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Question
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution 1
Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10–3 × 15.2 × 10–3
= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2
Modulus of elasticity, η = `"Stress"/"Strain" = (F/A)/"Strain"`
`:. "Strain" = F/(Aeta)`
`= 44500/(2.9xx10^(-4)xx42xx10^(9))`
`= 3.65 xx 10^(-3)`
Solution 2
A = 15.2 x 19.2 x 10-6 m2; F = 44500 N; η = 42 x 109 Nm-2
Strain = `"Stress"/"modulus of elasticity" = "F/A"/eta`
`=F/(Aeta) = 44500/((15.2xx19.2xx10^(-6))xx42xx10^9) = 3.65 xx 10^(-3)`
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