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!!!
If you show me an equation I can spot an error, but I dont 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 alot here.
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.
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.