Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm (0.31in.). A tensile force of 1000N (225 \(lb_{f}\)) produces an elastic reduction in diameter of \(2.8 x 10^{-4}\) mm (\(1.10 x 10^{-5}\) in.). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30.
50Cr: 4.34% A = 49.9460 amu
52Cr: 83.79% A = 51.9405 amu
53Cr: 9.50% A = 52.9407 amu
54Cr: 2.37% A = 53.9389 amu
Prove that \(A_{Cr}\) = 51.9963 amu
\(A = f_{1} * A_{1} + f_{2} * A_{2} + f_{3} * A_{3} + f_{4} * A_{4} \)
4.34% * 49.9460 + 83.79% * 51.9405 + 9.50% * 52.9407 + 2.37% * 53.9389 = 51.9963 amu
51.9963 amu = 51.9963 amu \( \implies \) correct