Forces on a beam, replace with force and couple

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Discussion Overview

The discussion revolves around the analysis of forces and moments acting on a beam, specifically focusing on how to replace these forces with a force-couple combination at a designated point. The scope includes technical reasoning, mathematical calculations, and conceptual clarifications related to statics in engineering mechanics.

Discussion Character

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

Main Points Raised

  • Some participants propose summing forces and moments to analyze the system, while others question the choice of reference point for moment calculations.
  • There is a discussion about the correctness of moment calculations about different points, with some participants asserting that moments can be calculated at any point, while others suggest using a specific point for efficiency.
  • Participants explore the need to establish a sign convention for moments, distinguishing between clockwise and counterclockwise directions.
  • There is confusion regarding the components of forces acting at an angle, with participants discussing how to break these forces into horizontal and vertical components for moment calculations.
  • Some participants express uncertainty about the direction of moments and how to visualize their effects based on the applied forces.
  • Participants correct earlier calculations and clarify the need to organize calculations to avoid errors in determining resultant forces and moments.
  • There is an ongoing effort to compute the equivalent force and moment at point A, with some participants questioning the relevance of certain calculations and the overall goal of the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach for calculating moments and forces, with multiple competing views on the reference points and methods for analysis remaining throughout the discussion.

Contextual Notes

Some calculations are dependent on the correct interpretation of angles and force components, and there are unresolved issues regarding the organization of calculations and the clarity of moment direction. The discussion reflects a learning process with evolving understanding among participants.

bnosam
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Homework Statement


http://oi58.tinypic.com/2a5d4w3.jpg

Homework Equations


[/B]
So I can just sum up all the forces then sum the moments?

The Attempt at a Solution



[/B]
FBx = 0 lb
FCx = 800 cos30 = 692.82 lb
FEx = 0 lb

FBy = -2500 lb
FCy = 800sin30 = 400 lb
FEy = -1200 lb

Resultant force:
FAx = 0 + 692.82 + 0 = 692.82 lb
FAy = -2500 + 400 - 1200 = -3300 lb

Moments about point C:
MOB = (2500)(2) = 5000 lb*ft counterclockwise
MOC = 0 lb*ft
MOE = having trouble with this oneDoes everything seem ok here so far?
 
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It's not clear why you are calculating moments about point C. The problem statement clearly states that the three forces are to be replaced by a force-couple combination acting at point A.
 
I thought I could calculate moments anywhere?
 
bnosam said:
I thought I could calculate moments anywhere?
Yes, you can calculate moments anywhere, but when you are asked to find them at a particular point, it saves time to use that point as the reference for your moment calculations.

BTW, your moment calculations about point C were incorrect, anyway. Remember, the force must act perpendicular to the moment arm in order to calculate the magnitude of the moment.
 
So moment about A:

B: 2500*4 = 10000 ft*lb

That's correct, right?

The reason why I wanted to take them about C is because of the angling on C and to simplify it more.
 
bnosam said:
So moment about A:

B: 2500*4 = 10000 ft*lb

That's correct, right?

The reason why I wanted to take them about C is because of the angling on C and to simplify it more.

The magnitude of this moment is correct, but you need to establish a sign convention for your moment directions. Moments can be either clockwise or counterclockwise, and failing to distinguish between these two directions will lead to erroneous results.

The angle of the arm BCD w.r.t. the horizontal only affects the moment due to the force at C.
 
Oh B should be counterclockwise, right?

so then C should be Moc = (800)(6)? If I get what you're saying?
 
bnosam said:
Oh B should be counterclockwise, right?

Right.

so then C should be Moc = (800)(6)? If I get what you're saying?

Point C is not 6' from point A. Remember the definition of what a moment is.

moment.PNG
 
So because it's on an angle we want only the horizontal component of where C is?
 
  • #10
bnosam said:
So because it's on an angle we want only the horizontal component of where C is?

Yes.

But it's not that simple. Because the force at C is also directed at an angle, it must be broken up into its components in order to calculate the moment acting about point A.
 
  • #11
so we want the cos of the angle? 2*cos(30) = 1.732 ft
 
  • #12
bnosam said:
so we want the cos of the angle? 2*cos(30) = 1.732 ft

The distance from A to C would be 4' + 2' * cos (30°), but the force at C is 800 lbs. acting at 60° to the horizontal. You've got to compute the components of that force. Each force component will create a moment acting about point A.
 
  • #13
4+ 2 cos(30) = 5.73

Moc = (5.73) * 800cos60 = (5.73)*400 = 2292 clockwise, right?

If I'm understanding you, that is.
 
  • #14
bnosam said:
4+ 2 cos(30) = 5.73

Moc = (5.73) * 800cos60 = (5.73)*400 = 2292 clockwise, right?

