DRAFT

STRUTTURE AERONAUTICHE1

course notes - Part 2 (P2)

Plates and Shells - Buckling instabilityStructural Optimization

Luigi Morino© Franco Mastroddi© Luigi Balis Crema©April 2020

1Realizzazione di parte delle figure a cura di Valerio Farano, Alessandro Settimi, Sabino Loiodice, Sara DiMario

Contents

III Plates and Shells – Buckling Instability 1

1 Thin plates with small deflections 3

1.1 Isotropic bending plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 Kinematic relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Stress description and constitutive relationships . . . . . . . . . . . . . . . 5

1.1.3 Equilibrium equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Orthotropic bending plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Isotropic laminated plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Analytic solutions with eigenfunctions methods . . . . . . . . . . . . . . . . . . . 14

1.4.1 Rectangular simply-supported isotropic plate . . . . . . . . . . . . . . . . 14

1.4.2 Rectangular simply-supported orthotropic plate . . . . . . . . . . . . . . . 17

1.5 Some issues on thermal stresses for plates . . . . . . . . . . . . . . . . . . . . . . 18

1.5.1 Solution for a rectangular simply supported plate loaded by a thermal field 22

2 Thin plates with large deflection - Buckling instability and response 25

2.1 General equations of isotropic plate with large deflection . . . . . . . . . . . . . . 25

2.1.1 Kinematic relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.1.2 Constitutive law for linearly isotropic materials . . . . . . . . . . . . . . . 26

2.1.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1.3.1 Rotations around the x and y axes . . . . . . . . . . . . . . . . . 26

i

2.1.3.2 Translations along x and y directions . . . . . . . . . . . . . . . 27

2.1.3.3 Translations along z direction . . . . . . . . . . . . . . . . . . . 28

2.1.3.4 Final Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Analysis of a two-edge buckled rectangular plate . . . . . . . . . . . . . . . . . . 30

2.2.1 Buckling stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.2 Response of a buckled plate to a static pressure load . . . . . . . . . . . 36

2.2.3 Response of a buckled plate having an initial deflection . . . . . . . . . . 37

2.3 Some general issues on buckling analysis for space-discretized structural models . 38

3 Shells with membrane behavior 41

3.1 General issues on membrane v.s. bending behavior of shell structures . . . . . . . 41

3.2 Axial symmetric shells undergone to axial symmetric loads . . . . . . . . . . . . 44

3.2.1 Spherical pressured shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.2.2 Conical shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2.3 Cylindrical shell of circular section . . . . . . . . . . . . . . . . . . . . . . 50

3.2.4 Spherical dome loaded by its own weight . . . . . . . . . . . . . . . . . . . 51

3.2.5 Open dome at the top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.2.6 Pressure Tanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.2.7 Toroidal tank with circular section with internal pressure∗ . . . . . . . . . 59

3.3 Axial-symmetric shells undergone to not axial-symmetric load . . . . . . . . . . . 61

3.3.1 Cylindrical shell of circular section . . . . . . . . . . . . . . . . . . . . . . 63

3.3.1.1 Cylindrical tank . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.3.2 Wind load on a spherical shell of revolution∗ . . . . . . . . . . . . . . . . 67

4 Shells with bending behavior 69

4.1 Curvature effects on bending stresses and strains . . . . . . . . . . . . . . . . . . 69

4.2 Circular cylindrical shell with axially symmetric load . . . . . . . . . . . . . . . . 73

4.2.1 Example of a cylindrical shell with axisymmetric load . . . . . . . . . . . 76

ii

4.2.2 Cylindrical shell subjected to forces and moments at the edges . . . . . . 79

4.2.3 Thermal stresses in cylindrical shells . . . . . . . . . . . . . . . . . . . . . 81

4.2.4 Pressurized cylindrical shell with a hemispherical end cap∗ . . . . . . . . 81

4.2.5 Pressurized cylindrical shell with end cap ellipsoid∗ . . . . . . . . . . . . . 84

4.3 Cylindrical bending shell with not axial symmetric load∗ . . . . . . . . . . . . . . 85

4.4 Buckling analysis for a cylindrical shell with circular transversal section . . . . . 88

4.4.1 Effect of pressurization∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

IV Structural Optimization 93

5 Some issues of Non-Linear Programming for structural optimization 95

5.1 Nonlinear programming: the statement of the problem . . . . . . . . . . . . . . . 95

5.2 Necessary conditions for a local solution of the problem of non-linear programming 96

5.3 Conditions for global optimal solution. Convex sets and convex functions . . . . 102

5.4 Some applications of Karush Khun Tucker conditions . . . . . . . . . . . . . . . . 104

5.4.1 Example 1∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.4.2 Example 2∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.4.3 Example 3∗ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5.4.4 An example of structural optimization of a wing-like-box structure . . . . 107

