Question

Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long rod changes to 1.02 m; (b) The length of a 2-m-long rod changes to 1.05 m; (c) The length of a 3-m-long rod changes to 3.07 m.

Answer 1

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

$\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}$

Here,

$\delta$= Change in final length $(l_0)$ and the initial length $l$

PART A)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{1.02-1}{1}$

$\epsilon = 0.01961$

PART B)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{2-1.05}{2}$

$\epsilon = 0.475$

PART C)

$\epsilon = \frac{\delta_1}{l}$

$\epsilon = \frac{l_0-l}{l_0}$

$\epsilon = \frac{3.07-3}{3}$

$\epsilon = 0.0233$

Therefore the rank of this deformation would be  B>C>A