An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that E 5 10.1 3 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.

Question

Chemistry

Michaela Garrison

9 May, 05:02

An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that E 5 10.1 3 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.

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Answers (1)

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Essence Hull · Answer 1

a) L = 36.07 in

b) A = 9.1 in^2

Explanation:

The length of the pipe will be calculated using the formula of the deformation under tensile load. The required area can be obtained by dividing the load that is applied over the area of the cross section. We have the following dа ta:

D = diameter = 5 mm

P = 127.5 ksi

E = 10.1 x 10^6 psi

δ = deformation = 0.05 in

σ = stress = 14 ksi

L = ?

A = ?

a)

δ = (P*L) / (A*E)

Clearing L:

L = (δ*A*E) / L = (δ*E) / σ = (0.05*10.1x10^6) / 14x10^3 = 36.07 in

b)

A = P/σ = 127.5x10^3/14x10^3 = 9.1 in^2