5.5 Deterministic numerical approaches for Non-Linear Programming problems . . . 110

5.5.1 Gradient method - Steepest descent . . . . . . . . . . . . . . . . . . . . . 110

5.5.2 Iterative process based on Newton’s method . . . . . . . . . . . . . . . . . 111

5.5.3 Examples of applications of the gradient method and Newton’s method∗ . 112

5.6 Nonlinear quadratic programming∗ . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6 Multi-objective optimization∗ 115

6.1 Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

6.2 State space and objective space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

iii

6.3 Efficiency and dominance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.4 Utopian point and nadir point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.5 Pareto optimality: local and global optimum . . . . . . . . . . . . . . . . . . . . 119

6.6 Pareto frontier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6.6.1 Numerical example: wing-box structure . . . . . . . . . . . . . . . . . . . 122

Bibliografia 124

iv

Part III

Plates and Shells – BucklingInstability

1

Chapter 1

Thin plates with small deflections

Plates can be defined as planar structural elements whose geometry is determined by the mean

plane, that can be divided in three categories:

� thin plates with small deflections

� thin plates with large deflections

� thick plates

Generally speaking, a plate is considered thin if the ratio between the thickness and the lowest

characteristic dimension is lower than tl ≤ 1

20 . Let us consider a plate like that in Figure 1.1,

with the main plane coinciding with the one identified by the co-ordinate plane with z = 0.

The components of the displacement vector are indicated with u, v and w. The fundamental

Figure 1.1: Reference plate geometry

hypotheses used in the theory of small deflections of thin plates are based on the so-called

Kirchhoff hypotheses: they are similar to those taken for the basic flexible beam theory; they

are summarized as follows:

� the deflection w in the mean plane of the plate is small with respect to the thickness of

the plate, even its derivative ∂w∂x or ∂w

∂y is thus small and the square power of the derivative

3

4

is negligible with respect to one;

� the membrane deformation of the mean plane is assumed to be zero;

� any straight material line normal to the plate mean plane before bending remains (1)

straight, (2) inextensible, and (3) normal to the beam axis after deformation. This implies

(see next Section) as a consequence that the shear strain components εxz, εyz, and εzz are

zero and the deflection of the plate is mainly due to the flexural deformations.

� the stress component σzz, orthogonal to the mean plane, is small with respect to the others

components of the stress tensor and so can be neglected.

If the deflection is not small, together with a vertical displacement of the plate, an in-plane

deformation of the mean plane may occur, the first two hypotheses would not be satisfied any

more and the theory of large deflection should be considered. (see Chapter 2).

Whether the plates are thick, the out-of-plane shear deformation would become important and

the last two hypotheses would not be valid. In this and a more general theory must be used.

1.1 Isotropic bending plates

1.1.1 Kinematic relationships

For a thin flat plate having a mean plane coincident with the z = 0 o-ordinate plane in a

Carthesian frame of reference, one can assume for the displacement solution the following Taylor

expansion in the z−direction

u(x, y, z) = u(x, y, 0) +∂u

∂z

∣∣∣∣z=0

z +O(z2)

v(x, y, z) = v(x, y, 0) +∂v

∂z

∣∣∣∣z=0

z +O(z2) (1.1)

w(x, y, z) = w(x, y, 0) +∂w

∂z

∣∣∣∣z=0

z +O(z2)

Considering the hypothesis previously mentioned of strain plane, i.e.,

εxz = 0 −→ ∂u

∂z= −∂w

∂xεyz = 0 −→ ∂v

∂z= −∂w

∂yεzz = 0 −→ ∂w

∂z= 0 (1.2)

one has

u(x, y, z) = u0(x, y)−∂w(x, y)

∂xz +O(z2) (1.3)

v(x, y, z) = v0(x, y)−∂w(x, y)

∂yz +O(z2) (1.4)

w(x, y, z) = w0(x, y) +O(z2)

5

where u0(x, y) and v0(x, y) stand for the membrane displacement of the mean plane of the plate

which, considering the initial hypothesis taken for the flexural problem, must be zero. Thus:

avoiding from now on the subscript ”0” for indicating mean plane quantitities, one has

u(x, y, z) = −z∂w(x, y)

∂x

v(x, y, z) = −z∂w(x, y)

∂y

w(x, y, z) = w(x, y)

the displacements u, v can be written in terms of vertical displacement w; replacing these

expressions of the displacements in the kinematic relations, one obtains:

εxx(x, y, z) =∂u(x, y)

∂x= −z

∂2w(x, y)

∂x2

εyy(x, y, z) =∂v(x, y)

∂y= −z

∂2w(x, y)

∂y2

γxy(x, y, z) =∂u(x, y)

∂y+

∂v(x, y)

∂x= −2z

∂w(x, y)

∂x∂y

these equations express the deformations in every point of the plate in terms of the vertical

displacement w(x, y).

