Answer:
The answers are as follow:
a) 10 mm
b) 12.730 N/
c) 127.307 N/
d) 0.25
Explanation:
d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9
Force = F =1000 N
let us find initial area first, A1 = pi* = 78.55
using reduction in area formula : 0.9 = (A1 - A2 ) / A1
solving it will give, A2 = 0.1 A1 = 7.855
a) The specimen elongation is final length - initial length
50 - 40 = 10 mm
b) Engineering stress uses the original area for all stress calculations,
Engineering stress = force / original area = F / A1 = 1000 / 78.55
Engineering stress = 12.730
c) True stress uses instantaneous area during stress calculations,
True fracture stress = force / final area = F / A2 = 1000 / 7.855
True Fracture stress = 127.30
e) strain = change in length / original length
strain = 10 / 40 = 0.25
