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So for part A using my FBD, I set up the following equations- ∑Fx = 0, -Lx = 0 therefore Lx = 0 ∑Fy = 0, Hy -P -Q + Ly = 0 ∑ML = 0, 8Hy - 4P - 2Q = 0 therefore Hy = P/2 + Q/4 so we sub Hy into ∑Fy equation, and get Ly = P/2 + 3Q/4 so those are the reactions I have calculated for the supports, and all my assumed directions are correct as all my "values" came out positive. Can someone please confirm if I have done this correctly?
And b has me totally stumped, would I have to use Sections method?
Yes method of sections will work for b)
where would I cut the section @surry99?
Suppose you want the force in BJ first. Then you must cut BJ, Now which other members would you cut at the same time? Try a few options and look at the resulting FBD's. Do the resulting FBD's allow you to determine the force in BJ?
Post your attempt and I would be happy to check it later.
so I had figured that DI = IE = KF = KG = 0 as they are all zero force members (I think) and if the above is correct, then since load at K is 2kN downward, from y equilibrium, KC should be 2kN upward and so CK would be 2 kN downward (action reaction forces) I then thought I'd make a cut down EC, JF and JK and use moments and equilibriums to solve for JC (JC = JF) and from doing this I found JF = √8 , JB = 1 (∑y equilibrium) and JK =2, but I am really uncertain of my answers @surry99 and thanks so much for your help! :)
And also Im unsure on how to determine if forces are under compression or tension..
If a member is being pulled on it is in tension. If a member is being pushed on, it is in compression. Please post your work and I will take a look.
So what I did was Mc = -(JK X 2) - (1 X 2) + (6 X 1) = 0 So -2JK -2 + 6 = 0 jK = 2 Then Mb = -(2 X JK) -( 2 X 1/sqrt(2) JC) + (4 X 1) = 0 So JC = 0 so JF = 0 and BJ = 1 (explained above) and CK = 2 (explained above) and for some reason I'm now getting JF = 0 instead of sqrt(8)
I would post a pic of my working out but my phone isn't letting me :/
@firekat97 have you solved the problem yet?
Not entirely... Still confused hahaha
@firekat97 If you have a chance to post your last version of your work (with your phone or otherwise), I'll be glad to look at it.
@mathmate okay 👍 thanks