Civil Engineering Reference
In-Depth Information
(1, 2)
3
300 psi
Y
1
(0, 0)
2
(1, 0)
X
100 psi
(a)
3
40
Y
2
1
20
10
10
X
(b)
Figure 9.4
(a) Distributed loads on a triangular element.
(b) Work-equivalent nodal forces.
Along edge 1-2,
y
=
0
,
p
x
=
0
,
p
y
=−
100
psi; hence, Equation 9.39 becomes
1
−
x
0
x
0
0 0
0
0
−
100
t
d
S
f
(
p
)
=
−
x
S
0
x
0
0
1
−
x
0
x
0
0 0
0
0
0
0
−
10
−
10
0
0
−
100
d
x
=
1
=
0
.
2
lb
−
x
0
0
x
0
0
