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7.2 Torsion
In this section, the geometry to be considered is that of a long slender circular bar and the load is one which twists the bar. Such problems are important in the analysis of twisting components, for example lug wrenches and transmission shafts.
7.2.1 Basic relations for Torsion of Circular Members
The theory of torsion presented here concerns torques^1 which twist the members but which do not induce any warping , that is, cross sections which are perpendicular to the axis of the member remain so after twisting. Further, radial lines remain straight and radial as the cross-section rotates – they merely rotate with the section.
For example, consider the member shown in Fig. 7.2.1, built-in at one end and subject to a torque T at the other. The x axis is drawn along its axis. The torque shown is positive, following the right-hand rule (see §7.1.4). The member twists under the action of the torque and the radial plane ABCD moves to ABC D^.
Figure 7.2.1: A cylindrical member under the action of a torque
Whereas in the last section the measure of deformation was elongation of the axial members, here an appropriate measure is the amount by which the member twists, the
rotation angle . The rotation angle will vary along the member – the sign convention is
that is positive in the same direction as positive T as indicated by the arrow in Fig.
7.2.1. Further, whereas the measure of strain used in the previous section was the normal
strain xx , here it will be the engineering shear strain xy (twice the tensorial shear strain
xy ). A relationship between (dropping the subscripts) and will next be established.
As the line BC deforms into BC ^ , Fig. 7.2.1, it undergoes an angle change . As defined in §4.1.2, the shear strain is the change in the original right angle formed by
BC and a tangent at B (indicated by the dotted line – this is the y axis to be used in xy ).
If is small, then
tan
CC R L
BC L
(^1) the term torque is usually used instead of moment in the context of twisting shafts such as those
considered in this section
T
x
B^ C
A C D
L
where L is the length, R the radius of the member and ( L )means the magnitude of
at L. Note that the strain is constant along the length of the member although is
not. Considering a general cross-section within the member, as in Fig. 7.2.2, one has
R ( ) x x
Figure 7.2.2: A section of a twisting cylindrical member
The shear strain at an arbitrary radial location r , 0 r R , is
x
r x r
showing that the shear strain varies from zero at the centre of the shaft to a
maximum R ( L )/ L R ( x )/ x on the outer surface of the shaft.
The only strain is this shear strain and so the only stress which will arise is a shear
stress . From Hooke’s Law
G (7.2.4)
where G is the shear modulus (the of Eqn. 6.1.5). Following the shear strain, the
shear stress is zero at the centre of the shaft and a maximum on the outer surface.
Considering a free-body diagram of any portion of the shaft of Fig. 7.2.1, a torque T acts on all cross-sections. This torque must equal the resultant of the shear stresses acting over the section, as schematically illustrated in Fig. 7.2.3a.
The elemental force acting over an element of area dA is dA and so the resultant
moment about r 0 is
dA
T r r dr (7.2.5)
But / r is a constant and so therefore also is / r (provided G is) and Eqn. 7.2.
can be re-written as
T
B E
A F
E
x
Example
Consider the problem shown in Fig.7.2.4, two torsion members of lengths L 1 (^) , L 2 ,
diameters d (^) 1 , d 2 and shear moduli G 1 (^) , G 2 , built-in at A and subjected to torques TB and
T C. Equilibrium of moments can be used to determine the unknown torques acting in
each member:
T 1 (^) TB TC 0 , T 2 TC 0 (7.2.11)
so that T 1 (^) TB TC and T 2 (^) TC.
Figure 7.2.4: A structure consisting of two torsion members; (a) subjected to torques T B and TC , (b,c) free-body diagrams
The shear stresses in each member are therefore
2
2 1
1 ,^ J
rT J
r TB TC C
where J 1 d 14 / 32 and J 2 d 24 / 32.
From Eqn. 7.2.10, the angle of twist at B is given by B T 1 L 1 / G 1 J 1. The angle of twist
at C is then
C GJ B
TL
2 2
Statically indeterminate problems can be solved using methods analogous to those used in the section 7.1 for uniaxial members.
A
x D
TB T C
L 1 B L 2 C
T 1
(a)
T 2
(b)
(c)
Example
Consider the structure in Fig. 7.2.5, similar to that in Fig. 7.2.4 only now both ends are built-in and there is only a single applied torque, TB.
Figure 7.2.5: A structure consisting of two torsion members; (a) subjected to a Torque T (^) B , (b) free-body diagram, (c) separate elements
Referring to the free-body diagram of Fig. 7.2.5b, there is only one equation of equilibrium with which to determine the two unknown member torques:
T 1 TB T 2 0 (7.2.14)
and so the deformation of the structure needs to be considered. A systematic way of dealing with this situation is to consider each element separately, as in Fig. 7.2.5c. The twist in each element is
2 2
2 2 2 1 1
1 1 1 ,^ GJ
TL
GJ
T L
The total twist is zero and so 1 2 0 which, with Eqn. 7.2.14, can be solved to obtain
B T B
LGJ LGJ
LGJ
T T
LGJ LGJ
LGJ
T
1 2 2 2 1 1
1 2 2 2 1 2 2 2 1 1
2 1 1 1 ,^
The rotation at B can now be determined, B 1 2.
A
x D
T B
L 1 B L 2 C
T 1
(a)
(b)
TB T^2
(c) T 1 T 1 T 2 T 2
