A cylindrical specimen of a metal alloy 48.0 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 368 MPa causes the specimen to plastically elongate to a length of 53.5 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 48.0 mm to a length of 57.4 mm.

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mariamtomori mariamtomori · Answer 1 · 2019-11-12T17:26:06+01:00

Answer:

the true stress for the second material is 629.19MPa

Explanation:

given for material 1,

l₀ = 48mm

l₁ = 53.5m

e = change in length = 53.5 - 48 = 5.5mm

true stress = 368MPa

strain = Δlength/original length = e/l₀

stress = force / area = F/A

young modulus, E = stress/strain = Fl₀/Ae

E = stress * l₀/e

E = 368 * 48/5.5 = 368 * 8.73 = 3212.64

for material 2 of the same properties, meaning E for both material is the same, therefore,

l₀ =48, l₁=57.4, e = 57.4-48

e=9.4mm

E = stress * l₀/e

3212.64 = stress * (48/9.4)

3212.64 = 5.106 * stress

stress for 3212.64/5.106 = 629.189MPa

the true stress for the second material is 629.19MPa