In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.
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Answer:
A) 1040 steel is not a possible candidate for this application
B) 35.94%
Explanation:
Initial length = 100 mm = 0.1 m
Initial diameter ( d ) = 7.5 mm = 0.0075 m
Tensile load ( p ) = 18,000 N
Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m
<u>A) would the 1040 steel be a possible candidate for this application</u>
<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>
stress = p / A0 = ( 18000 ) / ( ) * 0.0075^2
= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa
Yield strength of 1040 steel = 450 MPa
stress = 407.424 MPa
∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )
Therefore 1040 steel is not a possible candidate for this application
<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>
Area1 = ( ) ( 0.006 )^2 = 2.83 * 10^-5 m^2
therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K =
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators