2.081J/16.230J Plates and Shells

Professor Tomasz Wierzbicki

Contents

1 Strain-Displacement Relation for Plates 11.1 1-D Strain Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Engineering Strain . . . . . . . . . . . . . . . . . . . . . . . . 11.1.2 Green-Lagrangian Strain . . . . . . . . . . . . . . . . . . . . . 1

1.2 3-D Strain Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Derivation of Green-Lagrangian Strain Tensor for Plates . . . 11.2.2 Specification of Strain-Displacement Relation for Plates . . . 5

2 Derivation of Constitutive Equations for Plates 102.1 Definitions of Bending Moment and Axial Force . . . . . . . . . . . . 102.2 Bending Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.1 Bending Moment . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.2 Bending Energy Density . . . . . . . . . . . . . . . . . . . . . 112.2.3 Total Bending Energy . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Membrane Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.1 Axial Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Membrane Energy Density . . . . . . . . . . . . . . . . . . . 142.3.3 Total Membrane Energy . . . . . . . . . . . . . . . . . . . . . 16

3 Development of Equation of Equilibrium and Boundary ConditionsUsing Variational Approach 173.1 Bending Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 17

3.1.1 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . . 173.1.2 First Variation of the Total Potential Energy . . . . . . . . . 203.1.3 Equilibrium Equation and Boundary Conditions . . . . . . . 243.1.4 Specification of Equation for Rectangular Plate . . . . . . . . 24

3.2 Bending-Membrane Theory of Plates . . . . . . . . . . . . . . . . . . 293.2.1 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . . 29

i

3.2.2 First Variation of the Total Potential Energy . . . . . . . . . 293.2.3 Equilibrium Equation and Boundary Conditions . . . . . . . 31

4 General Theories of Plate 344.1 Bending Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 34

4.1.1 Derivation of the Plate Bending Equation . . . . . . . . . . . 344.1.2 Reduction to a System of Two Second Order Equations . . . 354.1.3 Exercise 1: Plate Solution . . . . . . . . . . . . . . . . . . . . 364.1.4 Exercise 2: Comparison between Plate and Beam Solution . . 404.1.5 Exercise 3: Finite Difference Solution of the Plate Bending

Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.2 Membrane Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . 47

4.2.1 Plate Membrane Equation . . . . . . . . . . . . . . . . . . . . 474.2.2 Plate Equation for the Circular Membrane . . . . . . . . . . 484.2.3 Example: Approximation Solution for the Clamped Membrane 48

4.3 Buckling Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 514.3.1 General Equation of Plate Buckling . . . . . . . . . . . . . . 514.3.2 Linearized Buckling Equation of Rectangular Plates . . . . . 524.3.3 Analysis of Rectangular Plates Buckling . . . . . . . . . . . . 544.3.4 Derivation of Raleigh-Ritz Quotient . . . . . . . . . . . . . . 624.3.5 Ultimate Strength of Plates . . . . . . . . . . . . . . . . . . . 654.3.6 Plastic Buckling of Plates . . . . . . . . . . . . . . . . . . . . 714.3.7 Exercise 1: Effect of In-Plane Boundary Conditions, δw = 0 . 744.3.8 Exercise 2: Raleigh-Ritz Quotient for Simply Supported Square

Plate under Uniaxial Loading . . . . . . . . . . . . . . . . . . 764.4 Buckling of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.4.1 Transition from Global and Local Buckling . . . . . . . . . . 774.4.2 Local Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5 Buckling of Cylindrical Shells 845.1 Governing Equation for Buckling of Cylindrical Shells . . . . . . . . 84

5.1.1 Special Case I: Cylinder under Axial Load P , q = 0 . . . . . 865.1.2 Special Case II: Cylinder under Lateral Pressure . . . . . . . 875.1.3 Special Case III: Hydrostatic Pressure . . . . . . . . . . . . . 885.1.4 Special Case IV: Torsion of a Cylinder . . . . . . . . . . . . . 89

5.2 Derivation of the Linearized Buckling Equation . . . . . . . . . . . . 895.3 Buckling under Axial Compression . . . . . . . . . . . . . . . . . . . 90

5.3.1 Formulation for Buckling Stress and Buckling Mode . . . . . 905.3.2 Buckling Coefficient and Batdorf Parameter . . . . . . . . . . 93

5.4 Buckling under Lateral Pressure . . . . . . . . . . . . . . . . . . . . 965.5 Buckling under Hydrostatic Pressure . . . . . . . . . . . . . . . . . . 995.6 Buckling under Torsion . . . . . . . . . . . . . . . . . . . . . . . . . 1005.7 Influence of Imperfection and Comparison with Experiments . . . . . 102

ii

1 Strain-Displacement Relation for Plates

1.1 1-D Strain Measure

1.1.1 Engineering Strain

Engineering strain ε is defined as the relative displacement:

ε =ds− ds0

ds0(1)

where ds0 is the increment of initial lenght and ds is the increment of current length.

1.1.2 Green-Lagrangian Strain

Instead of comparing the length, one can compare the square of lengths:

E =ds2 − ds202ds20

(2)

=ds− ds0

ds0

ds+ ds02ds0

Where ds → ds0, the second term is Eq. (2) tends to unity, and the Green strainmeasure and the engineering strain become identical. Equation (2) can be put intoan equivalnet form:

ds2 − ds20 = 2Eds20 (3)

which will now be generalized to the 3-D case.

1.2 3-D Strain Measure

1.2.1 Derivation of Green-Lagrangian Strain Tensor for Plates

Let define the following quanties:

• a = [ai]: vector of the initial (material) coordinate system

• x = [xi]: vector of the current (spatial) coordinate system

• u = [ui]: displacement vector

where the index i = 1, 2, 3. The relation between those quantities is:

xi= ai + ui (4)

dxi= dai + dui

1

O

a

u

x

du

da dx

O

a

u

x

du

da dx

Now, the squares of the initial and the current length increment can be writtenin terms of ai and ui:

ds20 = daidajδij (5)

ds2= dxidxjδij (6)

=(dai + dui) (daj + duj) δij

where the Kronecker tensor δij reads:

δij =

¯¯1 0 00 1 00 0 1

¯¯ (7)

The vector u can be considered as a function of:

• the initial (material) coordinate system, u (a), which leads to Lagrangiandescripion, or

• the current (spatial) coordinates, u (x), which leads to the Eulerian descrip-tion

In structural mechanics, the Lagrangian description is preferable:

ui= ui (ai) (8)

dui=∂ui∂ak

dak = ui,k dak

duj =∂uj∂al

dal = uj ,l dal

Let us calculated the difference in the length square:

ds2 − ds20 = (dai + dui) (daj + duj) δij − daidajδij (9)

2

Using Eq. (8) and the definition of δij , the difference in the length square can betransformed into:

ds2 − ds20= (dujdai + duidaj + duiduj) δij (10)

= (uj ,l dal dai + ui,k dak daj + ui,k dak uj ,l dal) δij

= [uj ,l (dajδjl) dai + ui,k (daiδik) daj + ui,k uj ,l (daiδik) (dajδjl)] δij

= (uj ,i+ui,j +ui,k uj ,k ) daidaj

=2Eij daidaj

where, by analogy with the 1-D case, the Lagrangian or Green strain tensor Eij isdefined:

Eij =1

2(ui,j +uj ,i+uk,i uk,j ) (11)

In the case of small displacement gradient (uk,i¿ 1), the second nonlinear termcan be neglected leading to the defintion of the infinitesimal strain tensor:

εij =1

2(ui,j +uj ,i ) (12)

From the defintion, the strain tensor is symmetric εij = εji, which can be seen byintechanign the indices i for j and j for i. In the moderately large deflection theoryof structures, the nonlinear terms are important. Therefore, Eq. (11) will be usedas a starting point in the development of the general theory of plates.

Components of Green-Lagrangian Strain Tensor Let define the followingrange convention for indices:

• Greek letters: α, β, ... = 1, 2

• Roman letters: i, j, ... = 1, 2, 3

With this range convention, the Roman letters are also written as:

i=α, 3 (13)

j = β, 3

The Lagrangian or Green strain tensor can be expressed:

11 12 13

3

21 22 23

31 32 33 3 33

ij

E E EE E

E E EE

E E E E E

αβ β

α

= =

where Eαβ is the in-plane component of strain tensor, Eα3 and E3β are out-of-plane

3

shear components of strain tensor, and E33 is the through-thickness component ofstrain tensor. Similarily, displacement vector can be divided into two components:

1

2

3

i

u uu

u vu

u w w

α

= = =

where uα is the in-plane components of the displacement vector, and u3 = w is theout-of-plane components of the displacement vector and also called as the trans-verse displacement.

Initial Undeformed Configuration

x

z

η

ξ

middlesurface

Deformed Configuration

a

x u

uα

3u

Assumptions of the von Karman Theory The von Karman thoery of mod-erately large deflection of plates assumes:

1. The plate is thin. The thickness h is much smaller than the typical platedimension, h¿ L.

2. The magnitude of the transverse deflection is of the same order as the thicknessof plate, |w| = O (h). In practice, the present theory is still a good engineeringapproximation for deflections up to ten plate thickness.

4

3. Gradients of in-plane displacements uα,β are small so that their product orsquare can be neglected.

4. Love-Kirchhoff hypothesis is satisfied. In-plane displacements are a linearfunction of the z−coordiate (3-coordinate).

uα = u◦α − z u3,α (14)

where u◦α is the displacement of the middle surface, which is independent ofz−coordinate, i.e. u◦α,3= 0; and u3,α is the slope which is negative for the"smiling" beam.

z

5. The out-of-plane displacement is independent of the z−coordiante, i.e. u3,3=0.

1.2.2 Specification of Strain-Displacement Relation for Plates

In the theory of moderately large deflections, the strain-displacement relation canbe specified for plates.

In-Plane Terms of the Strain Tensors From the general expression, Eq. (11),the 2-D in-plane componets of the strain tensor are:

Eαβ =1

2(uα,β +uβ,α+uk,α uk,β ) (15)

Here, consider the last, nonlinear term:

uk,α uk,β = u1,α u1,β +u2,α u2,β +u3,α u3,β (16)

= uγ,α uγ ,β +u3,α u3,β

In the view of the Assumption 3, the first term in the above equation is zero,uγ ,α uγ ,β ' 0. Therefore, the 2-D in-plane components of strain tensor reads:

Eαβ =1

2(uα,β +uβ,α+w,α w,β ) (17)

5

where w = u3. Introducing Eq. (14) into Eq. (17), i.e. applying Love-Kirchhoffhypothesis, one gets:

Eαβ =1

2

£(u◦α − z w,α ) ,β +

¡u◦β − z w,β

¢,α+w,α w,β

¤(18)

=1

2

¡u◦α,β +u

◦β,α−2 z w,αβ +w,α w,β

¢=1

2

¡u◦α,β +u

◦β,α¢− z w,αβ +

1

2w,α w,β

From the definiton of the curvature, one gets:

καβ = −w,αβ (19)

Now, Eq. (18) can be re-casted in the form:

Eαβ = E◦αβ + z καβ (20)

where the strain tensor of the middle surface E◦αβ is composed of a linear and anonlinear term:

E◦αβ =1

2

¡u◦α,β +u

◦β,α¢+1

2w,α w,β (21)

In the limiting case of small displacements, the second term can be neglected ascompared to the first term. In the classical bending theory of plate, the in-planedisplacements are assumed to be zero uα = 0 so that strains are only due to thecurvatue:

Eαβ = z καβ (22)

where

καβ =

¯κ11 κ12κ21 κ22

¯= −

¯¯ ∂2w

∂x2∂2w

∂x∂y∂2w

∂x∂y

∂2w

∂y2

¯¯ = −w,αβ (23)

In the above equation, κ11 and κ22 are curvatures of the cylindrical bending, andκ12 is the twist which tells how the slope in the x−direction changes with they−direction:

κ12 =∂

∂y

µ∂w

∂x

¶

for a cylinder

12 0κ =

6

Interpretation of the linear terms: 12

³u◦α,β +u

◦β,α

´Each component can

be expressed in the followings:

ε11 =1

2(u1,1+u1,1 ) = u1,1=

du1dx

(24)

ε22 =1

2(u2,2+u2,2 ) = u2,2=

du2dy

(25)

ε12=1

2(u1,2+u2,1 ) =

1

2

µdu1dy

+du2dx

¶(26)

ε12|if u2=0=1

2

du1dy

yx

11ε

22ε

yx

12ε

2 0u ≡

1u

yx

Therefore, ε11 and ε22 are the tensile strain in the two directions, and ε12 is thechange of angles, i.e. shear strain.

Interpretation of the nonlinear term: 12w,α w,β Let α = 1 and β = 1.

Then, the nonlienar term reads:

1

2w,α w,β

¯α=1,β=1

=1

2

dw

dx

dw

dx=1

2

µdw

dx

¶2(27)

One can also obtain the same quantity by the defintion of 1-D Green-Lagrangianstrain:

E =ds2 − ds202ds20

'¡ds20 + dw2

¢− ds20

2ds20=1

2

µdw

ds0

¶2=1

2

µdw

dx

¶2(28)

7

x

,z w

dx

0ds

dsdw

0ds dx=2 2 2

0ds ds dw= +

Thus, the conclusion is that the nonlinear term 12w,α w,β represents the change of

length of the plate element due to finite rotations.

