FBD Dilemma

[mechag]

Hey guys,
Im trying to work out a free body diagram for a suspended weight problem and am stuck as i cant seem to work out all the forces. Diagram is attached. I have called the bolt forces F1 and F2.

For the forces acting on the weight itself (what im trying to work out):

If i take Fx = 0 and Fy = 0 along with moments around either end and/or the centre of the weight i can easily find the forces in the 5 direction.
The only equation i can think of to find the x forces is Fx = 0 and this is not enough to solve for the forces as you get:

F1x + F2x = Wsin(theta)

How can i get another equation to find the x forces?
as any moment equation on the weight itself will simply result in the x forces having no effect.

Or is my FBD incorrect?

Cheers

 

[rb1957]

strictly speaking the problem is redundant, which doesn't mean unsolvable ... you have a three force body which tells you something specify about the forces ... look it up.

a reasonable solution is to assume the reactions are parallel with the applied load, now it is statically determinate (ie easily solveable)

 

[hydtools]

F1x + F2x = W*cos(theta), not W*sin(theta). Theta being the angle between W and the x-axis.

You could say F1x = F2x equally sharing the x-axis load.

Ted

 

[mechag]

rb1957 - imo having parallel loads is not a reasonably solution as the bolts will also be bearing a shear force as well, that would essentially be changing the problem to a simple 3 force supported beam problem.

hydtools - although i have not confirmed it i am pretty sure that is incorrect (given the angle of the applied force W the distribution will most likely not be equal).

I am pretty certain now i know how to do it. I think it will need to be solved as a statically indeterminant. I.e. i think moment equations need to be used with boundary conditions,
M(x) = E y'' I
and the differential equation solved.

 

[rb1957]

like i said, strictly speaking the problem is redundant (=indeterminant) 'cause two forces share the same line of action.

the two loads can react the load, 2 reactions parallel to the load ... each bolt would have tension and shear, Pcos(theta), Psin(theta).

i notice that you've only dimensioned the distance between the bolts ("a"), but not the offset distance (away from where "a" is measured). are you specifically looking for the reaction under the bolt head ? IMHO, 2 fasteners equally distanced from the load would react 1/2 the load each.