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:
a) W = 25.5 lbf
b) W = 150 lbf
Explanation:
Given data:
Mass of astronaut = 150 lbm
local gravity = 5.48 ft/s^2
a) weight on spring scale
it can be calculated by measuring force against local gravitational force which is equal to weight of body
W = mg
b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity
thus W= 150 lbf
Answer:
The detailed explanation of answer is given in attached file.
Explanation:
