Answer:

Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal
= 0.9
Radial load
= 2600 lbf
To find C10 value by using the formula:

where;


The Weibull parameters include:



∴
Using the above formula:


![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)

Recall that:
1 kN = 225 lbf
∴


OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.
It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
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