Out-Of-Plane Terms of the Strain Tensors Refering to the definition intro-duced in Section 1.2.1, there are three other componets of the strain tensor: E3β,Eα3 and E33. Using the general expression for the components of the strain tensor,Eq. (11), it can be shown that the application of Assumption 4 and 5 lead to thefollowing expressions:

E3β =1

2(u3,β +uβ,3+uk,3 uk,β ) (29)

=1

2[u3,β +uβ,3+(u1,3 u1,β +u2,3 u2,β +u3,3 u3,β )]

=1

2[u3,β −u3,β +(−u3,1 u1,β −u3,2 u2,β )]

=1

2(−u3,1 u1,β −u3,2 u2,β )

=−12w,γ uγ ,β

Eα3=1

2(uα,3+u3,α+uk,α uk,3 ) (30)

=1

2[uα,3+u3,α+(u1,α u1,3+u2,α u2,3+u3,α u3,3 )]

=1

2[−u3,α+u3,α+(−u1,α u3,1−u2,α u3,2 )]

=1

2(−u1,α u3,1−u2,α u3,2 )

=−12w,γ uγ ,α

8

E33=1

2(u3,3+u3,3+uk,3 uk,3 ) (31)

= u3,3+1

2

h(u1,3 )

2 + (u2,3 )2 + (u3,3 )

2i

=1

2

h(u1,3 )

2 + (u2,3 )2i

=1

2

h(−u3,1 )2 + (−u3,2 )2

i=1

2w,γ w,γ

The above are all second order terms which vanish for small deflection theory ofplates. In the theory of moderately larege deflection of plates, the out-of-plate shearstrains as well as the through-thickness strain is not zero. Therefore, an assumption"plane remains plane," expressed by Eq. (14), does not mean that "normal remainsnormal." The existance of the out-of-plane shear strain means that lines originallynormal to the middle surface do not remain normal to the deformed plate. However,the incremental work of these strains with the corresponding stresses is negligible:

E3βσ3β, Eα3σα3 and E33σ33, are small (32)

because the corresponding stress σ3β, σα3 and σ33 are small as compared to thein-plane stress σαβ. One can conclude that the elastic strain energy (and evenplastic dissipation) is well approximated using the plane strain assumption:Z

h

1

2σijεijdz '

Zh

1

2σαβεαβdz (33)

9

2 Derivation of Constitutive Equations for Plates

2.1 Definitions of Bending Moment and Axial Force

Hook’s law in plane stress reads:

σαβ =E

1− ν2[(1− ν) εαβ + ν εγγ δαβ] (34)

In terms of components:

σxx=E

1− ν2(εxx + ν εyy) (35)

σyy =E

1− ν2(εyy + ν εxx)

σxy =E

1 + νεxy

Here, strain tensor can be obtained from the strain-displacement relations:

εαβ = ε◦αβ + z καβ (36)

Now, define the tensor of bending moment:

Mαβ ≡Z h

2

−h2

σαβ z dz (37)

and the tensor of axial force (membrane force):

Nαβ ≡Z h

2

−h2

σαβ dz (38)

2.2 Bending Energy

2.2.1 Bending Moment

Let us assume that ε◦αβ = 0. The bending moment Mαβ can be calculated:

Mαβ =E

1− ν2

Z h2

−h2

[(1− ν) εαβ + ν εγγ δαβ] z dz (39)

=E

1− ν2£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤ Z h2

−h2

z dz

+E

1− ν2[(1− ν) καβ + ν κγγ δαβ]

Z h2

−h2

z2 dz

=Eh3

12 (1− ν2)[(1− ν) καβ + ν κγγ δαβ]

10

Here, we define the bending rigidity of a plate D as follows:

D =Eh3

12 (1− ν2)(40)

Now, one gets the moment-curvature relations:

Mαβ = D [(1− ν) καβ + ν κγγ δαβ] (41)

Mαβ =

¯M11M12

M21M22

¯(42)

where M12 =M21 due to symmetry.

M11=D (κ11 + ν κ22) (43)

M22=D (κ22 + ν κ11)

M12=D (1− ν) κ12

2.2.2 Bending Energy Density

One -Dimensional Case Here, we use the hat notation for a function of certainargument such as:

M11= M11 (κ11) (44)

=D κ11

Then, the bending energy density Ub reads :

Ub=

Z κ11

0M11 (κ11) dκ11 (45)

=D

Z κ11

0κ11 dκ11

=1

2D (κ11)

2

Ub =1

2M11 κ11 (46)

11κ

11M

11dκ

D

11

General Case General definition of the bending energy density reads:

Ub =

IMαβ dκαβ (47)

22κ

11κκ

0

Calculate the energy density stored when the curvature reaches a given value καβ.Consider a straight loading path:

καβ = η καβ (48)

dκαβ = καβ dη

αβκ0η=

1η=

η

αβκ

Mαβ

Mαβ

Mαβ = Mαβ (καβ) (49)

= Mαβ (η καβ)

= η Mαβ (καβ)

12

where Mαβ (καβ) is a homogeneous function of degree one.

Ub=

IMαβ (καβ) dκαβ (50)

=

Z 1

0η Mαβ (καβ) καβ dη

= Mαβ (καβ) καβ

Z 1

0η dη

=1

2Mαβ (καβ) καβ

=1

2Mαβ καβ

Now, the bending energy density reads:

Ub=D

2[(1− ν) καβ + ν κγγ δαβ] καβ (51)

=D

2[(1− ν) καβ καβ + ν κγγ καβ δαβ]

=D

2

h(1− ν) καβ καβ + ν (κγγ)

2i

The bending energy density expressed in terms of components:

Ub=D

2

n(1− ν)

h(κ11)

2 + 2 (κ12)2 + (κ22)

2i+ ν (κ11 + κ22)

2o

(52)

=D

2

n(1− ν)

h(κ11 + κ22)

2 − 2 κ11 κ22 + 2 (κ12)2i+ ν (κ11 + κ22)

2o

=D

2

nh(κ11 + κ22)

2 − 2 κ11 κ22 + 2 (κ12)2i− ν

h−2 κ11 κ22 + 2 (κ12)2

io=D

2

n(κ11 + κ22)

2 − 2 κ11 κ22 + 2 (κ12)2 − νh−2 κ11 κ22 + 2 (κ12)2

io=D

2

n(κ11 + κ22)

2 + 2 (1− ν)h−κ11 κ22 + (κ12)2

io

Ub =D

2

n(κ11 + κ22)

2 − 2 (1− ν)hκ11 κ22 − (κ12)2

io(53)

2.2.3 Total Bending Energy

The total bending energy is the integral of the bending energy density over the areaof plate:

Ub =

ZSUb dA (54)

13

2.3 Membrane Energy

2.3.1 Axial Force

Assume that καβ = 0. The axial force can be calculated:

Nαβ =E

1− ν2

Z h2

−h2

[(1− ν) εαβ + ν εγγ δαβ] dz (55)

=E

1− ν2

Z h2

−h2

£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤dz

+E

1− ν2

Z h2

−h2

[(1− ν) καβ + ν κγγ δαβ] z dz

=E

1− ν2£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤ Z h2

−h2

dz

+E

1− ν2[(1− ν) καβ + ν κγγ δαβ]

Z h2

−h2

z dz

=Eh

1− ν2£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤Here, we define the axial rigidity of a plate C as follows:

C =Eh

1− ν2(56)

Now, one gets the membrane force-extension relation:

Nαβ = Ch(1− ν) ε◦αβ + ν ε◦γγ δαβ

i(57)

Nαβ =

¯N11N12N21N22

¯(58)

where N12 = N21 due to symmetry.

N11=C (ε◦11 + ν ε◦22) (59)

N22=C (ε◦22 + ν ε◦11)

N12=C (1− ν) ε◦11

2.3.2 Membrane Energy Density

Using the similar definition used in the calculation of the bending energy density,the extension energy (membrane energy) reads:

Um =

INαβ dε◦αβ (60)

14

Calculate the energy stored when the extension reaches a given value ε◦αβ. Considera straight loading path:

ε◦αβ = η ε◦αβ (61)

dε◦αβ = ε◦αβ dη

Nαβ = Nαβ

¡ε◦αβ¢

(62)

= Nαβ

¡η ε◦αβ

¢= η Nαβ

¡ε◦αβ¢

where Nαβ

³ε◦αβ

´is a homogeneous function of degree one.

Um=

Z ε◦αβ

0Nαβ

¡ε◦αβ¢dε◦αβ (63)

=

Z 1

0η Nαβ

¡ε◦αβ¢ε◦αβ dη

=1

2Nαβ

¡ε◦αβ¢ε◦αβ

=1

2Nαβ ε◦αβ

Now, the extension energy reads:

Um=C

2

£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤ε◦αβ (64)

=C

2

h(1− ν) ε◦αβ ε◦αβ + ν

¡ε◦γγ¢2i

The extension energy expressed in terms of components:

Um=C

2

n(1− ν)

h(ε◦11)

2 + 2 (ε◦12)2 + (ε◦22)

2i+ ν (ε◦11 + ε◦22)

2o

(65)

=C

2

n(1− ν)

h(ε◦11 + ε◦22)

2 − 2 ε◦11ε◦22 + 2 (ε◦12)2i+ ν (ε◦11 + ε◦22)

2o

=C

2

n(ε◦11 + ε◦22)

2 − 2 ε◦11ε◦22 + 2 (ε◦12)2 − νh−2 ε◦11ε◦22 + 2 (ε◦12)2

io=C

2

n(ε◦11 + ε◦22)

2 + 2 (1− ν)h−ε◦11ε◦22 + (ε◦12)2

io

Um =C

2

n(ε◦11 + ε◦22)

2 − 2 (1− ν)hε◦11ε

◦22 − (ε◦12)

2io

(66)

15

2.3.3 Total Membrane Energy

The total membrane is the integral of the membrane energy density over the areaof plate::

Um =

ZSUm dS (67)

16

3 Development of Equation of Equilibrium and Boundary Condi-tions Using Variational Approach

3.1 Bending Theory of Plates

3.1.1 Total Potential Energy

The total potential energy of the system Π reads:

Π = Ub − Vb (68)

where Ub is the bending energy stored in the plate, and Vb is the work of externalforces.

Bending Energy

Ub=1

2

ZSMαβ καβ dS (69)

=−12

ZSMαβ w,

αβdS

where the geometrical relation καβ = −w,αβ has been used.Work of External Forces

Plate Loading Lateral load:

q (x) = q (xα) (70)

This is distributed load measured in [N/m2] or [lb/in2] force per unit area of themiddle surface of the plate.

( )q x( )q x

The distributed load contains concentrated load P as a special case:

P (x0, y0) = P0δ (x− x0) δ (y − y0) (71)

where δ is the Dirac delta function, [x0, y0] is the coordinate of the application ofthe concentrated force, and P0 is the load intensity.

17

x

Dirac -functionδ

0x

NOTE The shearing loads on the lateral surface of ice are normally not consid-ered in the theory of thin plates.

Load Classification

• Load applied at the horizontal surfaces.

transverse load

• Load applied at the lateral surfaces.

edge force

edge moment

Loads are assumed to be applied to the middle plane of the plateNOTE Other type of loading such as shear or in-plane tension or compression

do not deflect laterally the plate and therefore are not considered in the bendingtheory.

18

in-plane tension

in-plane shear

or compression

Potential Energy due to Lateral Load q Lateral (transverse) load doeswork on transverse deflection: Z

Sq w dS (72)

This is also called a work of external forces.

Potential Energy due to Edge Moment The conjugate kinematic variableassociated with the edge moment is the edge rotation dw/dxn.

edge moment

dl

Γ

t

n

We apply only the normal bending moment Mnn:

−ZΓMnn

dw

dxndl (73)

where the minus sign is included because positive bending moment results in anegative rotation and negative moment produces positive rotation.

0nnM < 0nnM >

nx0 a0

n

dwdx> 0

n

dwdx<

w

At the edge, Mtt = 0 and Mtn = 0.

19

Potential Energy due to Edge ForcesZΓVn w dl (74)

1Q 2Q

nV1x2x

Potential Energy due to All External Forces Now, the work of externalforces reads:

Vb =

ZSq w dS −

ZΓMnn

dw

dxndl +

ZΓVn w dl (75)

3.1.2 First Variation of the Total Potential Energy

The total potential energy reads:

Π=−12

ZSMαβ w,αβ dS (76)

−ZSq w dS +

ZΓMnn

dw

dxndl −

ZΓVn w dl

First variation of the total potential energy δΠ is expressed:

δΠ=−ZSMαβ δw,αβ dS (77)

−ZSq δw dS +

ZΓMnn δ

µdw

dxn

¶dl −

ZΓVn δw dl

We shall transform now the first integral with the help of the Gauss theorem.First note that from the rule of the product differentiation:

Mαβ δw,αβ = (Mαβ δw,α ) ,β −Mαβ,β δw,α (78)

then ZSMαβ δw,αβ dS =

ZS(Mαβ δw,α ) ,β dS −

ZSMαβ,β δw,α dS (79)

Now, the first integral on the right hand side of the above equation transforms tothe line integral:Z

SMαβ δw,αβ dS =

ZΓMαβ δw,α nβ dl −

ZSMαβ,β δw,α dS (80)

20

The integrand of the second integral on the right hand side of the above equationtransform to:

Mαβ,β δw,α= (Mαβ,β δw) ,α−Mαβ,αβ δw (81)

which results in:ZSMαβ δw,αβ dS =

ZΓMαβ δw,α nβ dl (82)

−ZS(Mαβ,β δw) ,α dS +

ZSMαβ,αβ δw dS

upon which the application of the Gauss rule gives:ZSMαβ δw,αβ dS =

ZΓMαβ δw,α nβ dl (83)

−ZΓMαβ,β δw nα dl +

ZSMαβ,αβ δw dS

We can return now to the expression for δΠ and substitute there the transformedfirst integral:

δΠ=

ZS(−Mαβ,αβ −q) δw dS (84)

+

ZΓMαβ,β δw nα dl −

ZΓVn δw dl

−ZΓMαβ δw,α nβ dl +

ZΓMnn δw,n dl

where δw,n= δ³

dwdxn

´. It is seen that integrals involving the prescribed forces Mnn

and Vn are written in a local coordinate system xγ {xn, xt} while the remainingtwo integrals over the contour Γ are written in the global coordinate system xα. Inorder to make comparison, we have to decide on one coordinate system. We choosethe local system.

