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Q.Prove that elastic energy density is 1/2 x (stress) x (strain).

Meghalaya MboseMBOSE Meghalaya 11th Board 2022Subjective· 2mImportance★★★★★
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The work done per unit volume in elastically deforming a wire, found using Hooke's law and integrating the force over the extension, equals (1/2) × stress × strain.

Step 1: Set up the wire.

Consider a wire of natural length L, cross-sectional area A, and volume V = AL. A force F is applied along its length, producing an extension x (up to a maximum extension l, within the elastic limit).

Step 2: Apply Hooke's law.

Within the elastic limit, F is proportional to the extension x:

F = (YA/L) x

where Y is the Young's modulus of the material (this comes from stress/strain = Y, i.e. (F/A)/(x/L) = Y, rearranged for F).

Step 3: Compute the work done in stretching the wire from 0 to l.

Since F varies linearly with x (starting at 0 when x=0), the work done is the area under the F-x graph, a triangle:

W = ∫₀^l F dx = ∫₀^l (YA/L)x dx = (YA/L) × [x^2/2]₀^l = (YA l^2)/(2L)

Step 4: Divide by the volume to get energy density (energy stored per unit volume).

u = W/V = [(YA l^2)/(2L)] / (AL) = (Y l^2)/(2L^2) = (1/2) Y (l/L)^2

Step 5: Recognize l/L as the strain.

strain = l/L, so:

u = (1/2) Y (strain)^2

Step 6: Express in terms of stress and strain together.

Since Y = stress/strain, substitute Y = (stress)/(strain): …

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