Chapter 6, Problem 6.29P

Interpretation Introduction

Interpretation:

The dimensions of the polymer bar are 1in× 2in×15in. The modulus of elasticity of bar is 600,000psi. Value of force which is required to stretch the bar from 15 into 15.25 in is to be determined.

Concept introduction:

The concept of strain, stress and Hooke's law will be used. The formula of strain and Hooke's law is given below -

  $e(engg.strain)=\frac{\Delta l}{l}$

The relation between true strain and engineering strain is given as-

$\epsilon =\ln (1+e)$

$\sigma (stress)=\frac{F}{A}$

And Hooke's law-

$\sigma \propto \epsilon$