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An aluminum pipe must not stretch more than 0.05 in. when
Chapter 2, Problem 2.5(choose chapter or problem)
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that \(E=10.1 \times 10^{6}\) 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.
Questions & Answers
QUESTION:
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that \(E=10.1 \times 10^{6}\) 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.
ANSWER:Step 1 of 4
We will find the allowable length of the pipe on the wire using the formula for deformation under tensile load. The required cross sectional area can be computed by dividing the applied load by the cross sectional area.
Given data are
Diameter of wire: D = 5 mm
Applied load: P = 127.5 ksi
Modulus of elasticity:
Deformation:
Allowable stress