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Find the distribution of deflection using collocation and Galerkin's
method, and compare the results with the exact solution.
7.16 Find the torsional properties of a rectangular cross-section of width 2
a
(
y
direction)
and height 2
b
(
z
direction). Use the Prandtl stress function in conjunction with the
principle of complementary virtual work.
Let
k
=
EI
=
1
.
Use the Ritz method. A possible trial solution is
ψ
=
ψ(
a
2
y
2
b
2
z
2
Hint:
−
)(
−
)
,
which satisfies
0 on the boundary (Chapter 1). From Eq. (4) of Chapter 2,
Example 2.9, represent the principle of complementary virtual work by
ψ
=
2
φ
)δψ
A
(
∇
ψ
+
2
G
dA
=
0
This gives
2
a
3
b
3
a
2
40
9
ψ
=
ψ
φ
(
a
2
b
2
φ
G
5
/
4
)/(
+
)
,
=
dy dz
=
b
2
G
t
+
A
φ
|
Ritz
=
φ
|
exact
=
Comparision for
b
/
a
=
1:
M
t
/
G
2
.
22
,M
t
/
G
2
.
25
7.17 Find the displacements and forces along the beam of Fig. P7.17 using the simple
Ritz method and then the extended
R
itz method. Represent the applied distributed
loading as a quadratic polynomi
al
p
z
=
α
1
x
2
1
,
2
,
3are
constants determined so that this
p
z
fits the loading distribution of the figure.
+
α
2
x
+
α
3
,
where
α
i
,i
=
FIGURE P7.17
L
2
,
Answer:
Loading:
α
1
=−
4
p
0
/
α
2
=
4
p
0
/
L,
α
3
=
0
.
For the trial solution,
2
3
4
5
w
=
(
a
1
+
a
2
ξ
+
a
3
ξ
+
a
4
ξ
+
a
5
ξ
+
a
6
ξ
)w
1
2
3
4
5
6
+
(
b
1
+
b
2
ξ
+
b
3
ξ
+
b
4
ξ
+
b
5
ξ
+
b
6
ξ
+
b
7
ξ
)w
2
using
a
1
=
a
2
=
b
1
=
b
2
=
0
,a
3
=−
a
4
=−
a
5
=
1
,b
3
=
b
4
=
1
,b
5
=−
2
,
and
b
6
=−
b
7
=−
3 to meet the displacement boundary conditions, the simple Ritz
method gives
p
0
L
4
29 370
EI
(
2
3
4
5
6
w
=
919
ξ
−
1277
ξ
−
740
ξ
+
1635
ξ
−
537
ξ
)
2
3
4
5
6
=
0
.
2163
ξ
−
0
.
3005
ξ
−
0
.
1742
ξ
+
0
.
3848
ξ
−
0
.
1264
ξ
For the trial solution,
2
3
4
5
6
w(
)
=
ξ
w
+
ξ
w
+
ξ
w
+
ξ
w
+
ξ
w
x
1
2
3
4
5
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