Consider the first integral: ZΓ(Mαβ,β nα) δw dl (85)

The term in the parenthesis is a scalar quantity and thus remain unchanged withrespect to the rotation of coordinate system. In the local system xγ , the lineintegral becomes: Z

Γ(Mγδ,δ nγ) δw dl (86)

where γ = 1 is the normal direction n, and γ = 2 is the tangential direction t. Thecoordinates of the unit normal vector is the local system are nγ {1, 0}. Hence,Z

Γ(Mγδ,δ nγ) δw dl=

ZΓ(M1δ,δ n1 +M2δ,δ n2) δw dl (87)

=

ZΓM1δ,δ δw dl

21

Furthermore, the integrand reads:

M1δ,δ =M11,1+M12,2 (88)

=∂M11

∂x1+

∂M12

∂x2=

∂Mnn

∂xn+

∂Mnt

∂xt

and we call it the shear force in the normal direction n and denote:

Qn ≡Mnδ,δ δ = {1, 0} or {n, t} (89)

Now, we can combine two line integrals in the equation of first variation of the totalpotential energy: Z

Γ

¡Qn − Vn

¢δw dl (90)

How the remaining integral is transformed?ZΓ(Mαβ nβ) δw,α dl =

ZΓ(Mγδ nδ) δw,γ dl (91)

Because it is a scalar quantity, we simply switch indices from global system (α andβ) to local (γ and δ). As before nδ {1, 0} so after summing with respect to δ, wehave: Z

Γ(Mγ1 n1 +Mγ2 n2) δw,γ dl=

ZΓMγ1 δw,γ dl (92)

=

ZΓMγn δw,γ dl

=

ZΓ(Mnn δw,n+Mtn δw,t ) dl

The first term can be absorbed with the line integral representing potential energyof bending moment:

−ZΓ

¡Mnn − Mnn

¢δw,n dl (93)

There remains though one integral which does not fit to anything. Since theboundary term must be equilibrated, it is suspected that this term might belong tothe shearing force term, at least partially:Z

ΓMtn δw,t dl transverse term (94)

In order to compare this term with the shearing force term, we have to make thisterm comparable as far as the kinematic quantity describing variation is concerned.One integral involves δw and the other one δw,t. Note that ∂w,t= ∂ (δw) /∂xt isthe derivative of the function δw in the tangential direction, i.e. direction along thecurve Γ. This means that we can integrate by parts along Γ. Thus,

Mtn δw,t= (Mtn δw) ,t−Mtn,t δw (95)

22

ZΓMtn δw,t dl =

ZΓ(Mtn δw) ,t dl −

ZΓMtn,t δw dl (96)

The first term in the right hand is equal to the value of the integrand calculated atthe beginning and end of the integration path:Z

Γ(Mtn δw) ,t dl = Mtn δw|endbeginning (97)

Consider now two cases.

• The contour Γ is a smooth closed curve, so the value at the beginning is equalto the value at the end:

Mtn δw|end − Mtn δw|beginning = 0 (98)

The term does not give any contribution.

Γend

begining

direction ofintegration

• The contour Γ is piece-wise linear or composed of a finite number, k, of smoothcurves with discontinuity. Therefore, the integration should be made in apiece-wise manner. Thus, the continuation of the beginning and end of eachshould be added: X

k

Mtn δw|endbeginning (99)

23

3.1.3 Equilibrium Equation and Boundary Conditions

Now, we can write the final expression for the first variation of δΠ:

δΠ=

ZS(−Mαβ,αβ −q) δw dS (100)

+

ZΓ

¡Qn − Vn

¢δw dl −

ZΓ

¡Mnn − Mnn

¢δw,n dl

−ÃX

k

Mtn δw|endbeginning −ZΓMtn,t δw dl

!

=

ZS(−Mαβ,αβ −q) δw dS

+

ZΓ

¡Vn − Vn

¢δw dl −

ZΓ

¡Mnn − Mnn

¢δw,n dl

−Xk

Mtn δw|endbeginning

where Vn = Qn +Mtn,t is the effective shear force.In order to make the functional Π stationary under arbitrary variation of the

displacement field δw, there must hold:

Equation of Equilibrium

Mαβ,αβ +q = 0 on S(101)

Boundary Conditions

Mnn − Mnn = 0 or δw,n= 0 on ΓVn − Vn = 0 or δw = 0 on Γ

Mnt = 0 or δw = 0at corner pointsof the contour Γ

(102)

3.1.4 Specification of Equation for Rectangular Plate

Consider a rectangular plate.

24

x t→

y n→

b

a

Boundary Conditions For edges parallel to x-axis, the normal direction is theydirection.

Myy − Myy =0 or∂w

∂y= 0 (103)

Vy − Vy =0 or w = 0

where

Vx=Qx +∂Myx

∂y(104)

Vy =Qy +∂Mxy

∂x

For edges parallel to y-axis, the normal direction is the x-direction.

Mxx − Mxx=0 or∂w

∂x= 0 (105)

Vx − Vx=0 or w = 0

where

Vy =Qy +∂Mxy

∂x(106)

Vx=Qx +∂Myx

∂y

Interpretation of Corner Points

25

1

23

4

[ ]segment k

[ ]segment 1k−

direction ofintegration

Boundary condition reads:

XMtn δw|endbeginning =M

[2]tn δw[2] −M

[1]tn δw[1] +M

[4]tn δw[4] −M

[3]tn δw[3] (107)

whereδw[3] = δw[2] (108)

thusXMtn δw|endbeginning = −M

[1]tn δw[1] +

³M[2]tn −M

[3]tn

´δw[2] +M

[4]tn δw[4] (109)

Consider the right angle.

x

y[ ]k

[ ]1k−

for the k − 1 side n= x, t = y (110)

for the k side n= y, t = x³Mtn δw|endbeginning

´at the right angle

= (Mxy −Myx) δw (111)

Interpretation of Corner Forces Plane stress:

τxy = τyx symmetry (112)

26

yσ

dSxσ xyτ

yxτ

surface element

Let us place the surface element at the corner.

Edge

Edge

[ ]k

[ ]1k−

y

x

yxM

xyM

yxτxyτ

z

The shearing stresses produce twisting moments which are in the opposite direction:

M [k−1]xy = −M [k]

yx (113)

Therefore, the boundary condition at the corner becomes:

Mtn δw|endbeginning =³M [k−1]

xy −M [k]yx

´δw = 2 Mxy δw = 0 (114)

Fcorner = 2 Mxy (115)

27

For the Entire Plate

2 xyM

2 xyM

2 xyM

2 xyM

Interpretation of the Effective Shear Vx

xQdy

xyM

dy

xyxy

MM dy

y∂+

∂

y

x

z

Equilibrium reads:

Qx dy +

µMxy +

∂Mxy

∂ydy

¶−Mxy (116)

=

µQx +

∂Mxy

∂y

¶dy

= Vx dy

Vx = Qx +∂Mxy

∂y(117)

28

3.2 Bending-Membrane Theory of Plates

3.2.1 Total Potential Energy

The total potential energy of the system Π reads:

Π = Ub + Um − Vb − Vm (118)

where Ub is the bending strain energy, Um is the membrane strain energy, Vb isthe potential energy of external loading causing flexural response, and Vm is thepotential energy of external loading causing membrane response.

Membrane Strain Energy The membrane strain energy reads:

Um =1

2

ZSNαβ ε◦αβ dS (119)

whereε◦αβ =

1

2(uα,β +uβ,α ) +

1

2w,α w,β (120)

Potential Energy of External Forces Evaluation of Boundary Terms

Normal in-plane loading, NnnZΓNnn un dl (121)

where un is normal in-plane displacement.

Shear in-plane loading, Ntt ZΓNtn ut dl (122)

where ut is shear component of the displacement vector.

Potential Energy of External Forces

Vm =

ZΓNnn un dl +

ZΓNtn ut dl (123)

3.2.2 First Variation of the Total Potential Energy

The first variation of the total potential energy reads:

δΠ = (δUb − δVb) + (δUm − δVm) (124)

The first parenthesis represent the terms considered already in the bending theoryof plates. All we have to do is to evaluate the term in the second parenthesis.Here, the first variation of the membrane energy reads:

δUm =

ZSNαβ δλαβ dS (125)

29

whereδλαβ =

1

2(δuα,β +δuβ,α ) +

1

2(δw,α w,β +δw,β w,α ) (126)

Because of the symmetry of the tensor of membrane forces:

Nαβ = Nβα (127)

by using the characteristics of dummy indices we obtain:

Nαβ δuβ,α= Nβα δuβ,α= Nαβ δuα,β (128)

Now, the first variation of the membrane strain energy reads:

δUm=

ZS

¿Nαβ

∙1

2(δuα,β +δuβ,α ) +

1

2(δw,α w,β +δw,β w,α )

¸ÀdS (129)

=

ZS(Nαβ δuα,β +Nαβ w,β δw,α ) dS

Note that the displacement vector has now three components:

{uα, w} (130)

so that there are three independent variations:

{δuα, δw} (131)

We expect those to end up with three independent equations of equilibrium. Thefirst term of δUm reads:Z

SNαβ δuα,β dS =

ZS(Nαβ δuα) ,β dS −

ZSNαβ,β δuα dS (132)

=

ZΓNαβ δuα nβ dl −

ZSNαβ,β δuα dS

=

ZΓNγδ δuγ nδ dl −

ZSNαβ,β δuα dS

=

ZΓNγ1 δuγ dl −

ZSNαβ,β δuα dS

=

ZΓNγn δuγ dl −

ZSNαβ,β δuα dS

=

ZΓ(Nnn δun +Ntn δut) dl −

ZSNαβ,β δuα dS

30

The second term of δUm reads:ZSNαβ w,β δw,α dS =

ZS(Nαβ w,β δw) ,α dS −

ZS(Nαβ w,β ) ,α δw dS (133)

=

ZΓNαβ w,β δw nα dl −

ZS(Nαβ w,β ) ,α δw dS

=

ZΓNγδ w,δ δw nγ dl −

ZS(Nαβ w,β ) ,α δw dS

=

ZΓN1δ w,δ δw dl −

ZS(Nαβ w,β ) ,α δw dS

=

ZΓ(N11 w,1+N12 w,2 ) δw dl −

ZS(Nαβ w,β ) ,α δw dS

=

ZΓ(Nnn w,n+Nnt w,t ) δw dl −

ZS(Nαβ w,β ) ,α δw dS

Now, the variation of external work reads:

δVm =

ZΓNnn δun dl +

ZΓNtn δut dl (134)

3.2.3 Equilibrium Equation and Boundary Conditions

The contribution of the term (δUm − δVm) then becomes:

δ (Um − Vm) =

ZΓ(Nnn δun +Ntn δut) dl −

ZSNαβ,β δuα dS (135)

+

ZΓ(Nnn w,n+Nnt w,t ) δw dl −

ZS(Nαβ w,β ) ,α δw dS

−ZΓNnn δun dl −

ZΓNtn δut dl

=−ZSNαβ,β δuα dS +

ZΓ

¡Nnn − Nnn

¢δun dl +

ZΓ

¡Ntn − Ntn

¢δut dl

−ZS(Nαβ w,β ) ,α δw dS +

ZΓ(Nnn w,n+Nnt w,t ) δw dl

The first three integrals involve independent variations of uα, i.e. δuα or {δun, δut}.This gives us two independent equations of equilibrium in the plane of the plate:

Equation of Equilibrium I

Nαβ,β = 0 on S(136)

31

and two additional boundary conditions:

Boundary Conditions I

Nnn − Nnn = 0 or δun = 0 on ΓNtn − Ntn = 0 or δut = 0 on Γ

(137)

The remaining two integrals involve variation in the out-of-plane displacementδw and thus should be combined with the equation of equilibrium and boundaryconditions governing the flexural response. The terms involving surface integralshould be added to the equation of equilibrium:

Equation of Equilibrium II

Mαβ,αβ +(Nαβ w,β ) ,α+q = 0 on S(138)

where the second term in the left hand is the new term arising from the finiterotation.

The term with the line integral should be added to the corresponding terminvolving variation δw:Z

Γ

¡Vn +Nnn w,n+Nnt w,t−Vn

¢δw dl = 0 (139)

The generalized boundary conditions reads:

Boundary Conditions II- (A)

Vn +Nnn w,n+Nnt w,t−Vn = 0 or δw = 0 on Γ(140)

where the second and third terms in the left hand side of the first equation are thenew terms arising from the finite rotation.

If the boundaries of the plate are kept undeformed w,t= 0 (simply supportedor clamped plate), then the boundary condition is satisfied:

Vn +Nnn w,n−Vn = 0 or δw = 0 on Γ (141)

Physically, the additional terms represent the contribution of the axial force to thevertical equilibrium. Using the in-plane equilibrium, Nαβ,β = 0, the out-of-planeequilibrium can be transformed to the form:

Mαβ,αβ +Nαβ,α w,β +Nαβ w,αβ +q = 0 on S (142)

32

Equation of Equilibrium II’

Mαβ,αβ +Nαβ w,αβ +q = 0 on S(143)

which is called as the von Karman equation. Note that Nαβ is related through theHook’s law with the gradient of the in-plane displacement uα, i.e. Nαβ = Nαβ (uα).Therefore, the new term Nαβ w,αβ represents in fact coupling between in-plane andout-out-plane deformation.

To make derivation complete, the final boundary conditions which do not changedfrom the bending theory of plate are presented:

Boundary Conditions II- (B)

Mnn − Mnn = 0 or δw,n= 0 on Γ

Mnt = 0 or δw = 0at corner pointsof the contour Γ

(144)

33

4 General Theories of Plate

4.1 Bending Theory of Plates

4.1.1 Derivation of the Plate Bending Equation

Then, groups of equations!

