Torsion occurs when an object, such as a bar with a cylindrical or square cross section (as shown in the figure), is twisted. The twisting force acting on the object is known as torque, and the resulting stress is known as Shear stress. If the object undergoes deformation as a result of and in the direction of the application of the force, the resulting deflection is known as Strain. Twisting a simple piece of blackboard chalk between ones fingers until it snaps is an example of a torsional force in action. A common example of torsion in engineering is when a transmission drive shaft (such as in an automobile) receives a turning force from its power source (the engine).
When a pure torque acts upon a shaft, shear stresses will be set up which act in directions
perpendicular to the radius at all points.
The complimentary shear stresses on longitudinal planes will cause the distortion of filaments
which were originally in the longitudinal direction. Assume that the points lying on a radius
before twisting will remain on that radius, and that the angle of twist is uniformly
over the length of the shaft.
The diagram shows the shear strain of elements which are at a distance r from the axis (
When T is constant is constant). The original line OA has twisted to OB, and the
angle ACB is now , the relative angle of twist for cross sections a distance l apart .
Compare the weights of equal lengths of hollow and solid shaft to transmit a given torque
for the same maximum shear stress, if the inside diameter is 2/3 of the outside.
From equation (7)
for a solid shaft:
This occurs frequently in practice when a shaft is subjected to both bending and inertia forces.
Stresses are set up due to gravity, torque, and shear forces, although the latter is usually
unimportant since its maximum value occurs at the neutral axis where the bending stress is zero.
For the purposes of design it is necessary to calculate all the principle stresses( maximum shear
stress, shear strain energy etc.) so that these can be used to compare them with the criterion of
If is the greatest bending stress and s is the greatest shear stress due to bending,
For loading in the vertical plane, these stresses will occur together at the end of a vertical
diameter, since there is no normal stress on longitudinal planes of the shaft, they are maximum
A flywheel weighing 1200lb. is mounted on a shaft 3in. in diameter and midway between two bearings
24in. apart. If the shaft is transmitting 40 h.p. (1 h.p. = 550 ft.lb./sec)Calculate the maximum
shear stress at the end of a vertical and horizontal diameter in a plane close to the flywheel.
thus on the "tension" side of the shaft the principle stresses are 3250 lb./sq.in. in tension and
528 lb./sq.in. compression. On the "Compression side the principle stresses are 3250lb./sq.in. in
compression and 528lb./sq.in. in tension.
From equation ( ) the maximum shear stress is given by:-