Civil Engineering Reference
In-Depth Information
Since we are assuming that the section rotates as a rigid body, it follows that
θ
is a
function of
x
only so that the above equation may be written
∂v
t
∂
x
=
d
θ
d
x
p
R
Substituting for
∂v
t
/∂
x
in Eq. (10.27) we have
∂
w
∂
s
d
θ
d
x
=
+
p
R
γ
Now
τ
G
=
q
s
Gt
γ
=
Thus
q
s
Gt
∂
w
∂
s
d
θ
d
x
Integrating both sides of this equation completely round the cross section of the beam,
i.e. from
s
=
+
p
R
=
0to
s
=
s
ι
(see Fig. 10.20(b))
q
s
Gt
d
s
∂
w
∂
s
d
s
p
R
d
s
d
θ
d
x
=
+
which gives
q
s
Gt
d
s
d
θ
d
x
2
A
The axial displacement,
w
, must have the same value at
s
[
w
]
s
=
s
ι
s
=
0
+
=
=
0 and
s
=
s
ι
. Therefore the
above expression reduces to
q
s
Gt
d
s
d
θ
d
x
=
1
2
A
(10.28)
For shear loads applied through the shear centre, d
θ/
d
x
=
0 so that
q
s
Gt
d
s
0
=
which may be written
1
Gt
(
q
b
+
0
=
q
s
,0
)d
s
Hence
(
q
b
/
Gt
)d
s
d
s
/
Gt
q
s
,0
=−
(10.29)
If
G
is constant then Eq. (10.29) simplifies to
(
q
b
/
t
)d
s
d
s
/
t
q
s
,0
=−
(10.30)