• EquilibriumMαβ,αβ +q = 0 on S (145)

• Geometryκαβ = −w,αβ (146)

• ElasticityMαβ = D [(1− ν) καβ + ν κγγ δαβ] (147)

Eliminating curvature καβ between Eq. (146) and (147), we obtain:

Mαβ = −D [(1− ν) w,αβ +ν w,γγ δαβ] (148)

Substituting Eq. (148) into Eq. (145) reads:

−D [(1− ν) w,αβ +ν w,γγ δαβ] ,αβ +q=0 (149)

−D [(1− ν) w,αβαβ +ν w,γγαβ δαβ] + q=0

Note that the components of the Kronecker "δαβ" tensor are constant and thus arenot subjected to differentiation:

δαβ =

¯1 00 1

¯or δαβ =

½1 if α = β0 if α 6= β

(150)

Also, note that only these components:

¤αβ δαβ = ¤αα (151)

survive in the matrix multiplication for which α = β. Therefore, Eq. (149) nowreads:

−D [(1− ν) w,αβαβ +ν w,γγαα ] + q = 0 (152)

Because "γγ" are "dummy" indices, they can be replaced by any other indices, forexample "ββ."

−D [(1− ν) w,αβαβ +ν w,ββαα ] + q = 0 (153)

The order of differentiation does not matter:

w,αβαβ = w,ααββ = w,ββαα

34

Thus, two terms in Eq. (153) can now be added to give the plate bending equation:

D w,ααββ = q for α, β = 1, 2 (154)

Here, the index notation can be expended:

w,ααββ =w,11ββ +w,22ββ (155)

=w,1111+w,2211+w,1122+w,2222

=w,1111+2 w,1122+w,2222

Now, letting "1→ x", "2→ y" leads:

D

µ∂4w

∂x4+ 2

∂4w

∂x2∂x2+

∂4w

∂y4

¶= q (x, y) (156)

Alternative notation can be:

D ∇4w = q (157)

where Laplacian ∇2w reads:

∇2w = ∂2w

∂x2+

∂2w

∂y2(158)

and bi-Laplican ∇4w reads:

∇4w=∇2¡∇2w

¢(159)

=∂2

∂x2¡∇2w

¢+

∂2

∂y2¡∇2w

¢=

∂2

∂x2

µ∂2w

∂x2+

∂2w

∂y2

¶+

∂2

∂y2

µ∂2w

∂x2+

∂2w

∂y2

¶=∂4w

∂x4+ 2

∂4w

∂x2∂x2+

∂4w

∂y4

4.1.2 Reduction to a System of Two Second Order Equations

DenoteD w,αα= −M (160)

Then, from the equilibrium equation:

[D w,αα ] ,ββ = q (161)

we obtain a system of two linear partial differential equations of the second order:

M,ββ = −qD w,αα= −M

(162)

35

or (∂2M∂x2 +

∂2M∂y2 = −q

D³∂2w∂x2

+ ∂2w∂y2

´= −M

(163)

What is "M" ? Let us calculate Mαα:

Mαα=M11 +M22 (164)

=D [(1− ν) κ11 + ν (κ11 + κ22) δ11]

+D [(1− ν) κ22 + ν (κ11 + κ22) δ22]

=D [(1 + ν) (κ11 + κ22)]

=D (1 + ν) καα

orMαα

1 + ν= D καα = −D w,αα=M (165)

Therefore,

Mαα=M (1 + ν) (166)

=D καα (1 + ν)

Now, moment sum reads:

M = D καα (167)

and in expanded notation it reads:

M = D [κxx + κyy] (168)

4.1.3 Exercise 1: Plate Solution

Consider a simply supported plate.

x

y

a

a

( )Square plate a a×

36

Boundary Condition General boundary condition reads:¡Mnn − Mnn

¢w,n= 0 on Γ¡

Vn − Vn¢w = 0 on Γ

(169)

Mnn = 0 ⇒Mnn = 0 on Γw = 0 on Γ

(170)

w = 0 at x = 0 and x = a , 0 ≤ y ≤ aw = 0 at y = 0 and y = a , 0 ≤ x ≤ a

(171)

Mxx = 0 at x = 0 and x = a , 0 ≤ y ≤ aMyy = 0 at y = 0 and y = a , 0 ≤ x ≤ a

(172)

Loading Condition Assume for simplicity the sinusoidal load distribution:

q (x, y) = q0 sin³π x

a

´sin³π y

a

´(173)

where q0 is a pressure intensity.

Solution of Problem The solution of the form

w (x, y) = w0 sin³π x

a

´sin³π y

a

´(174)

satisfy both the boundary conditions and the governing equations (see below).

Plate Bending Equation Substituting Eq. (173) and (174) into the plate bend-ing equation (156), one gets:½

D w0

∙³πa

´4+ 2

³πa

´4+³πa

´4¸− q0

¾sin³π x

a

´sin³π y

a

´=0 (175)½

4 D w0

³πa

´4− q0

¾sin³π x

a

´sin³π y

a

´=0

In order to satisfy the above equation for all values of x and y, the coefficient inthe bracket must vanish. This gives:

w0 =q04 D

³aπ

´4(176)

where D =¡Eh3

¢/£12¡1− ν2

¢¤.

Bending Moments The various bending moments are given by:

Mxx=−D∙∂2w

∂x2+ ν

∂2w

∂y2

¸= D (1 + ν)

³πa

´2w0 sin

³π x

a

´sin³π y

a(177)

Myy =−D∙∂2w

∂y2+ ν

∂2w

∂x2

¸= D (1 + ν)

³πa

´2w0 sin

³π x

a

´sin³π y

a

´Mxy =−D (1− ν)

∂2w

∂x∂y= −D (1− ν)

³πa

´2w0 cos

³π x

a

´cos³π y

a

´

37

Shear Components The shear components Qx and Qy are:

Qx=∂Mxx

∂x+

∂Mxy

∂y(178)

Qy =∂Myy

∂y+

∂Mxy

∂x

Now, using the previously obtained bending moments, we get the shear componentsin the interior of the plate:

Qx=2 D³πa

´3w0 cos

³π x

a

´sin³π y

a

´(179)

Qx=2 D³πa

´3w0 sin

³π x

a

´cos³π y

a

´Effective Shear Components Next, let us computer the effective shear com-

ponents:

Vx=Qx +∂Mxy

∂y(180)

Vy =Qy +∂Mxy

∂x

Using the previous results, we get:

Vx= (3− ν) D³πa

´3w0 cos

³π x

a

´sin³π y

a

´(181)

Vy = (3− ν) D³πa

´3w0 sin

³π x

a

´cos³π y

a

´We now need to evaluate the effective shear on the boundaries:

Vx/h(3− ν) D

¡πa

¢3iVy/

h(3− ν) D

¡πa

¢3ix = 0 w0 sin

¡π ya

¢0

x = a −w0 sin¡π y

a

¢0

y = 0 0 w0 sin¡π xa

¢y = a 0 −w0 sin

¡π xa

¢(182)

Because our sign convention is:

x

z

positive shear

38

in our case, shear along the boundary is:

x

y

From the above results, we can plot the shear distribution:

x

y

Force Balance Integrating the effective shear along the boundary, we get:

R =

ZLVn dxt = 4

Z a

0Vx|x=0 dy = 4 (3− ν) D

³πa

´3w0

Z a

0sin³π y

a

´dy (183)

Then, the reduction force due to effective shear on boundaries reads:

R = 2 (3− ν) q0¡aπ

¢2 (184)

Now, let us complete the total load acting on the plate:

P =

ZSq (x, y) dS =

Z a

0

Z a

0q0 sin

³π x

a

´sin³π y

a

´dx dy (185)

Then, the total external load acting on the plate reads:

P = 4 q0¡aπ

¢2 (186)

Notice that R and P do not balance! We did not include the corner forces. Theseare given by:

(Fcorner)x0,y0 = 2 (Mxy)|x=x0,y=y0 (187)

39

Because of the symmetry, all four forces are equal. So, compute the corner forceat x = y = 0, (Fcorner)0,0:

(Fcorner)0,0=2hcos³π x

a

´cos³π y

a

´i¯0,0

(188)

=− 2 D (1− ν)³πa

´2w0

Now, the vertical force balance is satisfied:

R+ 4 Fcorner = P (189)

2 (3− ν) q0

³aπ

´2− 8 D (1− ν)

³πa

´2w0=4 q0

³aπ

´2(190)

2 (3− ν) q0

³aπ

´2− 2 (1− ν) q0

³πa

´2=4 q0

³aπ

´24.1.4 Exercise 2: Comparison between Plate and Beam Solution

Plate Solution For a square simply supported plate under loading qplate (x, y)given by:

qplate (x, y) = (q0)plate sin³π x

a

´sin³π y

a

´(191)

we found that the plate deflection is:

wplate (x, y) = (w0)plate sin³π x

a

´sin³π y

a

´(192)

with:

(w0)plate=(q0)plate4 D

³aπ

´4(193)

=3¡1− ν2

¢(q0)plate

E h3

³aπ

´4For the plate, the total load is given by:

Pplate=

Z a

0

Z a

0(q0)plate sin

³π x

a

´sin³π y

a

´dy dx (194)

=4 (q0)plate

³aπ

´2

40

Wide Beam Solution For a wide beam under line loading given by:

qbeam (x, y) = (q0)beam sin³π x

a

´(195)

we need to compute the central deflection (w0)beam from:

E I w0000beam = qbeam (x) (196)

where I = ah3/12. Assuming the deflection wbeam (x):

wbeam (x) = (w0)beam sin³π x

a

´(197)

we get:

E I³πa

´4(w0)beam sin

³π x

a

´= (q0)beam sin

³π x

a

´(198)

Thus,

(w0)beam=(q0)beamE I

³aπ

´4(199)

=12 (q0)beamE a h3

³aπ

´4Now, let us compute the total forces:

Pbeam=

Z a

0(q0)beam sin

³π x

a

´dx (200)

=2 (q0)beama

π

Comparison For both total forces to be equal, we need to have:

Pplate=Pbeam (201)

4 (q0)plate

³aπ

´2=2 (q0)beam

a

π

(q0)beam = 2 (q0)platea

π(202)

With a concentrated load, the beam deflection now becomes:

(w0)beam=(q0)beamE I

³aπ

´4(203)

=24

π

(q0)plateE h3

³aπ

´4We now can compute the ratio of central deflections:

α=(w0)plate(w0)beam

=

3 (1−ν2) (q0)plateE h3

¡aπ

¢424π

(q0)plateE h3

¡aπ

¢4 (204)

=π

8

¡1− ν2

¢' 0.36

The above equation means that under the same total load, a plate is three timesstiffer than a wide beam. The ratio α will vary slightly depending on the loaddistribution (sinusoidal, uniform, concentrated load, etc.).

41

4.1.5 Exercise 3: Finite Difference Solution of the Plate Bending Prob-lem

Governing Equations read:∇2M = −q∇2w = −M

D

(205)

or in the component notation they read:

∂2M∂x2

+ ∂2M∂y2

= −q∂2w∂x2

+ ∂2w∂y2

= −MD

(206)

where M is the moment sum defined by:

M =Mαα

1 + ν= D καα (207)

Case of Simply Supported Plate The boundary condition of a simple sup-ported plate reads:

w=0 on Γ (208)

Mnn=0 on Γ

x

y

a

a

n

For sides parallel to x-axis (thick lines), one gets:

Mnn =Myy = 0 (209)

w = 0 → dw

dx= 0 → d2w

dx2= 0 → κxx = 0 (210)

From the general constitutive equations,

Myy =D [κyy + ν κxx] (211)

0=D [κyy + ν · 0] → κyy = 0

Therefore,M = D [καα + κββ] = D [0 + 0] = 0 (212)

42

Similar derivation can be performed for two edges parallel to y-axis. Then, M =0. It can be concluded that for a simply supported plate the following boundaryconditions hold: (

∂2M∂x2

+ ∂2M∂y2

= −q in S

M = 0 on Γ(213)

(∂2w∂x2

+ ∂2w∂y2

= −MD in S

w = 0 on Γ(214)

Therefore, the above two boundary value problems are uncoupled.

The Finite Difference Technique An approximation to the first and secondderivatives.

x

z

h

1nx +nx

nz1nz +

,n m 1n+1n−

1m+

1m−

h

x

y

dz

dx

¯backwardn

' zn − zn−1h

(215)

dz

dx

¯forwardn+1

' zn+1 − znh

43

d2z

dx2

¯n

=d

dx

∙dz

dx

¸(216)

=

¡dzdx

¢n+1−¡dzdx

¢n

h

=zn+1−zn

h − zn−zn−1h

h

=zn+1 − 2 zn + zn−1

h2

d2z

dy2

¯m

=zm+1 − 2 zm + zm−1

h2(217)

∇2z= ∂2z

∂x2+

∂2z

∂y2(218)

=1

h2(zn+1 − 2 zn + zn−1 + zm+1 − 2 zm + zm−1)

0 RightLeft

Bottom

Top

h

hh h

∇2z = 1

h2(zT + zB + zL + zR − 4 z0) (219)

Divide the plate into sixteen identical squares and distinguish six representativenodes: three in the interior and three at the boundary. Because of symmetry, it isenough to consider only an eighth of the plate.

44

a

a

4ah=1

23

456

Four axes of symmetry

Determination of Moment For each interior point (1, 2, 3), we write equation∇2M = −q. For each boundary point (4, 5, 6), we write boundary conditionM = 0 (uniform pressure).

