Civil Engineering Reference
In-Depth Information
which gives
M
F
P
CR
(1
−
cos
µ
L
)
sin
µ
L
C
2
=−
Hence Eq. (21.8) becomes
cos
µ
x
1
M
F
P
CR
(1
−
cos
µ
L
)
sin
µ
L
v
=−
+
sin
µ
x
−
(21.9)
Note that again
v
is indeterminate since
M
F
cannot be found. Also since d
v
/d
x
=
0at
x
=
L
we have from Eq. (21.9)
0
=
1
−
cos
µ
L
whence
cos
µ
L
=
1
and
µ
L
=
n
π
where
n
=
0, 2, 4,
...
For a non-trivial solution, i.e.
n
=
0, and taking the smallest value of buckling load
(
n
=
2) we have
4
π
2
EI
L
2
P
CR
=
(21.10)
BUCKLING LOAD FOR A COLUMN WITH ONE END FIXED
AND ONE END FREE
In this configuration the upper end of the column is free to move laterally and also to
rotate as shown in Fig. 21.6. At any section X the bending moment
M
is given by
M
=
P
CR
(
δ −
v
)or
M
=−
P
CR
v
+
M
F
x
P
CR
δ
EI
L
0
y
M
F
F
IGURE
21.6
Determination of buckling load for a
column with one end fixed and one end free
P
CR
