A Cold-Worked Columbium Member Under Tensile Load
Strings (SiPjAjk) = S7P5A51 Base Sequence = 12735 String Sequence = 12735 - 5 - 51
A member made of cold-worked columbium is 15 inches long and has a rectangular cross-section 1/4 in by 3/4 in. This member is under a tensile load of 5,000 lb.
Modulus of elasticity, E = 22.7 x 106
poisson's ratio μ = 0.28.
Assuming elastic behavior, determine:
(a) total change in Length
(b) lateral strain
(c) total change in Volume
S7P5A51 (Change - Physical Change).
Pj Problem of Interest is of type change (physical change).
Formulas of interest:
Stress, σ = Load/Area; Strain, ε = δ/Length; where δ is total elongation
Stress = modulus of elasticity x strain
Poisson's ratio, μ = lateral strain/longitudinal strain
Change in Volume, ΔV = Volume x longitudinal strain x (1 - 2μ).
(a) Cross-section area, A = (1/4) x (3/4) = 3/16.
So, stress, σ = 5,000(16/3) = 26,667 psi
Strain, ε = &sigma/E = 26,667/(22.7 x 106) = 1.135 x 10-3
So, change in length, δ = εL = 1.135 x 10-3 x 17 = 0.017 in.
(b) Lateral strain = -(0.28 x 0.00135) = -0.000318
(c) ΔV = V(ε)(1-2μ) = 15(3/16)(0.00135)(1-0.56) = 0.000167 in3
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