9

35

44π10

525

3π9

35

7π11

100

61π10

1050

7π11

100

176π10

1575

367π10

1575

47π10

420

61π10

1050

47π10

420

367π10

1575

31π11

350

7π11

100

31π11

350

223π12

3150

71π12

1050

223π12

3150

symm.

19π12

225

19π12

225

71π12

1050

97π13

1575

97π13

1575

194π14

3675

41π9

3153π9

3544π10

52547π10

420

367π11

31507π11

100

47π10

42041π9

315


and

π8

144π6

36f T=−qoπ9

240π10

400π9

240π7

72π7

72π8

120π8

120

The different shading represents a different level of approximation. The darkest shad-ing is the one-term approximation, the lighter shading is the four-term approximation,and the unshaded is the nine-term approximation. Clearly, all of the matrix componentscomputed for the lower-order approximations are apropos to the higher-order approxi-mation since the same base functions are reused.

The equations can be solved for the coefficients a. The MATHEMATICAT com-mands needed to form and solve the problem are given below for the nine-term approx-imation. To get the other approximations simply change the value of nterms.

g = x*y*(x - Pi)*(y - Pi)nterms = 4binom = List [ 1, x, y, x y, x^2, y^2, x^2 y, x y^2, x^2 y^2 ]h = g* Take[binom, nterms]h11 = D[h, x,2]h22 = D[h, y,2]h12 = D[h, x,1, y,1]k1 = Outer[Times, h11 + h22, h11 + h22]k2 = Outer[Times, h11, h22] + Outer[Times, h22, h11]k3 = 2 Outer[Times, h12, h12]ki = k1 - (1-n)(k2 - k3)K = Integrate[ki, x,0,Pi,y,0,Pi]f = Integrate[h , x,0,Pi,y,0,Pi]a = Simplify[ Inverse[K].f ]

The results are presented in the following table

− 588

a00

0

a10

0

− 4515155896

4515155896π2

− 514577948π4

− 514577948π2

4515155896π2

514577948π3

514577948π3

0

a01 a02 a22a11 a20 a21 a12

− 588

− 4515155896

− 366177948

1

4

9

N

where all the coefficients amn are normalized by qo∕D. Note that the coefficient amn isthe multiplier of the base function xm1 x

n2. As the table indicates, the four-term approxi-

mation yields the same result as the one-term approximation. The reason that the newcoefficients come out to be zero is that these are the values that best represent the sym-metry of the loading. The nine-term approximation allows the development of addi-tional symmetric displacement.

x1

x2−qo


(c) A mixed strategy for selecting the basis functions might also be fruitfullyemployed. Solve the problem with the four-term approximation

w(x1, x2) = sin x1 sin x2 a00+ a10x1+ a01x2+ a11 x1x2

We follow the same procedure as in part (a) with g(x1 , x2 )≡ sin x1 sin x2. The firstline of the MATHEMATICAT script needs to be changed to the line

g = Sin[x]*Sin[y]

also, the number of terms in the approximation must be specified as nterms=1 or asnterms=4 as appropriate. Carrying out the indicated integrals gives the componentsof the coefficient matrix K

K = D

π2

π26(9+2π2)

π32

π4

4

π32

π26(9+2π2)

π44

− 54π

2+π4+π6

9

symm.

π312 (9+2π

2)

π312 (9+2π

2)

and the right-side matrix as

π24f T=−qo 2π 2π

The results are presented in the following table

4π2

a00

0

a10

0 0

a01 a11

4π2

1

4

N

where, again, all the coefficients aij are normalized by qo∕D. As the table indicates,the four-term approximation again yields the same result as the one approximation.

214. Consider a square plate of depth h, sides of length ℓ,moduliλ* and m, subjected to a downward uniform load of intensity−qo. Assume that shear deformations are negligible (i.e., Kirch-hoff-Love plate theory is applicable). Find an approximate dis-placement for the following boundary conditions


fx

frfx

fx

ss

frss

ss

x2

x1ℓ

ℓ

0

x2

x1ℓ

ℓ

0

x2

x1ℓ

ℓ

0

(a) (b) (c)ss

ssfx

fx

where the designation ss indicates a simple support, fx indicates a fixed edge, and fr indi-cates a free edge.

The virtual work functional is given by the expression

G(w,w) ≡ Ω

h312λw,ααw,ββ+ 2mw,αβw,αβ − qwdA

Observing that λ= νC∕(1−ν2) and 2m= (1−ν)C∕(1−ν2), the virtual work func-tional can be put into the form

G(w,w) ≡ DΩ

w,ααw,ββ+ (1−ν) w,αβw,αβ− w,ααw,ββ dA−Ω

qw dA

where D= Ch3∕12(1−ν2). This form will prove convenient for organizing the com-putations for this problem.

For the purposes of this problem we shall use a one-term expansion. Thus, let usdefine the displacement as w= a g(x1 , x2 ) where g(x1 , x2 )= h1 (x1 )h2 (x2 ) is theproduct of one-dimensional functions. The derivatives can be computed as

w,11 = ag,11 w,22 = ag,22 w,12 = ag,12

It follows that

w,αβw,αβ− w,αα w,ββ = 2aa g,12 g,12− g,11 g,22

w,ααw,ββ = aa g,11+ g,22 2

By substituting these results into the virtual work expression we get the standard linearequations Ka= f, where the coefficient K is

K= DΩ

g,11+ g,22 2 − 2(1−ν) g,11 g,22 −g,12 2 dA

and the right-side f is

f ≡−Ω

qg dA


For part (a), the essential boundary conditions are

w(0, x2 )= w(π, x2 )= 0

w(x1 , 0 )= w(π, x2 )= 0

w′(x1 , 0 )= w′( 0, x2 )= 0

Thus, an appropriate choice of base functions is

h1 = ξ31− ξ21 h2 = ξ32− ξ22

where ξi≡ xi∕π. The MATHEMATICAT commands needed to form and solve the prob-lem are given below

h1 = (x1/Pi)^3 - (x1/Pi)^2h2 = (x2/Pi)^3 - (x2/Pi)^2g = h1*h2g11 = D[g,x1,2]g22 = D[g,x2,2]g12 = D[g,x1,1,x2,1]q = - qoki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2))K = Integrate [ ki, x1,0,Pi,x2,0,Pi]f = Integrate [q*g, x1,0,Pi,x2,0,Pi]a = Simplify[ f/K ]

It follows that

K=176D1575π2

f =−qoπ2

144⇒ a =−

175qoπ4

2816D

For part (b), the essential boundary conditions are

w(0, x2 )= 0

w(x1 , 0 )= w(π, x2 )= 0

Thus, an appropriate choice of base functions is h1 = x1∕π and h2 = sin x2. The firsttwo lines of the MATHEMATICAT script need to be changed to

h1 = x1/Pih2 = Sin[x2]

It follows that

K= Dπ26+1−ν f =−qoπ ⇒ a = −

6qoπD (π2+6−6ν)

For part (c), the essential boundary conditions are

w(0, x2 )= 0

w(x1 , 0 )= w(x1 ,π )= 0

w′(x1 , 0 )= w′(x1 , π )= w′( 0, x2 )= 0


Thus, an appropriate choice of base functions is h1 = x1∕π 2 and h2 = 1− cos2 x2.The first two lines of the MATHEMATICAT script need to be changed to

h1 = (x1/Pi)^2h2 = 1 - Cos[x2]^2

It follows that

K= D 43−2ν+ 3

2π2+ 2π2

5 f =−

qoπ2

6

The displacement coefficient is, therefore

a = −5qoπ4

D 45+40π2−60π2ν+12π4

215. Solve any of the above variations on the square-plate problemwith any combinationof the following differences(a) Consider a rectangular plate of dimension ℓ1×ℓ2.(b) Consider a load form different from a uniform load.

(a) Consider a rectangular plate of dimension ℓ1×ℓ2.Let us consider the plate shown in the figure below to be subjected to a uniform

downward load of magnitude qox2

x1ℓ1

ℓ2

0

ss

ssfx

fx

where the designation ss indicates a simple support and fx indicates a fixed edge. Forthe purposes of this problem we shall use a one-term expansion. Thus, let us define thedisplacement as w= a g(x1 , x2 ) where the function g(x1 , x2 )= h1 (x1 )h2 (x2 ) is theproduct of one-dimensional base functions. Now, the suitable base functions for thiscase are

h1 = x1ℓ13

−x1ℓ12

h2 = x2ℓ23

−x2ℓ22

Performing the integrals associated with K and f as in Problem 214 we get

K= 4D1575ℓ1ℓ2

15γ+ 1γ

2

−16 f =−qoℓ1ℓ2144


where γ≡ ℓ2∕ℓ1 is the aspect ratio of the plate. Solving Ka= f for a we get

a =−175qoℓ21ℓ22

64D15γ+ 1

γ 2

−16−1

Note the symmetry in the lengths ℓ1 and ℓ2. Note also that, when ℓ1 = ℓ2 = π, we getthe same result as we did in Problem 214(a). The MATHEMATICAT commands needed toform and solve the problem are given below

h1 = (x1/L1)^3 - (x1/L1)^2h2 = (x2/L2)^3 - (x2/L2)^2g = h1*h2g11 = D[g,x1,2]g22 = D[g,x2,2]g12 = D[g,x1,1,x2,1]q = - qoki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2))K = Integrate [ ki, x1,0,L1,x2,0,L2]f = Integrate [q*g, x1,0,L1,x2,0,L2]a = Simplify [f/K]

(b) Consider a load form different from a uniform load.

x1

x2

-- qmax-- qmin

Let us again consider the same plate, but define the loading as

q(x)=−qmax+ qmax−qmin x1 x24π21− x1

π1− x2π

which is a parabolic loading that takes on its minimum value qmin at the center of theplate. Thus,

f ≡−Ω

q(x)g(x)dA=−π2qmax144+qmin−qmax

225

Since K does not change, the solution is given by

a= fK=−

7π4

2816D9qmax+16qmin

Note that, if qmax= qmin = qo , then

a=−7π4

2816D9qo+16qo = −

175qoπ4

2816D


which is what we got in Problem 214(a). The MATHEMATICAT commands needed toform and solve this problem are as follows:

h1 = (x1/Pi)^3 - (x1/Pi)^2h2 = (x2/Pi)^3 - (x2/Pi)^2g = h1*h2g11 = D[g,x1,2]g22 = D[g,x2,2]g12 = D[g,x1,1,x2,1]q = -qmax + qmin*(4/Pi)^2*x1*x2*(1-x1/Pi)*(1-x2/Pi)ki = C1 ( (g11 + g22)^2 - const (g11 g22 - g12^2))K = Integrate [ ki, x1,0,Pi,x2,0,Pi]f = Integrate [q*g, x1,0,Pi,x2,0,Pi]a = Simplify[ f/K ]

216. Solve the square (π× π), simply supported plate under uniform load (shown in Fig.110) considering shearing deformations. Note: You can still neglect the in-plane problembecause it is uncoupled from the bending problem. Use the Ritz method. Note thatw(x1, x2) must vanish on the boundary, but that θ1(x1, x2) and θ2(x1, x2) are not restricted.Let w(x1, x2)= a1 sin x1 sin x2. Select the functions for θ1 and θ2 so that the shearing de-formation is linear, that is

θ1 = w,2+ b12x2−π , θ2 =−w,1+ b22x1−π

What is the result of making such an assumption? Noting that the tangential moment mustvanish on the boundary, is there a means of finding a better displacement function? Canwe use statical considerations to improve the approximation of the rotation field?

The virtual work functional is given by

G(w,w, θ, θ) ≡ Ω

Qα ⋅ εα +Mα ⋅ κα− q ⋅ w dA

Neglecting the in-plane actions, the first term is

Qα ⋅ εα = Qα w,α+εαηθη = hmw,α+εαβθβ w,α+εαη θη

However, for the present problem, we are given θβ = εβνw,ν+ηβ where the functionsηβ are given as η1 = b1 2x2−π and η2 = b2 2x1−π . Thus, since

w,α+εαβ θβ = w,α+εαβεβνw,ν+εαβηβ = w,α−δανw,ν+εαβηβ= εαβηβ

and noting that εαβ εανηβην= δβνηβην= ηβηβ, we get

Qα ⋅ εα = hmηβηβ


Similarly, neglecting the in-plane actions, the second term of the virtual work function-

al can be written as Mα ⋅ κα = Mαβθβ,α. The components of the bending moment

field are given by

Mαβ =h312λÁαβÁγηθη,γ+ mÁγβÁαηθη,γ+ θβ,α

= h312λ(θβ,α−θα,β )+ m (2θβ,α−δαβθν,ν )

where we have used the identity Áαβ Áγη= δαγδβη−δαηδβγ. Thus, the internal workassociated with bending is given by

Mαβ θβ,α = Dθβ,α θβ,α− ν θα,β θβ,α− (1−ν)θα,α θβ,β Let x1 ≡ 2x2−π e2, x2 ≡ 2x1−π e3, and aT= a1, b1, b2 so that ηβ= aTxβ.Further more let w≡ sin x1 sin x2 e1 and let y1 ≡ w,2+x1 and y2 ≡−w,1+x2 sothat the rotation field can be computed as θα= aTyα. The derivatives of the rotationfields can be computed as θα ,β= aTyα ,β. The principle of virtual work yields the stan-dard linear system of equations Ka= f where the matrix K can be constructed from abending component and a shear component as K = Kb+Ks where

Kb = DΩ

yβ,α yTβ,α− ν yα,β yTβ,α− ( 1−ν)yα,α yTβ,β dA

Ks = hmΩ

xαxTα dA

The load vector can be computed as

f=−Ω

qowdA

Evaluating the integrals, we get the system matrices

π2

π2312+γπ2

−8 (1+ν )

symm.

K = D f=−qo−4π2ν

4

0

0

8 (1+ν )

π2312+γπ2

where γ≡ mh∕D= 6(1−ν)∕h2 is the ratio of shear to bending modulus. Solving thelinear system of equations yields

a1 =−4π2qoD 12(1−ν)+ γπ2384(1+ν)2+ 12π2(1+ν)–γπ6

b1 =−

96qoD 1+ν384(1+ν)2+ 12π2(1+ν)–γπ6

= −b2


Note that in the thin plate limit, as γ→∞, we get

limγ→∞

a1=−4qoπ2D

and b1 = b2 = 0, which is the exact solution for a Kirchhoff-Love plate. The MATHE-MATICAT commands needed to form and solve this problem are as follows:

z1 = 0, 2 x2 - Pi , 0z2 = 0, 0, 2 x1 - Pi w = Sin[x1]Sin[x2], 0, 0 y1 = D[w, x2] + z1y2 = -D[w, x1] + z2y11 = D[y1,x1]y12 = D[y1,x2]y21 = D[y2,x1]y22 = D[y2,x2]Ks = M1 ( Outer[Times, z1, z1] + Outer[Times, z2, z2] )B11 = Outer[Times,y11,y11]B12 = Outer[Times,y12,y12]B21 = Outer[Times,y21,y21]B22 = Outer[Times,y22,y22]C12 = Outer[Times,y12,y21]C21 = Outer[Times,y21,y12]K1 = B11 + B12 + B21 + B22K2 = B11 + C12 + C21 + B22K3 = Outer[Times,y11 + y22, y11 + y22]Kb = M2 ( K1 - n K2 - (1-n) K3 )K = Integrate[Ks + Kb, x1,0,Pi,x2,0,Pi]f = Integrate[w, x1,0,Pi,x2,0,Pi]a = Simplify[Inverse[K].f]

Alternatively, we can try the displacement given by

w(x1, x2)= a1 sin x1 sin x2

θ1 = w,2+ b1 sin x1 cos x2

θ2 = –w,1+ b2 cos x1 sin x2

Which leads to the matrices

π2

π241+γ

π22

symm.

K = D f =−qo−π2

4

4

0

0

−π2

2

π241+γ

Solving these we get

a1 = − 4qoπ2D1+ 2

γ b1 =−8qoπ2mh

= −b2

ss

anyany

ss

x2

x1π

π

0


Note that, for both solutions, deflection at the center of the plate is w(π∕2,π∕2)= a1 .In the limit as the thickness of the plate goes to zero γ→∞ and

limγ→∞

a1=−4qoπ2D

and b1 = b2 = 0, which is the same as the center deflection of the exact solution of asimply supported Kirchhoff-Love Plate under uniform loading. The MATHEMATICAT

commands needed to form and solve this problem are the same as above except for thefirst two lines which should be replace by

z1 = 0, Sin[x1]Cos[x2] , 0z2 = 0, 0, Cos[x1]Sin[x2]

217. Consider a square (π× π) plate subjected to a uniformtransverse load q. The plate is simply supported along the edgesx2 = 0 and x2 = π with any boundary conditions along theother two edges. Assume that shear deformations are negligi-ble. According to the method of Kantorovich let us assume thatthe real and virtual displacement fields can be represented as

w(x1, x2)= W(x1) sin x2, w(x1, x2)= W(x1) sin x2

where W(x1) and W(x1) are unknown functions.(a) Substitute these functions into the virtual-work functional

G(w,w)≡ Ω

Dνw,ααw,ββ+ 1−ν w,αβw,αβ− qw dA

and show that W(x1) must satisfy the ordinary differential equation

WIV− 2W′′ +W = 4qπD

(b) Verify that the following function satisfies the above equation

W(x1) =4qπD+ a1+a2x1 cosh x1+ a3+a4x1 sinh x1

(c) What are the possible boundary conditions for W(x1)? Find the constants of integra-tion for the case where the plate is simply supported on all four edges.

(a) The second derivatives of the displacement function are given by

w,11 = W′′ sin x2 w,22 =−W sin x2 w,12 = W′ cos x2

with w,21 = w,12. Thus, we have

w,ααw,ββ = W′′−W W′′−W sin2 x2

w,αβ w,αβ = W′′W′′+WW sin2 x2+ 2W′W′ cos2 x2


Substituting these into the virtual work functional yields

h3

12Ω

λ W′′−W W′′−W + 2m W′′W′′+WW sin2 x2 dA

h3

12Ω

4mW′W′ cos2 x2 dA = Ω

qW sin x2dA

The integrals above are

π0

sin2 x2dx2= π

0

cos2 x2dx2=π2

π0

sin x2dx2 = 2

Thus, upon evaluating the integrals with respect to x2 we get

h3

12π0

λ W′′−W W′′−W + 2m W′′W′′+WW+2W′W′ dx1 = 4π

π

0

qW dx1

We can integrate by parts to shed the derivatives from the W terms as follows

π0

W′′−WW′′ dx1 = W′′−WW′−W′′′−W′W π

0+π

0

WIV−W′′ W dx1

π0

W′′W′′ dx1= W′′W′ −W′′′W π

0 +π

0

WIV W dx1

π0

W′W′ dx1= W′W π

0 –π

0

W′′W dx1

Using these results, noting that λ∕(λ+2m)= ν and 2m∕(λ+2m)= 1−ν, we get

Dπ0

W WIV−2W′′+W dx1+ Dν W′′−W W′−W′′′−W′ W π

0

D 1−ν W′′W′−W′′′W+2W′W π

0 =4π

π

0

qW dx1

where D= λ+2m h3∕12. Rearranging the boundary terms, we get

π0

W WIV−2W′′+W dx1+ (W′′−νW)W′–(W′′′+ν−2 W′ )W π

0

= 4πDπ0

qW dx1

Invoking the fundamental theorem of the calculus of variations, we can conclude thatthe function W must satisfy the differential equation


WIV− 2W′′ +W =4qπD

We shall see in detail in part (c) what are the implications of the boundary terms.

(b) Verify that the following function satisfies the above equation

W(x1) =4qπD+ a1+a2x1 cosh x1+ a3+a4x1 sinh x1

The derivatives of the displacement field are

W′ = a2 cosh x1+ a4 sinh x1+ a1+a2x1 sinh x1+ a3+a4x1 cosh x1

W′′ = 2a2 sinh x1+ a4 cosh x1 + W− 4qπD

W′′′ = 2a2 cosh x1+ a4 sinh x1 + W′

WIV = 2a2 sinh x1+ a4 cosh x1 + W′′

(*)

(* *)

Subtracting (*) from (**) we arrive at the result

WIV− 2W′′ +W =4qπD

(c) What are the possible boundary conditions for W(x1)? Find the constants of in-tegration for the case where the plate is simply supported on all four edges.

Since the normal vector at the boundaries at x1 = 0 and x1 = π are e1 and −e1respectively, the four possible boundary conditions are as follows (see page 259 in thetext):

w= W sin x2= 0 vanishing displacement

w,1 = W′ sin x2 = 0

w,11+ νw,22 = W′′ − νW sin x2 = 0

vanishing tangential rotation

vanishing tangential moment

vanishing effective shearw,11+ νw,22 = W′′′ + ν−2 W sin x2 = 0

Since these equations must hold for all x2 , we have

W= 0 vanishing displacement

W′ = 0

W′′ − νW= 0

W′′′ + ν−2 W= 0

vanishing tangential rotation

vanishing tangential moment

vanishing effective shear

Note that, on any edge, either the displacement or the effective shear is prescribed, butnot both; either the tangential rotation or the tangential moment is prescribed but not


both. Note that, the boundary terms in part (a) vanish from the functional if the thefunction W is made to satisfy the homogeneous essential boundary conditions.

For a plate that is simply supported on all four edges, we have the conditions

W(0)= W(π)= 0

W′′(0)− νW(0)= 0 ⇒ W′′(0)= 0

W′′(π)− νW(π)= 0 ⇒ W′′(π)= 0

For simplicity, let us write the displacement function as follows

W(x1) =4q

πD1+ a1+a2x1 cosh x1+ a3+a4x1 sinh x1

W′′(x1) =4q

πD2a2 sinh x1+ 2a4 cosh x1− 1 + W(x1)

Thus,

W′′(π)= 0

W(0)= 0

W′′(0)= 0

W(π)= 0

⇒ 1+ a1 = 0

⇒ 2a4− 1= 0

⇒ 2a2 sinh π+ 2a4 coshπ− 1= 0

⇒ 1+ −1+a2π coshπ+ a3− π2sinhπ= 0

These equations can be solved to give the constants

a1 =−1 a2 =−12tanh π

2

a3 =−14π− sinhπ sech2 π

2a4 =

12

218. There are many possibilities available within the context of the method of Kantoro-vich. In general, we use an approximation of the form w(x1, x2)= W(x1)φ(x2), whereφ(x2) is a known function. In Problem 217, the choicewas φ(x2)= sin x2. Amore generalapproach is to use the deflected shape of a simply supported beam subjected to the appliedloading (assuming that it does not varywith x2). Find the general expression for the coeffi-cients a, b, c, and q^ of the differential equation

aWIV− bW′′ + cW = q^

that results from applying the method of Kantorovich with the known function φ(x2). Ex-press the coefficients as integrals of φ(x2). CanKantorovich’smethod beapplied tobound-ary conditions other than simple supports along parallel edges? Howwould the approachchange to accommodate more general boundary conditions?


The derivatives of the displacement functions are given by

w,11 = W′′φ w,22 = Wφ′′ w,12 = W′φ′

Thus, we can compute

w,ααw,ββ = W′′φ−Wφ′′ W′′φ−Wφ′′

w,αβw,αβ = W′′W′′φ2+WW φ′′ 2+ 2W′W′ φ′

2

Substituting these into the virtual work functional yields

Ω

λ W′′ W′′φ2+Wφφ′′ +W W′′φφ′′+W φ′′ 2 dA

= 12h3Ω

qWφdA+Ω

2m W′′W′′φ2+WW φ′′ 2+2W′W′φ′ 2 dA

Since the function φ(x2) is specified a priori, the following integrals can be evaluated

α1 ≡ π

0

φ2dx2 α2 ≡ π

0

φ′ 2dx2

α3 ≡ π

0

φφ′′dx2 α4 ≡ π

0

φ′′ 2dx2 α5 ≡ π

0

φ dx2

Noting that λ∕(λ+2m)= ν and 2m∕(λ+2m)= 1−ν, and carrying out the integralswith respect to x2, we get

+ W α4W+να3W′′ − α5qW dx1 = 0

π0

DW′′ α1W′′+να3W +W′ 21−ν α2W′

Defining a new set of parameters γi≡ αi∕α1 and integrating by parts yields

π0

W WIV+2νγ3+(ν−1)γ2W′′+γ4W dx1+ B.T.=γ5Dπ0

qW dx1

where B.T. stands for the boundary terms. To wit,

B.T. ≡ W′′−ν γ3W W′ + W′′′ + νγ3+2(ν−1)γ2W′ W π

0

By identification, then, the coefficients of the problem statement have the form

a= 1, b=−2νγ3+ 2(1−ν)γ2 , c= γ4 , q^ = qDγ5


Observe that, for φ(x2)= sin x2

a= c= 1 , b=−2 , q^ =4qπD

and the boundary terms are

B.T. ≡ W′′−νW W′ + W′′′ + ν−2W′ W π

0

Substituting these results into the equation above, we get the same result as Problem217. Kantorovich’s method can be applied to other boundary conditions, but the func-tion φ(x2) must be chosen to satisfy the boundary conditions.

ℓ

x

b(x)

τouo

Chapter 9Energy Principlesand Static Stability

219. Consider a one-dimensional rod of length ℓ andmodulus C subjected to a body force b(x) and a tractionτo at the right end. The left end has a prescribed displace-ment of u(0)= uo. The Hu-Washizu energy functionalfor the rod is given in terms of the independent variablesσ(x) (stress), u(x) (displacement), and Á(x) (strain) as

H(u, σ, Á)≡ ℓ0

12CÁ

2−bu−σ Á−u′ dx− τou(ℓ)+ σ(0) u(0)−uo

Show that by taking the directional derivative of H in the direction of variations of eachof the variables, i.e.,

DH(u, σ, Á) ⋅ (u, σ, Á) = ddαH(u+αu, σ+ασ, Á+αÁ)

α=0

and setting the result equal to zero (to find the extremum), all of the classical governingequations for the one-dimensional bar result from applying the fundamental theorem ofthe calculus of variations

σ′ + b= 0

σ− CÁ = 0

Á− u′ = 0σ(ℓ)− τo = 0

u(0)− uo = 0x∈ (0, ℓ)

Note that the fields that appear in the functional are each functions of x themselves.

The directional derivatives of the functional in the directions of a variation in each ofthe argument functions are


D1H ⋅ u = ℓ

0

σu′−bu dx − τu(ℓ) + σ(0)u(0)

D2H ⋅ σ = ℓ

0

−σÁ−u′ dx + σ(0) u(0)−uo

D3H ⋅ Á = ℓ

0

CÁ−σ Á dx

= ℓ0

−σ′+b u dx + σ(ℓ)−τ u(ℓ)

where the notation D1H ⋅ u means take the directional derivative by considering u tobe the argument holding σ and Á constant (i.e., like the partial derivative). Setting eachof these expression equal to zero and invoking the fundamental theorem of the calculusof variations, considering the fields u, σ, and Á as independent gives the desired result.

220. Find the energy functional E(u,w, θ) for a Timoshenko beam.

The energy functional for the Timoshenko beam is

E(u,w, θ) = ℓ0

12EAu′ 2+ 1

2 EIθ′ 2+ 1

2GAw′−θ 2− pu− qw −mθ dx

plus boundary terms if there are applied loads at the end points. One can demonstratethat this is the correct functional by taking the directional derivative to get the virtualwork functional and then applying the Fundamental Theorem of the Calculus of Varia-tions to show that minimization of the energy is equivalent to the classical differentialequations.

221. Find a Hu-Washizu energy functional for a simply supported Bernoulli-Euler beamof length ℓ and modulus EI subject to a transverse load q(x). The appropriate field vari-ables are the transverse displacement w(x), the moment M(x), and the curvature κ(x).Show that the extremum of the energy functional with respect to the three field variablesgives the classical equations of Bernoulli-Euler beam theory. How must the functionalchange if the beam is fixed at x= 0 and pinned at x= ℓ?

The functional can be created by analogy with the little boundary value problem withthe following identifications u→ w, σ→ M, and Á → κ.

Chapter 9 Energy Principles and Static Stability 265

H(w,M, κ) = ℓ0

12EIκ2− qw −Mκ−w′′ dx +M′w ℓ

0

Differentiating with respect to w we get

D1H ⋅ w = ℓ

0

Mw′′ − qw dx +M′w ℓ0

= ℓ0

−M′w′ − qw dx +Mw′ ℓ0+M′w ℓ

0

= ℓ0

M′′−q w dx +Mw′ ℓ0

Setting the expression equal to zero to find the extremum we find, from the fundamen-tal theorem of the calculus of variations, that M′′−q= 0 in [0,ℓ] and that M(0)= 0and M(ℓ)= 0. Differentiating with respect to M we get

D2H ⋅M = −ℓ0

M κ−w′′ dx +M′w ℓ0

Setting the expression equal to zero to find the extremum we find, from the fundamen-tal theorem of the calculus of variations, κ= w′′ in [0,ℓ] and w(0)= 0 andw(ℓ)= 0. Finally, differentiating with respect to κwe get

D3H ⋅ κ = ℓ

0

EIκ−M κ dx

Setting the expression equal to zero to find the extremum we find, from the fundamen-tal theorem of the calculus of variations, M= EIκ in [0,ℓ].

For the beam that is fixed at x= 0 and pinned at x= ℓ we must add a term toaccount for the additional essential boundary condition at x= 0. To wit,

H(w,M, κ) = ℓ0

12EIκ2− qw −Mκ−w′′ dx +M′w ℓ

0−M(0)w′(0)

222. The Hellinger-Reissner energy functional for a three-dimensional hyperelastic solidbody B with boundary Ω is given by

R(u, S)= B

S ⋅ ∇u−b ⋅ u−U(S) dV −Ωt

τ^ ⋅ u dA −Ωu

Sn ⋅ u−u^ dA


where u is the displacement field, S is the stress field, τ^ is the prescribed traction over Ωt

(the portion of the boundary where tractions are prescribed), u^ is the prescribed displace-ment over Ωu (the portion of the boundary where displacements are prescribed), b is thebody force, and U(S) in the stress-energy function. What do the necessary conditions foran extremum imply? (Hint: take the directional derivative of the functional in the direc-tions of u and S, and apply the fundamental theorem of the calculus of variations.)

The directional derivative of R in the direction of u is

D1R ⋅ u = B

S ⋅ ∇u− b ⋅ u dV− Ωt

t^ ⋅ u dA −

Ωu

Sn ⋅ u dA

= B

div STu − divS ⋅ u− b ⋅ u dV

= −B

divS+ b ⋅ u dV + Ωt

Sn− t^ ⋅ u dA

− Ωt

t^ ⋅ u dA −

Ωu

n ⋅ STu dA

From the fundamental theorem of the calculus of variations we get the conclusions thatdivS+b= 0 in B and Sn= t

^on ∂Ωt. The directional derivative of R in the direc-

tion of S is

D2R ⋅ S = B

S ⋅ ∇u−∂U∂S ⋅ S dV − Ωu

Sn ⋅ u− u^ dA

= B

∇u−∂U∂S ⋅ S dV − Ωu

Sn ⋅ u− u^ dA

From the fundamental theorem of the calculus of variations we get the conclusions thatu= u^ on ∂Ωu, and that

E ≡ 12∇u+∇uT = ∂U

∂S

Note that because the expression in parentheses (i.e., ∇u−∂U∕∂S) is contracted withS, which is symmetric, the antisymmetric part of it is annihilated. Since S is symmetricone can conclude that it is appropriate to take only the symmetric part of ∇u. This is-sue can be eliminated if 1

2∇u+∇uT is used in the original functional R in place of

∇u.

223. Show that the energy functional for the Kirchhoff-Love plate, given in Eqn. (513),can be expressed in the equivalent form


E(w)=Ω

12 D w,11+w,22 2− 2 1−ν w,11w,22− w,212 − qw dA

where D is the plate modulus and ν is Poisson’s ratio. The term w,11 w,22−w,212 is anapproximation of the Gaussian curvature of the deformed reference surface of the plate.

The expression for the energy in Eqn. (513) is

E(w) = Ω

12Dν w,αα2− (1−ν)w,αβw,αβ− qwdA

Expanding the terms inside the parentheses we get

(⋅)≡ ν w,αα2+ (1−ν)w,αβw,αβ

= ν w,211+2w,11w,22+w,222 + (1−ν)w,211+2w,212+w,222

= w,211+w,222 + 2νw,11w,22+ 2(1−ν)w,212

= w,211+w,222 + 2w,11w,22− 2(1−ν)w,11w,22+ 2(1−ν)w,212

= w,11+w,222− 2(1−ν)w,11w,22− w,212

which demonstrates the equivalence.

224. Show that the energy functional for a three-dimensional linear elastic solid is

E(u) = B

12 λdivu 2+ 1

2 m∇uT+∇u ⋅ ∇u− b ⋅ u dV−

Ωt

t ⋅ u dA

Show also that the extremum of the energy gives the same equations as the principle ofvirtual work.

The virtual work functional has the form

G(u, u) = B

λdivudivu+ m ∇uT+∇u ⋅ ∇u− b ⋅ u dV−Ωt

t ⋅ u dA

First check symmetry of the virtual work functional

DG(u, u) ⋅ u^ = B

λdivu^ divu+ m ∇u^ T+∇u^ ⋅ ∇u dV

Noting that u^ j, i ui, j− uj,i u^i,j= 0 we have


DG(u, u) ⋅ u^ −DG(u, u^ ) ⋅ u = B

m ∇u^ T ⋅ ∇u−∇u T ⋅ ∇u^ dV= 0

Thus, an energy functional exists. Applying Vainberg’s theorem, noting the linearity ofdifferentiation, i.e., div(tu)= tdivu and ∇(tu)= t∇u, we find

E(u) = 10

G(tu, u) dt

Substituting into the virtual work functional we get

E(u) = 10

B

tλ divu 2+ tm ∇uT+∇u ⋅ ∇u− b ⋅ u dV−Ωt

t ⋅ u dA dtCarrying out the indicated integration with respect to the scalar t we obtain

E(u) = B

12 λdivu 2+ 1

2 m∇uT+∇u ⋅ ∇u− b ⋅ u dV−

Ωt

t ⋅ u dA

225. Show that the energy functional for a Bernoulli-Euler beam on an elastic foundationcan be expressed in the form

E(w)=ℓ0

12 EIw′′ 2+ 1

2 kw2− qw dx

where EI is the flexural modulus of the beam, k is the modulus of the foundation, and qis the transverse load.

The virtual work functional can be obtained by taking the directional derivative of theenergy functional in the direction of the function w

G(w,w)= ℓ0

EIw′′w′′ + kww− qw dx

= ℓ0

−EIw′′′w′ + kww− qw dx+ EIw′′w′ ℓ0

= ℓ0

EIwiv+ kw− q w dx+ EIw′′w′ ℓ0− EIw′′′w ℓ

0

Equilibrium is satisfied if G(w,w)= 0 for all w. Applying the fundamental theorem ofthe calculus of variations we obtain the classical equilibrium equations for the beam onelastic foundation


EIwiv+ kw− q= 0

Either EIw′′ = 0 or w′ = 0 at x= 0 and x= ℓ

Either EIw′′′ = 0 or w= 0 at x= 0 and x= ℓ

The classical equations can also be obtained from first principles using Cauchy-New-ton balance of momentum.

226. Find the Euler equation and boundary conditions for the functional

J(w) = ba

F(w,w′,w′′) dx

Use the Euler equation on the energy functional for a Bernoulli-Euler beam on an elasticfoundation to find the classical differential equation governing the beam.

We can find the Euler equations at the extremum of the functional J. Taking the direc-tional derivative in the direction of the function w we get

DJ(w) ⋅ w = ba

∂F∂ww+ ∂F∂w′w′ + ∂F∂w′′w′′ dx

= ba

∂F∂ww− ddx∂F∂w′w−

ddx∂F∂w′′w′ dx

= ba

∂F∂w− ddx∂F∂w′ +

d2dx2∂F∂w′′w dx

+ ∂F∂w′wba+ ∂F∂w′′w′ba

+ ∂F∂w′ − ddx∂F∂w′′w b

a+ ∂F∂w′′w′ba

The Euler equation for the functional J is, therefore,

∂F∂w−

ddx∂F∂w′ +

d2dx2∂F∂w′′ = 0

With boundary conditions indicated at a and b as

Either ∂F∂w′ −ddx∂F∂w′′ = 0

Either ∂F∂w′′ = 0

or w = 0 at x= a and x= b

or w′ = 0 at x= a and x= b


For the Bernoulli-Euler beam we can take

F(w,w′,w′′) = 12EIw′′ 2+ 1

2kw2− qw

The derivatives with respect to the various arguments are

∂F∂w = kw− q ∂F

∂w′ = 0 ∂F∂w′′ = EIw′′

Therefore, the Euler equation is

d2dx2

EIw′′ + kw− q = 0

One can observe that the boundary terms come out the same as the previous problem.

