First slide

Stress and strain

Question

A mild steel wire of length 2L and cross-sectional area A is stretched, well within elastic limit, horizontally between two pillars (figure ). A mass m is suspended from the mid-point of the wire. Strain in the wire is [NCERT Exemplar]

Moderate

A

$\frac{x^{2}}{2 L^{2}}$
B

$\frac{x}{L}$
C

$x^{2} / L$
D

$x^{2} / 2 L$

Solution

Consider the diagram below
Hence, change in length, $Δ L = B O + O C - (B D + D C)$
$= 2 B O - 2 B D (∵ B O = O C, B D = D C)$
$= 2 [B O - B D] = 2 [{(x^{2} + L^{2})}^{1 / 2} - L]$
$= 2 L [{(1 + \frac{x^{2}}{L^{2}})}^{1 / 2} - 1] \approx 2 L [1 + \frac{1}{2} \frac{x^{2}}{L^{2}} - 1] = \frac{x^{2}}{L} (∵ x << L)$
$∴ Strain = \frac{Δ L}{2 L} = \frac{x^{2} / L}{2 L} = \frac{x^{2}}{2 L^{2}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App