Civil Engineering Reference
In-Depth Information
or, if
G
, the shear modulus, is constant round the section
d
θ
d
x
=
d
s
t
T
4
A
2
G
(11.25)
E
XAMPLE
11.2
A thin-walled circular section beam has a diameter of 200mm and
is 2m long; it is firmly restrained against rotation at each end. A concentrated torque
of 30 kNm is applied to the beam at its mid-span point. If the maximum shear stress
in the beam is limited to 200N/mm
2
and the maximum angle of twist to 2
◦
, calculate
the minimum thickness of the beam walls. Take
G
25 000N/mm
2
.
=
The minimum thickness of the beam corresponding to the maximum allowable shear
stress of 200N/mm
2
is obtained directly using Eq. (11.22) in which
T
max
=
15 kNm.
Thus
10
6
15
×
×
4
t
min
=
200
=
1
.
2mm
200
2
2
×
π
×
×
The rate of twist along the beam is given by Eq. (11.25) in which
d
s
t
π
×
200
t
min
=
Hence
d
θ
d
x
=
T
4
A
2
G
×
π
×
200
t
min
(i)
Taking the origin for
x
at one of the fixed ends and integrating Eq. (i) for half the
length of the beam we obtain
T
4
A
2
G
×
200
π
t
min
θ
=
x
+
C
1
where
C
1
is a constant of integration. At the fixed end where
x
=
0,
θ
=
0 so that
C
1
=
0.
Hence
T
4
A
2
G
×
200
π
t
min
=
x
θ
The maximum angle of twist occurs at the mid-span of the beamwhere
x
=
1m. Hence
10
6
10
3
15
×
×
200
×
π
×
1
×
×
180
t
min
=
π
=
2
.
7mm
200
2
/
4)
2
4
×
(
π
×
×
25 000
×
2
×
The minimum allowable thickness that satisfies both conditions is therefore 2.7mm.
11.5 T
ORSION OF
S
OLID
S
ECTION
B
EAMS
Generally, by solid section beams, we mean beam sections in which the walls do not
form a closed loop system. Examples of such sections are shown in Fig. 11.13. An