227. Using the fundamental theorem of the calculus of variations, find the classical formof the governing differential equation for w(x) implied by the minimum of the energyfunctional

E(w)=ℓ0

12EIwivw− qw dx − 1

2EIw′′′w ℓ

0+ 1

2EIw′′w′ ℓ

0

Taking the directional derivative of the functional in the direction of the function w weget

− 12EIw′′′w+w′′′w ℓ

0+ 1

2EIw′′w′+w′′w′ ℓ

0

DE(w) ⋅ w=ℓ0

12EIwivw+wivw − qwdx

To apply the fundamental theorem of the calculus of variation we must integrate thefirst term by parts four times. The result is

+ 12EIw′′′w ℓ

0− 1

2EIw′′w′ ℓ

0+ 1

2EIw′w′′ ℓ

0− 1

2EIww′′′ ℓ

0

DE(w) ⋅ w=ℓ0

12EIwwiv+wivw − qwdx

− 12EIw′′′w+w′′′w ℓ

0+ 1

2EIw′′w′+w′′w′ ℓ

0

Simplifying we finally arrive at the result

DE(w) ⋅ w=ℓ0

EIwiv− q w dx− EIw′′′w ℓ0+ EIw′′w′ ℓ

0


Clearly, setting this functional equal to zero and applying the fundamental theorem ofthe calculus of variations results in the equations of the Bernoulli-Euler beam.

228. The potential energy of a simply supported, symmetrically loaded circular plate ofradius R is

E(w) = π R0

Dr w′′ 2+ 1r w′

2+ 2νw′w′′ − 2rqw drwhere the functionw(r) is the transverse deflection of the plate,D and ν are constants, andq is a known function of r. Find the variational (virtual work) form of the governing differ-ential equation. Find the classical (strong) form of the governing differential equation.What can you say, if anything, about the boundary conditions for the problem?

(a) Find the variational (virtual work) form of the governing differential equation.

DE(w) ⋅ w = πa0

D2rw′′w′′ + 2r w′w′ + 2ν w′w′′+w′w′′ − 2rqwdr

Integrating the term w′w′′ + w′w′′ by parts eliminates these terms to give

G(w,w) = 2πa0

Drw′′w′′ + 1r w′w′ − rqwdr+ 2πDνw′w′ a

0

(b) Find the classical (strong) form of the governing differential equation. Integratingby parts, using the above formulas, until all derivatives are shed from the w termsgives

DE(w) ⋅ w = 2πa0

Dwiv+ 2r w′′′ −

1r2w′′ + 1

r3w′ − qwrdr

+ 2πD rw′′+νw′ w′ a0+ 2πD1r w′−w′′−rw′′′w a

0

The classical equations can then be stated as

wiv+ 2r w′′′ −

1r2w′′ + 1

r3w′ − q

D= 0

(c) What can you say, if anything, about the boundary conditions for the problem?The boundary conditions can be determined from the boundary terms of the last ex-pression for DE ⋅ w. Because the loading is axisymmetric we have the conditionw′(0)= 0 as an essential boundary condition. The other boundary condition at r= 0is the natural condition (shear)


limr→01r w′−w′′−rw′′′ = 0

At the edge r= a a simply supported plate would have the conditions

w(a)= 0, aw′′(a)+ νw′(a) = 0

For a fixed edge at r= a the conditions would be

w′(a)= 0, 1aw′(a)− w′′(a)− aw′′′(a) = 0

229. Find an approximate solution to the problem of a simply supported, circular plateof radius R and modulus D, subjected to a uniform load of q. Use the Ritz method withthe energy functional given in Problem 228. Assume that the displacement is of the form

w(r)= a0 cosπr2Rwhere a0 is the, as yet, undetermined constant. If you had to pick additional terms in theapproximation, what would you choose? Why is the cosine function a good choice?


G(w,w) = a0

Drw′′w′′ + 1r w′w′ − rqwdr + Dνw′w′ a

0

The derivatives of w can be computed from the approximation as

w′ = −co π2a sinπr2a, w′′ = −co π24a2 cosπr2aLet ξ= πr∕2a and β≡ 4a2∕π2. The virtual work functional then becomes

G(w,w) = π2

0

Dβcocoξ cos2 ξ+ 1

ξsin2 ξ− qβcoξ cos ξdξ + Dν

βcoco

Carrying out the indicated integrals

G(co, co) = Dβcoco 1.191+ν − 0.5708qβco

Setting G equal to zero and solving for the constant we get

co = − qa4

D 0.093761.191+ν

ko

q(x)

ℓ

x

EI


The exact solution to this problem is

co = − qa4

64D5+ν1+ν

The approximation compares very favorably. For example, for ν= 0.3 the approxi-mate solution is co = 0.0629 and the exact is co = 0.0637.

Additional terms in the approximation. To increase the order of approximation onecould continue to chose cosine functions of higher wave number. The cosine functionis a good choice because it automatically satisfies the boundary condition w′(0)= 0,as required by symmetry. The term cos(3πr∕2a) would be the next choice as it satisfiesw(a)= 0.

230. A beam of length ℓ and modulus EI rests on anonlinearly elastic foundation that accrues trans-verse force in proportion to the cubeof the transversedisplacement, i.e., f (x)= kow3. The beam is sub-jected to downward transverse loading q(x). Axialand shear deformations are negligible. Take w(x) aspositive when it is upward. Find the virtual-work form of the equilibrium equations. Findthe energy functional E(w) for the system.

(a) Find the virtual work form of the equilibrium equations. The virtual work per unitlength done by the nonlinear force is simply −fw. Thus, the “external” work per unitlength is q−f )w. This term can be substituted directly into the expression for virtualwork to give

G(w,w) = ℓ0

EIw′′w′′ + kow3w− qw dx

(b) Find the energy functional E(w) for the system. First check the symmetry of thevirtual work functional

DG(w,w) ⋅ w^ = ℓ0

EIw^ ′′w′′ + 3kow2w^ w dx

The virtual work functional is clearly symmetric. Now the energy can be found fromVainberg’s theorem


E(w)=10

G(tw,w) dt

= 10

ℓ0

EI (tw′′)w′′ + ko (tw)3w− qw dxdt

= ℓ0

12 EIw′′ 2+ 1

4 kow4− qw dx

231. Consider a functional that takes scalar functions u(x) as input. The independent sca-lar variable is defined on the range x∈[0, 1]. The functional has the explicit form:

E(u)=10

11+x

u′(x) 2− 2u(x) dxwhere a prime indicates differentiation with respect to x. The functions are constrained atthe boundary to satisfy the conditions u(0)= 0 and u′(1)= 0. Find the classical differ-ential equation implied by stationarity (i.e., max, min, or saddle point) of the functional.Solve the classical differential equation. Compute the second derivative functional associ-ated with the given functional.

=−10

11+x u′ ′ + 2u dx + 1

1+x u′u1

0

DE(u ) ⋅ u= 10

11+x u′u ′−2u

dx

=−10

11+x u′ ′ + 2u dx − u′(0)u(0)

The classical differential equation can be deduced from the fundamental theorem of thecalculus of variations to be.

11+x u′ ′ + 2= 0

232. The virtual-work functional for a system is given by the following expression


G(w,w)= ℓ0

a w′′w+w′′w + bw3w dxwhere a, b, and ℓ are constants and w(x) andw(x) are functions of the independent variablex. Does this functional have an associated energy? Find the energy functional for the sys-tem, if it exists.

The existence of the energy functional depends upon the symmetry of the direc-tional derivative DG(w,w) ⋅ w^ = DG(w,w^ ) ⋅ w

DG(w,w) ⋅ w^ = ℓ0

a w^ ′′w+w′′w^ + 3bw2w^ w dx

= ℓ0

−2aw^ ′w′ + 3bw2w^ w dx+ a w^ ′w+w′w^ ℓ

0

This functional has an energy only for certain boundary conditions, but not in general.

233. Resolve Problem 232 with the functional

G(w,w)= ℓ0

−aw′′w+ bww dx − aw(ℓ)w′(ℓ)+ aw(0)w′(0)

The existence of the energy functional depends upon the symmetry of the direc-tional derivative DG(w,w) ⋅ w^ = DG(w,w^ ) ⋅ w

DG(w,w) ⋅ w^ = ℓ0

−aw^ ′′w+ bw^ w dx− aw^ (ℓ)w′(ℓ)+ aw^ (0)w′(0)

= ℓ0

aw^ ′w′ + bw^ w dx− a w^ ′w ℓ

0+ a w^ w′

ℓ

0

Again, this functional is not symmetric because of the presence of the asymmetricboundary terms.

It is worth noting that if the original functional had been

G(w,w)= ℓ0

−aw′′w+ bww dx − aw′(ℓ)w(ℓ)+ aw′(0)w(0)

then the functional would be symmetric because the boundary terms from integrationby parts would cancel the original boundary terms.


234. The deformation state of a particular system is characterized by the scalar functionθ(x), where the scalar variable x∈[0, ℓ]. The virtual-work functional for the system is giv-en by the following expression

G(θ, θ)= ℓ0

aθ′θ′ + bθ sin θ dx

where a, b, and ℓ are constants. Equilibrium of the system holds if G(θ, θ)= 0 for all θ.Does this functional have an associated energy? Find the energy functional for the system,if it exists.

(a) Does this functional have an associated energy?

DG(θ, θ ) ⋅ ψ= ℓ0

a ψ′θ′+bθ cos θψ dx

which is symmetric because swapping θ and ψ does not change the value. Therefore anenergy functional E(θ) exists.

(b) Find the energy functional for the system, if it exists. The energy functional canbe found by application of Vainberg’s theorem

E(θ)=10

G(t θ, θ) dt = 10

ℓ0

a t θ′ θ′ + b θ sin θ dx dt

= ℓ0

a t22θ′ 2+ −b cos tθ dx1

0

= ℓ0

12 aθ′ 2+ b 1− cos θ dx

235. The deformation state of a particular system is characterized by the scalar functionu(x), where the scalar variable x∈[0, 1]. The virtual-work functional for the system is giv-en by the following expression

G(u, u)= 10

au′u′ + bu′u+ g(u)u dx

where a and b are constants and g(u) is a given nonlinear function of the displacement func-tion u(x). Equilibrium of the system holds if G(u, u)= 0 for all u. For what values of theconstants a and b does this functional have an associated energy? Find the energy function-al for the system, if it exists.

k

P

ℓ

x


(a) For what values of the constants a and b does this functional have an associatedenergy? An energy functional exists if DG(u, u) ⋅ u^ = DG(u, u^) ⋅ u. Compute the di-rectional derivative as

DG(u, u) ⋅ u^ = ddε 1

0

a u+εu^ ′u′ + b u+εu^ ′u+ gu+εu^ u dxε=0

= 10

au^′u′ + b u^′u+ g′u^ u dx

Note: Symmetry holds only if b = 0, a can have any value.

(b) Find the energy functional for the system, if it exists. For b = 0 we can computethe energy by Vainberg’s theorem.

E(u)=10

G(t u, u) dt = 10

10

at(u′)2+g(t u)u dx dt

Define the function f (u) to be

f (u)≡10

g( t u )u dt

such that Df (u) ⋅ u= g(u)u. With this definition the energy functional can be writtenas

E(u)=10

12 au′ 2+ f (u) dx

236. A beam of length ℓ rests on an elastic founda-tion of modulus k (per unit length). It is pinned at theleft end and is subjected to a point load P at the rightend. The elastic foundation accrues a transverseforce in proportion to the transverse displacementw.The energy of the system can be expressed as:

E(w)=ℓ0

12EI(w′′)2+kw2 dx− Pw(ℓ)

Find the virtual-work form of the equilibrium equations. What are the essential boundaryconditions? Find the classical form of the equilibrium equations and the boundary condi-tions.Which of the three functions given beloware suitable for approximating the solutionwith the Ritz method? Explain why or why not in each case.


w(x)= x(ℓ−x)(a1+a2x), w(x)= x2(a1+a2x), w(x)= x(a1+a2x)

(a) Find the virtual work form of the equilibrium equations. What are the essentialboundary conditions?

Essential Boundary Condition: w(0)= 0, the rest are natural boundary condi-tions. The virtual work functional can be found by taking the directional derivative ofthe energy functional

G(w,w)= DE(w) ⋅ w= ddαE(w+αw) α=0

Carrying out this operation gives

G(w,w)= ℓ0

EIw′′w′′ + kww dx− Pw(ℓ)

G(w,w)= 0 for all virtual displacements that satisfy the essential boundary conditionis the weak form of equilibrium.

(b) Find the classical form of the equilibrium equations and the boundary conditions.To put the virtual work functional in a form suitable to apply the fundamental theoremof the calculus of variations, integrate G(w,w) by parts to get rid of w′′ terms:

G(w,w)= ℓ0

EIw′′′′+kw w dx+ EIw′′w′ ℓ0− EIw′′′w

ℓ0− Pw(ℓ)

From the Fundamental Theorem of the Calculus of Variations we can deduce

G(w,w)= 0 ∀w ⇒ EIw′′′′+kw= 0 for 0< x< ℓ

The virtual work functional tells us something about the possible boundary conditionstoo. In particular, the boundary terms vanish if

w(0)= 0

EIw′′(0)= 0

EIw′′(ℓ)= 0

EIw′′′(ℓ)+ P= 0

(essential BC)

(natural BC)

(natural BC)

(natural BC)

(c) Which of the three functions shown are suitable for approximating the solutionwith the Ritz method? Explain why or why not in each case.

The function w(x)= x ℓ−x a1+a2x is no good because w(ℓ)= 0 so the loadP does no work.

The function w(x)= x2 a1+a2x is no good because w′(0)= 0 implying anadditional essential boundary condition (fixed end).

h

ℓ P

WW

θ


The function w(x)= x a1+a2 x is fine.

237. Consider a rectangular (rigid) block of height handwidth ℓ andweightW.The block is prevented fromsliding by a small obstruction at the lower right cornerand is pushed by a force P at the upper left corner.Write the potential energy of the system in terms of theangle of rotation θ of the block. Find the force P as afunction of h, ℓ,W, and the angle θ needed to have stat-ic equilibrium. Find an expression for the angle atwhich equilibrium goes from being stable to being un-stable.

(a) Write the potential energy of the system in terms of the angle of rotation θ of theblock.

Datum for W

P

W

Datum for P

h sin θ

h2cos θ

ℓ2sin θ

ℓ2cos θ

Using terms defined in the sketch, we can write the energy of the system as

E(θ)= P ℓ cos θ−h sin θ + 12W ℓ sin θ+h cos θ

(b) Find the force P as a function of h, ℓ, W, and the angle θ needed to have staticequilibrium. Equilibrium holds when the energy is stationary, or

DE(θ) ⋅ θ= 0 ∀ θ

Computing the derivative of the energy, we get

E(θ) ⋅ θ= P −ℓ sin θ−h cos θ θ+ 12W ℓ cos θ−h sin θ θ= 0

This holds for values of the load that satisfy

P= W2ℓ cos θ− h sin θℓ sin θ+h cos θ

x1

x3

B

Ω

x2

n


(c) Find and expression for the angle at which equilibrium goes from being stable tobeing unstable. To establish the stability we must compute the second derivative

A(θ, θ)= d2dε2

E(θ+εθ)ε=0

Carrying out this computation for the system at hand we get

A(θ, θ)= P −ℓ cos θ+h sin θ θ 2+ 12W −ℓ sin θ−h cos θ θ 2

but P must satisfy equilibrium (i.e., it must satisfy the above expression)

−W2ℓ sin θ+h cos θ θ 2

A(θ, θ)=−W2ℓ cos θ−h sin θℓ sin θ+h cos θ

ℓ cos θ−h sin θ θ 2

=−W2 ℓ2+h2ℓ sin θ+h cos θ

θ 2Thus, we find that

A(θ, θ)≤ 0, 0< θ< π2

Therefore, equilibrium is always unstable.

238. Consider the solid spherical region shown in the sketch.As-sume that there exists a scalar field w(x) for which we can definethe functional

G(w, v) ≡ B

∇v ⋅ ∇w+ vw dV−Ω

p v dA

that has the property that if G(w, v)= 0for all (virtual) scalar functions v(x) then the clas-sical differential equations governing the real field w(x) are satisfied (i.e., G(w, v) is a“virtual-work” functional). Note that the scalar field p is defined on the surface of the solidregion. Show that an “energy” functional exists for this theory if the function p dependsonly upon the position vector x and not the function w(x), i.e., p= p(x). Determine theenergy functional in terms of the field w.

(a) Show that an “energy” functional exists for this theory if the function p dependsonly upon the position vector x and not the function w(x), i.e., p=p(x). An energy func-tional exists if the virtual work functional is symmetric in the sense ofDG(w, v) ⋅ u= DG(w, u) ⋅ v. Hence, we must compute the derivative


DG(w, v) ⋅ u= ddε

G(w+εu, v)ε=0

= B

∇u ⋅ ∇u+ vu dV

Therefore, an energy functional does exist.

(b) Determine the energy functional in terms of the field w. The energy functional canbe determined by Vainberg’s theorem

E(w)=10

G( tw,w) dt

Carrying out these operations we get

E(w)=10

B

∇w ⋅ ∇(tw)+ w(t w) dV−Ω

pw dA dt

= 12B

∇w ⋅ ∇w+ w2 dV−Ω

pw dA

239. Reconsider Problem 238 for the case where the function p depends upon the field wand the position x. Under what conditions would an “energy” functional exist in this case?(Hint: would an energy exist if p depends upon w itself? What if it depends upon deriva-tives of w, i.e., p= p(x,w(x),∇w(x), )?

The directional derivative of the function p can be carried out as

Dp(w) ⋅ u= ddε

p(w+εu)ε=0

This derivative takes the form

Dp(w) ⋅ u= ∂p∂w u+ ∂p∂∇w ⋅ ∇u+

Thus, last term of the given functional can be written as

Ω

∂p∂w uv+ ∂p∂∇w ⋅ ∇u v+

dA

k

P

ℓ∕2

x

ℓ∕2

EI


Hence, if terms in p involve w alone, then energy exists. Energy may not exist for high-er derivatives.

240. Abeam of length ℓ rests on an elastic founda-tion of modulus k (per unit length). It is fixed at theleft end, pinned at the right end, and is subjectedto a point loadP at midspan. The energy of the sys-tem can be expressed in terms of the transversedis-placement w(x) as:

E(w)=ℓ0

12EI(w′′)2+ kw2 dx+ Pw(ℓ∕2).

Find the virtual-work form of the equilibrium equations.What are the essential and naturalboundary conditions? Use the Ritz method to find a one-term approximation of the dis-placement field (use a polynomial approximation).

(a) Find the virtual work form of the equilibrium equations.

G(w,w)= DE(w) ⋅ w= ddεE(w+εw)

ε=0

= ℓ0

EIw′′w′′ + k ww dx+ P wℓ2


Essential: w(0)= 0, w′(0)= 0, w(ℓ)= 0

Natural: M(ℓ)= 0 (or w′′(ℓ)= 0)

(c) Use the Ritz method to find a one-term approximation of the displacement field(use a polynomial approximation). Start with a quartic polynomial and substitute thethree essential boundary conditions to get a one term polynomial approximation. Letξ≡ x∕ℓ and

w′ = 1ℓb+ 2cξ+ 3dξ2

w= a+ bξ+ cξ2+ dξ3

Substituting the boundary conditions we have

w′(0)= 1ℓb = 0

w(0)= a= 0

w(ℓ )= a+ b+ c+ d= 0 ⇒ c=−d


Thus, the approximation (and its derivatives) can be written as

w′ = cℓ2ξ−3ξ2 w= c ξ2−ξ3 w′ = c

ℓ22−6ξ

with similar expressions for the virtual displacement. Substituting this approximationinto the virtual work functional we have

G=10

EIℓ42−6ξ 2cc+ k ξ2−ξ3 2cc ℓdξ+ P c 14− 1

8 = 0 ∀ c

The integrals have the values

10

ξ2−ξ3 2 dξ= 10

ξ4−2ξ5+ξ6 dξ= 15− 26+

17=

42− 70+ 305 ⋅ 6 ⋅ 7 = 1

105

10

2−6ξ 2 dξ= 10

4−24ξ+36ξ2 dξ= 4−12+12= 4

Thus,

4EIℓ3 +

kℓ105c + P

8= 0

which allows the determination of the value of the constant c.

ℓ ℓ

P

kℓ

k

Chapter 10Fundamental Conceptsin Static Stability

241. The frame shown is composed of two rigid members con-nected by rotational springs. The moment developed by a springis related to the rotation by M= kφ, where φ is the rotationexpe-rienced by the spring. Both springs have modulus k. The frame issubjected to a load P acting downward. The motion of the struc-ture can be completely characterized by the rotation of theverticalmember from its original position. Examine the stability of thesystem. In particular, find the critical load and plot the bifurcation diagram. Note that thebifurcation diagram is not symmetric. Can you explain, in physical terms, why it is not?

P

θ βθ

ℓ(1− cos θ)

datum

Let θ be the rotation of the vertical member from its original position (see the figure).From geometry, the vertical drop of the hinge in the beam must be the same measuredfrom the left or from the right. Therefore, we get the relationship between the angles βand θ as

sin β = sin θ+ 1− cos θ

The potential energy of the frame is

E(θ) = 12kθ2+ 1

2k θ+β 2+ Pℓ cos θ


The virtual-work functional can be computed from the energy as

G(θ, θ) = DE(θ) ⋅ θ = kθ+ k θ+β 1+dβ∕dθ − Pℓ sin θ θ

where, up to the second-order of θ,

β≈ θ+ 12θ2,

dβdθ≈ 1+ θ, sin θ ≈ θ

The equilibrium condition G(θ, θ)= 0 for all θ yields two solutions:

θ= 0, P(θ) ≈ kℓ5+3θ

The value of P(θ) at θ= 0 is the critical load

Pcr = 5kℓ

The bifurcation diagram is asymmetric about θ because G(θ, θ) contains even terms inθ, which in turn causes P(θ) to have odd terms. A physical explanation is that for thesame amount of rotation of the vertical bar, the amount of rotation the horizontal barexperiences is larger for θ> 0 than for θ< 0, hence, more energy is stored in theframe in the former case. More energy leads to a higher bifurcation load. The secondderivative of the energy is

= k1+ θ+β + 2+θ 2− Pℓkcos θθ 2

A(θ, θ) = k1+ θ+β β′′ + 1+β′ 2− Pℓkcos θ θ 2

For the trivial path θ= 0

A(0, θ) = θ 2 5k− Pℓ

therefore, the trivial path is stable for P< Pcr and unstable for P> Pcr. For the bifur-cated path

A(θ, θ) = kθ 23θ+ O(θ2)

where O(θ2) denotes second and higher powers of θ. Therefore, the bifurcated path isstable for θ> 0 and unstable for θ< 0. The equilibrium paths are plotted in the at-tached figure.

53

1

θ

P(θ)ℓk

Chapter 10 Fundamental Concepts in Static Stability 287

242. Examine the effect of an imperfection in the system of Problem 241. Let the imper-fection be an initial value of the angle of rotation used to describe the motion, and assumethat the springs are such that they have no moment at this initial position. Plot the maxi-mum load versus the size of the initial imperfection.

The potential energy of the frame with an initial imperfection θo is

E(θ) = 12k θ−θo 2+ 1

2k θ−θo+β−βo 2+ Pℓ cos θ

Hence,

G(θ, θ) = DE(θ) ⋅ θ = kθ−θo + k θ−θo+β−βo 1+dβ∕dθ − Pℓ sin θ θ

From geometry, up to the second-order of θ

β≈ θ+ 12θ2, βo ≈ θo+ 1

2θ2o,

dβdθ≈ 1+ θ

G(θ, θ)= 0 for all θ gives

P(θ) ≈ kℓ51− θo

θ+ 2θ−θo +

(2+θ)θ2−θ2o 2θ

Setting P′(θ)= 0 gives the equation

3θ2+ 5θo = 0

Hence, there exists a limit load for θo < 0. It occurs at θcr=− −5θo∕3 . Substitut-ing θcr in the expression for P(θ), and retaining only the terms up to the first ordergives the maximum load as a function of the imperfection

Pmax ≈ Pcr 1− 2 − 35θo − 17

6 θowhere Pcr= 5k∕ℓ is the critical load for the perfect system. The maximum load as afunction of the imperfection is shown in the sketch.

θo 0−5o

1

PmaxPcr

ℓ

k

2ℓ

P

k

ℓ

k


243. Consider the two rigid bars hinged togeth-er and subjected to axial load P, as shown. Thebars have length ℓ and 3ℓ, and are restrained bythree elastic springs, with modulus k, that resistvertical motion. Find all equilibrium paths forthe system. Find the bifurcation loads of the system.Assess the stability of the straight andbent configurations.

θP

θ

datum

The system has one degree of freedom. The potential energy and virtual workfunctionals of the system are, respectively

E(θ) = 3kℓ2 sin2 θ+ 4Pℓ cos θ

G(θ, θ) = ℓ sin θ 6kℓ cos θ− 4P θ

G(θ, θ)= 0 for all θ gives the equilibrium paths for the system as

θ = 0, π, 2π,

P(θ) = 32kℓ cos θ

The bifurcation loads are Pcr= 32kℓ, corresponding to θ= 0 and θ= π, respec-

tively. The second derivative functional is

A(θ, θ) = ℓθ 2 6kℓcos2 θ− sin2 θ − 4P cos θ

For θ= 0, the second derivative functional is

A(0, θ) = ℓθ 2 6kℓ− 4P

Hence, the configuration θ= 0 is stable for P< 32kℓ, and unstable for P> 3

2kℓ. For

θ= π,

A( π, θ) = ℓθ 2 6kℓ + 4P

Hence, the straight configuration θ= π is stable for P>− 32kℓ, and it is unstable

for P<− 32kℓ. For the bent configuration P(θ)= 3

2kℓ cos θ,

A(θ, θ) = −6kℓ2θ 2 sin2 θ< 0

Hence, the bent configuration is always unstable.

ℓ

kk

ℓ ℓ

P


244. Consider the three-bar rigid linkageshown. The bars are hinged together and are re-strained by elastic springs that resist verticalmotion. The springs accrue force in proportionto their extension, with modulus k. The systemis subjected to an axial force P. Write an expression for the potential energy of the system.What are the equations of equilibrium governing the response of the system?Find the criti-cal loads and the buckling mode shapes of the system. Feel free to linearize the geometryof deformation as you see fit.

Pθ1θ2 β

datum

The system has two degrees of freedom. Let θ1, θ2, and β(θ) be the rotations ofthe bars (see sketch). For a linearized bifurcation analysis β≈ θ1+θ2. Let θ=θ1,θ2 be the degrees of freedom. The linearized potential energy functional of the systemis

E(θ) = 12kℓ2 θ21+ θ1+θ2 2 − 1

2Pℓθ21+ θ22+ θ1+θ2 2

and the linearized virtual-work functional of the system is

= 2θ1+θ2 kℓ−P ℓθ1+ kℓθ1+θ2 − Pθ1+2θ2 ℓθ2

G(θ, θ) = kℓ2 θ1θ1+ θ1+θ2 θ1+θ2

− Pℓθ1θ1+ θ2θ2+ θ1+θ2 θ1+θ2

G(θ, θ)= 0 for all θ gives the equations of equilibrium

kℓθ1+θ2 − P θ1+2θ2 = 0

kℓ − P 2θ1+θ2 = 0

Let m≡ P∕kℓ. The matrix form of the equations is

2−2m 1−m

1−m 1−2mθ1θ2 = 00kℓ

The solution is either θ= 0, or the determinant of the coefficient matrix is equal tozero

ℓ

kk

ℓ ℓ

P


2−2m 1−2m − 1−m 2 = 1−m 1−3m = 0

which yields the bifurcation loads as

P1cr = 13kℓ, P2cr = kℓ

The corresponding buckling mode shapes are θ1 = (1,−2), and θ2 = (1, 0), asshown in the sketch.

θ1 = (1,−2) θ2 = (1, 0)

245. Consider the three-bar rigid linkageshown. The bars are hinged together and arerestrained by elastic rotational springs. Thesprings accrue force in proportion to the rela-tive angle of distortion, with modulus k. Thesystem is subjected to an axial force P. Write an expression for the potential energy of thesystem.What are the equations of equilibrium governing the response of the system? Findthe critical loads and the buckling mode shapes of the system. Feel free to linearize thegeometry of deformation as you see fit.

β+θ2

Pθ1θ2 β

datum

The system has two degrees of freedom. Let θ= (θ1, θ2) be the degrees of freedom.Again, the linearized system has β≈ θ1+θ2 . The potential energy of the system is

E(θ) = 12k θ1−θ2 2+θ1+2θ2 2 − 1

2Pℓ θ21+ θ22+ θ1+θ2 2

The virtual work functional can be found as G(θ, θ)= DE(θ) ⋅ θ. To wit,

G(θ, θ) = k θ1−θ2 θ1−θ2 +θ1+2θ2 θ1+2θ2

− Pℓθ1θ1+θ2θ2+θ1+θ2 θ1+θ2


Rearranging we get

G(θ, θ) = 2kθ1+kθ2− Pℓθ1− Pℓθ1+θ2 θ1

+ kθ1+5kθ2− Pℓθ2− Pℓθ1+θ2 θ2The linearized equations of equilibrium are

k− Pℓθ1+ 5k− 2Pℓθ2 = 0

k− Pℓ 2θ1+ θ2 = 0

Let m≡ Pℓ∕k. The matrix form of the equations is

2−2m 1−m

1−m 5−2mθ1θ2 = 00k


21−m 5−2m − 1−m 2 = 31−m 3−m = 0

The bifurcation loads are

P1cr = kℓ P2cr = 3k

ℓ

The corresponding buckling mode shapes are θ1 = (1, 0), and θ2 = (1,−2).

246. Consider the rigid bar subjected to axial loadP and transverse load εP as shown. Thebar has length 2ℓ, is restrained against horizontal and vertical motion at the midpoint, andis supported by two elastic springs that resist vertical motion at the ends. The springs ac-crue force in proportion to their extension, withmodulus k. The deformationcan becharac-terized by the rotation of the bar relative to the horizontal position.

ℓ

k

ℓ

P

k

εP P

θ

π− πε= 0.1

Find all equilibrium paths P(θ) for the system (−π< θ< π). Determine the stabilityof these branches. Find the critical load of the system when ε= 0. Locate the limit pointon the bifurcation diagram plotted for ε= 0.1. Is the limit load at this point greater or lessthan the critical load?

PÁPθ


The system has a single degree of freedom. The potential energy, virtual work func-tional, and second derivative functional of the system are

E(θ) = kℓ2 sin2 θ+ Pℓcos θ−Á sin θ

G(θ, θ) = kℓ sin 2θ− P sin θ+Á cos θ ℓθ

A(θ, θ) = 2kℓ cos 2θ− P cos θ−Á sin θ ℓθ 2

G(θ, θ) = 0 for all θ gives the nonlinear equilibrium configurations

P(θ) = kℓ sin 2θsin θ+ Á cos θ

There are two equilibrium paths asymptotic to θo ≡− tan–1 Á where P(θ) goes topositive infinity at the left and negative infinity at the right, as shown in the figure. Thesecond derivative of the energy is

= Á cos3 θ− sin3 θÁ cos θ+ sin θ

kℓ2θ 2

A(θ, θ) = 2kℓ cos 2θ− kℓ sin 2θsin θ+ Á cos θ

cos θ−Á sin θ ℓθ 2

> 0

< 0

−π< θ<−π+θcr−π+θcr< θ< θo

> 0 θo < θ< θcr< 0 θcr< θ< π−θo> 0 π−θo < θ< π

(stable)

(unstable)

(stable)

(stable)

(unstable)

=

where θcr≡ tan–1 Á1∕3 . Note that P(θ) reaches an extremum at θcr. When Á = 0, theprinciple of virtual work, i.e., G(θ, θ) = 0 for all θ, yields the equilibrium configura-tions

θ = 0, π, 2π,

P(θ) = 2kℓ cos θ

The critical load is Pcr= 2kℓ. When Á = 0.1, the limit load is found to be 0.75Pcr,located at θcr= 0.43. Hence, the limit load for the imperfect system with Á = 0.1 is25% lower than that for the perfect system (Á = 0).

ℓ ℓ

P

k

ℓ


247. Consider the frame composed of rigid bars subjected to theload P as shown. The rigid members are hinged at the top rightcorner, with an elastic spring that resists relative rotation. Therotational spring accrues force in proportion to its relative anglechange, with modulus k. Find the buckling load for this system.Express the deformation of the system in terms of the angle ofrotation of the vertical member on the right side of the structure.What happens if you use the angle of rotation of the verticalmem-ber on the left?

βθ

P

θ

The system has a single degree of freedom. We shall take β as the measure ofdeformation, with the angle θ depending on β (see sketch). Based on the deformedgeometry, ℓ cos θ− 2ℓ sin θ = ℓ cos β, which implies that θ ≈ 1

4β2. The potential

energy of the system is

E(β) = 12k θ+ β 2+ Pℓcos θ− sin θ


G(β, β ) = k θ+β θ′+1 β− Pℓsin θ+ cos θ θ′β

≈ k14 β2+β 12 β+1β− Pℓ14 β2+1 12 ββ

≈ k− 12Pℓββ

Therefore, the linearized equation of equilibrium is

k− 12 Pℓβ = 0

The buckling load is then

Pcr = 2kℓ

If θ is used instead of β for the primary unknown, then the constraint θ> 0 must beaccounted for in the formulation. Consequently, the solution becomes messier.

ℓ

kk

ℓ

P

k

ℓ

kk

ℓ ℓ

P

ℓ

k

θ1

θ2


248. Consider the rigid bar subjected to axial load Pas shown. The bar has length 2ℓ and is supported byelastic springs that resist vertical motion. The springsaccrue force in proportion to their extension, withmo-dulus k. Find the critical loads and linearized bucklingmode shapes of the system. Note that this system hastwo degrees of freedom.

P

θ

u

The system has two degrees of freedom. Let u (vertical displacement at the center ofthe bar) and θ be the degrees of freedom of the system, as shown in the sketch. Then,the linearized potential energy of the system is

E(u, θ) = 12k u−ℓθ 2+ u2+ u+ℓθ 2 − Pℓθ2


G(u, θ, u, θ) = k u−ℓθ u−ℓθ + uu+ u+ℓθ u+ℓθ − 2Pℓθθ

= k3uu+ 2ℓ2θθ − 2Pℓθθ

Therefore, the linearized equations of equilibrium are

2kℓ2−Pℓθ = 0

3ku = 0

The trivial position (θ, u )=(0, 0) is an equilibrium configuration for any value of P.There is one critical load, Pcr = kℓ, with corresponding mode (θ, u )=(1, 0).

249. Consider the three-bar rigid linkageshown. The bars are hinged together andare restrained by elastic springs that resistvertical motion. The springs accrue forcein proportion to their extension, with mo-dulus k. The system is subjected to an axialforce P. Write an expression for the poten-tial energy of the system. What are the li-nearized equations of equilibrium governing the response of the system? Find the criticalloads of the system. A convenient choice of degrees of freedom is shown in the diagram.

kP

ℓ ℓ


Let θ= (θ1, θ2). The potential energy of the system is

E(θ) = 12kℓ2 2θ21+ θ1+θ2 2 − 1

2Pℓ3θ21+θ22


G(θ, θ) = kℓ2 2θ1θ1+θ1+θ2 θ1+θ2 − Pℓ 3θ1θ1+θ2θ2

= kℓ2 3θ1+θ2 − 3Pℓθ1+ kℓ2 θ1+θ2 − Pℓθ2

The linearized equations of equilibrium are

kℓ2 θ1+θ2 − Pℓθ2 = 0

kℓ2 3θ1+θ2 − 3Pℓθ1 = 0

Let m≡ P∕kℓ. The matrix form of the equations is

3−3m 1

1 1−mθ1θ2 = 00kℓ2


31−m 2− 1 = 3m2− 6m+ 2 = 0

The bifurcation loads are

P1cr = 1− 13 kℓ P2cr = 1+ 1

3 kℓ

The corresponding buckling mode shapes are θ1 = (1,− 3 ), θ2 = (1, 3 ).

