Answer:
B probably
Explanation:
Because the prompt doesn't specify what sort of violation it could be anything maybe when they release the metals during the day and so on.
Crazy Guy what do uh mean ?
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is 
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation

Here
- M is the moment which is to be calculated
- I is the moment of inertia given as

Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"



The yield moment is 400 k.in.
Part b:
The strain is given as

The stress at the station 2" down from the top is estimated by ratio of triangles as

Now the steel has the elastic modulus of E=29000 ksi

So the strain is 
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as

The plastic moment is 600 ksi.
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Explanation:
He would work on the thing like in the method you work on your question.