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A cylindrical specimen of some metal alloy 10.0 mm (0.3937 in.) in diameter is stressed elastically in tension. A force of 15500 N (3485 lbf) produces a reduction in specimen diameter of 9 × 10-3 mm (3.543 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).

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farhansiddiqui97

Answer:

Poisson ratio for this material is 196.56.

Explanation:

formula:

Poisson's ratio = lateral strain / axial strain

lateral strain = ΔD/D

axial strain = F/EA

Solution:

Poisson's ratio = (ΔD*E*A) / (F*D)

A = pi*( d / 2)∧2

A = 3.142*(0.39/2)∧2 = 0.12

so,

Poisson ratio = ( 0.3937*14.5*10∧6*0.12) / 3485

= 196.56 (ratio has no unit).

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