1. Units of N m

2. Torsion results from a loading that leads to a twisting action
A measure of the deformation of
the shaft and represents the shear
strain due to the shear stresses.
Shear strain  = (arc length B to B’) / l eqn A
This is the same value along the surface for
any value of l.

3. Amount of deformation (width of the pie slice) varies
with respect to the radial distance r measured from the
centerline of the shaft. This angle is called  and is the
angle of twist.
 = (arc length B to B’) / r0 eqn B
Now we can combine equations A and B above and solve
for the angle of twist in terms of the shear strain as:
 = l  / r0

4. Since the shaft as a whole is in static equilibrium
its individual parts have to be in static equilibrium. This
requires the presence of internal shearing forces that
are distributed over the cross-sectional area of the shaft.
The intensity of these internal forces per unit area is the
shear stress . Hence there exists in the object an
internal resisting torque.

5. Applied torque is T reaction Cut the object Shows the shear
stresses at some
distance  from
the axis. These
shear stresses
are acting on the
shaded area.

9. Example #1
If a twisting moment of lb in is impressed upon
a 1.75 inch diameter shaft what is the maximum shearing
stress that is developed? Also what is the angle of twist in a
4 ft length of the shaft?
Assume the material is steel for which G = 12 x 106 lb/in2.
Also assume elastic behavior.
max

10. Example #2
Consider a thin walled tube subject to torsion. Derive an
approximate expression for the allowable twisting moment
if the working stress limit in shear is given as w.
Assume elastic deformations.