1.1.2 Stress description and constitutive relationships

The constitutive equations, in the case of isotropic material, for a flexible plate with the condi-

tions:

εxz = εyz = εzz = 0

become:

σxx(x, y, z) =E

1− ν2(εxx + νεyy)

σyy(x, y, z) =E

1− ν2(εyy + νεxx)

σxy(x, y, z) = G γxy =E

2(1 + ν)γxy

using the above kinematic relations, these become:

σxx(x, y, z) = − E z

1− ν2

(∂2w(x, y)

∂x2+ ν

∂2w(x, y)

∂y2

)σyy(x, y, z) = − E z

1− ν2

(∂2w(x, y)

∂y2+ ν

∂2w(x, y)

∂x2

)σxy(x, y, z) = − E z

1 + ν

(∂w(x, y)

∂x∂y

)

6

It can be pointed out that stresses varies linearly inside the thickness of the plate and that they

are zero at the mean plane of the plate. These stresses bring to moments and shear forces per

unit length given by:1

Mx =

∫ t2

− t2

zσyy dz

My =

∫ t2

− t2

zσxx dz

Mxy =

∫ t2

− t2

zσxy dz

Qx =

∫ t2

− t2

σxz dz

Qy =

∫ t2

− t2

σyz dz

By expressing the stress in terms of vertical displacement w(x, y) one gets:

Mx = −D

(∂2w

∂y2+ ν

∂2w

∂x2

)(1.5)

My = −D

(∂2w

∂x2+ ν

∂2w

∂y2

)(1.6)

Mxy = −D (1− ν)∂2w

∂x∂y(1.7)

where D stands for the flexural stiffness of the plate given by the:

D =E t3

12 (1− ν2)

1.1.3 Equilibrium equations

When the plate is loaded the components of the stress tensor vary from point to point af the

plate itself and these variations are determined by the equilibrium equations.

Let us consider, as indicated in Figure 1.2 a dx dy element of plate with a load per unit surface

p(x, y). Moments and shear forces present some increases along dx or dy as shown in Figure 1.2.

From the equilibrium conditions with respect to the forces along the z axis one gets:

pdxdy −Qxdy −Qydx+

(Qx +

∂Qx

∂xdx

)dy +

(Qy +

∂Qy

∂ydy

)dx = 0

1Note that stress components σxz and σyz are generally present in this model although it is supposed that thematerial is not able to induce any consequent shear strain εxz and εyz.

7

Figure 1.2: Balance on the plate infinitesimal element

i.e., including also the inertial force,

∂Qx

∂x+

∂Qy

∂y+ p− ϱtw = 0

where w :=∂2w

∂t2.

From the condition of the moments equilibrium around the x axis one obtains:

∂Mxy

∂xdxdy +

∂Mx

∂ydxdy −Qydxdy = 0

hence:

∂Mxy

∂x+

∂Mx

∂y−Qy = 0 (1.8)

from the moments equilibrium condition around the y axis one gets:

∂Mxy

∂y+

∂My

∂x−Qx = 0 (1.9)

In the end combining these three equations together one gets:

∂2My

∂x2+ 2

∂2Mxy

∂x∂y+

∂2Mx

∂y2= −p(x, y) + ϱt w

8

which is the equilibrium equation for the flexural problem for thin plates.

Note that using Eqs. 1.8 and 1.9 together with 1.5, 1.6, and 1.7 one has for Qx, Qy

Qx = −D∂

∂x

(∂2w

∂x2+

∂2w

∂y2

)Qy = −D

∂

∂y

(∂2w

∂x2+

∂2w

∂y2

)Finally, the equilibrium can be expressed in terms of the displacement, using the constitutive

relationships given by 1.5, 1.6, and 1.7 in the homogeneous case

D

(∂4w

∂x4+ 2

∂4w

∂x2∂y2+

∂4w

∂y4

)+ ϱt w = p

which can be written as:

D∇4w + ϱt w = p(x, y; t)

in order to get the deflection w(x, y) one must integrate this equation in the field with the

boundary conditions of the problem considered.

1.1.4 Boundary conditions

A relevant point related to the boundary condition for a plate is given by the evaluation of the

global contribution in terms of vertical stress-force per unit length to be eventually imposed

to be zero in the case, for example, of free edge condition. Indeed, the difficulty is due to the

presence and contribution of the torsional moments Mxy.

If one consider, for example, a generic edge at x = const, the global contribution of the stress-

force per unit length is given by Qx but also a contribution of the local increment of Mxy along

the y direction has to be taken into account. Indeed, the local increment of this contribution

along the edge of an infinitesimal quantity dy is given by∂Mxy

∂ydy. This local incremental

contribution of moment is apparently given by an elementary vertical force∂Mxy

∂ymultiplied by

dy. Therefore, the global contribution in terms of vertical stress-force per unit length indicated

by Vx is given by

Vx = Qx +∂Mxy

∂y= −D

∂

∂x

(∂2w

∂x2+

∂2w

∂y2

)−D (1− ν)

∂3w

∂x∂y2

or

Vx = −D

[∂3w

∂x3+ (2− ν)

∂3w

∂x∂y2

]

9

In the same way:

Vy = Qy +∂Mxy

∂x= −D

[∂3w

∂y3+ (2− ν)

∂3w

∂x2∂y

]These relations are due to Kirchhoff: they set that the distribution of the torsional moments on

the free extreme is statically equivalent to a distribution of vertical forces.2 Now, some boundary

conditions can be considered in the following. In all cases reference is made to a rectangular

plate with its sides parallel to x and y axes and different boundary conditions are placed on the

x = a side.

