Computer Aided Engineering Design

FINITE ELEMENT METHOD 335

For Gauss point –^1
3

,^1

3

,^2

⎛

⎝

⎜

⎞

⎠

⎟ ke is

ke^26 =^16

15

10

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

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

0.5750 – 0.1396 0.3126 –0.1563 0.2791 –0.2875 –1.1667 0.5833

–0.2875 1.8198 –0.1563 0.5471 –0.1396 –0.3252 0.5833 –2.0417

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

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

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

1.0729 – 6.7915 0.5833 –2.0417 0.5208 1.2136 –21770 7.6196

×

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥

⎥

⎥⎥

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

For –^1
3

, –^1

3

,^3

⎛

⎝

⎜

⎞

⎠

⎟ ke is

ke^36 =^16

15

10

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

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

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

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

–1.1667 –0.5833 0.2791 0.2875 0.3126 0.1563 0.5750 0.1396

–0.5833 –2.0417 0.1396 –0.3252 0.1563 0.5471 0.2875 1.8198

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

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

×

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎥⎥

⎥

⎥

⎥

⎥

⎥

while for^1
3

, –^1

3

,^4

⎛

⎝⎜

⎞

⎠⎟

ke is

ke^46 =^16

15

10

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

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

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

0.5208 1.2136 –2.1770 7.6196 1.0729 –6.7915 0.5833 –2.0417

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

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

0.2791 –0.2875 –1.1667 0.5833 0.5750 –0.1396 0.3126 –0.1563

–0.1396 –0.3252 0.5833 –2.0417 –0.2875 1.8198 –0.1563 0.5471

×

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥

⎥

⎥

⎥

⎥

⎥⎥

⎥

⎥

⎥

⎥

⎥

Adding the four matrices above yields the element stiffness matrix, that is