Change In Dimensions Of A Columbium Member Under Tensile Load

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Change In Dimensions Of A Columbium Member Under Tensile Load

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 10^{6}

poisson's ratio μ = 0.28.

Assuming elastic behavior, determine:

(a) total change in Length

(b) lateral strain

(c) total change in Volume

**The strings**:
S_{7}P_{5}A_{51} (Change - Physical Change).
**The math**:

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 10^{6}) = 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 in^{3}

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The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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