Forces on a beam, replace with force and couple

  • Thread starter bnosam
  • Start date
  • #1
139
0

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 one


Does everything seem ok here so far?
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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.
 
  • #3
139
0
I thought I could calculate moments anywhere?
 
  • #4
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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.
 
  • #5
139
0
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.
 
  • #6
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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.
 
  • #7
139
0
Oh B should be counterclockwise, right?

so then C should be Moc = (800)(6)? If I get what you're saying?
 
  • #8
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
 
  • #9
139
0
So because it's on an angle we want only the horizontal component of where C is?
 
  • #10
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
so we want the cos of the angle? 2*cos(30) = 1.732 ft
 
  • #12
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
4+ 2 cos(30) = 5.73

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

If I'm understanding you, that is.
 
  • #14
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
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
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
I'm not sure what you mean?
 
  • #24
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,796
1,668
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
139
0
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.
 

Related Threads on Forces on a beam, replace with force and couple

Replies
3
Views
2K
Replies
7
Views
13K
Replies
0
Views
7K
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
9K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
686
  • Last Post
Replies
2
Views
2K
Top