Chapter 2

Q. 2.8

A high strength steel rod is compressed by axial force P. When there is no axial load, the diameter of the rod is 50 mm. In order to maintain certain clearances, the diameter of the rod must not exceed 50.02 mm. What is the largest permissible load P? Given E = 200 GPa, Poisson’s ratio = 0.3.


Verified Solution

This problem is almost similar to the previous one. We first calculate the lateral strain and longitudinal strain and then arrive at the value of load P.

\text { Lateral strain }=\varepsilon_{\text {lat }}=\frac{\text { change in diameter }}{\text { original diameter }}=\frac{50.02-50}{50}=4 \times 10^{-4}

\text { Longitudinal strain }=\varepsilon_{ long }=\frac{\sigma}{E}=\frac{P}{A \cdot E}=\frac{P}{\left(\pi 50^2 / 4\right) 200 \times 10^3}

\text { On simplification, } \varepsilon_{\text {long }}=2.546 \times 10^{-9} P

We know that,  \frac{1}{m}=\frac{\varepsilon_{\text {lat }}}{\varepsilon_{\text {long }}}, \quad \therefore \quad 0.3=\frac{4 \times 10^{-4}}{2.546 \times 10^{-9} P}

on solving,          P = 523598.78 N            or             523.6 kN.