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viktelen
2 months ago
5

A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f

or this specimen is 8,000 ksiand Poisson’s ratio is 0.35. Calculate the final length and the final diameter ofthis specimen under this load assuming that the material remains within thelinear elastic region.
Engineering
1 answer:
Viktor [391]2 months ago
8 0

Answer:

Final Length = 11.992 inches

Final Diameter = 6.001 inches

Explanation:

First, we compute the cross-sectional area:

Area = A = πr² = π(3 inches)² = 28.3 inches²

Next, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 inches²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 inches)/12 inches

Final Length = (-6.63 x 10⁻⁴)(12 inches) + 12 inches

Final Length = 11.992 inches

We know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 inches)/6 inches

Final Diameter = (2.32 x 10⁻⁴)(6 inches) + 6 inches

Final Diameter = 6.001 inches

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