Computer Aided Engineering Design

330 COMPUTER AIDED ENGINEERING DESIGN

For element 2

B 2 =^1

2

0010–10

0 –200 0 2

–20012–2

⎛

⎝

⎜

⎜

⎜

⎞

⎠

⎟

⎟

⎟

Hence

12 5 6 7 8

kBDBe AtT

sym

222 2

= =^167

15

10

0.84 0 0 – 0.42 – 0.84 0.42

2.24 – 0.28 0 0.28 – 2.24

0.56 0 – 0.56 0.28

0.21 0.42 – 0.21

1.40 – 0.7

2.45

1 2 5 6 7 8

×

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

The assembled matrix K is

12 34 5 678

K =^16

15

10

1.40 0.00 – 0.56 0.28 0.00 – 0.70 – 0.84 0.42

0.00 2.45 0.42 – 0.21 – 0.70 0.00 0.28 – 2.24

• 0.56 0.42 1.40 – 0.70 – 0.84 0.28 0.00 0.00
  0.28 – 0.21 – 0.70 2.45 0.42 – 2.24 0.00 0.00
  0.00 – 0.70 – 0.84 0.42 1.40 0.00 – 0.56 0.28


• 0.70 0.00 0.28 –2.24 0.00 2.45 0.42 – 0.21


• 0.84 0.28 0.00 0.00 – 0.56 0.42 1.40 – 0.70
  0.42 – 2.24 0.00 0.00 0.28 – 0.21 – 0.70 2.45

×^7

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎜⎜

⎜

⎜

⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

1 2 3 4 5 6 7 8

and the force vector F is [0 0 1000 –2000 0 0 0 0]T. After applying the displacement boundary
conditions that nodes 1 and 4 (degrees of freedom 1, 2, 7 and 8) are fixed, the relevant entries to
compute the unknown displacements [U(3)U(4)U(5)U(6)] are (in bold).

16

15

10

1.40 0.00 – 0.56 0.28 0.00 – 0.70 – 0.84 0.42

0.00 2.45 0.42 – 0.21 – 0.70 0.00 0.28 – 2.24

• 0.56 0.42 0.00 0.00
  0.28 – 0.21 0.00 0.00
  0.00 – 0.70 – 0.56 0.28


• 0.70 0.00 0.42 – 0.21


• 0.84 0.28 0.00 0.00 – 0.56 0.42 1.40 – 0.70
  0.42 – 2.24 0.00 0.00 0.28 – 0.21 – 0.70 2.45

×^7

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

1.40 – 0.70 – 0.84 0.28

• 0.70 2.45 0.42 – 2.24


• 0.84 0.42 1.40 0.00
  0.28 – 2.24 0.00 2.45

⎜⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

⎛

⎝

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟

0

0

(3)

(4)

(5)

(6)

0

0

=

1

2

7

8

U

U

U

U

2000

• 1000
  0
  0

F

F

F

F