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A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the true stress at the fracture point.

A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the true stress at the fracture point. Correct Answer 518.87 MPa

Engineering stress at maximum load = Pfracture/Afracture => Amax = π (fracture max.)2 => 35000 / π (5)2 *10-6 => 445 MPa So true stress is σ=s(e+1) Engineering strain= the Change in the diameter / original diameter => (6-5)/6 = 1/6 = 0.166 true stress is σ=445(1+0.166) = 518.87 MPa.

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