Chapter 20, Problem 20.1P

To determine

Circumferential stress developed in the walls of the pipe.

Answer to Problem 20.1P

   $S_{t_c}=14953.84psi$

Explanation of Solution

Given:

Diameter of water pipe $D_i=9ft$

Wall thickness of steel plate $t=\frac{13}{16}in=0.8125in$

Internal pressure of water in the pipe $p=225psi$

Calculation:

Thin wall theory is applicable if $t\le 0.1r_i$

   $\begin{array}{l}0.1r_i=0.1\left(\frac{9\times 12}{2}\right)in\\ r_i=5.4in\\ 0.8125in<5.4in(OK)\end{array}$

From the above values, the condition for thin wall theory is satisfied, therefore, the circumferential stress developed in the walls of the pipe is calculated by the formula:

   $\begin{array}{l}{S}_{t_c}=\frac{p{D}_i}{2t}\\ S_{t_c}=\frac{225\times \left(9\times 12\right)}{2\left(0.8125\right)}\\ S_{t_c}=14953.84psi\end{array}$

Conclusion:

Therefore, circumferential stress developed in the walls of the pipe is = $14953.84psi$.