A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
1 answer:
Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
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Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:
A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:
D = 35.5%.
Thus, the percentage ductility is 35.5%.