1. Clamped edge

The boundary conditions set the deflection and its derivative with respect to x to zero:

x = 0, x = a →

w = 0

∂w

∂x= 0

2. Simply-supported edge

In this case the boundary condition set the deflection and the Mx moment to zero:

x = 0, x = a →

w = 0

My = −D

(∂2w

∂x2+ ν

∂2w

∂y2

)= 0

the first condition on the deflection on the x = a side sets to zero the first and second

derivatives with respect to y, so the condition My = 0 becomes:

∂2w

∂x2= 0

3. Free edge

It means that moments and shear forces must be set to zero:

x = 0, x = a →

∂2w

∂x2+ ν

∂2w

∂y2= 0

∂3w

∂x3+ (2− ν)

∂3w

∂x∂y2= 0

the first expresses the Mx = 0 condition while the latter set the shear force to zero: Vx = 0

2If one applies the previous considerations to the case of to two free edges converging into a corner, one shouldhave at the corner point e concentrated load give by

Fc = 2Mxy = −2D (1− ν)∂2w

∂x∂y

10

4. Sliding edge

Here the condition is to set free deflection but bound rotation, moreover the constraint is

not able to suffer shear forces, so the conditions are:

x = 0, x = a →

∂w

∂x= 0

Vx = −D

[∂3w

∂x3+ (2− ν)

∂3w

∂x∂y2

]= 0

From the previous examples, it can be noted out that the boundary conditions can be of two

kinds:

a geometric boundary conditions (or kinematic), these are the ones relative to displacement and

its first derivatives;

b natural boundary conditions these are those one relative to forces and moments.

It is noticed that in the case of a locked side there will be only geometric conditions, in the case

of the free side there will be only natural boundary conditions and in the other two cases there

are mixed conditions.

Finally, it is also noticed that homogeneous boundary conditions have been adopted, but more

generally, non homogeneous boundary conditions may occour for the geometric or natural cases

as well.

1.2 Orthotropic bending plates

Let us consider a flexible plate made by ortotropic material, a modification in the constitutive

equations is needed.

For an orthotropic material the relations between stresses and deformations, are expressed in

terms of four independent constants by the following relations:

σxx =1

1− νLT νTL(ELεxx + νTLELεyy)

σyy =1

1− νLT νTL(νLTET εxx + ET εyy) (1.10)

σxy = GLTγxy

where the subscript L means quantity along the longitudinal direction (or fiber, for uni-directional

composite material) and T transversal one. Note that the further simmetry condition νTLEL =

11

νLTET reduce the number of the above quantities to be four. For the sake of simplicity the

previous equations can be rewritten as

σxx = Exεxx + Exyεyy

σyy = Exyεxx + Eyεyy (1.11)

σxy = GLTγxy

with Ex :=EL

1− νLT νTL, Ey :=

ET

1− νLT νTL, and Exy :=

νTLEL

1− νLT νTL=

νLTET

1− νLT νTL.

The kinematic relations are independent from the elastic characteristic of the material and so

they are written in the form:

εxx = −z∂2w

∂x2

εyy = −z∂2w

∂y2

γxy = −2z∂2w

∂x∂y

Taking these kinematic relations and imposing them inside the consitutive laws 1.11, one has:

σxx = −z

(Ex

∂2w

∂x2+ Exy

∂2w

∂y2

)σyy = −z

(Exy

∂2w

∂x2+ Ey

∂2w

∂y2

)σxy = −2zGLT

∂2w

∂x∂y

Imposing the stress expressions in the moments definitions:

My =

∫ t2

− t2

σxxzdz

Mx =

∫ t2

− t2

σyyzdz

Mxy =

∫ t2

− t2

σxyzdz

and integrating one obtains:

My = −(Dx

∂2w

∂x2+Dxy

∂2w

∂y2

)Mx = −

(Dxy

∂2w

∂x2+Dy

∂2w

∂y2

)Mxy = −2Gxy

∂2w

∂x∂y

12

with:

Dx =Ext

3

12; Dy =

Eyt3

12

Dxy =Exyt

3

12; Gxy =

GLT t3

12

these positions define the flexural stiffnesses, Dx, Dy and Dxy, and the torsional stiffness Gxy of

an orthotropic plate. Imposing the moments expressions in the equilibrium equations:

