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?

Asked by Abhisek | 1 year ago |  140

Solution :-

Area of the copper piece, A=19.1×10−3 ×15.2×10−3

=2.9×10−4m2

Tension force applied on the piece of copper, F=44,500 N

Modulus of elasticity of copper, Y=42×10Nm−2

Modulus of elasticity (Y) = $$\dfrac{Stress}{Strain}$$

=($$\dfrac{F}{A}.\dfrac{1}{Strain}$$)

Strain =$$\dfrac{F}{YA}$$

$$\dfrac{44500}{(2.9×10^{−4}×42×10^9)}$$

= 3.65×10−3

Answered by Abhisek | 1 year ago

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