Point 1: 4 M2 − 4 M1 = −q a2

16(220)

Point 2: M1 +M4 + 2 M3 − 4 M2 = −q a2

16

Point 3: 2 M5 + 2 M2 − 4 M3 = −q a2

16Point 4: M4 = 0

Point 5: M5 = 0

Point 6: M6 = 0

45

Substituting three last equations of Eq. (220) into the first three equations of Eq.(220), one ends up with the following system of linear algebraic equations:⎧⎪⎨⎪⎩

4 M2 − 4 M1 = − q a2

16

M1 + 2 M3 − 4 M2 = −q a2

16

2 M2 − 4 M3 = − q a2

16

(221)

whose solution is:

M1=9

128q a2 (222)

M2=7

128q a2

M3=11

256q a2

At the plate center, Mxx =Myy so that:

M =Mxx +Myy

1 + ν=2 Mxx

1 + ν(223)

Mxx =1

2(1 + ν) M (224)

At the center, M =M1, thus,

Mxx=1

2(1 + ν) M1 (225)

=1

2(1 + ν)

9

128q a2

=0.0457 q a2

This is 4.6% less than the exact solution which is (Mxx)exact = 0.0479 q a2 fromthe text book.

Determination of Deflection For each interior point (1, 2, 3), we write equation∇2w = −M/D. For each boundary point (4, 5, 6), we write boundary conditionw = 0.

Point 1: 4 w2 − 4 w1 = −M1

D

a2

16= −

µ9

128

qa2

D

¶a2

16(226)

Point 2: w1 +w4 + 2 w3 − 4 w2 = −M2

D

a2

16= −

µ7

128

q a2

D

¶a2

16

Point 3: 2 w5 + 2 w2 − 4 w3 = −M3

D

a2

16= −

µ11

256

q a2

D

¶a2

16

Point 4: w4 = 0

Point 5: w5 = 0

Point 6: w6 = 0

46

Similarly, ⎧⎪⎨⎪⎩4 w2 − 4 w1 = − 9

2048qa4

D

w1 + 2 w3 − 4 w2 = − 72048

q a4

D

2 w2 − 4 w3 = − 114096

q a4

D

(227)

Finally, the finite difference solution is:

w1=33

8196

q a4

D= 0.00403

q a4

D(228)

w2=3

1024

q a4

D= 0.00293

q a4

D

w3=35

16384

q a4

D= 0.00214

q a4

D

On the other hand, the exact deflection of the center point is:

(w1)exact = 0.00416q a4

D(229)

Thus, the error of the finite different solution is 3.1%.

4.2 Membrane Theory of Plates

4.2.1 Plate Membrane Equation

Assume that the bending rigidity is zero, D = 0. The plate becomes now amembrane.

• Equilibrium of in-plane equation

Nαβ,α= 0 on S (230)

Equilibrium of out-of-plane equation

Nαβ w,αβ +q = 0 on S (231)

• Strain-displacement relation

ε◦αβ =1

2(uα,β +uα,β ) +

1

2w,α w,β (232)

• Constitutive equation

Nαβ = C£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤(233)

where C = Eh/¡1− ν2

¢.

This is a non-linear system of equation which is difficult to solve. Note thatcorresponding system of equation for the plate bending was linear.

47

4.2.2 Plate Equation for the Circular Membrane

Cylindrical coordinate system is composed of ur, uθ, uz = w.

• Equilibrium of in-plane equation

r∂Nrr

∂r+Nrr −Nθθ = 0 on S (234)

Equilibrium of out-of-plane equation

∂

∂r

∙Nrr

∂w

∂rr

¸+ r q = 0 on S (235)

• Strain-displacement relation

λrr =∂ur∂r

+1

2

µ∂w

∂r

¶2(236)

λθθ =urr

• Constitutive equation

Nrr =C [λrr + ν λθθ] (237)

Nθθ =C [λθθ + ν λrr]

where C = Eh/¡1− ν2

¢.

4.2.3 Example: Approximation Solution for the Clamped Membrane

Consider a circular plate with the clamped support.

q

ra

( )w r

0w

48

Membrane Solution From the symmetry and clamped boundary condition, theradial displacement ur reads:

ur (r = 0)= 0 (238)

ur (r = a) = 0

Thus, as a first approximation, it is appropriate to assume:

ur ≡ 0 for 0 ≤ r ≤ a (239)

Then, the hoop strain vanishes:εθθ = 0

Now, the radial force and the radial strain component become:

Nrr = C εrr (240)

εrr =1

2

µ∂w

∂r

¶2(241)

With the assumption ur = 0, the in-plane equilibrium equation can not be satisfied.Consider out-of-plane equilibrium equation only. Substituting Eq. (240) and

(241) into Eq. (235), one gets:

∂

∂r

"C

2

µ∂w

∂r

¶2 ∂w

∂rr

#= − r q on S (242)

Integrating both sides once with respect to r reads:

C

2r

µ∂w

∂r

¶3= − r2 q

2+ c1 (243)

At the center of the membrane, the slope should be zero. Thus, one gets:

∂w

∂r= 0 at r = 0 (244)

⇒ c1 = 0

Then, Eq. (243) can be written:

∂w

∂r= − 3

rq r

C(245)

Integrating the above equation again reads:

w = −34

3

rq

Cr4/3 + c2 (246)

49

The integration constant c2 can be determined from the zero deflection conditionat the clamped edge:

w = 0 at r = a (247)

⇒ c2 =3

43

rq

Ca4/3

Recalling the definition of the axial rigidity C = Eh/¡1− ν2

¢, Eq. (246) can be

put into a final form:

w

a=3

4

3

r(1− ν2) q a

E h

∙1−

³ra

´4/3¸(248)

' 0.73 3

rq a

E h

∙1−

³ra

´4/3¸In particular, the central deflection w (r = 0) = w0 is related to the load intensityby:

w0a= 0.73 3

rq a

E h(249)

Bending Solution It is interesting to compare the bending and membrane re-sponse of the clamped circular plate. From the page 55 of Theory of Plates andShells (2nd Ed.) by Timoshenko and Woinowsky-Krieger, the central deflection ofthe plate is linearly related to the loading intensity:

w0a=

q a3

64 D(250)

=3¡1− ν2

¢q

16 E

³ah

´3' 0.17 q

E

³ah

´3Assume that a/h = 10, then Eq. (250) yields:

w0a= 17

q a

E h(251)

Comparison A comparison of the bending and membrane solution is shown inthe next figure.

50

It is seen that a transition from the bending to membrane response occurs atw0/a = 0.15 which corresponds to w0 = 1.5 h. When the plate deflection reachapproximately plate thickness, the membrane action takes over the bending actionin a clamped plate. If the plate is not restrained from axial motion, then theassumption ur = 0 is no longer valid, and a separate solution must be developed.

4.3 Buckling Theory of Plates

4.3.1 General Equation of Plate Buckling

• Equilibrium of in-plane equation

Nαβ,α= 0 on S (252)

Equilibrium of out-of-plane equation

Mαβ,αβ +Nαβ w,αβ +q = 0 on S (253)

• Strain-displacement relation

ε◦αβ =1

2(uα,β +uα,β ) +

1

2w,α w,β (254)

καβ =− w,αβ

• Constitutive equation for axial force and axial strain

Nαβ = C£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤(255)

where C = Eh/¡1− ν2

¢, and another one for moment and curvature

Mαβ = D [(1− ν) καβ + ν κγγ δαβ] (256)

where D = Eh3/£12¡1− ν2

¢¤.

51

By combining Eq. (254) and (256), one gets:

Mαβ,αβ = −D w,ααββ (257)

Substituting Eq. (257) into Eq. (253) leads:

−D w,ααββ +Nαβ w,αβ +q = 0 on S (258)

The buckling problem is specified by:

q ≡ 0 (259)

Now, changing signs leads the general out-of-plane equation for the buckling ofthe plates:

D w,ααββ −Nαβ w,αβ = 0 (260)

where the second term in the left hand is non-linear due to Nαβ which should beobtained from:

Nαβ,α= 0 (261)

4.3.2 Linearized Buckling Equation of Rectangular Plates

The nonlinear buckling equation can be separated into two linear equations: onefor in-plane equation for Nαβ and another one for w.

52

x

y

b

a

xPxP

xu

xu

xP

cP

Pre-buckling

Post-buckling

( )0wδ =

( )0wδ ≠

2k

k

Pre-Buckling Problem Recall that:

ε◦αβ =1

2

µ∂uα∂xβ

+∂uβ∂xα

¶+1

2

∂w

∂xα

∂w

∂xβ(262)

Nαβ =Eh

1− ν2£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤(263)

In the pre-buckling problem, the linear equilibrium equations are obtained byomitting the nonlinear terms in the governing equations Eq. (260) and (261). Theresulting equations are now:

D w,ααββ = 0

Nαβ,β = 0

For the pre-buckling trajectory, δw = 0, one gets the equilibrium equation:

Nαβ,β = 0 (264)

53

where

Nαβ =Eh

1− ν2£(1− ν) ε◦αβ + ν ε◦γγ δαβ

¤(265)

ε◦αβ =1

2

µ∂uα∂xβ

+∂uβ∂xα

¶(266)

and boundary condition: ¡Nnn − Nnn

¢δun = 0 on Γ (267)

Here, it is assumed that the unknown membrane force tensor Nαβ is equal tothe similar quantity known from the pre-buckling solution N◦

αβ:

Nαβ = −N◦αβ

where the compressive pre-buckling membrane force are defined as positive.

Post-Buckling Problem Now, the governing equation for buckling of platesreads:

D ∇4w + N◦αβ w,αβ = q +

BoundaryConditions

(268)

where membrane force tensor in the pre-buckling solution N◦αβ is defined as:

N◦αβ = λ Nαβ = λ

¯Nxx Nxy

Nyx Nyy

¯(269)

where Nαβ is the known direction from the pre-buckling analysis, and η is unknownload amplitude. Now, the nonlinear buckling equation becomes a linear eigenvalueproblem:

D ∇4w + λ Nαβ w,αβ = 0 (270)

where λ ≥ 0 is eigenvalues, and w is eigenfunctions.

4.3.3 Analysis of Rectangular Plates Buckling

Simply Supported Plate under In-Plane Compressive Loading Considera plate simply supported on four edges. The plate is subjected to an in-planecompressive load Px uniformly distributed along the edges x = [0, a].

54

x

y

b

a

xPxP

h

From equilibrium equations, one gets:

N◦αβ =

¯N◦xxN

◦xy

N◦yx N

◦yy

¯(271)

=λ

¯Nxx Nxy

Nyx Nyy

¯=Pxb

¯1 00 0

¯Introducing Eq. (271) into Eq. (268) leads:

D ∇4w + Pxb

w,xx= 0 (272)

Boundary condition for this simply supported plate are written as:

w=0 on Γ (273)

Mnn=0 on Γ

where the moment components read:

Mxx=−D (w,xx+ν w,yy ) = 0 (274)

Myy =−D (w,yy +ν w,xx ) = 0

Thus, one gets:

w=w,xx= 0 on x = [0, a] (275)

w=w,yy = 0 on x = [0, b]

Equation (272) is a constant-coefficient equation, and a solution of the followingform:

w = c1 sin³m π x

a

´sin³n π y

b

´for m,n = 1, 2 (276)

55

satisfies both the differential equation and the boundary conditions. Introductioninto Eq. (272) gives:

D

∙³m π

a

´4+ 2

³m π

a

´2 ³n π

b

´2+³n π

b

´4¸− Px

b

³m π

a

´2= 0 (277)

⇒ Pxb= D

³π a

m

´2 ∙³ma

´2+³nb

´2¸2(278)

where for the discrete values of Px Eq. (272) has nontrivial solutions. The criticalload can be determined by the smallest eigenvalue, i.e. n = 1 for all values of a:

Pxb=D

³π a

m

´2 "³ma

´2+

µ1

b

¶2#(279)

=π2 D

b2

µa b

m

¶2 "³ma

´2+

µ1

b

¶2#

=π2 D

b2

µm b

a+

a

m b

¶2Now, the critical load (Px)cr can be written as:

(Px)cr = kcπ2 D

b(280)

where

kc =

µm b

a+

a

m b

¶2(281)

where coefficient kc is a function of aspect ratio a/b and wavelength parameter m.

56

2 6 12

For a given a/b, m may be chosen to yield the smallest eigenvalue. In order tominimize kc in Eq. (281), treating m as a continuous variable produces:

∂kc∂m

= 2

µm b

a+

a

m b

¶µb

a− a

b m2

¶= 0 (282)

where the first bracket can not be zero, so the second bracket should be zero:

⇒ b

a− a

b

1

m2= 0 (283)

Now, one gets:

m =a

bkc = 4

(284)

Here, this is valid when a/b is integer and when considering a very long plates.Transition fromm tom+1 half-waves occurs when the two corresponding curves

have equal ordinates, i.e. from Eq. (281):

kc|m = kc|m+1 (285)

⇒ m b

a+

a

m b=(m+ 1) b

a+

a

(m+ 1) b(286)

⇒ a

b=pm (m+ 1)

57

a

b=pm (m+ 1) (287)

Example 1 For m = 1, a/b =√2

Example 2 For a very large m, i.e. a very long plate, a/b ' m. Thus, kc = 4 isnow independent of m.

A very long plate buckles in half-waves, whose lengths approach the width of theplate:

w = c1 sin³π x

b

´sin³n π y

b

´Thus, the buckled plate subdivides approximately into squares.

Various Boundary Conditions of Plate under In-Plane Compressive Load-ing The critical buckling load reads:

(Px)cr = kcπ2 D

b

Influence of boundary conditions on the buckling coefficients ofplates subjected to in-plane compressive loading

58

00

1 2

2

3 4

4

5

6

8

10

12

14

16

c

C

c c

c

A

D

E

B

ssss

ss

ss

free

free

Loaded edges simply supported.Loaded edges clamped.

ab

kc

Figure by MIT OCW.

Various Boundary Conditions of Plate under In-Plane Shear LoadingThe critical buckling load per unit length reads:

(Nxy)cr = kcπ2 D

b2

where the dimension of Nxy is [N/m].

