Question : A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in with tension of 44,500 N force, producing only elastic deformation. Calculate the resulting strain.
Doubt by Ananya
Solution :
Tension (F) = 44500 N
Area of Rectangular Cross Section of Wire
= l×b
= 15.2 mm × 19.1 mm
= (15.2×19.1)×10-6 m²
= l×b
= 15.2 mm × 19.1 mm
= (15.2×19.1)×10-6 m²
Young's Modulus of Elasticity of Copper (Y)
= 110×109 N/m2
= 110×109 N/m2
We know,
Y = Stress/Strain
Y = Stress/Strain
Y = (F/A)/Strain
Strain = F/AY
Strain = F/AY
= 44500/[(15.2×19.1×10-6)×(110×109)]
= 44500×10-3 / 31935.2
= (445000/319352) × 10-3
= 44500×10-3 / 31935.2
= (445000/319352) × 10-3
= 1.39×10-3 N/m2