250. Two rigid bars are hinged together and rest on a lin-early elastic foundation. The foundation accrues a forceper unit length proportional to the transverse displace-ment, i.e., f(x)=kw(x). The system is subjected to anaxialload P as shown. Find an expression for the energy functional for the system. Find an ex-pression for the virtual-work functional for the system. Find the buckling loads of the sys-tem by solving the linearized buckling eigenvalue problem.

(a) Find an expression for the energy functional for the system.

The system has two degrees of freedom. The displacement in the two rigid seg-ments can be expressed as w1(x)= x sin θ1 and w2(ξ)= ℓ sin θ1+ξ sin θ2. Thesequantities are shown in the sketch.


θ1

θ2

w1(x)

w2(ξ)

x ξ

E(θ1, θ2)= ℓ

0

12 kx sin θ1 2 dx+

ℓ

0

12 kℓ sin θ1+ξ sin θ2 2 dξ+ Pℓcos θ1+ cos θ2

= 12k sin2 θ1ℓ33 + 1

2k ℓ3 sin2 θ1+ℓ3 sin θ1 sin θ2+ℓ33 sin2 θ2

= 12kℓ3 4

3sin2 θ1+ sin θ1 sin θ2+

13sin2 θ2+ Pℓ cos θ1+ cos θ2

+ Pℓcos θ1+ cos θ2

(b) Find an expression for the virtual work functional for the system.

G(θ, θ)= 12 kℓ

3 83 sin θ1 cos θ1θ1+ sin θ2 cos θ1θ1

− Pℓ sin θ1θ1+ sin θ2θ2

+ 12 kℓ

3 sin θ1 cos θ2θ2+ 23 sin θ2 cos θ2θ2

≈ 12 kℓ3 83 θ1+ θ2 − Pℓ θ1 θ1+ 12 kℓ3 θ1+ 23 θ2 − Pℓ θ2 θ2

If we use a first order approximation of the nonlinear terms.

(c) Find the buckling loads of the system by solving the linearized buckling

eigenvalue problem. Let λ≡ 2P∕kℓ2. The equations of equilibrium can be written as

83− λ 1

1 23− λ θ2

θ1 =0

0

This system of equations as a nontrivial solution only if the determinant of the coeffi-cient matrix is zero. Thus,

83− λ 23− λ − 1= 0

or

ℓ

k

ℓ

P

k

kθ= 2kℓ2ℓ


9λ2− 30λ+ 7= 0

This quadratic equation has the solutions

λ1,2 =30 900− 4 * 7 * 9

2 * 9= 0.2525, 3.0809

Therefore, the critical loads are

P1 = 0.126 kℓ2 P2= 1.540 kℓ2

251. The vertical rigid bar is subjected to axial load P andis hinged to the horizontal rigid bar which has length 2ℓ. Arotational spring restrains the change in angle between thetwo bars. The horizontal bar is restrained against horizontaland vertical motion at the midpoint, and is supported by twoelastic springs that resist vertical motion at the ends. Thesprings accrue force in proportion to their extension, withmodulus k. Find an expression for the energy E of the sys-tem. Find the (nonlinear) equations of equilibrium of the system. Find the critical loadsof the system.

P

θ1

θ2

The system has two degrees of freedom. Let θ= (θ1, θ2). The potential energy of thesystem is

E(θ) = kℓ2 sin2 θ1+θ1+θ2 2 + Pℓ cos θ2

The virtual work functional can be found from G(θ, θ)= DE(θ) ⋅ θ to be

G(θ, θ) = 2kℓ2 sin θ1 cos θ1θ1+ θ1+θ2 θ1+θ2 − Pℓ sin θ2θ2

= 2kℓ2 sin θ1 cos θ1+ θ1+ θ2 θ1+ 2kℓ2 θ1+θ2 − Pℓ sin θ2 θ2

The nonlinear equations of equilibrium are

2kℓ2 θ1+θ2 − Pℓ sin θ2 = 0

2kℓ2θ1+θ2+ sin θ1 cos θ1 = 0

k

P

ℓ

k

k

ℓ ℓ

ℓ

ℓ P


Let m≡ P∕kℓ. The matrix form of the linearized equations is

4 2

2 2−mθ1θ2 = 00kℓ2


42−m − 4 = 41−m = 0

Since the characteristic equation is linear there is only one critical load. The criticalload is Pcr= kℓ. The buckling mode is θ= (1,−2).

252. Arigid bent of height 2ℓ and length 3ℓ rests on threeelastic springs, each with modulus k. The springs accrueforce in proportion to the amount by which they stretch.The bent is pinned at the corner end and is subjected to aload P at the top and a load of P at the right end. Find thevirtual-work form of the equilibrium equations. Find thesecond-derivative functional A for the system. Find allequilibrium configurations of the system and assess theirstability. Sketch the result on a bifurcation diagram.

P

θθ

P

The system has a single degree of freedom. The energy of the system is

E(θ) = 12k (ℓ sin θ)2+(ℓ sin θ)2+(2ℓ sin θ)2 + 2Pℓ cos θ− 3Pℓ cos θ

with virtual work functional

G(θ, θ) = 6kℓ2 sin θ cos θθ+ Pℓ sin θθ

and second-derivative functional

A(θ, θ) = 6kℓ cos 2θ+ P cos θ ℓθ 2

G(θ, θ)= 0 for all θ gives the equilibrium configurations of the system

ℓ

P

θ

P

αθ

k


θ = 0, π, 2π,

P(θ) = −6kℓ cos θ

Since A(0, θ)= 6kℓ+P ℓθ 2, the straight configuration θ= 0 is stable forP>−6kℓ, and unstable for P<−6kℓ. At θ= π we haveA( π, θ)= 6kℓ−P ℓθ 2. Thus, the straight configuration θ= π is stable forP< 6kℓ, and unstable for P> 6kℓ. For the bent configuration we have

A(θ, θ) = 6kℓ cos 2θ− 6kℓ cos2 θ ℓθ 2 = −6kℓ2θ 2 sin2 θ≤ 0

Thus, the bent configuration is always unstable.

253. A rigid bar of length ℓ is pinned and restrained by arotational spring ofmodulus k at the bottom. It is subjectedto a force P at the top. The force changes its direction withthe motion of the bar. If the bar rotates by an angle θ thenthe load rotates an angle αθ in the opposite sense (α is aknown constant). Find a suitable virtual work function forthe system? (Hint: start with a classical equilibrium equa-tion from a freebody diagram of the bar). Does an energyfunction exist? If so, then find it. Estimate the buckling load of the system.

(a) Find a suitable virtual work function for the system?

Consider a freebody diagram with moment M (and reaction forces) acting at thefixed end. Summing moments about the base gives

M= kθ= P cosαθℓ sin θ+ P sinαθℓ cos θ

= Pℓcosαθ sin θ+ sinαθ cos θ

= Pℓ sin(1+α)θ

Hence, kθ−Pℓ sin(1+α)θ= 0 is the classical equilibrium equation. We can developa virtual work functional by the method of weighted residuals

G(θ, θ)≡ kθ− Pℓ sin(1+α)θ θ

Then G(θ, θ)= 0 for all θ implies equilibrium by the fundamental theorem of thecalculus of variations. This is the principle of virtual work.

(b) Does an energy function exist? If so, then find it. Symmetry of the virtual workfunctional determines the existence of the energy functional. Taking the directionalderivative of the virtual work functional gives

DG(θ, θ) ⋅ θ^= kθ^− (1+α)Pℓ cos (1+α) θ

^ θ

= k− (1+α)Pℓ cos(1+α)θ θ^θ= DG(θ, θ

^) ⋅ θ

k

k

ℓ

P

ℓ

34


The symmetry is evident simply by swapping θ^and θ in the expression above. The

symmetry proves that an energy functional E(θ) exists. The energy functional can becomputed by Vainberg’s theorem.

E(θ)=10

k tθ− Pℓ sin (1+α) tθ θ dt

The second term in the integral can be integrated by letting u= 1+α tθ so thatdu= 1+α θdt. Carrying out the integral we get

E(θ)= 12kθ2+

Pℓ1+α cos(1+α)θ−

Pℓ1+α

Of course, this can be verified by taking the directional derivative of the energy andshowing that it gives the virtual work functional.

(c) Estimate the buckling load of the system. The governing equation iskθ− Pℓ sin(1+α)θ= 0. It is evident that one solution is θ= 0. For θ≠ 0 theequation is satisfied only for certain values of the load. In particular, for loads thatsatisfy

P= kℓ

θsin (1+α)θ

The critical load (buckling load) is the limit of this load as θ→ 0. To wit,

limθ→0

P= kℓ

1(1+α) cos(1+α)θ=

kℓ (1+α)

Hence, there is a bifurcation at

Pcr= kℓ (1+α)

We can check the stability of the straight configuration with the second derivative test.The second derivative of the energy is

A(θ, θ)= k− (1+α)Pℓ cos(1+α)θ θ 2

At θ= 0 we have A(0, θ)= k− (1+α)Pℓ θ 2, therefore the system is stable for

P< kℓ (1+α)

254. Two rigid bars, each of length ℓ are hinged together andattached to two linearly elastic springs of modulus k. The bot-tom end of the vertical member is on a roller that rolls on ahor-izontal plane. The right end of the horizontal member is on aroller that rolls on a slope. The column is subjected to a verti-cal force P. Find an expression for the energy of the system.Find the equilibrium configurations of the system. Find thecritical loads of the structure.


(a) Find an expression for the energy of the system.

∆

∆+ ℓ sin θ

ℓ cos θ

θ x

54 x

DOF : ∆ , θ

The energy functional can be established from the quantities shown in the sketch.For a linearized motion x≈ ∆+ℓ θ. Therefore,

E(∆ , θ)= 12 k ∆

2+ 12 k54∆+ℓ θ 2− Pℓ cos θ= 0

(b) Find the equilibrium configurations of the system. The virtual work functional canbe found by differentiating the energy functional. To wit,

G(∆, θ, ∆, θ ) = k ∆ ∆+ 2516k ∆+ℓ θ ∆+ℓ θ − Pℓθ θ = 0

Grouping terms with common virtual displacement measure we have

4116 k ∆+

2516 k ℓ θ ∆+ 2516 k ℓ ∆+ 25

16 k ℓ2 θ− P ℓ θ θ= 0 ∀ ∆, θ

Therefore, the equations of equilibrium are

2516k ℓ ∆+ 25

16k ℓ2 θ− P ℓ θ= 041

16k ∆+ 25

16kℓ θ= 0

(c) Find the critical loads of the structure. From the first equilibrium equation we candetermine

∆ =− 2541ℓ θ

Substituting this result into the second equation gives

2516 kℓ − 25

41 ℓ θ + 2516 k ℓ

2 θ− Pℓ θ= 0

1641kℓ2 θ− 16

25Pℓ θ= 0

One solution for this equation is θ= 0. A nontrivial solution is possible for the loadlevel

Pcr = 2541kℓ

which is the buckling load of the system.

k

W

4 ft

x


255. A ladder of length ℓ=20 ft leans against a wall with the base4 ft from the wall. Both ends are frictionless and the bottom end isrestrained by an elastic spring of modulus k=10 lb/ft. What is themaximumheight x (measured along the ladder as shown) that aper-son of weight W=200 lb can climb? The ladder can be assumedrigid, the rollers are very small relative to the length of the ladder,and the person climbs slowly enough to neglect dynamic effects.

x sin θ

W

xθ x cos θ

ℓ sin θ

sin θ0=35

The energy of the system can be established from the quantities given in thesketch. To wit,

E(θ)= 12kℓ2(sin θ− sin θ0)

2+Wx cos θ

The virtual work functional can be determined by differentiating the energy as

G(θ, θ) = DE(θ) ⋅ θ= kℓ2 sin θ− sin θ0 cos θ−Wx sin θ θ

The condition G(θ, θ)= 0 for all θ gives the equation

kℓ2 sin θ− sin θ0 cos θ−Wx sin θ= 0

Any (x, θ) pair that satisfies this equations is an equilibrium configurations. Hence, wecan write

x(θ)= kℓ2W1− sin θ0

sin θ cos θ

The value of θ that gives the greatest value of x can be determined from

dxdθ= kℓ2

W sin θ0sin θ

cos2 θ−1− sin θ0sin θ sin θ = 0

from which we find

sin θ0 cos2 θ+ sin2 θ − sin3 θ= 0 ⇒ sin θcr= sin1∕3 θ0

From trigonometry we know that

ℓ

k

P

ℓ ℓ ℓ

43


cos θcr= 1− sin2∕3 θ0 1∕2

Substituting these results into the equation for x(θ) we get

xmax = kℓ2W1− sin2∕3 θ0 3∕2

= (10)(20)2

(200)1− 0.6 2∕3 3∕2

= 3.1 ft< 20 ft

Therefore, the maximum occurs inside the range [0,20]. Check stability for allx Á 0, 20

A(θ, θ)= kℓ2 cos2 θ− sin2 θ+ sin θ sin θ0 −Wx cos θ θ 2

= kℓ2cos2 θ− sin2 θ+ sin θ sin θ0−1− sin θ0sin θ cos2 θ θ 2

= kℓ2 sin θ0− sin3 θsin θθ 2

Because A(θ, θ)> 0 for all θ< θc r it is unsafe to climb up the ladder past the dis-tance x = 3.1 ft!

256. Four rigid bars are hinged together and subjectedto the load P as shown. The two horizontal bars are re-strained by a linear, elastic rotational spring of modu-lus k. Find an expression for the energy of the system.Find an equation describing the equilibrium configu-rations of the system. Find the bifurcation load.

(a) Find an expression for the energy of the system. The geometry of the system canbe characterized by the angles α and θ(α). Note that the system has a single degree offreedom with relations shown

ℓ

P

ℓ

54 ℓ

α

2α

θ

2ℓ cos α+ 52ℓ cos θ= 4ℓ

−5 sin θ Dθ ⋅ α= 4 sin α α

5 cos θ= 8− 4 cos α

Dθ ⋅ α=− 4 sinα5 sin θ

α

ℓ

k

ℓ

P

k

kθ

ℓ ℓ


The energy of the system can be established from the quantities shown in the sketch as

E(α) = 12 k2α 2+ P 54 ℓ sin θ(α)

(b) Find an equation describing the equilibrium configurations of the system. Thevirtual work functional can be found by differentiating the energy as

G(α,α)= DE(α) ⋅ α= 4kα α+ 54Pℓ cos θ Dθ ⋅ α

The condition G(α,α)= 0 for all α (upon substituting the expression for Dθ ⋅ α)gives the equation

4kα− Pℓ cos θsin θ

sinα= 0

where, from the kinematic constraint we have cos θ= 8−4 sin α ∕5 and we can getsin θ= 1− cos2 θ −1∕2.(c) Find the bifurcation load. One solution for the equilibrium equation is α= 0. Anontrivial solution exists for certain values of the load P. Linearizing the equilibriumequation for α≪ 1, noting that

sinα≈ α, cos θ≈ cos θ0=45, sin θ≈ sin θ0 =

35

we find the linearized equation

4 k− 13 Pℓ α= 0

from which the critical buckling load can be found as

Pcr=3kℓ

257. Consider the linkage of two rigid bars subjected toaxial load P as shown. The linkage has length 4ℓ, ispinned at the left end, and has elastic springs that resistmotion. The translational springs accrue force in propor-tion to their extension, with modulus k. The rotationalspring accrues force in proportion to the its relative angle change, with modulus kθ= kℓ2.Find the critical loads and linearized buckling mode shapes of the system.

θ2

θ1


The energy can be established from the quantities defined in the sketch as

+ 12kℓ2 θ2−θ1 + 2Pℓcos θ1+ cos θ2−2

E(θ)= 12k ℓ sin θ1 2+

12k 2ℓ sin θ1+ℓ sin θ2 2

To solve the bifurcation problem we need only a quadratic energy functional. Using theTaylor series expansion we can quadratify the energy to

E(θ)= 12kℓ2 6θ21+ 2θ1θ2+ 2θ22 − Pℓθ21+θ22

The virtual work functional can be found by differentiating the energy as

G(θ, θ)= DE(θ) ⋅ θ= kℓ2 6θ1θ1+θ2θ1+θ1θ2+2θ2θ2 − 2Pℓθ1θ1+θ2θ2

The condition G(θ, θ)= 0 for all θ gives the equilibrium equations

kℓ2 θ1+2θ2 − 2Pℓθ2 = 0

kℓ2 6θ1+θ2 − 2Pℓθ1 = 0

Letting m≡ 2P∕kℓ we can write these equations in the form

6−m 1

1 2−m

θ1

θ2

0

0=

This system of equations has a nontrivial solution only if the determinant of the coeffi-cient matrix is equal to zero. This gives the equation

6−m 2−m − 1= 0

This equation can be simplified to m2−8m+11= 0, which has roots

m1 = 4− 5 , m2 = 4+ 5

The buckling mode shape can be found by substituting back into the governing equa-tion for the critical values, specifically for m1 = 4− 5 we have

a

1

1 −2+ 52+ 5 1

0

0=

a=−2− 5

φ1 = 1,−2− 5

and m2 = 4+ 5 we have

1 −2− 52− 5 1

a

1

0

0=a=−2+ 5

φ2 = 1,−2+ 5

3ℓ∕2

Pk

ℓ

ℓ


258. Consider the frame composed of rigid bars subjected to theload P as shown. The rigid members are hinged at the top left cor-ner, with an elastic spring that resists relative rotation. The rota-tional spring accrues force in proportion to the its relative anglechange, with modulus k. Find the buckling load for this system.Determine the post-buckling response of the system in terms ofthe rotation of the left column. Does it make a difference if theframe buckles to the left or to the right?

P

θ ψ

ψℓ + ℓ cos θ+ 3

2 ℓ sin ψ= 2ℓ cosψ

3ℓ∕2

ℓ 2ℓ

2ℓ 1− cosψ

The energy can be established in terms of the quantities shown in the sketch. Thesystem has one degree of freedom. Hence, the angles θ and ψ are related. In particular,the kinematic constraint has the form

2+ 2 cos θ+ 3 sinψ− 4 cosψ= 0

We can think of ψ(θ) as a dependent function. Then the energy can be written as

E(θ)= 12k θ+ψ(θ) 2+ 2Pℓ cosψ(θ)

The virtual work functional can be found by differentiating the energy

G(θ, θ)= k θ+ψ θ+Dψ ⋅ θ − 2Pℓ sinψ Dψ ⋅ θ

From the constraint −2 sin θθ+ 3 cosψ+4 sinψ Dψ ⋅ θ= 0. Hence,

Dψ ⋅ θ= 2 sin θ3 cosψ+4 sinψ θ≈

23θθ

Also, from the constraint equation we can determine from Taylors theorem, for smallvalues of the angle −θ2+ 3ψ= 0. Hence, the virtual work functional can be writtenapproximately as

G(θ, θ)= kθ+ 13 θ

2 1+ 23 θθ− 2Pℓ13 θ2 23 θθ = 0

The condition G(θ, θ)= 0 for all θ gives the equilibrium equations

θ k1+ 13 θ1+ 2

3 θ− 4

9Pℓθ2 = 0

ℓ

ℓ P

k, To ℓ

ℓ

ℓ

k

ℓ

ℓ P

W ℓ

ℓ

ℓ


The trivial solution θ= 0 satisfies this equation. For a nontrivial solution we musthave the load level of

P= k2ℓ3+θ 1+θ

θ2

θ= 0 is stable for any P. There is no bifurcation load.

259. Two rigid bars, each of length 2ℓ are connected by a linearelastic spring of length ℓ and modulus k. The right vertical bar issubjected to a force P as shown. The left bar is attached to a verti-cal spring of modulus k that has been stretched into place, givingit an initial tension force of To (i.e., when P=0). Write the energyfunctional for the system. Find the lowest buckling load Pcr of thesystem.

No solution available.

260. Two rigid bars, each of length 2ℓ are connected by a singlerigid bar of length ℓwhich is hinged at the ends. A weight of fixedvalueW hangs from the left bar while the right vertical bar is sub-jected to a force P as shown. Note: there are no elastic elements inthis system! Howmany degrees of freedom does the system have?Write the exact energy functional for the system. (Hint: You can de-scribe the deformation in terms of the rotation angles of eachmem-ber, but youmust write equations of constraint relating those anglesto your chosen degrees of freedom). Find the critical value of P atwhich buckling of the system takes place. Is the post-buckling behavior symmetric orasymmetric? Do you expect the post-buckling behavior to be stable or unstable?

(a) How many degrees of freedom does the system have?

System has 1 DOF. Because if we fix the rotation of any bar, we cannot rotate theremaining bars freely.

(b) Write the exact energy functional for the system. (Hint: You can describe thedeformation in terms of the rotation angles of each member, but you must writeequations of contraint relating those angles to your chosen degrees of freedom).


θ

β

α

2ℓ cos θ

2ℓ cosα

2ℓ sin α

ℓ sin β

ℓ cos β

The energy can be established from the quantities defined in the sketch. The systemhas one degree of freedom, θ. It is convenient to express the position in terms of twoadditional angles α(θ) and β(θ). The constraint equations relating these angles

3ℓ = ℓ cos θ+ℓ sin β+ 2ℓ cos αℓ = ℓ sin θ+ℓ cos β− 2ℓ sin α

The energy has the form

E(θ)=−2ℓW cos θ+ 2ℓP cosα(θ)

(c) Find the critical value of P at which buckling of the system takes place.

The virtual work functional can be determined from the energy by differentiation.To wit,

G(θ, θ)=−2ℓW sin θθ+ 2ℓP sin αDα ⋅ θ

For small angles we have sin θ≈ 2 sinα, θ≈ 2α, and θ≈ 2α. Hence, the workfunctional can be linearized to

G(α,α)=−2ℓ4Wα− Pα α= 0 ∀ α

Therefore, the critical load can be computed as

Pcr= 4W

(d) Is the post-buckling behavior symmetric or asymmetric? Do you expect thepost-buckling behavior to be stable or unstable?

Asymmetric. Stable in one direction, unstable in the other.

Chapter 11The Planar Bucklingof Beams

261. Vainberg’s theorem is simply a statement of integrability. This theorem can be ap-plied to the strain variations that we derive through a virtual work argument for a nonlinearplanar beam. Let u≡ u,w, θ and u≡ u,w, θ be the real and virtual displacementsand rotation. From the principle of virtual work, we have found that the virtual curvatureis given by κo(u, u)= θ′. Show that the symmetry condition holds for the virtual curva-ture, i.e., Dκo(u, u) ⋅ u^ = Dκo(u, u^ ) ⋅ u and, hence, that it is integrable. Show that thereal curvature is given by κo = θ′. Note that the directional derivative of κo is

Dκo(u, u) ⋅ u^ = ddεκo(u+εu^ , u)ε=0

and the integral of the virtual curvature can be computed by Vainberg’s formula as

κo = 10

κo( tu, u ) dt

where tu= tu, tw, tθ. Repeat the calculation for the virtual shear and axial strains

βo(u, u)= w′ cos θ− u′ sin θ− [w′ sin θ+ (1+u′) cos θ]θÁo(u, u)= u′ cos θ+ w′ sin θ+ [w′ cos θ− (1+u′) sin θ]θ

to get the real shear and axial strains

βo = w′ cos θ− (1+u′) sin θÁo = w′ sin θ+ (1+u′) cos θ− 1

Take the directional derivatives of the real strains to verify that these results are correct.

The virtual curvature is integrable because

Dθ(Ω,Ω) ⋅Ω^ = 0 = Dθ(Ω,Ω^) ⋅Ω


The integral of the virtual curvature is simply

κo = 10

κo(tΩ,Ω)dt = 10

θ′dt = θ′

It follows from the definition of the virtual shear strain that

− w^ ′ sin θ+ u^′ cos θ+ w′ cos θ− (1+u′) sin θ θ^θ

Dβo(Ω,Ω) ⋅ Ω^ = −(w′ sin θ+ u′ cos θ)θ

^

− w′ cos θ− (1+u′) sin θ θ^θ

= −w′θ^+w^ ′θ sin θ+ u′θ

^+u^′θ cos θ

which is clearly symmetric (just swap hats for bars and observe that the result does notchange). Hence,

Dβo(Ω,Ω) ⋅ Ω^ = Dβo(Ω,Ω

^) ⋅Ω

and the real shear strain is given by Vainberg’s theorem as

βo = 10

βo(tΩ,Ω)dt

= w′ cos θ− (1+u′) sin θ

= 10

w′ cos tθ− u′ sin tθ − tw′ sin tθ+ (1+tu′) cos tθ θ dt

= w′θsin tθ+ u′

θcos tθ − w′

θ sin tθ−tθ cos tθ − sin tθ− u′

θcos tθ−tθ sin tθ 1

0

The real axial strain can be obtained in exactly the same way. One can again demon-strate the required symmetry. Therefore, the axial strain can be computed from thevirtual axial strain by Vainberg’s theorem as

Áo = 10

Áo(tΩ,Ω)dt

= w′ sin θ+ (1+u′) cos θ− 1

= 10

w′ sin tθ+ u′ cos tθ + tw′ cos tθ− (1+tu′) sin tθ θ dt

= −w′θcos tθ+ u′

θsin tθ + w′

θcos tθ+tθ sin tθ + cos tθ− u′

θsin tθ−tθ cos tθ 1

0

P

ℓ

EI

Chapter 11 The Planar Buckling of Beams 311

Note that the following two integral identities have been used

x sin x = sin x− x cos x, x cos x = cos x+ x sin x

The directional derivatives of the real shear and axial strains are

Dβo(Ω) ⋅Ω = w′ cos θ− w′ sin θθ− u′ sin θ− (1+u′) cos θθ = βo(Ω,Ω)

DÁo(Ω) ⋅Ω = w′ sin θ+ w′ cos θθ+ u′ cos θ− (1+u′) sin θθ = Áo(Ω,Ω)

262. Consider the simply supported column of length ℓand flexural modulus EI. Assume that shear and axial de-formations are negligible, so that the constraints ofEuler’selastica are appropriate. Compute the critical loads for thiscolumn by solving the equation EIwiv+Pw′′ = 0 withthe appropriate boundary conditions. Carry out the stability analysis parallel to the analy-sis done for the cantilever model problem in the text.

The general solution to the the fourth-order differential equation is

w(x)= a0+ a1x+ a2 sin mx+ a3 cosmx

where m2 ≡ P∕EI, and a0, a1, a2, and a3 are integration constants. The boundary con-ditions at the left end, w(0)= 0 and w′′(0)= 0 immediately yield the constantsa0 = a3 = 0. At the right end, w(ℓ)= 0 and w′′(ℓ)= 0 give

a2 sinmℓ = 0 a1ℓ + a2 sin mℓ = 0

The first of these substituted into the second gives a1 = 0. The condition for existenceof a non-trivial solution (i.e., w(x)≠ 0) is, therefore,

sinmℓ = 0

This equation is the characteristic equation, and has solutions only for values of theload parameter mn = nπ∕ℓ, for n= 1, 2, 3, . The critical loads of the system are,therefore, Pn = n2π2EI∕ℓ2. Noting that a2 ≠ 0 only when m= mn we can substituteback into the expression for the displacement to get the buckling modes that corre-spond to the critical loads. The nth buckling mode is

wn(x)= a2 sinnπxℓ where the amplitude a2 is undetermined.

(b) Carry out the stability analysis parallel to the analysis done for the cantilevermodel problem in the text. Since the functions wn(x) form a complete basis, the testfunction w(x) can be written as a linear combination of them. To wit,

ℓ

P

EI

q(x)


w(x) = ∞n=1

anwn(x)

Then the second-derivative of E(w) for the trivial equilibrium is (see page 415 of thetext)

A(0,w) = ∞n=1

a2n (Pn−P)ℓ0

(wn′) 2dx

Therefore, the trivial equilibrium is stable for P< P1 and unstable for P> P1.

263. Consider the simply supported column of length ℓand flexural modulus EI. Assume that shear and axial de-formations are negligible so that constraints of Euler’selastica are appropriate. Can the classical elastica theorybe extended to accommodate the transverse load q(x)?What difficulties do you encounter when you attempt to do so? Can the virtual-work prin-ciple for the elastica be modified to account for the transverse load?

For this problem, p= m= 0. The horizontal and vertical forces are readilyavailable from overall equilibrium of the freebody diagram shown in the sketch. Towit,

H=−P, V(x)= ℓx

q(ξ)dξ

The moment equation (Eqn. 549 in the text) reduces to

EIθ ′′ + V(x) cos θ+ P sin θ = 0

with the boundary condition θ(0)= 0 and θ′(ℓ)= 0. The differential equation isharder to solve than that for Euler’s elastica because the coefficient of cos θ varies withx. The inextensibility constraints imply that Áo = βo = 0. Furthermore, we haveMθ|ℓ0 = 0 and Vw|ℓ0 = 0, either from the natural boundary conditions by imposinghomogeneous essential boundary conditions on w and θ. Consequently, the virtualwork functional for the problem reduces to

G(Ω,Ω) = ℓ0

EIθ′θ′ − qw dx+ Pu ℓ0

Using the inextensibility constraints w′ = sin θ and u′ = cos θ−1, we get

w′ = cos θθ, u′ = − sin θθ

ℓ

PEI


Noting that V′ = −q we can do the following computation

ℓ0

q(x)wdx = −ℓ0

V′(x)w dx

= ℓ0

V(x)w′dx− Vw ℓ0= ℓ

0

V(x) cos θθ dx

Also, we can compute

u ℓ0= ℓ

0

u′dx = −ℓ0

sin θθ dx

Substituting these results into the virtual work functional we get

G(θ, θ) = ℓ0

EIθ′θ′ − V(x) cos θ+ P sin θ θ dx

which, by the fundamental theorem of the calculus of variations, is equivalent to themoment equilibrium equation. If we had, for example, q(x)= qo then we can integratethe equation for the transverse force as V= qoℓ−x and the virtual work functionalin this particular case would be

G(θ, θ) = ℓ0

EIθ′θ′ − qo ℓ−x cos θ θ− P sin θ θ dx

264. Consider the bar of length ℓ with bending modulusEI, fixed at the left end, propped at the right end, and sub-jected to axial loadPas shown.Assume that shear andaxialdeformations are negligible. Compute the critical loads forthis column by solving the classical differential equation.Estimate the critical loads using the principle of virtual work in conjunction with the Ritzmethod. Use a polynomial basis.

(a) Compute the critical loads for this column by solving the classical differentialequation.

The linearized equation governing the column is

EIwiv+ Pw′′ = 0

The general solution to the equation is

w(x)= a0+ a1x+ a2 sinmx+ a3 cos mx


where m2 ≡ P∕EI. Substituting the boundary conditions w(0)= 0, w′(0)= 0,w(ℓ)= 0, and w′′(ℓ)= 0 into the general solution yields

a2 sinmℓ + a3 cos mℓ = 0

a0+ a3 = 0

a1+ ma2 = 0

a0+ a1ℓ+ a2 sinmℓ + a3 cos mℓ = 0

The first two equations give a0 =−a3 and a1 =−ma2. These can be substituted intothe second two equations to give the system

a2

a3

sinmℓ−mℓ

sinmℓ cosmℓ

cosmℓ−1=

0

0

Non-trivial solution w(x)≠ 0 exists only if the determinant of the coefficient matrixvanishes. Thus,

sinmℓ − mℓ cosmℓ = 0

The first solution to the transcendental equation is found to be m1 = 4.4934∕ℓ. Thecritical load is P1 = 20.16EI∕ℓ2. The corresponding buckling mode is

w1(x)= a2 sin m1x−m1x − a3 1− cos m1x

Observing that a2 sinmℓ+a3 cosmℓ = 0 we can express the buckling mode as

w1(x)= a^2 sinm1x−m1x cosm1ℓ+ 1− cos m1x sinm1ℓ

where a^2 = a2∕ cosm1ℓ. The buckling mode shape is shown below.

P

(b) Estimate the critical loads using the principle of virtual work in conjunction withthe Ritz method. Use a polynomial basis. The virtual work functional for the problemtakes the form

G(w,w) = ℓ0

EIw′′w′′ − Pw′w′ dx

Let the approximate displacement be

w(x)= b0ℓ+ b1x+ b2x2ℓ + b3

x3ℓ2+ b4

x4ℓ3

The function must satisfy the essential boundary conditions of the beam, i.e., it mustsatisfy w(0)= w′(0)= w(ℓ)= 0. The first two conditions give b0 = 0 and b1 = 0.The third condition gives b2+b3+b4 = 0. Substituting these conditions into the origi-nal function, letting a1 = b3, a2 = b4, and ξ≡ x∕ℓ, gives


w(ξ)= a1 ξ3−ξ2 ℓ + a2 ξ4−ξ2 ℓ

Thus, h1(ξ)= ξ3−ξ2 ℓ and h2(ξ)= ξ4−ξ2 ℓ are the base functions. Let the virtualdisplacement be w= a1h1+a2h2. The virtual work functional can be expressed in theform G(w,w)= a T Ka−λGa where the components of the coefficient matrices arecomputed as

Kij= 1

0

hi′′hj′′ dξ Gij= 1

0

hi′hj′ dξ

and λ≡ Pℓ2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional.We can perform the indicated integrals to get the equations

=a1

a2

4− 215λ 8− 7

30λ

845− 44

105λ

0

08− 730λ

The characteristic equation is obtained by setting the determinant of the coefficientmatrix equal to zero

1700λ2− 32

175λ+ 16

5= 0

which is readily solved to give λ1 = 20.919 and λ2 = 107.1. Therefore, the criticalloads are P1 = 20.919EI∕ℓ2 and P2 = 107.1EI∕ℓ2. The fundamental critical load ob-tained by the Ritz method is about 4% higher than the exact value found by the classi-cal method. The MATHEMATICAT commands needed to solve this problem are givenbelow

h = x^3 - x^2, x^4 - x^2hp = D[h,x]hpp = D[hp,x]K = Integrate[ Outer[Times,hpp,hpp],x,0,1]G = Integrate[ Outer[Times,hp ,hp ],x,0,1]F = K - m Gd = Det[F]Solve[d==0,m]

P

ℓ

k EI


265. The prismatic beam shown below has a cross sec-tion that is symmetric with respect to the plane of thepage. The cross section has flexural modulus EI. Axialand shear deformations can be neglected. The beam hasa deformable spring support at the left end that elastically restrains rotations. Themomentdeveloped by the spring is related to the rotation at that point by Ms= kθs, whereθs= w′(0) is the rotation experienced by the spring. The right end of the beam is free totranslate and to rotate. Solve the linearized buckling problem by the classical method, i.e.,by integrating the differential equation. What are the appropriate boundary conditions forthis problem? Solve the problem by integrating the differential equations and using theboundary conditions to find the constants of integration. What are the critical loads of thesystem? What is the smallest critical load as k→ 0? What is the smallest critical load ask→∞? Into what shapes does the beam deform at the critical loads? What are the shapesas k→ 0? What are the shapes as k→∞?

(a) What are the appropriate boundary conditions for this problem?

