Direct Variational and Weighted Residual Methods: Classical Trial Function Methods - Mechanics of Structures: Variational and Computational Methods - page 468

In contrast to the trial function of (1), here, none of the displacement boundary conditions
are satisfied. We find

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 ]

0
0
2
6

000 0 0
000 0 0
004 6 8
00612 18
00818 14 5

L 3 1
0

EI EI

L 3

2 ] d

k u = [0026 ξ 12 ξ ξ = (10)

ξ

2

12 ξ

Completion of the R and p u matrices leads to the equations

1 / 2

w 1
w 2
w 3
w 4
w 5

0000 4
0026 4
0

1 / 6

EI
L 3

2 4 12 32 = 1 / 12 p 0 L (11)

0 6 12 24 48

1 / 20

24 24 32 48 76 . 8

1 / 30

with the solution

p 0 L 4
120 EI

50] T w = [ − 2 . 17 5 . 00 4 . 00 − 8 . 00 2 . (12)

As a third case, use of a trial function with six free parameters

2 3 4 5 ][

6 ] T w( x ) = [1 ξξ ξ ξ ξ w w w w w w (13)

1 2 3 4 5

leads to the generalized displacements

p 0 L 4
120 EI [004

1] T

w = − 85 − (14)

In summary, the results are

A. Exact solution is equal to D .

B. First Trial Deflection:

p 0 L 4
120 EI ( −

2 3 4

w(ξ) = 0 . 4167 ξ + 4 ξ − 5 . 2916 ξ + 1 . 8229 ξ ) (15)

C. Second Trial Deflection:

p 0 L 4
120 EI ( −

2 3 4

w(ξ) = 2 . 17 + 5 ξ + 4 ξ − 8 ξ + 2 . 5 ξ ) (16)

D. Third Trial Deflection:

p 0 L 4
120 EI (

2 3 4 5

w(ξ) = 4 ξ − 8 ξ + 5 ξ − ξ ) (17)