Plates and Shells

2.081J/16.230J Plates and and Shells Professor  Tomasz  Wierzbicki 

Contents 1 Strain-Displacement Relation for Plates 1.1 1-D Strain Measure . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Engineering Strain . . . . . . . . . . . . . . . . . . . . . . 1.1.2 Green-Lagrangian Strain . . . . . . . . . . . . . . . . . . . 1.2 3-D Strain Measure . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Derivation Derivation of Green-Lagra Green-Lagrangian ngian Strain Strain Tensor Tensor for Plates Plates . 1.2.2 Specification of Strain-Disp Strain-Displaceme lacement nt Relation for Plates .

. . . . . . . . . . . .

1 1 1 1 1 1 5

2 Derivation of  Constitutive Equations for Plates 2.1 Definitions of Bending Moment and Axial Force . . . . . . . . . . . . 2.2 Bending Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Bending Moment . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Bending Energy Density . . . . . . . . . . . . . . . . . . . . . 2.2.3 Total Bending Energy . . . . . . . . . . . . . . . . . . . . . . 2.3 Membrane Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Axial Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Membrane Energy Density . . . . . . . . . . . . . . . . . . . 2.3.3 Total Membrane Energy . . . . . . . . . . . . . . . . . . . . .

10 10 10 10 11 13 14 14 14 16

3 Development of Equation of  Equation of Equilibrium of  Equilibrium and Boundary Conditions Using Variational Approach 3.1 Bending Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . . 3.1.2 First Variation of the Total Potential Energy . . . . . . . . . 3.1.3 Equilibrium Equation and Boundary Conditions . . . . . . . 3.1.4 Specification of Equation for Rectangular Plate . . . . . . . . 3.2 Bending-Membrane Theory of Plates . . . . . . . . . . . . . . . . . . 3.2.1 Total Potential Energy . . . . . . . . . . . . . . . . . . . . . .

17 17 17 20 24 24 29 29

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3.2.2 3.2.3

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

4 General Theories of  Plate 4.1 Bending Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Derivation of  the Plate Bending Equation . . . . . . . . . . . 4.1.2 Reduction to a System of  Two Second Order Equations . . . 4.1.3 Exercise 1: Plate Solution . . . . . . . . . . . . . . . . . . . . 4.1.4 Exercise 2: Comparison between Plate and Beam Solution . . 4.1.5 Exercise 3: Finite Diff erence erence Solution of  the Plate Bending Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Membrane Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Plate Membrane Equation . . . . . . . . . . . . . . . . . . . . 4.2.2 Plate Equation for the Circular Membrane . . . . . . . . . . 4.2.3 Example: Approximation Solution for the Clamped Membrane 4.3 Buckling Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 General Equation of Plate Buckling . . . . . . . . . . . . . . 4.3.2 Linearized Buckling Equation of  Rectangular Plates . . . . . 4.3.3 Analysis of  Rectangular Plates Buckling . . . . . . . . . . . . 4.3.4 Derivation of  Raleigh-Ritz Quotient . . . . . . . . . . . . . . 4.3.5 Ultimate Strength of Plates . . . . . . . . . . . . . . . . . . . 4.3.6 Plastic Buckling of Plates . . . . . . . . . . . . . . . . . . . . 4.3.7 Exercise 1: Eff ect ect of  In-Plane Boundary Conditions, δw = 0 . 4.3.8 Exercise 2: Raleigh-Ritz Quotient for Simply Supported Square Plate under Uniaxial Loading . . . . . . . . . . . . . . . . . . 4.4 Buckling of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Transition from Global and Local Buckling . . . . . . . . . . 4.4.2 Local Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Buckling of  Cylindrical Shells 5.1 Governing Equation for Buckling of  Cylindrical Shells . . . . 5.1.1 Special Case I: Cylinder under Axial Load P , P , q  = 0 . 5.1.2 Special Case II: Cylinder under Lateral Pressure . . . 5.1.3 Special Case III: Hydrostatic Pressure . . . . . . . . . 5.1.4 Special Case IV: Torsion of a Cylinder . . . . . . . . . 5.2 Derivation of  the Linearized Buckling Equation . . . . . . . . 5.3 Buckling under Axial Compression . . . . . . . . . . . . . . . 5.3.1 Formulation for Buckling Stress and Buckling Mode . 5.3.2 Buckling Coefficient and Batdorf Parameter . . . . . . 5.4 Buckling under Lateral Pressure . . . . . . . . . . . . . . . . 5.5 Buckling under Hydrostatic Pressure . . . . . . . . . . . . . . 5.6 Buckling under Torsion . . . . . . . . . . . . . . . . . . . . . 5.7 Influence of  Imperfection and Comparison with Experiments .

ii

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29 31 34 34 34 35 36 40 42 47 47 48 48 51 51 52 54 62 65 71 74 76 77 77 80 84 84 86 87 88 89 89 90 90 93 96 99 100 102

3.2.2 3.2.3

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

4 General Theories of  Plate 4.1 Bending Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Derivation of  the Plate Bending Equation . . . . . . . . . . . 4.1.2 Reduction to a System of  Two Second Order Equations . . . 4.1.3 Exercise 1: Plate Solution . . . . . . . . . . . . . . . . . . . . 4.1.4 Exercise 2: Comparison between Plate and Beam Solution . . 4.1.5 Exercise 3: Finite Diff erence erence Solution of  the Plate Bending Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Membrane Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Plate Membrane Equation . . . . . . . . . . . . . . . . . . . . 4.2.2 Plate Equation for the Circular Membrane . . . . . . . . . . 4.2.3 Example: Approximation Solution for the Clamped Membrane 4.3 Buckling Theory of Plates . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 General Equation of Plate Buckling . . . . . . . . . . . . . . 4.3.2 Linearized Buckling Equation of  Rectangular Plates . . . . . 4.3.3 Analysis of  Rectangular Plates Buckling . . . . . . . . . . . . 4.3.4 Derivation of  Raleigh-Ritz Quotient . . . . . . . . . . . . . . 4.3.5 Ultimate Strength of Plates . . . . . . . . . . . . . . . . . . . 4.3.6 Plastic Buckling of Plates . . . . . . . . . . . . . . . . . . . . 4.3.7 Exercise 1: Eff ect ect of  In-Plane Boundary Conditions, δw = 0 . 4.3.8 Exercise 2: Raleigh-Ritz Quotient for Simply Supported Square Plate under Uniaxial Loading . . . . . . . . . . . . . . . . . . 4.4 Buckling of Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Transition from Global and Local Buckling . . . . . . . . . . 4.4.2 Local Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Buckling of  Cylindrical Shells 5.1 Governing Equation for Buckling of  Cylindrical Shells . . . . 5.1.1 Special Case I: Cylinder under Axial Load P , P , q  = 0 . 5.1.2 Special Case II: Cylinder under Lateral Pressure . . . 5.1.3 Special Case III: Hydrostatic Pressure . . . . . . . . . 5.1.4 Special Case IV: Torsion of a Cylinder . . . . . . . . . 5.2 Derivation of  the Linearized Buckling Equation . . . . . . . . 5.3 Buckling under Axial Compression . . . . . . . . . . . . . . . 5.3.1 Formulation for Buckling Stress and Buckling Mode . 5.3.2 Buckling Coefficient and Batdorf Parameter . . . . . . 5.4 Buckling under Lateral Pressure . . . . . . . . . . . . . . . . 5.5 Buckling under Hydrostatic Pressure . . . . . . . . . . . . . . 5.6 Buckling under Torsion . . . . . . . . . . . . . . . . . . . . . 5.7 Influence of  Imperfection and Comparison with Experiments .

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29 31 34 34 34 35 36 40 42 47 47 48 48 51 51 52 54 62 65 71 74 76 77 77 80 84 84 86 87 88 89 89 90 90 93 96 99 100 102

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 of  initial lenght and ds is the increment of current of  current length. 1.1.2

Green-Lagrangian Strain

Instead of  comparing the length, one can compare the square of  lengths: ds2 ds20 2ds20 ds ds0 ds + ds0 = ds0 2ds0

−

E = E =

(2)

−

Where ds ds0 , the second term is Eq. (2) tends to unity, and the Green strain measure and the engineering strain become identical. Equation (2) can be put into an equivalnet form: ds2 ds20 = 2Eds 20 (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 dxi = dai + dui

1

(4)

du da

dx

u

x a

O

Now, the squares of the initial and the current length increment can be written in terms of  ai and ui : ds20 = daida j δ ij (5) ds2 = dxidx j δ ij

(6)

= (dai + dui ) (da j + du j ) δ ij where the Kronecker tensor δ ij reads:

¯¯ ¯¯

100 δ ij = 0 1 0 001 The vector

u

¯¯ ¯¯

(7)

can be considered as a function of:

• the initial (material) coordinate system, descripion, or  • the current (spatial) coordinates, tion

u

u

(a), which leads to Lagrangian

(x), which leads to the Eulerian descrip-

In structural mechanics, the Lagrangian description is preferable: ui = ui (ai ) ∂u i dui = dak = ui ,k dak ∂a k ∂u j du j = dal = u j ,l dal ∂a l

(8)

Let us calculated the diff erence in the length square: ds2

− ds20 = (dai + dui) (da j + du j ) δ ij − dai da j δ ij 2

(9)

Using Eq. (8) and the definition of  δ ij , the diff erence in the length square can be transformed into: ds2

− ds20 = (du j dai + duida j + duidu j ) δ ij

(10)

= (u j ,l dal dai + ui ,k dak da j + ui ,k dak u j ,l dal ) δ ij = [u j ,l (da j δ  jl ) dai + ui,k (dai δ ik ) da j + ui,k u j ,l (dai δ ik ) (da j δ  jl )] δ ij = (u j ,i +ui , j +ui ,k u j ,k ) dai da j = 2E ij daida j

where, by analogy with the 1-D case, the Lagrangian or Green strain tensor E ij is defined: 1 E ij = (ui , j +u j ,i +uk ,i uk , j ) (11) 2 In the case of  small displacement gradient (uk ,i ¿ 1), the second nonlinear term can be neglected leading to the defintion of  the infinitesimal strain tensor: εij =

1 (ui , j +u j ,i ) 2

(12)

From the defintion, the strain tensor is symmetric εij = ε ji , which can be seen by intechanign the indices i for j and j for i. In the moderately large deflection theory of  structures, the nonlinear terms are important. Therefore, Eq. (11) will be used as a starting point in the development of  the general theory of  plates. Components of  Green-Lagrangian Strain Tensor range convention for indices:

Let define the following

• 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:

E11

E12

#

#

E13

#

Eij = E21 "

E31

"

Eαβ 

E22

#

E23

"

#

"

E32

#

E33

#

E3β 

#

=

"

"

" #

"

Eα 3

#

E33

where E αβ is the in-plane component of strain tensor, E α3 and E 3β are out-of-plane 3

shear components of  strain tensor, and E 33 is the through-thickness component of  strain tensor. Similarily, displacement vector can be divided into two components:

u1

u uα

ui

=

u2

v

=

=

"

"

"

u3

w

w

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

Initial Undeformed Configuration

uα

a

z u x

x

u3

midd surfa

η

Deformed Configuration

ξ 

Assumptions of  the von Karman Theory erately large deflection of  plates assumes:

The von Karman thoery of  mod-

1. The plate is thin. The thickness h is much smaller than the typical plate dimension, h ¿ L. 2. The magnitude of the transverse deflection is of the same order as the thickness of plate, |w| = O (h). In practice, the present theory is still a good engineering approximation for deflections up to ten plate thickness. 4

3. Gradients of  in-plane displacements uα ,β are small so that their product or square can be neglected. 4. Love-Kirchho ff  hypothesis is satisfied. In-plane displacements are a linear function of  the z coordiate (3-coordinate).

−

uα = uα ◦

− z u3,α

(14)

where uα is the displacement of  the middle surface, which is independent of  z coordinate, i.e. uα ,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 can be 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: E αβ =

1 (uα ,β +uβ ,α +uk ,α uk ,β ) 2

(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: E αβ =

1 (uα ,β +uβ ,α +w,α w,β ) 2 5

(17)

where w = u3 . Introducing Eq. (14) into Eq. (17), i.e. applying Love-Kirchhoff  hypothesis, one gets: 1 (uα z w,α ) ,β + uβ z w,β ,α +w,α w,β 2 1 = u ,β +uβ ,α 2 z w,αβ +w,α w, β 2 α 1 1 = uα,β +uβ ,α z w,αβ + w,α w,β 2 2 From the definiton of  the curvature, one gets: E αβ =

£ ¡ ¡

◦

¡

−

◦

◦

◦

◦

− −

◦

¢

−

¢

¢

καβ =

−w,αβ

¤

(18)

(19)

Now, Eq. (18) can be re-casted in the form: E αβ = E αβ + z καβ

(20)

◦

where the strain tensor of  the middle surface E αβ is composed of  a linear and a nonlinear term: 1 1 E αβ = uα,β +uβ ,α + w,α w,β (21) 2 2 In the limiting case of  small displacements, the second term can be neglected as compared to the first term. In the classical bending theory of  plate, the in-plane displacements are assumed to be zero uα = 0 so that strains are only due to the curvatue: E αβ = z καβ (22) ◦

¡

◦

◦

where καβ

¯¯

¯¯

κ κ = 11 12 = κ21 κ22

¯¯ −¯ ¯

◦

¢

∂ 2 w ∂x 2 ∂ 2 w ∂x∂y

∂ 2 w ∂x∂y ∂ 2 w ∂y 2

¯¯ ¯¯

=

−w,αβ

(23)

In the above equation, κ11 and κ22 are curvatures of  the cylindrical bending, and κ12 is the twist which tells how the slope in the x direction changes with the y direction: ∂  ∂w κ12 = ∂y ∂x

−

−

µ ¶

 for  a cylinder 

κ12

6

=0

Interpretation of  the linear terms: be expressed in the followings:

