Correct Answer - Option 3 : 140
Explanation:
Circumferential stress of hoop stress σh
\({σ _1} ={σ _h} = \frac{{pd}}{{2t}} = \frac{{2 \times 14\;}}{{2 \times 0.05}} = 280MPa\)
Longitudinal stress σL
\({σ _2} ={σ _L} = \frac{{pd}}{{4t}} = \frac{{2 \times 14\;}}{{4 \times 0.05}} = 140MPa\)
As this is the case of a thin cylinder:
Radial stress \({σ _r} =0\)
Maximum shear stress \({τ _{max}} = \max \left\{ {\frac{{{\sigma _1} - {\sigma _2}}}{2},\frac{{{\sigma _1}}}{2},\frac{{{\sigma _2}}}{2}} \right\}\)
τmax = \(\frac{{σ _h}-{σ _r}}{{2}}=\frac{{σ _1} }{2} = \frac{280}{2}=140~ MPa\)
Maximum In-Plane shear stress/Surface shear stress:
\(τ_{max,inplane}=\frac{{σ _1}-{σ _2}}{{2}}\)
Maximum wall shear stress/Out plane shear stress/Absolute shear stress:
\(τ_{max,abs}=\frac{{σ _{max}}-{σ _{min}}}{{2}}=\frac{σ_1}{2}\)