A tensile specimen with a 12mm initial diameter and 50mm gage length reaches maximum load at 90KN and fractures at 70KN the minimum diameter at fracture is 10mm

determine the engineering stress at maximum load and the true fracture stress.

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 /  = 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

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