Question 3.3: A cylindrical water tank of height 25 m has inner diameter 2...

The pressure will be maximum at the base of the cylinder and that maximum pressure obviously forms the basis of design calculations.

p_{\max }=\text { Specific weight of water } \times \text { height }

=(10 × 10³) × 25

=0.25  MN / m ²

= 0.25 MPa

Considering circumferential stress, we can write

\sigma_1=\frac{p r}{t \eta}                (1)

where η is the longitudinal joint efficiency = 0.70. In this context, it may be mentioned that the circumferential joints are often assumed to be parallel, horizontal whereas longitudinal joints are vertically aligned (refer to Figure 3.9).

In the previous problem too, this specific concept was used. Now, we know from Chapter 1 that

\sigma_1=\text { Working stress }=\frac{\text { Yield stress }}{\text { Factor of safety }}=\frac{210}{3}=70  MPa

Putting this value in Eq. (1), we get

70=\frac{0.25 \times(2.2 \times 1000) / 2}{t \times 0.70}

or         t = 5.6 mm

Therefore, the thickness of steel tube is 5.6 mm.