P

V(ℓ)

M(0)

kw′(0)

The essential boundary condition for the beam is w(0)= 0, the natural boundaryconditions are M(0)= kw′(0), M(ℓ)= 0, and V(ℓ)= 0. These conditions can beexpressed in terms of w as follows:

⇒ w′′(0)− αw′(0)∕ℓ = 0

⇒ w′′(ℓ)= 0

⇒ w′′′(ℓ)+ m2w′(ℓ) = 0

M(0)= kw′(0)M(ℓ)= 0

V(ℓ)= 0

where m2 ≡ P∕EI and α≡ kℓ∕EI have been introduced.(b) Solve the problem by integrating the differential equations and using the boundaryconditions to find the constants of integration. The linearized equation governing thebeam is

EIwiv+ Pw′′ = 0

The general solution to the equation is


Substituting the general solution into the boundary conditions yields

a0+ a3 = 0

a2 sinmℓ + a3 cos mℓ = 0

m2a3+ α a1+ma2 ∕ℓ = 0

m a3 sin mℓ− a2 cosmℓ + a1+ ma2 cosmℓ− ma3 sinmℓ = 0


The first equation gives a0 =−a3 and the last equation gives a1 = 0. The remainingtwo equations can be expressed as

a2

a3αm∕ℓ

sinmℓ cosmℓ

m2=

0

0

A nontrivial solution exists only when the determinant of the coefficient matrix aboveis equal to zero. Hence, the characteristic equation is

mℓ sin mℓ− α cos mℓ = 0

There may be solutions to the characteristic equation for certain values of m (see part(c) below). However, observe that the four equations that result from the boundaryconditions must be used to determine five unknowns, the constants a0, a1, a2, a3 andthe load parameter m. If there exist solutions to the characteristic equation then wecannot completely determine the constants a0, a1, a2, and a3. The best we can hope todo is find three of them in terms of one that cannot be determined. If there is no solu-tion to the characteristic equation then the solution is a0=a1=a2=a3=0.One way to visualize the solutions of the characteristic equation is to let x≡ mℓ andrewrite the characteristic equation as

tan x =αx

The functions f (x)= tan x and f (x)= α∕x are easily visualized as shown in thesketch below. The solutions to the equation, xi, are the intersections of these twocurves. The first two solutions are shown in the sketch. Since the tangent function isasymptotic to the value π∕2 we can observe that x1 < π∕2 and thatπ∕2 < x2< 3π∕2. The hyperbolic function f (x)= α∕x is asymptotic to zero forlarge values of x. Since the tangent function crosses the axis at π, 2π, 3π, . . ., we canobserve that xn ≈ n−1 π for large values of n.

x

tan x

αx

x1 x2π2

f (x)tan x

(c) What are the critical loads of the system? What is the smallest critical load ask→ 0? What is the smallest critical load as k→∞? The characteristic equation frompart (b) can be solved numerically for specific values of the parameter α. For example,


if α= 2 the equation is satisfied for the following values of the load parameter:m1 = 1.077∕ℓ, m2 = 3.644∕ℓ, and m3 = 15.83∕ℓ. The fundamental critical load is,therefore, P(α=2)

1= 1.160EI∕ℓ2.

As k→ 0 the parameter α→ 0, and the characteristic equation reduces to theequation mℓ sin mℓ = 0. The first solution is m1 = 0, which yields a critical load ofzero. This is expected since, when k= 0, the beam can rotate about the left end withno resistance. A stability analysis will reveal that the straight configuration is unstablefor any value of P> 0. The higher critical values are given by mn = (n−1)π∕ℓ forn=2, 3, .... These values are essentially those of a pinned-pinned beam.

We can consider the other extreme, i.e., as k→∞, by dividing the characteristicequation by α and then taking the limit as α→∞. In the limit, the characteristicequation is cosmℓ = 0. The solution to the characteristic equation ismn = (2n−1)π∕2ℓ. The fundamental critical load is P(α→∞)

1= π2EI∕4ℓ2. This result

is again in agreement with our intuition since now the beam is cantilevered at the leftend.

(d) Into what shapes does the beam deform at the critical loads? What are the shapesas k→ 0? What are the shapes as k→∞? The buckling mode is

wn(x)= a^3 α 1− cos mn x + m2nℓ sin mn x

where a^3 = a3∕α. Since the mℓa3+αa2 = 0 we can determine that as k→∞wemust have a2 = 0 and the buckling mode shape is

w(α→∞)n = a3 1− cosmnℓ

The case where k→ 0 is more complicated because m→ 0 as α→ 0. Consider m1(α)to be a function of α. Because m≪ 1 we can expand the sine and cosine functionsapproximately as 1− cos θ≈ 1

2θ2 and sin θ≈ θ. Thus, we have

w1 = a^3 12 αm21x2+ m31ℓx

The characteristic equation reduces to

m2ℓ2− α 1− 12m

2ℓ2 = 0

which means that α∝ m2. Consequently, the term αm21 vanishes faster than the termm31 as α→ 0. In the limit, then, the mode shape for the case k→ 0 is simply

w(α→0)1

= a~x

The rigid body motion for the mode shape is the expected result.

266. For Problem 265, the virtual-work functional that accounts for the work done by thesprings and by the axial force is given by the expression


G(w,w) = ℓ0

EIw′′w′′ − Pw′w′ dx+ kw′(0)w′(0)

Estimate the buckling loads of the beam using a two-term polynomial expansion for thetransverse deflection. That is, assume the real and virtual transverse deflections to be

w(x)= a1x+ a2x2ℓ , w(x)= a1x+ a2

x2ℓ

Repeat the calculation with a three-term polynomial. The classical solution gives an infi-nite number of critical loads. How many did the two-term approximation give? Why?Were the critical loads higher or lower than the exact values? Why? Discuss what is goodand bad about the assumed shapes. Could the approximating functions be improved easi-ly? Suggest a better approximation.

The base functions are h1(ξ)= ξℓ and h2(ξ)= ξ2ℓ. The virtual work functionalcan be expressed as G(w,w)= a T Ka−λGa where the components of the coefficientmatrices are computed as

Kij= 1

0

hi′′hj′′ dξ+ αhi′(0)hj′(0) Gij= 1

0

hi′hj′ dξ

where α≡ kℓ∕EI and λ≡ Pℓ2∕EI. Notice that we divided out EI∕ℓ from the virtualwork functional. We can perform the indicated integrals to get the equation

=a1

a2

α−λ −λ

4− 43λ

0

0−λ


13λ2− 12+4α λ+ 12α = 0

which is readily solved to give

λ1,2 = 6+2α 2 9+3α+α2

Therefore, the critical loads are P1 = λ1(α)EI∕ℓ2 and P2 = λ2(α)EI∕ℓ2. In the limit asα→ 0 we have λ1 → α2 → 0 (show this by taking the Taylor expansion of λ1(α)about α= 0). Dividing the characteristic equation by α and taking the limit asα→∞ gives the equation 4λ−12= 0 giving λ1 = 3 (compare with the exact solu-tion of λ1 = 2.47).

(b) Repeat the calculation with a three-term polynomial. Let the three-term approxi-mation be

w(x)=3i=1

aihi(x), w(x)=3i=1

aihi(x)


with h3(ξ)≡ ξ3ℓ and h1, h2 unchanged. Following the principle of virtual work theequations for a1, a2 and a3 are

=

a3

a1

a2

−λα−λ −λ

4− 43λ

0

0

06− 32λ

12− 95λ6− 3

2λ

−λ

−λ

The characteristic equation is now cubic in the load parameter λ

λ3− 9α+8 λ2+ 2413α+30 λ− 720α = 0

The closed-form expressions for the roots of a cubic equation are very cumbersome.However, for a specific value of α one can numerically calculate the values of theroots. For example, for α= 2 the first three critical loads can be determined asλ1 = 1.160, λ2 = 17.36, and λ3 = 71.48. As usual, the corresponding critical loadsare given by Pi= λiEI∕ℓ2. The exact values of the critical loads were computed inProblem 265(c). The first two of were found to be λ1 = 1.160 and λ2 = 13.28. Ob-viously, the three-term approximation gives very good results for the fundamental criti-cal load.

(c) The classical solution gives an infinite number of critical loads. How many did thetwo-term approximation give? Why? Were the critical loads higher or lower than theexact values? Why?

The number of the critical loads one gets from a Ritz approximation is equal tothe number of unknown constants a1, a2, , aN in the expansions of the real and virtualdisplacement fields. The reason is that for an N-term approximation, the matrices Kand G have dimension N by N. Therefore, the characteristic equation detK−λG = 0is an Nth-order polynomial in λ with N roots. (Because K and G are symmetric, theroots are all real). For a two-term approximation, then, we get two critical loads. Theapproximate critical loads are always higher than the exact values due to theconstraints introduced by the approximating functions. In a qualitative sense one canview the base functions as containing errors in the form of deviations from the eigen-functions. Differentiation magnifies the errors. Thus, the product aTKa is augmentedmore than the product aTGa because the former has second derivatives and the latterhas first derivatives of the base functions. As a consequence, more constraint leads to astiffer system and hence higher buckling loads since λ≈ aTKa∕aTGa.

(d) Discuss what is good and bad about the assumed shapes. Could the approximatingfunctions be improved easily? Suggest a better approximation.

The good aspect of the assumed shape w(x)= a1x+ a2 x2∕ℓ is that it satisfies the

essential boundary condition of the problem w(0)= 0. The bad aspect is that it intro-duces a nonphysical constraint w′′(x)= 2a2∕ℓ, a constant curvature over the wholelength of the beam. We have improved the approximation by simply adding a cubicterm to the approximation.

k

P

ℓ

x

EI


267. The prismatic beam shown below has flexuralmodulus EI. Axial and shear deformations can be ne-glected. The beam has a spring of modulus k located atthe middle of the span. The force developed by thespring is related to the deflection at that point byf (ℓ∕2)= kw(ℓ∕2). Find a suitable expression for the virtual-work functional that ac-counts for the virtual work done by the spring. Estimate the critical loads of the columnusing the principle of virtual work in conjunction with the Ritz method. Use a polynomialbasis.

(a) Find a suitable expression for the virtual work functional that accounts for thevirtual work done by the spring.

The virtual work functional for the beam is simply

G(w,w) = ℓ0

EIw′′w′′−Pw′w′dx + kw(ℓ∕2)w(ℓ∕2)

(b) Estimate the critical loads of the column using the principle of virtual work inconjunction with the Ritz method. Use a polynomial basis.

Let ξ≡ x∕ℓ. The base functions h1 = ξ 1−ξ ℓ and h2 = ξ2 1−ξ ℓ satisfy theessential boundary conditions for the pinned-pinned beam. The virtual work functionalcan be expressed as G(w,w)= a T Ka−λGa where the components of the coefficientmatrices are computed as

Kij= 1

0

hi′′hj′′ dξ+12αhi (1∕2)hj (1∕2) Gij=

1

0

hi′hj′ dξ

where α≡ 2kℓ∕EI and λ≡ Pℓ2∕EI. Notice that we divided out EI∕ℓ from the virtualwork functional. We can perform the indicated integrals to get the equation

=a1

a2

4+ 132α− 1

3λ 0

02+ 164α− 1

6λ 4+ 1

128α− 2

15λ

2+ 164α− 1

6λ


13λ−60 32λ−3α−384 = 0

The critical loads are given by

λ1 = 12+ 332α, λ2 = 60

Therefore, the critical loads are P1 = λ1(α)EI∕ℓ2 and P2 = λ2(α)EI∕ℓ2. Notice thatthe second critical load does not depend upon the spring stiffness. The MATHEMATICAT

commands needed to solve this problem are given below


h = x (1-x) , x^2 (1-x)hp = D[h,x]hpp = D[hp,x]Kb = Integrate[ Outer[Times,hpp,hpp],x,0,1]G = Integrate[ Outer[Times,hp ,hp ],x,0,1]x = 1/2Ks = a Outer[Times,h,h]/2K = Kb + KsF = K - m Gd = Det[F]Solve[d==0,m]

268. Estimate the critical loads of the column in Problem 267 using theprinciple of virtualwork in conjunction with the Ritz method. Use the eigenbasis of the problem without thespring, that is wn(x)= sinmn x, where mn = nπ∕ℓ.

Let the real and virtual displacement fields be

w(x)=Nn=1

anwn(x), w(x)=Ni=1

anwn(x)

where N is the number of functions used in the approximation. The virtual work func-tional is

G(a, a)= a T Ka−λGa

where the ijth components of the coefficient matrices are give by

Kij = ℓ3ℓ0

wi′′wj′′dx+12αwi(ℓ∕2)wj(ℓ∕2), Gij = ℓℓ

0

wi′wj′dx

where α≡ 2kℓ3∕EI and λ≡ Pℓ2∕EI. Note that EI∕ℓ3 has been factored out of thevirtual work functional. Since wi (x) and wj (x) are eigenfunctions of the problem with-out the spring, they satisfy the orthogonality conditions

ℓ3ℓ0

wi′′wj′′dx =12 m

4i ℓ4δij , ℓℓ

0

wi′wj′dx =12m

2i ℓ2δij

where the usual summation convention is suspended. For the spring contribution to thematrix K we note that wi (ℓ∕2)= sin(iπ∕2). Thus,

wi (ℓ∕2)wj (ℓ∕2) =(−1)

12(i+j)−1

0

i, j

otherwise

both odd

Noting that mnℓ = nπ, the principle of virtual work yields

k

P

ℓx

EI


a1

a2

−α

0=

0

0

0 0

0 −α

⋮ ⋮ ⋮

π2 π2−λ +α

4π2 4π2−λ

9π2 9π2−λ +α⋅ ⋅ ⋅

a3⋮

0

⋮

Since the even rows and columns are not affected by the spring stiffness, the criticalloads associated with the even terms are the same as those for the beam without springsupport (by Gershgorin’s theorem). However, the critical loads associated with the oddterms increase due to the presence of the spring. For example, for a two-term approxi-mation (the relevant terms are shown shaded above), the characteristic equation re-duces to

4π2 4π2−λ π2 π2−λ +α = 0

The critical loads are

λ1 = π2+ απ2, λ2 = 4π2

The critical loads for the beam without the spring are λ1 = π2 and λ2 = 4π2. As αincreases λ1 increases and can pass λ2. In fact, for α> 3π4 we have λ2 > λ1 andbuckling is controlled by the “second” mode.

269. The prismatic beam shown below has flexu-ral modulus EI. Axial and shear deformations canbe neglected. The beam is supported on an elasticfoundation of modulus k. The force developed,per unit length, by the foundation is related to thedeflection at that point by f (x)= kw(x). Abeam on an elastic foundationwith axial thrustis governed by the following (linearized) differential equation and boundary conditions

EIwiv+ Pw′′ + kw = 0

w(0)= 0, w′′(0)= 0, w(ℓ)= 0, w′′(ℓ)= 0

The eigenfunctions of the beamwithout the elastic foundation are wn = sin nπx∕ℓ. Verifythat the virtual-work functional, accounting for the elastic foundation, is

G(w,w)= ℓ0

EIw′′w′′ − Pw′w′ + kww dx

Does the presence of the elastic foundation affect the boundary conditions? Find the buck-ling loads of the system using the Ritz method, assuming that the real and virtual displace-ments have the shape of the nth eigenfunction

w(x)= a sin nπxℓ , w(x)= a sin nπxℓ


How does the buckling load vary with the elastic properties of the system, namely EI andk? Express your result in terms of P1 ≡ π2EI∕ℓ2 and the ratio of foundation stiffness tobeam stiffness, given by the dimensionless parameter β≡ kℓ4∕π4EI. (Hint: the criticalbuckling mode depends upon β.) Is your answer exact?

Consider the following weighted residual of the differential equation

G(w,w) = ℓ0

EIwiv+ Pw′′ + kw w dx

By the fundamental theorem of the calculus of variations, G(w,w)= 0 for all w isclearly equivalent to the original classical differential equation. Integrating the firstterm by part twice and the second term once we obtain

G(w,w) = ℓ0

EIw′′w′′−Pw′w′+kww dx

+ EIw′′′+Pw′ w ℓ0− EIw′′w′ ℓ

0

Note that the classical boundary conditions for M= 0 and V= 0 can be expressed asEIw′′ = 0 and EIw′′′+Pw′ = 0, respectively. If we insist that our virtual displace-ments satisfy homogeneous essential boundary conditions, then the virtual work func-tional is the one given.

(b) Does the presence of the elastic foundation affect the boundary conditions?

The elastic foundation does not affect the boundary conditions because it is a dis-tributed effect. The elastic foundation affects only the domain terms of both the classi-cal differential equation and the virtual work functional.

(c) Find the buckling loads of the system using the Ritz method, assuming that thereal and virtual displacements have the shape of the nth eigenfunction

w(x)= a sin nπxℓ , w(x)= a sin nπxℓ

Let mn = nπ∕ℓ. Then w′ = amn cos mn x and w′′ = −am2n sin mn x. The deriva-tives of the virtual displacement field is similar. Thus, the virtual work functional canbe expressed as

G(a, a) = ℓ0

aaEIm4n sin2mn x− Pm2n cos2 mn x+ k sin2 mn x dx

= 12EIℓ5aa m4nℓ4− λm2nℓ2+ π4β

where λ≡ Pℓ2∕EI and β≡ kℓ4∕π4EI. Setting G(a, a)= 0 for any a implies thateither a= 0 (no displacement) or


λn = m2nℓ2+π4βm2nℓ2

= π2n2+ βn2

Note that this approach is valid since the base functions used are eigenfunctions are areorthogonal to each other. If we started the approximation by taking the sum of eigen-functions times unknown constants the equations would uncouple and the results wouldbe the same. Note that Pn = λnEI∕ℓ2.

(d) How does the buckling load vary with the elastic properties of the system, namelyEI and k? Express your result in terms of P1 ≡ π2EI∕ℓ2 and the ratio of foundationstiffness to beam stiffness, given by the dimensionless parameter β≡ kℓ4∕π4EI.(Hint: the critical buckling mode depends upon β.) Is your answer exact?

The buckling loads are a function of the ratio of flexural stiffness to foundationstiffness. In particular, they are a function of the defined parameter β. In the limit asβ→ 0 the buckling loads reduce to those of the pinned-pinned beam without elasticfoundation. For non-zero values of β, the change from the foundationless case is mostapparent for n= 1 and reduces for the higher modes. Physically, the higher modes aremore tortuous but the net deflection from the initial position is less. Consequently,more energy is vested in flexure and less in deforming the foundation. Ordinarily, λ1would be the fundamental buckling mode, but in this case the lowest buckling loaddepends upon the foundation stiffness. For each value of β we must find the smallestλn over all values of n. This process is shown graphically in the sketch below. Thevalue λ1 controls for 0≤ β≤ 4, after which the value λ2 is minimal. In general, thevalue λn is minimal for n−1≤ β ∕n≤ n+1.

1+β

4+ 14 β

9+ 19 β

β

λmin(β)

40 9

4

9

The solution is exact because the base functions are eigenfunctions of the classicalproblem. One can observe that these functions are eigenfunctions by substituting theminto the differential equation, noting that wiv= am4n sinmn x, we have, substituting intoEIwiv+Pw′′+kw= 0, with λ and β defined previously,

aEIℓ4 m4n− λm2n+ β sinmn x = 0

Thus, the classical differential equation is satisfied for all x if λ= λn.

ℓ

P

qo


270. The prismatic beam shownhas a cross section thatis symmetric with respect to the plane of the page. Thecross section has flexural modulus EI. Axial and sheardeformations can be neglected. The beam is fixedagainst transverse deflection and rotation at both ends,but the supports provide no resistance to the axial force P. The beam is also subjected toa uniform transverse load of magnitude qo. The linearized buckling theory for a beamwithtransverse load and axial thrust gives rise to the following differential equations andboundary conditions for the present configuration

EIwiv+ Pw′′ = q(x)

w(0)= 0, w′(0)= 0, w(ℓ)= 0, w′(ℓ)= 0

Solve the governing differential equations by the classical method to find an expressionfor the transverse deflection w(x) and the bending moment M(x).

In the design of beams subjected to transverse load and thrust, sometimes called beam-columns, the concept of the magnification factor is often used. The idea behind themagni-fication factor is that the influence of the axial thrust is to magnify the values of displace-ment and moment that would be present if the axial thrust were not (i.e., the solution ifP= 0). Show that the maximum deflection and moment can be expressed as

w(ℓ∕2)= wo 11−P∕P1, M(0)= Mo1−0.4P∕P1

1−P∕P1

where wo and Mo are the maximum deflection and moment that would occur if P= 0,and P1 would be the fundamental critical load of the column if qo= 0.

(a) Solve the governing differential equations by the classical method to find an ex-pression for the transverse deflection w(x) and the bending moment M(x).

A particular solution for q(x)=−qo is

wp =− 12

qom2EI

x2

where m2 = P∕EI. Therefore, the general solution to the given differential equation is

w(x)= a0ℓ+ a1x+ a2ℓ sin mx+ a3ℓ cosmx−12

qom2EI

x2

Substituting the general solution into the boundary conditions yields

a0+ a3 = 0

a1+ mℓa2 = 0

a0+ a1+ a2 sinmℓ + a3 cos mℓ− qoℓ∕2m2EI = 0

a1+ a2mℓ cosmℓ− a3mℓ sinmℓ − qoℓ∕m2EI = 0

The first two equations give a0 =−a3 and a1 =−mℓa2. Substituting these into thethird and fourth equations gives


a2

a3mℓ cos mℓ−mℓ

sinmℓ−mℓ cosmℓ−1

−mℓ sin mℓ= qoℓ

2m2EI

1

2

These equations can be solved to give

a2 =−12qom3EI

, a3 =−12qom3EIcos mℓ+1

sinmℓ Observing that tan θ= sin 2θ∕(1+ cos 2θ), and letting α≡ mℓ∕2, the displacementcan be expressed as

w(x)= 24wo

α3 1− cos 2αξ

tanα + 2αξ−ξ2 − sin 2αξwhere wo ≡ qoℓ4∕384EI is the maximum deflection that the beam would experience ifit had no axial load and ξ≡ x∕ℓ. The bending moment field is

M(x)= 4Moα cos 2αξtanα −

1α+ sin 2αξ

where Mo ≡ qoℓ2∕12 is the maximum moment that the beam would experience if ithad no axial load.

(b) Show that the maximum deflection and moment for this problem can be expressedin the form

w(ℓ∕2)= wo 11− P∕P1, M(0)= Mo1− 0.4P∕P1

1− P∕P1

where wo and Mo are the maximum deflection and moment that would occur if P= 0,and P1 would be the fundamental critical load of the column if qo=0.

The maximum deflection is at ξ= 1∕2 and the maximum moment is at ξ= 0.Thus, the exact solution for the maxima are given by

wmaxwo= 122 cos α−2+α sin α

α3 sinα Mmax

Mo= 3 tanα−α

α2 tan α

where α= mℓ∕2. To compare the approximate and exact expressions we can expandboth in Taylor series and examine the difference. Noting that the load ratio can be ex-pressed as P∕P1 = α2∕π2, we can expand the approximate maximum deflection as

wapproxmaxwo

= 11− α2∕π2 = 1+ α

2

π2+ α

4

π4+ α

6

π6+ α

8

π8+⋅⋅⋅

The Taylor expansion for the exact midspan deflection is

wmaxwo= 1+ 1

10α2+ 17

1680α4+ 31

30240α6+ 691

6652800α8+⋅⋅⋅

P

P

x ℓ

ℓ


The difference between the approximate and the exact expressions is

wapproxmaxwo−wmax

wo= 0.01304β2+ 0.01431β4+ 0.01445β6+⋅⋅⋅

where β≡ P∕P1. The moments can be treated similarly. We can expand the approxi-mate maximum moment as

Mapproxmax

Mo=

1− 0.4α2∕π21− α2∕π2

= 1+ 35α2π2+ α

4

π4+ α

6

π6+ α

8

π8+⋅⋅⋅

The Taylor expansion for the exact end moment is

Mmax

Mo= 1+ 1

15α2+ 2

315α4+ 1

1575α6+ 2

31185α8+⋅⋅⋅

The difference between the approximate and the exact expressions is

Mapproxmax

Mo−Mmax

Mo= 0.0578β2+ 0.0185β4+ 0.0104β6+⋅⋅⋅

As the axial load approaches the buckling load α→ π and P→ P1. The approximationcontains the most error in this vicinity. However, even at P= 0.99P1 the error in dis-placement is 1.5% and the error in moment is 1.4%.

271. The column shown is subjected to axial forces at the midpoint andtop, both of magnitude P. The column has variable flexural modulusgiven by the expression EI(x)= EIo 2−x∕2ℓ. Estimate the bucklingload of the system using the Ritz method with a one-term polynomialbasis. Is your estimate higher or lower than the actual buckling load?Explain your answer. Propose a function for a one-term Ritz approxi-mation that will give better results than you got in the first part. Whydo you think it is better? Estimate the buckling load of the system usingthe Ritz method with a two-term polynomial basis. Repeat with a three-term polynomial basis.

(a) Estimate the buckling load of the system using the Ritz method with a one-termpolynomial basis.

The function w(x)= ax2 satisfies the essential boundary conditions of zero dis-placement and rotation at x= 0. Then w′ = 2ax and w′′ = 2a. The virtual displace-ment field and its derivatives are similar. Letting ξ≡ x∕ℓ, the virtual work functionalcan be expressed as


G(a, a) = 10

aa2EIoℓ4−ξ − 8Pℓ3ξ 2 dξ+21

aa2EIoℓ4−ξ − 4Pℓ3ξ2 dξ

= aa7EIoℓ − 83 Pℓ

3+ 5EIoℓ− 283 Pℓ

3

= aa12EIoℓ − 12Pℓ3

Setting G(a, a)= 0 for any a implies that either a= 0 (no displacement) or

Pcr =EIoℓ2

(b) Is your estimate higher or lower than the actual buckling load? Explain your an-swer.

The estimate is higher than the actual buckling load because the Ritz method al-ways provides an upper bound on buckling loads.

(c) Propose a function for a one-term Ritz approximation that will give better resultsthan you got in part (a). Why do you think it is better?

The function w(x)= a x3−6x2ℓ satisfies the essential boundary conditions ofzero displacement and rotation at x= 0. It also satisfies the condition of vanishingmoment at x= 2ℓ. It should, therefore, give better results than the previous one-termpolynomial approximation. The derivatives of the displacement function arew′ = 3a x2−4xℓ and w′′ = 6a x−2ℓ. The virtual displacement field and its deriv-atives are similar. Thus, the virtual work functional can be expressed as

G(a, a) = 10

aa 18EIoℓ3 4−ξ ξ−2 2− 18Pℓ5 ξ2−4ξ 2 dξ

= aa 3032 EIℓ3− 3185Pℓ5+ 33

2 EIℓ3− 609

5Pℓ5

= aa 168EIℓ3− 9275Pℓ5

+ 21

aa 18EIoℓ3 4−ξ ξ−2 2− 9Pℓ5 ξ2−4ξ 2 dξ


Pcr = 280309

EIℓ2

The coefficient is 0.906, which is less than the previous value of 1.0.

(d) Estimate the buckling load of the system using the Ritz method with a two-termpolynomial basis. Repeat with a three-term polynomial basis.

The base functions h1(ξ)= ξ2ℓ, h2(ξ)= ξ3ℓ, h3(ξ)= ξ4ℓ, etc. satisfy the essen-tial boundary conditions for the cantilever beam. The virtual work functional can beexpressed as G(w,w)= a T Ka−λGa where the coefficients are computed as

k

P

ℓ

x

EI Rigid

ℓ


Kij= 2

0

124−ξ hi′′hj′′ dξ Gij=

1

0

2hi′hj′ dξ+2

1

hi′hj′ dξ

where λ≡ Pℓ2∕EIo. Notice that we divided out EIo∕ℓ from the virtual work function-al. For the two-term approximation we can perform the indicated integrals to get theequation

=a1

a2

12−12λ 0

032− 512λ 120− 297

5λ

32− 512λ


125120λ2− 2604

5λ+ 416 = 0


λ1 = 0.89497, λ2 = 7.43117

Therefore, the critical loads are P1 = λ1EIo∕ℓ2 and P2 = λ2EIo∕ℓ2. Notice that thefirst critical load is smaller than the ones computed with the one-term expansions. Thethree-term expansion is similar to the two-term expansion. The MATHEMATICAT com-mands needed to solve this problem are given below

h = x^2 ,x^3, x^4 hp = D[h,x]hpp = D[hp,x]EI = (4 - x)/2K = Integrate[ EI Outer[Times,hpp,hpp],x,0,2]G1 = Integrate[ 2 Outer[Times,hp ,hp ],x,0,1]G2 = Integrate[ Outer[Times,hp ,hp ],x,1,2]G = G1 + G2d = Det[K - m G]Solve[d==0,m]

The critical loads for the three-term expansion are given by

λ1 = 0.88749, λ2 = 5.85419, λ3 = 27.1855

272. A flexible beam of length ℓ and modulusEI is welded to a rigid beam of length ℓ and restson an elastic foundation of modulus k (per unitlength). It is pinned at the left end and is sub-jected to a compressive axial load P at the rightend. The elastic foundation accrues a transverseforce in proportion to the transverse displacement w. Shear and axial deformations in thebeam are negligible.Write the expression for the energy of the system.What are the essen-tial and natural boundary conditions for the flexible beam? Find an approximate solutionfor the buckling loads and mode shapes using a two-term polynomial Ritz basis.


(a) Write the expression for the energy of the system.

Let us designate the deflection of the beam as w(x), applicable over the entirelength [0, 2ℓ]. Let ∆ℓ ≡ w(ℓ) be the deflection at the end of the flexible portion of thebeam and let θℓ ≡ w′(ℓ) be the rotation at the end of the flexible portion of the beam.The potential energy of the system using the linearized theory is

E(w) = ℓ0

12EI (w′′) 2+kw2−P(w′)2 dx +2ℓ

ℓ

12 kw

2dx − 12 Pℓθ

2ℓ

The deflection of the rigid bar can be expressed as

w(x)= ∆ℓ + x−ℓθℓ, ℓ ≤ x≤ 2ℓ

Therefore, the second integral can be evaluated explicitly as

2ℓℓ

12 kw

2dx = 12 kℓ∆2

ℓ +13 ℓ

2θ2ℓ + ℓ∆ℓθℓ

The virtual work functional can be obtained by taking the directional derivative of theenergy expression

G(w,w) = ℓ0

EIw′′w′′+kww−Pw′w′ dx +2ℓℓ

kwwdx − Pℓθℓθℓ

where θℓ = w′(ℓ) and w= ∆ℓ + x−ℓθℓ in the rigid portion of the beam. The termassociated with the elastic foundation under the rigid bar can be obtained from theabove expression as

2ℓℓ

kwwdx = 12 kℓ2∆ℓ∆ℓ + 2

3 ℓ2θℓθℓ + ℓ∆ℓθℓ + ℓ∆ℓθℓ

The virtual work functional can be written as

G(w,w) = ℓ0

EIw′′w′′ + kww− Pw′w′ dx− Pℓw′(ℓ)w′(ℓ)

+ 12 kℓ2w(ℓ)w(ℓ)+ 2

3 ℓ2w′(ℓ)w′(ℓ)+ ℓw(ℓ)w′(ℓ)+ ℓw′(ℓ)w(ℓ)

(b) What are the essential and natural boundary conditions for the flexible beam?Integrating the virtual work functional by parts until all of the derivatives are shedfrom the w terms we can find an equivalent expression for the virtual work functional

G(w,w) = ℓ0

EIwiv+kw+Pw′′ wdx−M(0)w′(0)+Q(0)w(0)

+ −EIw′′′(ℓ)−Pw′(ℓ)+kℓw(ℓ)+ 12 kℓ

2w′(ℓ) w(ℓ)+ EIw′′(ℓ)−Pℓw′(ℓ)+ 1

2 kℓ2w(ℓ)+ 1

3 kℓ3w′(ℓ) w′(ℓ)


By the fundamental theorem of the calculus of variation we can determine that thegoverning differential equation for the beam is

EIwiv+ Pw′′ + kw = 0

and that the appropriate boundary conditions for this problem are

EIw′′(ℓ)− Pℓw′(ℓ)+ 12kℓ2w(ℓ)+ 1

3kℓ3w′(ℓ) = 0

EIw′′′(ℓ)+ Pw′(ℓ)− kℓw(ℓ)− 12kℓ2w′(ℓ) = 0

w(0)= 0

w′′(0)= 0

The first boundary condition is essential and the last three are natural.

(c) Find an approximate solution for the buckling loads and mode shapes using a two-term polynomial Ritz basis. The base functions h1(ξ)= ξℓ, h2(ξ)= ξ2ℓ satisfy theessential boundary conditions for the cantilever beam. Let w= a1h1+a2h2 andw= a1h1+a2h2. The virtual work functional can be expressed asG(w,w)= a T Ka−λGa where the coefficients are computed as

Kij= 1

0

hi′′hj′′+αhihj dξ

Gij= 1

0

hi′hj′ dξ + hi′(1)hj′(1)

+ 12α2hi(1)hj(1)+ 2

3 hi′(1)hj′(1)+ hi(1)hj′(1)+ hi′(1)hj(1)

where λ≡ Pℓ2∕EI and α≡ kℓ3∕EI. Notice that we divided out EI∕ℓ from the virtualwork functional. For the two-term approximation we can perform the indicated inte-grals to get the equation

=a1

a2

83α−2λ 0

04+ 6815α− 16

3λ

4112α−3λ

4112α−3λ


53 λ

2− 8+ 25190 αλ+ 323 α+ 299

720 α2 = 0

The critical loads can be computed numerically for specific values of α. For example,for the value α= 1 we get

λ1 = 1.2804, λ2 = 5.1929

Therefore, the critical loads are P1 = λ1EI∕ℓ2 and P2 = λ2EI∕ℓ2. The MATHEMATI-CAT commands needed to solve this problem are given below

ℓ p,EI


h = x, x^2hp = D[h,x]hpp = D[hp,x]K1 = Integrate[Outer[Times,hpp,hpp],x,0,1]K2 = Integrate[Outer[Times,h ,h ],x,0,1]G1 = Integrate[Outer[Times,hp ,hp ],x,0,1]x = 1K3 = 2 Outer[Times,h,h]K4 = 2 Outer[Times,hp,hp]/3K5 = Outer[Times,hp,h] + Outer[Times,h,hp]G2 = Outer[Times,hp,hp]K = K1 + a ( K2 + (K3 + K4 + K5)/2 )G = G1 + G2d = Det[K - m G]Solve [ d==0 , m ]

In this script, a stands for α and m stands for λ.

273. The column shown has modulus EI and weight per unitlength p. It is fixed at one end and free at the other. Shear and axialdeformations can be neglected. Find the (classical) governing dif-ferential equations and boundary conditions for the transversedeflection w(x). Express the governing equations in virtual-workform. Estimate by the Ritz method the maximum length the col-umn can have before it buckles under its own weight.

(a) Find the (classical) governing differential equations and boundary conditions forthe transverse deflection w(x).

x

H

V

M

ℓ

p(x)= po

EI

po

From equilibrium of forces we have H(x)=−po ℓ−x and V(x)= 0. The (li-nearized) classical equations are, therefore,

EIwiv+ po ℓ−x w′ ′ = 0


w(0)= 0, w′(0)= 0, M(ℓ)= 0, V(ℓ)= 0

(b) Express the governing equations in virtual work form.

The virtual work functional is given by

G(w,w) = ℓ0

EIw′′w′′ − po ℓ−x w′w′ dx

ℓ3P

P

ℓ

EI

2EI


(c) Using a polynomial approximation for w, estimate the maximum length the col-umn can have before it buckles under the axial force. Use the Ritz method.

The function w(x)= ax2 satisfies the essential boundary conditions of zero dis-placement and rotation at x= 0. Then w′ = 2ax and w′′ = 2a. The derivatives ofthe virtual displacement field are similar. Letting ξ≡ x∕ℓ, the virtual work functionalcan be expressed as

G(a, a) = 10

aa 4EIℓ− 4poℓ41−ξ ξ2 dξ

= 4aa EIℓ − 112 poℓ

4


ℓcr = 12EIpo1∕3

274. The stepped column shown has a variable modulus and is sub-jected to vertical forces at two points. It is fixed at one end and freeat the other. Shear and axial deformations can be neglected. Find the(classical) governing differential equations and boundary condi-tions for the transverse deflectionw(x). Express the governing equa-tions in virtual-work form. Using a two-term polynomial approxi-mation for w, estimate the critical load using the Ritz method.