∂Mxy

∂x+

∂Mx

∂y−Qy = 0

∂My

∂x+

∂Mxy

∂y−Qx = 0

thus:

Qx = − ∂∂x

(Dx

∂2w∂x2 +H ∂2w

∂y2

)Qy = − ∂

∂y

(Dy

∂2w∂y2

+H ∂2w∂x2

)where:

H = Dxy + 2Gxy

Starting from the equilibrium equation in terms of moments including inertial force:

∂2My

∂x2+ 2

∂2Mxy

∂x∂y+

∂2Mx

∂y2= −p(x, y) + ϱtw (1.12)

and from the moments expressions in terms of deflection w(x, y) one gets the equation for the

flexion of an orthotropic plate:

Dx∂4w

∂x4+ 2H

∂4w

∂x2∂y2+Dy

∂4w

∂y4+ ϱtw = p(x, y; t) (1.13)

1.3 Isotropic laminated plates

Let us consider a plate made by a laminate, generally speaking every lamina can have different

geometric and elastic properties such as thickness and elastic parameters; for simplicity let us

consider the case of a plate made by several isotropic laminae, Fig. 1.3. The kinematic relations

must be written with respect to every ith layer:

ε(i)xx = −zi∂2w

∂x2

ε(i)yy = −zi∂2w

∂y2

γ(i)xy = −2zi∂2w

∂x∂y

13

Figure 1.3: Geometry of an isotropic laminated plate

and the constitutive laws of isotropic layers for the ith layer are:

σ(i)xx =

Ei

1− ν2i

[ε(i)xx + νiε

(i)yy

]σ(i)yy =

Ei

1− ν2i

[ε(i)xx + νiε

(i)yy

]σ(i)xy =

Ei

2 (1 + νi)γ(i)xy

Therefore, for the ith layer the stress tensor components can be written in terms of the w(x, y):

σ(i)xx = −zi

Ei

1− ν2i

(∂2w

∂x2+ νi

∂2w

∂y2

)σ(i)yy = −zi

Ei

1− ν2i

(∂2w

∂y2+ νi

∂2w

∂x2

)σ(i)xy = −zi

Ei

1 + νi

∂2w

∂x∂y

Note that the equilibrium equation in terms of moments are the same as in the homogeneous

case, equation 1.12; for the plate equation on gets:

Dt∇4w = p(x, y)

where with Dt is the equivalent flexural stiffness of the laminate plate. It can be obtained

considering the previous definitions and the laminated z−symmetry

My =

∫ t/2

−t/2zσxxdz = −2

n∑i=1

∫ ti

ti+1

z2Ei

1− ν2i

(∂2w

∂x2+ νi

∂2w

∂y2

)dz − 2

∫ tn+1

0

z2En+1

1− ν2n+1

(∂2w

∂x2+ νn+1

∂2w

∂y2

)dz

≃ −

(2

n∑i=1

z3

3

∣∣∣∣titi+1

Ei

1− ν2i+ 2

z3

3

∣∣∣∣tn+1

0

En+1

1− ν2n+1

)(∂2w

∂x2+ ν

∂2w

∂y2

)= −Dt

(∂2w

∂x2+ ν

∂2w

∂y2

)

14

where

Dt =2

3

[n∑

i=1

Ei

1− ν2i

(t3i − t3i+1

)+

En+1

1− ν2n+1

t3n+1

]

Note that a averaged value ν for the coefficients νi has been assumed for combining the partial

derivatives of w since this values are very close to be equal to 0.3 for most of the isotropic

materials. Moreover, the laminate is assumed symmetric, the last n + 1 lamina is assumed to

be the central one and all the other are numbered, in one side, starting by the more external.

Note that, once set the boundary conditions, the applied load p(x, y) and the elastic and geomet-

ric characteristics of every single layer, the model is quite similar to that for the homogeneous

case.

1.4 Analytic solutions with eigenfunctions methods

In this Section some analytic solutions for statics plate structures are proposed using the eigen-

functions approach.

1.4.1 Rectangular simply-supported isotropic plate

Let us consider a rectangular plate, with a and b edges supported on the border, stressed with

a distributed load shown in Figure 1.4.

Figure 1.4: Geometry of simply-supported rectangular isotropic plate

In 1820 Navier proposed the solution for this problem based on an expansion with Fourier series

15

of the load applied and of the unknown function w(x, y) by posing:

p(x, y) =∞∑

m=1

∞∑n=1

pmn sinmπ

ax sin

nπ

by

(1.14)

w(x, y) =

∞∑m=1

∞∑n=1

wmn sinmπ

ax sin

nπ

by

where pmn and amn are the known and unknown coefficients, respectively, to be determined.

