A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 lb. Under this force, the bar elastically deforms so that the gauge length increases to 2.007 in. and the cross section decreases to 0.599 in. x 0.599 in. Determine the modulus of elasticity and Poisson’s ratio for this metal.

modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

Explanation:

Modulus is the ratio of tensile stress to tensile strain

Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain within the direction of the stretching force

And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in

answer:

no man

explanation:

no i dont want it