Review of Strain Energy Methods and Introduction to...

Duke UniversityDepartment of Civil and Environmental Engineering

CEE 421L. Matrix Structural AnalysisHenri P. Gavin

Fall, 2012

Review of Strain Energy Methods andIntroduction to Stiffness Matrix Methods

of Structural Analysis

1 Strain Energy

Strain energy is stored within an elastic solid when the solid is deformed under load.In the absence of energy losses, such as from friction, damping or yielding, the strain energyis equal to the work done on the solid by external loads. Strain energy is a type of potentialenergy.

Consider the work done on an elastic solid by a single point force F . When the elasticsolid carries the load, F , it deforms with strains (ε and γ) and the material is stressed (σand τ).

D is a displacement in the same location and in the same direction as a point force, F .D and F are colocated. The work done by the force F on the elastic solid is the area underthe force vs. displacement curve.

W =∫F dD (1)

This work is stored as strain energy U within the elastic solid.

U = 12

∫V

(σxxεxx + σyyεyy + σzzεzz + τxyγxy + τxzγxz + τyzγyz) dV. (2)

This is a very general expression for the strain energy, U , and is not very practical forstructural elements like bars, beams, trusses, or frames.

2 CEE 421L. Matrix Structural Analysis – Duke University – Fall 2012 – H.P. Gavin

1.1 Bars

For a bar in tension or compression, we have internal axial force, N , only,

ε

Nx

Nx

x

dl

σxx

xx dl

��

��

so σyy = 0, σzz = 0, τxy = 0, τxz = 0, and τyz = 0, and

U = 12

∫Vσxxεxx dV ,

where σxx = N/A and εxx = N/EA. Substituting dV = A dx we get

U = 12

∫L

N(x)2

E(x) A(x) dx , (3)

and if N , E, and A are constantU = 1

2N2 L

E A.

Alternatively, we may express the strain as a function of the displacements along the barux(x), εxx = ∂ux(x)/∂x, and σxx = E ∂ux(x)/∂x. Again substituting dV = A dx ,

U = 12

∫LE(x) A(x)

(∂ux(x)∂x

)2

dx , (4)

and if E, A and ∂ux/∂x = (u2 − u1)/L are constants,

U = 12EA

L(u2 − u1)2

CC BY-NC-ND H.P. Gavin

http://creativecommons.org/licenses/by-nc-nd/3.0/

Strain Energy and Matrix Methods of Structural Analysis 3

1.2 Beams

For a beam in bending we have internal bending moments, M , and internal shear forces,V . For slender beams the effects of shear deformation are usually neglected.

��

��

��

��

Mzzzz

M

y

x

σxx

dl

v" dl

As in the axially loaded bar, σyy = 0, σzz = 0, τxy = 0, τxz = 0, and τyz = 0, and

U = 12

∫Vσxxεxx dV .

For bending, σxx = My/I and εxx = My/EI. Substituting dV = dA dx,

U = 12

∫L

∫A

M(x)2 y2

E(x) I(x)2 dA dx ,

where∫A y

2dA = I, so

U = 12

∫L

M(x)2

E(x) I(x) dx . (5)

Alternatively, we may express the moment in terms of the curvature of the beam, φ ≈∂2uy/∂x

2,

M(x) = E(x) I(x) ∂2uy(x)∂x2 ,

from which σxx = E (∂2uy/∂x2) y and εxx = (∂2uy/∂x

2) y, so that

U = 12

∫L

∫AE(x)

(∂2uy(x)∂x2

)2

y2 dA dx

where, again,∫A y

2dA = I, so

U = 12

∫LE(x) I(x)

(∂2uy(x)∂x2

)2

dx . (6)




1.3 Summary

External work is done by a set of forces, Fi, on a linear elastic solid, producing a set ofdisplacements, Di, in the same locations and directions.

