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
Answer:
See attached images for the diagrams and tables
Answer:
The percentage of the remaining alloy would become solid is 20%
Explanation:
Melting point of Cu = 1085°C
Melting point of Ni = 1455°C
At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.
The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:
60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy
At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:
40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.
The percentage of remaining alloy in solid is equal to
Solid = (10/50) * 100 = 20%
Answer:21.3%
Explanation:
Given
80 % reduction in tool life
According to Taylor's tool life
=c
where V is cutting velocity
T=tool life of tool
80 % tool life reduction i.e. New tool Life is 0.2T
Thus
=1.213V
Thus a change of 21.3 %(increment) is required to reduce tool life by 80%
