The Finite Difference Method - Mechanics of Structures: Variational and Computational Methods - page 522

where

1 ν 0

∂ 0

x

Et

u y ] T

ν 10

00 1 − ν
2

D = 0 ∂ E = and u = [ u x

y

1 − ν

2

∂ ∂

y x

in which u x and u y are the displacements in the x and y directions, and t is the thickness of
the plate. Expand the first term on the right-hand side of (2)

u T u D T ED u u dA

A δ

x u x

u y

dA

− ν
2

− ν
2

1 1

Et ∂∂ + ∂ ∂ ∂ν ∂ + ∂ ∂

x y y x y y x

= [ δ u x δ u y ] (3)

1 − ν

2 1 − ν

2

1 − ν

2

∂ν ∂ x + x ∂ ∂ y ∂∂ y + x ∂ ∂ x

A y

where x ∂ means that the derivative ∂/∂ x is taken of the variable to the left and ∂ x is the

usual derivative x of the variable to the right.

Employ the central finite difference scheme, representing the first derivatives of u x with
respect to x and y for the triangle corresponding to Fig. 8.11a as

∂/∂

− 1

1
0

∂ u x

∂

u x 2 − u x 1 1

h x =

= [ u x 1 u x 2 u x 3 ]

h

(4)

0

∂ u x

∂

u x 3 − u x 2 1

h

y =

= [ u x 1 u x 2 u x 3 ] − 1

1

h

Similar formulas apply for ∂ u y /∂ x and ∂ u y /∂ y . Substitute (4) and the formulas for ∂ u y /∂ x

and ∂ u y /∂ y into (3). This leads to

v T E ∗ k 11

v

k 12

u T u D T ED u u dA v T kv

A δ

= δ = δ

k 21 k 22

where A is the area of the triangle, E ∗ =

2

h 2 , and tE A /( 1 − ν )

u y 3 ] T v = [ u x 1 u x 2 u x 3 u y 1 u y 2

− 1

1
0

0

1 − ν

2

[ [0

k 11 = − 110] + − 1

1

− 11]

1 − 10

= − 1 γ − β

0 − ββ

− 1

1
0

0

1 − ν

2

ν [0 [

k 12 = − 11] + − 1

1

− 110]

(5)

ν − ν
β − βν
− ββ

0

=

0

0
−

− 1

1
0

1 − ν

2

ν [ [0

k 21 = 1

1

− 110] + − 11]