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

  • Thread starter Thread starter shreddinglicks
  • Start date Start date
  • Tags Tags
    Force
Click For Summary

Discussion Overview

The discussion revolves around determining the shear force and moment at a specific point (Point C) in a beam subjected to a distributed load. Participants are exploring the calculations necessary to find these values, including the total load on the beam and the application of static equilibrium equations. The context is primarily homework-related, with participants seeking clarification on their approaches and calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants are attempting to calculate the force at "FAY" and the moment at point B using the equation M = FAY(18) - ?(6) = 0, but are uncertain about the value to replace the question mark.
  • There is a discussion about the total load on the beam, with some participants calculating it as 27 kip using the formula 1/2 * 3 * 18.
  • Participants express confusion about how to determine the correct load at point C and the significance of the load distribution along the beam.
  • Some participants suggest that using point A as a reference for calculations may be more beneficial than using point C.
  • There is mention of needing to calculate the load distribution at intermediate points between A and B, with suggestions to use ratios or linear interpolation.
  • One participant expresses feeling "clueless" about the calculations and the application of the distributed load concept.

Areas of Agreement / Disagreement

Participants generally agree that the total applied load acting on the beam is 27 kip, but there is disagreement and confusion regarding how to calculate the shear force and moment at point C. Multiple competing views on the correct approach to the problem remain unresolved.

Contextual Notes

Some participants highlight the importance of clear equations of static equilibrium and the need to accurately account for the distributed load, which varies as a function of position from point A. There are unresolved questions about the correct reference points for calculations and the interpretation of the load distribution diagram.

shreddinglicks
Messages
225
Reaction score
7

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?
 

Attachments

  • media_23b_23b57840-0912-48fe-ade5-c350d7ef588c_php1gMEoH.png
    media_23b_23b57840-0912-48fe-ade5-c350d7ef588c_php1gMEoH.png
    4.3 KB · Views: 563
Physics news on Phys.org
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!
 

Similar threads

How Do I Draw This Shear and Moment Diagram?
  • · Replies 12 ·
Replies
12
Views
2K
Loader Boom maximum bending moment
  • · Replies 2 ·
Replies
2
Views
2K
Given force, need to determine what bearing to use for a crane
  • · Replies 69 ·
3
Replies
69
Views
5K
Engineering Analyzing Shear & Bending on Idealized Cross Section
  • · Replies 4 ·
Replies
4
Views
2K
Shear force and bending moment
  • · Replies 16 ·
Replies
16
Views
4K
Engineering Statics - Determine the Reactions on this bent bar levering between two surfaces
  • · Replies 27 ·
Replies
27
Views
4K
Shear force diagram and bending moment diagram
  • · Replies 10 ·
Replies
10
Views
3K
Engineering Statics, Finding Reaction forces for A and H
  • · Replies 6 ·
Replies
6
Views
3K
Engineering 2nd Degree Inderteminacy for Structure Using Force Method
  • · Replies 3 ·
Replies
3
Views
3K
Engineering Calculate shear force and bending moments with a UDL and a point load
  • · Replies 2 ·
Replies
2
Views
5K