Answer:
hello some part of your question is missing below is the complete question
answer :
A) 162750 Ib.ft
B) - 64950 Ib.ft
Explanation:
Applying Muller-Breslau's law
we will make assumptions which include assuming an imaginary hinge at G
therefore the height of I.LD for B.M at G = ( 12 * 8 ) / 20 = 4.8
height of I.L.D at C = 2.4 ( calculated )
height of I.L.D at F = 1.5 ( calculated )
A) Determine Maximum positive moment produced at G
= [ (1/2 * 20 * 4.8 ) ( 600 + 300 ) ] + [ ( 25 * 4.8 * 10^3 ) ] - [ ( 1/2 *2.4*20 ) * 300 ] + [ (1/2 * 1.5 * 10 ) ( 600 + 300 ) ]
= 162750 Ib.ft
B) Determine the maximum negative moment produced at G
= [ ( 1/2 * 20 * 4.8 ) * 300 ] - [ ( 1/2 * 2.4 * 20 ) ( 600 + 300 ) ] - [ (2.5 * 10^3 * 2.4 ) ] + [ ( 1/2 * 1.5 * 10) * 300 ]
= - 64950 Ib.ft
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the full step by step explanation to your question.
Answer:
<u>Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)</u>
Explanation:
<u>Step 1.</u>
Taking derivative of the equation with respect to 'r' we get:
d/dr(EN) = - A/r² - nB/r^(n+1)
Setting this equation to zero:
<u>Step 2.</u>
Solving for r:
- A/r² - nB/r^(n+1) = 0
A/r² + nB/r^(n+1) = 0
Ar^(n+1) + nBr² = 0
Ar^(n+1) = - nBr²
[r^(n+1)]/r² = - nB/A
r^(n+1-2) = - nB/A
r^(n-1) = - nB/A
Taking power 1/(n-1) on both sides:
r = [-nB/A]^(1/n-1)
This is the value of ro:
ro = [-nB/A]^(1/n-1)
<u>Step 3.</u>
Substituting value of ro in eqn we get value of Eo
<u>Eo = A/[-nB/A]^(1/n-1) + B/[-nB/A]^(n/n-1)</u>
Answer:
Net discharge per hour will be 3.5325
Explanation:
We have given internal diameter d = 25 mm
Time = 1 hour = 3600 sec
So radius
We know that area is given by
We know that discharge is given by , here A is area and V is velocity
So
So net discharge in 1 hour =
Answer:
option B.
Explanation:
The correct answer is option B.
Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.
In-plane shear stress are the shear stress which is acting on the plane.
The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.