1 2

³

uα ,β +uβ ,α ◦

◦

1 (u1,1 +u1,1 ) = u1,1 = 2 1 ε22 = (u2,2 +u2,2 ) = u2,2 = 2 ε11 =

1 1 ε12 = (u1 ,2 +u2 ,1 ) = 2 2 1 du1 ε12|if  u2=0 = 2 dy

µ

´

Each component can

du1 dx du2 dy

(24) (25)

du1 du2 + dy dx

¶

(26)

u2 ≡ 0

ε11 ε12

x

x

x

y

y

y

ε22

u1

Therefore, ε11 and ε22 are the tensile strain in the two directions, and ε12 is the change of  angles, i.e. shear strain. Interpretation of  the nonlinear term: Then, the nonlienar term reads: 1 w, α w,β 2

¯¯ ¯

α=1,β=1

1 2 w,α

w, β

1 dw dw 1 = = 2 dx dx 2

Let α = 1 and β  = 1.

dw dx

2

µ ¶

(27)

One can also obtain the same quantity by the defintion of  1-D Green-Lagrangian strain: ds20 + dw2 ds2 ds20 E  = ' 2ds20 2ds20

−

¡

¢−

7

ds20

1 = 2

dw ds0

2

1 = 2

dw dx

2

µ ¶ µ ¶

(28)

dx

x

ds0 = dx ds 2 = ds02 + dw2

ds0 dw

ds

z, w

Thus, the conclusion is that the nonlinear term 12 w,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 introduced in Section 1.2.1, there are three other componets of  the strain tensor: E 3β , E α3 and E 33. 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 the following expressions: 1 (u3 ,β +uβ ,3 +uk ,3 uk ,β ) 2 1 = [u3 ,β +uβ ,3 + (u1 ,3 u1,β +u2 ,3 u2 ,β +u3 ,3 u3 ,β )] 2 1 = [u3 ,β u3 ,β + ( u3 ,1 u1 ,β u3 ,2 u2 ,β )] 2 1 = ( u3 ,1 u1 ,β u3 ,2 u2 ,β ) 2 1 = w,γ  uγ ,β 2

E 3β =

−

−

−

(29)

−

−

−

1 (uα ,3 +u3 ,α +uk ,α uk ,3 ) 2 1 = [uα ,3 +u3 ,α + (u1 ,α u1 ,3 +u2 ,α u2 ,3 +u3 ,α u3 ,3 )] 2 1 = [ u3 ,α +u3,α + ( u1 ,α u3,1 u2,α u3 ,2 )] 2 1 = ( u1 ,α u3 ,1 u2 ,α u3 ,2 ) 2 1 = w,γ  uγ ,α 2

E α3 =

− −

−

−

−

−

8

(30)

1 (u3,3 +u3,3 +uk ,3 uk ,3 ) 2 1 = u3 ,3 + (u1 ,3 )2 + (u2,3 )2 + (u3 ,3 )2 2 1 = (u1 ,3 )2 + (u2,3 )2 2 1 = ( u3 ,1 )2 + ( u3 ,2 )2 2 1 = w,γ  w,γ  2

E 33 =

h h−

h

i

−

(31)

i

i

The above are all second order terms which vanish for small deflection theory of  plates. In the theory of moderately larege deflection of plates, the out-of-plate shear strains 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 remains  normal."  The existance of the out-of-plane shear strain means that lines originally normal 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: E 3β σ 3β , E α3 σ α3 and E 33σ 33, are small

(32)

because the corresponding stress σ 3β , σ α3 and σ 33 are small as compared to the in-plane stress σ αβ . One can conclude that the elastic strain energy (and even plastic dissipation) is well approximated using the plane strain assumption:

Z

h

1 σ ij εij dz ' 2

9

Z  h

1 σ αβ εαβ dz 2

(33)

2

Derivation of  Constitutive Equations for Plates

2.1

Definitions of  Bending Moment and Axial Force

Hook’s law in plane stress reads: E  [(1 1 ν 2

σ αβ =

−

− ν )

εαβ + ν εγγ  δ αβ ]

(34)

In terms of  components: E  (εxx + ν εyy ) 1 ν 2 E  σ yy = (εyy + ν εxx) 1 ν 2 E  σxy = εxy 1 + ν 

σxx =

(35)

− −

Here, strain tensor can be obtained from the strain-displacement relations: εαβ = εαβ + z καβ

(36)

◦

Now, define the tensor of  bending moment: h

Z  ≡

2

M αβ

−

h

σ αβ z dz

(37)

2

and the tensor of  axial force (membrane force): h

N αβ

Z  ≡

2

−

2.2 2.2.1

h

σ αβ dz

(38)

2

Bending Energy Bending Moment

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

E  M αβ = 1 ν 2

−

E  = 1 ν 2

−

h

Z  − £− 2

−

h

[(1

2

(1

E  + [(1 1 ν 2

−

=

Eh 3 12 (1

ν ) εαβ + ν εγγ  δ αβ ] z dz

− ν 2)

h

ν ) εαβ + ν εγγ  δ αβ

− ν ) [(1

◦

◦

¤ Z  Z  2

−

h

z dz

2

h 2

καβ + ν κγγ  δ αβ ]

−

− ν ) 10

καβ + ν κγγ  δ αβ ]

h 2

z 2 dz

(39)

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

Eh 3 12 (1 ν 2 )

(40)

−

Now, one gets the moment-curvature relations: M αβ = D [(1

− ν )

καβ + ν κγγ  δ αβ ]

¯¯

M 11 M 12 = M 21 M 22

M αβ where M 12 = M 21 due to symmetry.

¯¯

(42)

M 11 = D (κ11 + ν κ22) M 22 = D (κ22 + ν κ11) M 12 = D (1 2.2.2

− ν )

(41)

(43)

κ12

Bending Energy Density

One -Dimensional Case Here, we use the hat notation for a function of certain argument such as: ˆ 11 (κ11) M 11 = M 

(44)

= D κ11 ¯b reads : Then, the bending energy density U  ¯b = U 

κ ¯11

Z  0

=D

(45)

κ ¯11

Z  0

=

ˆ 11 (κ11) dκ11 M  κ11 dκ11

1 D (¯ κ11 )2 2 ¯b = 1 M 11 κ U  ¯ 11 2

(46)

 M 11

D κ11

dκ11

11

General Case General definition of  the bending energy density reads: ¯b = U 

I 

M αβ dκαβ

(47)

κ

κ11

κ22

0

Calculate the energy density stored when the curvature reaches a given value κ ¯ αβ . Consider a straight loading path: καβ = η κ ¯ αβ

(48)

dκαβ = κ ¯ αβ dη

 M αβ 

η=1

 M αβ  η η=0

καβ 

ˆ αβ (καβ ) M αβ = M  ˆ αβ (η κ = M  ¯ αβ ) ˆ αβ (¯ = η M  καβ )

12

καβ 

(49)

ˆ αβ (καβ ) is a homogeneous function of  degree one. where M  ¯b = U 

I  Z 

ˆ αβ (καβ ) dκαβ M 

1

=

0

(50)

ˆ αβ (¯ η M  καβ ) κ ¯ αβ dη

ˆ αβ (¯ = M  καβ ) κ ¯ αβ

1

Z 

η dη

0

1 ˆ M αβ (¯ καβ ) κ ¯αβ 2 1 = M αβ κ ¯ αβ 2 =

Now, the bending energy density reads: ¯b = D [(1 U  2 D = [(1 2 D = (1 2

h

− ν ) κ¯αβ + ν κ¯γγ  δ αβ ] κ¯αβ − ν ) κ¯αβ κ¯αβ + ν κ¯γγ  κ¯αβ δ αβ ] − ν ) κ¯αβ κ¯αβ + ν  (¯κγγ )2

(51)

i

The bending energy density expressed in terms of  components: ¯b = D U  2 D = 2 D = 2 D = 2 D = 2

n− h n− h nh n n

ν ) (¯κ11)2 + 2 (κ ¯ 12 )2 + (κ ¯ 22)2 + ν  (¯ κ11 + ¯κ22 )2

(1

ν ) (¯κ11 + ¯κ22)2

i i − h− h − − io

− 2 κ¯11 κ¯22 + 2

(¯κ11 + ¯κ22)2

o

(52)

(κ ¯ 12)2 + ν  (¯ κ11 + ¯κ22)2

o

io io

− 2 κ¯11 κ¯22 + 2 (κ¯12)2 ν  2 κ¯11 κ¯22 + 2 (κ¯12)2 (¯κ11 + ¯κ22)2 − 2 κ ¯ 11 κ ¯22 + 2 (κ ¯ 12)2 ν  2κ ¯11 κ ¯ 22 + 2 (κ ¯ 12)2 (¯κ11 + ¯κ22)2 + 2 (1 − ν ) −κ ¯ 11 κ ¯ 22 + (κ ¯12 )2 ¯b = D U  2

2.2.3

i

(1

h

n

2

(¯ κ11 + ¯κ22)

− 2 (1 − ν )

h

κ ¯ 11 κ ¯ 22

2

− (¯κ12)

io

(53)

Total Bending Energy

The total bending energy is the integral of the bending energy density over the area of  plate: ¯b dA U b = U  (54)

Z 

S 

13

2.3 2.3.1

Membrane Energy Axial Force

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

E  N αβ = 1 ν 2 =

Z  − Z  £ − Z  − £ − − 2

−

2

E  1 ν 2

2

−

(1

h 2

2

−

◦

2

−

Eh (1 1 ν 2

[(1

h

E  + [(1 1 ν 2

£

− ν ) εαβ + ν εγγ  δ αβ ◦

h

E  = (1 1 ν 2

−

(55)

h

E  + 1 ν 2

=

− ν ) εαβ + ν εγγ  δ αβ ] dz

[(1

h

− ν )

¤

dz

καβ + ν κγγ  δ αβ ] z dz h

ν ) εαβ + ν εγγ  δ αβ ◦

◦

¤ Z  Z  ¤ 2

−

h

dz

2

h

− ν )

2

καβ + ν κγγ  δ αβ ]

−

− ν ) εαβ + ν εγγ  δ αβ ◦

◦

h

z dz

2

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

Eh 1 ν 2

(56)

−

Now, one gets the membrane force-extension relation:

h

N αβ = C  (1

N αβ

− ν )

εαβ + ν εγγ  δ αβ ◦

¯¯ ¯

◦

N  N  = 11 12 N 21 N 22

where N 12 = N 21 due to symmetry.

¯¯ ¯

N 11 = C  (ε11 + ν ε22) ◦

◦

i

(57)

(58)

(59)

N 22 = C  (ε22 + ν ε11) ◦

N 12 = C  (1 2.3.2

◦

− ν ) ε11 ◦

Membrane Energy Density

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

I 

Nαβ dεαβ

14

◦

(60)

Calculate the energy stored when the extension reaches a given value ¯εαβ . Consider a straight loading path: ◦

εαβ = η ¯εαβ ◦

(61)

◦

dεαβ = ¯εαβ dη ◦

◦

ˆαβ εαβ N αβ = N  ˆαβ η ¯ε = N  αβ ˆαβ ¯ε = η N 

¡ ¢ ¡ ¢ ¡ ¢

(62)

◦

◦

◦

αβ

³ ´

ˆαβ ε where N  αβ is a homogeneous function of  degree one. ◦

¯m = U 

¯εαβ

Z  Z 

◦

¡ ¢ ¡ ¢ ¡ ¢

0

dεαβ

◦

0

1

=

ˆαβ ε N  αβ

(63)

◦

ˆαβ ¯ε η N  εαβ dη αβ ¯ ◦

◦

1 ˆ N αβ ¯εαβ ¯εαβ 2 1 = N αβ ¯εαβ 2 =

◦

◦

◦

Now, the extension energy reads: ¯m = C  (1 U  2 C  = (1 2

£ h

¤

− ν ) ¯εαβ + ν  ¯εγγ  δ αβ ¯εαβ − ν ) ¯εαβ ¯εαβ + ν  ¯εγγ  2 ◦

◦

◦

(64)

◦

¡ ¢i

◦

◦

The extension energy expressed in terms of  components: ¯m = C  U  2 C  = 2 C  = 2 C  = 2

n n n n

h h

i

(¯ ε11)2 + 2 (¯ ε12 )2 + (¯ε22)2 + ν  (¯ ε11 + ¯ε22)2

− ν ) (1 − ν ) (1

◦

◦

◦

(¯ ε11 + ¯ε22)2

◦

i

◦

o

(65)

o io

− 2 ¯ε11¯ε22 + 2 (ε¯12)2 + ν (¯ε11 + ¯ε22)2 (¯ ε11 + ¯ε22)2 − 2 ¯ε11¯ε22 + 2 (ε¯12)2 − ν −2 ¯ε11¯ε22 + 2 (ε¯12)2 (¯ ε11 + ¯ε22)2 + 2 (1 − ν ) −¯ε11¯ε22 + (ε¯12 )2 ◦

◦

◦

◦

¯m = C  U  2

◦

◦

◦

◦

◦

2

(¯ ε11 + ¯ε22 ) ◦

◦

◦

h

◦

h

n

◦

◦

◦

− 2 (1 − ν ) 15

◦

¯ε11¯ε22 ◦

◦

◦

2

− (¯ε12) ◦

◦

◦

io

◦

h

◦

io

(66)

2.3.3

Total Membrane Energy

The total membrane is the integral of  the membrane energy density over the area of  plate:: ¯m dS  U m = U  (67)

Z 

S 

16

3

Development of  Equation of  Equilibrium and Boundary Conditions Using Variational Approach

3.1 3.1.1

Bending Theory of  Plates Total Potential Energy

The total potential energy of  the system Π

Π

= U b

reads:

− V b

(68)

where U b is the bending energy stored in the plate, and V b is the work of  external forces. Bending Energy 1 U b = 2 =

Z  Z  − S 

1 2

where the geometrical relation καβ = Work of  External Forces

M αβ καβ dS 

S 

(69)

M αβ w,αβ dS 

−w,

αβ

has been used.

