Civil Engineering Reference
In-Depth Information
4
3
25 mm
2
1
25 mm
1
2
(b)
(a)
Figure 9.14
Finite element model of Example 9.7.
■
Solution
Observing the symmetry conditions, we model one-fourth of the cross section using
three-node linear triangular elements, as in Figure 9.14b. For simplicity of illustration, we
use only two elements and note that, at nodes 2, 3, and 4, the value of the stress function
is specified as zero, since these nodes are on the surface. Also note that the planes of sym-
metry are such that the partial derivatives of the stress function across those planes are
zero. These conditions correspond to zero normal heat flux (perfect insulation) in a con-
duction problem.
The element stiffness matrices are given by
k
(
e
)
=
∂
d
A
T
∂
∂
T
∂
N
∂
y
N
∂
y
N
∂
z
N
∂
z
+
A
and element nodal forces are
f
(
e
)
=
2
G
[
N
]
T
d
A
A
(Note the use of
y
,
z
coordinates in accord with the coordinate system used in the preced-
ing developments.)
These relations are obtained by analogy with Equation 7.35 for heat conduction. The
interpolation functions are as defined in Equation 9.28.
Element 1
1
2
A
(625
−
25
y
)
∂
N
1
∂
y
25
2
A
∂
N
1
∂
yz
=
0
N
1
=
=−
1
2
A
(25
y
−
25
z
)
∂
N
2
∂
y
25
2
A
∂
N
2
∂
z
25
2
A
N
2
=
=
=−
1
2
A
(25
z
)
∂
N
3
∂
y
∂
N
3
∂
z
25
2
A
N
3
=
=
0
=
Since the partial derivatives are all constant, the stiffness matrix is
−
25
25
0
0
−
25
25
k
(1)
=
1
4
A
1
4
A
[
−
25
25
0 ]
+
[ 0
−
25
25 ]
