A brass rod with a length of 1.15 m and a cross-sectional area of 2.39 cm2 is fa

Question

A brass rod with a length of 1.15 m and a cross-sectional area of 2.39 cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 1.22 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 3.05×104 N at its ends.

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

b)What is the stress in the brass rod?

c)What is the stress in the nickel rod?

d)What is the strain in the brass rod?

e)What is the strain in the nickel rod?

Can you explain and place equations used. Thanks.

Explanation / Answer

A.

elongation in both rods is equal

e1 = e2

F*L1/Y1*A1 = F*L2/Y2*A2

1.15/(9*10^10*2.39*10^-4) = L2/(21*10^10*1.22*10^-4)

L2 = 1.15*21*10^10*1.22*10^-4/(9*10^10*2.39*10^-4)

L2 = 1.37 m

B.

stress in brass rod = F/A1 = 3.05*10^4/(2.39*10^-4)

stress = 1.28*10^8 N/m^2

C.

Stress in Nickel rod = F/A2 = 3.05*10^4/(1.22*10^-4)

Stress = 2.5*10^8 N/m^2

D.

Strain = Stress/Y

for brass

strain = 1.28*10^8/(9*10^10) = 1.42*10^-3

E.

strain = 2.5*10^8/(21*10^10) = 1.19*10^-3

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