Q.A piece of copper having a rectangular cross-section of 15.2 mm x 19.1 mm is pulled in tension with 44,500 N force producing only elastic deformation. Calculate the resulting strain. Given Shear modulus of elasticity of copper is 42 x 10^9 Nm^-2.
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Start your 14-day free trial to unlock the full solution →Strain = Stress/Modulus of elasticity; computing stress as force/area from the given rectangular cross-section and dividing by the given modulus (42x10^9 Pa) gives strain approximately 3.65 x 10^-3.
Given:
- Cross-section dimensions: 15.2 mm x 19.1 mm
- Tensile force, F = 44,500 N
- Modulus of elasticity relating this tensile stress and strain (as given), E = 42 x 10^9 N m^-2
- Deformation is purely elastic (so Hooke's law applies: stress proportional to strain)
Step 1 -- Convert cross-section to SI units and find area:
Width w = 15.2 mm = 15.2 x 10^-3 m
Thickness t = 19.1 mm = 19.1 x 10^-3 m
A = w x t = (15.2 x 10^-3) x (19.1 x 10^-3) = 290.32 x 10^-6 m^2 = 2.9032 x 10^-4 m^2
Step 2 -- Calculate the tensile stress:
Stress, sigma = F / A = 44500 / (2.9032 x 10^-4)
sigma is approximately 1.5328 x 10^8 N m^-2 (approximately 153.3 MPa)
Step 3 -- Apply Hooke's law to find strain:
For small elastic deformation, stress is directly proportional to strain, with the modulus of elasticity as the constant of proportionality:
Strain = Stress / E
Strain = (1.5328 x 10^8) / (42 x 10^9)
Strain is approximately 3.65 x 10^-3
Step 4 -- Sanity check: …
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