Question

I really need help with this question...

Answer 1

this is how I labelled the reaction forces

Answer 2

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?

Answer 3

And b has me totally stumped, would I have to use Sections method?

Answer 4

Yes method of sections will work for b)

Answer 5

where would I cut the section @surry99?

Answer 6

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?

Answer 7

Post your attempt and I would be happy to check it later.

Answer 8

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! :)

Answer 9

And also Im unsure on how to determine if forces are under compression or tension..

Answer 10

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.

Answer 11

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)

Answer 12

I would post a pic of my working out but my phone isn't letting me :/

Answer 13

@firekat97 have you solved the problem yet?

Answer 14

Not entirely... Still confused hahaha

Answer 15

@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.

Answer 16

@mathmate okay 👍 thanks