(a) Since the modulus and axial load are discontinuous we must consider each seg-ment separately and relate the two segments with continuity conditions. Let w1(x) andw2(x) be the transverse displacement in the first and second segments so that

w(x) =w1(x1)

w2(x2)

0≤ x1≤ ℓ

0≤ x2≤ ℓ

The segments are governed by the differential equations

wivi + m2i wi′′ = 0 i = 1, 2

where m21 = 2m22 = P∕EI. These equations have the solution

wi = ai+ bixi+ ci sin mi xi+ di cosm i xi


w1(0)= 0

w2′′(ℓ)= 0

w1′(0)= 0

w2′′′(ℓ)+ m22w2′(ℓ)= 0


The continuity conditions are

w1(ℓ)= w2(0)

2w1′′(ℓ)= w2′′(0)w1′(ℓ)= w2′(0)

2 w1′′′(ℓ)+ m21w1′(ℓ) = w2′′′(0)+ m22w2′(ℓ)

(b) Express the governing equations in virtual work form. Assume that the real andvirtual displacement functions are defined on the entire range [0,2ℓ]. Then the virtualwork functional can be expressed as

G(w,w) = ℓ0

2EIw′′w′′−4Pw′w′ dx + 2ℓℓ

EIw′′w′′−Pw′w′ dx

(c) Using a two-term polynomial approximation for w, estimate the critical load usingthe Ritz method. The base functions h1(ξ)= ξ2, h2(ξ)= ξ3, h3(ξ)= ξ4, etc. satisfythe essential boundary conditions for the cantilever beam. Note that the derivatives arewith respect to x so that, for example, h1′ = 2ξ∕ℓ and h1′′ = 2∕ℓ2. The virtual workfunctional can be expressed as G(w,w)= a T Ka−λGa where the coefficients arecomputed as

Gij= 1

0

4hi′hj′ ℓdξ +2

1

hi′hj′ ℓdξ

Kij= 1

0

2EIhi′′hj′′ ℓdξ +2

1

EIhi′′hj′′ ℓdξ

where λ≡ Pℓ2∕EI. Notice that we divided out EI∕ℓ from the virtual work functional.For the two-term approximation we can perform the indicated integrals to get the equa-tion

=a1

a2

12− 443λ 0

030− 572λ 108−63λ

30− 572λ


4474λ2− 630λ+ 396 = 0


λ1 = 0.72071, λ2 = 4.91688

Therefore, the critical loads are P1 = λ1EIo∕ℓ2 and P2 = λ2EIo∕ℓ2. The MATHEMATI-CAT commands needed to solve this problem are given below

ℓ

PEI,GA


h = x^2 ,x^3 hp = D[h,x]hpp = D[hp,x]K1 = Integrate[2 Outer[Times,hpp,hpp],x,0,1]K2 = Integrate[ Outer[Times,hpp,hpp],x,1,2]G1 = Integrate[4 Outer[Times,hp ,hp ],x,0,1]G2 = Integrate[ Outer[Times,hp ,hp ],x,1,2]K = K1 + K2G = G1 + G2d = Det[K - m G]Solve[d==0,m]

275. Consider the bar of length ℓwith bending modulus EIand shear modulusGA, subjected to axial load P as shown.Show that the linearized virtual-work functional for thebuckling of a beam with shear deformation is given by

G(w, θ,w, θ) = ℓ0

EIθ′θ′ +GA(w′−θ) (w′−θ)− P w′θ+w′θ−θθ dx

Make the assumption that the (generalized) shear strain in the beam is constant. Estimatethe critical loads of the beam using a polynomial approximation with theRitz method. Forexample, a three-parameter approximation would have the expression

θ= a0xℓ + a1

x2ℓ2 , w= a1x+ 1

2 a0x2ℓ +

13 a1

x3ℓ2

Describe at least two ways of improving the approximation, and rank them according towhich is likely to give the most improvement (no calculations necessary). If EI is verylarge in comparison to GAℓ2, what will the buckled shape of the beam look like?

Derivation. The inextensibility constraint Á2o+ β2o = 1 can be maintained withoutsacrificing the shearing deformations if we introduce the angle φ such that

w′ = sinφ, 1+u′ = cos φ

Noting that βo = sinφ cos θ− cos φ sin θ, and using a trigonometric identity, theconstitutive equation for shear can be written as

Q = V cos θ−H sin θ = GA sin(φ−θ)

The equation can be linearized to give

Hθ+GA φ−θ − V = 0, (*)

Equilibrium of moments, Eqn. (549), can now be put into the (linearized) form

M′ + V−Hφ = 0, (**)

Let us construct a weighted residual expression for the moment equilibrium equation(**) and the transverse force equation (*). To wit, multiply the moment equation by θ


and the transverse force equation by φ−θ and integrate over the length of the beam toget

G(φ, θ, φ, θ)=ℓ0

−M′+V−Hφ θ+ Hθ+GA φ−θ − V φ−θ dxIntegrating by parts and rearranging terms we get

G(φ, θ, φ, θ) = ℓ0

Mθ′ − Vφ+H φθ+φθ−θθ + GA φ−θ φ−θ dxFinally, noting that V= 0, H=−P , and M= EIθ′ we get the expression

G(φ, θ, φ, θ) = ℓ0

EIθ′θ′ +GA φ−θ φ−θ − P φθ+φθ−θθ dxNoting that φ= w′ we can see the difference in the proposed expression and the cor-rect expression. Note that this form of the virtual work functional reduces to the onefor the Bernoulli-Euler theory when φ= θ= w′.

(a) Make the assumption that the (generalized) shear strain in the beam is constant.Estimate the critical loads of the beam using a polynomial approximation with the Ritzmethod. For example, a three-parameter approximation would have the expression

θ= a0xℓ + a1

x2ℓ2 , w= a2x+ 1

2 a0x2ℓ +

13 a1

x3ℓ2

Note that φ= w′ = a2+a0 ξ+a1ξ2 where ξ≡ x∕ℓ. The shear deformation canbe computed as φ−θ= a2, which is constant, as desired. Assuming that the virtualdisplacements and rotations are interpolated the same way as the real ones we canwrite the virtual work functional as

G(a, a) = 10

a0+2a1ξ a0+2a1 ξ + βa2a2

+ λ a0ξ+a1ξ2 a0ξ+a1ξ2 dx

− λ a2+a0ξ+a1 ξ2 a0ξ+a1ξ2

− λ a2+a0ξ+a1 ξ2 a0ξ+a1ξ2

where β≡ GAℓ2∕EI and λ= Pℓ2∕EI. Note that the virtual work functional has beendivided through by EI∕ℓ. Carrying out the integrations we arrive at the system of equa-tions


=

a0

a1

1− 14λ 0

043− 1

5λ − 1

3λ

− 12λ

a2 0β

1− 13λ

1− 14λ

− 12λ − 1

3λ


8λ3+ 9β−240 λ2− 312βλ+ 720β = 0

The critical loads can be computed numerically for specific values of β. Some valuesare given in the table below

λ1 λ2 λ3β

1 1.232 30.07 --2.429 1.14910 2.117 30.55 --13.918 2.048100 2.437 31.67 --116.60 2.4091000 2.481 32.12 --1129.6 2.461

λEXACT1

The exact value of the buckling load is given by the equation

λEXACT1 = π2

2+2 1+π2∕β

Notice that as β gets large the fundamental buckling load approaches the buckling loadfor the Bernoulli-Euler beam, that is λ1 → π2∕4 = 2.467.

It is interesting to observe that λ3 < 0, suggesting the existence of a critical loadfor tensile loads. An explanation can be constructed as follows. The classical differen-tial equation can be obtained by adding (*) and (**) to eliminate V and by subsequentlyeliminating φ= 1+λ∕β θ by (*) with V= 0. Dividing the result by EI we finallyobtain the classical equation

ℓ2θ′′+λ 1+λ∕β θ = 0

where λ and β are as previously defined. For the load λ=−β this equation reducesto θ′′ = 0 along with φ= 0. Thus, θ= a0+a1x. The boundary condition θ(0)= 0gives a0 = 0 and the boundary condition M(ℓ)= 0 gives a1 = 0. Thus, θ(x)= 0 forthis value of the load parameter. The negative root to the characteristic equation is thebest effort the approximation can make to finding this “other” solution in the context ofthe given functional. Note that, to within the roughness of the approximation, the nu-merical solution gives λ3 ≈−β.(b) Describe at least two ways of improving the approximation, and rank them ac-cording to which is likely to give the most improvement (no calculations necessary).The approximation can be improved by increasing the order of the basis and by elimi-nating the constant shear assumption. Actually, you cannot do the latter without doing

ℓ

x wo(x)

P


the former. The computation can be organized conveniently by defining the followingmatrices

aT = a1, a2, a3, a4 gT(ξ) = 0, 0, ξ, ξ2 hT(ξ) = ξ, ξ2 , 0, 0

Now θ= aTh(ξ), φ= aTg(ξ), θ= a Th(ξ), φ= a Tg(ξ). The virtual work functionalcan be computed as G= a T K−λG awhere the components of the coefficient ma-trices are given by

Kij = 1

0

hi′hj′+β gi−hi gj−hj dξ Gij = 1

0

hi gj+gi hj−hihj dξ

The MATHEMATICAT commands needed to solve this problem are given below

h = x, x^2, 0, 0g = 0, 0, x, x^2hp = D[h,x]gmh = g - hK1 = Integrate[Outer[Times,hp ,hp ],x,0,1]K2 = Integrate[Outer[Times,gmh,gmh],x,0,1]G1 = Integrate[Outer[Times,g ,h ],x,0,1]G2 = Integrate[Outer[Times,h ,g ],x,0,1]G3 = Integrate[Outer[Times,h ,h ],x,0,1]K = K1 + b K2G = G1 + G2 - G3d = Det[K - m G]

The result for the four-term solution for β= 100 is

λ1 = 2.427, λ2 = 25.62, λ3 = −102.4, λ4 = −125.6

Observe that now we get two negative eigenvalues that are roughly equal to β.

(c) If EI is very large in comparison to GAℓ2, what will the buckled shape of thebeam look like?

The deformation will be dominated by shear, not flexure. Therefore, the shape ofthe beam should be a straight line in the limit.

276. Consider the beam of length ℓ, fixed at both ends,with constant modulus EI shown in the sketch. The beamis subjected to a compressive axial load P. When thebeam is not loaded, the initial shape can be described as

wo(x)= co3 x2ℓ2− 2 x3

ℓ3

where co% 1 is known as given data. Assume that shear and axial deformations are negli-gible. Find an expression for the energy functional E and the virtual-work functionalG forthis problem. Estimate the deflection of the beam as a function of load P using the Ritzmethod and a one-term approximation as follows

w(x)= wo(x)+ ax2ℓ2− 2 x3

ℓ3+x4ℓ4


assuming that the displacements are small enough to use the linearized buckling theory.

(a) Find an expression for the energy functional E and the virtual work functional Gfor this problem.

The energy functional E for the imperfect system is

E(w) = 12ℓ0

EI w′′−wo′′ 2− P w′ 2 dx

The virtual work functional can be found by taking the directional derivative of theenergy to get

G(w,w) = ℓ0

EI w′′−wo′′ w′′ − Pw′w′ dx

(b) Estimate the deflection of the beam as a function of load P using the Ritz methodand a one-term approximation as follows

w(x)= wo(x)+ ax2ℓ2− 2 x3

ℓ3+x4ℓ4

assuming that the displacements are small enough to use the linearized bucklingtheory.

Let the base function h(ξ) and wo(ξ) be given by

h(ξ) = ξ2−2ξ3+ξ4, wo(ξ) = co 3ξ2−2ξ3

so that w(ξ)= wo(ξ)+ah(ξ) and w(ξ)= ah(ξ). Then the virtual work functional canbe expressed as

10

aa EIℓh′′ 2− Pℓ (h′) 2 dξ = Pℓ10

wo′h′dξ

Observe that the integral on the right side of the equation vanishes, i.e.,

10

wo′h′dξ = 10

co 6ξ−6ξ2 2ξ−6ξ2+4ξ3 dξ = 0

Thus, we have

aa4EI5ℓ3−

2P105ℓ = 0

hence the solution is either a= 0, the trivial solution, or a≠ 0 when the load reachesa critical value

k x

EI

ℓ

P

ℓ

2P 2P


Pcr = 42EIℓ2

277. A flexible beam of length2ℓ andmodulusEI restson an elastic foundation of modulus k. The beam iscompressed by a known fixed force 2P and is subjectedto a transverse load P at its midpoint. The propertieshave values such that kℓ4 = EI. Axial and shear de-formations of the flexible beam can be neglected. Esti-mate the deflection at the middle and ends of the beam using virtual work and the Ritzmethod. (Note: due to symmetry odd base functions need not be included.)

The appropriate virtual work functional is

G(w,w)= ℓ−ℓ

EIw′′w′′−2Pw′w′+kww dx− Pw(0)w(0)= 0

Using base functions h= [1 , ξ2 ], h′ = [0 , 2ξ], h′′ = [0 , 2] we get the followingstiffness matrix and force vector

K = 210

EIℓ3 h′′ h′′ − 2P

ℓ h′ h′ + kℓh h dξ, f= Ph(0)

Carrying out the integrals,

K =2EIℓ3

0 0

0 4 f= P1

0− 4Pℓ

0 0

0 4∕3 + 2kℓ1 1∕3

1∕3 1∕5

Noting that kℓ4= EI, and letting λ≡ 40Pℓ2∕EI, we get

K =2EI15ℓ3

15 5

5 63−λf=

2EI15ℓ3

3λℓ16

1

0

The coefficients of the displacement approximation can be found by solving Ka= f.To wit,

15a0+ 5a1 = 3λℓ∕16

5a0+ 63−λ a1= 0

from which we find

a0 =3λℓ63−λ 80 184−λ

, a1=−15λℓ

80 184−λ

P

ℓ

ℓ

2P

EIRigidx


Note that the system will buckle when the denominator of these coefficients becomesinfinite or when λ= 184. This condition occurs at a critical load of

Pcr= 1.53 EIℓ2

The displacement is

w(ξ)=3λℓ

80 184−λ 63−λ − 5ξ2

from which it is easy to find w(0) and w(1).

278. A square frame of dimension ℓ is composed of twocolumns connected together by a beam (the beam can beconsidered rigid). The left column, which is rigid andpinned at both ends, is subject to a force 2P. The rightcolumn, which has flexural modulus EI, is subjected toa force P. What are the essential and natural boundaryconditions on the flexible beam. Express all boundaryconditions in terms of the transverse displacement w(x)of the flexible beam. Find the classical characteristicequation that determines the buckling load of the system.Find the exact valueof theprima-ry buckling load from the characteristic equation. Recall that the classical differentialequation wiv+ m2w′′ = 0 has the general solution in the form

w(x)= a0+a1x+a2 sin mx+a3 cosmx.

Estimate the buckling capacity of the structure using the Ritz method in conjunction withthe principle of virtual work. Compare the classical and variational solutions. [Note: Theleft column is often referred to as a “leaner” because it leans on the right column to findresistance to sway. By itself the left column has no lateral stiffness, but it carries a destabi-lizing force.]

(a) What are the essential and natural boundary conditions on the flexible beam.Express all boundary conditions in terms of the transverse displacement w(x) of theflexible beam.

Equilibrium of the rigid link on the left gives

2Pw(ℓ)= Nℓ ⇒ N= 2Pw(ℓ)ℓ

This force acts, through the horizontal rigid link, on the flexible beam. Therefore theboundary conditions are

Essential: w(0)= 0, w′(0)= 0


P2P

x

N N N N

w(ℓ) w(ℓ)

Natural: w′′(ℓ)= 0, − EIw′′′(ℓ)+ Pw′(ℓ)+ 2Pw(ℓ)∕ℓ = 0

(b) Find the classical characteristic equation that determines the buckling load of thesystem. Find the exact value of the primary buckling load from the characteristicequation. Recall that the solution to the classical differential equation wiv+ m2w′′ = 0is given by


Substituting the boundary conditions gives equations to determine the coefficientsof the solution

w(0)= a0+ a3 = 0

w′(0)= a1+ ma2 = 0

w′′(ℓ)=−m2 a2 sinmℓ+ a3 cosmℓ = 0

+ Pℓa1ℓ+ mℓa2 cosmℓ − a3 sin mℓ

V=−EIm3 a2 cosmℓ− a3 sinmℓ

+ 2 a0+ a1ℓ + a2 sin mℓ+ a3 cosmℓ = 0

Let S≡ sin mℓ and C≡ cos mℓ and T≡ S∕C. Substituting the first equations into thelast two give, noting that m2 = P∕EI, and dividing through by P∕ℓ

a2S+ a3C= 0 ⇒ a3 =−a2T

−mℓa2C− a3S + mℓ−a2+ a2C− a3S + 2 −a3−mℓa2+ a2S+ a3C = 0

Substituting the first into the second gives−mℓa2+ 2 a2T−mℓa2 = 0 or

2 tan mℓ = 3mℓ

This equation has a solution at mℓ = 0.968. Therefore, the critical load is

Pcr= 0.937 EIℓ2

(c) Estimate the buckling capacity of the structure using the Ritz method inconjunction with the principle of virtual work. Use a two term polynomialapproximation. Compare the solutions from parts (b) and (c).

The appropriate virtual work functional is


G(w,w)= ℓ0

EIw′′w′′ − Pw′w′ dx− 2Pw(ℓ)w(ℓ)∕ℓ = 0

Using base functions h= [ξ2, ξ3 ] we get the following stiffness and geometric stiff-ness matrices

K =ℓ0

EIℓ3 h′′ h′′ dx, G= P

ℓ ℓ

0

h′ h′ dξ+ 2Pℓ h(ℓ) h(ℓ)

Carrying out the integrals gives the results

K = EIℓ3

4 6

6 12G = P

ℓ

72

195

103

72

Letting λ= Pℓ2∕EI we have the equation K−λG a= 0 governing the response ofthe system this equation has a nontrivial solution only if the determinant of the coeffi-cient matrix is zero. Hence, we can determine the value of the critical load from thecondition

6− 72λ

12− 195λ

4− 103λ

6− 72λ

det = 0

or

4− 103 λ12− 19

5λ − 6− 7

2 λ 2 = 0

This equation can be simplified to

0.416667λ2− 13.2λ+ 12= 0

which has solution λ= 0.936. Therefore, the critical load is

Pcr= 0.936 EIℓ2

which is nearly identical to the classical solution found previously.

k

P

ℓ

x

EI

ℓ


279. A flexible beam of length 2ℓ andmodulusEI is stuck on an elastic foundation of modulusk (per unit length) over half of its length. It ispinned at the left end and is subjected to a com-pressive axial loadP at the right end. Theelasticfoundation accrues a transverse force inpropor-tion to the transverse displacement w. Shear and axial deformations in the beam are negli-gible. Write the expression for the energy of the system.What are the essential and naturalboundary conditions for the beam? Find an approximate solution for the buckling loadsusing a polynomial Ritz basis.

(a) Write the expression for the energy of the system. This problem is essentially astandard beam buckling problem but with and elastic foundation only over half of thelength. Therefore, the energy is

E(w)=2ℓ0

12EI(w′′)2− P(w′)2 dx+ℓ

0

12 kw

2 dx

(b) What are the essential and natural boundary conditions for the beam?

The displacement and moment are zero at the left end and the moment and shearare zero at the right end. Therefore, the boundary conditions are

Essential: w(0)= 0

Natural: w′′(0)= w′′(2ℓ)= 0 (Moment),

w′′′(2ℓ)+ m2w′(2ℓ)= 0 (Shear)

(c) Find an approximate solution for the buckling loads using a two-term polynomialRitz basis. Let ξ= x∕ℓ so that dx= ℓ dξ. A two-term polynomial that satisfies theessential boundary condition w(0)= 0 is w= a1ξ+ a2 ξ

2. The derivatives of thedisplacement (with respect to x) is

w′ = 1ℓa1+ a22ξ w′′ = 1

ℓ22a2

with similar expressions for the virtual displacement. The virtual work functional is

G(w,w)= 2ℓ0

EIw′′w′′−Pw′w′ dx+ℓ0

kww dx= 0

Substituting the approximation into the virtual work functional

G(w,w)= 20

EI a2a2 2∕ℓ2 2− Pℓ2a1+2a2ξ a1+2a2ξ ℓdξ

+10

k a1ξ+a2ξ2 a1ξ+a2ξ2 ℓdξ= 0

k

ℓ

x

EI Rigid

ℓ

P

ℓ

k

Rigid


Carrying out the integrals suggests defining the matrices

K = EIℓ3

0 0

0 8G =− P

ℓ

4

4 323

2+kℓ

14

14

15

13

Let β≡ kℓ4∕EI and λ≡ Pℓ2∕EI. The equation governing the response of the systemis K−λG a= 0, which has nontrivial solution only if the determinant of the coeffi-cient matrix is zero. Hence, we can find the critical loads from the condition

det

β3−2λ

β4−4λ

β4−4λ 8+

β5− 32λ

3

= 0

This equation can be solved for the critical load as a function of the stiffness.

280. A flexible beam of length ℓ and modu-lus EI is connected to rigid beams of length ℓat both ends. The beams are supported by twolinear springs with modulus k= βEIℓ3,where β is a given constant. The beam is sup-ported as shown and is subjected to an endloadP. Shear and axial deformations in the beam are negligible.Write the (quadratic) ener-gy functional E and the virtual-work functional G for the system. What are the essentialand natural boundary conditions for the flexible beam? Find an approximate solution forthe lowest buckling load using a two-term polynomial Ritz basis. Express the result interms of β, i.e. Pcr(β).What is the buckling load for very large spring stiffnesses (i.e., asβ→1)? Does the approximation appear to make sense in the limit? Explain.

(a) Write the (quadratic) energy functional E and the virtual-work functional G for thesystem. The figure below shows the deformed position of the system (with all end dis-placements shown positive).

w(0)

w(x)

x

P

θ1

θ2

The geometry of the rigid bars gives the relations

θ2 = w′(ℓ) θ1 = w′(0)= w(0)ℓ


The energy of the system is the elastic energy (with the axial load term) for the flexiblebeam in the middle plus the energy of the two rigid portions. The axial load is constantand equal to P along the entire length. Hence,

E(w)=ℓ0

12EIw′′ 2− Pw′ 2 dx+ 1

2 kw(0) 2+ℓw′(ℓ) 2

− Pℓ1− cos θ1 − Pℓ1− cos θ2

Or, recognizing that θ1 = w′(0)= w(0)∕ℓ and θ2 = w′(ℓ) we have

E(w)=ℓ0

12EIw′′ 2− Pw′ 2 dx+ 1

2 kw(0) 2+ℓw′(ℓ) 2

− 12Pℓ w′(0) 2+ w′(ℓ) 2

The virtual work functional can be found by taking the directional derivative of theenergy in the direction of a virtual displacement w(x). Hence,

G(w,w)= ℓ0

EIw′′w′′ − Pw′w′ dx

+ kℓ2w′(0)w′(0)+w′(ℓ)w′(ℓ) − Pℓw′(0)w′(0)+w′(ℓ)w′(ℓ)

(b) What are the essential and natural boundary conditions for the flexible beam? Toobtain the natural boundary conditions we must take freebody diagrams of the tworigid segments, representing the section forces with the positive sign convention andincluding the forces in the springs.

P

M(0)

kℓw′(ℓ)

ℓV(0)+M(0)−ℓ kw(0)+ Pw(0)= 0

V(0)

R1

PM(ℓ)

V(ℓ)

R2kw(0)

−M(ℓ)+ Pℓw′(ℓ)−kℓw′(ℓ)= 0

PP

Hence the boundary conditions are

Essential: w(0)−ℓw′(0)= 0, w(ℓ)= 0

Natural: −EIℓw′′′(0)+Pℓw′(0)+ EIw′′(0)− kℓw(0)+ Pw(0)= 0

−EIw′′(ℓ)+ Pℓw′(ℓ)− kℓw′(ℓ)= 0

Note that the first essential boundary condition is the result of the kinematic constrainimposed by the rigid part on the left. It is the manifestation at the left end of the flex-ible beam of the pin at the left end of the rigid part.


(c) Find an approximate solution for the lowest buckling load using a two-termpolynomial Ritz basis. Express the result in terms of β, i.e., Pcr(β).

Let ξ= x∕ℓ so that dx= ℓ dξ. A two-term polynomial that satisfies the essentialboundary condition can be found by starting with a quartic polynomial and substitutingthe two essential boundary conditions. To wit,

w′ = 1ℓb+ 2cξ+ 3dξ2

w= a+ bξ+ cξ2+ dξ3

The essential boundary conditions are

EBC 1: w(ℓ)= 0 ⇒ a+ b+ c+ d= 0

EBC 2: w(0)−ℓw′(0) ⇒ a− b= 0

Therefore, we get a= b and c=−2b−d. Substitution these into the original expres-sion gives

w= b+ bξ+ −2b−d ξ2+ dξ3

= b1+ ξ− 2ξ2 + d ξ3− ξ2 ≡ bh1(ξ)+ dh2(ξ)

The base functions and their derivatives are

h1 = 1+ ξ− 2ξ2 h2 = ξ3− ξ2

h1′′ = −4ℓ2

h1′ =1ℓ1−4ξ h2′ =

1ℓ3ξ2−2ξ

h2′′ =1ℓ26ξ−2

Thus, the components of the stiffness and geometric stiffness matrices can be com-puted as

Kij= ℓ

0

EI hi′′hj′′ dx+ kℓ2 hi′(0)hj′(0)+ hi′(ℓ)hj′(ℓ)

Gij= P ℓ0

hi′hj′ dx+ Pℓ hi′(0)hj′(0)+ hi′(ℓ)hj′(ℓ)

Substituting the base functions yields

--4( 6ξ--2 ) ( 6ξ--2)2

16 −4 6ξ−2 dξ+ kℓ2K = EI

ℓ ℓ

0--3 1

10 --3

Let λ≡ Pℓ2∕EI and noting that β≡ kℓ3∕EI we have

−4 4

16 −4K = EI

ℓ+ βEIℓ −3 1

10 −3

k

ℓ

x

EI Rigid

ℓ

P


Similarly, for the geometric stiffness we get

1−4ξ 3ξ2−2ξ 3ξ2−2ξ 21−4ξ 2 1−4ξ 3ξ2−2ξ

G = Pℓ ℓ0

dξ+ Pℓ−3 1

10 −3

or, carrying out the integrals

−1∕3 2∕15

7∕3 −1∕3G = Pℓ

−3 1

10 −3+ Pℓ

−10∕3 17∕15

37∕3 −10∕3= Pℓ

− 4− 3β+ 103 λ 4+ β− 17

15λ

16+ 10β− 373 λ − 4− 3β+ 10

3 λEIℓK−G = EI

ℓ ≡ EIℓ A λ, β

Critical loads are computed from detA= 0 detA= 0 ⇒ Pcr(β).

(d) What is the buckling load for very large spring stiffnesses (i.e., as β→1)? Doesthe approximation appear to make sense in the limit? Explain.

From a physical point of view, if the springs are infinitely rigid then the bucklingload is exactly the same as a beam fixed at both ends

Pcr β⇒∞ =P⇒ Pcr= 4π2EI

ℓ2

The numerical approximation does not make sense in the limit because quadraticapproximation is needed?

281. A flexible beam of length ℓ and modulus EI iswelded to a rigid beam of length ℓ which rests on aspring ofmodulus k=2EI/ℓ3. The beam is supportedas shown and is subjected to an end loadP. Shear andaxial deformations in the beam are negligible. Writethe energy functional E and the virtual-work func-tionalG for the system.What are the essential and natural boundary conditions for the flex-ible beam? Find an approximate solution for the displacementw(x) using a two-term poly-nomial Ritz basis.

(a) Write the energy functional E and the virtual work functional G for the system.The energy functional of the system includes the standard part for the flexible beam,the energy of the spring, and the potential energy of the load for the rigid part

E(w)= 12ℓ0

EI(w′′)2−P(w′)2 dx+ 12 kℓ

2 w′(ℓ) 2− 12Pℓw′(ℓ)

2


The corresponding virtual work functional can be found by taking the directional deriv-ative of the energy in the direction of a virtual displacement

G(w,w)= ℓ0

EIw′′w′′−Pw′w′ dx+ kℓ2w′(ℓ)w′(ℓ)− Pℓw′(ℓ)w′(ℓ)

(b) What are the essential and natural boundary conditions for the flexible beam? Thenatural boundary conditions can be found by considering a freebody diagram of therigid piece as shown below.

PM(ℓ)

kℓw′(ℓ)M(ℓ)= Pℓw′(ℓ)− kℓ2w′(ℓ)

R

Hence, the boundary conditions are

Essential: w(0)= 0, w(ℓ)= 0

Natural: w′′(0)= 0, EIw′′(ℓ)− Pℓw′(ℓ)+ kℓ2w′(ℓ)= 0

(c) Find an approximate solution for the displacement w(x) using a two-termpolynomial Ritz basis. Let ξ= x∕ℓ so that dx= ℓ dξ. A two-term polynomial thatsatisfies the essential boundary condition is w= a1ξ1−ξ +a2ξ21−ξ . The basefunctions are, therefore,

h1 = ξ−ξ2 h2 = ξ2−ξ3

h1′′ = −2ℓ2 h2′′ =

1ℓ22−6ξ

h1′ =1ℓ1−2ξ h2′ =

1ℓ2ξ−3ξ2

Note that h1′(ℓ)=−1 and h2′(0)=−1. Substituting the base functions into thevirtual work functional gives

K =10

EIℓℓ2

4 −22−6ξ

−22−6ξ 2−6ξ 2

1 −1−1 1dξ+ kℓ2

Observing that k= 2EI∕ℓ3 and carrying out the integrals we get

K =EIℓ

4 2

2 4 + kℓ21 −1

−1 1 =EIℓ

6 0

0 6


2ξ−3ξ2 21−2ξ 2ξ−3ξ2

G =10

Pℓ1−2ξ 2 1 −1

−1 1dξ+ Pℓ1−2ξ 2ξ−3ξ2

k

ℓ

x

EIRigid

ℓ

P


or

4

5G = Pℓ

30

10+ Pℓ

305 30

−3030

−30 = Pℓ30 34

−2540

−25

Let λ≡ Pℓ2∕30EI. The governing equations K−G a= 0 has a nontrivial solutiononly if the determinant of the coefficient matrix is zero, Thus,

det6−34λ−25λ6−40λ

−25λ= 6−40λ 6−34λ −25λ2 = 0

which can be simplified to give the quadratic equation

36−444λ+1360λ2−25λ= 1335λ2−444λ+36= 0

that has the solution(s)

λ= 444 (444)2− (4)(1335)(36)(2)(1335)

= 0.192, 0.140

The displacement of the structure is only possible in a buckling mode. The first buck-ling mode is associated with the lowest buckling eigenvalue 0.140. Take a1 = 1 andcompute a2 from the equilibrium equations to give

6−40(0.140)−25(0.140)a2 = 0 ⇒ 0.4− 3.5a2 = 0 ⇒ a2= 0.114

Therefore the first mode shape is [ 1.000, 0.114].

282. A flexible beam of length ℓ and modu-lus EI is welded to a rigid beam of length ℓand rests on an elastic foundation ofmodulusk=20EI/ℓ4. The beam is simply supportedand is subjected to an end load P. The elasticfoundation accrues a transverse force in pro-portion to the transverse displacement w. Shear and axial deformations in the beam arenegligible. Write the energy functionalE and the virtual-work functionalG for the system.What are the essential and natural boundary conditions for the flexible beam? Find anapproximate solution for the displacement w(x) using a polynomial Ritz basis.

(a) Write the energy functional E and the virtual work functional G for the system.The energy can be established by considering the sketch below, which includes theforces due to the elastic foundation.


θ≡ w′(ℓ)

P

ξ

P

H(x) kθξ

V(x)

E(w)=ℓ0

12EIw′′ 2+ 1

2 k w2− 1

2 Pw′ 2 dx+ℓ

0

12 kθξ 2 dξ− 1

2Pℓθ2

= 12ℓ0

EI w′′ 2+ k w2− Pw′ 2 dx+ 16 kℓ

3θ2− 12 Pℓθ

2

The virtual work functional can be found by taking the directional derivative of theenergy in the direction of a virtual displacement. To wit,

G(w,w)= ℓ0

EIw′′w′′ + k ww− Pw′w′ dx+ 13 kℓ

3θθ− Pℓθθ

(b) What are the essential and natural boundary conditions for the flexible beam?

M(ℓ)= Pℓθ−ℓ0

ξ ⋅ kθξ dξ

= Pℓθ− 13 kθℓ

3

P

kθξ

V

M

Essential: w(0)= 0, w(ℓ)= 0

Natural: w′′(0)= 0, EIw′′(ℓ)+ 13kℓ3w′(ℓ)− Pℓw′(ℓ)= 0

Note that the shear boundary condition includes the effects of the elastic foundationunder the rigid part.

(c) Find an approximate solution for the displacement w(x) using a two-term polyno-mial Ritz basis. Let ξ= x∕ℓ so that dx= ℓ dξ. A two-term polynomial that satisfiesthe essential boundary condition is w= a1ξ1−ξ + a2ξ

21−ξ . The base functionsare, therefore,

h1 = ξ−ξ2 h2 = ξ2−ξ3

h1′′ = −2ℓ2 h2′′ =

1ℓ22−6ξ

h1′ =1ℓ1−2ξ h2′ =

1ℓ2ξ−3ξ2

The stiffness matrix can be computed from the base functions as

ℓ

P

EI

ℓ rigid

x


K =10

EIh′′ h′′ + kh h ℓdξ+ 13 kℓ

3h′(1) h′(1)

10

2K = EI

ℓ34

+ 20EIℓ32 1∕105

1∕601∕30

1∕30+ 20EIℓ3 1∕3

1∕31∕3

1∕3

16.857

9.000= EIℓ3

11.333

9.000

Similarly, the geometric stiffness can be computed as

G =10

Ph′ h′ ℓdξ + Pℓ h′(1) h′(1)

−2ξ 2

2ξ−3ξ2 2−2ξ 2ξ−3 ξ2

−2ξ 2ξ−3 ξ2 G = Pℓ10

ℓ dξ+ Pℓ1

11

1

4∕3

1

1+ Pℓ

1G = Pℓ

12∕15

1∕61∕3

1∕6= Pℓ

17∕15

7∕6

7∕6

The governing equations for the system are K−λG a= 0 which has a nontrivialsolution only if the determinant of the coefficient matrix is zero.

283. A flexible bar of length ℓ and bending modulus EI is weldedto a rigid bar of length ℓ. The structure is fixed at the bottom andsubjected to a compressive axial load P at the top as shown. Whatare the appropriate essential and natural boundary conditions forthis problem? Find an appropriate energy functional E(w) for thesystem, where w(x) is the transverse deflection of the flexible bar.Compute an approximation of the critical load of the system usingthe Ritz method and a polynomial basis function.

(a) What are the appropriate essential and natural boundary conditions for this prob-lem?

Essential: w(0)= 0, w′(0)= 0

Natural: V(ℓ)=−EIw′′′(ℓ)+ Pw′(ℓ)= 0

M(ℓ)= EIw′′(ℓ)= Pℓ w′(ℓ)

(b) Find an appropriate energy functional E(w) for the system, where w(x) is the trans-verse deflection of the flexible bar.

ℓ

P

ℓ

2ℓ

EI EI

EI

4EI

roller


E(w)= 12ℓ0

EI w′′ 2− P w′ 2 dx− 12 Pℓ w′(ℓ)

2

(c) Compute an approximation of the critical load of the system using the Ritz meth-od and a one-term polynomial basis function.

G(w,w)= ℓ0

EI w′′w′′ − P w′w′ dx− p ℓ w′(ℓ)w′(ℓ)

Let w(x)= a x2. Carrying out the computation we get

G(a, a)= a ℓ0

EI(4)− P(4x2) dx− P ℓ(4ℓ2) a = 0 ∀ a

Carrying out the integrals and noting the fundamental theorem of the calculus of varia-tions we have

4EIℓ − Pℓ343+ 4 a= 0

This equation has a trivial solution for the value

Pcr=3EI4ℓ2

which is (an approximation of) the critical buckling load of the system.

284. A beam of flexural modulus 4EI carries theload P to the frame as shown in the figure. Theframe is made of two columns pinned together bya beam atmidheight. The framemembers all havelength 2ℓ and flexural modulus EI as shown. Theforce P is applied directly above the left column.The members have axial modulus EA≫EI/ℓ2.Estimate the smallest buckling capacityPcr of thestructure using the Ritz method. Resolve theproblem assuming that the load can be placedanywhere along the top beam.

