Solving Strange Structures: Platt Truss Issues

  • Thread starter Thread starter physicsCU
  • Start date Start date
  • Tags Tags
    Strange Structures
AI Thread Summary
The discussion centers on issues with calculating forces in a Platt truss, specifically regarding discrepancies between the user's calculations and the book's answers. The user calculated a diagonal force of approximately 240 kN, while the book states it should be 96.2 kN, suggesting a significant error in the user's approach. Key equations for summing forces and moments were shared, indicating that the user’s methodology might be correct but could contain an error in the moment calculations. The conversation also highlights the importance of verifying support reactions and the potential for errors in the textbook. Overall, the user is encouraged to revisit their calculations and consider the possibility of a mistake in the book.
physicsCU
Messages
198
Reaction score
1
OK, I am having some issues with some trusses.

The first one is a platt truss, h = 8 m, length of horizontal beams = 8 m

F = 340 kN occurs at each of the 5 pts between the supports, one is a pin, the other is a roller.

I get the force in a diagonal to be ~240 kN, but the answer in the book is 96.2 kN, a factor of 2.5.

I did a bunch of searches on google, and I think my procedure is right, even according to the examples. So where does that 2.5 come from? Or is the book wrong? Part of me thinks that the book is wrong, but this isn't the first problem in this chapter that would be true.

Thanks!
 
Physics news on Phys.org
If you show me an equation I can spot an error, but I don't feel like doing the problem from scratch right now because its 1:13 am, sorry. Just make sure you sumed the forces in the y and x directions correctly and found the correct support reactions when you sum the moments about one of the supports. The factor of 2.5 comes from you getting the wrong anwser. A picture would help a lot here.
 
Last edited:
OK, we are to find the axial force in member EK.

Point A is a pin support, H is a roller.

F_x = A_x = 0
F_y = H_y + A_y - 5F = 0
M_a = 6LH_y - LF - 2LF - 3LF - 4LF - 5LF = 0

These give A_x = 0, H_y = 850 kN, A_y = 850 kN

Now I cut through DE, EK, and KL.

F_y = H_y - 2F - EKsin(45 degrees) = 0

This is the procedure I did.

I can't get a pic up, but imagine this:
B C D E G
__________
/|\ |\ | /| /|\
/ | \| \|/ |/ | \
A ---------------- H
I J K L M

F acts downward at I,J,K,L,M

Again L = 8 m, F = 340 kN

L is the vertical and horizontal distances, but not diagonals.

Hope this helps.

The bridge won't work out, but you get the idea.
 
your work seems right, book might be wrong. Is the book by hibbeler 9th ed.?
 
No, it is Bedford and Fowler.

I will look at it tomorrow and see what I can do with it.
 
Your work does seem correct, to me at least.
 
Thanks.

I had another issue earlier today that was resolved by an incorrect moment equation. I will work on that tomorrow, and report back here.
 
Back
Top