In a tension test of steel, the ultimate load was 13,100 lb and the elongation was 0.52 in. The original diameter of the specimen was 0.50 in. and the gage length was 2.00 in. Calculate (a) the ultimate tensile stress (b) the ductility of the material in terms of percent elongation

Answer:

a) the ultimate tensile stress is 66717.8 psi

b) the ductility of the material in terms of percent elongation is 26%

Explanation:

Given the data in the question;

ultimate load P = 13,100 lb

elongation δl = 0.52 in

diameter of specimen d = 0.50 in

gage length l = 2.00 inch

First we determine the cross-sectional area of the specimen

A = [tex]\frac{\pi }{4}[/tex] × d²

we substitute

A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²

A = 0.1963495 in²

a) the ultimate tensile stress σ[tex]_u[/tex]

tensile stress σ[tex]_u[/tex] = P / A

we substitute

tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495

tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi

Therefore, the ultimate tensile stress is 66717.8 psi

b) ductility of the material in terms of percent elongation;

percentage elongation of specimen = [change in length / original length]100

% = [ δl / l ]100

we substitute

% = [ 0.52 in / 2.00 in ]100

= [ 0.26 ]100

= 26

Therefore, the ductility of the material in terms of percent elongation is 26%

Assuming you determine the required section modulus of a wide flange beam is 200 in3, determine the lightest beam possible that will satisfy this condition.