Limiting Case: Wide Plates Consider a wide plate for which a/b¿ 1. Fromthe diagram, we see that if a/b < 1, then m is set to be equal to unity, i.e. just onewavelength in the x-direction.

59

15

13

11

9

7

5

30 1 2 3 4 5

Clamped Edges

Simply Supported Edges

ab

kc

rcr =η2Db2

hkc

Critical values of shear stress for plates subjected to in-plane loading.

Figure by MIT OCW.

b

ax

y

The buckling formula thus becomes:

(Nx)cr = kc|m=1π2 D

b2(288)

=π2 D

b2

µb

a+

a

b

¶2=π2 D

a2

³ab

´2µ b

a+

a

b

¶2=π2 D

a2

∙1 +

³ab

´2¸2If a/b ¿ 1, then the second term in the bracket can be neglected so that thebuckling load per unit length becomes:

(Nx)cr =π2D

a2

which is called Sezawa’s formula for wide plates.

Example 3 Here, relative merits of stiffening a large panel are investigated in thelongitudinal or in the transverse direction. It is assumed that the stiffeners providefor a simply supported boundary conditions.Consider the case of longitudinal stiffeners.

60

L

crσ

s

From the von Karman formula, the buckling load per unit length for each dividedpart reads:

(Nx)cr =4π2D

s2

Now, the buckling stress can be calculated:

(σcr)longitudinal =(Nx)cr

h

=4π2D

s2h

Consider the case of transverse stiffeners.

crσ s

L

From the Sezawa’s formula, the buckling load per unit length along the loaded edgesreads:

(Nx)cr =π2D

s2

Now, the buckling stress can be calculated:

(σcr)transverse=(Nx)cr

h

=π2D

s2h

Thus, it is concluded that

(σcr)longitudinal = 4 (σcr)transverse

This shows advantages of longitudinal stiffeners over transverse stiffeners.

61

4.3.4 Derivation of Raleigh-Ritz Quotient

Recall the total potential energy of system and other corresponding definitions:

Π = (Ub − Vb) + (Um − Vm) (289)

where each term for buckling problems will be discussed in the following.

Term Relating to Plate Bending Response In the buckling problem, thework done by external load causing bending response considered as zero:

Vb = 0 (290)

The bending energy can be expressed:

Ub=1

2

ZSMαβ καβ dS (291)

=D

2

ZS[(1− ν) καβ + ν κγγ δαβ] καβ dS

=D

2

ZS

h(1− ν) καβκαβ + ν (κγγ)

2idS

=D

2

ZS

h(1− ν) (κ11κ11 + κ12κ12 + κ21κ21 + κ22κ22) + ν (κ11 + κ22)

2idS

=D

2

ZS

n(κ11 + κ22)

2 − 2 (1− ν)hκ11 κ22 − (κ12)2

iodS

Here, the term in the square bracket is called Gaussian curvature:

κ11 κ22 − (κ12)2 = κI κII (292)

where κI and κII are the principal curvatures. For plates with straight edges,Gaussian curvature vanishes, so one gets:

Ub =D

2

ZS(κ11 + κ22)

2 dS (293)

The integrand of the above equation can be written in terms of the transversedisplacement:

(κ11 + κ22)2=

¡−∇2w

¢2(294)

=∇2w ∇2w

Now, the bending energy reads:

Ub =D

2

ZS∇2w ∇2w dS (295)

62

Term Relating to Plate Membrane Response The work done by externalload causing membrane response reads:

Vm =

ZΓNnn un dl +

ZΓNtn ut dl (296)

In the buckling problem, the axial force N◦αβ = λNαβ is determined from the

pre-buckling solution and is considered as constant, so the membrane energy reads:

Um=−λZSNαβ ε◦αβ dS (297)

=−λZSNαβ

∙1

2(uα,β +uα,β ) +

1

2w,α w,β

¸dS

=−λZSNαβ uα,β dS − λ

2

ZSNαβ w,α w,β dS

Here, the first term can be extended in a similar way shown in Eq. (132):

−λZSNαβ uα,β dS =−λ

ZΓ

³Nnn un + Ntn ut

´dl (298)

−λZSNαβ,β uα dS

=−λZΓ

³Nnn un + Ntn ut

´dl

where the in-plane equilibrium is applied:

−λ Nαβ,β = 0 (299)

Now, the membrane energy can be expressed:

Um = −λZΓ

³Nnn un + Ntn ut

´dl − λ

2

ZSNαβ w,α w,β dS (300)

Thus, the term relating to membrane response can be summarized:

Um − Vm=−λZΓ

³Nnn un + Ntn ut

´dl − λ

2

ZSNαβ w,α w,β dS (301)

−ZΓNnn un dl −

ZΓNtn ut dl

=

ZΓ

³−λ Nnn − Nnn

´un dl −

ZΓ

³−λ Ntn − Ntn

´ut dl

−λ2

ZSNαβ w,α w,β dS

=−λ2

ZSNαβ w,α w,β

where the boundary conditions on Γ are applied.

63

Total Potential Energy and Its Variations Now, one gets the total potentialenergy:

Π=(Ub − Vb) + (Um − Vm) (302)

=D

2

ZS∇2w ∇2w dS − λ

2

ZSNαβ w,α w,β dS

The first variation of the potential energy can be obtained:

δΠ=D

2

ZS

£δ¡∇2w

¢∇2w +∇2w δ

¡∇2w

¢¤dS (303)

−λ2

ZSNαβ (δw,α w,β +w,α δw,β ) dS

=D

ZS∇2w δ

¡∇2w

¢dS − λ

ZSNαβ w,α δw,β dS

=D

ZS∇2w ∇2δw dS − λ

ZSNαβ w,α δw,β dS

where Nαβ is considered as constant under the variation. Similarly, the secondvariation of the potential energy reads:

δ2Π=D

ZSδ¡∇2w

¢∇2δw dS − λ

ZSNαβ δw,α δw,β dS (304)

D

ZS∇2δw ∇2δw dS − λ

ZSNαβ δw,α δw,β dS

Raleigh-Ritz Quotient Application of the Trefftz condition for invertability,δ2Π = 0, determines the load intensity:

λ =DRS∇

2δw ∇2δw dSRS Nαβ δw,α δw,β dS

(305)

Here, choose a trial function for w:

w = A φ (306)

where A is the undetermined magnitude, and φ = φ (x, y) is a normalized shapefunction. Then, the variation of the trial function reads:

δw = δA φ (307)

Now, the load intensity reads:

λ=DRS ∇

2 (δA φ) ∇2 (δA φ) dSRS Nαβ (δA φ) ,α (δA φ) ,β dS

(308)

=DRS δA ∇2φ δA ∇2φ dSR

S Nαβ δA φ,α δA φ,β dS

=DRS ∇

2φ ∇2φ dSRS Nαβ φ,α φ,β dS

64

The Raleigh-Ritz quotient is defined as:

λ =DRS ∇

2φ ∇2φ dSRS Nαβ φ,α φ,β dS

(309)

Example 4 As a special case, consider 1-D case:

Nαβ =

¯1 00 0

¯then, the Raleigh-Ritz quotient becomes:

λ =DRS ∇

2φ ∇2φ dSRS (φ,x )

2 dS

Example 5 Similarly, consider 2-D compression case:

Nαβ = δαβ =

¯1 00 1

¯then, the Raleigh-Ritz quotient becomes:

λ=DRS ∇

2φ ∇2φ dSRS φ,α φ,α dS

=DRS ∇

2φ ∇2φ dSRS

h(φ,x )

2 + (φ,y )2idS

4.3.5 Ultimate Strength of Plates

The onset of buckling stress σcr does not necessarily means the total collapse of theplate. Usually, there is redistribution of stresses, and the plate takes additionalload until the ultimate strength σu is reached.

Von Karman Analysis of the Effective Width For a simply supported plate,the buckling load is:

Pcr =4π2D

b(310)

=4π2E

12 (1− ν2)

h3

b

and the corresponding buckling stress is:

σcr =Pcrbh

(311)

=4π2E

12 (1− ν2)

µh

b

¶2=

π2E

3 (1− ν2)

µh

b

¶265

The normalization of buckling stress by the yield stress reads:

σcrσy=

π2

3 (1− ν2)

E

σy

µh

b

¶2(312)

=(1.9)2

σyE

¡bh

¢2σcrσy

=

µ1.9

β

¶2(313)

where β is a non-dimensional parameter defined by:

β =

rσyE

b

h(314)

The relation between the normalized buckling stress versus β is plotted in the nextfigure.

1.9

On further loading the plate beyond σcr, a greater proportion of the load istaken by the regions of the plate near the edge. Von Karman assumed that theseedge regions, each of the width beff/2, carry the stress up to the yield while thecenter is stress free.

66

σ

Before Buckling

( )yσ

After Buckling

x

y

b

yσ

( )yσ

yσ2effb

b

2effb

Actual Stress von Karman Model

The edge zones are at yield, i.e. σcr/σy = 1, but the width of the effectiveportion of the plate is unknown:

σcrσy

=(1.9)2

σyE

³beffh

´2 = 1 (315)

from which one obtains:

beff = 1.9 h

sE

σy(316)

Taking, for example, E/σy = 900 for mild steel, one gets:

beff = 1.9√900 = 57h (317)

This is somehow high, but there is not much difference from the empirically deter-mined values of beff = 40h ∼ 50h.

67

The total force at the point of ultimate load is:

P = σy beff h (318)

Now, the average ultimate stress can be calculated:

σu=P

bh(319)

= σybeffb

=1.9h

b

pEσy

The average ultimate stress can be normalized by the yield stress:

σuσy= 1.9

hb

pEσy

σy(320)

σuσy

=1.9

β(321)

The average ultimate stress is plotted with respect to β in the next figure.

1.9

Comparison of the ultimate and buckling load solution is shown in the nextfigure.

68

1.9

Thick plates yield before buckle.

Thin plates buckle before reaching ultimate load.

Under the uniaxial loading, the relation between an applied load and the cor-responding displacement is schematically shown all the way to collapse in the nextfigure.

xu

xP

crP

pre-buckling

post-buckling

uP

collapseor crash

x

y

xPxP

xu

Empirical Formulas

• Foulkner correctionσuσy=2

β− 1

β2(322)

69

• Gerard (Handbook of elastic stability)

σuσy= 0.56

Ãgh2

A

sE

σy

!0.85(323)

where g is the sum of the number of cuts and the number of flanges after thecuts, A is the cross sectional area A = bh, and the coefficients 0.56 and 0.85are empirical constants.

Example 6 Consider a plate which has one cut and two flanges.

b

h

Then,g = 1 + 2 = 3

Now,

σuσy=0.56

Ã3h2

bh

sE

σy

!0.85

=1.42

Ãh

b

sE

σy

!0.85=1.42

β0.85

Modifications in Codes In the original von Karman formula, the effective widthratio reads:

beffb=

rσcrσy

(324)

In the ANSI specification, imperfection is considered:

beffb=

rσcrσy

µ1− 0.218

rσcrσy

¶(325)

In UK specification, the effective width ratio is defined as:

beffb=

"1 + 14

µrσcrσy− 0.35

¶4#−0.2(326)

70

A comparison of theoretical (von Karman) and experimental predictions for the effective width in compressed steel plates

y

eff

4.3.6 Plastic Buckling of Plates

Stocky plates with low b/h ratio will yield before buckling at the point B. Afteradditional load, the plate will deform plastically on the path BC until conditionsare met for the plate to buckle in the plastic range.

1.9

B

A

C

Stowell’s Theory for the Buckling Strain Stowell developed the theory ofplastic buckling for simply-supported square plates loaded in one direction.

71

x

y

b

b

thickness h

The critical buckling strain εcr was derived by him in the form:

εcr =π2

9

µh

b

¶2 h2 +

p1 + 3 (Et/Es)

i(327)

where the tangent modulus Et and the secant modulus Es are defined by:

Et =dσ

dε; Es =

σ

ε(328)

ε

σ

yσ

sE

tE

For the materials obeying the power hardening low:

σ = σr

µε

εr

¶n

(329)

where σr and εr are the reference stress and strain. Now, the tangent and secantmodulus are:

Et=dσ

dε= n

σrεr

µε

εr

¶n−1(330)

Es=σ

ε=

σrεr

µε

εr

¶n−1

72

Substituting these expression back into the buckling equation (327), one gets:

εcr =π2

9

µh

b

¶2 ¡2 +√1 + 3n

¢(331)

The exponent n varies usually between n = 0 (perfectly plastic material) and n = 1(elastic material). This makes the coefficient

¡2 +√1 + 3n

¢vary in the range

3 ∼ 4. For a realistic value of n = 0.3, the buckling strain becomes:

εcr = 3.7

µh

b

¶2(332)

Having determined εcr, the corresponding buckling stress is calculated from thepower law.

Approximate Solution for the Buckling Strain Consider an elastic planestress relation:

σxx=E

1− ν2(εxx + νεyy) (333)

σyy =E

1− ν2(εyy + νεxx)

Using the solution for the pre-buckling state, i.e. σxx = σcr and σyy = 0 leads:

σcr =E

1− ν2(εxx + νεyy) (334)

0=E

1− ν2(εyy + νεxx)

from which one gets:

εyy =−νεxx (335)

σcr =Eεxx = Eεcr

The critical elastic buckling stress is:

σcr = kcπ2E

12 (1− ν2)

µh

b

¶2(336)

So, the corresponding critical elastic buckling strain read:

εcr = kcπ2

12 (1− ν2)

µh

b

¶2(337)

Equation (337) is more general than a similar expression Eq. (327) given bythe Stowell theory because it applies to all type of boundary conditions. At the

73

same time, Stowell’s equation was derived only for the simply supported boundaryconditions. In particular, for kc = 4, Eq. (337) predicts:

εcr = 3.6

µh

b

¶2(338)

which should be compared with the coefficient 3.7 of Eq. (332) in the Stowell’stheory. For a plastic material or very high hardening exponent, the prediction ofboth method are much closer.

