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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)
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 :/
Not entirely... Still confused hahaha