Question 2.13: A cylindrical pressure tank of mean diameter 1.6 m, is fabri...

(a) Cylindrical tank:

First consider the equilibrium of forces at a transverse cut for evaluating longitudinal stress \sigma _1 as shown in Fig. 2.20.

Inward force,  p \times \pi r^2 = Outward force, \sigma_1 \times \pi r \times t

or, \sigma_1=\frac{p r}{2 t}=\frac{0.8 \times 800}{2 \times 8}=40  MPa

Similarly consider the section of a unit length of tank as shown in Fig. 2.20 for circumferential stress \sigma_2,

or, \sigma_2=\frac{p r}{t}=\frac{0.8 \times 800}{8}=80  MPa
Let stresses across and along the weld be \sigma_{x^{\prime}}  and  \tau_{x^{\prime}y^{\prime}}  \text{Angle between}  \sigma_{x^{\prime}}  and  \sigma_1=20^{\circ}. Using Eqs (2.1) and (2.2),

\sigma _{x^{\prime}}=\frac{\sigma _x+\sigma _y}{2}+\frac{\sigma _x-\sigma _y}{2}\cos 2\theta -\tau _{xy}\sin 2 \theta           (2.1b)

\tau _{x^{\prime}y^{\prime}}=\frac{\sigma_x- \sigma_y}{2}\sin 2\theta +\tau _{xy}\cos 2\theta                  (2.2)

\sigma_{x^{\prime}}=\frac{40+80}{2}+\frac{40-80}{2} \cos 40^{\circ} = 44.68 MPa (tension)

\tau_{x^{\prime}y^{\prime}}  =\frac{40-80}{2} \sin 40^{\circ} = -12.86 MPa (Anticlockwise)

(b) Spherical cap:

The stress in spherical cap is same in all directions due to symmetry. A section of a sphere along any diameter will be similar to the transverse section of the cylinder in Fig. 2.20.

∴     \sigma_1=\sigma_2=\frac{p r}{2 t}=\frac{0.8 \times 800}{2 \times 10}=32  MPa

Shear stress in the plane of the sphere is zero. The shear stress on the walls of the vessel is greater if third dimension is also considered.