Correct Answer - Option 3 : pd / 8t

__Concept:__

At any point in the material of the thin cylinder shell, there are two principal stresses.

Longitudinal stress: \({σ _L} = \frac{{pd}}{{4t}}\)

Hoop stress: \({σ _h} = \frac{{pd}}{{2t}} = 2{σ _L}\)

These two stresses are tensile and perpendicular to each other.

__Calculation:__

__Given:__

Internal pressure, pi = p , diameter = d, thickness = t

Maximum shear stress on the surface is maximum in-plane shear stress and is given by:

\({\tau _{max}} = \frac{{{σ _1} - {σ _2}}}{2} = \frac{{{σ _h} - {σ _L}}}{2} = \frac{\frac{pd}{2t}-\frac{pd}{4t}}{2}=\frac{\frac{pd}{4t}}{2}=\frac{pd}{8t}\)