Acylindrical specimen of some hypothetical alloy is stressed in compression. if its original and final diameters are 30.00 and 30.04mm respectively, and its final length is 130.00 mm, calculate its original length if the deformation is totally elastic. the elastic and shear moduli for this material are 65.5 and 25.4 gpa respectively.

Lin = 136.28 mm

Explanation:

Given

Din = 30.00 mm

Dfin = 30.04 mm

Lfin = 130.00 mm

Lin = ?

E = 65.5 GPa

G = 25.4 GPa

The deformation is totally elastic

Knowing that

G = E / (2*(1+υ))    ⇒  υ = (E-2G) / (2G)

⇒  υ = (65.5 GPa-2*25.4 GPa) / (2*25.4 GPa)

⇒  υ = 0.2893

then we apply

υ = - εtransv / εlong  ⇒  εlong = - εtransv / υ

If

εtransv = (Dfin-Din) / Din = (30.04 mm-30.00 mm) / (30.00 mm) = 0.0133

then

εlong = - εtransv / υ = - 0.0133 / 0.2893 = - 0.046

Finally, we can get Lin as follows

εlong = ΔL / Lin = (Lfin - Lin) / Lin

⇒  Lin = Lfin / (1+εlong) = (130.00 mm) / (1-0.046)

⇒  Lin = 136.28 mm

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explanation: