**Branch :**First Year-Engineering Syllabus

**Subject :**Basic Manufacturing Process

## Torsion

**Torsion:**** **Many cutting processes induce failure by shear, such as cutting of sheet metal (also called shearing), cutting paper with a pair of scissors, punching processes that cut holes and slots in sheet metal, etc. Therefore it is useful to study how a material behaves under pure shearing stresses. This is done by the use of a torsion test.

Torsion is a twisting force, called a moment, applied (usually) to a bar. It introduces a shear stress in the material. Consider a solid cylindrical bar at equilibrium and subjected to torsion as shown in figure 9a below. A torque T is applied to it, causing it to twist by an angle of θ over the length L. Initially, the bar remains in the elastic deformation range. Then, due to symmetry, every planar cross-section parallel to the end-faces in the bar gets twisted, but remains planar. So there is a pure shear in the interface between any imaginary cross-section plane. Also, every radial line in the solid is turns around the axis, but remains as a straight line. For such a bar, the shear stress is maximum at the surface and zero at the center along the axis. The stress increases linearly along the radius, and the following relations hold:

Angle of twist: θ = TL/GJ ....................(1)

Shear stress: τ = Tr/J ......................................(2)

Maximum shear stress = τmax = TR/J..................... (3)

Shear strain = γ = rθ/L .................................................(4)

where

T = torque,

L = cylinder length,

r = distance from axis of the cylinder, and R = radius of the cylinder

J = polar moment of inertia of the cross section of the bar, and is computed as J = ∫ r2 dA

For a solid cylinder, J = πd4/32;

For a hollow cylinder with outer diameter D and inner diameter d, J = π( D4−d4)/32

G is a constant called the modulus of rigidity.

Because the stress and strain vary along the radius, failure begins at the outer surface. It is conventional to measure torsion properties using a thin, hollow cylinder, since for such a part, the shear stress across the cross section is nearly constant. In this case, the above formulas (1), (3) and (4) remain valid by using the correct value of J. The formula for shear stress can be simplified as:

Shear stress for thin wall tube (Figure 9b) = τ = T/(2 π r2 t) ..............................(5)

where t = thickness of the tube = (D-d)/2, and r = mean radius = (D d)/2.

Figure 1 (a). Solid cylinder under torsion (b) Thin walled cylinder under torsion