Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combination is subjected to a tensile force with magnitude 4000 N.

A) Forthe steel rod, what is the strain?

B) Forthe steel rod, what is the elongation?

C) For the copper rod, what is thestrain?

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied) / Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π (D^2)) / 4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257 / (210 * (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl) / (original length, lo)

Elongation, dl = strain * original length

dl = 0.0001078 * 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257 / (120 * (10^9)) = 0.000189