Answer:
18.75in
Explanation:
Modulus of elasticity = Stress/Strain
Since stress = Force/Area
Given
Force = 1000lb
Area = 0.75sqin
Stress = 1000/0.75
Stress = 1333.33lbsqin
Strain
Strain = Stress/Modulus of elasticity
Strain = 1333.33/5,000,000
Strain = 0.0002667
Also
Strain = extension/original length
extension = 0.005in
Original length = extension/strain
Original length = 0.005/0.0002667
Original length = 18.75in
Hence the original length of the rectangular bar is 18.75in
Answer:
I would say do it at an even pace
Explanation:
Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast
Answer:

Explanation:
Given data:
Diffusion constant for nitrogen is 
Diffusion flux 
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law

Take C_A as point on which nitrogen concentration is 2 kg/m^3

Assume X_A is zero at the surface


Complete question:
A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Answer:
Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection
so that critical flow is subject to detection
Explanation:
We are given:
Plane strain fracture toughness K 
Yield strength Y = 860 MPa
Flaw detection apparatus = 3.0mm (12in)
y = 1.0
Let's use the expression:

We already know
K= design
a = length of surface creak
Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.
Therefore
![a= \frac{1}{pi} * [\frac{k}{y*a}]^2](https://tex.z-dn.net/?f=%20a%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7Bk%7D%7By%2Aa%7D%5D%5E2%20)
Substituting figures in the expression above, we have:
![= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2](https://tex.z-dn.net/?f=%20%3D%20%5Cfrac%7B1%7D%7Bpi%7D%20%2A%20%5B%5Cfrac%7B98.9%20MPa%20%5Csqrt%7Bm%7D%7D%20%7B10%20%2A%20%5Cfrac%7B860MPa%7D%7B2%7D%7D%5D%5E2)
= 0.0168 m
= 17mm
Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection