Answer:
The right solution is "2625 kN".
Explanation:
According to the question,
The average pressure will be:
= 
By putting values, we get
= 
= 
= 
hence,
The average force will be:
= 
= 
= 
Or,
= 
Answer:
a) 2∪p/lb (l+b)dH
b) po exp( 4∪x/l)
Explanation:
please check the attachment for proper explanation and proper sign notations thanks.
Answer:
Step 1 of 3
Case A:
AISI 1018 CD steel,
Fillet radius at wall=0.1 in,
Diameter of bar
From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel
Tensile strength
Yield strength
The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.
At the critical stress elements on the top and bottom surfaces transverse shear is zero
Explanation:
See the next steps in the attached image
Answer:
your answer is correct
Explanation:
You have the correct mapping from inputs to outputs. The only thing your teacher may disagree with is the ordering of your inputs. They might be written more conventionally as ...
A B Y
0 0 1
0 1 0
1 0 0
1 1 1
That is, your teacher may be looking for the pattern 1001 in the last column without paying attention to what you have written in column B.