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So we started doing structural analysis in my engineering class lately, and are doing trusses. However, I'm not quite sure how method of sections works conceptually. I'm getting the right numbers, but I am not sure how the directions in my FBD relate to whether a member is in tension or compression. I found on slide 5 of the first Google result that

When a member force points toward the joint it is attached to, the member is in compression. If that force points away from the joint it is attached to, the member is in tension.

Taking my FBD, FBD of a Pratt bridge truss. Members ML, CL, and CD have been cut. A 100kN force acts at point A, the origin, a 50kN force acts at point M at {3mi+4mj}. The force of ML acts as {4mj}. The force of CL acts at {6mi}. The force of CD acts along the x-axis. does this mean that forces ML, CL, and CD should be considered in tension if those are the members I have cut?

The reason I am confused is because the original system looks like thisscreenshot of the original truss so ML, CL, and CD are all acting "compressively" relative to the bridge.

My prof hasn't really gone into detail about this in class yet, nor have the videos provided been much help, so an explanation/link to an explanation of how to determine the "direction" (compression/tension) of the members would be a great help.

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From the looks of it, you are cutting members CD, CL, and ML and analyzing the remaining section to the left. Next, I assume you correctly calculated the reaction force (100 KN) at the pin support A. Finally, I assume you chose the directions of the unknown forces acting on members CD, CL, and ML arbitrarily as being in tension (away from the joints).

It's ok to assume the directions of the forces, but in order to determine what the directions actually are you have to calculate their actual values. You do this by (1) summing the vertical forces and setting them equal to zero, (2) summing the horizontal forces and setting to zero and (3) summing the moments of the forces about A and setting to zero. If the magnitudes of a calculated force is positive and it is pointing away from the joint, you have tension. If it is negative and pointing away from the joint you have compression.

You only have one unknown vertical force, CL. Its value and therefore direction will be obvious. You have two unknown horizontal forces, ML and CD. In order for them to sum to zero, one of them has to be chosen in the wrong direction. In other words, one of them has to be in compression. CD contributes no moment about point A, but ML and CL does. Since you now know CL, summing the moments of all the forces about A will give you ML. From there you can find CD. The signs of the results will tell you if they are in tension or compression.

Hope this helps.

p.s.

When you become more familiar with statics you will in some cases be able to determine whether members are in tension or compression simply by inspection. But until you get more familiar with the subject, it is best to do the detailed calculations.

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  • $\begingroup$ Your assumptions were all correct, sorry for not specifying that in my original question! I suppose I should also have mentioned that I did calculate the various numbers as well. Oops, hit enter! Will type more, just wait a mo' $\endgroup$ – Tropingenie Mar 6 at 23:36
  • $\begingroup$ @Tropingenie Great. So what did you find out. Which members are in tension and compression. Let me know and I will let you know what I believe are the answers. Hint: Member CL is not in tension. $\endgroup$ – Bob D Mar 6 at 23:43
  • $\begingroup$ After reading over your reply, I think I understand how determining the direction works. Determining T or C in method of sections is based on the direction relative to the joint, but for whatever reason, I thought it was relative to the member, hence my confusion. Since it was a practice question, I already have the answers (hence how I knew my directions were the only thing wrong). Thanks for your time :D $\endgroup$ – Tropingenie Mar 6 at 23:46
  • $\begingroup$ @Tropingenie OK, so if you found that CL is in compression, CD is in tension, and ML is in compression, you got it right. $\endgroup$ – Bob D Mar 7 at 0:53

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