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or
−
4110
1
ψ
−
0
.
2222
2
,
2
=
401
10
−
ψ
−
0
.
2222
2
,
3
ψ
3
,
2
ψ
3
,
3
41
011
−
−
0
.
2222
(3b)
−
4
−
0
.
2222
K
U
=
P
The solution of (3) provides the values of
ψ
at the nodal points
ψ
0
.
11111
(
8%
)
2
,
2
ψ
2
,
3
ψ
3
,
2
ψ
3
,
3
=
0
.
11111
(
8%
)
(4)
0
.
11111
(
8%
)
0
.
11111
(
8%
)
The percentage in parenthesis following the values of
ψ
indicates the error of the finite
difference solution.
Refine the grid mesh to that shown in Fig. 8.9c, and continue to use the finite difference
scheme of Eq. (8.26). This leads to the system of linear equations
−
410100000
1
ψ
2
,
2
ψ
2
,
3
ψ
2
,
4
ψ
3
,
2
ψ
3
,
3
ψ
3
,
4
ψ
4
,
2
ψ
4
,
3
ψ
4
,
4
−
0
.
125
41010000
01
−
−
0
.
125
4001000
100
−
−
0
.
125
410100
0101
−
−
0
.
125
41010
00101
−
=
−
0
.
125
(5)
4001
000100
−
−
0
.
125
410
0000101
−
−
0
.
125
41
00000101
−
−
0
.
125
−
4
−
0
.
125
K
U
=
P
Note that the matrix is banded and symmetric. The solution to the linear equations is
ψ
2
,
2
ψ
2
,
3
ψ
2
,
4
ψ
3
,
2
ψ
0
.
0859
(
5
.
15%
)
0
.
10937
(
4
.
3%
)
0
.
0859
(
5
.
15%
)
0
.
10937
(
4
.
3%
)
=
0
.
140625
(
4
.
5%
)
(6)
3
,
3
ψ
0
.
10937
(
4
.
3%
)
3
,
4
ψ
0
.
0859
(
5
.
15%
)
4
,
2
ψ
0
.
10937
(
4
.
3%
)
4
,
3
ψ
0
.
0859
(
5
.
15%
)
4
,
4
It can be seen that the accuracy improves as the mesh of the grid is refined.
After the stress function is calculated, the shear stresses on the cross-section can be ob-
tained. From Chapter 1, Eq. (1.155) or, better still, from Eq. (1) of Chapter 7, Example 7.8,
φ
∂ψ
∂
φ
∂ψ
∂
τ
xy
=
G
τ
xz
=−
G
(7)
z
y
From an appropriate finite difference scheme, the nodal stresses can be calculated. For
example, with the simple central difference formulas, the shear stresses at point 3,3 of
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