Statics question - Method of joints

[Patdon10]

The problem is attached. I have to solve for the force in the member BC. The thing I need help with is solving for the two attachments. I already broke it down into the forces and directions, but what do I do with the lower attachment? How do I find the x displacement of point G from E if I'm trying to find the moment at point E about G. I understand the ratio is supposed to help me, but how? This problem would be much simpler if truss FG wasn't there, and the support was at point F.

 

[nvn]

Patdon10: You know the force in member FG is axial, because it is a truss member. You have three equations, and three unknowns (Ex, Ey, and G). Write an expression for Gx in terms of G. Write an expression for Gy in terms of G. Then write equilibrium equations, and solve for the three unknowns.

 

[nvn]

It is interesting to note that if your only objective is to find the force in member BC, then using method of sections, using a vertical section cut through points D and C, you could find the force in member BC, without knowing the reaction forces at points E and G.

Alternately, without doing any of the above, you could apply method of joints at point A, then apply method of joints at point B, and solve for the force in member BC, without knowing the reaction forces at points E and G.

 

[Patdon10]

Patdon10: You know the force in member FG is axial, because it is a truss member. You have three equations, and three unknowns (Ex, Ey, and G). Write an expression for Gx in terms of G. Write an expression for Gy in terms of G. Then write equilibrium equations, and solve for the three unknowns.

To be honest. I'm not sure I understand what an 'axial force' is. A web search didn't really help me too much either. Are you saying I can find both supports without using a moment?

So in other words, it needs to be...
G_x = G*cos(53.13)
G_y = G*sin(53.13)

Summation_F_x = -E_x + G*cos(53.13) = 0
Summation_F_y = (G*sin(53.13) - 900 - 1200 - E_y) = 0

I have 3 unknowns, and only 2 equations. What am I missing?

 

[SteamKing]

An Axial Force is one whose direction coincides with the length of the member in which it occurs.

 

[Patdon10]

F_BD = 1000 lbs (C). using the method of joints in the way you described I found it to be 562.5 lbs. Maybe you'd be able to spot my mistake? it's a lot of work, but I can try...

Joint A
-----------
F_x = F_AB*cos(36.87) - F_AC = 0
F_y = -900 + F_AB*sin(36.87) = 0
F_AB = 1500 lb (T)
F_AC = 1200 lb (C)

Joint B (There is lots of solving for angles in here)
----------
F_x = -1500*cos(36.87) + F_BD*cos(36.87) + F_BC*cos(36.87) = 0
F_y = -1200 - F_BC*sin(36.87) + F_BD*sin(36.87) - 1500*sin(36.87)

I had 2 equations and 2 unknowns. Used F_x to solve for F_BD = 1500 - F_BC
Plugged F_BD into the F_y to solve for F_BC. F_BC = -562.5 so it'd be 562.5 (C).

 

[nvn]

Are you saying I can find both supports without using a moment?

No.

Regarding post 6, you did excellent work. It appears you made some algebra mistake (not shown). Try again, and watch your algebra.

 

[Patdon10]

I can't find the mistake. I looked everything over and it looks correct. I'll worry about that later.

Can you give me a nudge in the right direction for solving for those supports? What am I missing?

 

[nvn]

Write a summation of moment equation.

 

[Patdon10]

Write a summation of moment equation.

Sure.

I have:
summation of M_E = 1200(14) + 900(18) - (G_y)(x_displacement) + Gx (y displacement) = 0

I simply don't know how to find those displacements.

 

[nvn]

No, it should not be displacement. Change those to x and y distance from point E.

 

[Patdon10]

yeah. that's what i meant. How would I go about finding those distances though?

 

[nvn]

Patdon10: Go ahead and assume the length of member FG is 1.

 

[Patdon10]

That would make the problem fairly easy, but can you do that? I literally posted all the information I was given in the problem.

 

[pongo38]

In statically determinate frames like this one, the lengths themselves are unimportant. It's the ratios of lengths that determine the force distribution. (That is why it doesn't matter if the units are feet or metres or anything else you want, the forces are the same).

 

[Patdon10]

In statically determinate frames like this one, the lengths themselves are unimportant. It's the ratios of lengths that determine the force distribution. (That is why it doesn't matter if the units are feet or metres or anything else you want, the forces are the same).

I guess that makes sense. So as long as the ratio corresponds to the length of that member, everything will check out.