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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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