Question

A bar of 30 mm diameter is subjected to a pull of 60KN. The measured extension on gauge length of 200mm is 0.09mm and the change in diameter is 0.0039. calculate the Poisson’s ratio and the value of the three moduli. ​

Answer 1

Answer:

Data: d=30 mm, L=200 mm, P =60 kN, δL=0.09 mm, δd = 0.0039 mm Calculate: μ and E

A=πd24=π×3024=706.858mm2

E=PLAδL=60×103×200706.858×0.09=188628.08N/mm2

E=1.89N/mm2

μμμ= Lateral Strain Linear Strain =(δdd)(δLL)=(0.003930)(0.09200)=0.29=0.29

Answer 2

POISSON'S RATIO= 0.29

Modulus of elasticity = 1.89N/$mm^{2}$

D = 30mm

L = 200mm

P = 60kN

δL = 0.09mm

δd = 0.0039mm

μ = ?

E = ?

Area = $\pi (D)(24) = 3024\pi = 706.858m^{2}$

E = PLAδL = (60)(103)(200)(706.858)(0.09) = 188628.08N/$mm^{2}$

E = 1.89N/$mm^{2}$

μ = Lateral strain

Linear strain = (LδL)(Dδd) = (0.003930)(0.09200) = 0.29.]

Therefore, the poisson's ratio is 0.29 and the modulus of elasticity is 1.89N/$mm^{2}$.