Question

A brass rod with a length of 1.23 m and a cross-sectional area of 1.53 cm2 is fastened end to end to a nickel rod with length Land cross-sectional area 1.29 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 3.57×10⁴ Nat its ends.

a- Find the length L of the nickel rod if the elongations of the two rods are equal. (in m)

b- What is the stress in the brass rod? (in Pa)

c- What is the stress in the nickel rod? (in Pa)

d- What is the strain in the brass rod?

e- What is the strain in the nickel rod?

Answer 1

Young's Modulus is given by:

Y = F*L/(A*dL)

Change in length or elongation will be:

dL = F*L/(A*Y)

Part A.

Given that elongation in both rod is same, So

dL_brass = dL_Nickel

F*Lb/(Ab*Yb) = F*Ln/(An*Yn)

Lb/(Ab*Yb) = Ln/(An*Yn)

Ln = Length of Nickel Rod = Lb*An*Yn/(Ab*Yb)

Yn = Young's modulus of Nickel = 21*10^10 Pa

Yb = Young's modulus of Brass = 9*10^10 Pa

An = 1.29 cm^3

Ab = 1.53 cm^2

Lb = 1.23 m

So,

Ln = 1.23*1.29*21*10^10/(9*10^10*1.53)

Ln = length of Nickel rod = 2.42 m

Part B.

Stress In brass Rod will be

Stress = Force/Area

Area = 1.53 cm^2 = 1.53*10^-4 m^2

Stress = 3.57*10^4/(1.53*10^-4)

Stress = 2.33*10^8 N/m^2 = 2.33*10^8 Pa

Part C.

Stress In Nickel Rod will be

Stress = Force/Area

Area = 1.29 cm^2 = 1.29*10^-4 m^2

Stress = 3.57*10^4/(1.29*10^-4)

Stress = 2.77*10^8 N/m^2 = 2.77*10^8 Pa

Part D.

Since we know that

Y = Stress/Strain

Strain = Stress/Y

For Brass rod

Strain = 2.33*10^8/(9*10^10) = 2.58*10^-3 m = 2.58 mm

Part E.

For Brass rod

Strain = 2.77*10^8/(21*10^10) = 1.32*10^-3 m = 1.32 mm

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