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FIGURE P2.15
A bar structure. Each bar has the same modulus of elasticity and cross-sectional area.
10 7
0
.
6 .
2.16 Resolve Problem 2.15 if the material is nonlinearly elastic with
σ =
20
×
2.17 Solve Problem 2.15 using the principle of stationary potential energy.
2.18 Use the principle of stationary potential energy to find the equation of equili br ium
for a flexible string of length L with tension N and transverse loading intensity p z (
x
)
.
be the transverse displacement.
2.19 An elastic string of length L , modul us E , and cross-sectional area A is fixed at its ends
and subjected to a transverse force p z .
Let
w(
x
)
1
2
The strain is given by
x =
u ,x +
2 w
,x , where
u is the axial component of displacement and
is the transverse displacement. Use
the principle of stationary potential energy to show that the governing equations are
w
EA u ,x +
d
dx
1
2 w
2
,x
=
0
and
u
| x = 0 =
u
| x = L =
0
EA u ,x +
w ,x
d
dx
1
2 w
2
,x
+
p z =
0
w | x = 0 = w | x = L =
0
Hint:
1
2 EA u ,x +
2
dx
1
2
1
2 w
2
,x
=
σ x x dV
p z w
dx
=
p z w
L
L
δ =
0 leads to the governing equations.
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