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expression leads to
=
EAL
0
0
C
1
C
2
C
3
0
0
0
L
3
L
3
0
(
/
3
)(
k
s
GA
−
P
)
−
(
/
6
)
k
s
GA
(3)
L
3
L
3
0
−
(
/
6
)
k
s
GA
EIL
+
(
/
12
)
k
s
GA
From det
K
=
0, the critical load is found to be
12
EI
/
L
2
P
cr
=
(4)
1
1
+
k
s
GA
12
EI
/
L
2
This value is lower than when shear deformation is not considered. In fact,
P
cr
|
No Shear
P
cr
|
Shear
=
(5)
1
+
P
cr
|
No Shear
/
k
s
GA
EXAMPLE 11.13 Variable Axial Force and Variable Cross-Section
Consider a beam element with moment of inertia and axial force that vary linearly along
the beam (see Fig. 11.41).
The moment of inertia is assumed to vary as
EI
0
x
L
β
=
−
ξβ)
=
−
(
EI
EI
0
EI
0
1
(1)
with
ξ
=
x
/
L
, and similarly, the axial force is taken to be
P
(ξ )
=
P
0
(
1
−
ξβ)
(2)
where
β
and
β
are prescribed. The virtual work relationship can be written as
v
T
G
T
N
T
u
G
T
N
T
u
L
EI
0
dx
G
N
u
EI
0
dx
G
x
−
β
N
u
−
δ
W
=
δ
G
T
N
u
N
u
(
dx
G
G
T
N
u
N
u
dx
G
v
x
L
(
−
P
0
)
+
β
P
0
)
=
0
(3)
FIGURE 11.41
Beam element of variable cross-section with a variable axial force.
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