The concepts of shearing stress and shearing strain have an important application in torsion problems and find an important place in GATE Mechanical Engineering and other engineering exams. Such problems arise in shafts transmitting, heavy torques, in eccentrically loaded beams, in aircraft wings and fuselages, and many other instances.

The simplest torsion problem is that of the twisting of a uniform thin circular tube; the tube shown in figure below is of thickness t, and the mean radius of the wall is r, L is the length of the tube. Shearing stresses τ are applied around the circumference of the tube at each end, and in opposite directions.

If the stresses τ are uniform around the boundary, the total torque T at each end of the tube is,

Thus the shearing stress around the Circumference due to an applied torque T is,

We consider next the strains caused by these shearing stresses. We note firstly that complementary shearing stresses are set up in the wall parallel to the longitudinal axis of the tube. If  is a small length of the circumference then an element of the wall ABCD, figure given above, is in a state of pure shearing stress. If the remote end of the tube is assumed not to twist, then the longitudinal element ABCD is distorted into the parallelogram ABC'D', the angle of shearing strain being.

if the material is elastic, and has a shearing (or rigidity) modulus G. But if  is the angle of twist of the near end of the tube we have

Hence,

It is sometimes more convenient to define the twist of the tube as the rate of change of twist per unit length; this is given by , and from equation (5) this is equal to,

Torsion on solid circular shafts

The torsion of a thin circular tube is a relatively simple problem as the shearing stress may be assumed constant throughout the wall thickness. The case of a solid circular shaft is more complex because the shearing stresses are variable over the cross-section of the shaft. The solid circular shaft of figure given below has a length L and radius a in the cross-section.

When equal and opposite torques T are applied at each end about a longitudinal axis we assume that

1. the twisting is uniform along the shaft, that is, all normal cross-sections the same distance apart suffer equal relative rotation;
2. cross-sections remain plane during twisting; and
3. radiii remain straight during twisting.

If  is the relative angle of twist of the two ends of the shaft, then the shearing strain  of an elemental tube of thickness  and at radius is

If the material is elastic, and has a shearing modulus G, then the circumferential shearing stress on this elemental tube is,

The thickness of the elemental tube is , so the total torque on this tube is

The total torque on the shaft is then,

Substitute the value of  from equation (8), we will get

Now,

The above given value is the polar moment of area of the cross section about an axis through the centre, and is usually denoted by J. Then equation (9) can be written as,

We can combine equations (8) and (11) as,

From equation (8), we can see that shear stress varies linealy with radius, from zero at the centre of the shaft to  at the circumference. Along any radius of the cross section, the shearing stresses are normal to the radius and in the plane of the cross section, as depicted the figure below.

.

For hollow shafts, we can use the same equation as equation (12) for mathematical calculations.

In the next article, we will discuss about the principal stresses in a twisted shaft.