Computer Aided Engineering Design

334 COMPUTER AIDED ENGINEERING DESIGN

where the weights wi = 1, i = 1,... , 4 and ξηii = =^1
3

± are the four Gauss points.

Example 11.5.Solve the rectangular plate example above using a quadrilateral element.
The nodal coordinates are (0, 0), (2, 0), (2, 1) and (0, 1). The Jacobian matrix in Eq. (11.11l) is

J =^1

4

2 (1 – ) + 2 (1 + ) 0

0 (1 – ) + (1 + )

=^1

4

40

02

ηη

ξξ

⎡

⎣

⎢

⎤

⎦

⎥

⎡

⎣

⎢

⎤

⎦

⎥

It turns out for this example that the Jacobian matrix is a constant which may not be so in general.
The matrix A in Eq. (11.11j) is

A =^1

0.5

0.5 0 0 0

0001

0 1 0.5 0

=

1000

0002

0210

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

From Eqs. (11.11i) and (11.11k), the strain displacement matrix B can be determined as

B =^1

4

–1 + 0 1 – 0 1 + 0 –1 – 0

0 – 2 + 2 0 – 2 – 2 2 + 2 0 2 – 2

• 2 + 2 – 1 + – 2 – 2 1 – 2 + 2 1 + 2 – 2 – 1 –

ηηηη

ξξξξ

ξη ξη ξηξη

0

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

and the elasticity matrix from Example 11.4 is

D =^16

15

224 10

1 1

4

0

1

4

10

003

8

××^7

⎛

⎝

⎜

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

⎟

For thickness t = 10–2 m, and for Gauss point^1
3

,^1

3

⎛

⎝

⎜

⎞

⎠

⎟, the stiffness matrix k^1 e is given as

ke^16 =^16

15

10

0.3126 0.1563 0.5750 0.1396 –1.1667 –0.5833 0.2791 0.2875

0.1563 0.5471 0.2875 1.8198 –0.5833 –2.0417 0.1396 –0.3252

0.5750 0.2875 2.7375 –0.5833 –2.1458 –1.0729 –1.1667 1.3687

0.1396 1.8198 –0.5833 7.0134 –0.5208 –6.7915 0.9646 –2.0417

–1.1667 –0.5833 –2.1458 –0.5208 4.3541 2.1770 –1.0416 –1.0729

–0.5833 –2.0417 –1.0729 –6.7915 2.1770 7.6196 –0.5208 1.2136

0.2791 0.1396 –1.1667 0.9646 –1.0416 –0.5208 1.9292 –0.5833

0.2875 –0.3252 1.3687 –2.0417 –1.0729 1.2136 –0.5833 1.1533

×

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥

⎥

⎥⎥

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