# Torsion (mechanics)

File:Twisted bar.png
Torsion of a square section bar.

In solid mechanics, torsion is the twisting of an object due to an applied torque, therefore is expressed in N·m or ft·lbf. In sections perpendicular to the torque axis, the resultant shear stress in this section is perpendicular to the radius.

For solid shafts of uniform circular cross-section or hollow circular shafts with constant wall thickness, the torsion is:

$T=\frac{J_{T}}{r}\tau=\frac{J_{T}}{\ell}G\theta$

where:

• $\tau$ is the maximum shear stress at the outer surface.
• JT is the torsion constant for the section. It is identical to the second moment of area Jzz for concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and finite element analysis (FEA) may be the best method.
• r is the distance between the rotational axis and the furthest point in the section (at the outer surface).
• is the length of the object the torque is being applied to or over.
• θ is the angle of twist in radians.
• G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2.
• The product JT G is called the torsional rigidity wT.

## Properties

The shear stress at a point within a shaft is:

$\tau_{{\varphi_{{z}}}}={Tr\over J_{T}}$

Note that the highest shear stress is at the point where the radius is maximum, the surface of the shaft. High stresses at the surface may be compounded by stress concentrations such as rough spots. Thus, shafts for use in high torsion are polished to a fine surface finish to reduce the maximum stress in the shaft and increase its service life.

The angle of twist can be found by using:

$\varphi={T\ell\over w_{T}}.$

## Sample calculation

Calculation of the steam turbine shaft radius for a turboset:

Assumptions:

• Power carried by the shaft is 1000 MW; this is typical for a large nuclear power plant.
• Yield stress of the steel used to make the shaft (τyield) is: 250 x 106 N/m².
• Electricity has a frequency of 50 Hz; this is the typical frequency in Europe. In North America the frequency is 60 Hz.

The angular frequency can be calculated with the following formula:

$\omega=2\pi f$

The torque carried by the shaft is related to the power by the following equation:

$P=T\omega$

The angular frequency is therefore 314.16 rad/s and the torque 3.1831 x 106 N·m.

The maximal torque is:

$T_{\max}=\frac{{\tau}_{\max}J_{{zz}}}{r}$

After substitution of the polar moment of inertia the following expression is obtained:

$d=\sqrt[3]{\frac{16T_{\max}}{\pi{\tau}_{\max}}}$

The diameter is 40 cm. If one adds a factor of safety of 5 and re-calculates the radius with the maximal stress equal to the yield stress/5 the result is a diameter of 69 cm, the approximate size of a turboset shaft in a nuclear power plant.

## Failure mode

The shear stress in the shaft may be resolved into principal stresses via Mohr's circle. If the shaft is loaded only in torsion then one of the principal stresses will be in tension and the other in compression. These stresses are oriented at a 45-degree helical angle around the shaft. If the shaft is made of brittle material then the shaft will fail by a crack initiating at the surface and propagating through to the core of the shaft fracturing in a 45-degree angle helical shape. This is often demonstrated by twisting a piece of blackboard chalk between one's fingers.