Question 4.4: Draw the Mohr’s circle for the biaxial stress element shown ...

The given element, subjected to biaxial stress is in ‘pure shear’. No normal stress acts on the surfaces.
In case of circular shaft subjected to torsional moment such elements of pure shear are found. In this case, \sigma_{\text {average }}=0 \text { and } \tau_{\max }=20  MPa . So, we can draw the Mohr’s circle with centre at (0,0) and radius 20 MPa as shown in Figure 4.13.

From the Mohr’s circle, we find \sigma_{\max }=20  MPa \text { and } \sigma_{\min }=-20  MPa . These are the principal stresses.
To locate the principal planes, that is, the planes on which these stresses act, one has to rotate.

A by 90° in clockwise direction from x plane to reach point A′ on the Mohr’s plane. Similarly, to get B′, one has to rotate A in counterclockwise direction by 90°. On Mohr’s circle, if these angles are 90° each, in case of the body the angles should be 45° each. Thus, the planes are making 45° angles with x-axis as shown in Figure 4.13(b).
Note that we have already known that cases of pure shear occur in circular shafts subjected to torsional moment. Now we find that maximum tensile or compressive stress develops on the planes making 45° angles with positive x-axis, that is, axis of the shaft (Figure 4.14).

If the shaft is made up of ductile material and the stress developed is high enough, it will suffer failure along the plane with maximum shear and in this case, that plane is vertical or horizontal. If the shaft is made up of ductile materials, like aluminium or mild steel, failure occurs perpendicular to the x-axis, that is, axis of the shaft as they are weak in shear. For brittle materials, failure occurs along the plane with maximum tensile stress that is making an angle 45° with axis of the shaft. In case of wood, a natural composite, resistance to shear in axial direction is low. So while subjected to torsional moment, they are more susceptible to failure due to shear along axial direction, parallel to the fibres.