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A 0.82-in-diameter aluminum rod is 5.5 ft long and carries a load of 3000 lbf. Find the tensile stress, the total deformation, the unit strains, and the change in the rod diameter.

User SRobertJames
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1 Answer

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7 votes

Answer:

Tensile stress = 0.1855Kpsi

Total deformation = 0.0012243 in

Unit strain = 1.855 *10^-5 or 18.55μ

Change in the rod diameter = 5.02 * 10^ -6 in

Step-by-step explanation:

Data given: D= 0.82 in

L = 5.5 ft * 12 = 66 in

load (p) = 3000 (Ibf) /32.174 = 93.243 Ibm

Area = (π/4) D² = (π/4) 0.82² = 0.502655 in²

∴ Tensile stress Rt = P/A = 93.243/0.502655 = 185.50099 pound/in²

Rt = 0.1855 Kpsi

∴ Total deformation = PL / AE = Rt * L/ Eal

= 0.1855 * 10³ * 66 / 10000 * 10³

= 0.0012243 in

∴the unit strains = total deformation / L = 0.0012243/ 66

=0.00001855 = 1.855 *10^-5

= 18.55μ

∴ Change in rod Δd/ d = μ ΔL/L

= (0.33) 1.855 *10^-5 * 0.82

= 5.02 * 10^ -6 in

User Brinda Rathod
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