The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
Answer:
Stress corrosion cracking
Explanation:
This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.
Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.
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- Telephone
- Face-to-face discussions
- Contact with others
- Importance of being exact or accurate.
O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:
- Workforce development professionals
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On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.
According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:
1. Telephone
2. Face-to-face discussions
3. Contact with others
4. Importance of being exact or accurate.
Read more on work contexts here: brainly.com/question/22826220
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, 
= 0.51
Void ratio in the loosest state, 
= 0.87
Now,
Dry density, 
= 17.14 kN/m³
Also,
here, G = Specific gravity = 2.7 for sand
or
e = 0.545
Relative density = 
= 
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%
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