Note that the right column has no axial load on it and hence only adds lateral stiff-ness to the left column. The stiffness of the right beam is essentially equivalent to an


equivalent spring of modulus k= 3EI∕ℓ3 (i.e., the stiffness of a cantilever beam).Theappropriate virtual work functional is

G(w,w)= 2ℓ0

EIw′′w′′ − Pw′w′ dx+ 3EIℓ3 w(ℓ)w(ℓ)

Let ξ= x∕ℓ and consider a cubic approximation of the displacement that satisfies theessential boundary conditions

w= a1ξ2+ a2ξ

3 , w′ = 1ℓ2a1 ξ+3a2ξ2 , w′′ = 1

ℓ22a1+6a2 ξ

4

36ξ212ξ

12ξK = EI

ℓ320

dξ+ 3EIℓ3 1

11

1


11

3

3+ EIℓ3

3K = EI

ℓ3 396

248

24= EIℓ3 99

27

27


4ξ2

9ξ46ξ3

6ξ3G = 1

ℓ2

0

dξ= 1ℓ

2432∕324 288∕5

Let λ≡ Pℓ2∕EI

K− λG = EIℓ3

11− 323 λ 27−24λ

27−24λ 99− 2885λ

The condition for a nontrivial solution to the governing equation is

det K−λG = 11− 323λ99− 288

5λ− 27− 24λ 2= 0

which simplifies to λ2−10.25λ+9.375= 0, which gives λ= 1.015, 9.235. There-fore, the lowest critical load is

Pcr= 1.015EIℓ2

If the load can be placed at any point along the top, the both columns have axialload and we must formulate the problem a bit differently. Assume that the load isplaced a distance βℓ from the right column line. Then the axial force in the left col-umn is βP and the axial force in the right column is 1−β P. We can again take apolynomial approximation for the displacement in the two columns, wL= a1ξ

2+a2ξ3and wR= a3ξ

2+a4ξ3. The constraint on the two displacement fields is that


wL(1)= wR(1). Thus, a1+a2 = a3+a4. We can, therefore, eliminate one variablewith this constraint, a4 = a1+a2−a3. This allows us to writewR= a1ξ

3+a2ξ3+a3 ξ2−ξ3 .Let a= a1, a2, a3 and define base functions as follows

hL′ = 2ξ, 3ξ2, 0 ∕ℓ hR′ = 3ξ2, 3ξ2, 2ξ−3ξ2 ∕ℓ

hL= ξ2, ξ3, 0 hR= ξ3, ξ3, ξ2−ξ3

hL′′ = 2, 6ξ, 0 ∕ℓ2 hR′′ = 6ξ, 6ξ, 2−6ξ ∕ℓ2

Now, wL= a ⋅ hL(ξ) and wR= a ⋅ hR(ξ) and the displacement functions automatical-ly satisfy the kinematic constraint. The virtual work functional is

G(wL,wR,wL,wR)=2ℓ

0

EIwL′′wL′′ − βPwL′wL′ dx

+2ℓ0

EIwR′′wR′′ − 1−β PwR′wR′ dx

Substituting the Ritz approximations into the virtual work functional yields the follow-ing stiffness matrix

K =20

EI hL′′ hL′′+ hR′′ hR′′ ℓdξ

and geometric stiffness matrices

GL= 2

0

P hL′ hL′ ℓdξ, GR= 2

0

P hR′ hR′ ℓdξ

4+36ξ2 12ξ 1+3 ξ

K = EIℓ320

dξ

12ξ 1−3 ξ

72ξ2

2−6ξ 212ξ 1−3 ξ 12ξ 1+3 ξ

12ξ 1−3 ξ 12ξ 1−3 ξ

and G = βGL+1−β GR with

4ξ2 6ξ3

GL= Pℓ20

dξ

0

9ξ4

0

06ξ3

0 0

9ξ4

GR= Pℓ20

dξ

3ξ3 2−3ξ

ξ22−3ξ 2

9ξ4

9ξ4 9ξ4 3ξ3 2−3ξ

3ξ3 2−3ξ 3ξ3 2−3ξ


Now, the critical load Pcr(β) can be found by solving the equation

det K−G(P, β) = 0

285. Resolve Problem 284 by solving the classical differential equations and boundaryconditions.

No solution available.

Chapter 12Numerical Computationsfor Nonlinear Problems

286. Modify the program NEWTON to account for initial geometric imperfections in thetwo-bar rigid linkage connected by rotational springs.

The governing equations for the rigid linkage with imperfections is given as

g1(θ) ≡ 2 θ1−θo1 − θ2−θo2 − λ sin θ1 = 0

g2(θ) ≡ −θ1−θo1 + θ2−θo2 − λ sin θ2 = 0

where θo = θo1, θo2 represent the imperfections. Because g(θ, λ)= 0 is linear inthe imperfections, the Hessian matrix is identical to the one for the system withoutimperfections, given in the text. The modification to the program Newton are sketchedbelow:

The input data must be altered. We read in xo(1) and xo(2) as the values of theimperfections. The initial value of the load xo(3) is assumed to be zero.

c.... Read problem parametersread(5,1000) tol,alpharead(5,1001) maxsteps,maxitread(5,1000) xo(1),xo(2)read(5,1000) x(1), x(2)xo(3) = 0.0write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3)

Before starting the march along the load path the imperfection values are stored in thearray because the array is subsequently used as an estimate of the next equilibriumstate.

c.... Initialize values for zero load,set direction to next trial state[code that is omitted is unchanged]y(1) = xo(1)y(2) = xo(2)write(6,2001) n,xo(1),xo(2),xo(3)

ℓ

kk

ℓ ℓ

P


c.... Compute MAXSTEPS points along the Equilibrium Path

Finally, the expression for the residual of the equilibrium equations must be altered toaccount for the imperfections.

c.... Compute equilibrium and constraint at current stateb(1) = 2.0*(x(1)-y(1)) - (x(2)-y(2)) - x(3)*dsin(x(1))b(2) = -(x(1)-y(1)) + (x(2)-y(2)) - x(3)*dsin(x(2))b(3) = 0.0

The following example is indicative of the revised program.

Convergence tolerance : .1000E-07Arc length parameter : .5000Number of load steps : 10Maximum number of iterations : 50Initial theta 1, theta 2, load : .1000 .1000 .0000Next theta 1, theta 2, load : .2000 .2000 .2000

n Theta 1 Theta 2 Load EV 1 EV 2 NU || b ||0 .10000 .10000 .000001 .31675 .44218 .29314 .11324 2.34325 6 .1412E-132 .58101 .85997 .36811 .12271 2.32940 5 .2429E-083 .85255 1.27355 .44032 .20617 2.37511 3 .6349E-084 1.13333 1.67414 .54372 .35113 2.47461 4 .2429E-115 1.42284 2.04864 .70474 .57000 2.65020 4 .3419E-096 1.70943 2.37312 .95490 .88476 2.93375 5 .9155E-157 1.96366 2.61807 1.30898 1.30065 3.33413 5 .4957E-138 2.16156 2.77878 1.73910 1.78270 3.81187 5 .1212E-139 2.30716 2.87932 2.20675 2.29141 4.32193 4 .4459E-0810 2.41533 2.94387 2.69062 2.80765 4.84222 4 .8173E-10

287. Modify the programNEWTON to analyzethe three-bar rigid linkage shown. The bars arehinged together and are restrained by elasticsprings that resist vertical motion. The springsaccrue force in proportion to their extension,with modulus k. The system is subjected to anaxial force P.

ℓ

ℓℓ

Pφ1φ2

ℓ cosφ1+ cosφ2+ cos γ

γ

Let us describe the motion of the linkage with the independent angles φ1 and φ2 asshown in the sketch. The angle γ(φ) depends upon φ1 and φ2 through the constraintrelationship

sin γ = sinφ1+ sinφ2

Chapter 12 Numerical Computation for Nonlinear Problems 361

The energy of the system is

E(φ) = 12kℓ2 sin2φ1+ sin2 γ + Pℓ cosφ1+ cosφ2+ cos γ

The virtual work functional can be computed by taking the directional derivative of theenergy functional. The result, after dividing by kℓ2 and defining λ≡ P∕kℓ is

G(φ,φ) = sinφ1 cosφ1φ1+ sin γ cos γDγ ⋅ φ

− λ sinφ1φ1+ sinφ2φ2+ sin γDγ ⋅ φ

where Dγ ⋅ φ can be computed from the constraint equation as

cos γDγ(φ) ⋅ φ = cosφ1φ1+ cos φ2φ2

Thus, the equilibrium equations are

g1(φ) ≡ 2 sinφ1+ sinφ2 cos φ1− λ sinφ1+ tan γ cos φ1 = 0

g2(φ) ≡ sinφ1+ sinφ2 cosφ2− λ sinφ2+ tan γ cosφ2 = 0

The Hessian A(φ, λ)= ∇φg(φ, λ) can be computed by taking the derivatives of thefunction g(φ) the components are

A11 = cos 2φ1− sinφ1 sinφ2−λ

cos γcos(φ1+γ)+ cos2 φ1 sec

2 γ

A12 = cosφ1 cosφ2−λ

cos3 γcos φ1 cosφ2 = A21

A22 = cos 2φ2− sinφ1 sinφ2−λ


2 γ

The components of ∇λg(φ, λ) can be computed easily from the equilibrium equationsas

A13 = − sinφ1− tan γ cosφ1A23 = − sinφ2− tan γ cosφ2

The equations associated with the arc-length constraint are identical to the exampleproblem in the text and need not be modified. These equations can be directly imple-mented into the program NEWTON.

program Newtonimplicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3),xo(3)

C *-----------------------------------------------------------*C | Problem 287. Fundamentals of Structural Mechanics |C *-----------------------------------------------------------*

c.... Read problem parametersread(5,1000) tol,alpharead(5,1001) maxsteps,maxitread(5,1000) xo(1),xo(2),xo(3)read(5,1000) x(1), x(2), x(3)write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3)

c.... Initialize values for load step zero,set next trial staten = 0


sum = 0.0do j = 1,3b(j) = x(j) - xo(j)sum = sum + b(j)*b(j)

end dosum = alpha/dsqrt(sum)do j = 1,3x(j) = xo(j) + sum*b(j)

end dowrite(6,2001) n,xo(1),xo(2),xo(3)

c.... Compute MAXSTEPS points along the Equilibrium Pathdo 2 n = 1,maxsteps

c.... Perform Newton iteration at each load stepnu = 0

1 nu = nu + 1

c.... Compute equilibrium and constraint at current states1 = dsin(x(1))c1 = dcos(x(1))s2 = dsin(x(2))c2 = dcos(x(2))sgamma = s2 + s1if (dabs(sgamma).lt.1.d0) thengamma = dasin(sgamma)

elsestop ’Arcsine of value greater than 1’

end iftang = dtan(gamma)b(1) = 2.0*s1*c1 + s2*c1 - x(3)*(s1 + tang*c1)b(2) = s1*c2 + s2*c2 - x(3)*(s2 + tang*c2)b(3) = 0.0do k = 1,3

b(3) = b(3) + (x(k) - xo(k))**2end dob(3) = b(3) - alpha**2

c.... Compute norm of residual for convergence testtest = dsqrt(b(1)**2 + b(2)**2 + b(3)**2)

c.... Compute eigenvalues of Tangent stiffness matrixc2p1 = dcos(2.d0*x(1))c2p2 = dcos(2.d0*x(2))secg = 1.d0 + tang**2cosg = dcos(gamma)f1 = dcos(x(1) + gamma)f2 = dcos(x(2) + gamma)aa = 2.0*c2p1 - s1*s2 - x(3)*(f1 + c1*c1*secg)/cosgbb = c2p2 - s1*s2 - x(3)*(f2 + c2*c2*secg)/cosgab = c1*c2 - x(3)*c1*c2/(cosg**3)det = aa*bb - ab*abeig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det))eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det))

c.... Compute Hessian matrixA(1,1) = aaA(1,2) = abA(1,3) = -s1 - tang*c1A(2,1) = abA(2,2) = bbA(2,3) = -s2 - tang*c2A(3,1) = 2.0*(x(1) - xo(1))A(3,2) = 2.0*(x(2) - xo(2))A(3,3) = 2.0*(x(3) - xo(3))

c... Update state vectorcall invert(A,3,3)do j=1,3do k=1,3x(j) = x(j) - A(j,k)*b(k)

end doend do


c.... Test for convergence, if successful output values and updateif ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test

c.... Set values for previous converged state and guess at next statedo j=1,3temp = xo(j)xo(j) = x(j)x(j) = 2.0*x(j) - temp

end do

2 continuestop ’Maximum number of steps exhausted’

1000 format(3f10.0)1001 format(2i10)2000 format(’ Problem 287. Fundamentals of Structural Mechanics’//

* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Initial Phi 1, Phi 2, load : ’,3f12.4/* 5x, ’ Next Phi 1, Phi 2, load : ’,3f12.4//* ’ n’ 5x,’ Phi 1’,5x,’ Phi 2’,8x,’Load’,* 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’)

2001 format(i5,5f12.5,i5,e12.4)end

The first branch can be computed starting at the point (0, 0, 13) and moving in the di-rection of the first mode (1, --2).

Problem 287. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .2000Number of load steps : 15Maximum number of iterations : 50Initial Phi 1, Phi 2, load : .0000 .0000 .3333Next Phi 1, Phi 2, load : .1000 -.2000 .3000

n Phi 1 Phi 2 Load EV 1 EV 2 NU || b ||0 .00000 .00000 .333331 .08914 -.17899 .32933 -.00957 1.64334 4 .9411E-102 .17638 -.35856 .31730 -.03789 1.57813 4 .5885E-143 .25966 -.53927 .29714 -.08382 1.48476 4 .1491E-134 .33663 -.72164 .26856 -.14541 1.38399 4 .6271E-135 .40446 -.90600 .23096 -.21979 1.30067 4 .3693E-126 .45984 -1.09215 .18321 -.30276 1.26028 4 .2486E-117 .49935 -1.27893 .12361 -.38780 1.28566 4 .1457E-108 .52030 -1.46382 .05029 -.46444 1.39372 4 .5250E-109 .52208 -1.64335 -.03783 -.51836 1.59201 4 .8426E-1010 .50661 -1.81458 -.14001 -.53690 1.87669 4 .5235E-1011 .47733 -1.97650 -.25370 -.51619 2.23466 4 .1345E-1012 .43761 -2.13024 -.37531 -.46183 2.64809 4 .1611E-1113 .38979 -2.27830 -.50097 -.38421 3.09761 4 .9048E-1314 .33504 -2.42396 -.62661 -.29412 3.56292 4 .3292E-1415 .27347 -2.57097 -.74742 -.20121 4.02101 4 .6629E-14

The computation of the equilibrium path emanating from the second bifurcation pointcan fail under the previous parameterization because φ2 = 0 (by symmetry) and γeventually exceeds π∕2 making the arcsine function ambiguous. We can parameterizethe problem to be more favorable for this case, but then the algorithm fails when com-puting the equilibrium path emanating from the first bifurcation point.


ℓ

ℓℓ

Pφ1φ2

ℓ cosφ1+ cosφ2+ cos γ

γ

Let us describe the motion of the linkage with the independent angles φ1 and φ2 . Theangle γ(φ) depends upon φ1 and φ2 through the constraint relationship

sin γ = sinφ2− sinφ1

The energy of the system is

E(φ) = 12kℓ2 sin2φ1+ sin2φ2 + Pℓcos φ1+ cos φ2+ cos γ

The virtual work functional can be computed by taking the directional derivative of theenergy functional. The result, after dividing by kℓ2 and defining λ≡ P∕kℓ is

G(φ,φ) = sinφ1 cosφ1φ1+ sinφ2 cosφ2φ2− λ sinφ1φ1+ sinφ2φ2+ sin γDγ ⋅ φ

where Dγ ⋅ φ can be computed from the constraint equation as

cos γDγ(φ) ⋅ φ ≡ cosφ2φ2− cos φ1φ1

Thus, the equilibrium equations are

g1(φ) ≡ sinφ1 cosφ1− λ sinφ1− tan γ cosφ1 = 0

g2(φ) ≡ sinφ2 cosφ2− λ sinφ2+ tan γ cosφ2 = 0

The Hessian A(φ, λ)= ∇φg(φ, λ) can be computed by taking the derivatives of thefunction g(φ) the components are

A11 = cos 2φ1−λ

cos γcos(φ1−γ)+ cos2 φ1 sec

2 γ

A12 =λ

cos3 γcos φ1 cos φ2 = A21

A22 = cos 2φ2−λ


2 γ

The components of ∇λg(φ, λ) can be computed easily from the equilibrium equationsas

A13 = − sinφ1+ tan γ cosφ1A23 = − sinφ2− tan γ cosφ2

The equations associated with the arc-length constraint are identical to the exampleproblem in the text and need not be modified. These equations can be directly imple-


mented into the program NEWTON. The differences from the previous parameterizationcan be observed in the code segment given below. All of the remaining code is thesame.

c.... Compute equilibrium and constraint at current states1 = dsin(x(1))c1 = dcos(x(1))s2 = dsin(x(2))c2 = dcos(x(2))sgamma = s2 - s1if (dabs(sgamma).lt.1.d0) thengamma = dasin(sgamma)

elsestop ’Arcsine of value greater than 1’

end iftang = dtan(gamma)b(1) = s1*c1 - x(3)*(s1 - tang*c1)b(2) = s2*c2 - x(3)*(s2 + tang*c2)b(3) = 0.0do k = 1,3



c.... Compute eigenvalues of Tangent stiffness matrixc2p1 = dcos(2.d0*x(1))c2p2 = dcos(2.d0*x(2))secg = 1.d0 + tang**2cosg = dcos(gamma)aa = c2p1 - x(3)*(dcos(x(1)-gamma) + c1*c1*secg)/cosgbb = c2p2 - x(3)*(dcos(x(2)+gamma) + c2*c2*secg)/cosgab = x(3)*c1*c2/(cosg**3)det = aa*bb - ab*abeig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det))eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det))

c.... Compute Hessian matrixA(1,1) = aaA(1,2) = abA(1,3) = -s1 + tang*c1A(2,1) = abA(2,2) = bbA(2,3) = -s2 - tang*c2A(3,1) = 2.0*(x(1) - xo(1))A(3,2) = 2.0*(x(2) - xo(2))A(3,3) = 2.0*(x(3) - xo(3))

The first branch can be computed starting at the point (0, 0, 13) and moving in the di-rection of the first mode (--1, 1). Note that the eigenvectors of the linearized problemare slightly different for this parameterization. As you can see, the solution fails whenthe center bar goes to vertical because the constraint causes a computational error try-ing to find the arcsine of a number greater than one.

Problem 287. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .0500Number of load steps : 40Maximum number of iterations : 50Initial Phi 1, Phi 2, load : .0000 .0000 .3333Next Phi 1, Phi 2, load : -.1000 .1000 .3000

n Phi 1 Phi 2 Load EV 1 EV 2 NU || b ||0 .00000 .00000 .333331 -.03535 .03535 .33271 -.00376 .66583 4 .1263E-082 -.07068 .07068 .33082 -.01519 .66336 3 .2531E-093 -.10597 .10597 .32766 -.03474 .65931 3 .2754E-09


4 -.14118 .14118 .32318 -.06325 .65378 3 .3174E-095 -.17629 .17629 .31732 -.10206 .64693 3 .3902E-096 -.21127 .21127 .30999 -.15330 .63898 3 .5179E-097 -.24606 .24606 .30110 -.22026 .63022 3 .7562E-098 -.28060 .28060 .29046 -.30826 .62102 3 .1247E-089 -.31482 .31482 .27786 -.42611 .61190 3 .2409E-0810 -.34857 .34857 .26298 -.58924 .60353 3 .5721E-0811 -.38166 .38166 .24538 -.82657 .59687 4 .1250E-1312 -.41376 .41376 .22440 -1.19748 .59328 4 .1398E-1213 -.44427 .44427 .19914 -1.83910 .59472 4 .2833E-1114 -.47217 .47217 .16843 -3.12288 .60402 4 .1378E-0915 -.49571 .49571 .13113 -6.30658 .62467 5 .6830E-1416 -.51267 .51267 .08726 -17.51073 .65927 6 .4016E-10

STOP Arcsine of value greater than 1

The second branch is easily computed with this parameterization as shown below. Notethat the solution starts at the bifurcation point (0,0,1) and moves in the direction of thesecond mode (1,1).

Problem 287. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .2000Number of load steps : 15Maximum number of iterations : 50Initial Phi 1, Phi 2, load : .0000 .0000 1.0000Next Phi 1, Phi 2, load : .1000 .1000 .9000

n Phi 1 Phi 2 Load EV 1 EV 2 NU || b ||0 .00000 .00000 1.000001 .14125 .14125 .99004 -1.96066 -.01982 6 .8184E-092 .28114 .28114 .96074 -1.85054 -.07698 4 .5193E-113 .41859 .41859 .91366 -1.69064 -.16522 4 .2933E-114 .55289 .55289 .85101 -1.50841 -.27578 4 .1198E-115 .68376 .68376 .77520 -1.33076 -.39906 4 .3751E-126 .81124 .81124 .68860 -1.17886 -.52583 4 .9459E-137 .93559 .93559 .59335 -1.06573 -.64794 4 .1975E-138 1.05722 1.05722 .49130 -.99580 -.75862 4 .3361E-149 1.17659 1.17659 .38407 -.96580 -.85249 3 .8605E-0810 1.29423 1.29423 .27306 -.96616 -.92544 3 .2620E-0811 1.41063 1.41063 .15948 -.98268 -.97457 3 .5739E-0912 1.52633 1.52633 .04446 -.99820 -.99802 3 .6039E-1013 1.64182 1.64182 -.07096 -.99496 -.99425 3 .3481E-1214 1.75763 1.75763 -.18575 -.96550 -.95268 3 .2276E-1115 1.87427 1.87427 -.29884 -.91069 -.85732 3 .1135E-09

288. Implement the constraint c(θ, λ)= detA(θ, λ)= 0 into the programNEWTON to lo-cate bifurcation points exactly. At the bifurcation point, compute the eigenvectors of thetangent stiffness matrix and switch to another equilibrium branch.

Let us implement the constraint for the example problem in the text. The determinantof the Hessian matrix can be computed as

c(θ, λ) ≡ detA(θ, λ) = 2−λ cos θ1 1−λ cos θ2 − 1

The derivatives with respect to θ1, θ2, and λ are, respectively,

A31 = λ sin θ1 1−λ cos θ2

A33 =− cos θ1 1−λ cos θ2 − cos θ2 2−λ cos θ1

A32 = λ sin θ2 2−λ cos θ1


These equations can be directly implemented into the program NEWTON. The programlisting is given below. Note that we do not march along a load path, but rather searchdirectly for a bifurcation point.

program Newtonimplicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3)

C *-----------------------------------------------------------*C | Problem 288. Fundamentals of Structural Mechanics |C *-----------------------------------------------------------*

c.... Read problem parametersread(5,1000) tolread(5,1001) maxitread(5,1000) x(1),x(2),x(3)write(6,2000) tol,maxit,(x(j),j=1,3)


1 nu = nu + 1

c.... Compute equilibrium and constraint at current stateaa = 2.0 - x(3)*dcos(x(1))bb = 1.0 - x(3)*dcos(x(2))det = aa*bb - 1.0b(1) = 2.0*x(1) - x(2) - x(3)*dsin(x(1))b(2) = -x(1) + x(2) - x(3)*dsin(x(2))b(3) = det


c.... Compute eigenvalues of Tangent stiffness matrixeig1 = 0.5*((aa+bb) - dsqrt((aa+bb)**2-4*det))eig2 = 0.5*((aa+bb) + dsqrt((aa+bb)**2-4*det))

c.... Compute Hessian matrixA(1,1) = aaA(1,2) = -1.0A(1,3) = -dsin(x(1))A(2,1) = -1.0A(2,2) = bbA(2,3) = -dsin(x(2))A(3,1) = x(3)*dsin(x(1))*bbA(3,2) = x(3)*dsin(x(2))*aaA(3,3) = - dcos(x(1))*bb - dcos(x(2))*aa

c.... Update state vectorcall invert(A,3,3)do j=1,3do k=1,3x(j) = x(j) - A(j,k)*b(k)

end doend do

c.... Test for convergence, if successful output values and updateif ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,teststop ’Critical point found’


* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Initial theta 1, theta 2, load : ’,3f12.4//* ’ n’ 5x,’Theta 1’,5x,’Theta 2’,8x,’Load’,* 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’)

2001 format(i5,5f12.5,i5,e12.4)end

P

k

θ1

ℓ

ℓ

ℓ( cos θ1+ cos θ2

P

k

θ2

ℓk

θ3

+ cos θ3)


The program can, of course, be used to find the critical points on the trivial equilibriumpath. However, there are other critical points that are less obvious. Using the startingvalues given below, we find one of those critical points.

Problem 288. Fundamentals of Structural Mechanics

Convergence tolerance : .1000E-07Maximum number of iterations : 50Initial theta 1, theta 2, load : .2000 -1.5000 5.0000

n Theta 1 Theta 2 Load EV 1 EV 2 NU || b ||0 .84928 4.50308 -3.73532 .00000 4.69111 10 .7174E-14

This solution corresponds to the position shown in the sketch. The top bar has rotatedaround nearly three quarters of a complete circle.

λ= 3.73

θ1 = 0.85

θ2 = 4.5

289. Modify the program NEWTON toanalyze the three-bar rigid linkageshown below. This structure has threedegrees of freedom. The bars are hingedtogether and are restrained by elasticsprings that resist relative rotation. Thesprings accrue force in proportion totheir extension, with modulus k. Thesystem is subjected to an axial force P.

Lateral loads can be viewed as im-perfections to a purely axial loadingsystem. Modify the equations of equi-librium to allow the applications of the lateral loads Á1P, Á2P, and Á3P at locations ℓ, 2ℓ,and 3ℓ respectively, where Ái is a fixed value recording the ratio of the lateral load to theaxial load. Implement the load imperfections in the program.

The energy of the system, including the transverse loads (acting to the left), is

E(θ) = 12k θ21+ θ2−θ1 2+ θ3−θ2 2 + Pℓcos θ1+ cos θ2+ cos θ3

− Pℓ Á1+Á2+Á3 sin θ1+ Á2+Á3 sin θ2+ Á3 sin θ3

The virtual work functional can be computed by taking the directional derivative of theenergy functional. The result, after dividing by k and defining λ≡ Pℓ∕k is


G(θ, θ) = θ1θ1+ θ2−θ1 θ2−θ1 + θ3−θ2 θ3−θ2

− λ sin θ1θ1+ sin θ2θ2+ sin θ3θ3

− λ Á1+Á2+Á3 cos θ1θ1+ Á2+Á3 cos θ2θ2+ Á3 cos θ3θ3

The equilibrium equations can be obtained as

g1(θ) ≡ 2θ1− θ2− λ sin θ1+ Á123 cos θ1 = 0

g2(θ) ≡ −θ1+ 2θ2− θ3− λ sin θ2+ Á23 cos θ2 = 0

g3(θ) ≡ −θ2+ θ3− λ sin θ3+ Á3 cos θ3 = 0

where Á123 ≡ Á1+Á2+Á3 and Á23 ≡ Á2+Á3. The constraint is similar to the example inthe text, except that there are three degrees of freedom

c(θ, λ) ≡ ‖ θ−θn ‖2+ λ−λn 2+ α2 = 0

The components of the matrix A are given by

A11 = 2−λ cos θ1−Á123 sin θ1

A22 = 2−λ cos θ2−Á23 sin θ2

A33 = 1−λ cos θ3−Á3 sin θ3

The off-diagonal terms are A12 = A23 =−1 and A13 = 0, with corresponding sym-metric terms below the diagonal. The derivatives with respect to λ give

A14 =−sin θ1+Á123 cos θ1



The terms associated with the constraint are the same as before, extended to one moredegree of freedom. These equations can be directly implemented into the programNEWTON. The program listing is given below.

program Newtonparameter (ndm=4)implicit double precision (a-h,o-z)dimension A(ndm,ndm),b(ndm),x(ndm),xo(ndm),sn(3),cn(3)dimension H(3,3),V(3,3),e(3)

C *-----------------------------------------------------------*C | Problem 289. Fundamentals of Structural mechanics |C *-----------------------------------------------------------*

c.... Read problem parametersread(5,1000) tol,alpharead(5,1001) maxsteps,maxitread(5,1000) eps1, eps2, eps3read(5,1000) xo(1),xo(2),xo(3),xo(4)read(5,1000) x(1), x(2), x(3), x(4)write(6,2000) tol,alpha,maxsteps,maxit,eps1,eps2,eps3,* (xo(j),j=1,4),(x(j),j=1,4)


c.... Initialize values for load step zero,set next trial staten = 0sum = 0.0do j = 1,ndmb(j) = x(j) - xo(j)sum = sum + b(j)*b(j)

end do

sum = alpha/dsqrt(sum)do j = 1,ndmx(j) = xo(j) + sum*b(j)

end do

eps123 = eps1 + eps2 + eps3eps23 = eps2 + eps3write(6,2001) n,(xo(i),i=1,4)



1 nu = nu + 1

c.... Compute equilibrium and constraint at current statedo j=1,3sn(j) = dsin(x(j))cn(j) = dcos(x(j))

end dob(1) = 2.0*x(1) - x(2) - x(4)*(sn(1)+ eps123*cn(1))b(2) = -x(1) + 2.0*x(2) - x(3) - x(4)*(sn(2)+ eps23*cn(2))b(3) = - x(2) + x(3) - x(4)*(sn(3)+ eps3*cn(3))b(4) = 0.0do k = 1,ndmb(4) = b(4) + (x(k)-xo(k))**2

end dob(4) = b(4) - alpha**2

c.... Compute norm of residual for convergence testtest = dsqrt(b(1)**2 + b(2)**2 + b(3)**2 + b(4)**2)

c.... Compute eigenvalues of Tangent stiffness matrixA(1,1) = 2.0 - x(4)*(cn(1)- eps123*sn(1))A(1,2) = -1.0A(1,3) = 0.0A(2,1) = -1.0A(2,2) = 2.0 - x(4)*(cn(2)- eps23*sn(2))A(2,3) = -1.0A(3,1) = 0.0A(3,2) = -1.0A(3,3) = 1.0 - x(4)*(cn(3)- eps3*sn(3))do j=1,3do k=1,3H(j,k) = A(j,k)

end doe(j) = 1.0

end docall eigens(H,V,e,3,6,nr)

c.... Compute border of Hessian matrixA(1,4) = -(sn(1) + eps123*cn(1))A(2,4) = -(sn(2) + eps23*cn(2))A(3,4) = -(sn(3) + eps3*cn(3))A(4,1) = 2.0*(x(1)-xo(1))A(4,2) = 2.0*(x(2)-xo(2))A(4,3) = 2.0*(x(3)-xo(3))A(4,4) = 2.0*(x(4)-xo(4))

c... Update state vectorcall invert(A,4,4)do j=1,4do k=1,4x(j) = x(j) - A(j,k)*b(k)

end doend do


c.... Test for convergence, if successful output values and updateif ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,(x(i),i=1,4),(e(i),i=1,3),nu,test

c.... Set values for previous converged state and guess at next statedo j=1,4temp = xo(j)xo(j) = x(j)x(j) = 2.0*x(j) - temp

end do



* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Eps 1, Eps 2, and Eps 3 : ’,3f12.4/* 5x, ’ Initial theta 1, 2, 3, and load : ’,4f12.4/* 5x, ’ Next theta 1, 2, 3, and load : ’,4f12.4//* 4x, ’n’,5x,’Theta 1’,5x,’Theta 2’,5x,’Theta 3’,8x,’Load’,* 8x, ’EV 1’,8x,’EV 2’,8x,’EV 3’,’ NU’,4x,’|| b ||’)

2001 format(i5,7f12.5,i5,e12.4)end

Moving from a two by two system to a three by three system requires that we get alittle more general with our solution of the eigenvalue problem at each step. The sub-routine EIGENS solves the eigenvalue problem

Hφ = mEφ

where H is a symmetric N× Nmatrix and E is a diagonal N× Nmatrix. The diago-nal elements of E are stored as a vector. For our purposes E= I. The eigenvalue prob-lem is solved by the Jacobi algorithm wherein off-diagonal elements are zeroed byperforming a rotation. The eigenvectors are returned in the array U and the eigenvaluesare returned in the array E.

subroutine eigens(h,u,e,n,ns,nr)C *---------------------------------------------------------------*C | Eigenvalues and eigenvectors by Jacobi’s method |C | solve H.u = m.E.u where E is a diagonal matrix |C | the eigenvectors are stored in u, eigenvalues in E |C | n is the dimension of the matrices H and u |C | ns is the number of digits of accuracy required (input) |C | nr is the number of Jacobi rotations performed (output) |C *---------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension h(n,n),u(n,n),e(n)

test=1.0/10.**(2*ns)nn=n-1nr=0nrlm=5*n**2toler=0.1

c.... Normalize to standard eigenvalue problemdo 10 i=1,n

10 e(i)=1.0/dsqrt(e(i))do 30 i=1,ndo 20 j=1,nh(i,j)=e(i)*h(i,j)*e(j)

20 u(i,j)=0.030 u(i,i)=1.0


c.... Reduce matrix50 xmax=0.0

do 700 ii=1,nnjl=ii+1do 600 jj=jl,n

c.... Check if rotation is requiredhii=h(ii,ii)hij=h(ii,jj)hjj=h(jj,jj)d=dabs(hii*hjj)h2=hij*hijif(h2.gt.xmax*d) xmax=h2/dif(h2.lt.toler*d) go to 600

c.... Compute tan, sin and cosnr=nr+1ht=.5*(hii-hjj)/hijtn=-ht-dsign(dsqrt(ht*ht+1.0),ht)cs =1.0/dsqrt (1.+tn **2)sn=cs*tnc2=cs**2s2=sn**2

c.... Reduce ii,jj element to zeroht=2.*hij*cs*snh(ii,jj)=0.0h(ii,ii)= hii*c2 + ht + hjj*s2h(jj,jj)= hii*s2 - ht + hjj*c2do 530 i=1,nif(i-ii) 370,530,420

370 ht=h(i,ii)h(i,ii)= cs*ht + sn*h(i,jj)h(i,jj)=-sn*ht + cs*h(i,jj)go to 530

420 if(i-jj) 430,530,480430 ht=h(ii,i)

h(ii,i)= cs*ht + sn*h(i,jj)h(i,jj)=-sn*ht + cs*h(i,jj)go to 530

480 ht=h(ii,i)h(ii,i)= cs*ht + sn*h(jj,i)h(jj,i)=-sn*ht + cs*h(jj,i)

530 continue

c.... Operate on eigenvectors540 do 550 i=1,n

ht=u(i,ii)u(i,ii)= cs*ht + sn*u(i,jj)

550 u(i,jj)=-sn*ht + cs*u(i,jj)600 continue700 continue

c.... Test for end iteration set new toleranceif(nrlm.lt.nr) go to 1000if(xmax.lt.test) go to 710toler=0.10*xmaxgo to 50

c.... Normalize and order eigenvectors710 do 800 i=1,n

do 750 j=1,n750 u(i,j)=u(i,j)*e(i)800 e(i)=h(i,i)

c.... Order eigenvalues and eigenvectorsdo 900 i=1,nnjl=i+1ht=e(i)im=ido 850 j=jl,nif(ht.lt.e(j)) go to 850ht=e(j)im=j


850 continuee(im)=e(i)e(i)=htdo 900 j=1,nht=u(j,i)u(j,i)=u(j,im)

900 u(j,im)=htreturn

1000 write (not,2000)return

2000 format(’ **** warning: iteration terminated without convergence’/)end

The program is demonstrated by finding the equilibrium path emanating from the criti-cal load λ1 = 0.198 moving in the direction of the buckling mode φ1 =0.328, 0.591,0.737. The imperfections are zero for the bifurcation problem.

Problem 289. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .2500Number of load steps : 10Maximum number of iterations : 20Eps 1, Eps 2, and Eps 3 : .0000 .0000 .0000Initial theta 1, 2, 3, and load : .0000 .0000 .0000 .1980Next theta 1, 2, 3, and load : .3300 .5900 .7400 .1980

n Theta 1 Theta 2 Theta 3 Load EV 1 EV 2 EV 3 NU || b ||0 .0000 .0000 .0000 .19801 .0820 .1478 .1842 .1989 .0018 1.3578 3.0498 3 .881E-112 .1643 .2956 .3682 .2016 .0071 1.3605 3.0525 3 .395E-103 .2471 .4437 .5518 .2062 .0161 1.3650 3.0571 3 .656E-104 .3306 .5921 .7348 .2129 .0288 1.3715 3.0639 3 .112E-095 .4151 .7408 .9168 .2218 .0454 1.3801 3.0730 3 .213E-096 .5010 .8900 1.0977 .2333 .0663 1.3910 3.0850 3 .433E-097 .5887 1.0398 1.2771 .2479 .0916 1.4046 3.1003 3 .920E-098 .6786 1.1901 1.4545 .2662 .1220 1.4213 3.1197 3 .203E-089 .7712 1.3411 1.6295 .2890 .1581 1.4416 3.1442 3 .467E-0810 .8672 1.4925 1.8014 .3173 .2007 1.4665 3.1752 4 .576E-15

The problem was also run with the imperfections values Á1 = Á2 = Á3= 0.01. Notethat in this case one can specify the direction in which to move as 0, 0, 0, 1. Theresults of both cases are shown in the figure below.

