Civil Engineering Reference
In-Depth Information
Generally, radial body force arises from rotation of an axisymmetric body
about the
z
axis. For constant angular velocity
,
the radial body force compo-
nent
R
B
is equal to the magnitude of the normal acceleration component
r
2
and
directed in the positive radial direction.
EXAMPLE 9.6
The axisymmetric element of Figure 9.11 is part of a body rotating with angular velocity
10 rad/s about the
z
axis and subjected to gravity in the negative
z
direction. Compute the
equivalent nodal forces. Density is 7.3(10)
−
4
lb-s
2
/in.
4
■
Solution
For the stated conditions, we have
2
100
r
in./s
2
R
B
=
r
=
=−
g
=−
386
.
4 in./s
2
Z
B
Using the interpolation functions as given in Example 9.5,
4
4
−
r
(4
−
r
−
z
)
r
2
d
z
d
r
f
r
1
=
2
N
1
R
B
r
d
r
d
z
=
2
(100)
=
0
.
84 lb
A
3
0
4
4
−
r
(
r
−
3)
r
2
d
z
d
r
f
r
2
=
2
N
2
R
B
r
d
r
d
z
=
2
(100)
=
0
.
98 lb
A
3
0
4
4
−
r
zr
2
d
z
d
r
f
r
3
=
2
N
3
R
B
r
d
r
d
z
=
2
(100)
=
0
.
84 lb
A
3
0
4
4
−
r
f
z
1
=
2
N
1
Z
B
r
d
r
d
z
=−
2
(386
.
4)
(4
−
r
−
z
)
r
d
z
d
r
=−
1
.
00 lb
A
3
0
4
4
−
r
f
z
2
=
2
N
2
Z
B
r
d
r
d
z
=−
2
(386
.
4)
(
r
−
3)
r
d
z
d
r
=−
1
.
08 lb
A
3
0
4
4
−
r
f
z
3
=
2
N
3
Z
B
r
d
r
d
z
=−
2
(386
.
4)
zr
d
z
d
r
=−
1
.
00 lb
A
3
0
The integrations required to obtain the given results are straightforward but algebraically
tedious. Another approach that can be used and is increasingly accurate for decreasing
element size is to evaluate the body forces and the integrand at the centroid of the cross
section of the element area as an approximation. Using this approximation, it can be
shown that
rA
3
N
i
(
r
,
z
)
r
d
z
d
r
=
i
=
1, 3
A
