Civil Engineering Reference
In-Depth Information
Substituting for
M
in the first of these expressions from Eq. (13.3) (equally we could
use the second) we obtain
d
2
v
d
x
2
P
CR
EI
=
(
δ −
v
)
(21.11)
which, on rearranging, becomes
d
2
v
d
x
2
P
CR
EI
P
CR
EI
δ
+
v
=
(21.12)
The solution of Eq. (21.12) is
v
=
C
1
cos
µ
x
+
C
2
sin
µ
x
+ δ
(21.13)
where
µ
2
=
P
CR
/
EI
. When
x
=
0,
v
=
0 so that
C
1
=−δ
. Also when
x
=
L
,
v
= δ
so that
from Eq. (21.13) we have
δ =−δ
cos
µ
L
+
C
2
sin
µ
L
+ δ
which gives
cos
µ
L
sin
µ
L
C
2
= δ
Hence
cos
µ
x
1
cos
µ
L
sin
µ
L
sin
µ
x
v
=−δ
−
−
(21.14)
Again
v
is indeterminate since
δ
cannot be determined. Finally we have d
v
/d
x
=
0at
x
=
0. Hence from Eq. (21.14)
cos
µ
L
=
0
whence
n
π
2
µ
L
=
where
n
=
1, 3, 5,
...
Thus taking the smallest value of buckling load (corresponding to
n
=
1) we obtain
π
2
EI
4
L
2
P
CR
=
(21.15)
BUCKLING OF A COLUMN WITH ONE END FIXED AND
THE OTHER PINNED
The column in this case is allowed to rotate at one end but requires a lateral force,
F
,
to maintain its position (Fig. 21.7).
At any section X the bending moment
M
is given by
M
=−
P
CR
v
−
F
(
L
−
x
)
