Civil Engineering Reference
In-Depth Information
Substituting for
τ
from Eq. (11.1) we have
R
o
2
π
r
3
G
θ
T
=
L
d
r
R
i
from which
π
2
(
R
o
−
R
i
)
G
θ
T
=
L
The polar second moment of area,
J
, is then
π
2
(
R
o
−
R
i
)
J
=
(11.5)
STATICALLY INDETERMINATE CIRCULAR SECTION BARS
UNDER TORSION
Inmany instances bars subjected to torsion are supported in such away that the support
reactions are statically indeterminate. These reactions must be determined, however,
before values of maximum stress and angle of twist can be obtained.
Figure 11.6(a) shows a bar of uniform circular cross section firmly supported at each
end and subjected to a concentrated torque at a point B along its length. From
equilibrium we have
T
=
T
A
+
T
C
(11.6)
A second equation is obtained by considering the compatibility of displacement at B
of the two lengths AB and BC. Thus the angle of twist at B in AB must equal the angle
of twist at B in BC, i.e.
θ
B(AB)
=
θ
B(BC)
T
T
A
T
C
A
C
B
L
AB
L
BC
(a)
T
C
T
F
IGURE
11.6
Torsion of a
circular section bar with built-in
ends
T
A
(b)
