External Reaction of Truss

[TwelveRox_12]

Hey guys,

I'm new here and sorry for this newbie question, I would to ask how to I obtain the reaction of the pinned connection of A, B, and C (See attachment)? Do I really have to use Force-Method? I actually attempted to use Moment at different places and do simultaneously (Idk if it's the right thing to do) but apparently I got Ay = 38.07kN, By = 37.39kN and Cy = 0, which I think it is wrong, can anybody help? After getting this, then I can proceed to calculate the internal force and pick the type of section to use.

Thanks.

-TR

 

[jayrod12]

That structure is indeterminate. You can't solve it with just the normal 3 statics equations.

For what it's worth, this feels like homework.

 

[TwelveRox_12]

jayrod12: Thanks for the reply, Jay! Well, uhm yeah I get that it's indeterminate as it has more unknowns than equation. So I was finding for solution on whether if Force-Method is needed to be used. It's a real design (structured designed by STAAD Pro), but just that I had totally forgotten what I've learnt in my university and only a week into consultant design, now slowly touching back my basics using manual calculation.

-TR

 

[techiestruc]

For indeterminate structures, I'd go for matrix analysis since its general solution is more convenient for me.

 

[TwelveRox_12]

techiestruc : Thanks for the reply, Techie. Totally forgotten about that, it's a good recommendation! I really forgotten about this stiffness method, I will try and compare to the results from STAAD. Pro, thanks again!

-TR

 

[IDS]

but just that I had totally forgotten what I've learnt in my university and only a week into consultant design

After completing my degree I spent 18 months doing highway work then moved to site supervision on a viaduct project. I still remember the struggle I had trying to remember reinforced concrete basics when a senior engineer asked me to check the reinforcement in a small retaining wall.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[jayrod12]

The stiffness method is the only way to go really in my eyes.

Alternately, you could look at the actual support connections and see if you can't rationalize it into a determinate structure. Perhaps at the low end of the truss on the left and the high connection on the right they become rollers. Do you actually have two connections on the right side of the truss? That seems odd. Generally we would bear the truss at the bottom only, and perhaps connect the top of the truss for out of plane loads only.

 

[rb1957]

I thought of Ay, By, and Cx as reactions, though with the loading, Cx = 0

then I thought Ay, Bx, Cx; or Bx, By, Cx and add Ay (as a propped cantilever, unit force method).

as I see it there are several members (well 2) with zero load (including the member BC).

this is (I think) triply redundant, I'd use FEA (maybe play with different react sets to see the changes).
Maybe too instead of hard constraints support the truss on springs (ie finite stiffness constraints)

another day in paradise, or is paradise one day closer ?

 

[bhiggins]

My question: is the truss truly pinned, pinned, pinned? Most trusses are designed as pinned roller, could you idealize A & C as roller connections to make it statically determinate?

 

[BAretired]

My question: is the truss truly pinned, pinned, pinned? Most trusses are designed as pinned roller, could you idealize A & C as roller connections to make it statically determinate?

No, that would still be indeterminate.

You could make A or B a horizontal roller and remove C to make it determinate. Or you could remove A and make B a vertical roller (or any direction other than horizontal). Or you could remove B and make C a vertical or horizontal roller (or any direction other than normal to AC).

Most would likely choose the first option.

BA

 

[bhiggins]

I was thinking of making A a horizontal roller and C a vertical roller. That would leave 4 reactions, have 4 equilibrium equations, (Fx, Fy, M(a), M(b)) shouldn't that be enough to make it statically determinate?

 

[BAretired]

I was thinking of making A a horizontal roller and C a vertical roller. That would leave 4 reactions, have 4 equilibrium equations, (Fx, Fy, M(a), M(b)) shouldn't that be enough to make it statically determinate?

If B remains hinged and C becomes a vertical roller, the structure is stable and determinate. Making A a horizontal roller adds a redundant reaction.

Alternatively, if B remains hinged and A becomes a horizontal roller, the structure is stable and determinate. Making C a vertical roller adds a redundant reaction.

M(a) and M(b) are not independent equations. For equilibrium, the sum of moments about any point must be zero. If that is so, the moment about any other point must also be zero.

BA

 

[klaus.]

in 2-D you have 3 Equilibrium equations
Sum X =0
Sum Y = 0
Sum M = 0
that's it.... no more

So 4 unknown is one too much
Maybe it would help to read one on the basic text books on mechanics




best regards
Klaus

 

[hokie66]

BAretired and klaus are correct, but in this case, with only vertical loading, I think the vertical roller at C will behave just as if it were released. I get a vertical reaction at A of 38.3, and 37.5 at B. Close enough for me.

 

[BAretired]

BAretired and klaus are correct, but in this case, with only vertical loading, I think the vertical roller at C will behave just as if it were released. I get a vertical reaction at A of 38.3, and 37.5 at B. Close enough for me.

Perhaps, but an equal and opposite horizontal reaction at B and C will reduce the vertical reaction at A and increase the vertical reaction at B as the truss attempts to act as fixed ended at BC.

BA

 

[rb1957]

a propped cantilever is different to a simply supported beam ...

interestingly I think you can solve Bx, By, Cx, and Cy (because of the relationship between Cx and Cy); I thought the problem triply redundant but maybe it is doubly redundant ?
Cx and Cy can be determined from sum moments about B (> Cx) which defines the load in the truss member going into C (member BC sees no load, as both B and C are constrained) and so Cy can be determined. Then you can determine Bx and By (from sum forces). Then you could plot the deflection of point A, and probably see Ay is the significant reaction here.

another day in paradise, or is paradise one day closer ?

 

[BAretired]

A propped cantilever is what we have. It cannot be solved by statics alone.
BA