Name: Properties of Solid
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: Do you know which are the important mechanical properties of solids? In this video, we will discuss the solid state of matter and its different properties.

Question

Define strain energy and derive the equation for the same.

Accepted Answer

Consider a wire which subjected to tensile stress. We know that work is done against inter-molecular forces in this instance and this work is stored in the wire as elastic potential energy or stain energy.

When a wire of original length L and area of cross-section A is subjected to a deforming force F along the length of the wire, let the length of the wire be elongated by $x$ .

We have, Young's modulus, $Y = \frac{\frac{F}{A}}{\frac{x}{L}} \Rightarrow F = \frac{Y A x}{L}$

The work done for an additional small increase $d x$ in length,

$d W = F d x \Rightarrow \frac{Y A}{L} x . d x$

Hence, total work done in increasing the length by $x$ ,

$W = \int_{0}^{x} F . d x \Rightarrow W = \int_{0}^{x} \frac{Y A}{L} x . d x \Rightarrow W = \frac{Y A x^{2}}{2 L} \Rightarrow W = \frac{1}{2} \times Y \times {(\frac{x}{L})}^{2} \times A L \Rightarrow W = \frac{1}{2} \times Young' s modulus \times {strain}^{2} \times volume of wire \Rightarrow W = \frac{1}{2} \times stress \times strain \times volume of wire$

This work is stored in the wire in the form of elastic potential energy $(U)$ . Therefore, the elastic potential energy per unit volume of the wire $(u)$ is

$u = \frac{1}{2} \times σ \times ε$

Telangana Board Solutions for Chapter: Mechanical Properties of Solids, Exercise 1: EXERCISES

Telangana Board Physics Solutions for Exercise - Telangana Board Solutions for Chapter: Mechanical Properties of Solids, Exercise 1: EXERCISES

Questions from Telangana Board Solutions for Chapter: Mechanical Properties of Solids, Exercise 1: EXERCISES with Hints & Solutions

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