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:\(τ_{max,abs}=\frac{{σ _{max}}-{σ _{min}}}{{2}}=\frac{σ_1}{2}\)