Q. A wire of cross section A is stretched horizontally between two clamps located 2/ m apart. A weight W kg is suspended from the mid-point of the wire. If the mid-point sags vertically through a distance $x < < l$ the strain produced is
Mechanical Properties of Solids
Solution:
Here change in length is
$\Delta l = [AC + BC ] - 2l$
= $2 (I^2 + x^2)^{1/2} , - 2 l = 2l \left( 1 + \frac{x^2}{l^2} \right)^{1/2}$
= $2 l \left( 1 + \frac{1}{2} \frac{x^2}{l^2} \right) - 2l = \frac{x^2}{l}$
$\therefore $ strain = $\frac{\Delta l}{2 l} = \frac{x^2}{2 l^2}$
