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Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain.

1 Answer

6 votes

Answer:


strain = 1.4 * 10^(-3)

Step-by-step explanation:

As we know by the formula of elasticity that


E = (stress)/(strain)

now we have


E = 110 GPA


E = 110 * 10^9 Pa

Area = 15.2 mm x 19.1 mm


A = 290.3 * 10^(-6)

now we also know that force is given as


F = 44500 N

here we have

stress = Force / Area


stress = (44500)/(290.3 * 10^(-6))


stress = 1.53 * 10^8 N/m^2

now from above formula we have


strain = (stress)/(E)


strain = (1.53 * 10^8)/(110 * 10^9)


strain = 1.4 * 10^(-3)

User Jonah
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