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  the middle surface of  the plate.

q ( x)

The distributed load contains concentrated load P  as a special case: P  (x0 , y0 ) = P 0 δ (x

− x0) δ (y − y0)

(71)

where δ  is the Dirac delta function, [x0, y0 ] is the coordinate of  the application of  the concentrated force, and P 0 is the load intensity.

17

Dirac δ -function

x

x0

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

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 plate NOTE Other  type  of  loading  such  as  shear  or  in-plane  tension  or  compression  do not  de  fl ect  laterally  the  plate  and  therefore  are  not  considered  in  the  bending  theory. 18

in-plane tension or compression

in-plane shear

Potential Energy due to Lateral Load q  Lateral (transverse) load does work on transverse deflection: q w dS  (72)

Z 

S 

This is also called a work of  external forces. Potential Energy due to Edge Moment The conjugate kinematic variable associated with the edge moment is the edge rotation dw/dxn.

dl

t n

edge moment Γ

We apply only the normal bending moment M nn : ¯ nn dw dl M  dxn Γ

Z  −

(73)

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

 M nn < 0

 M nn > 0

0 dw dxn

a dw dxn

>0

w At the edge, M tt = 0 and M tn = 0.

19

<0

xn

Potential Energy due to Edge Forces

Z 

¯n w dl V 

(74)

Γ

Q2

Q1

x2

x1 V n

Potential Energy due to All External Forces Now, the work of  external forces reads: ¯ nn dw dl + V  ¯n w dl V b = q w dS  M  (75) dxn S  Γ Γ 3.1.2 First Variation of  the Total Potential Energy

Z

Z  −

Z 

The total potential energy reads: Π=

1 2

Z  − Z−

S 

M αβ w,αβ dS 

(76)

¯ nn dw dl q w dS + M  dxn S  Γ

Z 

Z  −

¯n w dl V 

Γ

First variation of  the total potential energy δ Π is expressed: δ Π =

Z  − Z  −

S 

M αβ δw, αβ dS  q δw dS +

S 

(77) dw dxn

Z  µ ¶ − Z  ¯ nn δ  M 

Γ

dl

¯n δw dl V 

Γ

We shall transform now the first integral with the help of  the Gauss theorem. First note that from the rule of  the product diff erentiation: M αβ δw, αβ = (M αβ δw, α ) ,β then

Z

S

M αβ δw, αβ dS  =

Z 

S 

−M αβ ,β

(M αβ δw, α ) ,β dS 

Z  −

S 

δw, α M αβ ,β δw, α dS 

(78) (79)

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

Z

S 

M αβ δw, αβ dS  =

Z 

M αβ δw, α nβ dl

Γ

20

Z  −

S 

M αβ ,β δw, α dS 

(80)

The integrand of  the second integral on the right hand side of  the above equation transform to: M αβ ,β δw, α = (M αβ ,β δw) ,α M αβ ,αβ δw (81)

−

which results in:

Z

S 

M αβ δw, αβ dS =

Z  Z− Z  Z−

M αβ δw, α nβ dl

(82)

Γ

S

(M αβ ,β δw) ,α dS +

upon which the application of  the Gauss rule gives:

Z

S 

M αβ δw, αβ dS =

Z 

S 

M αβ ,αβ δw dS 

M αβ δw, α nβ dl

(83)

Γ

M αβ ,β δw nα dl +

Γ

Z 

S 

M αβ ,αβ δw dS 

We can return now to the expression for δ Π and substitute there the transformed first integral: δ Π =

Z  − Z Z  − S 

( M αβ ,αβ

−q )

δw dS 

M αβ ,β δw nα dl

+

Γ

(84)

Z  − Z 

M αβ δw, α nβ dl +

Γ

³ ´

¯n δw dl V 

Γ

¯ nn δw, n dl M 

Γ

dw ¯ nn where δw, n = δ  dx . It is seen that integrals involving the prescribed forces M  n ¯n are written in a local coordinate system xγ  {xn , xt } while the remaining and V  two integrals over the contour Γ are written in the global coordinate system xα . In order to make comparison, we have to decide on one coordinate system. We choose the local system. Consider the  fi rst  integral:

Z 

(M αβ ,β nα ) δw dl

(85)

Γ

The term in the parenthesis is a scalar quantity and thus remain unchanged with respect to the rotation of  coordinate system. In the local system xγ , the line integral becomes:

Z 

(M γδ ,δ nγ ) δw dl

(86)

Γ

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

Z

(M γδ ,δ nγ ) δw dl =

Γ

Z  Z 

(M 1δ ,δ n1 + M 2δ ,δ n2 ) δw dl

Γ

=

M 1δ ,δ δw dl

Γ

21

(87)

Furthermore, the integrand reads: M 1δ ,δ = M 11,1 +M 12,2 ∂M 11 ∂M 12 ∂M nn ∂M nt = + = + ∂x 1 ∂x 2 ∂x n ∂x t

(88)

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

≡ M nδ ,δ

δ  = {1, 0} or {n, t}

(89)

Now, we can combine two line integrals in the equation of  first variation of the total potential energy: ¯n δw dl Qn V  (90)

Z ¡ Γ

−

¢ Z 

How the remaining  integral is transformed?

Z

(M αβ nβ ) δw, α dl =

Γ

(M γδ nδ ) δw, γ  dl

(91)

Γ

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

Z

(M γ 1 n1 + M γ 2 n2 ) δw, γ  dl =

Γ

Z  Z  Z 

M γ 1 δw, γ  dl

(92)

Γ

=

M γn δw, γ  dl

Γ

=

(M nn δw, n +M tn δw, t ) dl

Γ

The first term can be absorbed with the line integral representing potential energy of  bending moment: ¯ nn δw, n dl M nn M  (93)

Z  − ¡ Γ

−

¢

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

Z 

M tn δw, t dl

transverse term

(94)

Γ

In order to compare this term with the shearing force term, we have to make this term 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) /∂x t is the derivative of the function δw in the tangential direction, i.e. direction along the curve Γ. This means that we can integrate by parts along Γ. Thus, M tn δw, t = (M tn δw) ,t M tn ,t δw

−

22

(95)

Z

M tn δw, t dl =

Γ

Z 

(M tn δw) ,t dl

Γ

Z  −

M tn,t δw dl

(96)

Γ

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

Z  Γ

(M tn δw) ,t dl = M tn δw|end beginning

(97)

Consider now two cases. • The contour Γ is a smooth closed curve, so the value at the beginning is equal to the value at the end: M tn δw|end

− M tn δw|beginning = 0

(98)

The term does not give any contribution.

direction of integration

end begining

Γ

• The contour Γ is piece-wise linear or composed of a finite number, k, of smooth curves with discontinuity. Therefore, the integration should be made in a piece-wise manner. Thus, the continuation of  the beginning and end of  each should be added: M tn δw|end (99) beginning

X k

23

3.1.3

Equilibrium Equation and Boundary Conditions

Now, we can write the final expression for the first variation of  δ Π: δ Π =

Z  − − Z¡ − ¢ ÃX − Z  − − Z¡ − ¢ X − S 

( M αβ ,αβ

(100)

Z  − ¡ Z  −

¯n δw dl V 

Qn

+

q ) δw dS  M nn

Γ

Γ

M tn δw |end beginning

k

=

S 

( M αβ ,αβ

Γ

¢

M tn,t δw dl

Γ

δw, n dl

!

q ) δw dS 

¯n δw dl V 

V n

+

− M ¯ nn

Z  − ¡

M nn

Γ

− M ¯ nn

M tn δw|end beginning

¢

δw, n dl

k

where V n = Qn + M tn,t is the eff ective shear force. In order to make the functional Π stationary under arbitrary variation of  the displacement field δw, there must hold: Equation of Equilibrium

M αβ ,αβ +q  = 0

on S 

(101)

Boundary Conditions

¯ nn = 0 M nn M  ¯n = 0 V n V 

or or

δw, n = 0 δw = 0

M nt = 0

or

δw = 0

−

3.1.4

−

on Γ on Γ at corner points of  the contour Γ

Specification of  Equation for Rectangular Plate

Consider a rectangular plate.

24

(102)

a

x→t

b

y→n

Boundary Conditions ydirection.

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

− M ¯ yy = 0 ¯y = 0 V y − V 

M yy

or or

∂w =0 ∂y w=0

(103)

where ∂M yx ∂y ∂M xy V y = Qy + ∂x

V x = Qx +

(104)

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

− M ¯ xx = 0 ¯x = 0 V x − V 

M xx

or or

∂w =0 ∂x w=0

(105)

where ∂M xy ∂x ∂M yx V x = Qx + ∂y V y = Qy +

Interpretation of  Corner Points

25

(106)

1 segment [ k − 1 ]

2

segment [ k ]

direction of integration

3

4 Boundary condition reads:

X

[2]

[2] M tn δw|end beginning = M tn δw

− M tn[1] δw [1] + M tn[4] δw [4] − M tn[3] δw [3]

(107)

where

δw [3] = δw [2]

(108)

thus

X

M tn δw|end beginning

=

[1] M tn

−

δw

[1]

+

Consider the right angle.

³

[2] M tn

−

[3] M tn

x

´

[4]

δw [2] + M tn δw [4]

(109)

[ k − 1]

y

[k]

for the k 1 side for the k side

−

³

M tn δw|end beginning

´

n = x, t = y n = y, t = x

at the right angle

= (M xy

− M yx )

(110)

δw

(111)

Interpretation of  Corner Forces Plane stress: τ xy = τ yx

symmetry

26

(112)

σy

surface element

τ xy

dS

σx

τ yx Let us place the surface element at the corner.

x y

z τ xy

Edge [ k − 1]

τ yx

Edge [ k]

 M xy  M yx

The shearing stresses produce twisting moments which are in the opposite direction: [k M xy

1]

=

−

−M yx[k]

(113)

Therefore, the boundary condition at the corner becomes: M tn δw|end beginning

=

³

[k−1] M xy

−

[k] M yx

´

F corner = 2 M xy

27

δw = 2 M xy δw = 0

(114)

(115)

For the Entire Plate

2 M xy 2 M xy 2 M xy 2 M xy

Interpretation of  the Eff ective Shear V x

x dy

y

Qx

z

dy  M xy  M xy +

∂ M xy ∂y

dy

Equilibrium reads: ∂M xy Qx dy + M xy + dy ∂y ∂M xy = Qx + dy ∂y = V x dy

µ

µ

¶

V x = Qx +

28

∂M xy ∂y

¶−

M xy

(116)

(117)

3.2 3.2.1

Bending-Membrane Theory of  Plates Total Potential Energy

The total potential energy of  the system Π

Π

= U b + U m

reads:

− V b − V m

(118)

where U b is the bending strain energy, U m is the membrane strain energy, V b is the potential energy of  external loading causing flexural response, and V m is the potential energy of  external loading causing membrane response. Membrane Strain Energy The membrane strain energy reads: 1 U m = 2

Z 

S 

where

N αβ εαβ dS  ◦

1 1 (uα ,β +uβ ,α ) + w,α w,β 2 2 Potential Energy of  External Forces Evaluation of  Boundary Terms εαβ = ◦

(119)

(120)

Normal in-plane loading, N nn

Z 

¯nn un dl N 

(121)

Γ

where un is normal in-plane displacement. Shear in-plane loading, N tt

Z 

¯tn ut dl N 

(122)

Γ

where ut is shear component of  the displacement vector. Potential Energy of  External Forces V m =

Z 

¯nn un dl + N 

Γ

3.2.2

Z 

¯tn ut dl N 

(123)

Γ

First Variation of  the Total Potential Energy

The first variation of  the total potential energy reads: δ Π = (δU b

− δV b) + (δU m − δV m)

(124)

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

Z 

S 

N αβ δλ αβ dS  29

(125)

where

1 1 (δu α,β +δuβ ,α ) + (δw, α w,β +δw, β w, α ) 2 2 Because of  the symmetry of  the tensor of  membrane forces: δλαβ =

(126)

N αβ = N βα

(127)

by using the characteristics of  dummy indices we obtain: N αβ δu β ,α = N βα δuβ ,α = N αβ δuα ,β

(128)

Now, the first variation of  the membrane strain energy reads: δU m = =

Z  ¿ ∙ Z  S 

S 

N αβ

1 1 (δu α,β +δu β ,α ) + (δw, α w,β +δw, β w,α ) 2 2

¸À

dS  (129)

(N αβ δu α ,β +N αβ 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. The  fi rst  term of  δU m reads:

Z

S

N αβ δu α ,β dS = =

Z  Z Z  Z  Z  Z 

S 

(N αβ δu α ) ,β dS  N αβ δuα nβ dl

Γ

=

N γ 1 δu γ  dl

Γ

=

Γ

=

N γn δu γ  dl

S 

S 

N γδ δu γ  nδ dl

Γ

=

Z  − Z  − Z  − Z  − Z  − S 

S 

S 

N αβ ,β δu α dS 

N αβ ,β δu α dS 

N αβ ,β δu α dS 

N αβ ,β δuα dS  N αβ ,β δu α dS 

(N nn δu n + N tn δu t ) dl

Γ

30

Z  −

S 

N αβ ,β δu α dS 

(132)

The second  term of  δU m reads:

Z

S

N αβ w,β δw, α dS = =

Z  Z  Z  Z  Z  Z 

S 

(N αβ w,β δw) ,α dS  N αβ w,β δw nα dl

Γ

=

N 1δ w,δ δw dl

Γ

=

S 

S 

N γδ w,δ δw nγ  dl

Γ

=

Z  − Z  − Z  − Z  − S 

S 

(N αβ w,β ) ,α δw dS 

(N αβ w, β ) ,α δw dS 

(N αβ w,β ) ,α δw dS 

(N αβ w,β ) ,α δw dS 

(N 11 w,1 +N 12 w,2 ) δw dl

Γ

=

(133)

(N nn w,n +N nt w,t ) δw dl

Γ

Z  − Z  − S 

(N αβ w,β ) ,α δw dS 

S 

(N αβ w,β ) ,α δw dS 

Now, the variation of  external work reads: δV m =

Z 

¯nn δu n dl + N 

Γ

3.2.3

¯tn δu t dl N 

(134)

