How Do I Determine Shear Force and Moment at Point C?

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To determine the shear force and moment at point C in the beam, it is essential to calculate the total applied load using the distributed load value of 3 kip/ft at point C. The total load from point A to point C can be calculated as 1/2 * base * height, resulting in 9 kip for a triangular load distribution. The correct approach involves using point A as the reference for calculating xbar and ensuring that the load distribution is accurately represented. It is crucial to apply the correct values for w, as it is given in kip/ft, not kip. Understanding these principles will help in accurately solving for the shear force and moment at point C.
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Homework Statement


determine the sheer force and the moment acting at a section passing through point C in the beam. w = 3 kip/ft

Homework Equations


M = FR

The Attempt at a Solution


I am trying to find the force at, "FAY." So I tried to get the moment at B.

M = FAY(18) - ?(6) = 0

I am trying to figure out what goes in place of the question mark. I got the length of six by finding my xbar = 18/3.
The solution manual I saw gives the force that goes in place of the question mark to be 27 kip. How do I get that?
 

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shreddinglicks said:

The Attempt at a Solution


I am trying to find the force at, "FAY." So I tried to get the moment at B.

M = FAY(18) - ?(6) = 0

I am trying to figure out what goes in place of the question mark. I got the length of six by finding my xbar = 18/3.
The solution manual I saw gives the force that goes in place of the question mark to be 27 kip. How do I get that?

What's the total load on the beam, given the distribution as shown in the diagram? You know the loading is 0 kip/ft at point A and 3 kip/ft at point C.

BTW, this same problem has been ventilated at length recently in this forum. Scroll back about 10 days and you should find the other threads which cover this exact problem.
 
SteamKing said:
What's the total load on the beam, given the distribution as shown in the diagram? You know the loading is 0 kip/ft at point A and 3 kip/ft at point C.

BTW, this same problem has been ventilated at length recently in this forum. Scroll back about 10 days and you should find the other threads which cover this exact problem.

I did 1/2*3*18 = 27 Is that correct?

I tried to apply that to get torque about C.

M = 9(6) + 3(2) = 0

found xbar = 6/3
I tried the same thing, 1/2*6*3 = 9. How do I get the 3 that is shown in the solutions?
 
shreddinglicks said:
I did 1/2*3*18 = 27 Is that correct?

I tried to apply that to get torque about C.

M = 9(6) + 3(2) = 0

found xbar = 6/3
I tried the same thing, 1/2*6*3 = 9. How do I get the 3 that is shown in the solutions?

Yes, 27 kip is the total applied load acting on the beam. (Please show units on your calculations)

It's not clear what reference you are using to calculate xbar.

As always, write clear equations of static equilibrium for these problems. It saves a lot of time and guess work.

Remember, you have a distributed load which varies as a function of position from point A. Once you calculate the ordinate of the load distribution at point C, then you can calculate the total applied load acting on the beam between points A and C.
 
SteamKing said:
Yes, 27 kip is the total applied load acting on the beam. (Please show units on your calculations)

It's not clear what reference you are using to calculate xbar.

As always, write clear equations of static equilibrium for these problems. It saves a lot of time and guess work.

Remember, you have a distributed load which varies as a function of position from point A. Once you calculate the ordinate of the load distribution at point C, then you can calculate the total applied load acting on the beam between points A and C.

I calculated xbar using point C as my reference.

Since I know FAY = 9 kip
To calculate the load at point C do I just do 9 + FC = 0

so FC = -9 kip?

But if I take that value of -9 and get the total load from A to C, 1/2*9*6 = 27. Not the 3 kip the solutions has.

How do I know the total applied load acting on point A to B is 3 kip?
 
Last edited:
Like I said, you have to calculate the applied load between points A and C using the load distribution as shown in the OP. You cannot ignore this step and expect to get the correct value of the shear force at point C.

BTW, referring xbar to point C is not helpful here. It's better to use point A as the reference.
 
SteamKing said:
Like I said, you have to calculate the applied load between points A and C using the load distribution as shown in the OP. You cannot ignore this step and expect to get the correct value of the shear force at point C.

BTW, referring xbar to point C is not helpful here. It's better to use point A as the reference.

I really am clueless. If I just make a free body diagram from point A to C I have a triangle with a length of 6 feet. I have 3 kip/ft applied to the triangle.

I do 1/2*6*3 = 9 kip. Not the 3 kip I should have as the total applied load.

Regardless of where I take my xbar from, I have a center of 4 ft from point A or 2 feet from point C.
 
shreddinglicks said:
I really am clueless. If I just make a free body diagram from point A to C I have a triangle with a length of 6 feet. I have 3 kip/ft applied to the triangle.

You have a distributed load value of w = 3 kip/ft applied only at x = 18 feet from point A. Is the diagram not clear to you on this point?

If you want to find values of w at intermediate points between A and B, you have to calculate them. (Hint: think ratios or linear interpolation)
 
SteamKing said:
You have a distributed load value of w = 3 kip/ft applied only at x = 18 feet from point A. Is the diagram not clear to you on this point?

If you want to find values of w at intermediate points between A and B, you have to calculate them. (Hint: think ratios or linear interpolation)
SteamKing said:
You have a distributed load value of w = 3 kip/ft applied only at x = 18 feet from point A. Is the diagram not clear to you on this point?

If you want to find values of w at intermediate points between A and B, you have to calculate them. (Hint: think ratios or linear interpolation)

So you mean I do a [(change in y) / (change in x)] sort of thing?

I was thinking since it is 3 kip at 18 ft, so 12 ft is 2 kip, and 6 ft is 1 kip? Is that the idea here?
 
Last edited:
  • #10
shreddinglicks said:
So you mean I do a [(change in y) / (change in x)] sort of thing?

I was thinking since it is 3 kip at 18 ft, so 12 ft is 2 kip, and 6 ft is 1 kip? Is that the idea here?

Yes, exactly. You cannot calculate the correct applied load between A and C unless you use the correct values for w. But w is in kip/ft, not kip.
 
  • #11
SteamKing said:
Yes, exactly. You cannot calculate the correct applied load between A and C unless you use the correct values for w. But w is in kip/ft, not kip.

Thank you, That was the 1st problem of that nature I had to solve. Thanks again!
 

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