Truss axial forces ~ Section size 1

[dgkhan]

Manually when I used to analyse a truss (Whether by Joint, Section or graphical method), I never used to consider the section sizes, as far as I remember. For long, now I use STAAD or SAP. By modifying the section sizes, I see slight change in results (axial forces). I am not changing any geometry and ignoring self weight. What caused the results to be changed?

 

[patswfc]

Are you loading the truss at node points only and is every member pinned at both ends including the top and bottom chord members. If so then the forces (there will only be axial forces) shouldn't change. The deflection is the only thing that will change.

Galambos
Even if he has pinned-pinned instead of pinned-roller supports the forces in the pinned-pinned models should still be the same although not correct.

 

[Stillerz]

If you are using STAAD, are you analyzing a "plane" frame or a "Space" frame? This is of PARAMOUNT importance in specifying member releases.
In a space frame relative stiffness will matter and will change the load paths within the truss members as there is no way to specify all members as "Truss" members in a space frame without runnign into instability problems. In a space frame truss design the larger the capacity of the section, the more load it will "suck" from surrounding members. Truss design using STAAD is can be an iterative process.

 

[Stillerz]

STAAD will absolutely allow you to analyze a plane frame truss and specify ALL members as "truss" members. No "pin" connections need to be specified so long as you have a plane model. Can't get any simpler.
Pin support one end, roller on the other, and specify every single damn memeber as a "truss".

 

[Stillerz]

Member sizes should not matter in a theoretical analysis that ignores selfweight.

 

[dgkhan]

1. Let: m = number of members in the truss
r = number of external supports for the truss
j = number of nodes in the truss
a) m + r = 2j is statically determinate
b) m + r < 2j is statically unstable
c) m + r > 2j is statically indeterminate

Case c applies for me
72+3 > 2 x35
soeven with roller support and all pin joints, truss can be indeterminate

 

[dgkhan]

here is the truss. supports are only at top chords

 

[civeng80]

Just a thought here.

The software doesn't distinguish structures from statically determinate or indeterminate. It just calulates member stiffness and assembles the stiffness matrix of the structure. Thats why it asks for member sizes. It then inverts the stiffness matrix to solve for the diplacements and then solves for actions.

Maybe when it inverts the matrix it just comes up with slightly different values with different member stiffnesses.

Its just an error due to the nuber crunching.

Maybe!

Cheers!!

 

[Galambos]

patswfc, i disagree completely.

let's just see a screenshot...that should solve it.

 

[frv]

I second galambos, with a caveat..

If only some members are changing, then the axial force would be different with a pin-pin connection; if all members are changing to something else, then yes, the axial force should be the same.

 

[haynewp]

 

[apsix]

I second galambos, no caveat.

 

[DRC1]

Civeng80 has the answer. For a pure detrminate truss(axial forces, pinned joints, loads at joints) hand solutions by method of sections or joints is exact. FE is ALWAYS an aproximate solution. It does not distnguish between determinate or indeterminate. Changes in A change the stiffness matrix and change the residual errror. That is why there ar MINOR differences bet

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member sizes. When you do an FE problem, ever notice the static check is not quite zero? Same concept. Know the limitation of your tools.

 

[Galambos]

I'm not sure you will be able to solve a pin-pin truss by using hand solutions methods of sections or joints.

a pin-roller truss on the other hand is a very different story.

 

[IFRs]

Did anyone ask what percentage the results were changing by? It may be small enough to ignore, given the inherent inaccuracy of computers and the algorithms used to solve matrix math. Also, if the truss is nowhere near 0,0,0 there may be other round-off errors in simply the coordinate system. It also looks like some of the members are coming together at fairly acute angles, presenting ill-conditioned equations to be solved - depending on the algorithms used there may be errors there. Did the OP ever run the same analysis twice to see if there were small differences between them?

 

[civeng80]

DRC1

Exactly my line of thinking. For a critical form structures for example determinant is never zero, but very small. Matrix structural analysis is always an approximate method(albeit with high accuracy!) Thats why with different member sections it comes up with slightly different solutions. However I would expect these to be insignificantly small, with the hand solution which doesn't depend on section properties being the exact solution.

Again an example of computers going about solving simple problems the hard way and getting close to the correct answers. Fortunately we humans know a quicker and in this case exact way of solving the problem.

What this proves is that no computer program is perfect.

 

[apsix]

I don't use FEA but using my classical matrix analysis software resulted in identical results to 6 decimal places for a range of relative section sizes.
This was for a determinate, pin jointed, node loaded truss.