Γ

Equilibrium Equation and Boundary Conditions

The contribution of  the term (δU m δ (U m

Z 

− V m) =

Z  Z  Z  − Z  − Z  −

− δV m) then becomes:

(N nn δu n + N tn δut ) dl

Γ

+

Z  −

S 

(N nn w, n +N nt w, t ) δw dl

Γ

Γ

=

S 

S 

¯nn δu n dl N 

Z  −

¯tn δu t dl N 

N αβ ,β δu α dS 

Z  −

S 

(135)

(N αβ w,β ) ,α δw dS 

Γ

N αβ ,β δu α dS +

Z ¡

N nn

Γ

(N αβ w,β ) ,α δw dS +

− N ¯nn

Z 

¢

δu n dl +

Z  ¡ Γ

N tn

− N ¯tn

(N nn w,n +N nt w,t ) δw dl

¢

δut dl

Γ

The first three integrals involve independent variations of uα , i.e. δuα or {δu n, δu t }. This gives us two independent equations of  equilibrium in the plane of  the plate: Equation of Equilibrium I

N αβ ,β = 0

31

on S 

(136)

and two additional boundary conditions: Boundary Conditions I

N nn N tn

− N ¯nn = 0 − N ¯tn = 0

δu n = 0 δu t = 0

or or

on on

(137)

Γ Γ

The remaining two integrals involve variation in the out-of-plane displacement δw and thus should be combined with the equation of  equilibrium and boundary conditions governing the flexural response. The terms involving surface integral should be added to the equation of  equilibrium: Equation of Equilibrium II

M αβ ,αβ + (N αβ w,β ) ,α +q  = 0

(138)

on S 

where the second term in the left hand is the new term arising from the finite rotation. The term with the line integral should be added to the corresponding term involving variation δw:

Z ¡

¯n δw dl = 0 V n + N nn w,n +N nt w, t V 

−

Γ

The generalized boundary conditions reads:

¢

(139)

Boundary Conditions II- (A)

¯n = 0 V n + N nn w,n +N nt w,t V 

−

or

δw = 0

on

Γ

(140)

where the second and third terms in the left hand side of  the first equation are the new terms arising from the finite rotation. If  the boundaries of  the plate are kept undeformed w,t = 0 (simply supported or clamped plate), then the boundary condition is satisfied: ¯n = 0 V n + N nn w,n V 

−

or

δw = 0 on

Γ

(141)

Physically, the additional terms represent the contribution of  the axial force to the vertical equilibrium. Using the in-plane equilibrium, N αβ ,β = 0, the out-of-plane equilibrium can be transformed to the form: M αβ ,αβ +N αβ ,α w,β +N αβ w, αβ +q  = 0

32

on S 

(142)

Equation of Equilibrium II’

M αβ ,αβ +N αβ w,αβ +q  = 0

on S 

(143)

which is called as the von Karman equation. Note that N αβ is related through the Hook’s law with the gradient of the in-plane displacement uα , i.e. N αβ = N αβ (uα ). Therefore, the new term N αβ w,αβ represents in fact coupling between in-plane and out-out-plane deformation. To make derivation complete, the final boundary conditions which do not changed from the bending theory of  plate are presented: Boundary Conditions II- (B)

M nn

− M ¯ nn = 0

M nt = 0

or

δw, n = 0

or

δw = 0

33

on Γ at corner points of  the contour Γ

(144)

4

General Theories of  Plate

4.1 4.1.1

Bending Theory of  Plates Derivation of  the Plate Bending Equation

Then, groups of  equations! • Equilibrium M αβ ,αβ +q  = 0

on S 

(145)

• Geometry καβ =

−w,αβ

(146)

• Elasticity M αβ = D [(1

− ν ) καβ + ν κγγ  δ αβ ]

(147)

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

−D

[(1

− ν )

w,αβ +ν w,γγ  δ αβ ]

(148)

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

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

(149)

Note that the components of  the Kronecker "δ αβ " tensor are constant and thus are not subjected to diff erentiation: δ αβ

¯¯

10 = 01

¯¯

δ αβ =

or

½

1 0

if  α = β  if  α 6 = β 

(150)

Also, note that only these components: ¤αβ

δ αβ = ¤αα

(151)

survive in the matrix multiplication for which α = β . Therefore, Eq. (149) now reads: D [(1 ν ) w,αβαβ +ν w,γγαα ] + q  = 0 (152)

−

−

Because "γγ " are "dummy" indices, they can be replaced by any other indices, for example "ββ ." D [(1 ν ) w,αβαβ +ν w,ββαα ] + q  = 0 (153)

−

−

The order of  diff erentiation 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

∂ 4 w ∂ 4 w ∂ 4 w +2 2 2 + ∂x 4 ∂x ∂x ∂y 4

¶

= q (x, y)

(156)

Alternative notation can be: D where Laplacian

∇2w reads:

∇4w = q  2

(157)

2

∇2w = ∂ ∂xw2 + ∂ ∂yw2 and bi-Laplican

∇4w reads: ∇4w = ∇2 ∇2w ∂ 2 = 2 ∇2 w ∂x

¡ ¢ ¡ ¢ ¡∇ ¢ µ ¶ µ

∂ 2 2 w ∂y 2 ∂ 2 ∂ 2 w ∂ 2 w ∂ 2 = 2 + + ∂x ∂x 2 ∂y 2 ∂y 2 ∂ 4 w ∂ 4w ∂ 4 w = +2 2 2 + ∂x 4 ∂x ∂x ∂y 4

4.1.2

(158)

(159)

+

∂ 2 w ∂ 2 w + ∂x 2 ∂y 2

¶

Reduction to a System of  Two Second Order Equations

Denote D w,αα = Then, from the equilibrium equation:

−M 

[D w,αα ] ,ββ = q 

(160)

(161)

we obtain a system of  two linear partial diff erential equations of  the second order: M, ββ = q  D w,αα = M 

−

35

−

(162)

or

2 ∂ 2 M  + ∂ ∂yM  2 ∂x2 ∂ 2 w ∂ 2 w + ∂y 2 ∂x 2

( ³ D

What is "M " ? Let us calculate M αα:

=

q 

´ −− =

M 

M αα = M 11 + M 22 = D [(1 ν ) κ11 + ν  (κ11 + κ22) δ 11] +D

(163)

(164)

− [(1 − ν ) κ22 + ν  (κ11 + κ22) δ 22 ]

= D [(1 + ν ) (κ11 + κ22)] = D (1 + ν ) καα or

M αα = D καα = 1 + ν 

−D w,αα = M 

(165)

Therefore, M αα = M  (1 + ν )

(166)

= D καα (1 + ν ) Now, moment sum reads: M  = D καα

(167)

M  = D [κxx + κyy ]

(168)

and in expanded notation it reads:

4.1.3

Exercise 1: Plate Solution

Consider a simply supported plate.

Square plate ( a × a) a

a y

36

x

Boundary Condition General boundary condition reads: ¯ nn w, n = 0 M nn M  ¯n w = 0 V n V 

¡¡

¢¢

− −

¯ nn = 0 M 

on on

⇒ M nn = 0

x = 0 and x = a , 0 y = 0 and y = a , 0

at at

M xx = 0 M yy = 0

x = 0 and x = a , y = 0 and y = a ,

at at

(169)

Γ

on Γ on Γ

w=0

w =0 w =0

Γ

(170)

≤y≤a ≤x≤a 0≤y≤a 0≤x≤a

(171) (172)

Loading Condition Assume for simplicity the sinusoidal load distribution: q (x, y) = q 0 sin where q 0 is a pressure intensity.

πx a

πy a

³ ´ ³ ´ sin

(173)

Solution of  Problem The solution of  the form πx πy w (x, y) = w0 sin sin a a

³ ´ ³ ´

(174)

satisfy both the boundary conditions and the governing equations (see below). Plate Bending Equation Substituting Eq. (173) and (174) into the plate bending equation (156), one gets: π a

π a

π 4 a π 4 a

πx a πx sin a

πy =0 a πy sin =0 a

½ ∙³ ´ ³ ´ ³ ´ ¸ − ¾ ³ ´ ³ ´ ½ ³´− ¾ ³ ´ ³ ´ D w0

4

+2

4

+

4 D w0

q 0 sin q 0

sin

(175)

In order to satisfy the above equation for all values of  x and y, the coefficient in the bracket must vanish. This gives: q 0 w0 = 4D where D = Eh 3 / 12 1

ν 2 .

¡ ¢ £ ¡ − ¢¤ ∙ ¸ − ∙ ¸ −

Bending Moments

a π

4

³´

The various bending moments are given by:

∂ 2w ∂ 2 w π + ν  = D (1 + ν  ) ∂x 2 ∂y 2 a 2 2 ∂  w ∂  w π M yy = D + ν  = D (1 + ν  ) ∂y 2 ∂x 2 a ∂ 2 w π M xy = D (1 ν ) = D (1 ν ) ∂x∂y a

M xx = D

−

−

(176)

−

−

37

πx πy sin (177) a a 2 πx πy w0 sin sin a a 2 πx πy w0 cos cos a a 2

³´ ³´ ³´

³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´

w0 sin

Shear Components

The shear components Qx and Qy are: ∂M xx ∂M xy + ∂x ∂y ∂M yy ∂M xy Qy = + ∂y ∂x

Qx =

(178)

Now, using the previously obtained bending moments, we get the shear components in the interior of  the plate: Qx = 2 D Qx = 2 D

π a π a

3

³´ ³´

3

πx a πx w0 sin a w0 cos

Eff ective Shear Components ponents:

πy a πy cos a

³ ´ ³ ´ ³ ´ ³ ´ sin

(179)

Next, let us computer the eff ective shear com-

∂M xy ∂y ∂M xy V y = Qy + ∂x

V x = Qx +

(180)

Using the previous results, we get: V x = (3 V y = (3

π a π a

3

πx w0 cos a πx w0 sin a

πy sin a πy cos a

³ ´ ³ − ³ ´ ³ − h − ¡ ¢i h − ¡¡ ¢¢ − ¡¡ ν ) D ν ) D

³´ ³´

3

We now need to evaluate the eff ective shear on the boundaries: V x / (3 x=0 x=a y=0 y=a

ν ) D

w0 sin w0 sin 0 0

π 3 a

π y a π y a

V y / (3

ν ) D

0 0 w0 sin w0 sin

−

Because our sign convention is:

positive shear

x

z 38

´ ´

¡ ¢i ¢¢

π x a π x a

(181)

π 3 a

(182)

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 eff ective shear along the boundary, we get: R=

Z  L

a

V n dxt = 4

Z  0

V x |x=0 dy = 4 (3

− ν ) D

π a

3

a

πy a

³ ´ Z  ³ ´ w0

sin

0

dy (183)

Then, the reduction force due to eff ective shear on boundaries reads: R = 2 (3

− ν ) q 0

a 2 π

(184)

¡¢

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

Z 

S 

a

q (x, y) dS  =

a

Z  Z  0

0

q 0 sin

πx a

πy a

³ ´ ³ ´ sin

dx dy

(185)

Then, the total external load acting on the plate reads: P  = 4 q 0

a 2 π

¡¢

(186)

Notice that R and P  do not balance! We did not include the corner forces. These are given by: (F corner )x0 ,y0 = 2 (M xy )|x=x0 ,y=y0 (187) 39

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

=

πx a

πy a

h ³ ´ ³ ´i¯¯ ³ ´ − −

(F corner )0,0 = 2

cos

2 D (1

cos π a

ν )

2

(188)

0,0

w0

Now, the vertical force balance is satisfied:

R + 4 F corner = P 

− ν ) q 0 2 (3 − ν )

2 (3

4.1.4

a π

2

³´− ³´− a π

q 0

8 D (1

2

− ν ) (1 − ν )

2

(189)

π a

2

³´ ³´

w0 = 4 q 0

π a

q 0

2

= 4 q 0

a π a π

2

³´ ³´

(190)

2

Exercise 2: Comparison between Plate and Beam Solution

Plate Solution given by:

For a square simply supported plate under loading q  plate (x, y) q  plate (x, y) = (q 0 ) plate sin

we found that the plate deflection is:

πx a

sin

πy a

(191)

πx a

sin

πy a

(192)

³ ´ ³ ´ ³ ´ ³ ´

w plate (x, y) = (w0 ) plate sin with:

(q 0 ) plate a 4 (w0 ) plate = 4D π 2 3 1 ν  (q 0 ) plate a = E h3 π

³´ ¡− ¢

a

P  plate =

a

0

0

(q 0 ) plate sin

= 4 (q 0 ) plate

a π

2

³´

40

πx a

4

³´

For the plate, the total load is given by:

Z  Z 

(193)

πy a

³ ´ ³ ´ sin

dy dx

(194)

Wide Beam Solution

For a wide beam under line loading given by: πx q beam ( q 0 )beam sin beam (x, y ) = (q  a we need to compute the central deflection (w0 )beam from:

³ ´

0000

E I wbeam = q beam beam (x)

(196)

where I  = ah3/12 12.. Assuming the deflection wbeam (x): πx wbeam (x) = (w0 )beam sin a we get: π 4 πx πx E I  (w0 )beam sin = (q 0 )beam sin a a a Thus,

³ ´

³´

³ ´

³ ´

(q 0 )beam a 4 (w0 )beam = E I  π 12 (q 0 )beam a 4 = E a h3 π Now, let us compute the total forces: a πx P beam = (q 0 )beam sin dx beam a 0 a = 2 (q 0 )beam π Comparison For both total forces to be equal, we need to have:

³´ ³´ ³ ´

Z 

P  plate = P beam beam 2 a a 4 (q 0 ) plate = 2 (q 0 )beam π π a (q 0 )beam = 2 (q 0 ) plate π With a concentrated load, the beam deflection now becomes:

³´

(q 0 )beam a 4 (w0 )beam = E I  π 24 (q 0 ) plate a = π E h3 π We now can compute the ratio of  central deflections:

³´ ³´

α=

(w0 ) plate (w0 )beam

=

(195)

(197) (198)

(199)

(200)

(201)

(202)

(203)

4

3 (1−ν 2 ) (q0 )plate a 4 π E h3 ( q ) 4 0 plate 24 a π E h3 π

¡¢ ¡¢

(204)

π 1 ν 2 ' 0.36 8 The above equation means that under the same total load, a plate is three times stiff er er than a wide beam. The ratio α will vary slightly depending on the load distribution (sinusoidal, uniform, concentrated load, etc.). =

¡− ¢

41

4.1.5

Exercise 3: Finite Diff erence erence Solution of  the Plate Bending Problem

Governing Equations read:

∇2M  = −q  ∇2w = − M D

(205)

or in the component notation they read:

2 ∂ 2 M + ∂ ∂yM  q  2 = ∂x 2 2 2 ∂  w + ∂ ∂yw2 = M  D ∂x 2

− −

(206)

where M  is the moment sum defined by: M  =

M αα αα = D καα 1 + ν 

(207)

Case of  Simply Supported Plate The boundary condition of  a simple supported plate reads: w=0

on

Γ

M nn nn = 0

on

Γ

(208)

a x

a

n y

For sides parallel to x-axis (thick lines), one gets: M nn nn = M yy yy = 0 dw =0 dx From the general constitutive equations, w=0

→

→

d2 w =0 dx2

(209)

→

M yy yy = D [κyy + ν κxx ] 0 = D [κyy + ν · ν  · 0]

κxx = 0

(210)

(211)

→

κyy = 0

Therefore, M  = D [καα + κββ ] = D [0 + 0] 0] = 0 42

(212)

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 boundary conditions hold: 2 ∂ 2 M + ∂ ∂yM  q  in S  2 = ∂x2 (213) M  = 0 on Γ

( (

−

∂ 2 w ∂x2

2

+ ∂ ∂yw2 = w=0

− M D

in S  on Γ

(214)

Therefore, the above two boundary value problems are uncoupled. The Finite Diff erence erence Technique derivatives.