��

��

��

��

��

��

��

��

��

��

��

��

��

��

i

i

D

F

D

Fj

D j

n

1

F1

D

F

n

F

D D

FF

D D

D

F

D

1

1

1

1

i

i

n

Fj

F j

Dj

ji

i

Fn

Fn

Dn

The work done by these forces is

W = 12F1D1 + 1

2F2D2 + 12F3D3 + · · ·

The external forces are resisted by internal moments, M , and axial forces, N . The totalstrain energy stored within the solid is

U = 12

∫L

M2

E Idx + 1

2∑j

N2j Lj

Ej Aj(7)

where the first term is the integral over all lengths of all the beams and the second term isthe sum over all the bars. If torsion and shear are included, then two additional terms are

12

∫L

T 2

G Jdx and 1

2

∫L

V 2

G A/αdx .

Alternatively, we can think of external forces producing curvatures (∂2uy/∂x2) by bend-

ing, and axial stretches (∂ux/∂x). In this case

U = 12

∫LE I

(∂2uy∂x2

)2

dx + 12∑j

Ej AjLj

(u2j − u1j)2 (8)

If torsion and shear are included, then two additional terms are

12

∫LG J

(∂uxθ∂x

)2

dx , and 12

∫LG A/α

(∂uy∂x

)2

dx ,

where uxθ is the torsional rotation about the x-axis, ∂uxθ/∂x is the torsional shear strain,γxθ, (on the face perpendicular to the x-axis and in the θ-direction) and ∂uy/∂x is the shearstrain, γxy, (on the face perpendicular to the x-axis and in the y-direction).

Analyses using expressions of the form of equations (3), (5), or (7) are called forcemethod or flexibility method analyses.

Analyses using expressions of the form of equations (4), (6), or (8) are called displace-ment method or stiffness method analyses.




2 Castigliano’s Theorems

2.1 Castigliano’s Theorem - Part I

U =∫F dD ... strain energy

��

��

0

Dj

∆

jD∆D+j

Dj

U(D)

jF

∆U

Fi = ∆U∆Di

= ∂U

∂Di

A force, Fi, on an elastic solid is equal to the derivative of the strain energy with respect tothe displacement, Di, in the direction and location of the force, Fi.

2.2 Castigliano’s Theorem - Part II

U∗ =∫D dF ... complementary strain energy

��

��

0

∆ jFF∆ jF+j

jF

Dj

U*(F)

∆U*

Di = ∆U∗

∆Fi= ∂U∗

∂Fi

A displacement, Di, on an elastic solid is equal to the derivative of the complementary strainenergy with respect to the force, Fi, in the direction and location of the displacement, Di.

If the solid is linear elastic, then U∗ = U .




3 Superposition

Superposition is an extremely powerful idea that helps us solve problems that are stati-cally indeterminate. To use the principle of superposition, the system must behave in a linearelastic fashion.

The principle of superposition states:

Any response of a system to multiple inputs can be represented as the sum of the re-sponses to the inputs taken individually.

By “response” we can mean a strain, a stress, a deflection, an internal force, a rotation,an internal moment, etc.

By “input” we can mean an externally applied load, a temperature change, a supportsettlement, etc.

4 Detailed Example of Castigliano’s Theorem and Superposition

An example of a statically indeterminate system with external loads w(x) and threeredundant reaction forces, RB, RC , and RD, is shown below.

��

��

��

��

��

��

��

x

A B C

y

w(x)

EA

L

H

D

EI

In general, the displacements at the locations of the unknown reaction forces are known,and, in this example these displacements will be taken as zero: DB = 0, DC = 0, DD = 0.

Invoking the principle of superposition, we may apply the external loads, (w(x)) andthe unknown reactions (RB, RC , and RD) individually, and then sum-up the responses toeach individual load. Further, we may represent the response to a reaction force, (e.g., RB) asthe response to a unit force co-located with the reaction force, times the value of the reactionforce. Note that all four systems to the right of the equal sign in the following figure arestatically determinate. Expressions for Mo(x), m1(x), m2(x), m3(x), No(x), n1(x), n2(x),and n3(x) may be found from static equilibrium alone.




