A brass rod with a length of 1.40m and a cross-sectional area of 2.37cm2 is fast

Question

A brass rod with a length of 1.40m and a cross-sectional area of 2.37cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 0.720cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 4.28*10^4N at its ends.

Part A

Find the length L of the nickel rod if the elongations of the two rods are equal.

Part B

What is the stress in the brass rod?

Part C

What is the stress in the nickel rod?

Part D

What is the strain in the brass rod?

Part E

What is the strain in the nickel rod?

Explanation / Answer

dL = L*alpha*dT

a) elongations are same

alpha_brass = 18.7x10^-6

alpha_nickel = 13x10^-6

L1*alpha_brass*dT = L2*alpha_nickel*dT

L2 = 1.4*18.7x10^-6/13x10^-6 = 2.014 m

b) stress_brass = F/A1 = 4.28x10^4/2.37x10^-4 = 1.806x10^8 Pa

c) stress_nickel = F/A2 = 4.28x10^4/0.72x10^-4 = 5.94x10^8 Pa

d) Y_brass = 125 x10^9 Pa

strain = stress/Y_brass = 1.806x10^8/125 x10^9 = 1.448x10^-3

e) Y_brass = 125 x10^9 Pa

strain = stress/Y_brass = 5.94x10^8/200 x10^9 = 2.97x10^-3

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