16.
Derive an expression for energy stored in a wire? Find the work done in stretching a wire of cross-section
1mm and length 2m through Omm. Young's modulus for the material of wire is 2 x 10' N/m?
Answers
Answer:
In stretching a wire work is done against internal restoring forces. This work is stored as elastic potential energy or strain energy. If a force F acts along the length L of the wire or cross section A and stretches it by x then: Young′sModulus(Y)=StressStrain=F/Ax/L=FLAx⇒F=YALx.
Explanation:
In stretching a wire work is done against internal restoring forces. This work is stored as elastic potential energy or strain energy.
If a force F acts along the length L of the wire or cross section A and stretches it by x then:
Young′sModulus(Y)=StressStrain=F/Ax/L=FLAx⇒F=YALx.
So, the work done for an additional small increase dx in length,
dW=Fdx=YALx.dx.
Hence, total work done in increasing the length (l):
W=∫0ldW=∫0lF.dx=12YALl2, this work done is stored in the wire
Therefore, Energy stored in wire, U=12YAl2L=12Fl(As,F=YAlL).
Now, dividing both sides by volume of the wire we get energy stored in unit volume of wire.
UV=12FlAL = ½ x Stress x Strain = ½ x Y x (Strain)²
UV=12Y(Stress)2 (As AL = Volume of Wire)
If the force on the wire is increased from F₁ to F₂ and the elongation in wire is l then energy stored in the wire, U=12(F1+F22)l.
Thermal energy density = Thermal energy per unit volume = ½ x Thermal Stress x Strain
=12FlAL = ½ (Y α Δθ) (α Δθ) = ½ Y α² (Δθ)².
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