If I'm understanding you, that is.

Nope. You should draw the components of the force at C on your picture and see how they produce their moments.
 
  • #15
I made a mistake:

4+ 2 cos(30) = 5.73

Moc = (5.73) * 800sin60 = (5.73)*692.82 = 3969.86 clockwise, right?

Does that seem better?
 
  • #16
bnosam said:
I made a mistake:

4+ 2 cos(30) = 5.73

Moc = (5.73) * 800sin60 = (5.73)*692.82 = 3969.86 clockwise, right?

Does that seem better?

The magnitude looks OK. Are you sure the direction of this moment is clockwise w.r.t. point A?

What about the moment due to the horizontal component of the force at C?
 
  • #17
No it should be counterclockwise I believe. Is there a better way to tell despite intuition?

In the case of C:
10.46 * 1200 = 12552 lb*ft
 
  • #18
bnosam said:
No it should be counterclockwise I believe. Is there a better way to tell despite intuition?

In the case of C:
10.46 * 1200 = 12552 lb*ft
This is actually the moment due to the force applied at point E.

What about the moment due to the horizontal component of the force applied at point C?

Examination of the direction of the force in relation to the reference point is fine. There are different methods of calculating moments which take care of having to visualize which direction the moment will take, but you will learn those later.
 
  • #19
SteamKing said:
This is actually the moment due to the force applied at point E.

What about the moment due to the horizontal component of the force applied at point C?

Examination of the direction of the force in relation to the reference point is fine. There are different methods of calculating moments which take care of having to visualize which direction the moment will take, but you will learn those later.
Sorry, I mistyped and meant E :p

Wouldn't the horizontal component of C cause a counterclockwise rotation also? Since it would push against the angle and push it upwards.
 
  • #20
bnosam said:
Wouldn't the horizontal component of C cause a counterclockwise rotation also? Since it would push against the angle and push it upwards.

Yes, it would.
 
  • #21
So all the data I have:

FBx = 0 lb
FCx = 800 cos30 = 692.82 lb
FEx = 0 lb

FBy = -2500 lb
FCy = 800sin30 = 400 lb
FEy = -1200 lb

Resultant force:
FAx = 0 + 692.82 + 0 = 692.82 lb
FAy = -2500 + 400 - 1200 = -3300 lb

Moments about point A: (anticlockwise positive)
MOB = (2500)(4) = 10000 lb*ft

MOC = Vertical + Horizontal Components
MOC = 3969.86 + 5(400) lb*ft after squaring and adding then square rooting, magnitude is 4445.20 lb*ft
MOE = 12552 lb*ft

Result of adding moments: 26997.2 lb*ft

I'm not sure what I'd do next to complete the problem to replace it with a force and couple at A, assuming I did the above correctly.
 
  • #22
bnosam said:
FCx = 800 cos30 = 692.82 lb
FCy = 800sin30 = 400 lb

For some reason, you have ignored your previous calculation of the components of the force at C.
 
  • #23
I'm not sure what you mean?
 
  • #24
bnosam said:
I'm not sure what you mean?
If you'll take a moment and review your previous posts in this thread, it should become apparent what I mean.
 
  • #25
SteamKing said:
If you'll take a moment and review your previous posts in this thread, it should become apparent what I mean.
I read it over twice and I'm still not understanding what you mean.
 
  • #26
bnosam said:
I read it over twice and I'm still not understanding what you mean.
Review post Nos. 15 & 16 in this thread.

A big part of obtaining a successful solution to problems is organizing your calculations and keeping them straight in your head or on paper.
 
  • #27
SteamKing said:
Review post Nos. 15 & 16 in this thread.

A big part of obtaining a successful solution to problems is organizing your calculations and keeping them straight in your head or on paper.

Ohh did you mean:

FCx = 800 cos60 = 400 lb
FCy = 800sin60 = 692.82 lb
 
  • #28
bnosam said:
Ohh did you mean:

FCx = 800 cos60 = 400 lb
FCy = 800sin60 = 692.82 lb

Yeeessss. ;)
 
  • #29
SteamKing said:
Yeeessss. ;)
So our resultant:

0 = FAx + 400
FAx = -400 lb

0 = FAy -2500 - 1200 + 692.82
FAy = -3007.18 lb
 
  • #30
bnosam said:
So our resultant:

0 = FAx + 400
FAx = -400 lb

0 = FAy -2500 - 1200 + 692.82
FAy = -3007.18 lb

The magnitudes of the forces look OK, but the OP isn't looking for the reactions at point A, just the equivalent force and moment.

BTW, where is your moment calculation for point A?
 

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