An approximation to the first and second

z

zn

+1

zn

h xn x n

x +1

m−1 h n−1

x

n+1

n, m m+1 y

dz dx

backward

¯¯ ¯¯ ¯

dz dx

'

n forward

'

n+1

43

zn

− zn

1

−

h

zn+1 zn h

−

(215)

d2 z dx2

¯¯ ∙ ¸ ¡ ¢ −¡ ¢ =

n

= = =

d2 z dy2 2

∇

¯¯

=

d dz dx dx

dz dx n+1

zn+1−zn h

zn+1

(216)

dz dx n

h

−z

n −zn−1

h

h

− 2 zn + zn

1

−

h2

zm+1

− 2 zm + zm

1

−

(217)

h2

m

∂ 2 z ∂ 2 z z= 2 + 2 ∂x ∂y 1 = 2 (zn+1 2 zn + zn h

−

(218) 1

−

+ zm+1

− 2 zm + zm

1)

−

Top h h

h

Left h

0

Right

Bottom

∇2z = h12 (zT  + zB + zL + zR − 4 z0)

(219)

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

44

6

5

4

3

2

a

1

h=

a 4

a

Four axes of symmetry

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

∇

−

2

Point 1:

4 M 2

− 4 M 1 = − q16a

Point 2:

M 1 + M 4 + 2 M 3

Point 3:

2 M 5 + 2 M 2

Point 4:

M 4 = 0

Point 5:

M 5 = 0

Point 6:

M 6 = 0

(220)

− 4 M 2 = −

2

45

− 4 M 3 = − q16a

q a2 16

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: 2

1

q a2 16

3

2 q a2 16

3

2

whose solution is:

1

⎧⎪⎨ 4 M  − 4 M  = − ⎪⎩ M 2 M +−2 M 4 M − =4 M − = −

q a2 16

9 q a2 128 7 M 2 = q a2 128 11 M 3 = q a2 256 M 1 =

(221)

(222)

At the plate center, M xx = M yy so that: M  =

M xx + M yy 2 M xx = 1 + ν  1 + ν 

M xx =

1 (1 + ν ) M  2

(223) (224)

At the center, M  = M 1 , thus, 1 (1 + ν ) M 1 2 1 9 = (1 + ν ) q a2 2 128 2 = 0.0457 q a

M xx =

(225)

This is 4.6% less than the exact solution which is (M xx)exact = 0.0479 q a2 from the text book. Determination of Deflection For each interior point (1, 2, 3), we write equation 2 w = M/D. For each boundary point (4, 5, 6), we write boundary condition w = 0.

∇

−

Point 4:

9 qa2 a2 4 w2 4 w1 = (226) 128 D 16 M 2 a2 7 q a2 a2 w1 + w4 + 2 w3 4 w2 = = D 16 128 D 16 M 3 a2 11 q a2 a2 2 w5 + 2 w2 4 w3 = = D 16 256 D 16 w4 = 0

Point 5: Point 6:

w5 = 0 w6 = 0

Point 1: Point 2: Point 3:

−

−

M1 a2 = D 16

−

µ −

−

−

−

46

¶ µ ¶ − µ ¶ −

Similarly, 9 qa4 2048 D

⎧⎪⎨ 4 w − 4 w = − ⎪⎩ w2 w+−2 w4 w− =4 w− = − 2

1

1

q a4 7 2048 D 11 q a4 4096 D

3

2

(227)

2

3

Finally, the finite diff erence solution is:

33 q a4 q a4 = 0.00403 8196 D D 4 3 qa q a4 w2 = = 0.00293 1024 D D 4 35 q a q a4 w3 = = 0.00214 16384 D D w1 =

(228)

On the other hand, the exact deflection of  the center point is: (w1)exact

q a4 = 0.00416 D

(229)

Thus, the error of  the finite diff erent solution is 3.1%.

4.2 4.2.1

Membrane Theory of  Plates Plate Membrane Equation

Assume that the bending rigidity is zero, D = 0. membrane.

The plate becomes now a

• Equilibrium of  in-plane equation N αβ ,α = 0

on S 

(230)

Equilibrium of  out-of-plane equation N αβ w,αβ +q  = 0

on S 

(231)

• Strain-displacement relation εαβ = ◦

1 1 (uα ,β +uα,β ) + w,α w,β 2 2

(232)

• Constitutive equation

£

N αβ = C  (1 ν 2 .

¡− ¢

where C  = Eh/ 1

− ν )

εαβ + ν εγγ  δ αβ ◦

◦

¤

This is a non-linear system of  equation which is difficult to solve. corresponding system of  equation for the plate bending was linear. 47

(233)

Note that

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

∂N rr + N rr ∂r

− N θθ = 0

on S 

(234)

on S 

(235)

Equilibrium of  out-of-plane equation ∂  ∂w N rr r + r q  = 0 ∂r ∂r

∙

¸

• Strain-displacement relation ∂u r 1 λrr = + ∂r 2 ur λθθ = r

∂w ∂r

2

µ ¶

(236)

N rr = C  [λrr + ν λθθ ]

(237)

• Constitutive equation

N θθ = C  [λθθ + ν λrr ] ν 2 .

¡− ¢

where C  = Eh/ 1 4.2.3

Example: Approximation Solution for the Clamped Membrane

Consider a circular plate with the clamped support. q

r 

a

w0

()

w r 

48

Membrane Solution From the symmetry and clamped boundary condition, the radial 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: N rr = C εrr εrr

1 = 2

(240) 2

∂w ∂r

µ ¶

(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

"µ ¶ C  ∂w 2 ∂r

#

2

∂w r = ∂r

− r q 

on S 

(242)

Integrating both sides once with respect to r reads: C r 2

∂w ∂r

3

µ ¶

=

−

r 2 q  + c1 2

(243)

At the center of  the membrane, the slope should be zero. Thus, one gets: ∂w =0 at r = 0 ∂r c1 = 0

(244)

⇒

Then, Eq. (243) can be written: ∂w = ∂r

r  −

qr C 

(245)

q  4/3 r + c2 C 

(246)

3

Integrating the above equation again reads: w=

−

3 4

r  3

49

The integration constant c2 can be determined from the zero deflection condition at the clamped edge: w=0

⇒

at r = a 3 3 q  4/3 c2 = a 4 C 

(247)

r 

w 3 = a 4

ν 2 , Eq. (246) can be

¡− ¢ r  − ∙ ³ ´ ¸ − r  ∙ − ³ ´ ¸

Recalling the definition of  the axial rigidity C  = Eh/ 1 put into a final form: 3

' 0.73

(1 3

ν 2 ) q a 1 Eh qa r 1 Eh a

r a

4/3

(248)

4/3

In particular, the central deflection w (r = 0) = w0 is related to the load intensity by: w0 qa = 0.73 3 (249) a Eh

r 

Bending Solution It is interesting to compare the bending and membrane response of  the clamped circular plate. From the page 55 of  Theory  of  Plates  and  Shells  (2nd  Ed.) by Timoshenko and Woinowsky-Krieger, the central deflection of  the plate is linearly related to the loading intensity: w0 q a3 = a 64 D 3 1 = 16 q ' 0.17 E

(250) ν 2 q  a E h a 3 h

¡− ¢ ³´ ³´

3

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

w0 qa = 17 a Eh

(251)

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

50

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

4.3 4.3.1

Buckling Theory of  Plates General Equation of  Plate Buckling

• Equilibrium of  in-plane equation N αβ ,α = 0

on S 

(252)

Equilibrium of  out-of-plane equation M αβ ,αβ +N αβ w,αβ +q  = 0

on S 

(253)

• Strain-displacement relation 1 1 (uα ,β +uα ,β ) + w,α w,β 2 2 καβ = w,αβ εαβ = ◦

(254)

−

• Constitutive equation for axial force and axial strain

£

N αβ = C  (1

(255)

− ν ) καβ + ν κγγ  δ αβ ]

(256)

εαβ + ν εγγ  δ αβ ◦

◦

ν 2 , and another one for moment and curvature

¡− ¢ £ ¡ − ¢¤

where C  = Eh/ 1

¤

− ν )

M αβ = D [(1

where D = Eh 3 / 12 1

ν 2 . 51

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

−D w,ααββ

(257)

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

−D w,ααββ +N αβ 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 of  the plates: D w,ααββ

−N αβ w,αβ = 0

(260)

where the second term in the left hand is non-linear due to N αβ which should be obtained from: N αβ ,α = 0 4.3.2

(261)

Linearized Buckling Equation of  Rectangular Plates

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

52

a

x Px

Px b ux y

Px

Post-buckling (δ w ≠ 0) k 2

Pc

Pre-buckling (δ w = 0) k

ux

Pre-Buckling Problem εαβ ◦

Recall that: 1 = 2

µ

∂u α ∂u β + ∂x β ∂x α

¶

+

1 ∂w ∂w 2 ∂x α ∂x β

(262)

Eh (1 ν ) εαβ + ν εγγ  δ αβ (263) 1 ν 2 In the pre-buckling problem, the linear equilibrium equations are obtained by omitting the nonlinear terms in the governing equations Eq. (260) and (261). The resulting equations are now: D w,ααββ = 0 N αβ =

−

£

−

◦

◦

¤

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

53

(264)

where N αβ =

Eh (1 1 ν 2

−

εαβ ◦

£

1 = 2

− ν ) εαβ + ν εγγ  δ αβ

µ

◦

◦

∂u α ∂u β + ∂x β ∂x α

¶

δun = 0

on

¤

(265) (266)

and boundary condition:

¡

N nn

− N ¯nn

¢

(267)

Γ

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

N αβ =

−N αβ ◦

where the compressive pre-buckling membrane force are defined as positive. Post-Buckling Problem Now, the governing equation for buckling of  plates reads: D

∇4 w

+ N αβ w,αβ = q  + ◦

Boundary Conditions

(268)

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

˜αβ N αβ = λ N  ◦

˜ N  ˜ N  = λ ˜xx ˜xy N yx N yy

¯¯

¯¯

(269)

¯αβ is the known direction from the pre-buckling analysis, and η is unknown where N  load amplitude. Now, the nonlinear buckling equation becomes a linear eigenvalue problem: D where λ 4.3.3

∇4w + λ N ˜αβ w,αβ = 0

(270)

≥ 0 is eigenvalues, and w is eigenfunctions.

Analysis of  Rectangular Plates Buckling

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

54

a

x Px

Px b y h

From equilibrium equations, one gets:

¯¯ ¯¯ ¯ ¯¯

¯¯ ¯¯ ¯ ¯¯

N  N  N αβ = xx xy N yx N yy ˜ N  ˜ N  = λ ˜xx ˜xy N yx N yy ◦

=

◦

◦

◦

◦

P x b

10 00

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

∇4w + P bx w,xx = 0

(271)

(272)

Boundary condition for this simply supported plate are written as: w=0

on

Γ

M nn = 0

on

Γ

(273)

where the moment components read: M xx = D (w,xx +ν w,yy ) = 0

− M yy = −D

(274)

(w,yy +ν w,xx ) = 0

Thus, one gets: w = w,xx = 0

on x = [0, a]

w = w,yy = 0

on x = [0, b]

(275)

Equation (272) is a constant-coefficient equation, and a solution of the following form: mπx nπy w = c1 sin sin for m, n = 1, 2 (276) a b

³ ´ ³ ´ 55

satisfies both the diff erential equation and the boundary conditions. Introduction into Eq. (272) gives: D

∙³ m π ´

4

a

mπ a

nπ b

nπ b

P x b

mπ a

³ ´ ³ ´ ³ ´ ¸− ³ ´ ∙³ ´ ³ ´ ¸ ³ ´ ⇒

+2

2

P x =D b

2

+

πa m

4

m a

2

2

n + b

2 2

2

=0

(277)

(278)

where for the discrete values of  P x Eq. (272) has nontrivial solutions. The critical load can be determined by the smallest eigenvalue, i.e. n = 1 for all values of  a: P x =D b

³ ´ "³ ´ µ ¶ # µ ¶ "³ ´ µ ¶ # µ ¶ πa m

π2 D = 2 b π2 D = 2 b

m a

2

ab m

2

2

1 b

+

m a

2

mb a + a mb

2

+

2

1 b

(279)

2

Now, the critical load (P x )cr can be written as:

(P x )cr = kc

π2 D b

where kc =

µ

mb a + a mb

¶

(280)

2

(281)

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

56

6

2

12

For a given a/b, m may be chosen to yield the smallest eigenvalue. In order to minimize kc in Eq. (281), treating m as a continuous variable produces: ∂k c =2 ∂m

µ

mb a + a mb

¶µ

b a

−

a b m2

¶

=0

(282)

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

⇒

b a

− ab m12 = 0

(283)

Now, one gets: a b kc = 4

m=

(284)

Here, this is valid when a/b is integer and when considering a very long plates. Transition from m to m+1 half-waves occurs when the two corresponding curves have equal ordinates, i.e. from Eq. (281): kc |m = kc|m+1

⇒ ⇒

mb a (m + 1) b a + = + a mb a (m + 1) b a = m (m + 1) b

p 

57

(285)

(286)

a = b Example 1 For  m = 1, a/b =

p 

m (m + 1)

(287)

√ 2

Example 2 For  a  very  large  m, i.e. a  very  long  plate, a/b ' m. Thus, kc = 4 is  now  independent  of  m. A very  long  plate  buckles  in  half-waves, whose  lengths approach  the  width  of  the  plate: πx nπy w = c1 sin sin b b Thus, the  buckled  plate  subdivides  approximately into squares.

³ ´ ³ ´

Various Boundary Conditions of Plate under In-Plane Compressive Loading The critical buckling load reads: (P x )cr

π2 D = kc b

16

14

12

10

k c 8

6

c

A

 ss

C

c  ss  ss

 free E

B

 free D

c

c

 ss

4 Loaded edges clamped. Loaded edges simply supported.

2

0

0

1

2

a  b

3

4

5

Figure by MIT OCW. Influence of boundary conditions on the buckling coefficients of plates subjected to in-plane compressive loading

58

Various Boundary Conditions of  Plate under In-Plane Shear Loading The critical buckling load per unit length reads: (N xy )cr = kc

π2 D b2

15

r cr  =

13

2 � D  b2h

k 

c

Clamped Edges 11

k c 9 Simply Supported Edges

7

5

3

0

1

2

3

a  b

4

5

Critical values of shear stress for plates subjected to in-plane loading. Figure by MIT OCW.

Limiting Case: Wide Plates Consider a wide plate for which a/b ¿ 1. From the diagram, we see that if  a/b < 1, then m is set to be equal to unity, i.e.  just one wavelength in the x-direction. 59

a x

b

y

The buckling formula thus becomes: (N x )cr = kc |m=1 π2 D = 2 b π2 D = 2 a =

π2 D a2

π2 D b2 b a + a b

(288) 2

µ ¶ ³´µ ¶ ∙ ³ ´¸ a b

b a + a b

2

1+

a b

2

2 2

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

(N x )cr

π2D = a2

which is called Sezawa’s formula for wide plates. Example 3 Here, relative  merits  of  sti  ff ening a  large panel  are investigated in the  longitudinal or  in  the  transverse  direction. It  is  assumed  that  the  sti  ff eners  provide   for  a  simply  supported  boundary  conditions. Consider  the  case  of  longitudinal  sti  ff eners.

60

L σcr 

s From  the  von  Karman  formula, the  buckling  load  per  unit  length  for  each  divided  part  reads: 4π 2 D (N x )cr = s2 Now, the  buckling  stress  can be  calculated: (N x )cr h 2 4π D = 2 s h

(σcr )longitudinal =

Consider  the case of  transverse  sti  ff eners.

L s

σcr 

From  the  Sezawa’s  formula, the  buckling load  per unit  length  along  the  loaded  edges  reads: π2D (N x )cr = 2 s Now, the  buckling  stress  can be  calculated: (N x )cr h 2 π D = 2 s h

(σ cr )transverse =

Thus, it  is  concluded  that  (σ cr )longitudinal = 4 (σ cr )transverse This  shows  advantages  of  longitudinal  sti  ff eners  over  transverse  sti  ff eners. 61

4.3.4

Derivation of  Raleigh-Ritz Quotient

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

− V b) + (U m − V m)

= (U b

(289)

where each term for buckling problems will be discussed in the following. Term Relating to Plate Bending Response In the buckling problem, the work done by external load causing bending response considered as zero: V b = 0

(290)

The bending energy can be expressed: 1 U b = 2 D = 2 D = 2 D = 2 D = 2

Z  Z  − Z  h − Z  h − Z  n S 

M αβ καβ dS  [(1

S 

(1

S 

(1

S 

S 

(291)

ν ) καβ + ν κγγ  δ αβ ] καβ dS  2

ν ) καβ καβ + ν  (κγγ )

i

dS  2

ν ) (κ11κ11 + κ12 κ12 + κ21κ21 + κ22 κ22) + ν  (κ11 + κ22) 2

(κ11 + κ22)

− 2 (1 − ν )

h

κ11 κ22

− (κ12)

2

io

dS 

i

dS 

Here, the term in the square bracket is called Gaussian curvature: κ11 κ22

− (κ12)2 = κI  κII 

where κI  and κII  are the principal curvatures. Gaussian curvature vanishes, so one gets: D U b = 2

Z 

S 

(292)

For plates with straight edges,

(κ11 + κ22)2 dS 

(293)

The integrand of  the above equation can be written in terms of  the transverse displacement: (κ11 + κ22)2 = =

2

2

¡−∇ ¢ w

(294)

∇2w ∇2w

Now, the bending energy reads: D U b = 2

Z  ∇

2

S 

62

w

∇2w dS 

(295)

Term Relating to Plate Membrane Response The work done by external load causing membrane response reads: V m =

Z

Z 

¯nn un dl + N 

Γ

¯tn ut dl N 

(296)

Γ

˜αβ is determined from the In the buckling problem, the axial force N αβ = λN  pre-buckling solution and is considered as constant, so the membrane energy reads: ◦

Z  − ∙ Z  − Z  −

U m = λ

˜αβ ε dS  N  αβ

(297)

◦

S 

= λ

S 

= λ

S

˜αβ N  ˜αβ N 

1 1 (uα ,β +uα ,β ) + w, α w,β dS  2 2 λ ˜αβ w,α w, β dS  uα ,β dS  N  2 S 

¸

Z 

−

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

Z  − λ

S 

Z  ³ − Z  − Z  ³ −

˜αβ uα ,β dS = λ N 

˜nn un + N  ˜tn ut N 

Γ

λ

S 

˜αβ ,β uα dS  N  ˜nn un + N  ˜tn ut N 

= λ

Γ

where the in-plane equilibrium is applied:

´

dl

´

dl

(298)

−λ N ˜αβ ,β = 0

(299)

Now, the membrane energy can be expressed: U m =

Z  ³ −

˜nn un + N  ˜tn ut N 

λ

Γ

´

dl

λ 2

−

Z 

S 

˜αβ w,α w,β dS  N 

(300)

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

− V m =

Z  ³ − Z  − Z  ³− Z  − Z  − λ

˜nn un + N  ˜tn ut N 

Γ

Z  − ´ −

¯nn un dl N 

Γ

=

˜nn λ N 

Γ

=

λ 2 λ 2

S 

S 

´

dl

−

¯tn ut dl N 

Z 

S 

˜αβ w,α w,β dS  N 

Γ

¯nn N 

un dl

Z  ³ − −

˜αβ w,α w, β N  Γ

˜tn λ N 

Γ

˜αβ w,α w, β dS  N 

where the boundary conditions on

λ 2

are applied. 63

−

¯tn N 

´

ut dl

(301)

Total Potential Energy and Its Variations Now, one gets the total potential energy: V b ) + (U m

Π = (U b

=

D 2

− V m) λ 2 w ∇2 w dS  − 2

(302)

Z − ∇ S 

Z 

S 

˜αβ w,α w,β dS  N 

The first variation of  the potential energy can be obtained: D δ Π = 2 λ 2

Z  £ ¡∇ ¢ ∇ ∇ ¡∇ ¢¤ Z  − Z  ∇ ¡∇ ¢ − Z  Z  ∇ ∇ − Z  2

δ 

2

w

2

w+

w δ 

2

w

dS 

S 

˜αβ (δw, α w,β +w,α δw, β ) dS  N  2

=D

2

w δ 

w dS 

λ

S 

S 

2

=D

2

w

δw dS 

λ

S 

S 

˜αβ w,α δw, β dS  N 

˜αβ w,α δw, β dS  N 

˜αβ is considered as constant under the variation. where N  variation of  the potential energy reads: 2

δ  Π = D

Z  ¡∇ ¢ ∇ Z  ∇ ∇ 2

δ 

2

w

δw dS 

S 

D

(303)

S 

2

2

δw

δw dS 

S

−λ

−λ

Z 

S 

Z 

S 

Similarly, the second

˜αβ δw, α δw, β dS  N 

(304)

˜αβ δw, α δw, β dS  N 

Raleigh-Ritz Quotient Application of  the Treff tz condition for invertability, δ 2 Π = 0, determines the load intensity: λ=

D

2

R  ∇ S 

δw

∇2δw dS 

˜ S  N αβ δw, α δw, β dS 

R 

(305)

Here, choose a trial function for w: w=Aφ

(306)

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

(307)

Now, the load intensity reads: λ= = =

2

R  ∇ R  R  ∇ R  R  ∇ ∇ R  D

S 

(δA φ)

∇2 (δA φ) dS 

˜ S  N αβ (δA φ) ,α (δA φ) ,β dS 

D

S 

˜

S  N αβ

D

S 

˜

2

δA

S  N αβ

φ δA

∇2φ dS 

δA φ,α δA φ,β dS  2

φ

2

φ dS 

φ,α φ,β dS  64

(308)

The Raleigh-Ritz quotient is defined as: D

λ=

2

R  ∇ S 

φ

∇2φ dS 

˜ S  N αβ φ,α φ,β dS 

R 

(309)

Example 4 As  a  special  case, consider  1-D  case: 10 = 00

˜αβ N  then, the  Raleigh-Ritz  quotient  becomes: λ=

D

¯¯

2

R  ∇ R  S 

φ

S  (φ,x )

¯¯

∇2φ dS 

2

dS 

Example 5 Similarly, consider  2-D  compression  case: ˜αβ = δ αβ N 

¯¯

10 = 01

then, the  Raleigh-Ritz  quotient  becomes: λ= =

φ 2 φ dS  S  φ,α φ,α dS 

R R  ∇ ∇ R  R  h ∇ ∇ i D

S 

D

S 

4.3.5

2

¯¯

2

S 

2

φ

φ dS 

(φ,x )2 + (φ,y )2

dS 

Ultimate Strength of  Plates

The onset of buckling stress σcr does not necessarily means the total collapse of the plate. Usually, there is redistribution of  stresses, and the plate takes additional load until the ultimate strength σ u is reached. Von Karman Analysis of the Eff ective Width For a simply supported plate, the buckling load is: 4π 2 D b 4π 2 E  h3 = 12(1 ν 2 ) b

P cr =

(310)

−

and the corresponding buckling stress is: P cr σ cr = bh 4π 2 E  = 12(1 ν 2 )

−

π 2 E  = 3 (1 ν 2 )

−

65

(311) 2

h b

µ¶ µ¶ h b

2

The normalization of  buckling stress by the yield stress reads: σ cr π2 E  = σy 3 (1 ν 2 ) σ y

−

=

(1.9)2 σy E 

h b

2

µ¶

(312)

b 2 h

¡¢ µ ¶

σ cr = σy

1.9 β 

2

(313)

where β  is a non-dimensional parameter defined by: β  =

r 

σy b E h

(314)

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

1.9

On further loading the plate beyond σ cr , a greater proportion of  the load is taken by the regions of  the plate near the edge. Von Karman assumed that these edge regions, each of  the width beff  /2, carry the stress up to the yield while the center is stress free.

66

σ (y)

σ

x

b y Before Buckling

After Buckling

σy

beff 

beff 

2

2

σy σ (y )

b Actual Stress

von Karman Model

The edge zones are at yield, i.e. σ cr /σy = 1, but the width of  the eff ective portion of  the plate is unknown: σ cr = σy from which one obtains:

(1.9)2 σy E

2

³ ´ beff  h

=1

s 

beff  = 1.9 h

E  σy

(315)

(316)

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

√ 

beff  = 1.9 900 = 57h

(317)

This is somehow high, but there is not much diff erence from the empirically determined values of  beff  = 40h 50h.

∼

67

The total force at the point of  ultimate load is: P  = σ y beff  h

(318)

Now, the average ultimate stress can be calculated: σu =

P  bh

(319)

beff  b h = 1.9 Eσ y b = σy

p 

The average ultimate stress can be normalized by the yield stress: h Eσ y σu = 1.9 b σy σy

p 

(320)

σu 1.9 = σy 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 next figure.

68

1.9

Thin plates buckle before reaching ultimate load.

Thick plates yield before buckle.