θ3

λ

λ1

Á1 = Á2 = Á3= 0.01


290. Modify the program ELASTICA to incorporate a distributed transverse loading q(x)on the cantilever column in addition to the load P. Examine the case where the transverseload is proportional to the axial load, as well as the case where the transverse load is fixedand the axial load is increased.

The expression for the virtual work functional for the elastica with transverse load wasderived in Problem 263 in Chapter 11. The result for q(x)= qo is

G(λ, θ, θ) = ℓ0

θ′θ′ − λ sin θ+γ 1−ξ cos θ θ dξ

where ξ≡ x∕ℓ, λ≡ P∕EI, and γ≡ qo∕P. The directional derivatives of the virtualwork functional can be computed as

DG ⋅ ∆θλo, θo

= ℓ0

∆θ′θ′ − λo cos θo−γ 1−ξ sin θo ∆θθ dx

DG ⋅ ∆λ λo, θo

= ℓ0

− ∆λo sin θo+γ 1−ξ cos θo θ dx

Interpolating the functions θ, θ, and ∆θ with base functions hi(x) we can put the linea-rized virtual work functional in the form G

^(λ, a, a)= aT Kν∆a+kν∆λ+gν where the

components of the coefficient matrices are given by

Kνij = ℓ

0

hi′hj′ − λν cos θν−γ 1−ξ sin θν hihj dx

kνi = ℓ

0

−sin θν+γ 1−ξ cos θν hi dx

gνi = ℓ

0

θν ′hj′ − λν sin θν−γ 1−ξ cos θν hi dx

The arc-length constraint is exactly the same as the example in the text. The modifica-tions to the program ELASTICA are manifested in the following changes to the program.First, the ratio of the transverse load to the axial load must be input. The input segmentof the code is modified as follows

c.... Read problem parametersread(5,1000) tol,alpha,xlength,gamma

The output of the input values is also appropriately modified. The call to the subroutineFCN is changed to pass the value of γ as

call fcn(A,b,x,z,xlength,gamma,factor)

Finally, the main changes are to the subroutine FCN to incorporate the new definitionsof the matrices Kν, kν, and gν as given above. Recall, that this routine is called for


each integration point z and is accumulated in the arrays. It should be clear how toalter the functional form of the transverse loading in this subroutine.

subroutine fcn(A,b,x,z,xlength,gamma,factor)C *------------------------------------------------------------------*C | Compute Simpson contribution to Hessian and residual matrices |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3)data ev1/1.570796327d0/,ev2/4.712388981d0/

c.... Basis functions and their derivativesh1 = dsin(ev1*z)h2 = dsin(ev2*z)dh1 = ev1*dcos(ev1*z)/xlengthdh2 = ev2*dcos(ev2*z)/xlength

c.... Transverse loading functionq = (1.0 - z)*gamma

c.... Compute rotation and first derivative at current pointtheta = x(1)*h1 + x(2)*h2dtheta = x(1)*dh1 + x(2)*dh2c1 = dcos(theta)- dsin(theta)*qc2 = dsin(theta)+ dcos(theta)*q

c.... Compute integral part of residual vectorb(1) = b(1) + (dtheta*dh1 - x(3)*c2*h1)*factorb(2) = b(2) + (dtheta*dh2 - x(3)*c2*h2)*factor

c.... Compute integral part of Hessian matrixA(1,1) = A(1,1) + (dh1*dh1 - x(3)*c1*h1*h1)*factorA(1,2) = A(1,2) + (dh1*dh2 - x(3)*c1*h1*h2)*factorA(1,3) = A(1,3) - (c2*h1)*factorA(2,1) = A(2,1) + (dh2*dh1 - x(3)*c1*h2*h1)*factorA(2,2) = A(2,2) + (dh2*dh2 - x(3)*c1*h2*h2)*factorA(2,3) = A(2,3) - (c2*h2)*factor

returnend

The program we run with γ= 0.1 with the following results. Note that the directionfor the next state is simply 0, 0, 1 because the imperfection can drive the problem.

Problem 290. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .5000Number of load steps : 10Maximum number of iterations : 30Number of integration intervals : 10Column Length : 1.0000Transverse load ratio to P : .1000Initial a1, a2, load : .0000 .0000 .0000Next a1, a2, load : .0000 .0000 1.0000

n a1 a2 P/EI EV 1 EV 2 NU || b ||0 .00000 .00000 .000001 .01175 .00118 .49986 .98385 10.85346 4 .3886E-152 .03149 .00242 .99947 .73450 10.60411 3 .4089E-083 .07132 .00372 1.49788 .48733 10.35669 4 .2939E-134 .18546 .00505 1.98468 .25795 10.12403 4 .3451E-085 .52441 .00659 2.35225 .18865 10.02147 7 .5684E-146 .93486 .01045 2.63776 .34207 10.08710 5 .2003E-137 1.29972 .01843 2.97953 .60593 10.22891 4 .9325E-098 1.60012 .03019 3.37906 .91886 10.40964 4 .3600E-089 1.84204 .04453 3.81640 1.24719 10.60985 4 .6387E-0910 2.03768 .06039 4.27627 1.57544 10.81942 4 .5783E-10

The problem was also run with γ= 0 to compute the equilibrium path emanating fromthe critical load λ1 = 2.467 with the direction input as 1, 0, 2.467. Both results areplotted in the following figure.


a1

λ

γ= 0.1

λ1

The case of transverse loading that is not proportional to the axial load is very simpleto implement. The components of the coefficient matrices are given by

Kνij = ℓ

0

hi′hj′ − λν cos θν−γ 1−ξ sin θν hihj dx

kνi = ℓ

0

− sin θνhi dx

gνi = ℓ

0

θν ′hj′ − λν sin θν−γ 1−ξ cos θν hi dx

where γ≡ qo∕EI. The modifications to the program are trivial.

291. Modify the program ELASTICA to incorporate N base functions. Examine the perfor-mance of the system as the number of base functions is increased.

The exercise is primarily one in generalizing the program. The program from the pre-vious problem (including the transverse load) is generalized to more base functions.The base functions used are the eigenfunctions

hi(ξ) = sin (2i−1)πξ

The main program and the subroutine FCN are the only code segments that change.These are listed below.

program ElasticaC *-----------------------------------------------------------*C | Problem 291. Fundamentals of Structural Mechanics |C | NDM is the number of base functions |C *-----------------------------------------------------------*

parameter(ndm=4,nn=ndm+1)implicit double precision (a-h,o-z)dimension A(nn,nn),b(nn),x(nn),xo(nn),h(ndm),dh(ndm)dimension AA(ndm,ndm),U(ndm,ndm),e(ndm)

data zero/0.d0/,one/1.d0/,two/2.d0/

c.... Read problem parametersread(5,1000) tol,alpha,xlength,gamma


read(5,1001) maxsteps,maxit,interread(5,1000) (xo(i),i=1,nn)read(5,1000) (x(i),i=1,nn)write(6,2000) tol,alpha,maxsteps,maxit,inter,xlength,gammawrite(6,2004) (j,xo(j),x(j),j=1,nn)

c.... Initialize constants of integrationdz = one/(2*inter)npoints = 2*inter + 1

c.... Initialize values for load step zero,set next trial staten = 0sum = zerodo j = 1,nnb(j) = x(j) - xo(j)sum = sum + b(j)*b(j)

end dosum = alpha/dsqrt(sum)do j = 1,nnx(j) = xo(j) + sum*b(j)

end dowrite(6,2001) n,nn,sumwrite(6,2002) (i,xo(i),i=1,nn)



1 nu = nu + 1

c.... Execute Simpson integration of Hessian and residual componentscall zerovec(b,nn)call zerovec(A,nn*nn)z = zerodo m=1,npointscall simpson(m,npoints,dz,xlength,factor)call fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm)z = z + dz

end do

c.... Finish computation of Hessian and residualdo k=1,nn

b(nn) = b(nn) + (x(k) - xo(k))**2A(nn,k) = two*(x(k) - xo(k))

end dob(nn) = b(nn) - alpha**2

c.... Compute norm of residual for convergence testtest = 0.d0do k=1,nntest = test + b(k)**2

end dotest = sqrt(test)

c.... Compute eigenvalues of Tangent stiffness matrixdo j=1,ndmdo k=1,ndmAA(j,k) = A(j,k)

end doe(j) = 1.d0

end docall eigens(AA,U,e,ndm,8,NR)

c... Update state vectorcall invert(A,nn,nn)do j=1,nndo k=1,nnx(j) = x(j) - A(j,k)*b(k)

end doend do

c.... Test for convergence, if successful output values and updateif ((test.gt.tol).and.(nu.lt.maxit)) go to 1


if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,nu,testwrite(6,2005) ’ Displacement Coefficients’write(6,2002) (i,x(i),i=1,nn)write(6,2005) ’ Eigenvalues of tangent stiffness matrix’write(6,2002) (i,e(i),i=1,ndm)

c.... Set values for previous converged state and guess at next statedo j=1,nntemp = xo(j)xo(j) = x(j)x(j) = two*x(j) - temp

end do



* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Number of integration interval s : ’,i6/* 5x, ’ Column Length : ’,f12.4/* 5x, ’ Transverse load ratio to P : ’,f12.4/* 5x, ’ Initial - Next values of a1,..., a(ndm), load : ’,/)

2004 format(15x,i3,5x,2f12.4)2001 format(/’ Step number > ’,i4/,

* ’ No. Iterations > ’,i4/,* ’ || b || > ’,e12.4)

2002 format(5x,i3,5x,f12.5)2005 format(/a)

end

C----------------------------------------------------------------------FCNsubroutine fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm)

C *------------------------------------------------------------------*C | Compute Simpson contribution to Hessian and residual matrices |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension A(nn,nn),b(nn),x(nn),h(ndm),dh(ndm)data zero,one/0.d0,1.d0/

c.... Basis functions and their derivativesev = 2.d0*atan(1.d0)do i=1,ndmcev = (2*i-1)*evh(i) = dsin(cev*z)dh(i) = cev*dcos(cev*z)/xlength

end do

c.... Transverse loading functionq = (one - z)*gamma

c.... Compute rotation and first derivative at current pointtheta = zerodtheta = zerodo i=1,ndmtheta = theta + x(i)* h(i)dtheta = dtheta + x(i)*dh(i)

end doc1 = dcos(theta)- dsin(theta)*qc2 = dsin(theta)+ dcos(theta)*q

c.... Compute integral part of Hessian matrix and residual vectordo i=1,ndmdo j=1,ndmA(i,j) = A(i,j) + (dh(i)*dh(j) - x(nn)*c1*h(i)*h(j))*factor

end doA(i,nn) = A(i,nn) - c2*h(i)*factorb(i) = b(i) + (dtheta*dh(i) - x(nn)*c2*h(i))*factor

end do


returnend

The program was run for the load case of Problem 290 with γ= 0.1 using a one, two,three, and four term expansion of the displacement function. The results at the end of10 steps are summarized below

Displacement Coefficients Eigenvalues of tangent stiffness1 Term 2 terms 3 terms 4 terms 1 Term 2 terms 3 terms 4 terms

a1 2.01914 2.03768 2.03806 2.03807 1.60159 1.57544 1.57438 1.57435a2 .06039 .06073 .06074 10.81942 10.79868 10.79777a3 .00480 .00482 30.53156 30.51467a4 .00087 60.13066Load 4.28929 4.27627 4.2760 4.27603 4.28929 4.27627 4.2760 4.27603

It is clear that the response is dominated by the first mode for this structure. The con-tributions of the higher modes decreases with mode number.

292. Modify the program ELASTICA to use polynomial base functions. Examine the per-formance of the system as the number of base functions is increased. Compare the perfor-mance of the polynomial functions with the eigenfunctions.

The modification to the program in Problem 291 involves only the definition of thebase functions in the subroutine FCN. We have implemented the base functions

hn(ξ) = ξn

where ξ= x∕ℓ. The modified segment of subroutine FCN is given below.c.... Base functions and their derivatives

do i=1,ndmh(i) = z**idh(i) = i*z**(i-1)/xlength

end do

The program was run for the load case of Problem 290 with γ= 0.1 using a one, two,three, and four term expansion of the displacement function. The results at the end of10 steps are summarized below.

Displacement Coefficients Eigenvalues of tangent stiffness1 Term 2 terms 3 terms 4 terms 1 Term 2 terms 3 terms 4 terms

a1 1.90046 2.77087 2.79818 2.63404 .84553 .10743 .01261 .00083a2 -1.31778 -.81545 -.25533 1.84271 .24839 .03437a3 -.41310 -1.27835 3.66093 .37639a4 .43191 5.69782Load 4.30798 3.19684 3.37669 3.31565 4.30798 3.19684 3.37669 3.31565Rot 1.90046 1.45309 1.56963 1.53227

One can clearly see the differences between the polynomial basis and the eigenbasis.At four terms the solution still changes quite a bit with the addition of another basefunction. The last row, called Rot, is the value θ(ℓ) computed from the approximation.It is evident that the load and rotation are converging as the number of base functionsincreases. For a point of reference we can interpolate the values from Problem 290 togive a load value of 3.289 at a displacement of 1.532, an error of less than 1%. Theconditioning of the system of equations (as measured by the ratio of the largest to the


smallest eigenvalues of the tangent stiffness matrix) is considerably worse for the poly-nomial basis.

293. Modify the programELASTICA to account for a nonlinearmoment curvature relation-ship of the form

M(κo)=EIoκo

1+moκ2o

where EIo and mo are material constants. Note that for small values of mo, the constitutivemodel reduces to the linear model originally used. The material is hyperelastic becausean energy function exists such that M= ∂W(κo)∕∂κo. What is the strain energy functionW? Plot the bifurcation diagrams for various values of the material constants.

The strain energy function can be computed by finding the anti-derivative (indefiniteintegral) of the moment

W(κo) = M(κo) dκo = EIomo 1+moκ2o − 1

where the constant of integration has been set to give zero energy when the curvature iszero. The validity of this expression can be demonstrated by differentiating the resultto give the original relationship between moment and curvature.

The curvature is related to the rotation as κo = θ′. The virtual work functionalfrom Problem 290 can be modified to read

G(λ, θ, θ) = ℓ0

m(θ′)θ′ − λ sin θ+γ 1−ξ cos θ θ dξ

where the function m(θ′)= M(θ′)∕EIo and the parameter λ≡ P∕EIo. We can com-pute the directional derivative of G in the direction of ∆θ as

DG ⋅ ∆θ = ℓ0

m′(θ′)∆θ′θ′ − λ cos θ−γ 1−ξ sin θ ∆θθ dξ

where the derivative of the function m(θ′) is given by

dm(κo)dκo

= 1+moκ2o −3∕2

The components of the matrices Kν, kν, and gν are modified from those of Problem290 to be


Kνij = ℓ

0

m′(κνo)hi′hj′ − λν cos θν−γ 1−ξ sin θν hi hj dx

kνi = ℓ

0

−sin θν+γ 1−ξ cos θν hi dx

gνi = ℓ

0

m(κνo)hj′ − λν sin θν−γ 1−ξ cos θν hi dx

The modifications are shown in the following code segment. The remaining parts ofthe code are unchanged from the previous problem. In the main program the value ofmo must be input (it is called rmo) and passed to the subroutine FCN.

c.... Read problem parametersread(5,1000) tol,alpha,xlength,gamma,rmo

[ code between is unchanged ]call fcn(A,b,x,z,xlength,gamma,rmo,factor,h,dh,nn,ndm)

The main changes are in the subroutine FCN as shown below.subroutine fcn(A,b,x,z,xlength,gamma,rmo,factor,h,dh,nn,ndm)

C *------------------------------------------------------------------*C | Compute Simpson contribution to Hessian and residual matrices |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension A(nn,nn),b(nn),x(nn),h(ndm),dh(ndm)data zero,one/0.d0,1.d0/

c.... Basis functions and their derivativesev = 2.d0*atan(1.d0)do i=1,ndmcev = (2*i-1)*evh(i) = dsin(cev*z)dh(i) = cev*dcos(cev*z)/xlength

end do


c.... Compute rotation and first derivative at current pointtheta = zerodtheta = zerodo i=1,ndmtheta = theta + x(i)* h(i)dtheta = dtheta + x(i)*dh(i)

end doc1 = dcos(theta)- dsin(theta)*qc2 = dsin(theta)+ dcos(theta)*qf = dsqrt(one + rmo*dtheta**2)fmu = dtheta/ffdmu = 1.0/f**3

c.... Compute integral part of Hessian matrix and residual vectordo i=1,ndmdo j=1,ndmA(i,j) = A(i,j)+(fdmu*dh(i)*dh(j)-x(nn)*c1*h(i)*h(j))*factor

end doA(i,nn) = A(i,nn) - c2*h(i)*factorb(i) = b(i) + (fmu*dh(i) - x(nn)*c2*h(i))*factor

end do

returnend

The problem was also run with γ= 0 to compute the equilibrium path emanating fromthe critical load λ1 = 2.467 with the direction input as 1, 0, 2.467. Both results areplotted in the following figure.


The program was run for the problem with γ= 0.1 and mo = 5 with two basefunctions (eigenfunctions) starting at the state (0,0,0) and moving in the direction(0,0,1). The arc length was set at α= 0.05.Problem 293. Fundamentals of Structural Mechanics

Convergence tolerance : .1000E-07Arc length parameter : .0500Number of load steps : 10Maximum number of iterations : 50Number of integration intervals : 10Column Length : 1.0000Transverse load ratio to P : .1000Ratio (EIo/Mo)^2 : 5.0000Initial - Next values : Initial Next

a1 .0000 .0000a2 .0000 .0000Load .0000 1.0000

The problem was run for 100 steps (10 of which are given below). The results areplotted in the figure below. (Note: the program writes the output in a slightly differentform).

n a1 a2 P/EIo EV1 EV20 .00000E+00 .00000E+00 .00000E+001 .95678E-03 .11608E-03 .49991E-01 .12087E+01 .11078E+022 .19540E-02 .23269E-03 .99981E-01 .11836E+01 .11053E+023 .29943E-02 .34985E-03 .14997E+00 .11585E+01 .11027E+024 .40807E-02 .46758E-03 .19996E+00 .11334E+01 .11001E+025 .52162E-02 .58589E-03 .24994E+00 .11081E+01 .10975E+026 .64044E-02 .70481E-03 .29993E+00 .10829E+01 .10948E+027 .76492E-02 .82436E-03 .34991E+00 .10575E+01 .10921E+028 .89548E-02 .94458E-03 .39990E+00 .10320E+01 .10894E+029 .10326E-01 .10655E-02 .44988E+00 .10065E+01 .10866E+0210 .11768E-01 .11872E-02 .49986E+00 .98086E+00 .10837E+02

0 1 2

1

2

λ

a1

γ= 0.1mo = 5.0α= 0.05

One can observe that the nonlinear moment-curvature relationship causes the columnto exhibit a limit point under the applied loading.

294. Modify the program ELASTICA to use piecewise linear finite element base functions.Examine the performance of the system as the number of base functions is increased.Compare the performance of the finite element base functions with the polynomial func-tions.

The modification to the program involves only the definition of the base functions inthe subroutine FCN. We have implemented the base functions given in the text, special-ized for equidistant spacing of the nodes. To wit,


hn(x) =

Nℓ x−xn−1 xn−1 ≤ x≤ xn

xn ≤ x≤ xn+1Nℓ xn+1−x

0 otherwise

hn′(x) =

Nℓ xn−1 ≤ x≤ xn

xn ≤ x≤ xn+1−Nℓ

0 otherwise

where xn is the coordinate of node n. The change requires only the modification of thesubroutine FCN, which computes the basis functions.

c.... Base functions (finite element) and their derivativesdo i=1,ndmza = dfloat(i-1)/dfloat(ndm)zb = dfloat(i)/dfloat(ndm)zc = dfloat(i+1)/dfloat(ndm)if ((z.ge.za).and.(z.lt.zb)) thenh(i) = (z-za)/(zb-za)dh(i) = dfloat(ndm)/xlength

elseif ((z.ge.zb).and.(z.le.zc)) thenh(i) = (zc-z)/(zc-zb)dh(i) = -dfloat(ndm)/xlength

elseh(i) = 0.0dh(i) = 0.0

end ifend do

The program was run for the problem with γ= 0.1 with from two to nine base func-tions starting at the state a= 0 and λ= 0 moving in the direction a= 0 and λ= 1.The arc length was set at α= 0.5 and the problem was run for 10 steps. The numberof integration points was set to 200 to be certain that the integration was accurate forthe finest discretization. (See Problem 300 for more discussion on this point.) The val-ues of the input parameters are given below.

Problem 294. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .5000Number of load steps : 10Maximum number of iterations : 30Number of integration intervals : 200Column Length : 1.0000Transverse load ratio to P : .1000

The results are summarized below. The values of the displacement coefficients andload at the end of step 10 are presented for each approximation. Since the solution iscontrolled by the arc-length, the place on the path that the solutions falls is different foreach case. One can assess the accuracy by interpolating the results given in Problem290.

Displacement Coefficients2 Terms 3 Terms 4 Terms 5 Terms 6 Terms 7 Terms 8 Terms 9 Terms

a1 1.39358 .95416 .70394 .54302 .43433 .35197 .30091 .25640a2 1.88422 1.58194 1.25495 1.00551 .81769 .68215 .58129 .50175a3 1.78957 1.59065 1.34622 1.13719 .96381 .83174 .71951a4 1.70350 1.55441 1.36803 1.19218 1.04387 .91666a5 1.62413 1.50611 1.35576 1.21005 1.07958a6 1.55289 1.45565 1.33166 1.21023a7 1.48843 1.40501 1.30163a8 1.42966 1.35838a9 1.37718

Load 4.00767 3.79835 3.61925 3.47365 3.35279 3.25393 3.16795 3.09819Cond. No. 6.3 16.6 33.0 56.9 89.2 131.3 186.1 251.8


The condition number (ratio of maximum to minimum eigenvalues of the tangent stiff-ness matrix) at step 10 is presented for each approximation. One can observe that thefinite element equations are very well conditioned. For the 4-term approximation wecan compare with the polynomial basis (Problem 292) and the eigenbasis (Problem291). The polynomial basis had a condition number of 6855, the eigenbasis had 38, andthe finite element basis had 33.

295. Describe a method for using the program NONLINEARBEAM to locate the bifurcationpoints of a system without imperfections.

Technically, the constraint c(x)≡ detK(x)= 0, where K(x) is given in the text andx≡ (λ, a ), would be a suitable constraint for homing in on a critical point. However,observing that there are three fields in the approximation, the dimension of the matrixK(x) gets large as the order of the approximation (i.e., the number of base functions)increases. Computation of the determinant is not very practical for large systems ofequations. Linearization of the constraint is also difficult.

We can observe that an eigenvalue of the tangent stiffness matrix goes to zero at acritical point let the eigenvalue problem be stated as

K(x)φ = mφ

Because the tangent stiffness matrix is a function of the state x so are the eigenvalue mand the eigenvector φ (implicitly). The constraint equation can be stated as

c(x) = m(x) = 0

The eigenvalue problem at any given state can be solved by any of a variety of algo-rithms. Thus, let us assume that m(xo) and φ(xo) can be computed at the state xo. Theconstraint equation can be linearized to be

c^(x) = ∇m(xo)∆x+ m(xo) = 0

where [∇m]i= ∂m∕∂xi. The gradient of the eigenvalue with respect to x can be com-puted from the original eigenvalue problem. Taking the derivative we find

∇K−∇mI φ+ K−mI ∇φ = 0

Multiplying the above equation by φT annihilates the second term, and hence obviatesthe need to compute the rate of change of the eigenvector. Rearranging terms we arriveat the result

∇m ∆x = φT ∇K∆x φ

if φ is normalized such that φTφ= 1. The above expression can be computed in com-ponents to be φi Kij,kφj∆xk. This equation gives us what we need to form the bordered


Hessian matrix. The main issue in this computation is how to compute ∂K∕∂ai and∂K∕∂λ. It is possible, albeit tedious, to compute these terms analytically. The maindifficulty is that K depends implicitly on the load parameter through the internal mo-ment, shear, and axial force. One can also compute these derivatives numerically byaugmenting x and taking the difference, essentially replacing ∇K with

∂K∂xi≈K(x+δxiei )− K(x )

δxi

where ei is a vector with a one in the ith slot and zeros everywhere else. The value ofδxi must be chosen small enough to give a good approximation but large enough toavoid an ill-conditioned operation.

296. Explore the features of the programNONLINEARBEAMby using it to solve the cantile-ver beam problem under a variety of loading scenarios.

This problem is open-ended and, as such, has no identifiable “solution” to present.Some suggestions on loading scenarios include investigation of the effects of propor-tional versus non-proportional transverse loading on the buckling behavior of the beamand the effects of different transverse load distributions. One can make comparisonswith some of the classical results that are available in the literature.

297. Modify the programNONLINEARBEAM to account for initial geometric imperfectionsin the column. Is it sufficient to specify imperfections only in the field w(x)?

It is not sufficient to consider only an imperfection in the transverse displacement w(x).One also needs to consider an imperfection in the rotation field. It is not really neces-sary to consider imperfections in the axial displacement u, but we can. Let the imper-fection fields be called uo(x), wo(x), and θo(x). The strains associated with the initialimperfections are

κoi = θo′

βoi = wo′ cos θo− 1+uo′ sin θo

Áoi = wo′ sin θo+ 1+uo′ cos θo− 1

The initial strains are accounted for in the constitutive equations

M = EI κo− κoi

Q = GA βo− βoi

N = EA Áo− Áoi


Since the initial strains are not a function of the subsequent displacements, which aremeasured from a nominal straight configuration, there are no further complicationsfrom initial imperfections.

We must be specific about the variation of the initial imperfection fields. We can,for example consider a shear-free initial imperfection by taking

wo(ξ) = co1− cosπξ2 θo(ξ) = π

2ℓ co sinπξ2

uo(ξ) = 0

where ξ≡ x∕ℓ, so that wo′−θo = 0. If the constant co is small there should be onlyan effect on the moment equation. The changes to the program are minor. First, thevalue of co must be input. In the main program we can read in this value as cimperf

c.... Read problem parametersread(5,1000) tol,alpha,xlength[ code between is unchanged ]read(5,1000) (dpo(i),i=1,3)read(5,1000) cimperf

The call to subroutine FCN must pass the value of co. The functions given above havebeen implemented in the subroutine FCN as follows

subroutine fcn(A,b,D,x,z,xlength,cimperf,factor,ndm,nbasis)C *------------------------------------------------------------------*C | Compute contribution to K and g at current integration point |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)

dimension A(ndm,ndm),b(ndm),x(ndm),D(3)dimension G(4,4),GB(4,3),BGB(3,3)

data zero/0.d0/,one/1.d0/

c.... Compute displacements, derivatives, and current load factor clfdu = zerodw = zerodtheta = zerotheta = zerodo i=1,nbasis

call basis(i,z,xlength,h,dh,nbasis)du = du + x(3*i-2)*dhdw = dw + x(3*i-1)*dhdtheta = dtheta + x(3*i)*dhtheta = theta + x(3*i)*h

end doct = dcos(theta)st = dsin(theta)

c.... Compute axial strain, shear strain, and curvatureepsilon = dw*st + (one + du)*ct - onebeta = dw*ct - (one + du)*stcurv = dtheta

c.... Compute initial axial strain, shear strain, and curvaturetwopi = 2.0*atan(1.0)dw0 = (cimperf*twopi/xlength)*dsin(twopi*z)du0 = 0.0th0 = dw0ct0 = dcos(th0)st0 = dsin(th0)eps0 = dw0*st0 + (one + du0)*ct0 - onebeta0 = dw0*ct0 - (one + du0)*st0curv0 = (cimperf*(twopi/xlength)**2)*dcos(twopi*z)

c.... Compute axial force, shear force, bending moment, and other forcesbend = D(1)*(curv-curv0)shear = D(2)*(beta-beta0)axial = D(3)*(epsilon-eps0)


[ Remaining part of subroutine is unchanged from text ]

The program was run for a cantilever beam subjected to a compressive end load withan imperfection of co = 0.1. The input values are given below

Problem 297. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .5000Number of load steps : 40Maximum number of iterations : 40Number of integration intervals : 10Number of points in plotted shape : 20Output shape every N points, N = : 10Number of basis functions : 6

Beam properties, end load and dist. load amplitudesColumn Length : 10.00EI, GA, EA : 1000. .1000E+07 .1000E+07Mo, Qo, Po : .0000E+00 .0000E+00 -2.460mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00Imperfection : .1000

The problem was run for 40 steps. The output is given below for selected steps (theoutput of the program has been edited)

Step Load u(L) w(L) theta(L) NU || b ||1 .48855E+00 -.76802E-04 .10512E+00 .16512E-01 8 .2498E-082 .98851E+00 -.16657E-03 .11093E+00 .17425E-01 6 .8651E-083 .14885E+01 -.27032E-03 .11743E+00 .18445E-01 4 .2784E-094 .19884E+01 -.39166E-03 .12472E+00 .19592E-01 4 .4601E-095 .24883E+01 -.53539E-03 .13299E+00 .20890E-01 4 .1660E-0910 .49874E+01 -.19457E-02 .19887E+00 .31240E-01 5 .1699E-0915 .74787E+01 -.90667E-02 .39230E+00 .61647E-01 6 .2205E-0920 .95113E+01 -.16797E+00 .16425E+01 .25994E+00 9 .6065E-0925 .10322E+02 -.10142E+01 .39219E+01 .64369E+00 5 .2608E-0830 .11317E+02 -.24660E+01 .58131E+01 .10184E+01 6 .1000E-0835 .12762E+02 -.42030E+01 .70895E+01 .13524E+01 6 .2343E-0840 .14639E+02 -.58404E+01 .77553E+01 .16192E+01 6 .3906E-08

The results are plotted below for two values of the imperfection. One can observe thatthe equilibrium path does not start at the origin because of the imperfection.

0 10

15

λ

w(ℓ)0 10

15

θ(ℓ)

co = 0.1

co = 0.3

co = 0.1

co = 0.3


298. Modify the subroutine basis in the programNONLINEARBEAM to use piecewise linearfinite-element base functions. Examine the performance of the system as the number ofbase functions is increased. Compare the performance of the finite element base functionswith the sinusoids used in the original program.

The finite element base functions are simple to implement and are identical to thoseimplemented in Problem 294. The program NONLINEARBEAM was already structured tomake the change to a new basis simple. The new subroutine BASIS is given as

subroutine basis(i,z,xlength,h,dh,nbasis)C *------------------------------------------------------------------*C | Evaluate ith basis function h and derivative dh at point z |C *------------------------------------------------------------------*


za = dfloat(i-1)/dfloat(nbasis)zb = dfloat(i) /dfloat(nbasis)zc = dfloat(i+1)/dfloat(nbasis)if ((z.ge.za).and.(z.le.zb)) thenh = (z-za)/(zb-za)dh = dfloat(nbasis)/xlength

elseif ((z.gt.zb).and.(z.le.zc)) thenh = (zc-z)/(zc-zb)dh = -dfloat(nbasis)/xlength

elseh = 0.0dh = 0.0

end ifreturnend

One can observe that the solution with the linear finite element base functions shows atendency to shear lock, that is, displacements are constrained to very small values be-cause the contribution to the tangent stiffness from the shear term does not have theappropriate rank. This phenomenon is well-known and can be cured by “reduced in-tegration” of the system matrices. One point integration, with the sampling point at theelement center, is known to cure the shear-locking problem. We can trick the programinto doing one-point integration by making a modification to the SIMPSON subroutine.We have adjusted the weights in the Simpson rule to give all of the weight to the mid-point of a segment and none to the ends. If the number of integration points is equal tothe number of base functions then we achieve the appropriate reduced integrationscheme.

subroutine simpson(m,npoints,dz,xlength,factor)C *------------------------------------------------------------------*C | Evaluate the weight factor for the current integration point m |C *------------------------------------------------------------------*


c = xlength*dz/3.d0n = mod(m,2)if((m.eq.1).or.(m.eq.npoints)) then

factor = 0.0else if (n.eq.0) then

factor = 6.0*celse

factor = 0.0endifreturnend


The program was run on the example presented in the text (cantilever beam subjectedto a uniform transverse load and a compressive axial load). The properties of the sys-tem are given below.


Beam properties, end load and dist. load amplitudesColumn Length : 10.00EI, GA, EA : 1000. .1000E+07 .1000E+07Mo, Qo, Po : .0000E+00 .0000E+00 -2.460mo1, qo1, po1 : .0000E+00 .2000 .0000E+00mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The results for the first 10 steps are present below. One can compare the result at step10 with the example in the text from the output. Note that the transverse tip displace-ment with 6 eigenfunctions was w(ℓ)= 0.478 compared to w(ℓ)= 0.463 for 8 finiteelement functions.

Step Load u(L) w(L) theta(L) NU || b ||1 .27270E+00 -.37735E-02 .25655E+00 .34614E-01 7 .5542E-092 .77227E+00 -.41948E-02 .26975E+00 .36665E-01 7 .7040E-093 .12717E+01 -.46873E-02 .28444E+00 .38950E-01 4 .4582E-094 .17711E+01 -.52683E-02 .30088E+00 .41510E-01 4 .1080E-085 .22702E+01 -.59610E-02 .31940E+00 .44398E-01 4 .4007E-086 .27691E+01 -.67964E-02 .34041E+00 .47679E-01 5 .4267E-097 .32677E+01 -.78173E-02 .36445E+00 .51438E-01 5 .3071E-098 .37658E+01 -.90833E-02 .39222E+00 .55786E-01 5 .3231E-099 .42632E+01 -.10680E-01 .42464E+00 .60868E-01 5 .4285E-0910 .47595E+01 -.12731E-01 .46296E+00 .66882E-01 5 .4386E-09

0 10

50

λ

w(ℓ)


299. Modify the basis subroutine in the programNONLINEARBEAM to use the polynomialshi (x)∈ x, x2,. . ., xN . Examine the performance of the system as the number of basefunctions is increased using the pure bending problem. Are these functions able to capturethe exact solution, which is a circular shape, as shown in the text example?Why is conver-gence so difficult with a large number of base functions? Implement the orthogonal basefunctions described in Chapter 6. Do these base functions work better than the originalpolynomials?

The modification of the subroutine BASIS is straightforward. For the polynomials higiven above the routine is



h = z**idh = i*z**(i-1)/xlengthreturnend

The orthogonal polynomials can also be easily implemented, but are a bit morecumbersome. The expressions for the first four are on page 212 of the text. The fifthand sixth orthogonal polynomials are given by

g5(ξ) = 210ξ5− 504 ξ4+ 420ξ3− 140ξ2+ 15ξ

g6(ξ) = 792ξ6− 2310 ξ5+ 2520ξ4− 1260ξ3+ 280ξ2− 21ξ

The orthogonal polynomials up to order six are implemented in the routine as follows:



if (nbasis.gt.6) Stop ’Not enough basis functions in BASIS’go to (1,2,3,4,5,6),i

c.... Basis function h1(z) = z1 h =zdh=1./xlengthreturn

c.... Basis function h2(z) = (4z-3)z2 h =(4.*z-3.)*zdh=(8.*z-3.)/xlengthreturn

c.... Basis function h3(z) = ((15z-20)z+6)z3 h =((15.*z-20.)*z+6.)*zdh=((45.*z-40.)*z+6.)/xlengthreturn

c.... Basis function h4(z) = (((56z-105)z+60)z-10)z4 h =((( 56.*z-105.)*z+ 60.)*z-10.)*zdh=(((224.*z-315.)*z+120.)*z-10.)/xlengthreturn

c.... Basis function h5(z) = ((((210z-504)z+420)z-140)z+15)z5 h =(((( 210.*z- 504.)*z+ 420.)*z-140.)*z+15.)*zdh=((((1050.*z-2016.)*z+1260.)*z-280.)*z+15.)/xlengthreturn


c.... Basis function h6(z) = (((((792z-2310)z+2520)z-1260)z+280)z-21)z6 h =((((( 792.*z- 2310.)*z+ 2520.)*z-1260.)*z+280.)*z-21.)*zdh=(((((4752.*z-11550.)*z+10080.)*z-3780.)*z+560.)*z-21.)dh=dh/xlengthreturnend

The program was run on the example problem in the text wherein a beam is subjectedto an end moment of magnitude sufficient to curl it into a circle with circumferenceequal to its original length of 10. The input for the problem is

Problem 299. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : 10.0000Number of load steps : 10Maximum number of iterations : 40Number of integration intervals : 10Number of points in plotted shape : 20Output shape every N points, N = : 10Number of basis functions : 6

Beam properties, end load and dist. load amplitudesColumn Length : 10.00EI, GA, EA : 1000. .1000E+07 .1000E+07Mo, Qo, Po : 6.280 .0000E+00 .0000E+00mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The output for the case of orthogonal polynomials with 6 base functions is given belowas an example.