��

��

��

��

��

��

��

��

��

��

��

��

��

��

w(x)

A B C

C

B

RB

RC

R

11

1

+

+

+

=

N(x) N (x)o

M(x) M (x)o

m (x)3

m (x)2

m (x)1

n (x)3

n (x)2

n (x)1

w(x)

CBA

D

D

* R

* RB

C

D* R

D

D

In equation form, the principle of superposition says:

M(x) = Mo(x) +m1(x)RB +m2(x)RC +m3(x)RD (9)N = No + n1RB + n2RC + n3RD (10)

(Note that in this particular example, No(x) = 0, n1 = 0, n2 = 0, n3 = 1, m1(x) = 0 forx > xB, and m2(x) = 0 for x > xC .)




The total strain energy, U , in systems with bending strain energy and axial strainenergy is,

U = 12

∫ L

0

M(x)2

EIdx + 1

2∑ N2 H

EA(11)

We are told that the displacements at points B, C, and D are all zero and we need to assumethe structure behaves linear elastically in order to invoke superposition in the first place.Therefore, from Castigliano’s Second Theorem,

Di = ∂U∗

∂Fi= ∂U

∂Fi,

we obtain three expressions for the facts that DB = 0, DC = 0, and DD = 0.

DB = 0 = ∂U

∂RB

DC = 0 = ∂U

∂RC

DD = 0 = ∂U

∂RD

Inserting equation (11) into the three expressions for zero displacement at the fixed reactions,noting that EI and EA are constants in this problem, and noting that the strain energy, U ,depends on the reactions R, only through the internal forces, M and N , we obtain

DB = 0 = 1EI

∫ L

0M(x)∂M(x)

∂RB

dx + H

EAN∂N

∂RB

DC = 0 = 1EI

∫ L

0M(x)∂M(x)

∂RC

dx + H

EAN∂N

∂RC

DD = 0 = 1EI

∫ L

0M(x)∂M(x)

∂RD

dx + H

EAN∂N

∂RD

Now, from the superposition equations (9) and (10), ∂M(x)/∂RB = m1(x), ∂M(x)/∂RC =m2(x), ∂M(x)/∂RD = m3(x), ∂N(x)/∂RB = n1, ∂N(x)/∂RC = n2, and ∂N(x)/∂RD = n3.Inserting these expressions and the superposition equations (9) and (10) into the aboveequations for DB, DC , and DD,

DB = 0 = 1EI

∫ L

0[Mo +m1RB +m2RC +m3RD] m1 dx+ H

EA[No +n1RB +n2RC +n3RD] n1

DC = 0 = 1EI

∫ L



DD = 0 = 1EI

∫ L



These three expressions contain the three unknown reactions RB, RC , and RD. Everythingelse in these equations (m1(x), m2(x) ... n3) can be found without knowing the unknown




reactions. By taking the unknown reactions out of the integrals (they are constants), we canwrite these three equations in matrix form.∫ L

0m1m1EI

dx+ n1n1HEA

∫ L0

m1m2EI

dx+ n1n2HEA

∫ L0

m1m3EI

dx+ n1n3HEA∫ L

0m2m1EI

dx+ n2n1HEA

∫ L0

m2m2EI

dx+ n2n2HEA

∫ L0

m2m3EI

dx+ n2n3HEA∫ L

0m3m1EI

dx+ n3n1HEA

∫ L0

m3m2EI

dx+ n3n2HEA

∫ L0

m3m3EI

dx+ n3n3HEA

RB

RC

RD

= −

∫ L

0Mom1EI

dx+ Non1HEA∫ L

0Mom2EI

dx+ Non2HEA∫ L

0Mom3EI

dx+ Non3HEA

(12)

This 3-by-3 matrix is called a flexibility matrix, F. The values of the terms in the flexibilitymatrix depend only on the responses of the structure to unit loads placed at various points inthe structure. The flexibility matrix is therefore a property of the structure alone, and doesnot depend upon the loads on the structure1. The vector on the right-hand-side dependson the loads on the structure. Recall that this matrix looks a lot like the matrix from thethree-moment equation. All flexibility matrices share several properties:

• All flexibility matrices are symmetric.

