A master stud pattern is laid out somewhat like a?
1 answer:
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Answer:
a) diameter available = 0.0384 nm
b)The space is smaller than the carbon atom which has a radius of 0.077 nm and this simply means that the carbon atom will not conquer these sites
Explanation:
For BCC iron
From Appendix B given,select the lattice parameter ( a ) as = 0.2866 nm
The BCC iron has 4 atomic radii and therefore the body diagonal length =
expressing the atomic radius of the BCC iron
4r =
insert the value of (a) from appendix B which is = 0.2866 nm
4r =
therefore r = 0.4964 nm / 4 = 0.1241 nm
Refer again to appendix C given select the atomic radius of the BCC iron as = 0.1241 nm assuming the atomic radius of the iron are the same
then the radius ratio = 0.62
Refer to the Figure 3.2 given, the amount of space required for an interstitial at the BCC position is between the atoms at the FCC position and also in this space there are two atoms that are equal to a radius of 0.2482 nm
The diameter of the minimum space available
= 0.2482 nm
a = 0.2666 nm
therefore
d = 0.2866 nm - 0.2482 nm = 0.0384 nm
comparing this to the diameter of a carbon atom
The space is smaller than the carbon atom which has a radius of 0.077 nm and this simply means that the carbon atom will not conquer these sites
Answer:
Explanation:
The formula for critical stress is
K is the plane strain fracture toughness
Y is dimensionless parameters
We are to Determine the Critical stress
Now replacing the critical stress with 54.8
a with 0.2mm = 0.2 x 10⁻³
Y with 1
The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa
Therefore, the specimen does not failure for surface crack of 0.2mm