Under the uniaxial loading, the relation between an applied load and the corresponding displacement is schematically shown all the way to collapse in the next figure. Px x Pu

Pcr 

post-buckling

Px

Px

collapse or crash

ux pre-buckling

y ux

Empirical Formulas • Foulkner correction

σu 2 = σy β 

69

− β 12

(322)

• Gerard (Handbook of  elastic stability) σu = 0.56 σy

0.85

à s  ! gh 2 A

E  σy

(323)

where g is the sum of  the number of  cuts and the number of  flanges after the cuts, A is the cross sectional area A = bh, and the coefficients 0.56 and 0.85 are empirical constants. Example 6 Consider  a  plate  which  has  one  cut  and  two  fl anges. b h

Then, g=1+2=3 Now, σu = 0.56 σy = 1.42 =

0.85

à s  ! à s  ! 3h2 bh h b

E  σy

E  σy

0.85

1.42 β 0.85

Modifications in Codes In the original von Karman formula, the eff ective width ratio reads: beff  σ cr = (324) b σy

r 

In the ANSI specification, imperfection is considered: beff  = b

σ cr σy

σ cr σy

r  µ − r  ¶ 1

0.218

(325)

In UK specification, the eff ective width ratio is defined as:

" µr 

beff  = 1 + 14 b

σ cr σy

70

4

− 0.35

0.2

−

¶#

(326)

eff 

y

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

4.3.6

Plastic Buckling of  Plates

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

C 

B

 A

1.9

Stowell’s Theory for the Buckling Strain Stowell developed the theory of  plastic buckling for simply-supported square plates loaded in one direction. 71

b

x

b

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

π2 = 9

h b

2

µ ¶ h p 

i

1 + 3 (E t /E s )

2+

(327)

where the tangent modulus E t and the secant modulus E s are defined by: E t =

dσ dε

E s =

;

σ ε

(328)

σ

Et σy

Es ε For the materials obeying the power hardening low: σ = σr

ε εr

n

µ¶

(329)

where σ r and εr are the reference stress and strain. Now, the tangent and secant modulus are: dσ σr E t = =n dε εr σ σr E s = = ε εr

ε εr

n−1

µ¶ µ¶

72

ε εr

n−1

(330)

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

π2 = 9

h b

2

µ ¶¡

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

h b

µ¶

¢

2

(332)

Having determined εcr , the corresponding buckling stress is calculated from the power law. Approximate Solution for the Buckling Strain stress relation:

Consider an elastic plane

E  (εxx + νε yy ) 1 ν 2 E  σ yy = (εyy + νε xx) 1 ν 2

σxx =

−

(333)

−

Using the solution for the pre-buckling state, i.e. σ xx = σ cr and σyy = 0 leads: E  (εxx + νε yy ) 1 ν 2 E  0= (εyy + νε xx) 1 ν 2

σ cr =

−

(334)

−

from which one gets: εyy = νε xx

(335)

−

σcr = Eε xx = Eε cr The critical elastic buckling stress is: σ cr

π2E = kc 12(1 ν 2 )

−

2

h b

µ¶

(336)

So, the corresponding critical elastic buckling strain read: εcr

π2 = kc 12(1 ν 2 )

−

h b

2

µ¶

(337)

Equation (337) is more general than a similar expression Eq. (327) given by the 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 boundary conditions. In particular, for kc = 4, Eq. (337) predicts: εcr = 3.6

h b

µ¶

2

(338)

which should be compared with the coefficient 3.7 of  Eq. (332) in the Stowell’s theory. For a plastic material or very high hardening exponent, the prediction of  both method are much closer. 4.3.7

Exercise 1: Eff ect of  In-Plane Boundary Conditions, δw = 0

No Constraint in In-Plane Displacement displacement in y-direction, N yy = 0.

Consider no constraint in in-plane

◦

expansion due to Poison's effect uy a

x

N D

N D a ux y The membrane force tensor reads: N αβ ◦

N  0 = 0 0

From the constitutive equation, one gets:

N yy = =

⇒

¯¯ ¯

◦

¯¯ ¯

Eh ε + ν εxx = 0 = 1 ν 2 yy εyy = ν εxx

−

◦

£

−

◦

◦

◦

¤

(339)

−N yy ◦

(340)

By applying the geometric equation between strain and the displacement and considering δw = 0, here, one gets the relation between ux and uy : ∂u y = ∂y

x −ν ∂u ∂x

74

(341)

Integrating both sides over the plate length leads: a

Z  0

a ∂u y ∂u x dy = ν  dx ∂y 0 ∂x uy = ν ux

Z  −

(342)

−

Constraint in In-Plane Displacement the y-direction, uy = 0.

Consider a plate fully constrained in D N yy

a

D

= ν N 

x D D = N  N xx

N D

N D a ux y Consequently, one also gets:

εyy = ◦

∂u y =0 ∂y

(343)

Under the uniform compression, εxx reads: ◦

εxx = ◦

∂u x ux = ∂x a

(344)

From the constitutive relation, the membrane forces reads: Eh 1 ν 2 Eh N yy = 1 ν 2

N xx =

− −

£ £

◦

◦

εyy + ν εxx ◦

◦

Finally, the membrane force tensor reads: N αβ ◦

where

¤ ¤

Eh ux = N xx 1 ν 2 a Eh ux = ν  = N yy 2 1 ν  a

εxx + ν εyy =

¯¯

−

−

−

¯¯

◦

−

¯¯

¯¯

(345)

◦

N  0 10 = xx =λ 0 N yy 0 ν 

(346)

Eh ux 1 ν 2 a

(347)

λ=

◦

◦

−

75

4.3.8

Exercise 2: Raleigh-Ritz Quotient for Simply Supported Square Plate under Uniaxial Loading

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

a

x

N D

N D a y

Then, the membrane force tensor reads: N αβ = N  ◦

◦

¯¯ ¯

¯¯ ¯

¯¯ ¯

10 10 =λ 00 00

¯¯ ¯

(348)

The plate will deform into a dish, so for the trial function take the following: ˆ (x, y) = sin φ=φ

πx πy sin a a

³ ´ ³ ´

Now, in order to obtain the load intensity, we first calculate φ,x and π πx πy cos sin a a a π πx πy φ,y = sin cos a a a

³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ − ³´ ³ ´ ³ ´ −

φ,x =

φ,xx = φ,yy =

π a π a

πx πy sin a a 2 πx πy sin sin a a 2

sin

∇2φ = φ,xx +φ,yy π 2 = −2 sin a ∇2φ ∇2φ = 4

4

sin2

76

∇2φ : (350)

(351)

(352) πx πy sin a a

³´ ³ ´ ³ ´ ³´ ³ ´ ³ ´ π a

(349)

πx πy sin2 a a

(353)

Now, the Raleigh-Ritz quotient is calculated:

=

2

2

R  ∇ ∇ R  ¡ ¢ R R  ¡¡ ¢¢ ¡¡ ¢¢ ¡¢ ³´

N  = λ = ◦

4D

= 4D

D

S 

φ

φ dS 

(354)

2 S  (φ,x ) dS 

π 4 2 π x a S sin a π x 2 2 π S cos a a π 2

sin2 sin2

π y a π y a

dS  dS 

a

N cr = λ = 4D ◦

2

µ ¶ π a

(355)

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

4.4 4.4.1

Buckling of  Sections Transition from Global and Local Buckling

Euler buckling load of  a simply-supported column reads: (P cr )column

π2 EI  = 2 l

(356)

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

( Pcr )column

l

EI 

77

On the other hands, local buckling force of  a simply-supported plate reads: kc π2 D N cr = b2

(357)

where D = Eh 3 / 12 1 ν 2 and kc = [(mb) /a + a/ (mb)]2 . Thus, the total local buckling load can be obtained:

£ ¡ − ¢¤

(P cr ) plate =

kc π 2 D b

(358)

N cr 

h a

b

Transition from global to local buckling can be calculated by (P cr )column = (P cr ) plate: π2 D 2 EI  π 2 = kc l b =

(359)

kc π 2E h3 12 (1 ν 2 ) b

−

I kc = bh3 12 (1 ν 2)

−

Example 7 Consider  a  square  box column.

78

l b

2

µ¶

(360)

h

b

b

l

Then, one  obtains: kc = 4 2 I = hb3 3 Now, the global  buckling load reads: (P cr )column

I  2π 2 E hb3 = π E  2 = l 3 l2 2

and  the  local  buckling  load  from  four  plates  can  be calculated: (P cr )four  plates =

4kc π 2 E h3 12(1 ν 2 ) b

−

Then, applying  (P cr )column = (P cr )four  plates, one  gets: b2 hl

2

µ¶

' 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 with lips 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 both sides of Eq. (357) by the plate thickness b gives the expression of the buckling stress σ cr : N cr kc π2 E h 2 σ cr = = (361) h 12 (1 ν 2) b

µ¶

−

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

Plate 1 h1

b1

h2 b2

Plate 2

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

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: k1 π 2 E (σ cr )1 = 12 (1 ν 2 )

−

k2 π 2 E (σ cr )2 = 12 (1 ν 2 )

−

h1 b1

2

h2 b2

2

µ ¶ µ ¶

(362)

Before buckling, stresses in the entire cross-section are the same. So, at the point of  buckling, one gets: (σ cr )1 = (σ cr )2 (363) from which the buckling coefficient k2 is relating to k1 : k2 = k1

h1 b2 h2 b1

2

µ ¶

(364)

The total buckling load on the angle  element is: P cr = (σ c )1 h1b1 + (σ c )2 h2 b2 π2E = k1 12 (1 ν 2 )

−

(365)

"µ ¶ µ ¶ # " µ ¶ # µ ¶ h1 b1

2

h1 b1 + k2

π 2 E  (h1 )3 = k1 + k1 12 (1 ν 2 ) b1

−

π2 Ek 1 = 12 (1 ν 2 )

−

h1 b1

h1 b2 h2 b1

h2 b2 2

2

h2 b2

(h2 )3 b2

2

A

where A is the sectional area of two plate A = b1 h1 +b2 h2 . From this derivation, the conclusion is that only one buckling coefficient is needed to calculate the buckling load of  the section consisting of  several plates. Determination of  the buckling coefficient is a bit more complicated because of  the existence of  the edge bending moment. This can be illustrated in an example of  a box column with a rectangular cross section with the same thickness h. b1

b2

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 long before. Thus, there is an interaction between left plates, and a compromise must be established because left plates must buckle at the same time. The buckling coefficient as a function of  the ratio b2 /b1 is plotted in the figure next.

h h

Buckling coefficients for box sections

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

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

82

83

5 5.1

Buckling of  Cylindrical Shells Governing Equation for Buckling of  Cylindrical Shells

The starting point of  the analysis is the strain-displacement relation for plates: 1 1 (uα ,β +uβ ,α ) + w,α w,β 2 2 καβ = w,αβ εαβ =

(366)

◦

−

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

a x

x

y

aθ

Can the strain-displacement relation for a cylinder be derived from similar relation for a flat plate? x x ; dx dx (367)

→ y → aθ

→ ∂  1 ∂  → ∂y a ∂θ

;

(368)

Consider component by component. Use the notation: u1 = ux u2 = uy u3

→u →v →w

Then, one gets:

(369)

1 εxx = u,x + (w,x )2 (370) 2 u,θ 1 w,θ 2 w εθθ = + + (371) a 2 a a 1 u,θ 1 w,θ εxθ = + u,x + w,x (372) 2 a 2 a There is a new term w/a in the expression for the hoop strain. The physical meaning of  this new term becomes clear if  we consider axisymmetric deformation with v = 0 and w independent of  θ. Define the hoof  strain as a relative change ◦

◦

◦

³

³ ´ ´

84

in the length of  circumference when the original circle has a radius of  "a" before deformation and a new circle has a radius of  "a + w" after deformation: εθθ = ◦

2π (a + w) 2πa

− 2πa = w

(373)

a

a θ

w

Mathematically, Eq. (370)-(372) can be derived from its counterparts by transforming the rectangular coordinate system into the curvilinear coordinate system. x = r sin θ y = r cos θ

(374)

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 are transformed in a similar way: κxx = w,xx w, κθθ = 2θθ a w, xθ κxθ = a

(375)

Using the variational approach explained in details for the plate problem, one can see that the only new term in the expression for δ Π = 0 is

Z 

S 

N θθ

δw dS  a

(376)

Therefore, the new term should be added in the equation for out-of-plane equilibrium: 1 2 1 D 4 w + N θθ N xxw,xx + N xθ w,xθ + 2 N θθ w,θθ = q  (377) a a a

∇

where

µ −

¶

∇4w = w,xxxx + a22 w,xxθθ + a14 w,θθθθ 85

(378)

The above equations are the nonlinear equilibrium equations for quasi-shallow cylindrical shells. The linear equilibrium equations are obtained by omission the nonlinear terms, i.e. terms in the parenthesis. The resulting equations are: aN xx,x +N xθ ,θ = 0

(379)

aN θθ ,x +N θθ ,θ = 0 1 D 4 w + N θθ = q  a

∇

with w N xx = C  u,x +ν  a w0 N θθ = C  + ν u,x a

³ ´ ³ ´ ¡− ¢

(380)

where C  is the axial rigidity, C  = Eh/ 1 ν 2 . The pre-buckling solution should satisfy the system Eq. (379) and (380). These solutions will be denoted by N αβ = 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  1 0 = 2πa 0 0

and the corresponding displacement field reads: u (x) = w=

¯¯ ¯

(381)

P  − 2πahE  x

(382)

ν  P l E 2πah

u (l)

l

w

x aθ

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 = N θθ ◦

¯¯ ¯

00 01

¯¯ ¯

(383)

where from the constitutive equations Eq. (380): N θθ = Eh

w a

(384)

Substituting Eq. (384) into Eq. (379) leads the following linear forth order inhomogeneous ordinary diff erential equation for w (x): Dw or in a dimensionless form:

0000

+

Ehw = q  a2

(385)

d4 w q  4 + 4β  w = dx4 D

where

(386)

3 1 ν 2 Eh 4 β  = 2 = (387) a D a2 h4 The dimension of β  is [L 1 M 0 T 0], so βx is dimensionless. There are four boundary conditions for a simply supported cylinder:

¡− ¢

−

M xx = w = 0

at

x=0

M xx = w = 0

at

x=l

(388)

where one gets from the moment-curvature relation: d2 w M xx = D 2 dx M θθ = 0

(389)

−

The general solutions of  the above boundary value problem is: w (x) = e

βx

−

[c1 sin βx + c2 cos βx] + eβx [c3 sin βx + c4 cos βx]

− D4q β 4

(390)

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

87

A

It can be conducted that the curvature and bending is confined to a narrow boundary zone of  the width xb = π/ (2β ). The remainder of  the shell undergoes a uniform radial contraction: q  w0 = (391) D 4β 4

−

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 pressure by N θθ = qa and the pre-buckling solution is:

→

N αβ ◦

5.1.3

¯¯

00 = qa 01

¯¯

(392)

Special Case III: Hydrostatic Pressure

For a cylinder subjected to the hydrostatic pressure, the total axial compressive force is: P  = qπa 2 (393)

The pre-buckling solution is: N αβ ◦

¯¯

1/2 0 = qa 0 1

¯¯

which is a classical membrane stress state in a thin cylinder. 88

(394)

5.1.4

Special Case IV: Torsion of  a Cylinder

The pre-buckling stress in a cylinder subjected to the total torque of the magnitude T  is: M  0 1 N αβ = (395) 2πa 1 0

¯¯

◦

¯¯

x aθ N θx  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 of  buckling from the pre-buckling value. Thus, N αβ =

−N αβ

(396)

◦

and Eq. (377) becomes: 1 2 1 D 4 w + N θθ + N xxw,xx + N xθ w,xθ + 2 N θθ w,θθ a a a

∇

∙

◦

◦

◦

¸

= q 

(397)

where the hoop membrane force N θθ , the second term in Eq. (397), depends linearly on three component of  the displacement vector (u, v, w): 1 w N θθ = C  v,θ + + νu, x a a

µ

89

¶

(398)

Therefore, the out-of-plane equilibrium equation is coupled with the in-plane displacement (u, v) through the presence of  the term N θθ . Note that in the plate buckling problem the in-plane and out-of-plane response was uncoupled. It is possible to eliminate the terms involving in-plane components using the full set of  equilibrium and constitutive equations in the in-plane direction. By doing this, the order of  the governing equation has to be raised by four to give eight: D

8

∇ w+

1

− ν 2 Cw, a2

xxxx +

4

∇

∙

2 1 N xxw,xx + N xθ w, xθ + 2 N θθ w,θθ a a ◦

◦

◦

¸

=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 over and above those produced by the pre-buckling solution. The total deflection is a sum of  the two.