Step Load u(L) w(L) theta(L) NU || b ||0 0 0 0 0 0 01 .96915E+01 -.60604E+00 .29503E+01 .60863E+00 7 .3324E-092 .19390E+02 -.22944E+01 .53723E+01 .12177E+01 5 .1436E-083 .29101E+02 -.47075E+01 .68613E+01 .18275E+01 5 .9216E-094 .38830E+02 -.73477E+01 .72304E+01 .24384E+01 5 .5750E-085 .48583E+02 -.96970E+01 .65572E+01 .30492E+01 6 .2784E-086 .58364E+02 -.11360E+02 .51758E+01 .36555E+01 7 .1646E-087 .68174E+02 -.12192E+02 .35257E+01 .42540E+01 8 .5047E-088 .78013E+02 -.12265E+02 .19918E+01 .48407E+01 8 .2023E-089 .87884E+02 -.11817E+02 .85711E+00 .54015E+01 8 .1992E-0810 .97790E+02 -.11171E+02 .22152E+00 .59092E+01 8 .3493E-08

The problem was run with 3, 4, 5, and 6 base functions for both the original polyno-mials h and the orthogonal polynomials g. Below we plot the final deflected shape ofthe beam for each of these approximations along with the exact shape and the shapeobtained with the 6-eigenfunction basis (presented in the text). The approximationusing the 6 eigenfunctions is labeled e6. The exact solution is shown as a dotted line.

4

e6

5

3 4

e65

3

6

polynomials orthogonalpolynomials

xn

xo

xn+1

xν

x

α

do

g(x)= 0


We also present a plot of the load factor versus the end moment for each approxima-tion. The exact solution is the diagonal line.

5

100

0 2π

λ

θ(ℓ)

4 e63 5

100

0 2π

λ

θ(ℓ)

4e63

6

polynomials orthogonalpolynomials

One can clearly see the difference between the original and orthogonal polynomials.For the 3-term approximation the results are essentially identical. As the order in-creases the original polynomials give out while the orthogonal polynomials do verywell. The solution with 6 original polynomial base functions failed to converge on the6th load step out of 10. Ill-conditioning is the culprit for the failure to converge. Thesolution for the orthogonal polynomials is still robust at 6 terms. It is interesting toobserve the reluctance to rotate beyond a certain point in the low-order polynomials(both original and orthogonal).

300. The arc-length constraint forces the next equi-librium configuration to be a fixed distance from theprevious converged state. Therefore, all iteratesmust lie on a sphere of radius α centered on the con-verged state xn. One of the problemswith this strate-gy is that the equilibrium path pierces the sphere attwo points (at least). For a highly nonlinear equilib-rium path the Newton iterations can converge to theother point on the sphere, which causes the loadingdirection to change. We can observe this phenome-non in the program NONLINEARBEAM if the step size is not judiciously chosen. Once theloading direction has turned around, it is not likely to change back. Consider another pos-sible constraint pictured on the following page.

The difference between the estimate xν and xo is forced to lie in a hyperplane normalto the tangent direction do ≡ xo− xn. The normal plane is set a distance α from xn.Therefore, ‖ do ‖= α. How can one compute the tangent direction do? Show that thenormality condition is dν ⋅ do = α2, where the νth increment in state is dν≡ xν− xn.Develop amethod based upon a secant directionwhere do ≡ xn− xn−1 (the previous twoconverged states). How would you start a method based on this definition? Implement


these constraints in the programNONLINEARBEAM and assess their performance. Are thereadvantages over the arc-length constraint? Are there disadvantages?

(a) How can one compute the tangent direction do?

Observe that x= (λ, a ), where a is the vector of displacement coefficients and λ isthe load parameter. The tangent direction can be obtained by examining the linearizedequilibrium equations g^(λ, a)= 0. We have

g^(λ, a) = K(a)∆a+ k(λ)∆λ + g(λ, a) = 0

At the state xn = (λn, an ) equilibrium is satisfied. Therefore g(λn, an)= 0. To find thetangent direction we need only assume a unit increment in the load parameter and com-pute the corresponding change in displacement. To wit, let do ≡ (1, ∆ao ). The incre-ment in displacement can then be computed by solving the system of equations

K(an)∆ao = −k(λn)

The vector do can subsequently be normalized to have length ‖ do ‖= α. Note that ifthe current state is on a downhill branch of the equilibrium path, as evidenced by nega-tive eigenvalues of the tangent stiffness K(an), then the choice do ≡ (−1, ∆ao ) isappropriate. Clearly, the computation will fail if K(an) is singular, as it is at a limitpoint or bifurcation point.

(b) Show that the normality condition is dν ⋅ do = α2, where the νth increment instate is dν≡ xν−xn. By definition of the method the increment xν−xo is orthogonalto the tangent line do. Thus,

0 = do ⋅ xν−xo

0 = do ⋅ xν−xn+ xn−xo

0 = do ⋅ xν−xn − do ⋅ xo−xn

0 = do ⋅ xν−xn − do ⋅ do

Since do ⋅ do = α2 we get the result do ⋅ xν−xn = α2. Letting dν≡ xν−xn thengives the desired result.

(c) Develop a method based upon a secant direction where do ≡ xn−xn−1 (the pre-vious two converged states). How would you start a method based on this definition?The method is identical to the previous one except that it avoids the computation of thetangent direction from the condition given in part (b). Since there is no state x−1 themethod cannot be started with a secant. Because most problems start out linear, onemight start the secant method with the tangent direction do ≡ (1, ∆ao ) with ∆ao com-puted from the equation

K(0)∆ao = −k(0)

The secant method can commence on the second step. As an alternative, a guess at thenext direction can be an input to the algorithm (as it is for most of the programs in the


chapter). This approach is necessary if starting from a bifurcation point since the tan-gent computation fails there.

(d) Implement these constraints in the program NONLINEARBEAM and assess theirperformance. Are there advantages over the arc-length constraint? Are there disadvan-tages?

The tangent line algorithm appears to have little advantage over the regular arc-lengthconstraint.

301. When using finite element base functions in the programsELASTICA andNONLINEAR-BEAM, most of the integration points contribute nothing to the integrals because the basefunctions are zero over much of the region. Restructure the order of the programs to makethem more efficient by putting the loop over integration points inside the loop over basefunctions.

The modification to the program in Problem 294 involves a change in the way thecomputations are organized. In the main program the dimension of the arrays for thebase functions is changed to 2 because only two base functions will be non-zero at anyone time.

dimension A(nn,nn),b(nn),x(nn),xo(nn),h(2),dh(2)

Also in the main program, the integration of the Hessian and residual components ischanged to have a loop over “elements” with an inner loop over integration pointswithin that element.

c.... Execute Simpson integration of Hessian and residual componentscall zerovec(b,nn)call zerovec(A,nn*nn)do nel=1,ndmz = dfloat(nel-1)/dfloat(ndm)do m=1,npointscall basis(z,xlength,h,dh,ndm,nel)call simpson(m,npoints,dz,xlength,factor)call fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm,nel)z = z + dz

end doend do

The subroutine BASIS is modified to return the values of the base functions and deriv-atives for only the ith and (i--1)th base functions. All of the remain functions are knownto be zero at this point.

subroutine basis(z,xlength,h,dh,ndm,i)C *------------------------------------------------------------------*C | Compute the ith and i-1th basis functions at point z |C | h(1) is i-1th (ramping down), h(2) is ith (ramping up) |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension h(2),dh(2)

c.... Compute basis function (finite element) and its derivativeza = dfloat(i-1)/dfloat(ndm)zb = dfloat(i)/dfloat(ndm)h(1) = (zb-z)/(zb-za)


h(2) = (z-za)/(zb-za)dh(1) = -dfloat(ndm)/xlengthdh(2) = dfloat(ndm)/xlengthreturnend

The subroutine FCN is modified to recognize that only two of the base functions arenon-zero. Hence the computation of θ and θ′ only involve two of the coefficients. Fur-thermore, the contributions to A and b come only from the two base functions.

subroutine fcn(A,b,x,z,xlength,gamma,factor,h,dh,nn,ndm,nel)C *------------------------------------------------------------------*C | Compute Simpson contribution to Hessian and residual matrices |C *------------------------------------------------------------------*

implicit double precision (a-h,o-z)dimension A(nn,nn),b(nn),x(nn),h(2),dh(2)data zero,one/0.d0,1.d0/


c.... Compute rotation and first derivative at current pointif(nel.eq.1) thenx0 = 0.0

elsex0 = x(nel-1)

endifx1 = x(nel)theta = x0*h(1) + x1*h(2)dtheta = x0*dh(1) + x1*dh(2)c1 = dcos(theta)- dsin(theta)*qc2 = dsin(theta)+ dcos(theta)*q

c.... Compute integral part of Hessian matrix and residual vectordo i=1,2ii = nel+i-2do j=1,2jj = nel+j-2if((ii.gt.0).and.(jj.gt.0))

* A(ii,jj)= A(ii,jj)+(dh(i)*dh(j)-x(nn)*c1*h(i)*h(j))*factorend doif(ii.gt.0) thenA(ii,nn) = A(ii,nn) - c2*h(i)*factorb(ii) = b(ii) + (dtheta*dh(i) - x(nn)*c2*h(i))*factor

end ifend do

returnend

The program was run for the problem with γ= 0.1 with from two to nine base func-tions starting at the state a= 0 and λ= 0 moving in the direction a= 0 and λ= 1,exactly as was done in Problem 294. The arc length was set at α= 0.5 and the prob-lem was run for 10 steps. The number of integration points were set to 2 (per element).The values of the input parameters are given below.

Problem 183. Fundamentals of Structural MechanicsConvergence tolerance : .1000E-07Arc length parameter : .5000Number of load steps : 10Maximum number of iterations : 10Number of integration intervals : 2Column Length : 1.0000Transverse load ratio to P : .1000

The results are summarized below. The values of the displacement coefficients andload at the end of step 10 are presented for each approximation. Since the solution iscontrolled by the arc-length the place on the path that the solutions falls is different for


each case. Again, one can assess the accuracy by interpolating the results in Problem290.

Displacement Coefficients2 Terms 3 Terms 4 Terms 5 Terms 6 Terms 7 Terms 8 Terms 9 Terms

a1 1.39142 .95682 .70144 .54050 .43216 .35538 .29878 .25569a2 1.88316 1.58142 1.25254 1.00296 .81944 .68286 .57910 .49855a3 1.78954 1.59063 1.34658 1.13578 .96545 .82956 .72071a4 1.70316 1.55483 1.36665 1.19188 1.04176 .91591a5 1.62427 1.50615 1.35597 1.21024 1.07966a6 1.55271 1.45501 1.33192 1.20903a7 1.48807 1.40530 1.30227a8 1.42980 1.35847a9 1.37723

Load 4.00905 3.79776 3.62091 3.47457 3.35367 3.25335 3.16951 3.09879

The condition number (ratio of maximum to minimum eigenvalues of the tangent stiff-ness matrix) at step 10 was the same as those reported in Problem 294, as expected.The small differences in the values reported here and those reported in Problem 294 aredue to the accuracy of numerical integration. Ironically, the numerical integration ismore accurate in the present implementation than in the former implementation. Thefinite element base functions are not smooth. When we do Simpson quadrature, whichapproximates the curve as a quadratic, over the entire segment (Problem 294) we sufferunder the inaccuracy of the method trying to integrate over the peaks of the base func-tions and the jumps in their derivatives (and the ambiguity if a sampling point is rightat the jump). The numerical integration can be made arbitrarily accurate by takingmore segments, but that number is unacceptably large. In the present implementationwe integrate within each element and sum over all elements. In doing so we explicitlyaccount for the peaks and jumps in the functions. A two-segment (5 sampling points)Simpson quadrature is very accurate in this case. Also, with this organization, the num-ber of integration points increases as the order of approximation increases (i.e., twosegments per element times nine elements is 18 segments, far below the 200 used inProblem 294).

302. Add a subroutine to find the eigenvalues and eigenvectors of K(a) at each point onthe equilibrium path in the programNONLINEARBEAM to examine the stability of equilibri-um. Implement a procedure for branch switching so that the programwill trace thebifurca-tion diagram when there are no imperfections.

Finding the eigenvalues and eigenvectors of the tangent stiffness matrix is simply amatter of a call to the subroutine EIGENS that we have used in the other programs.The implementation involves providing storage space, as indicated in these commandsin the main program,

parameter (nn=ndm-1)dimension AA(nn,nn),U(nn,nn),e(nn)

and preparing for the eigenvalue problem, and calling the subroutine, again in the mainprogram

c.... Compute eigenvalues of Tangent stiffness matrixdo j=1,nn


do k=1,nnAA(j,k) = A(j,k)

end doe(j) = 1.d0

end docall eigens(AA,U,e,nn,8,NR)enext = e(1)

To get the solution to switch from the trivial branch to a buckled configuration we sim-ply watch for a change in sign of the first eigenvalue. When detected we provide aguess at the next state in the direction of the first eigenvector, scaled to have length α,as the following code segment indicates (again in the main program).

c.... Update values for previous converged state and guess at next stateif ((elast.gt.zero).and.(enext.lt.zero)) thensum = zerodo j = 1,nnsum = sum + U(j,1)*U(j,1)

end dosum = alpha/dsqrt(sum)do j = 1,nn

x(j) = x(j) + sum*U(j,1)end do

elsedo j=1,ndmtemp = xo(j)xo(j) = x(j)x(j) = two*x(j) - temp

end doend ifelast = enext

The problem was run for the same beam as used in the example in the text. The inputproperties are given below


Beam properties, end load and dist. load amplitudesColumn Length : 10.00EI, GA, EA : 1000. .1000E+07 .1000E+07Mo, Qo, Po : .0000E+00 .0000E+00 -2.470mo1, qo1, po1 : .0000E+00 .0000E+00 .0000E+00mo2, qo2, po2 : .0000E+00 .0000E+00 .0000E+00mo3, qo3, po3 : .0000E+00 .0000E+00 .0000E+00

The computation was run for 50 steps, some of which are included below. The resultsare shown in the accompanying figure.

Step Load u(L) w(L) theta(L) NU || b ||1 .50000E+00 -.12233E-04 .00000E+00 .00000E+00 2 .7486E-102 .10000E+01 -.24466E-04 .00000E+00 .00000E+00 1 .1692E-103 .15000E+01 -.36699E-04 .00000E+00 .00000E+00 1 .2108E-10

[ steps omitted for brevity ]19 .95000E+01 -.23243E-03 .00000E+00 .00000E+00 1 .3641E-1020 .10000E+02 -.24466E-03 .00000E+00 .00000E+00 1 .2708E-1121 .10500E+02 -.25690E-03 .00000E+00 .00000E+00 1 .1348E-1022 .10488E+02 -.96550E-03 .10771E+00 .16919E-01 15 .6174E-1023 .10985E+02 -.18518E-02 .16095E+00 .25284E-01 8 .6422E-0824 .11483E+02 -.27403E-02 .20061E+00 .31514E-01 6 .6705E-0925 .11982E+02 -.36297E-02 .23366E+00 .36707E-01 6 .6114E-1026 .12481E+02 -.45194E-02 .26258E+00 .41252E-01 5 .4401E-0927 .12981E+02 -.54094E-02 .28863E+00 .45346E-01 5 .5623E-10


28 .13480E+02 -.62994E-02 .31251E+00 .49099E-01 5 .9440E-1029 .13980E+02 -.71895E-02 .33470E+00 .52586E-01 5 .6043E-1030 .14479E+02 -.80797E-02 .35550E+00 .55856E-01 4 .7212E-08

[ steps omitted for brevity ]

Step Load EV1 EV2 EV31 .50000E+00 .28233E+01 .21087E+05 .44830E+052 .10000E+01 .26760E+01 .21087E+05 .44830E+053 .15000E+01 .25287E+01 .21087E+05 .44830E+05

[ steps omitted for brevity ]19 .95000E+01 .17229E+00 .21086E+05 .44830E+0520 .10000E+02 .25003E-01 .21085E+05 .44830E+0521 .10500E+02 -.12228E+00 .21085E+05 .44830E+0522 .10488E+02 .23747E+00 .21086E+05 .44831E+0523 .10985E+02 .53030E+00 .21086E+05 .44833E+0524 .11483E+02 .82388E+00 .21086E+05 .44835E+0525 .11982E+02 .11177E+01 .21086E+05 .44837E+05

[ steps omitted for brevity ]

One can see that the first occurrence of a negative first eigenvalue is step 21. Thus, thebifurcation path is sought in step 22. The actual bifurcation point is λ= 10, which isapproximately reached in step 20. However, in step 20 the first eigenvalue is still posi-tive (albeit very near zero). This worst case scenario gives us about a two step lag ingetting on the right path. We show the jump onto the path with a dotted line in the fig-ure.

10

0 0.2 0.4

λ

w(ℓ)

20


Appendix A. Fortran ProgramsThe solutions to the problems in this chapter were done for the first edition of

the text, inwhich the programswere written inFortran. The original Fortran codesare listed here for reference because the solutions only indicate small code seg-ments that need to be inserted into this code. Perhaps I will get the energy to redothese solutions in MatLab, but for now these will have to do.

The Program NEWTON

program NEWTON

implicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3),xo(3)

c.... Read problem parameters

read(5,1000) tol,alpharead(5,1001) maxsteps,maxitread(5,1000) xo(1),xo(2),xo(3)read(5,1000) x(1), x(2), x(3)write(6,2000) tol,alpha,maxsteps,maxit,(xo(j),j=1,3),(x(j),j=1,3)

c.... Initialize values for load step zero,set next trial state

n = 0sum = 0.0do j = 1,3b(j) = x(j) - xo(j)sum = sum + b(j)*b(j)




do 2 n = 1,maxsteps

c.... Perform Newton iteration at each load step

nu = 01 nu = nu + 1

c.... Compute equilibrium and constraint at current state

b(1) = 2.0*x(1) - x(2) - x(3)*dsin(x(1))b(2) = -x(1) + x(2) - x(3)*dsin(x(2))b(3) = 0.0do k = 1,3


c.... Compute Hessian matrix A

A(1,1) = 2.0 - x(3)*dcos(x(1))A(1,2) = -1.0A(1,3) = -dsin(x(1))A(2,1) = -1.0A(2,2) = 1.0 - x(3)*dcos(x(2))A(2,3) = -dsin(x(2))A(3,1) = 2.0*(x(1) - xo(1))A(3,2) = 2.0*(x(2) - xo(2))A(3,3) = 2.0*(x(3) - xo(3))

c.... Compute norm of residual for convergence test

test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2)


c.... Compute eigenvalues of Tangent stiffness matrix

tr = A(1,1) + A(2,2)det = A(1,1)*A(2,2) - A(2,1)*A(1,2)eig1 = 0.5*(tr - dsqrt(tr**2 - 4*det))eig2 = 0.5*(tr + dsqrt(tr**2 - 4*det))

c... Update state vector

call invert(A,3,3)do j=1,3do k=1,3x(j) = x(j) - A(j,k)*b(k)

end doend do

c.... Test for convergence, if successful output values and update

if ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test

c.... Set values for previous converged state and guess at next state

do j=1,3temp = xo(j)xo(j) = x(j)x(j) = 2.0*x(j) - temp

end do


The Program ELASTICA

program Elasticaimplicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3),xo(3)



read(5,1000) tol,alpha,xlengthread(5,1001) maxsteps,maxit,interread(5,1000) xo(1),xo(2),xo(3)read(5,1000) x(1), x(2), x(3)write(6,2000) tol,alpha,maxsteps,maxit,inter,xlength,* (xo(j),j=1,3),(x(j),j=1,3)

c.... Initialize constants of integration

dz = one/(2*inter)npoints = 2*inter + 1

c.... Initialize values for load step zero,set next trial state

n = 0sum = zerodo j = 1,3b(j) = x(j) - xo(j)sum = sum + b(j)*b(j)




do 2 n = 1,maxsteps


nu = 01 nu = nu + 1


c.... Execute Simpson integration of Hessian and residual components

call zerovec(b,3)call zerovec(A,9)z = zerodo m=1,npointscall simpson(m,npoints,dz,xlength,factor)call fcn(A,b,x,z,xlength,factor)z = z + dz

end do

c.... Finish computation of Hessian and residual

do k=1,3b(3) = b(3)+(x(k)-xo(k))**2A(3,k) = two*(x(k)-xo(k))

end dob(3) = b(3) - alpha**2


test = dsqrt(b(1)**2 + b(2)**2 + b(3)**2)

c.... Compute eigenvalues of Tangent stiffness matrix

tr = A(1,1) + A(2,2)det = A(1,1)*A(2,2) - A(2,1)*A(1,2)eig1 = 0.5d0*(tr - dsqrt(tr**2 - 4.d0*det))eig2 = 0.5d0*(tr + dsqrt(tr**2 - 4.d0*det))

c... Update state vector

call invert(A,3,3)do j=1,3do k=1,3x(j) = x(j) - A(j,k)*b(k)

end doend do

c.... Test for convergence, if successful output values and update

if ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’write(6,2001) n,x(1),x(2),x(3),eig1,eig2,nu,test

c.... Set values for previous converged state and guess at next state

do j=1,3temp = xo(j)xo(j) = x(j)x(j) = two*x(j) - temp

end do


c.... IO Formats

1000 format(3f10.0)1001 format(3i10)2000 format(’ Equilibrium Path for Cantilever Elastica’//

* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Number of integration intervals : ’,i6/* 5x, ’ Column Length : ’,f12.4/* 5x, ’ Initial a1, a2, load : ’,3f12.4/* 5x, ’ Next a1, a2, load : ’,3f12.4//* ’ n’ 10x,’a1’,10x,’a2’,8x,’P/EI’,* 8x, ’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’)

2001 format(i5,5f12.5,i5,e12.4)end

C---Compute Simpson contribution to Hessian and residual matrices--------FCN

subroutine fcn(A,b,x,z,xlength,factor)

implicit double precision (a-h,o-z)dimension A(3,3),b(3),x(3)data ev1/1.570796327d0/,ev2/4.712388981d0/


c.... Base functions and their derivatives

h1 = dsin(ev1*z)h2 = dsin(ev2*z)dh1 = ev1*dcos(ev1*z)/xlengthdh2 = ev2*dcos(ev2*z)/xlength

c.... Compute rotation and first derivative at current point

theta = x(1)*h1 + x(2)*h2dtheta = x(1)*dh1 + x(2)*dh2c1 = x(3)*dcos(theta)c2 = x(3)*dsin(theta)

c.... Compute integral part of residual vector

b(1) = b(1) - (c2*h1 - dtheta*dh1)*factorb(2) = b(2) - (c2*h2 - dtheta*dh2)*factor

c.... Compute integral part of Hessian matrix

A(1,1) = A(1,1) + (dh1*dh1 - c1*h1*h1)*factorA(1,2) = A(1,2) + (dh1*dh2 - c1*h1*h2)*factorA(1,3) = A(1,3) - (dsin(theta)*h1)*factorA(2,1) = A(2,1) + (dh2*dh1 - c1*h2*h1)*factorA(2,2) = A(2,2) + (dh2*dh2 - c1*h2*h2)*factorA(2,3) = A(2,3) - (dsin(theta)*h2)*factor

returnend

C----Evaluate Simpson integration weight factors---------------------SIMPSON

subroutine simpson(m,npoints,dz,xlength,factor)


c = xlength*dz/3.d0n = mod(m,2)

if((m.eq.1).or.(m.eq.npoints)) thenfactor = c

else if (n.eq.0) thenfactor = 4.d0*c

elsefactor = 2.d0*c

endif

returnend

C---Initialize the array V to zero-----------------------------------ZEROVEC

subroutine zerovec(v,n)

implicit double precision (a-h,o-z)dimension v(n)

do i=1,nv(i) = 0.d0

end do

returnend

c.... Output formats

1000 format(3f10.0)1001 format(2i10)2000 format(’ Equilibrium Path for Chapter 10 Example Problem’//

* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Initial theta 1, theta 2, load : ’,3f12.4/* 5x, ’ Next theta 1, theta 2, load : ’,3f12.4//* 10x,’Theta 1’,5x,’Theta 2’,8x,’Pl/k’,


* 8x,’EV 1’,8x,’EV 2’,’ NU’,4x,’|| b ||’)2001 format(i5,5f12.5,i5,e12.4)

end

C---- Invert matrix A (nmax,nmax) when array dimension is NDM ---INVERT

subroutine invert(a,nmax,ndm)

implicit double precision (a-h,o-z)dimension a(ndm,ndm)

do 200 n = 1,nmaxd = a(n,n)do 100 j = 1,nmaxa(n,j) = -a(n,j)/d

100 continue

do 150 i = 1,nmaxif(n.eq.i) go to 150do 140 j = 1,nmaxif(n.ne.j) a(i,j) = a(i,j) + a(i,n)*a(n,j)

140 continue

a(i,n) = a(i,n)/d150 continue

a(n,n) = 1.0/d200 continue

returnend

The Program NONLINEARBEAM

program NonlinearBeam


parameter (nbasis=6)parameter (ndm=3*nbasis+1)

dimension A(ndm,ndm),b(ndm),x(ndm),xo(ndm),D(3)common /loads/ dmo(3),dqo(3),dpo(3)common /out/ npts,nnstep



read(5,1000) tol,alpha,xlengthread(5,1001) maxsteps,maxit,interread(5,1001) npts,nnstepread(5,1000) D(1),D(2),D(3)read(5,1000) Cm,Cq,Cpread(5,1000) (dmo(i),i=1,3)read(5,1000) (dqo(i),i=1,3)read(5,1000) (dpo(i),i=1,3)write(6,2000) tol,alpha,maxsteps,maxit,inter,npts,nnstep,nbasiswrite(6,2001) xlength,D(1),D(2),D(3),Cm,Cq,Cp,* (dmo(i),dqo(i),dpo(i),i=1,3)

c.... Open output files and write header for summary output

open(3,file=’disp.beam’,status=’unknown’)open(4,file=’coef.beam’,status=’unknown’)write(6,2002)

c.... Initialize integration increment, change intervals to Simpson points

dz = one/(2*inter)npoints = 2*inter + 1

c.... Initialize values for load step zero,set first load level

call zerovec(x,ndm)call zerovec(xo,ndm)x(ndm) = alpha



do 2 nstep = 1,maxsteps


nu = 01 nu = nu + 1

c.... Execute numerical integration of Hessian and residual components

call zerovec(b,ndm)call zerovec(A,ndm*ndm)z = 0.d0

do mpoint = 1,npointscall simpson(mpoint,npoints,dz,xlength,factor)call fcn(A,b,D,x,z,xlength,factor,ndm,nbasis)z = z + dz

end do

C.... Add end load terms to the residual and coefficient matrix

do i=1,nbasiscall basis(i,one,xlength,h,dh,nbasis)mm = 3*(i-1)b(mm+1) = b(mm+1) - h*Cp*x(ndm)b(mm+2) = b(mm+2) - h*Cq*x(ndm)b(mm+3) = b(mm+3) - h*Cm*x(ndm)A(mm+1,ndm) = A(mm+1,ndm) - h*CpA(mm+2,ndm) = A(mm+2,ndm) - h*CqA(mm+3,ndm) = A(mm+3,ndm) - h*Cm

end do

c.... Add arc-length constraint terms to Hessian and residual

do k=1,ndmb(ndm) = b(ndm) + (x(k) - xo(k))**2A(ndm,k) = two*(x(k) - xo(k))

end dob(ndm) = b(ndm) - alpha**2


test = zerodo k=1,ndm

test = test + b(k)**2end dotest = dsqrt(test)

c.... Update state vector

call invert(A,ndm,ndm)do j=1,ndmdo k=1,ndmx(j) = x(j) - A(j,k)*b(k)

end doend do

c.... Test for convergence, if successful output values

if ((test.gt.tol).and.(nu.lt.maxit)) go to 1if (nu.ge.maxit) stop ’Iteration limit exceeded’call results(x,xlength,nstep,nu,test,ndm,nbasis)

c.... Update values for previous converged state and guess at next state

do j=1,ndmtemp = xo(j)xo(j) = x(j)x(j) = two*x(j) - temp

end do

2 continueclose(3)close(4)stop ’Maximum number of steps exhausted’

c.... IO Formats

1000 format(3f10.0)1001 format(3i10)


2000 format(/’ Fully Nonlinear Beam Analysis’//* 5x, ’ Convergence tolerance : ’,e12.4/* 5x, ’ Arc length parameter : ’,f12.4/* 5x, ’ Number of load steps : ’,i6/* 5x, ’ Maximum number of iterations : ’,i6/* 5x, ’ Number of integration intervals : ’,i6/* 5x, ’ Number of points in plotted shape : ’,i6/* 5x, ’ Output shape every N points, N = : ’,i6/* 5x, ’ Number of basis functions : ’,i6/)

2001 format(/’ Beam properties, end load and dist. load amplitudes’/* 5x, ’ Column Length : ’,g12.4/* 5x, ’ EI, GA, EA : ’,3g12.4/* 5x, ’ Mo, Qo, Po : ’,3g12.4/* 5x, ’ mo1, qo1, po1 : ’,3g12.4/* 5x, ’ mo2, qo2, po2 : ’,3g12.4/* 5x, ’ mo3, qo3, po3 : ’,3g12.4//)

2002 format(’ Step’,8x,’Load’,8x,’u(L)’,8x,’w(L)’,4x,’theta(L)’,* ’ NU’,4x,’|| b ||’)end

C----Compute contribution to A and b at current integration point-------FCN

subroutine fcn(A,b,D,x,z,xlength,factor,ndm,nbasis)


dimension A(ndm,ndm),b(ndm),x(ndm),D(3)dimension G(4,4),GB(4,3),BGB(3,3)

data zero/0.d0/,one/1.d0/

c.... Compute displacements and derivatives

du = zerodw = zerodtheta = zerotheta = zero

do i=1,nbasiscall basis(i,z,xlength,h,dh,nbasis)du = du + x(3*i-2)*dhdw = dw + x(3*i-1)*dhdtheta = dtheta + x(3*i)*dhtheta = theta + x(3*i)*h

end do

ct = dcos(theta)st = dsin(theta)

c.... Compute axial strain Áo, shear strain βo, and curvature κoepsilon = dw*st + (one + du)*ct - onebeta = dw*ct - (one + du)*stcurv = dtheta

c.... Compute axial force, shear force, bending moment, and other forces

bend = D(1)*curvshear = D(2)*betaaxial = D(3)*epsilonHor = axial*ct - shear*stVer = axial*st + shear*ctXi = (one + du)*Hor + dw*VerYi = dw*Hor - (one + du)*Ver

c.... Compute components of [ETDE+G] store in matrix G

G(1,1) = D(2)*st*st + D(3)*ct*ctG(1,2) = ct*st*(D(3)-D(2))G(1,3) = D(2)*st*(one+epsilon) + D(3)*ct*beta - VerG(1,4) = zeroG(2,2) = D(2)*ct*ct + D(3)*st*stG(2,3) = D(3)*st*beta - D(2)*ct*(one+epsilon) + HorG(2,4) = zeroG(3,3) = D(2)*(one+epsilon)**2 + D(3)*beta**2 - XiG(3,4) = zeroG(4,4) = D(1)


c.... Compute the rest of [ETDE+G] by symmetry

do i=1,3do j=i+1,4G(j,i) = G(i,j)

end doend do

c.... Form stiffness matrix K and store it in matrix Ado i=1,nbasiscall basis(i,z,xlength,hi,dhi,nbasis)do j=1,nbasiscall basis(j,z,xlength,hj,dhj,nbasis)

c.... Compute BTi [ETDE+G]B j noting the sparse structure of B

do k=1,4GB(k,1) = dhj*G(k,1)GB(k,2) = dhj*G(k,2)GB(k,3) = hj*G(k,3) + dhj*G(k,4)

end do

do k=1,3BGB(1,k) = dhi*GB(1,k)BGB(2,k) = dhi*GB(2,k)BGB(3,k) = hi*GB(3,k) + dhi*GB(4,k)

end do

c.... Assemble the result into the matrix Ado m=1,3mm = 3*(i-1)do n=1,3nn = 3*(j-1)A(mm+m,nn+n) = A(mm+m,nn+n)+ BGB(m,n)*factor

end doend do

end do

c.... Form integral part of residual force and assemble into matrix bmm = 3*(i-1)call applied(z,dm,dq,dp,x(ndm),1)b(mm+1) = b(mm+1) + (dhi*Hor - hi*dp)*factorb(mm+2) = b(mm+2) + (dhi*Ver - hi*dq)*factorb(mm+3) = b(mm+3) + (dhi*bend + hi*(Yi - dm))*factor

c.... Form the integral part of load factor part of matrix A

call applied(z,dm,dq,dp,x(ndm),2)A(mm+1,ndm) = A(mm+1,ndm) - hi*dp*factorA(mm+2,ndm) = A(mm+2,ndm) - hi*dq*factorA(mm+3,ndm) = A(mm+3,ndm) - hi*dm*factor

end doreturnend

C---- Evaluate ith basis function h and derivative dh----------------BASIS

subroutine basis(i,z,xlength,h,dh,nbasis)


n = mod(i,2)m = i/2 + na = dble(2*m-1)*2.d0*datan(1.d0)

if (n.eq.0) thenh = dsin(a*z)dh = a*dcos(a*z)/xlength

elseh = 1.d0 - dcos(a*z)dh = a*dsin(a*z)/xlength

end ifreturnend


C----Evaluate the distributed load functions at point z--------------APPLIED

subroutine applied(z,dm,dq,dp,clf,n)

implicit double precision (a-h,o-z)common /loads/ dmo(3),dqo(3),dpo(3)

c.... Compute the nominal values of the applied forces at point z

f = 1.d0 - zgo to (1,2), n

c.... Compute total transverse loads

1 dm = dmo(1) + (dmo(2) + dmo(3)*f)*clfdq = dqo(1) + (dqo(2) + dqo(3)*f)*clfdp = dpo(1) + (dpo(2) + dpo(3)*f)*clfreturn

c.... Compute transverse loads associated with load factor only

2 dm = dmo(2) + dmo(3)*fdq = dqo(2) + dqo(3)*fdp = dpo(2) + dpo(3)*freturnend

C----Print results of current step to various files----------------RESULTS

subroutine results(x,xlength,nstep,nu,test,ndm,nbasis)

implicit double precision (a-h,o-z)logical plot

dimension x(ndm),disp(3)common /out/ npts,nnstep

c.... Determine if current step is an output step

if(mod(nstep,nnstep).eq.0) thenplot = .true.nplot = npts

elseplot = .false.nplot = 1

end if

c.... Write coefficients for current step

if (plot) thenwrite(4,2001) nstep, x(ndm), nu, testdo i=1,nbasisii = 3*(i-1)write(4,2002) i,(x(ii+k),k=1,3)

end doend if

c.... Compute and print current geometry of beam

if (plot) write(3,3001) nstep,x(ndm)z = 0.d0dz = 1.0/nplot

do ii=1,nplot+1do j=1,3disp(j) = 0.d0do i=1,nbasiscall basis(i,z,xlength,h,dh,nbasis)disp(j) = disp(j) + h*x(3*(i-1)+j)

end doend do

if (plot) write(3,3000) z*xlength,z*xlength+disp(1),disp(2)z = z + dz

end do

write(6,2003) nstep,x(ndm),(disp(j),j=1,3),nu,testreturn

2001 format(/’ Load Step :’,i4,5x,’ Load factor :’,e12.5,* /’ Iterations :’,i4,5x,’ Norm of Residual :’,e12.5/


* /’ i’,15x,’ a(i)’,15x,’ b(i)’,15x,’ c(i)’)2002 format(i5,3e20.5)2003 format(i5,4e12.5,i5,e12.4)3000 format(4e15.5)3001 format(’ Step : ’,i5,5x,’ Load : ’,e15.5)

end

Solution

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Transcript of Solution