• No diagonal terms are negative.

• Flexibility matrices for structures which can not move or rotate without deforming arepositive definite. This means that all of the eigenvalues of a flexibility matrix describinga fixed structure are positive.

• The unknowns in a flexibility matrix equation are forces (or moments).

• The number of equations (rows of the flexibility matrix) equals the number of unknownforces (or moments).

1There are some fascinating cases in which the behavior does depend upon the loads, but that is a storyfor another day!




It is instructive to now examine the meaning of the terms in the matrix, F

F11 =∫ L

0

m1m1

EIdx+ n1n1H

EA= δ11

displacement at “1” due to unit force at “1”

��

��

A B C

D

1

11F

F21 =∫ L

0

m2m1

EIdx+ n2n1H

EA= δ21


��

��

A B C

D

1

21F

F12 =∫ L

0

m1m2

EIdx+ n1n2H

EA= δ12


��

��

A B C

D

F12

1

F31 =∫ L

0

m3m1

EIdx+ n3n1H

EA= δ31


��

��

A B C

D

1

31

F

The fact that F12 = F21 is called Maxwell’s Reciprocity Theorem.




5 Introductory Example of the Stiffness Matrix MethodIn this simple example, elements are springs with stiffness k. A spring with stiffness

k > 0 connecting point i to point j, will have a force f = k(dj−di) where di is the displacementof point i and dj is the displacement of point j. (Tension is positive so di points “into” thespring and dj points “away” from the spring.)

The stiffness matrix for this structure can be found using equilibrium and force-deflectionrelationships (f = kd) for the springs.

#1: ∑Fx = 0 : f1 − k1d1 + k2(d2 − d1) = 0

#2: ∑Fx = 0 : f2 − k2(d2 − d1)− k4d2 + k3(d3 − d2) = 0

#3: ∑Fx = 0 : f3 − k3(d3 − d2)− k5d3 = 0

In matrix form these three equations may be written: k1 + k2 −k2 0−k2 k2 + k3 + k4 −k3

0 −k3 k3 + k5

d1d2d3

=

f1f2f3

The displacements are found by solving the stiffness matrix equation for d, d = K−1 f .

• The matrix K is called a stiffness matrix.

• All stiffness matrices are symmetric.




• All diagonal terms of all stiffness matrices are positive.

• Stiffness matrices are diagonally dominant. This means that the diagonal terms areusually larger than the off-diagonal term.

• If the structure is not free to translate or rotate without deforming, then the stiffnessmatrix is positive definite. This mathematical property guarantees that the stiffnessmatrix is invertible, and a unique set of displacements, d, can be found by solvingK d = f .

• The total potential energy, U , in this system of springs is

U = 12k1d

21 + 1

2k2(d2 − d1)2 + 12k3(d3 − d2)2 + 1

2k4d22 + 1

2k5d23 .

You should be able to confirm that this is equal to

U = 12 dT K d

Also, note that no matter what the values of the displacements, d, may be, the energyU is always positive. The statement 1

2dTKd > 0 ∀d 6= 0 is another way of saying thatK is positive-definite.

• The set of forces required to deflect coordinate “i” by a deflection of “1 unit” equals the“i-th” column of the stiffness matrix. For example consider the case in which d1 = 1,d2 = 0, and d3 = 0, k1 + k2 −k2 0

−k2 k2 + k3 + k4 −k30 −k3 k3 + k5

1

00

=

k1 + k2−k2

0

,

which is equal to the first column of the stiffness matrix.




