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surface area
S
. Note that
r
2
b
h
V
=
π
3
+
r
2
h
r
2
b
2
S
=
π
+
+
19.
3 m
4 m
2
1
P
= 200 kN
3
P
= 200 kN
The equilibrium equations of the truss shown are
4
5
σ
2
A
2
=
3
5
σ
2
A
2
+
σ
3
A
3
=
σ
1
A
1
+
P
P
where
σ
i
is the axialstress in member
i
and
A
i
are the cross-sectional areas.
The third equationissuppliedbycompatibility (geometricalconstraints on the
elongations of the members):
16
5
σ
1
−
9
5
σ
3
=
5
σ
2
+
0
Find the cross-sectional areas of the membersthat minimize the weight of the
truss without the stresses exceeding 150MPa.
20.
B
1
y
1
L
1
y
2
H
2
W
1
L
2
3
L
3
W
2
A cable supportedat the endscarries the weights
W
1
and
W
2
. The potentialenergy
of the systemis
=−
W
1
y
1
−
V
W
2
y
2
=−
W
1
L
1
sin
θ
−
W
2
(
L
1
sin
θ
+
L
2
sin
θ
2
)
1
1
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