Direct Variational and Weighted Residual Methods: Classical Trial Function Methods - Mechanics of Structures: Variational and Computational Methods - page 472

Numerical expressions for R u and R p are computed as

0
0
6
24

0
2
0
0

0
0
6
0

EI
L 3

R u = [1111] + [1000] − [0000]

+ − =

0000
0000
6666
24

0000
2000
0000
0000

0000
2000
6666
24

EI
L 3

EI
L 3

= 0 (6)

24 24 24 24 24 24

1
2
3
4

026 2
041224
061836
082448

EI
L 3

EI
L 3

= =

R p [0 2 6 12]

Thus, the matrix on the left-hand side of Eq. (7.76) is

026 4
2 4 12 32
6 2 4 8
24

EI
L 3

k = k u + R u + R p = (7)

32 48 76 . 8

We find that the w are the same as obtained in Example 7.9 for the same trial function i.e.,

case B.
As a second case, we choose to use the trial function of case C of Example 7.9. Here,
w =

2 3 4 ][

5 ] T

w w w w w

[1 ξξ ξ ξ (8)

1 2 3 4

which satisfies none of the boundary conditions. We find

2 3 4 ]

N u = [1 ξξ ξ ξ

1
L [012

N u =

2 3 ]

ξ 3 ξ 4 ξ

1
L 2 [0026

N u =

2 ]

ξ 12 ξ (9)

1
L 3 [000624

N u = ξ ]

1
L 4 [000024]

N i u =

Then compute

1
ξ
ξ

000024
000012
00008
00006
000024

L 3 1
0

EI EI

L 3

2

k u = [000024] d ξ = (10)

3

ξ

/ 5

4

ξ

1 / 2

1
ξ
ξ

p 0 L 1
0

1 / 6

2

p u = ( 1 − ξ) d ξ = 1 / 12 p 0 L

3

ξ

1 / 20

4

ξ

1 / 30

(11)