Answer to Question #98139 in Mechanics | Relativity for Anirudh

Question

A cylindrical tank with diameter d = 300 mm is subjected to internal gas pressure p = 2 MPa. The tank is constructed of steel sections that are welded circumferentially. The heads of the tank are hemispherical. The allowable tensile and shear stress are 60 MPa and 24 MPa, respectively. Also, the allowable tensile stress perpendicular to a weld is 40 MPa. Determine the thickness tmin of
(i) cylindrical part of the tank and
(ii) the hemispherical heads.

Answer 1

Determine the thickness t_min of cylindrical part of the tank and the hemispherical heads.

We need to determine the thickness of t_min of cylindrical part of the tank and the hemispherical heads.

Answer:

Given Information is

Diameter of the Tank = d = 300 mm

So, radius = r = 150 mm

(i). Minimum Thickness of Cylinder:

Let t be the thickness of the Cylindrical part of the Tank

Maximum tensile =

$$= \frac { p . r}{t}$$

$$t _{min}= \frac { p . r}{\sigma_{allow}} = \frac {21 \times 150}{60} = 5 mm$$

Shear =

$$Max = \frac {p \times r}{2t}$$

$$t _{min}= \frac { p . r}{2 \sigma_{allow}} =\frac {21 \times 150}{2 \times 24} = 6.2 mm$$

(ii). Minimum thickness of hemisphere:

Tension:

max =

$$= \frac { p . r}{2t}$$

$t_{min} =$

$$= \frac { p . r}{{2 \sigma_{allow}}}=\frac {21 \times 150}{2 \times 60} = 2.5 mm$$

Shear:

$\tau _{max}$ =

$$= \frac { p . r}{4t}$$

$$t_{min} = \frac {pr}{4 τ_{max} } = \frac {21 \times 150}{4 \times 24} =3.125mm$$