Civil Engineering Reference
In-Depth Information
while for element 2, we have
5
A
0
8
A
2
E
L
2
5
A
0
E
8(
L
/
2)
=
5
A
0
E
4
L
A
1
=
and
k
2
=
=
Since no load is applied at the center of the bar, the equilibrium equations for the
system of two elements is
=
−
k
1
k
1
0
U
1
U
2
U
3
F
1
0
P
−
k
1
k
1
+
k
2
−
k
2
0
−
k
2
k
2
Applying the constraint condition
U
1
=
0
results in
k
1
+
k
2
U
2
U
3
0
P
−
k
2
=
−
k
2
k
2
Adding the two equations gives
4
PL
7
A
0
E
and substituting this result into the first equation results in
P
k
1
=
U
2
=
k
1
+
k
2
k
2
48
PL
35
A
0
E
=
PL
A
0
E
U
3
=
=
1
.
371
(c)
To obtain the exact solution, we refer to Figure 2.7d, which is a free-body diagram of
a section of the bar between an arbitrary position
x
and the end
x
=
L
.
For equilibrium,
A
0
1
x
2
L
and since
x
A
=
P
A
=
A
(
x
)
=
−
the axial stress variation along the length of the bar is described by
P
A
0
1
−
x
=
x
2
L
Therefore, the axial strain is
x
E
=
P
EA
0
1
−
ε
x
=
x
2
L
Since the bar is fixed at
x
=
0
, the displacement at
x
=
L
is given by
L
L
P
EA
0
d
x
1
−
=
ε
x
d
x
=
x
2
L
0
0
ln(2
L
−
x
)]
2
PL
EA
0
[
2
PL
EA
0
[ln(2
L
)
2
PL
EA
0
ln 2
PL
A
0
E
L
0
=
−
=
−
ln
L
]
=
=
1
.
386
