For a wire of length L and cross-section area A subjected to a deforming force F, show that $U = \dfrac{1}{2} imes stress imes strain$.

Question

For a wire of length L and cross-section area A subjected to a deforming force F, show that $U = \dfrac{1}{2} 	imes stress 	imes strain$.

Accepted Answer

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> [!TLDR]
> The elastic potential energy stored per unit volume of a stretched wire works out to exactly $	frac12$ stress $	imes$ strain.
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> Consider a wire of natural length $L$ and cross-sectional area $A$, stretched by a deforming force that increases gradually from $0$ to $F$, producing a total extension $\Delta L$. Within the elastic limit, the restoring force is proportional to extension, so at an intermediate extension $x$ the applied force is $f(x) = \dfrac{F}{\Delta L}x$ (increasing linearly from 0 to $F$).
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> The work done in stretching the wire (which is stored as elastic potential energy $U_{total}$) is
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> $$U_{total} = \int_0^{\Delta L} f(x)\,dx = \int_0^{\Delta L}\dfrac{F}{\Delta L}x\,dx = \dfrac{F}{\Delta L}\cdot\dfrac{(\Delta L)^2}{2} = \dfrac12 F\,\Delta L$$
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> Now write $F = 	ext{stress}	imes A$ (since stress $=F/A$) and $\Delta L = 	ext{strain}	imes L$ (since strain $=\Delta L/L$):
> …

Q.For a wire of length L and cross-section area A subjected to a deforming force F, show that $U = \dfrac{1}{2} \times stress \times strain$ .

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.For a wire of length L and cross-section area A subjected to a deforming force F, show that U=12×stress×strainU = \dfrac{1}{2} \times stress \times strainU=21​×stress×strain.

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Q.For a wire of length L and cross-section area A subjected to a deforming force F, show that $U = \dfrac{1}{2} \times stress \times strain$ .

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