Answer to Question #219906 in Mechanical Engineering for moham

Question

A steel thick cylinder of external diameter 150 mm has two strain gauges fixed

externally, one along the longitudinal axis and the other at right angles to read the hoop strain. The cylinder is subjected to an internal pressure of 75 MN/m²

and this causes the following strains:

(a) hoop gauge: 455 × 10^-6 tensile.

(b) longitudinal gauge: 124 × 10^-6 tensile.

Find the internal diameter of the cylinder assuming that Young’s modulus for steel is 208 GN/m² and Poisson’s ratio is 0.283.

Answer 1

Since ,  = 0 at the outside surface of the cylinder for zero external pressure

Hoop's strain  and longitudinal strain  can be written as

$\epsilon_H= \frac{1}{E}(\sigma_H-v\sigma_L)$  --------------------------- (1)

$\epsilon_L= \frac{1}{E}(\sigma_L-v\sigma_H)$  ---------------------------- (2)

where E = Young's modulus for steel,  = stress,  = Poisson's ratio

Putting values in equation (1) & (2), we obtain

 $455*10^{-6}*208*10^9 = \sigma_H-0.283 \sigma_L= 94640*10^3$  ------------------ (3)

And,

  $124*10^{-6}*208*10^9 = \sigma_L-0.283 \sigma_H= 25792*10^3$ -------------------(4)

Solving equation $0.283 * (3) + (4)$ , we get

$\sigma_H= 111MN/m^2\\ \sigma_L=57.2MN/m^2$

Now, we have

$\sigma_H= 111MN/m^2$  at r = 0.075m and   $\sigma_r$ = 0 at the outside surface

$\therefore \sigma_H= A+\frac{B}{r^2}\\ 111=A+177.8B \& 0 =A-177.8B\\ A=55.5\\ B=0.312\\$

Therefore for the internal radius R₁ ,

$-75=55.5-\frac{0.321}{R_1^2}\\ \implies R_1=49 mm$

Or, Internal Diameter = 2R₁ = 98 mm