Question 1.1: A cylindrical pressure vessel has an internal diameter of 2 ...

The expressions for the longitudinal and circumferential stresses produced by the internal pressure may be found in any text on stress \text { analysis }{ }^{1} and are

Longitudinal stress \left(\sigma_{x}\right)=\frac{p d}{4 t}=1.5 \times 2 \times 10^{3} / 4 \times 20=37.5 N / mm ^{2}

Circumferential stress \left(\sigma_{y}\right)=\frac{p d}{2 t}=1.5 \times 2 \times 10^{3} / 2 \times 20=75 N / mm ^{2}

The direct stress due to the axial load will contribute to \sigma_{x} and is given by

A rectangular element in the wall of the pressure vessel is then subjected to the stress system shown in Fig. 1.9. Note that no shear stresses act on the x and y planes; in this case, \sigma_{x} and \sigma_{y} form a biaxial stress system. The direct stress, \sigma_{n}, and shear stress, \tau,, on the plane AB, which makes an angle of 60^{\circ} with the axis of the vessel, may be found from first principles by considering the equilibrium of the triangular element ABC or by direct substitution in Eqs. (1.8) and (1.9). Note that, in the latter case, \theta=30^{\circ} \text { and } \tau_{x y}=0. Then,

\sigma_{n}=\sigma_{x} \cos ^{2} \theta+\sigma_{y} \sin ^{2} \theta+\tau_{x y} \sin 2 \theta  (1.8)

\tau=\frac{\left(\sigma_{x}-\sigma_{y}\right)}{2} \sin 2 \theta-\tau_{x y} \cos 2 \theta  (1.9)

The negative sign for \tau indicates that the shear stress is in the direction BA and not AB. From Eq. (1.9), when \tau_{x y}=0,

\tau=\left(\sigma_{x}-\sigma_{y}\right)(\sin 2 \theta) / 2  (i)

The maximum value of \tau therefore occurs when \sin 2 \theta is a maximum, that is, when \sin 2 \theta=1 \text { and } \theta=45^{\circ} \text { . } Then, substituting the values of \sigma_{x} and \sigma_{y} in Eq. (i),