Forces on a Truss Homework: Find Tension/Compression

  • Thread starter Thread starter Bradracer18
  • Start date Start date
  • Tags Tags
    Forces Truss
AI Thread Summary
The discussion revolves around calculating tension and compression in a truss system, where specific forces and distances are provided. The user, Brad, attempts to find the forces at various points using equilibrium equations but encounters discrepancies with the book's answers. Key errors identified include a math mistake in calculating the vertical reaction force at point C and neglecting the horizontal reaction force at point A. After receiving feedback, Brad acknowledges the mistakes and expresses intent to review his calculations. The conversation highlights the importance of careful verification in structural analysis problems.
Bradracer18
Messages
203
Reaction score
0

Homework Statement



Ok, you'll have to wait for my picture. Basically we are doing trusses, and I'm not sure where I'm going wrong here.

We are to find the tension/compression in each member(line) of the truss.

Ok, from my picture. F1 is directly at B(in neg y direction), and is 2800N. F2 is at point D(in the neg x direction) and is 2100N.

The distances between A&B, and B&C are 4m. A is a pin system(basically has x and y forces)...and C is held up by a cable(I believe only supports in the y direction).

I also drew another picture, which consists of little circles on each joint, and arrows pointing in the directions of the tensions(and forces acting on the "circles").

Homework Equations





The Attempt at a Solution



Ok, I'm only going to post some of this, to see if I started correctly(where I think I messed up).

So, from my picture, I took overall equilibrium. So, I chose to sum moments about point A=0(with a counter-clockwise rotation I chose).

moments at pt A = 0= 8Cy -2800(4) - 2100(3.4641)
Cy = 1259.33N up

Sum Forces in +x direction=0

Ax - 2100 = Ax=2100N to the right

Sum forces in +y direction = 0

Ay + Cy - 2800 = Ay = 1540.67N up

If this is correct, then I'll post more. Next, I drew that other picture, and started at points and found the tension(as I draw them all in tension to begin with) at each point.

Thanks,
Brad

View attachment problem 6.bmp
 
Physics news on Phys.org
Looks okay to me.
 
For point A:

Sum Forces in +x=0:

Tab + Tadcos(60) =0 Tab = -3558.02N
Sum Forces in +y=0:

Ay(or 1540.67) - Tadsin(60) = 0 Tad = 1779.01N

I'll stop here again too.

The book says Tab = 2.38kN and Tad = 0.567kN

So, what did I do wrong?
 
Is this portion correct?
 
trying to do another one of these now...so any help would be nice:)
 
Is my drawing or the last post not clear...I'd like to know where I've went wrong. I'll do the calculations, I can't find any other way to do them, as of now...need to know if I'm right up to this point.
 
Bradracer18 said:
Is my drawing or the last post not clear...I'd like to know where I've went wrong. I'll do the calculations, I can't find any other way to do them, as of now...need to know if I'm right up to this point.
You have a math error up front when you calculated C_y. Right equation, careless math.

Then you left out the A_x reaction when you summed forces in x direction at A.

And one of those members at A will not be in tension.
 
Ok...I figured it out...I guess I didn't go over my work as carefully as I thought I did. Thanks for the help...I'm sure by the end of the day, I'll have another one of these posted...ha ha...Thanks again!
 
Back
Top