# Don't know where to start on this truss

1. Feb 17, 2013

### dreamliner

1. The problem statement, all variables and given/known data

You are given truss ABC. In joint C there is a horisontal force F(13,2kN). a is the measurement between A and B, and b is the measurement between A and C. Calculate forces Nac and Nbc.

2. Relevant equations
ƩMa=0
ƩFy=0
ƩFx=0

3. The attempt at a solution

My main problem is that I don't see where to start in order to eliminate some of the unknowns.

I first tried to calculate the angle at joint C by taking invers tan 2,8/5,4 - which gave me an angle of 44 degrees.

I then looked at joint C and tried the method of joints. Which gave me:
ƩFy= 0 --> -AC*sinv-BC*sinv=0
ƩFx=0 --->F-AC*cosv-BC*cosv=0

No way to eliminate unknowns, so no go.

I then looked at the construction as a whole and started with calculating moment in A:

ƩMa=0--> By*2,8=0
ƩFx=0 --> F-Ax+Bx...And I guess forces on the beams would come into play here as well, but that only gives me more unknowns. The same would happen with ƩFy=0, so I don't think this is the way to go either.

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2. Feb 17, 2013

### voko

You can ignore the moments equation. Here is why: compute moments around C. They are all zero. So you have only the forces equation to consider.

And you don't care about the angles, either. Use vectors. The normal forces are along the rods AC and BC. Can you represent AC and BC as vectors?

Last edited: Feb 17, 2013
3. Feb 17, 2013

### dreamliner

Thank you for replying so quickly.
Unfortunately I suck when it comes to vectors, but I suppose you mean constructing a triangle of forces?
If so it would have to consist of rods AC and BC, as well as force F. All forces meet in joint C so equilibrium would be present. But in order to construct a triangle I would have to move BC and that, I'm guessing, is a big no no...

(I'm sure it can be done without moving anything, but as I said I suck when it comes to vector construction. I appreciate your attempt at helping me though. I didn't know you could solve trusses with vectors. My course book only does the moment and force equations.)

4. Feb 17, 2013

### voko

Force $\vec{N}_{AC}$ acts along $\vec{AC}$, just like $\vec{N}_{BC}$ does along $\vec{AC}$. What does "act along" mean? It means that $\vec{N}_{AC}$ is parallel to $\vec{AC}$, and that in turn means that $\vec{N}_{AC} = a \vec{AC}$, where $a$ is some number you need to find out. Likewise, $\vec{N}_{BC} = b \vec{BC}$. We have the equilibrium equation $\vec{N}_{AC} + \vec{N}_{AC} + \vec{F} = 0$, from which it follows that $a\vec{AC} + b\vec{AC} + \vec{F} = 0$.

Can you continue from here?

5. Feb 17, 2013

### dreamliner

Hm. I will certainly give it a go. Thank you.

Another question when it comes to trusses(I only started learning about them on Friday, so my knowledge is very limited.)
In the figure both A and B are fixed nodes but are they also considered to be joints in the truss?

6. Feb 17, 2013

### voko

I think they can be, but I am not completely fluent with this terminology, so perhaps someone could clarify.