Civil Engineering Reference
In-Depth Information
y
S
BA
u
B
B
S
AB
M
BA
M
AB
EI
υ
B
A
υ
A
F
IGURE
16.32
Slope and deflection
of a beam
u
A
x
moment at the right-hand end would induce a positive bending moment at all sections
of the beam. We shall therefore adopt a sign convention such that themoment at a point
is positive when it is applied in a clockwise sense and negative when in an anticlockwise
sense; thus in Fig. 16.32 both moments
M
AB
and
M
BA
are positive. We shall see in the
solution of a particular problem how these end moments are interpreted in terms of
the bending moment distribution along the length of a beam. In the analysis we shall
ignore axial force effects since these would have a negligible effect in the equation
for moment equilibrium. Also, the moments
M
AB
and
M
BA
are independent of each
other but the shear forces, which in the absence of lateral loads are equal and opposite,
depend upon the end moments.
From Eq. (13.3) and Fig. 16.32
EI
d
2
v
d
x
2
=
M
AB
+
S
AB
x
Hence
S
AB
x
2
EI
d
v
d
x
=
M
AB
x
+
2
+
C
1
(16.17)
and
M
AB
x
2
S
AB
x
3
EI
v
=
2
+
6
+
C
1
x
+
C
2
(16.18)
When
d
v
d
x
x
=
0
=
θ
A
v
=
v
A
Therefore, from Eq. (16.17)
C
1
=
EI
θ
A
and from Eq. (16.18),
C
2
=
EI
v
A
. Equations
(16.17) and (16.18) then, respectively, become
S
AB
x
2
EI
d
v
d
x
=
M
AB
x
+
2
+
EI
θ
A
(16.19)
and
M
AB
x
2
S
AB
x
3
EI
v
=
2
+
6
+
EI
θ
A
x
+
EI
v
A
(16.20)
