Question

A tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN

Answer 1

Answer:

i) 796.18 N/mm^2

ii) 1111.11 N/mm^2

Explanation:

Initial diameter ( D ) = 12 mm

Gage Length = 50 mm

maximum load ( P ) = 90 KN

Fractures at =  70 KN

minimum diameter at fracture = 10mm

Calculate the engineering stress at Maximum load and the True fracture stress

i) Engineering stress at maximum load = P/ A

= P / $\pi \frac{D^2}{4}$  = 90 * 10^3 / ( 3.14 * 12^2 ) / 4

= 90,000 / 113.04 = 796.18 N/mm^2

ii) True Fracture stress =  P/A

= 90 * 10^3 / ( 3.24 * 10^2) / 4

= 90000 / 81  =  1111.11 N/mm^2