Plates - Mechanics of Structures: Variational and Computational Methods - page 772

Then the shear forces are

= ∂

M
∂

= ∂

M
∂

q x q y (13.43)

x y

Finally, the fourth order governing relationship of Eq. (13.41) is replaced by two equations
of similar structure

2 M 2 2

∇ = ∂

x M

+ ∂

y M

=− p z

(13.44)

2 2

x

2
y

∇ w = ∂ w + ∂ w =− M / K

Stresses
The stress components σ

x ,

σ

y , and

τ

xy of Eq. (13.38) were found using the material law.

The stress components

yz cannot be obtained from the material law, since they
cannot be related to the corresponding strains,

σ

z ,

τ

xz , and

τ

yz , which are neglected as a
result of the basic assumptions. However, the equations of equilibrium will perm it these
stresses to be determined. The three-dimensional equilibrium equation D T σ

z ,

γ

xz , and

γ

+

p V =

0

without body forces is (Chapter 1, Eq. 1.51)

∂ x σ x + ∂ y τ yx + ∂ z τ zx = 0

∂ x τ xy + ∂ y σ y + ∂ z τ zy = 0

(13.45)

∂ x τ xz + ∂ y τ yz + ∂ z σ z = 0

Upon integration, the first equation gives

τ xz =− (∂ x σ x + ∂ y τ xy ) dz + C

where C is an arbitrary integration constant. Equivalently,

z
a (∂ x σ x + ∂ y τ xy )

τ xz =− dz

where a is an arbitrary point along z and z is the coordinate of a point along the thickness
of the plate. Let a = t / 2, then using Eqs. (13.38) and (13.31a)

t / 2
z (∂

t / 2

t 3 t / 2

12 z
t 3 (∂

12

τ = σ + ∂ τ ) dz =

x m x

+ ∂

y m xy

) dz = zq x dz

xz x x y xy

z z

3 q x
2 t

2 ]

= [1 − ( 2 z / t ) (13.46a)

Similarly,

t / 2
z (∂ y σ y + ∂ x τ xy )

3 q y
2 t

2 ]

τ yz = dz = [1 − ( 2 z / t ) (13.46b)

Integration of the final relation of Eq. (13.45) leads to the distribution of the transverse
normal stress. From Eqs. (13.45), (13.46a), (13.46b), and (13.31a)

1 2 z

t

2 1 2 z

t

2 p z

3
2 t

3
2 t

∂ z σ z =− ∂ x τ zx − ∂ y τ zy =− − (∂ x q x + ∂ y q y ) = −