# How do I find the total force?

1. Nov 15, 2014

### shreddinglicks

1. The problem statement, all variables and given/known data
determine the sheer force and the moment acting at a section passing through point C in the beam. w = 3 kip/ft

2. Relevant equations
M = FR

3. 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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2. Nov 15, 2014

### SteamKing

Staff Emeritus
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.

3. Nov 15, 2014

### shreddinglicks

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?

4. Nov 15, 2014

### SteamKing

Staff Emeritus
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.

5. Nov 15, 2014

### shreddinglicks

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: Nov 15, 2014
6. Nov 15, 2014

### SteamKing

Staff Emeritus
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.

7. Nov 15, 2014

### shreddinglicks

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.

8. Nov 15, 2014

### SteamKing

Staff Emeritus
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)

9. Nov 15, 2014

### shreddinglicks

So you mean I do a [(change in y) / (change in x)] sorta 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: Nov 15, 2014
10. Nov 15, 2014

### SteamKing

Staff Emeritus
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. Nov 15, 2014

### shreddinglicks

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