Note also that the basis functions

ϕmn(x, y) = sinmπ

ax sin

nπ

by

are eigenfunctions of the structural operator of the plate simply supported in a Cartesian rect-

angular domain as it will be shown later. Note that, since by eigenfunctions definition one has

Lϕmn = λmnϕ, then in the present case

λmn = D

[(mπ

a

)2+(nπ

b

)2]2The support condition on the border becomes:

x = 0, x = a →

w = 0

∂2w

∂x2= 0

y = 0, y = b →

w = 0

∂2w

∂y2= 0

The evaluation of the coefficients pmn of the loaded condition is obtained proceeding a la Fourier

so multiplying for sin mπa x sin nπ

b and integrating on the plate:

⟨p(x, y), ϕmn⟩ = pmn

∫ b

0

∫ a

0sin2

mπ

ax sin2

nπ

by dxdy = pmn∥ϕmn(x, y)∥2

with ∥ϕmn(x, y)∥2 = ab/4, one gets for load components:

pmn =4

ab

∫ b

0

∫ a

0p(x, y) sin

mπ

ax sin

nπ

by dxdy (1.15)

Thus, the coefficients of the deflection wmn are obtained by applying the eigenfunction method

for statics, or

wmn =pmn

λmn

16

This implies that

wmn =pmn

Dπ4(m2

a2+ n2

b2

)2Replacing the wmn coefficient in the displacement expansion, Eq. 1.14 one gets:

w(x, y) =

∞∑m=1

∞∑n=1

pmn

λmnϕ(x, y) =

1

π4D

∞∑m=1

∞∑n=1

pmn[(ma

)2+(nb

)2]2 sin mπ

ax sin

nπ

by

where the coefficients pmn are determined a la Fourier with respect to the load applied, Eq.

1.15.

If the load is equally distributed on the structure:

p(x, y) = p0

one gets:

pmn =16p0π2mn

; m,n = 1, 3, 5, · · ·

these coefficients are zero for even m or n indexes and imposing this expression in the expansion

which gives the deflection, one gets:

w(x, y) =16p0π6D

∞∑m=1

∞∑n=1

1

mn

sin mπa x sin nπ

b y[(ma

)2+(nb

)2]2 ; m,n = 1, 3, 5, · · ·

From the deflection it is possible to evaluate the series expansion related to the moments Mx,

Mxy, My. For example, one gets:

My =16p0π6D

∞∑m=1

∞∑n=1

(ma

)2+ ν

(nb

)2mn

[(ma

)2+(nb

)2]2 sin mπ

ax sin

nπ

by

and does it for Mx and Mxy. It can be noted out that Mx and My are respectively zero on the

edges for x = 0, x = a, y = o, y = b, while the torsional moment does not vanish on the border

of the plate.

In the case of square plate, a = b = a∗, supported on the border and with equally distributed

load, considering only the first terms of the series expansion of the deflection and referring to

the center of the plate, one gets:

w

(a∗

2,a∗

2

)= 0.00416

p0a∗4

D

17

otherwise if the first four terms of the series expansion are considered, one has:

w

(a∗

2,a∗

2

)= 0.00406

p0a∗4

D

(m = 1, n = 1, 3m = 3, n = 1, 3

)with a very slight difference between the two approximations. However if the values of the

moments is considered using only the first terms of the series expansion, one gets:

Mxmax = Mymax = 0.0534 p0a∗2

while if the first four terms of the series expansion are considered, one gets:

Mx = My = 0.0469 p0a∗2

it can easily noted out that the convergence to the right value of the approximated moments is

slower than the case of the deflection.

1.4.2 Rectangular simply-supported orthotropic plate

The eigenfunction (or Navier for this specific plate appication) approach used for the problem of

the isotropic plate can be used also for the case of orthotropic plate. Referring to a rectangular

plate with a and b long edges charged with a load per unit surface p(x, y) and with simple support

constraint boundary conditions, it is possible to look for the solution with the expansion used

in the 1.14, which once put in the plate equation 1.13, gives:

∞∑m=1

∞∑n=1

[wmn

(m4π4

a4Dx + 2

m2n2π4

a2b2H +

n4π4

b4Dy

)− pmn

]sin

mπ

ax sin

nπ

by = 0

This relation must be satisfied for every x, y and so if one set the terms in the squared brackets

to zero then gets the relation between the unknown coefficients of the w(x, y) expansion to those,

this time known, of the expansion of the given load:

wmn =pmn(

m4π4

a4

)Dx + 2

(m2n2π4

a2b2

)H +

(n4π4

b4

)Dy

the coefficient of the expansion of the given load follow from the:

pmn =4

ab

∫ b

0

∫ a

0p(x, y) sin

mπ

ax sin

nπ

by dxdy

for the expansion of the unkown function w(x, y) one gets:

w(x, y) =4

ab

∞∑m=1

∞∑n=1

∫ b

0

∫ a

0

p(x, y) sin mπa x sin nπ

b y(m4π4

a4

)Dx + 2

(m2n2π4

a2b2

)H +

(n4π4

b4

)Dy

dxdy

sinmπ

ax sin

nπ

by

18

in the particular case in which the given load is equally distributed on the plate surface p(x, y) =

p0 one gets the previously used expression for the expansion coefficients:

pmn =16p0π2mn

; m,n odd

and so one gets the following expression for the vertical displacement:

w(x, y) =16p0π6

∞∑m=1

∞∑n=1

sin mπa x sin nπ

b y

mn[(

m4

a4

)Dx + 2

(m2n2

a2b2

)H +

(n4

b4

)Dy

]once the w(x, y) is evaluated, it is possible to derive the moments and all the stresses.

