Civil Engineering Reference
In-Depth Information
Substituting this expression for
τ
in Eq. (13.20) we obtain
y
2
S
y
A
¯
2
y
A
S
y
β
=
b
0
d
y
b
0
I
z
y
1
which gives
y
2
(
A
¯
y
)
2
b
0
A
I
z
β
=
d
y
(13.21)
y
1
Suppose now that
v
s
is the deflection due to shear in the elemental length of beam of
Fig. 13.16. The work done by the shear force
S
y
(assuming it to be constant over the
length
δ
δ
x
and gradually applied) is then
1
2
S
y
δ
v
s
which is equal to the strain energy stored. Hence
S
A
2
1
2
S
y
δ
v
s
=
β
2
G
×
×
A
δ
x
which gives
S
A
β
G
δ
v
s
=
δ
x
The total deflection due to shear in a beam of length
L
subjected to a vertical shear
force
S
y
is then
S
y
A
d
x
β
G
v
s
=
(13.22)
L
E
XAMPLE
13.17
A cantilever beam of length
L
has a rectangular cross section of
breadth
B
and depth
D
and carries a vertical concentrated load,
W
, at its free end.
Determine the deflection of the free end, including the effects of both bending and
shear. The flexural rigidity of the cantilever is
EI
and its shear modulus
G
.
Using Eq. (13.21) we obtain the form factor
β
for the cross section of the beamdirectly.
Thus
D
/
2
B
D
y
1
2
D
2
+
y
2
BD
(
BD
3
/
12)
2
1
B
β
=
2
−
d
y
(see Ex. 10.1)
−
D
/
2
which simplifies to
D
4
16
−
y
4
d
y
D
/
2
D
2
y
2
2
36
D
5
β
=
+
−
D
/
2
Integrating we obtain
D
4
y
16
D
/
2
D
2
y
3
6
y
5
5
36
D
5
β
=
−
+
−
D
/
2
