Civil Engineering Reference
In-Depth Information
or using Eq. (11.3)
T
A
L
AB
GJ
T
C
L
BC
GJ
=
whence
L
BC
L
AB
T
A
=
T
C
Substituting in Eq. (11.6) for
T
A
we obtain
T
C
L
BC
1
T
=
L
AB
+
which gives
L
AB
L
AB
+
T
C
=
T
(11.7)
L
BC
Hence
L
BC
L
AB
+
T
A
=
T
(11.8)
L
BC
The distribution of torque along the length of the bar is shown in Fig. 11.6(b). Note
that if
L
AB
>
L
BC
,
T
C
is the maximum torque in the bar.
E
XAMPLE
11.1
A bar of circular cross section is 2.5m long (Fig. 11.7). For 2m of
its length its diameter is 200mm while for the remaining 0.5m its diameter is 100mm.
If the bar is firmly supported at its ends and subjected to a torque of 50 kNm applied
at its change of section, calculate the maximum stress in the bar and the angle of twist
at the point of application of the torque. Take
G
80 000N/mm
2
.
=
In this problem Eqs (11.7) and (11.8) cannot be used directly since the bar changes
section at B. Thus from equilibrium
T
=
T
A
+
T
C
(i)
and from the compatibility of displacement at B in the lengths AB and BC
θ
B(AB)
=
θ
B(BC)
50 kN m
Diameter
200 mm
C
Diameter
100 mm
A
B
F
IGURE
11.7
Bar of
Ex. 11.1
2.0 m
0.5 m
