Plates - Mechanics of Structures: Variational and Computational Methods - page 767

FIGURE 13.5
Positive stress resultants.

Substitute the kinematic relations of Eq. (13.22) and the material law of Eq. (13.25) into
the stress resultants of Eq. (13.27). We find, for example,

t
2

t
2

t
2

E E

m x =

2 σ x zdz

=

2 ( x + ν y )

zdz =

2 (

z κ x + ν z κ y ) zdz

1 − ν 1 − ν

t t

2

t
2

− − −

t
2

Et 3

E

z 2 =

2 (κ

+ νκ ) dz =

) (κ

+ νκ ) (13.28)

x y x y

1 − ν 12 ( 1 − ν

2

−

t
2

In matrix notation, the relations for the moments are

m x
m y
m xy

1 ν 0 κ x

κ y
2

Et 3

=

ν 10

00 1 − ν
2

(13.29)

12 ( 1 − ν

2

)

κ xy

13.2.3 Conditions of Equilibrium

To derive the dif fe rential equations of equilibrium, consider an element tdxdy subject to
an applied load p z (Fig. 13.6). The condition that the sum of the vertical forces is equal to
zero gives

∂ q x

∂

+ ∂

q y
∂

dx dy dx dy + p z dx dy = 0 (13.30a)

x y

or

∂ q x

∂ x + ∂

q y
∂

y +

p z = 0

The summation of moments about the x axis set to zero yields

∂ m xy

∂

+ ∂

m y
∂

dx dy dx dy − q y dx dy = 0 (13.30b)

x y

or

∂ m xy

∂

+ ∂

m y
∂

y −

q y = 0

x

Similarly, the equilibrium of moments about the y axis provides the relationship

∂ m yx

∂

+ ∂

m x
∂

x −

q x = 0 (13.30c)

y