Forces on a beam, replace with force and couple

AI Thread Summary
The discussion revolves around calculating the resultant forces and moments acting on a beam, specifically replacing three forces with a force-couple combination at point A. Participants clarify the need to calculate moments about a specific point for efficiency, emphasizing the importance of breaking forces into components to determine their respective moments correctly. There is a focus on establishing a consistent sign convention for moment directions, with counterclockwise moments considered positive. The calculations for resultant forces and moments are iteratively refined, leading to a net moment of 26,997.2 lb*ft, although the final verification of these calculations remains pending. Understanding the components and their effects on moments is crucial for accurately solving the problem.
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?
 
  • #31
SteamKing said:
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?
You mean the sum of all the moments about A?
moment about A: 26997.2 lb*ft
 
  • #32
bnosam said:
Moments about point A: (anticlockwise positive)
MOB = (2500)(4) = 10000 lb*ft
Is this moment CCW about point A?
MOC = Vertical + Horizontal Components
MOC = 3969.86 + 5(400) lb*ft after squaring and adding then square rooting, magnitude is 4445.20 lb*ft

We just had to straighten out the components of the force applied at C. Is this calculation for the moment due to the force at C correct?

BTW: Unlike adding two force vectors together, if you want to find the net moment acting about a point, you just add the magnitudes of the moments together algebraically.

MOE = 12552 lb*ft
Is this moment CCW about point A?
Result of adding moments: 26997.2 lb*ft
Net moment still subject to verification. See comments above.
 

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