1.5 Some issues on thermal stresses for plates

The presence of non-uniform thermal field in a plates requires a new set of relations between

stress and deformation and these can be obtained by adding the deformation due to thermal

loads to those ones due to mechanical loads.

In the case of an isotropic and homogeneous material suffering a variation of temperature:

∆Θ = Θ−Θ0

in which Θ0 stands for a reference temperature, this causes a deformation in every direction and

takes the tensor form:

ET = α ∆Θ I

i.e., in terms of strain axial components (the only not equal to zero)

εT = α ∆Θ

where α, coefficient of thermal expansion, is a characteristic property of the material; α depends

on temperature, but for limited variations of temperature it can be considered constant.

Obviously the elastic properties of a material ( E and ν in the case of mechanically isotropic

one) also depend on the temperature, but still for limited variations of temperature, they can

be considered constant.

The direct effects of the temperature variation are deformations; to these a stress status can

follow if the original thermal deformations are prevented, for example, from the constraint

conditions on the structure. The stresses due to thermal load are indicated with thermal stresses.

In the case of a free plate, made by isotropic material, uniformly heated, the thermal effect

on the plate is given by deformations but not by stresses; if the plate suffers a non-uniform

temperature field, generally speaking there would be thermal stresses too.

19

The boundary conditions could be such to prevent thermal deformations and so one could have

thermal stresses even if the temperature field suffered by an isotropic made plate is uniform.

In the general case of a plate made by an anisotropic material there could be thermal stresses

even in the case of a uniform temperature field for a free plate.

Considering the deformation due to the presence of a thermal field, in the case of isotropic

material, one gets the following relations:

εxx =1

E(σxx − νσyy) + α ∆Θ

εyy =1

E(σyy − νσxx) + α ∆Θ

γxy =σxyG

these relations can be recast in a stiffness form as

σxx =E

(1− ν2)[εxx + νεyy − (1 + ν)α∆Θ]

σyy =E

(1− ν2)[εyy + νεxx − (1 + ν)α∆Θ]

σxy = Gγxy

For the hypotheses of the plate theory generally including membrane effects (see Chapter 2) one

has

u(x, y, z) = u0(x, y)− z∂w

∂x

v(x, y, z) = v0(x, y)− z∂w

∂y

w(x, y, z) = w0(x, y)

one can obtain the following expressions for the deformations field:

εxx =∂u0∂x

− z∂2w

∂x2

εyy =∂v0∂y

− z∂2w

∂y2

γxy =

(∂u0∂x

+∂v0∂y

)− 2z

∂2w

∂x∂y

stresses can be expressed in terms of deformations:

σxx =E

1− ν2

[∂u0∂x

+ ν∂v0∂y

− z

(∂2w

∂x2+ ν

∂2w

∂y2

)− (1 + ν)α∆Θ

]σyy =

E

1− ν2

[∂v0∂y

+ ν∂u0∂x

− z

(∂2w

∂y2+ ν

∂2w

∂x2

)− (1 + ν)α∆Θ

]σxy = G

[∂u0∂y

+∂v0∂x

− 2z∂2w

∂x∂y

]

20

from these stress expressions one can get the resultant in terms of forces per unit length, Nx,

Ny, Nxy and moments per unit length Mx, Mx, Mxy. Respectively

Nx =Et

1− ν2

(∂u0∂x

+ ν∂v0∂y

)− N∗

1− ν

Ny =Et

1− ν2

(∂v0∂y

+ ν∂u0∂x

)− N∗

1− ν(1.16)

Nxy =Et

2(1− ν)

(∂u0∂y

+ ν∂v0∂x

)with:

N∗ = αE

∫ t2

− t2

∆Θ(z)dz

and

My = −D

(∂2w

∂x2+ ν

∂2w

∂y2

)− M∗

1− ν

Mx = −D

(∂2w

∂y2+ ν

∂2w

∂x2

)− M∗

1− ν(1.17)

Mxy = −(1− ν)D∂2w

∂x∂y

with:

M∗ = αE

∫ t2

− t2

∆Θ(z)zdz

Note that N∗ and M∗ define those contributions due to thermal effects.