This fact may be used to derive the stiffness matrix:d1 = 1, d2 = 0, d3 = 0

f1 = k1 + k2 , f2 = −k2 , f3 = 0 ... 1st columnd1 = 0, d2 = 1, d3 = 0

f1 = −k2 , f2 = k2 + k3 + k4 , f3 = −k3 ... 2nd columnd1 = 0, d2 = 0, d3 = 1

f1 = 0 , f2 = −k3 , f3 = k3 + k5 ... 3rd column




6 Basic Concepts of the Stiffness Matrix Method

The previous example illustrates some of the basic concepts needed to apply the stiffnessmatrix method to structures made out of bars and beams. There are, however, a few addi-tional complications. Displacements in structures can be vertical, horizontal, or rotational,and structural bars and beams have a more complicated force-displacement relationshipsthan those of simple springs.

In applying the matrix stiffness method of structural analysis, structures are describedin terms of elements that connect nodes which can move in certain coordinate directions.

6.1 Elements

In the stiffness matrix method, structures are modeled as assemblies of elements such asbars, beams, cables, shafts, plates, and walls. Elements connect the nodes of the structuralmodel. Like the simple springs in the previous example, structural elements have clearlydefined, albeit more complicated, force-displacement relationships. The stiffness propertiesof structural elements can be determined from equilibrium equations, Castigliano’s Theorems,the principle of minimum potential energy, and/or the principle of virtual work. Structuralelements can be mathematically assembled with one another (like making a structural systemusing a set of tinker-toys), into a system of equations for the entire structure.

6.2 Nodes

The force-displacement relationship of a structural element is defined in terms of theforces and displacements at the nodes of the element. Nodes define the points where elementsmeet. The nodes in the model of a truss are at the joints between the truss bars. The nodesin the model of a beam or a frame are at the reaction locations, at locations at which elementsconnect to each other, and possibly at other intermediate locations.

6.3 Coordinates

Coordinates describe the location and direction at which forces and displacements acton an element or on a structure. Trusses are loaded with vertical and horizontal forces atthe joints. The joints of a 2D truss can move vertically and horizontally; so there are twocoordinates per node in a 2D truss. Beams and frames carry vertical and horizontal loadsas well as bending moments. The nodes of a 2D frame can move vertically, horizontally, andcan rotate; so there are three coordinates per node in a 2D frame.

Structural coordinates can be classified into two sets. Displacement coordinates haveunknown displacements but know forces. Reaction coordinates have unknown forces butknown displacements (usually zero).




6.4 Elements, Nodes and Coordinates

Planar (2D) truss bar elements have two nodes and four coordinates, two at each end.

Space (3D) truss bar elements have two nodes and six coordinates, three at each end.

Planar (2D) frame elements have two nodes and six coordinates, three at each end.

Space (3D) frame elements have two nodes and twelve coordinates, six at each end.




6.5 Structural Nodes and Coordinates

Planar (2D) truss nodes and coordinates

Planar (2D) frame nodes and coordinates




7 Relate the Flexibility Matrix to the Stiffness Matrix• Column “j” of the stiffness matrix: The set of forces at all coordinates required to

produce a unit displacement at coordinate “j”

• Column “j” of the flexibility matrix: The set of displacements at all coordinates result-ing from a unit force at coordinate “j”

• Stiffness matrix equation:

K11 K12 K13 · · · K1nK21 K22 K23 · · · K2nK31 K32 K33 · · · K3n

... ... ... . . . ...Kn1 Kn2 Kn3 · · · Knn

d1d2d3...dn

=

f1f2f3...fn

• Flexibility matrix equation:

F11 F12 F13 · · · F1nF21 F22 F23 · · · F2nF31 F32 F33 · · · F3n

... ... ... . . . ...Fn1 Fn2 Fn3 · · · Fnn

f1f2f3...fn

=

d1d2d3...dn

• K = F−1 and K−1 = F

• A useful fact for 2-by-2 matrices ...[a bc d

]−1

= 1ad− bc

[d −b−c a

]

... you should be able to prove this fact to yourself.



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