Direct Variational and Weighted Residual Methods: Classical Trial Function Methods - Mechanics of Structures: Variational and Computational Methods - page 453

As a second possibility, choose a trial solution with

2 3

N u 1 (ξ ) = (ξ − ξ )

3 4

N u 2 (ξ ) = (ξ − ξ ) (14)

4 5

N u 3 (ξ ) = (ξ − ξ )

Here, N u 1 is the same as the previous case, the combination b 1 = b 2 = b 3 = 0 ,b 4 = 1 , and

b 5 =− 1 substituted in (4) leads to N u 2 , and N u 3 is a higher order polynomial which satisfies

the required displacement boundary conditions. Our trial deflection is

w

1

2 3 3 4 4 5

w

w(ξ) = [ (ξ − ξ )ξ − ξ )ξ − ξ ) ] (15)

2

w

3

For this assumption, we obtain

2 2 3 2 3 4

4 − 24 ξ + 36 ξ 12 ξ − 60 ξ + 72 ξ 24 ξ − 112 ξ + 120 ξ

L 4 1
0

EI

2 3 4 3 4 5

k u = 36 ξ − 144 ξ + 144 ξ 72 ξ − 264 ξ + 240 ξ dx (16)

4 5 6

Symmetric 144 ξ − 480 ξ + 400 ξ

dx

/

1 1 1 30

ξ

2

− ξ

3

N u dx

ξ

3

− ξ

4

/

p u = p z ( x ) = p 0 ( 1 − ξ) = p 0 L 1 60 (17)

0 0 4 5

ξ − ξ

1 / 105

The equations for w are

44 4 w

1 / 30

1

=

424 / 5 6 / 5 w

p 0 L 1 / 60

EI
L 3

2

(18)

w /

426 / 5 208 / 35 1 105

3

k u w = p u

which provides the solution

p 0 L 4
120 EI [4

T

w = − 41 (19)

Let us compare both of the above results with the exact solution.

A. Exact solution:

p 0 L 4
120 EI (

2 3 4 5

w(ξ) = ξ − ξ + ξ − ξ )

4 8 5

B. Trial Deflection with m = 2:

p 0 L 4
120 EI (

2 3 4

w(ξ) = 3 . 5 ξ − 6 ξ + 2 . 5 ξ )

C. Trial Deflection with m = 3:

p 0 L 4
120 EI (

2 3 4 5

w(ξ) = 4 ξ − 8 ξ + 5 ξ − ξ )