From the equilibrium in transversal direction and not including the membrane effects one has

∂2My

∂x2+ 2

∂2Mxy

∂x∂y+

∂2Mx

∂y2= −p(x, y)

one gets:

D∂4w

∂x4+ νD

∂4w

∂x2∂y2+ 2(1− ν)D

∂4w

∂x2∂y2+D

∂4w

∂y4+ νD

∂4w

∂x2∂y2+

1

(1− ν)∇2M∗ = p

Thus:

D

(∂4w

∂x4+

∂4w

∂y4

)+ 2D

∂4w

∂x2∂y2= p− 1

(1− ν)∇2M∗

and therefore

D∇4w = p− 1

(1− ν)∇2M∗ = p+ p∗ (1.18)

21

where p∗ indicates the equivalent load due to the effect of thermal field:

p∗ = − 1

(1− ν)∇2M∗

the effect due to the thermal field is added to that of the load p(x, y).

The solution of a thermo-elastic problem is obtained by the equation 1.18 on the deflection

w(x, y) and other equations on the forces lying on the plane for the membrane problem given by

Eqs. 2.11 and 2.14 together with Eq. 1.16. The solutions to these two problems can be obtained

independently and the stress can be added together. Between these two elastic problems, those

defined by the equation 1.18 is characteristic of the plate theory while the other is a problem of

membrane theory.

For what it concerns the effect of the thermal field on the boundary conditions, one must consider

the expressions obtained for the moments (Eq. 1.17) and forces; from those one on gets for the

edges with x = const:

My = −D

(∂2w

∂x2+ ν

∂2w

∂y2

)− M∗

(1− ν)

Vx = −D

[∂3w

∂x3+ (2− ν)

∂3w

∂x∂y2

]− 1

(1− ν)

∂M∗

∂x

Therefore, for the tipical conditions still on the x = a side, one gets:

(i) cantilevered edge (x = const)

w = 0∂w

∂x= 0

(ii) simply supported edge (x = const)

w = 0

My = D∂2w

∂x2+

M∗

(1− ν)= 0

(iii) free edge (x = const)

My = 0

Vx = 0

22

or

D

(∂2w

∂x2+ ν

∂2w

∂y2

)= − M∗

(1− ν)

D

[∂3w

∂3x+ (2− ν)

∂3w

∂x∂2y

]= − 1

(1− ν)

∂M∗

∂x

Note that some boundary conditions become non homogeneous and the term due to the thermal

effects appears in the static conditions.

1.5.1 Solution for a rectangular simply supported plate loaded by a thermalfield

It is considered a simply supported rectangular plate with a non-uniform temperature distribu-

tion. The boundary conditions are:

x = 0, x = a w = 0, My = 0

y = 0, y = b w = 0, Mx = 0

from the expression of the moments and by this boundary conditions one can write an equation

which is effective on the plate boundary:

w = 0 D∇2w = − M∗

(1− ν)(1.19)

Moreover, from the equation of the plate with thermal load, once put equal to zero the p(x, y)

load, one can get the following relation on the whole field:

D∇4w = − ∇2M∗

(1− ν)

Thus, this field equation can be rewritten, considering also its natural boundary conditions, as: D∇2w +M∗

(1− ν)=: f(x, y) = 0

∇2f = 0(1.20)

Thus, since the function f(x, y) must be equal to zero on the plate boundary and the Laplace

operator be equal to zero in the plate domain, this necessarily implies f(x, y) = 0 in the plate

domain. The first of the 1.20 becomes:

D∇2w +M∗

(1− ν)= 0

with the boundary conditions:

x = 0, x = a w = 0

y = 0, y = b w = 0

23

The problem so defined can be solved with the eigenfunction approach by setting:

M∗[x, y] =

∞∑m=1

∞∑n=1

Mmn sinmπx

asin

nπy

b

w(x, y) =∞∑

m=1

∞∑n=1

wmn sinmπx

asin

nπy

b

where the Mmn are coefficients to be found by projection from:

Mmn =4

ab

∫AM∗(x, y) sin

mπx

asin

nπy

bdA

Thus, by the eigenfunction approach one obtains the relation between the unknown coefficients

wmn and those already known Mmn by the algebraic relationship

wmn =Mmn

(1− ν)π2D[(

ma

)2+(nb

)2]By using this equation in the expansion of the unknown function w(x, y), one obtains:

w(x, y) =∞∑

m=1

∞∑n=1

Mmn

(1− ν)π2D[(

ma

)2+(nb

)2] sin mπx

asin

nπy

b

which represents the deflection due to the thermal load defined by M∗(x, y). In the case in

which heating is such that the temperature varies only along the thickness of the plate, but is

constant in the plane, one gets:

M∗(x, y) = αE

∫ t2

− t2

∆Θ(z)zdz = cost = M∗

and thus one can obtain for the expression of the displacement:

Mmn =16M∗

π2mn(for odd m, n)

w(x, y) =16M∗

(1− ν)Dπ4

∞∑m=1

∞∑n=1

sin mπxa sin nπy

b

mn

[(ma

)2+

(n

b

2)] (m = n = 1, 3, ....)

24