# Homework Help: Analysis Of Truss, Method Of Joints.

1. Sep 12, 2011

### slain4ever

1. The problem statement, all variables and given/known data
Hi so I have the following truss:
http://screencast.com/t/BIdagRt5abg
and I need to calculate the forces in every member by method of joints

3. The attempt at a solution

So I got the Reaction forces:
F(BG) = 25N and F(AC) = 35N

Now I'm not exactually sure what to do next, I know you calculate the x and y forces at each joint, I started at C
and then I got F(y) = 0 =35N + F(CJ) sin(45)
F(CJ) = -49.497

Im fairly sure that that is right
but when I get to the horizontal my equation is like this 0=F(CJ) * Cos(45)
which doesnt seem right since it means F(CJ) is zero

2. Sep 12, 2011

### slain4ever

the equation should be 0=F(CJ) * Cos(45) + F(CD) right?

how do you tell if it is + or - ??
like is it what I wrote above or 0=F(CJ) * Cos(45) - F(CD)??

3. Sep 12, 2011

### PhanthomJay

After calculating reactions, you should start with the easiest joint first, a joint with only 2 members framing into it, like Joint H. This will assure you that when you get to joint C, there is no force in member CH. When you isolate joint C, look at the direction of the member forces required for equilibrium in the x and y directions. If a member force points in toward a joint, it is a compression member., and if it points away from the joint, it is a tension member. Don't forget this when you move to your next joint.

Th magnitude of your forces are thus far correct.

4. Sep 12, 2011

### slain4ever

if it is in compression does that mean its negative? or is it the other way around?

also we were told that CH, HJ, EK, GM and LM are zero force members

5. Sep 12, 2011

### slain4ever

i've calculated F(CD) to be 35 and F(DJ) = 65 F(JK) =35 and F(DK) = 91.924

i have no idea if they are negative or what

6. Sep 12, 2011

### slain4ever

i've decided to just write expressions for each of the members can you please tell me which ones I need to change to negative.

BG =25
AC = 35
CJ = -49.497
CD = 35
DJ = 65
JK = 35
DK = -91.924
DE = -(35 + -91.924* sin(45))
0 = DK sin(45) + KF sin (45) - 20
0 = JK + KL + DK sin(45) + KF sin(45)
EF = -DE
0 = GL sin(45) + BG
FG = -LG sin(45)
FL = -LG sin (45) + 10
KL = -LG sin(45)
FL = -KFsin(45)

thank you

7. Sep 12, 2011

### PhanthomJay

By convention, a compression force in a member is considered a negative force, and a tension force in a member is considered a positive force. But the direction of the x and y components of these forces are determined from Newton 1, and you have to be careful about signage. This then determines whether a member is in compression or tension. If the member force (resultant of x and y components of force in that member) points toward the joint, it's a compression member...if it pulls away from the joint, it is a tension member.

If you get one wrong then the others most all become wrong. Please look at Joint C first. There is a support force of 35 N acting up, and there are unknown forces CD and CJ acting along the member's long axis in an unknown direction. First realize that truss members are 2 force members....they can only take axial loads in compression or tension.

Since there is 35 N up from the support, and since no part of the vertical force can be carried by CD, then the vert comp in CJ must be 35 N down, per Newton 1. That means the horizontal comp in CJ must be 35 N to the right. The resultant force is sq rt sum of squares = 49.5 N, pulling away from the member. So CJ is in tension , call it a plus. And since since for equilibrium the force in CD must be 35 N acting left, that is a force pushing toward the joint C, so that is a compression force in CD, call it a minus.

Now move on to the other joints, tackling the joints with the least amount of unknowns first. Don't forget Newtons 3rd Law!!