4.3.7 Exercise 1: Effect of In-Plane Boundary Conditions, δw = 0

No Constraint in In-Plane Displacement Consider no constraint in in-planedisplacement in y-direction, N◦

yy = 0.

x

y

a

ayu

xu

NN

expansion due toPoison's effect

The membrane force tensor reads:

N◦αβ =

¯N◦ 00 0

¯(339)

From the constitutive equation, one gets:

Nyy =Eh

1− ν2£ε◦yy + ν ε◦xx

¤= 0 = −N◦

yy (340)

=⇒ ε◦yy = −ν ε◦xx

By applying the geometric equation between strain and the displacement and con-sidering δw = 0, here, one gets the relation between ux and uy:

∂uy∂y

= −ν ∂ux∂x

(341)

74

Integrating both sides over the plate length leads:Z a

0

∂uy∂y

dy=−νZ a

0

∂ux∂x

dx (342)

uy =−ν ux

Constraint in In-Plane Displacement Consider a plate fully constrained inthe y-direction, uy = 0.

x

y

a

a

xu

NN xxN N=

yyN Nν=

Consequently, one also gets:

ε◦yy =∂uy∂y

= 0 (343)

Under the uniform compression, ε◦xx reads:

ε◦xx =∂ux∂x

=uxa

(344)

From the constitutive relation, the membrane forces reads:

Nxx=Eh

1− ν2£ε◦xx + ν ε◦yy

¤=

Eh

1− ν2uxa= −N◦

xx (345)

Nyy =Eh

1− ν2£ε◦yy + ν ε◦xx

¤=

Eh

1− ν2νuxa= −N◦

yy

Finally, the membrane force tensor reads:

N◦αβ =

¯N◦xx 00 N◦

yy

¯= λ

¯1 00 ν

¯(346)

where

λ =Eh

1− ν2uxa

(347)

75

4.3.8 Exercise 2: Raleigh-Ritz Quotient for Simply Supported SquarePlate under Uniaxial Loading

Consider a simply supported square plate subjected to uniform compressive load inthe x-direction.

x

y

a

a

N N

Then, the membrane force tensor reads:

N◦αβ = N◦

¯1 00 0

¯= λ

¯1 00 0

¯(348)

The plate will deform into a dish, so for the trial function take the following:

φ = φ (x, y) = sin³π x

a

´sin³π y

a

´(349)

Now, in order to obtain the load intensity, we first calculate φ,x and ∇2φ :

φ,x=π

acos³π x

a

´sin³π y

a

´(350)

φ,y =π

asin³π x

a

´cos³π y

a

´

φ,xx=−³πa

´2sin³π x

a

´sin³π y

a

´(351)

φ,yy =−³πa

´2sin³π x

a

´sin³π y

a

´

∇2φ=φ,xx+φ,yy (352)

=−2³πa

´2sin³π x

a

´sin³π y

a

´∇2φ ∇2φ = 4

³πa

´4sin2

³π x

a

´sin2

³π y

a

´(353)

76

Now, the Raleigh-Ritz quotient is calculated:

N◦= λ =DRS ∇

2φ ∇2φ dSRS (φ,x )

2 dS(354)

=4D

¡πa

¢4¡πa

¢2RS sin

2¡π xa

¢sin2

¡π ya

¢dSR

S cos2¡π xa

¢sin2

¡π ya

¢dS

=4D³πa

´2

N◦cr = λ = 4D

µπ

a

¶2(355)

This is the classical buckling solution, and it is exact because of the right guess ofthe displacement field.

4.4 Buckling of Sections

4.4.1 Transition from Global and Local Buckling

Euler buckling load of a simply-supported column reads:

(Pcr)column =π2EI

l2(356)

where I is the bending rigidity of the column. Consider a section column which iscomposed of several thin plates, then the Euler buckling load can be considered asa global buckling load of the column.

EIl

( )cr columnP

77

On the other hands, local buckling force of a simply-supported plate reads:

Ncr =kcπ

2D

b2(357)

where D = Eh3/£12¡1− ν2

¢¤and kc = [(mb) /a+ a/ (mb)]2. Thus, the total local

buckling load can be obtained:

(Pcr)plate =kcπ

2D

b(358)

a

crN

b

h

Transition from global to local buckling can be calculated by (Pcr)column =(Pcr)plate:

π2EI

l2= kc

π2D

b(359)

=kcπ

2E

12 (1− ν2)

h3

b

I

bh3=

kc12 (1− ν2)

µl

b

¶2(360)

Example 7 Consider a square box column.

78

bb

l

h

Then, one obtains:

kc=4

I =2

3hb3

Now, the global buckling load reads:

(Pcr)column = π2EI

l2=2π2E

3

hb3

l2

and the local buckling load from four plates can be calculated:

(Pcr)four plates =4kcπ

2E

12 (1− ν2)

h3

b

Then, applying (Pcr)column = (Pcr)four plates, one gets:µb2

hl

¶2' 2.20

Thus, the local and global buckling loads become same when

b2 ' 1.5hl

For example, if b = 40h, then l = 60b.

Transition from the local to global buckling for an open channel section withlips is shown in the figure below.

79

Buckling coefficients and modes for a hat section

4.4.2 Local Buckling

The remainder of this section deals only with the local buckling. Dividing bothsides of Eq. (357) by the plate thickness b gives the expression of the buckling stressσcr:

σcr =Ncr

h=

kcπ2E

12 (1− ν2)

µh

b

¶2(361)

Consider two adjacent plates of a section of the prismatic column.

Compatibility and equilibrium conditions at junction of adjoining walls of a section

1h

1b

2h

2b

1Plate

2Plate

In general, there will be a restraining moment acting at the corner line between

80

Plate 1 and Plate 2. The buckling stresses for those two plates are:

(σcr)1=k1π

2E

12 (1− ν2)

µh1b1

¶2(362)

(σcr)2=k2π

2E

12 (1− ν2)

µh2b2

¶2Before buckling, stresses in the entire cross-section are the same. So, at the pointof buckling, one gets:

(σcr)1 = (σcr)2 (363)

from which the buckling coefficient k2 is relating to k1:

k2 = k1

µh1h2

b2b1

¶2(364)

The total buckling load on the angle element is:

Pcr =(σc)1 h1b1 + (σc)2 h2b2 (365)

=π2E

12 (1− ν2)

"k1

µh1b1

¶2h1b1 + k2

µh2b2

¶2h2b2

#

=π2E

12 (1− ν2)

"k1(h1)

3

b1+ k1

µh1h2

b2b1

¶2 (h2)3b2

#

=π2Ek1

12 (1− ν2)

µh1b1

¶2A

where A is the sectional area of two plate A = b1h1+b2h2. From this derivation, theconclusion is that only one buckling coefficient is needed to calculate the bucklingload of the section consisting of several plates.

Determination of the buckling coefficient is a bit more complicated because ofthe existence of the edge bending moment. This can be illustrated in an exampleof a box column with a rectangular cross section with the same thickness h.

1b

2b

uniform thickness h

The wider flange will be ready to buckling first while the narrow plate is not ready

81

to buckle. When the second plate buckles, the first plate would have buckled longbefore. Thus, there is an interaction between left plates, and a compromise mustbe established because left plates must buckle at the same time. The bucklingcoefficient as a function of the ratio b2/b1 is plotted in the figure next.

Buckling coefficients for box sections

h

h

In the limiting case of a square box (b1 = b2 = b), k1 = 4 and the edge interactivemoment between adjacent plates is zero.

Some useful graphs and formulas for typical sections are given next.

82

83

5 Buckling of Cylindrical Shells

5.1 Governing Equation for Buckling of Cylindrical Shells

The starting point of the analysis is the strain-displacement relation for plates:

ε◦αβ =1

2(uα,β +uβ,α ) +

1

2w,αw,β (366)

καβ =− w,αβ

Consider a flat plate (x, y) and a segment of a cylinder (x, aθ), where a is the radiusof a cylinder and θ is the hoop coordinate.

aθ

a

y

x x

Can the strain-displacement relation for a cylinder be derived from similar re-lation for a flat plate?

x→ x ; dx→ dx (367)

y → aθ ;∂

∂y→ 1

a

∂

∂θ(368)

Consider component by component. Use the notation:

u1 = ux→ uu2 = uy→ v

u3 →w(369)

Then, one gets:

ε◦xx = u,x+1

2(w,x )

2 (370)

ε◦θθ =u,θa+1

2

³w,θa

´2+

w

a(371)

ε◦xθ =1

2

³u,θa+ u,x

´+1

2w,x

w,θa

(372)

There is a new term w/a in the expression for the hoop strain. The physicalmeaning of this new term becomes clear if we consider axisymmetric deformationwith v = 0 and w independent of θ. Define the hoof strain as a relative change

84

in the length of circumference when the original circle has a radius of "a" beforedeformation and a new circle has a radius of "a+ w" after deformation:

ε◦θθ =2π (a+ w)− 2πa

2πa=

w

a(373)

a

wθ

Mathematically, Eq. (370)-(372) can be derived from its counterparts by trans-forming the rectangular coordinate system into the curvilinear coordinate system.

x= r sin θ (374)

y= r cos θ

The step-by-step derivation can be found, for example, in the book by Y.C. Fung,"First Course in the Continuum Mechanics." The expression for curvature aretransformed in a similar way:

κxx=w,xx (375)

κθθ =w,θθa2

κxθ =w,xθa

Using the variational approach explained in details for the plate problem, one cansee that the only new term in the expression for δΠ = 0 isZ

SNθθ

δw

adS (376)

Therefore, the new term should be added in the equation for out-of-plane equilib-rium:

D∇4w + 1aNθθ −

µNxxw,xx+

2

aNxθw,xθ+

1

a2Nθθw,θθ

¶= q (377)

where∇4w = w,xxxx+

2

a2w,xxθθ+

1

a4w,θθθθ (378)

85

The above equations are the nonlinear equilibrium equations for quasi-shallow cylin-drical shells. The linear equilibrium equations are obtained by omission the non-linear terms, i.e. terms in the parenthesis. The resulting equations are:

aNxx,x+Nxθ,θ =0 (379)

aNθθ,x+Nθθ,θ =0

D∇4w + 1aNθθ = q

with

Nxx=C³u,x+ν

w

a

´(380)

Nθθ =C³w0a+ ν u,x

´where C is the axial rigidity, C = Eh/

¡1− ν2

¢. The pre-buckling solution should

satisfy the system Eq. (379) and (380). These solutions will be denoted by Nαβ =−N◦

αβ.

5.1.1 Special Case I: Cylinder under Axial Load P , q = 0

The membrane forces in the pre-buckling state are:

N◦αβ =

P

2πa

¯1 00 0

¯(381)

and the corresponding displacement field reads:

u (x) =− P

2πahEx (382)

w=ν

E

Pl

2πah

lw

( )u l

aθ

x

It is easy to prove that the above solution satisfies all field equation.

86

5.1.2 Special Case II: Cylinder under Lateral Pressure

N◦αβ = Nθθ

¯0 00 1

¯(383)

where from the constitutive equations Eq. (380):

Nθθ = Ehw

a(384)

Substituting Eq. (384) into Eq. (379) leads the following linear forth order inho-mogeneous ordinary differential equation for w (x):

Dw0000 +Ehw

a2= q (385)

or in a dimensionless form:d4w

dx4+ 4β4w =

q

D(386)

where

β4 =Eh

a2D=3¡1− ν2

¢a2h4

(387)

The dimension of β is [L−1M0T 0], so βx is dimensionless. There are four boundaryconditions for a simply supported cylinder:

Mxx=w = 0 at x = 0 (388)

Mxx=w = 0 at x = l

where one gets from the moment-curvature relation:

Mxx=−Dd2w

dx2(389)

Mθθ =0

The general solutions of the above boundary value problem is:

w (x) = e−βx [c1 sinβx+ c2 cosβx] + eβx [c3 sinβx+ c4 cosβx]−q

D4β4(390)

The four integration constants can be found from the boundary conditions. Atypical term of the solution is a rapidly decaying function of x.

87

It can be conducted that the curvature and bending is confined to a narrowboundary zone of the width xb = π/ (2β). The remainder of the shell undergoes auniform radial contraction:

w0 = −q

D 4β4(391)

For the sake of simplicity, this localized bending can be neglected (D → 0).Then, from Eq. (379), the hoop membrane force is related to the lateral pressureby Nθθ = qa and the pre-buckling solution is:

N◦αβ = qa

¯0 00 1

¯(392)

5.1.3 Special Case III: Hydrostatic Pressure

For a cylinder subjected to the hydrostatic pressure, the total axial compressiveforce is:

P = qπa2 (393)

The pre-buckling solution is:

N◦αβ = qa

¯1/2 00 1

¯(394)

which is a classical membrane stress state in a thin cylinder.

88

5.1.4 Special Case IV: Torsion of a Cylinder

The pre-buckling stress in a cylinder subjected to the total torque of the magnitudeT is:

N◦αβ =

M

2πa

¯0 11 0

¯(395)

xNθ

aθ

x

M

5.2 Derivation of the Linearized Buckling Equation

We are now in the position to linearize the nonlinear buckling equation Eq. (377).It is assumed that the state of membrane forces does not change at the point ofbuckling from the pre-buckling value. Thus,

Nαβ = −N◦αβ (396)

and Eq. (377) becomes:

D∇4w + 1aNθθ +

∙N◦xxw,xx+

2

aN◦xθw,xθ+

1

a2N◦θθw,θθ

¸= q (397)

where the hoop membrane force Nθθ, the second term in Eq. (397), depends linearlyon three component of the displacement vector (u, v, w):

Nθθ = C

µ1

av,θ+

w

a+ νu,x

¶(398)

89

Therefore, the out-of-plane equilibrium equation is coupled with the in-plane dis-placement (u, v) through the presence of the term Nθθ. Note that in the platebuckling problem the in-plane and out-of-plane response was uncoupled. It ispossible to eliminate the terms involving in-plane components using the full set ofequilibrium and constitutive equations in the in-plane direction. By doing this, theorder of the governing equation has to be raised by four to give eight:

D∇8w + 1− ν2

a2Cw,xxxx+∇4

∙N◦xxw,xx+

2

aN◦xθw,xθ+

1

a2N◦θθw,θθ

¸= 0 (399)

The above equation is called the Donnell stability equation in the uncoupled form.Not that w (x, θ) in the above equation represents additional lateral deflection overand above those produced by the pre-buckling solution. The total deflection is asum of the two.

