Question

If my material has an ultimate tensile strength of 52 KSU, yield strength of 35 ksi, elastic modulus of 6.5 msi, and a poisssons ratio of 0.33. What is the maximum force that can be reached till failure?

Answer 1

Ans) Given,

   Ultimate tensile strength = 52000 psi

Yield strength = 35000 psi

Elastic modulus(E) = 6.5 x 10⁶ psi

   Poisson ratio = 0.33

We know,

Ultimate Force = Ultimate stress x area

Stress = E x strain

35000 = 6.5 x 10⁶ x strain

   Strain = 0.00538

Strain = P / AE

0.00538 = P / A x 6.5 x 10⁶..................(1)

E = 2G ( 1 + $\nu$)

  6.5 x 10⁶ = 2 G x 1.33

   G = 2.44 x 10⁶ psi

Poisson ratio = Lateral strain / longitudinal strain

0.33 = Lateral strain / 0.00538

Lateral strain = 0.00177

=> 0.0177 = P / A' 2.44 x 10⁶ ...............(2)

A' x 2.44 x 0.0177 = A x 6.5 x 0.00538

A' / A= 0.81

2$\pi$rL / 2$\pi$r² = 0.81   

L/r = 0.81

From 1,

0.00538 = 35000 x L  / 6.5 x 10⁶

L = 1 in

r = 1.23in

=> A = $\pi$r²= 1.187 in²

=> ultimate force = 1.187 x 52000

= 61756.5 lb or 61.75 kips