Civil Engineering Reference
In-Depth Information
Writing
P
EI
µ
2
=
Equation (21.52) becomes
d
2
v
d
x
2
Wa
EIL
x
µ
2
v
+
=−
the general solution of which is
Wa
PL
x
v
=
C
1
cos
µ
x
+
C
2
sin
µ
x
−
(21.54)
Similarly the general solution of Eq. (21.53) is
W
PL
(
L
v
=
C
3
cos
µ
x
+
C
4
sin
µ
x
−
−
a
)(
L
−
x
)
(21.55)
where
C
1
,
C
2
,
C
3
and
C
4
are constants which are found from the boundary conditions
as follows.
When
x
=
0,
v
=
0, therefore from Eq. (21.54)
C
1
=
0. At
x
=
L
,
v
=
0 giving, from
Eq. (21.55),
C
3
=−
C
4
tan
µ
L
. At the point of application of the load the deflection
and slope of the beam given by Eqs (21.54) and (21.55) must be the same. Hence
equating deflections
Wa
PL
(
L
Wa
PL
(
L
C
2
sin
µ
(
L
−
a
)
−
−
a
)
=
C
4
[ sin
µ
(
L
−
a
)
−
tan
µ
L
cos
µ
(
L
−
a
)]
−
−
a
)
and equating slopes
Wa
PL
=
Wa
PL
(
L
C
2
µ
cos
µ
(
L
−
a
)
−
C
4
µ
[ cos
µ
(
L
−
a
)
+
tan
µ
L
sin
µ
(
L
−
a
)]
+
−
a
)
Solving the above equations for
C
2
and
C
4
and substituting for
C
1
,
C
2
,
C
3
and
C
4
in
Eqs (21.54) and (21.55) we have
W
sin
µ
a
P
µ
sin
µ
L
sin
µ
x
Wa
PL
x
v
=
−
for
x
≤
L
−
a
(21.56)
W
sin
µ
(
L
−
a
)
W
PL
(
L
=
sin
µ
(
L
−
x
)
−
−
a
)(
L
−
x
)
for
x
≥
L
−
a
v
(21.57)
P
µ
sin
µ
L
These equations for the beam-column deflection enable the bending moment and
resulting bending stresses to be found at all sections.
A particular case arises when the load is applied at the centre of the span. The
deflection curve is then symmetrical with a maximum deflection under the load of
2
P
µ
tan
µ
L
W
WL
4
P
v
max
=
−
2
Finally we consider a beam column subjected to endmoments,
M
A
and
M
B
, in addition
to an axial load,
P
(Fig. 21.15). The deflected form of the beam column may be found
