Strain Energy Stored In An Elastic Material
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Strain Energy Stored In An Elastic Material

A 15 in member is to be designed using a safety factor of 1.50, to withstand a tensile load of 6000 lb. The three choices of material available are:
(a) Aluminum Alloy, for which modulus of Elasticity, E = 10 x 106 psi and tensile stress σy = 52, 0000 psi
(b) Magnesium Alloy, for which modulus of Elasticity, E = 6.5 x 106 psi and tensile stress σy = 28,500 psi
(c) Molded Nylon, for which modulus of Elasticity, E = 410,000 psi and tensile stress σy = 8000 psi.

Calculate the total amount of strain energy stored by each member at the 6000 lb load.

The strings: S7P3A31 (Force - Pull).

The math:
Pj Problem of Interest is of type force (pull).

Formulas of interest:
Stress, σ = Load/Area; Strain, ε = δ/Length; where δ is total elongation
Stress = (modulus of elasticity)(strain)
Poisson's ratio, μ = lateral strain/longitudinal strain
Offset yield strength σy = (working stress σw)(safety factor)
Total strain energy, U = ∫0δ P(x) dx.
Where x is elongation and P(x) = kx is force as a function of elongation.
Strain energy per unit volume = σ2/(2E)

(a) yield strength, σy = [Load(safety factor)]/(Cross-section Area)
So, 52,000 = 6,000(1.50)/A
So, A = 6,000(1.5)/52,000 = 0.173 in2.
Total strain energy, U at Load can be calculated by integrating P(x) = kx from 0 to δ
Where k, (load/δ) is the slope of the graph relating load and elongation when material is elastic.
So, U = 6,000[(σy)/(1.5E)](length/2)
So, total strain energy, U = 6,000[(52,000/(1.5 x 10 x 106)](15/2)
= 156 lb-in.

(b) A = 6,000(1.5)/28,500 = 0.316 in2.
So, total strain energy, U = 6,000[(28,500/(1.5 x 6.5 x 106)](15/2)
= 132 lb-in.

(c) A = 6,000(1.5)/8000 = 1.125 in2
So, total strain energy, U = 6,000[(8,000/(1.5 x 410,000)](15/2)
= 586 lb-in.

Total strain energy can also be calculated by first calculating strain energy per unit volume and then multiplying by total volume.
So, total strain energy = [σ2/(2E)]V.

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