The Finite Difference Method - Mechanics of Structures: Variational and Computational Methods - page 523

β − β
ν − ββ
− ν

0

=

ν 0

0
−

− 1

1
0

1 − ν

2

[0 [

k 22 = 1

1

− 11] + − 110]

β − β 0

= − βγ − 1

0 − 1 − 1

with β = ( 1 − ν)/ 2 and γ = ( 3 − ν)/ 2.

Then k becomes

1 − 100 ν − ν

− 1 γ − ββ − βν

0 − ββ − ββ 0

E ∗ k = (6)

0 β − ββ − β 0

ν − ββ − βγ − 1

− ν ν 00 − 10

Note that k is symmetric.
For the triangle in Fig. 8.11b, the central difference expressions for ∂ u x /∂ x and ∂ u x /∂ y

would be

0

∂ u x

∂

u x 3 − u x 2 1

h

x =

= − 1

1

[ u x 1 u x 2 u x 3 ]

h

(7)

− 1

1
0

∂ u x

∂

u x 2 − u x 1 1

h y =

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

h

Similar expressions can be obtained for ∂ u y /∂ x and ∂ u y /∂ y . Substitute these expressions

into (3), and obtain k as

ββ 00 β − β

− βγ − 1 ν − ββ

0 − 11 − ν ν 0

E ∗

k = (8)

0 ν − ν 1 − 10

β − βν − 1 γ − β

− ββ 00 − ββ

This matrix is also symmetric.
Now calculate the loading vector. The tractions are applied on the side of the plate
connecting nodes 7, 8, and 9 and can be lumped to these nodes. For triangle 5, the traction
distribution on the side connecting nodes 7 and 8 is shown in Fig. 8.13a. From the conditions
of equilibrium, the equivalent resultant force F 5 is hp 0 3 away from node 7.

The nodal reaction forces, which can be viewed as the loading on these nodes, are obtained

/ 4 acting 2 h /