Static equilibrium of a beam, 2 force and 3 force member?

[J-dizzal]

Homework Statement

Homework Equations

ΣF=0
ΣM=0

The Attempt at a Solution

I think i correctly identified two 2 force members at A and B. But would the system be a 3 force member? If it is i don't quite know how to solve for the missing angles.
Is the problem asking for the sum of the forces at A and B including moments at each point? Thanks.

 

[Dr. Courtney]

You usually only need to pick one axis and properly sum the torques to be zero, along with the equilibrium force equations.

You can pick any axis, but making a good choice always makes the math easier.

 

[J-dizzal]

You usually only need to pick one axis and properly sum the torques to be zero, along with the equilibrium force equations.

You can pick any axis, but making a good choice always makes the math easier.

Would the sum of torques about the y-axis be the only choice?

 

[Dr. Courtney]

Summing the torques about any axis into or out of the page would work.

But some choices are easier than others.

 

[J-dizzal]

Summing the torques about any axis into or out of the page would work.

But some choices are easier than others.

I don't understand how to sum the torques about the entire axis, wouldn't you need to split it up at the points given?

edit: would the sum of moments about the axis be the equal to sum of the equations i have? ΣM_A+M_B+M_C

edit; this is not a homework problem, i would like to figure this out before my exam tomorrow though. I think this one has an easy way to solve it because of the 2 force and 3 force members but i can't remember how to apply that technique.

I have 5 equations and I am trying to find the angle α, do my equilibrium equations look right?

 

[J-dizzal]

I don't think the problem has enough information to solve for numerical values.

 

[haruspex]

I have 5 equations and I am trying to find the angle α, do my equilibrium equations look right?

For a rigid body, in a 2D problem, there are only three equations available. Yes, you can write down as many as you like by taking moments about different points, and by resolving linear forces in different directions, but it will turn out that the equations are not independent. There will be three from which all others can be deduced. If you resolve in two different directions and take moments about some point, that will give you all the equations you can usefully get. As Dr Courtney posts, some selections for the reference axis may be more convenient than others.

I don't think the problem has enough information to solve for numerical values.

The 5 equations you posted involve the three unknowns as follows:
1. alpha, F_B
2. alpha,F_A, F_B
3. alpha, F_B
4. alpha, F_A
5. F_A, F_B
Pick n equations involving n of the unknowns, including alpha.

 

[SteamKing]

I don't understand how to sum the torques about the entire axis, wouldn't you need to split it up at the points given?

edit: would the sum of moments about the axis be the equal to sum of the equations i have? ΣM_A+M_B+M_C

You can only write one moment equation for this beam, using either point A, B, or C as the reference. (Hint: there are two choices of moment reference which will simplify the solution of the equations of statics, so that you can determine angle α.)

You cannot write three moment equations, as you have done, and expect to solve this problem.

edit; this is not a homework problem, i would like to figure this out before my exam tomorrow though. I think this one has an easy way to solve it because of the 2 force and 3 force members but i can't remember how to apply that technique.

This is a simple statics problem. All you need to solve it are the equations of statics, ΣF = 0 and ΣM = 0.

2-force or 3-force members are completely irrelevant to obtaining the solution.

Remember, for ΣF = 0, ΣF_x = 0 and ΣF_y = 0.

The support links at A and B are pinned connections. Only forces can be developed there, no moments. If you draw the FBD of the beam CAB, like you started to do in your calculations, there is enough information here to solve for α.

I have 5 equations and I am trying to find the angle α, do my equilibrium equations look right?

Pick one point about which to take moments, and the other two moment equations can be discarded.

 

[J-dizzal]

For a rigid body, in a 2D problem, there are only three equations available. Yes, you can write down as many as you like by taking moments about different points, and by resolving linear forces in different directions, but it will turn out that the equations are not independent. There will be three from which all others can be deduced. If you resolve in two different directions and take moments about some point, that will give you all the equations you can usefully get. As Dr Courtney posts, some selections for the reference axis may be more convenient than others.

The 5 equations you posted involve the three unknowns as follows:
1. alpha, F_B
2. alpha,F_A, F_B
3. alpha, F_B
4. alpha, F_A
5. F_A, F_B
Pick n equations involving n of the unknowns, including alpha.

equations 1. and 5. have all the unknowns. so then I could sum them and they will equal 0
edit; actually i will need 3 equations with 3 unknowns, so 1,2, and 5

 

[haruspex]

edit; actually i will need 3 equations with 3 unknowns, so 1,2, and 5

I didn't make clear that I was concentrating on the first question - finding alpha. For that, the obvious choice is equations 1 and 3. Neither involves F_A, so it's very quick.
More generally, picking three at random might not work. You might happen to pick one that can be derived from the other two. Safest is to use the two linear equations and anyone moment equation.

 

[J-dizzal]

I didn't make clear that I was concentrating on the first question - finding alpha. For that, the obvious choice is equations 1 and 3. Neither involves F_A, so it's very quick.
More generally, picking three at random might not work. You might happen to pick one that can be derived from the other two. Safest is to use the two linear equations and anyone moment equation.

ok, using 1 and 3 I've got it down to cosα-sinα=(F_Bsin30 - 2F_Bcos30)/4
edit: i don't know of any trig identies to solve this

 

[haruspex]

ok, using 1 and 3 I've got it down to cosα-sinα=(F_Bsin30 - 2F_Bcos30)/4

That doesn't help because you now have only one equation but still two unknowns.
The basic procedure in solving simultaneous equations is to use one equation to express one unknown in terms of others, then use that to substitute for that unknown in all of the other equations.

 

[J-dizzal]

That doesn't help because you now have only one equation but still two unknowns.
The basic procedure in solving simultaneous equations is to use one equation to express one unknown in terms of others, then use that to substitute for that unknown in all of the other equations.

ok so let me attempt to solve 1 for alpha and then substitute alpha into eq 3.
edit: actually just cosα from the eq 1. then sub that into eq 3
edit2: on closer inspection I am going to solve for F_B then sub that into eq 3

 

[J-dizzal]

That doesn't help because you now have only one equation but still two unknowns.
The basic procedure in solving simultaneous equations is to use one equation to express one unknown in terms of others, then use that to substitute for that unknown in all of the other equations.

does this answer look correct; α=arctan(2 cos30/sin30)

 

[haruspex]

does this answer look correct; α=arctan(2 cos30/sin30)

Yes.