5.3 Buckling under Axial Compression

5.3.1 Formulation for Buckling Stress and Buckling Mode

We are now in a position to develop solutions to the buckling equations, Eq. (399)for four different loading cases discussed in the previous section. Consider firstCase I of a simply-supported cylindrical shell in which the pre-buckling solution isgiven by Eq. (381). In this case, Eq. (399) reduces to:

D∇8w + 1− ν2

a2Cw,xxxx+

P

2πa∇4w,xx= 0 (400)

The buckling deflection of the shell is assumed in the following form:

w (x, θ) = c1 sin³mπx

l

´sin (nθ) (401)

where c1 is the magnitude, and the integer numbers (m, n) denote the numberof half-waves respectively in the axial and circumferential direction. The abovedeformation satisfies both simply-supported boundary conditions at the ends, x = 0and x = L, and periodicity conditions along the circumference.

Here, the half-length of the buckling wave is defined:

λ =l

m(402)

It is convenient to introduce a dimensionless buckling number, m:

mπx

l=³mπa

l

´ x

a= m

x

a(403)

⇒ m =mπa

l

90

Using the dimensionless buckling number, substituting the solution Eq. (401)into the governing equation Eq. (400) leads:∙D

a2¡m2 + n2

¢4+ m4

¡1− ν2

¢C − P

2πa

¡m2 + n2

¢2m2

¸c1a6sin³mx

a

´sin (nθ) = 0

(404)By setting the coefficient in the square bracket to zero, the critical buckling mem-brane force per unit length becomes:

Ncr =Pcr2πa

(405)

=D

a2

¡m2 + n2

¢2m2

+¡1− ν2

¢C

m2

(m2 + n2)2

Here, by introducing the dimensionless parameter χ:

χ =

¡m2 + n2

¢2m2

(406)

Eq. (405) reads:

Ncr =D

a2χ+

¡1− ν2

¢C1

χ(407)

The dependence of the buckling force on the parameter χ is shown in the figurebelow.

91

Treating χ as a continuous variable, one can find an analytical minimum:

dNcr

dχ=

D

a2−¡1− ν2

¢C1

χ2= 0 (408)

from which the optimum value of the parameter χ is:

χopt=

r(1− ν2)Ca2

D(409)

=a

h

p12 (1− ν2)

' 3.3ah

Introducing the expression for the optimum parameter χ into Eq. (407) and usingdefinitions of bending and axial rigidity, one gets:

(Ncr)min =Ep

3 (1− ν2)

h2

a(410)

The buckling stress is obtained by dividing the critical membrane fore by the shellthickness h:

σcr =(Ncr)min

h(411)

=Ep

3 (1− ν2)

h

a

σcr ' 0.605 Eh

a(412)

This is the classical solution for the buckling stress of a cylindrical shell subjectedto axial compression. While the buckling load is unique and does not depend on(m, n), the buckling mode is not unique as:

χopt =

¡m2 + n2

¢2m2

= 3.3a

h(413)

There are infinity of combinations of m and n that give the same expression.

Example 8 Let a/h = 134, then

χopt =

¡m2 + n2

¢2m2

= 3.3× 134 ' 442

⇒ m2 + n2

m' 21

92

n =p21m− m2

5.3.2 Buckling Coefficient and Batdorf Parameter

Let us introduce the dimensionless buckling stress or the buckling coefficient kc:

kc = σcrhl2

π2D(414)

Additionally, the Batdorf parameter Z is defined:

Z =p1− ν2

l2

ah(415)

Using Eq. (411), the buckling coefficient becomes:

kc=Ep

3 (1− ν2)

h

a× hl2

π2D(416)

=12√3π2

p1− ν2

l2

ah

=12√3π2

Z

kc ' 0.702 Z (417)

This relation between kc and Z are shown in the figure below together with twolimiting cases of very short and very long cylindrical shells. These two limitingcases are discussed below.

93

Limiting Cases: Short Cylinders, a À l Consider a case of the followingconditions:

aÀ l

⇒ a

l→∞ (418)

Then, it is natural to assume that the number of half-waves in the axial directionis unity:

m = 1

Consequently, one gets:m = π

a

l→∞ (419)

Since m is much larger than n, one also gets:

m2 + n2 ' m2 (420)

94

Now, from Eq. (405), the membrane force reads:

Ncr =P

2πa(421)

=D

a2m2 +

¡1− ν2

¢C1

m2

=m2D

a2

=π2D

l2

The buckling stress reads:

σcr =Ncr

h=

π2D

hl2

Now, from the definition of the buckling coefficient, Eq. (414), the buckling coeffi-cient for the very short cylindrical shells reads:

kc = 1

This solution indicated in previous figure as the lower bound cut-off value. Theupper bound cut-off values is given by the Euler buckling load.

Limiting Cases: Very Long Cylinders, lÀ a The cylindrical shell is becom-ing the Euler column. The buckling load of the column is:

Pcr =π2EI

l2(422)

where I = πa3h for cylinder sections.

95

The buckling stress reads:

σcr =Pcr2πah

(423)

=π2

2E³al

´2For this very long cylindrical shells, the buckling coefficient can be written:

kc = 6¡1− ν2

¢ ³ah

´2(424)

kc = 6

µa

l

¶4Z2 (425)

From Eq. (411) and (423), the transition between the local shell buckling andglobal thin-walled column buckling occurs when

(σcr)shell = (σcr)column (426)

which gives:Ep

3 (1− ν2)

h

a=

π2

2E³al

´2(427)

⇒ hl2

a3=

π2p3 (1− ν2)

2' 8.15 (428)

5.4 Buckling under Lateral Pressure

From the pre-buckling solution, Eq. (392), the governing equation Eq. (399) be-comes:

D∇8w + 1− ν2

a2Cw,xxxx+

q

a∇4w,θθ= 0 (429)

Assuming the double sine buckling deflection function, similar to the case of axialcompression, the governing equation becomes:∙

D

a2¡m2 + n2

¢4+ m4

¡1− ν2

¢C − qa n2

¡m2 + n2

¢2¸ c1a6sin³mx

a

´sin (nθ) = 0

(430)By setting the coefficient in the square bracket to zero, the equation for the bucklingpressure becomes:

qa =D

a2

¡m2 + n2

¢2n2

+¡1− ν2

¢C

m4

n2 (m2 + n2)2(431)

96

It can be shown that the smallest eigenvalue is obtained when m = 1 or m = πa/l.With this observation, the solution becomes a function of the parameter n, ordimensionless parameter n:

n =nl

πa(432)

Additionally, we define the dimensionless buckling pressure, q:

q = ql2a

π2D(433)

Now, substituting m, n and p into Eq. (431) leads:

q =

¡1 + n2

¢2n2

+1

n2 (1 + n2)2

¡1− ν2

¢C

a2

µl

a

¶2(434)

By introducing the Batdorf parameter Z, one gets:

q =

¡1 + n2

¢2n2

+1

n2 (1 + n2)212

πZ2 (435)

For any value of the geometrical parameter Z, there exists a preferred n whichminimize the buckling pressure. Treating n as a continuous variable, the optimumn can be found analytically from dp/dn = 0. Substituting this back into Eq. (432),there will be a unique relation between the buckling pressure and the Batdorf pa-rameter. The solution is shown graphically in the figure below.

Limiting Cases: Very Long Cylinders, l À a In the limiting case of a longtube (lÀ a), one gets:

m = mπa

l→ 0 (436)

97

In this case, the last term in Eq. (431) vanishes and the buckling pressure becomes:

q =n2 D

a3(437)

Imagine a long cylinder consisting of a change of ring, each of the height b. Themoment of inertia of the ring along the axial axis reads:

I =bh3

12(438)

From Eq. (437) and (438), the intensity of the line load Q = qb can be written:

Q= qb (439)

=n2

a3Eh3

12 (1− ν2)b

=n2EI

a3 (1− ν2)

The above approximation is due to Donnell. The smallest integer value is n = 1which gives the following buckling mode.

98

The Donnell solution should be compared with more exact solution of the ringbuckling problem which take into account a more complex incremental displacementfield with both w (θ) and v (θ). Here, a distinction should be made between thecentrally directed pressure (as in all preceding analysis) and the field -pressureloading where pressure is always directed normal to the deformed surface.

In the later case, the ring buckling occurs at:

Qc =¡n2 − 1

¢ EIa3

(440)

where the smallest integer n = 2 so that Q = 3EI/a3. The buckling mode has noweight nodal points rather than four.

The solution of centrally directed pressure loading case is:

Qc =

¡n2 − 1

¢2n2 − 2

EI

a3(441)

where again the smallest n = 2. Thus the smallest buckling load intensity is:

Qc = 4.5EI

a3(442)

It is seen that for a realistic value n=2, the present solution, Eq. (439) givesthe buckling pressure between the two cases of field-pressure loading and centrallydirected loading, 3 < 4 < 4.5.

5.5 Buckling under Hydrostatic Pressure

Special case of the combined loading in which the total axial load P is:

P = πa2q (443)

99

The pre-buckling solution is given by Eq. (394). The solution of the buckling equa-tion still can be sought through the sinusoidal function, Eq. (401). The optimumsolution can be found by a trial and error method varying parameters m and n. Agraphical representation of the solution is shown in figure below.

5.6 Buckling under Torsion

A twist applied to one end of a cylindrical shell process a twisting force Nxθ:

N◦xθ =

M

2πa(444)

The other two components of the pre-buckling membrane forces vanishN◦xx = N◦

θθ =0. Moreover, the force is constant.

100

Under these conditions, the governing equation reduces to:

D∇8w + 1− ν2

a2C w,xxxx+

2

aN◦xθ ∇4w,xθ= 0 (445)

In view of the presence of odd-ordered derivatives in the above equation, the sep-arable form of the solution for w (x, θ), assumed previously, does not satisfy theequation.

Under torsional loading, the buckling deformation consists of circumferentialwaves that spiral around the cylindrical shell from one end to the other. Suchwaves can be represented by a deflection function of the form:

w (x, θ) = C sin³mx

a− nθ

´(446)

withm =

mπa

L(447)

where m and n are integers. The alone displacement field satisfy the differentialequation and the periodical condition is the circumferential direction, but does notsatisfy any commonly used boundary at the cylinder ends. Consequently, thissimple examples can be used only for long cylinders.

For such cylinders, introduction of the additional displacement field into thegoverning equation, yields:

N◦xθ =

¡m2 + n2

¢22 m n

D

a2+

m3

2 (m2 + n2)2 n

¡1− v2

¢C (448)

For sufficiently long cylinders, the shell buckles in two circumferential waves, n = 2.Also, the term m2 is small compared with 4. Then, the approximate expression is:

N◦xθ =

4

m

D

a2+

m3

64

¡1− v2

¢C (449)

An analytical minimization of the alone experiment with repeat to m gives:

m4 =64

9 (1− ν2)

µh

a

¶2(450)

101

Upon substitution, the final expression for the buckling force, or better, criticalshear strain causing buckling is:

τ cr =N◦xθ

h=

0.272 E

(1− ν2)3/4

µh

a

¶3/2(451)

The above solution was given by Donnell. As noted, the above solution is invalidfor short shells due to the difficulties in satisfying boundary condition. A morerigorous analytical-numerical solution is shown in the figure below.

As the radius of the shell approaches infinity, the critical stress coefficient forsimply supported and clamped edge approaches respectively the value 5.35 and 8.98corresponding to plates under the shear loading.

5.7 Influence of Imperfection and Comparison with Experiments

Because of the presence of unavoidable imperfection in real shells, the experimen-tally measured buckling load are much smaller than the ones found theoretically.

Reduction in the buckling strength due to imperfection

Comparison of theoretical and experimental results for four different type of loads:

102

• Axial compression

• Torsion

• Lateral pressure

• Hydrostatic pressureare shown in the subsequent two pages. Note that the graphs were presented

in log − log scale. Replotting the results for axially loaded plate yields the graphshown below.

The differences are shown to be very large. Design curves for cylindrical andother shells are based on the theoretical solution modified by empirical predictionfactors called "know-down factors". For example, the solid curve shown in thereceding page is a "90 percent probability curve." For a/h = 50, the reductionfactor predicted by this probability curve is 0.24. Thus, the theoretical solutionσcr = 0.605 Eh/a should be multiplied by 0.24 for the design stress σ = 0.15 Eh/a.

Cylindrical shells loaded in different way are been sensitive to imperfections andthe resulting knock-down factors are smaller.

Most industrial organization establish their own design criteria. They are fre-quently loose-leaf and are continuously updated on the volume of the experimentedevidence increase. The situation can be converted to the firmly established manual

103

Figure by MIT OCW.

1.1

1.0

0.8

0.6

0.4

0.2

00 500 1,000 1,500 2,000 2,500 3,000 3,500

Theory

A design recommendation

Distribution of test data for cylinders subjected to axial compression.

σ cr =

0.6

05 E

h/a

σ exp

ah

of steel constructions for the design of columns and beam. The reason is that col-umn are not sensitive to imperfection as far as the ultimate strength is concerned.

104