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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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