a. Derive an expression for elastic potential energy of a strained body.

**OR**

b. Show that the relation between the three coefficients of thermal expansion is given by $\gamma = \dfrac{3\beta}{2} = 3\alpha$.

Question

a. Derive an expression for elastic potential energy of a strained body.

**OR**

b. Show that the relation between the three coefficients of thermal expansion is given by $\gamma = \dfrac{3\beta}{2} = 3\alpha$.

Accepted Answer

>

> [!TLDR]
> When a body is elastically strained (e.g. a wire stretched within its elastic limit), the work done by the external force is stored as elastic potential energy; this is found by integrating the work done against the internal restoring force as the strain builds up from zero.
>
> Consider a wire of natural length $L$ and cross-sectional area $A$, being stretched by an external force. Within the elastic limit, Young's modulus $Y = \dfrac{	ext{stress}}{	ext{strain}} = \dfrac{F/A}{\Delta l/L}$, so at an extension $x$ (where $0\le x\le \Delta L$), the restoring (and hence applied) force is:
>
> $F(x) = \dfrac{YAx}{L}$
>
> The work done in stretching the wire by a further small amount $dx$ is $dW = F(x)\,dx$. The total work done (= elastic PE stored) in stretching from $x=0$ to $x=\Delta L$ is:
>
> $U = \int_0^{\Delta L} F(x)\,dx = \int_0^{\Delta L} \dfrac{YAx}{L}\,dx = \dfrac{YA}{L}\cdot\dfrac{(\Delta L)^2}{2}$
>
> $U = \dfrac12 \left(\dfrac{Y\Delta L}{L}ight)\left(\dfrac{\Delta L}{L}ight)(AL) = \dfrac12\,(	ext{stress})(	ext{strain})\,(	ext{volume})$
> …

Q.a. Derive an expression for elastic potential energy of a strained body. OR b. Show that the relation between the three coefficients of thermal expansion is given by $\gamma = \dfrac{3\beta}{2} = 3\alpha$ .

Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

The Intuition: Stiffness per Unit Size

The Precise Definition

What the Number Tells You

A Worked Example

Q.a. Derive an expression for elastic potential energy of a strained body. OR b. Show that the relation between the three coefficients of thermal expansion is given by γ=3β2=3α\gamma = \dfrac{3\beta}{2} = 3\alphaγ=23β​=3α.

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Q.a. Derive an expression for elastic potential energy of a strained body. OR b. Show that the relation between the three coefficients of thermal expansion is given by $\gamma = \dfrac{3\beta}{2} = 3\alpha$ .

Unlock everything free for 14 days