Civil Engineering Reference
In-Depth Information
Then, noting that
d
r
d
z
=
d
A
, we have
2
L
i
L
j
r
d
r
d
z
=
2
L
i
L
j
(
L
1
r
1
+
L
2
r
2
+
L
3
r
3
)d
A
A
A
which is of the appropriate form for application of the integration formula. For
i
=
1,
=
2,
the integral becomes
j
2
L
1
L
2
r
d
r
d
z
=
2
L
1
L
2
(
L
1
r
1
+
L
2
r
2
+
L
3
r
3
)d
A
A
A
r
1
A
r
2
A
r
3
A
L
1
L
2
d
A
+
L
1
L
2
=
2
2
d
A
+
2
L
1
L
2
L
3
d
A
Applying the integration formula to each of the three integrals on the right,
2
L
1
L
2
r
d
r
d
z
A
A
r
1
(2!)(1!)(0!)
(2
+
1
+
0
+
2)!
+
r
2
(1!)(2!)(0!)
(1
+
2
+
0
+
2)!
+
r
3
(1!)(1!)(1!)
(1
+
1
+
1
+
2)!
=
4
A
2
r
1
2
r
2
120
+
r
3
120
=
A
30
=
4
120
+
(2
r
1
+
2
r
2
+
r
3
)
The integration technique used in Example 6.5 is also applicable to higher-
order, straight-sided triangular elements, as shown in the next example.
EXAMPLE 6.6
For an axisymmetric element based on the six-node, quadratic triangular element having
interpolation functions given by Equation 6.47, evaluate the integral
I
=
N
2
N
4
d
V
V
■
Solution
Using Equation 6.93,
I
=
N
2
N
4
d
V
=
2
N
2
N
4
r
d
r
d
z
=
2
L
2
(2
L
2
−
1)(4
L
1
L
2
)
r
d
r
d
z
V
A
A
Now observe that, even though the interpolation functions vary quadratically over the
element area, the area coordinates, by definition, vary linearly. Since the element sides
are straight, the radial coordinate can still be expressed as
r
=
L
1
r
1
+
L
2
r
2
+
L
3
r
3
Therefore, we have
I
=
2
L
2
(2
L
2
−
1)(4
L
1
L
2
)(
L
1
r
1
+
L
2
r
2
+
L
3
r
3
)d
A
A
