Civil Engineering Reference
In-Depth Information
3
p
z
p
z
p
r
z
d
S
p
r
2
1
r
(a)
(b)
Figure 9.10
(a) Axisymmetric element. (b) Differential length
of the element edge.
force) and gravity. In each case, the external influences are reduced to nodal
forces using the work equivalence concept previously introduced.
The triangular axisymmetric element shown in Figure 9.10a is subjected to
pressures
p
r
and
p
z
in the radial and axial directions, respectively. The equivalent
nodal forces are determined by analogy with Equation 9.39, with the notable
exception depicted in Figure 9.10b, showing a differential length d
S
of the ele-
ment edge in question. As d
S
is located a radial distance
r
from the axis of sym-
metry, the area on which the pressure components act is
2
r
d
S
.
The nodal
forces are given by
f
(
p
)
=
f
(
p
)
r
f
(
p
)
z
[
N
]
T
p
r
p
z
r
d
S
=
2
(9.101)
S
and the path of integration
S
is the element edge. In this expression,
[
N
]
T
is as
defined by Equation 9.40.
EXAMPLE 9.5
Calculate the nodal forces corresponding to a uniform radial pressure
p
r
=
10 psi acting
as shown on the axisymmetric element in Figure 9.11.
■
Solution
As we have pressure on one face only and no axial pressure, we immediately observe that
f
r
2
=
f
z
1
=
f
z
2
=
f
z
3
=
0
The nonzero terms are
f
r
1
=
2
N
1
p
r
r
d
S
S
f
r
3
=
2
N
3
p
r
r
d
S
S