5.3 5.3.1

Buckling under Axial Compression Formulation for Buckling Stress and Buckling Mode

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

8

∇ w+

1

− ν 2 Cw, a2

xxxx +

P  2πa

∇4w,xx = 0

(400)

The buckling deflection of  the shell is assumed in the following form: w (x, θ) = c1 sin

mπx sin(nθ) l

³ ´

(401)

where c1 is the magnitude, and the integer numbers (m, n) denote the number of  half-waves respectively in the axial and circumferential direction. The above deformation satisfies both simply-supported boundary conditions at the ends, x = 0 and 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 mπa x x = =m ¯ l l a a mπa m ¯ = l

⇒

³ ´ 90

(403)

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

∙D ¡

c1 x sin m ¯ sin (nθ) = 0 a2 a6 a (404) By setting the coefficient in the square bracket to zero, the critical buckling membrane force per unit length becomes: 2

2 4

m ¯ +n

¢

4

+m ¯

2

¡− ¢ 1

ν  C 

−

P  m ¯ 2 + n2 2πa

P cr 2πa ¯ 2 + n2 D m = 2 a m ¯2

¡

¢ ¸ ³ ´ 2

m ¯2

N cr =

¡

(405) 2

m ¯2 ν  C  (m ¯ 2 + n2 )2

¢ ¡ ¢ − ¡ ¢ + 1

2

Here, by introducing the dimensionless parameter χ: m ¯ 2 + n2 χ= m ¯2 Eq. (405) reads:

2

(406)

D 1 χ + 1 ν 2 C  (407) 2 a χ The dependence of  the buckling force on the parameter χ is shown in the figure below. N cr =

¡− ¢

91

Treating χ as a continuous variable, one can find an analytical minimum: dN cr D cr = 2 dχ a

− 1 − ν 2

¡ ¢

1 C  2 = 0 χ

(408)

from which the optimum value of  the parameter χ is: χopt =

r  − p  (1

ν 2 ) C a2 D

a 12 (1 h a ' 3.3 h

=

(409)

− ν 2)

Introducing the expression for the optimum parameter χ into Eq. (407) and using definitions of  bending and axial rigidity, one gets: (N cr cr )min =

E h2 3 (1 ν 2 ) a

p 

−

(410)

The buckling stress is obtained by dividing the critical membrane fore by the shell thickness h: (N cr cr )min h E h = 3 (1 ν 2 ) a

σ cr =

p 

(411)

−

σ cr ' 0.605 E 

h a

(412)

This is the classical solution for the buckling stress of  a cylindrical shell subjected to axial compression. While the buckling load is unique and does not depend on (m ¯ , n), the buckling mode is not unique as: m ¯ 2 + n2 χopt = m ¯2

¡

2

¢

= 3.3

a h

There are infinity of  combinations of  m ¯ and n that give the same expression. Example 8 Let  a/h = 134 134,, then  χopt

2

m ¯ 2 + n2 = = 3.3 × 134 ' 442 m ¯2 m ¯ 2 + n2 ' 21 m ¯

¡

¢

⇒

92

(413)

n=

5.3.2

p 

21m ¯

− m¯ 2

Buckling Coefficient and Batdorf  Parameter

Let us introduce the dimensionless buckling stress or the buckling coefficient kc : kc = σ cr

hl2 π 2D

(414)

Additionally, the Batdorf  parameter Z  is defined: Z  =

p  − 1

ν 2

l2 ah

(415)

Using Eq. (411), the buckling coefficient becomes: E  h hl2 × 2 3 (1 ν 2 ) a π D 12 l2 = 1 ν 2 ah 3π 2 12 = Z  3π 2

kc =

p 

√  √ 

−

(416)

p  −

kc ' 0.702 Z 

(417)

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

93

Limiting Cases: Short Cylinders, a À l conditions: aÀl a l

⇒

→∞

Consider a case of  the following

(418)

Then, it is natural to assume that the number of  half-waves in the axial direction is unity: m=1 Consequently, one gets:

a l Since m ¯ is much larger than n, one also gets: m ¯ =π

→∞

(419)

m ¯ 2 + n2 ' m ¯2

(420)

94

Now, from Eq. (405), the membrane force reads: P  2πa D 2 = 2m ¯ + 1 a m ¯ 2D = 2 a π2 D = 2 l

N cr =

(421) 1 ν 2 C  2 m ¯

¡− ¢

The buckling stress reads:

N cr π2D σ cr = = h hl2 Now, from the definition of  the buckling coefficient, Eq. (414), the buckling coefficient for the very short cylindrical shells reads: kc = 1 This solution indicated in previous figure as the lower bound cut-off  value. upper bound cut-off  values is given by the Euler buckling load.

The

Limiting Cases: Very Long Cylinders, l À a The cylindrical shell is becoming the Euler column. The buckling load of  the column is: π 2 EI  P cr = 2 l where I  = πa3 h for cylinder sections.

95

(422)

The buckling stress reads: P cr 2πah π2 a = E  2 l

σ cr =

(423) 2

³´

For this very long cylindrical shells, the buckling coefficient can be written: a h

2

¡ − ¢³ ´ µ ¶ ν 2

kc = 6 1

a l

kc = 6

(424)

4

Z 2

(425)

From Eq. (411) and (423), the transition between the local shell buckling and global thin-walled column buckling occurs when (σ cr )shell = (σ cr )column which gives:

E  h π2 a = E  2 l 3 (1 ν 2 ) a

p 

⇒ 5.4

2

³´

−

π2 hl2 = a3

(426)

p 

3 (1 2

− ν 2) ' 8.15

(427) (428)

Buckling under Lateral Pressure

From the pre-buckling solution, Eq. (392), the governing equation Eq. (399) becomes: 1 ν 2 q  4 D 8w + Cw, + w,θθ = 0 (429) xxxx a2 a Assuming the double sine buckling deflection function, similar to the case of  axial compression, the governing equation becomes:

−

∇

∙D ¡

∇

c1 x sin m ¯ sin (nθ) = 0 a2 a6 a (430) By setting the coefficient in the square bracket to zero, the equation for the buckling pressure becomes: 2

2 4

m ¯ +n

¢

4

+m ¯

2

¡− ¢

ν  C 

1

− qa n

¯ 2 + n2 D m qa = 2 a n2

¡

2

2

¡

2

m ¯ +n

m ¯4 ν 2 C  n2 (m ¯ 2 + n2 )2

¢ ¡ ¢ − + 1

96

¢¸ ³ ´

2 2

(431)

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, or dimensionless parameter n ¯: nl n ¯= (432) πa Additionally, we define the dimensionless buckling pressure, q  ¯: l2 a q  ¯ = q  2 π D

(433)

Now, substituting m ¯, n ¯ and  p¯ into Eq. (431) leads: 1+n ¯2 q  ¯= n ¯2

¡ ¢

2

1 ν 2 C  l 1 + a2 a n ¯ 2 (1 + n ¯ 2 )2

2

¡− ¢ µ¶

(434)

By introducing the Batdorf  parameter Z , one gets: 1+n ¯2 q  ¯= n ¯2

¡ ¢

2

+

1 12 2 Z  2 n ¯ 2 (1 + n ¯2) π

(435)

For any value of  the geometrical parameter Z , there exists a preferred n ¯ which minimize the buckling pressure. Treating n ¯ as a continuous variable, the optimum n ¯ can be found analytically from d¯  p/d¯ n = 0. Substituting this back into Eq. (432), there will be a unique relation between the buckling pressure and the Batdorf  parameter. The solution is shown graphically in the figure below.

Limiting Cases: Very Long Cylinders, l À a In the limiting case of  a long tube (l À a), one gets: a m ¯ = mπ 0 (436) l

→

97

In this case, the last term in Eq. (431) vanishes and the buckling pressure becomes: n2 D q  = a3

(437)

Imagine a long cylinder consisting of  a change of  ring, each of  the height b. The moment 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 n2 Eh 3 = 3 b a 12 (1 ν 2 ) n2 EI  = 3 a (1 ν 2)

(439)

−

−

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

98

The Donnell solution should be compared with more exact solution of  the ring buckling problem which take into account a more complex incremental displacement field with both w (θ) and v (θ). Here, a distinction should be made between the centrally directed pressure (as in all preceding analysis) and the field -pressure loading where pressure is always directed normal to the deformed surface. In the later case, the ring buckling occurs at: Qc = n2

¡ −¢ 1

EI  a3

(440)

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

The solution of  centrally directed pressure loading case is: n2 Qc = n2

¡ −¢ −

2

1 EI  2 a3

(441)

where again the smallest n = 2. Thus the smallest buckling load intensity is: Qc = 4.5

EI  a3

(442)

It is seen that for a realistic value n=2, the present solution, Eq. (439) gives the buckling pressure between the two cases of  field-pressure loading and centrally directed 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  = πa2 q 

99

(443)

The pre-buckling solution is given by Eq. (394). The solution of the buckling equation still can be sought through the sinusoidal function, Eq. (401). The optimum solution can be found by a trial and error method varying parameters m ¯ and n ¯. A graphical 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 N xθ : N xθ = ◦

M  2πa

(444)

The other two components of the pre-buckling membrane forces vanish N xx = N θθ = 0. Moreover, the force is constant. ◦

100

◦

Under these conditions, the governing equation reduces to: 2

D

∇8w + 1 −a2ν  C w,xxxx + a2 N xθ ∇4w,xθ = 0 ◦

(445)

In view of  the presence of  odd-ordered derivatives in the above equation, the separable form of  the solution for w (x, θ), assumed previously, does not satisfy the equation. Under torsional loading, the buckling deformation consists of  circumferential waves that spiral around the cylindrical shell from one end to the other. Such waves can be represented by a deflection function of  the form: x w (x, θ) = C sin m ¯ nθ (446) a

³

with

´

−

mπa (447) L where m and n are integers. The alone displacement field satisfy the diff erential equation and the periodical condition is the circumferential direction, but does not satisfy any commonly used boundary at the cylinder ends. Consequently, this simple examples can be used only for long cylinders. For such cylinders, introduction of  the additional displacement field into the governing equation, yields: m ¯ =

m ¯ 2 + n2 N xθ = 2m ¯ n ◦

¡

¢

2

D m ¯3 + 1 a2 2 (m ¯ 2 + n2 ) 2 n

v 2 C 

¡− ¢

(448)

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

4 D m ¯3 + 1 m ¯ a2 64

v 2 C 

¡− ¢ µ¶

(449)

An analytical minimization of  the alone experiment with repeat to m ¯ gives: 64

4

m ¯ =

9 (1

− ν 2)

101

h a

2

(450)

Upon substitution, the final expression for the buckling force, or better, critical shear strain causing buckling is: N  0.272 E τ cr = xθ = h (1 ν 2 )3/4 ◦

−

h a

3/2

µ¶

(451)

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

As the radius of  the shell approaches infinity, the critical stress coefficient for simply supported and clamped edge approaches respectively the value 5.35 and 8.98 corresponding 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 experimentally 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 diff erent type of  loads: 102

• Axial compression • Torsion • Lateral pressure • Hydrostatic pressure are 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 graph shown below.

−

1.1

Theory

1.0

  a 0.8    /    h    E   p    5   x    0   e    6

   �

 .    0   = 0.6

  r   c

A design recommendation

   �

0.4

0.2

0

0

500

1,000

1,500 a 2,000

2,500

3,000

3,500

h

Distribution of test data for cylinders subjected to axial compression.

Figure by MIT OCW. The diff erences are shown to be very large. Design curves for cylindrical and other shells are based on the theoretical solution modified by empirical prediction factors called "know-down  factors" . For example, the solid curve shown in the receding page is a " 90 percent  probability  curve."  For a/h = 50, the reduction factor predicted by this probability curve is 0.24. Thus, the theoretical solution σ cr = 0.605 Eh/a should be multiplied by 0.24 for the design stress σ = 0.15 Eh/a. Cylindrical shells loaded in diff erent way are been sensitive to imperfections and the resulting knock-down factors are smaller. Most industrial organization establish their own design criteria. They are frequently loose-leaf and are continuously updated on the volume of the experimented evidence increase. The situation can be converted to the firmly established manual 103