Units of N m

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.

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

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.

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.

Example #1 If a twisting moment of 10000 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

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.