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Maximum shear stress at any point in a thin cylinder of diameter (d) and thickness (t) subjected to an internal fluid pressure (p) is given by-
1. pd / 2t
2. pd / 4t
3. pd / 8t
4. pd / 6t

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Correct Answer - Option 3 : pd / 8t


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.



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

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

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

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