Determine the modulus of elasticity, Poisson's ratio and the shear modulus.

[texasfight]

Homework Statement

A rod with a diameter of 1.00 in and length of 6.0 ft undergoes an axial deformation of 0.150 in when subjected to an axial force of 52.0 kip. The diameter of the rod decreases by 0.0007 in at this load. Determine the modulus of elasticity, Poisson's ratio and the shear modulus for the rod's material.

Homework Equations

E = PL/Aδ
√=-ΔDL/DΔL
G=E/2(1+√)

The Attempt at a Solution

E = (52.0 kip)(72.0 in)/(((∏(1.0in)²)/4) = 31800Ksi
√ = (-(-0.0007 in)(2.00 in))/((1.00 in)(0.150in)) = 0.009
G = 31800 ksi/(2(1+0.0009)) = 15900 Ksi

Somehow the correct answer for the modulus of elasticity is about 30 Ksi, the Poisson's ratio is 0.3 and the shear modulus is about 10000 ksi

 

[Chestermiller]

Homework Statement

A rod with a diameter of 1.00 in and length of 6.0 ft undergoes an axial deformation of 0.150 in when subjected to an axial force of 52.0 kip. The diameter of the rod decreases by 0.0007 in at this load. Determine the modulus of elasticity, Poisson's ratio and the shear modulus for the rod's material.

Homework Equations

E = PL/Aδ
√=-ΔDL/DΔL
G=E/2(1+√)

The Attempt at a Solution

E = (52.0 kip)(72.0 in)/(((∏(1.0in)²)/4) = 31800Ksi
√ = (-(-0.0007 in)(2.00 in))/((1.00 in)(0.150in)) = 0.009
G = 31800 ksi/(2(1+0.0009)) = 15900 Ksi

Somehow the correct answer for the modulus of elasticity is about 30 Ksi, the Poisson's ratio is 0.3 and the shear modulus is about 10000 ksi

In your calculation for E, you forgot to divide by the $\Delta$L. Also, the correct answer for E is not 30 Ksi, it is 30 Msi. The elastic properties in this problem are those for steel.