Mass hanging - equilibrium equations using rod length

[Pawel27]

I am looking for equation describing situation like on the picture. It should be equilibrium equation using Rod Length and Rod Young modulus and concerns gravity center.
The problem is that I have ready quite long Visual Basic code made by other author but I cannot understand it. Let's start my question without this code.

 

[desertfox]

Hi Pawel27

If all the rods are of the same material which I assume they are and the rods were symmetrically spaced around the centre of gravity then each rod would carry a load relative to its stiffness however you indicted in your first post that the centre of gravity wasn't central to the supporting rods.

 

[Pawel27]

Yes, gravity center is usually misaligned (not central to the supporting rods). I just try to write down relations (equations) to describe such kind of structures looking at VBA code.

 

[rb1957]

then i have no idea what ML/EA represents.

PL/EA is the extension due to a force P.

i think Doug(IDS) posts a reasonable solution strategy. if the load CG aligns with the rods' centroid then the rods will be extending an equal amount (the weight will simply translate). if not the weight will rotate (slightly) and the off-set moment will be reacted by varying loads in the rods. As the rods see tension from the weight, and some will see compression (-ve tension if you will) from the moment, you need to check that the compression from the moment doesn't exceed the tension from the weight ... if it does then the rod becomes ineffective in both calcs (it can't carry a resultant compression load). I can't see this check in your posts. if the rods have some I, now i see only A, then you could react some compression (as an euler column).

Quando Omni Flunkus Moritati

 

[dhengr]

Pawel27:
I can’t see your sketches unless they are pdf files, so I’m guessing a bit here, but based on the posts above.... I think you have to find the C.G. of the mass, and write a set of equations which compare its location to the locations of all of the supporting rods. The rods are support springs with a stiffness or elongation of PL/AE and they will carry the load in proportion to their distance from the C.G. and their section properties, mainly ‘A & L’, and they should sum-up to the total weight. But, then a set of additional equations will relate these spring support forces to the stiffness of the mass structure. As if it were an inverted spread footing/a plate (some structure) supported by randomly located springs. It’s a beam or plate supported by springs. The rods are probably not going to change their vertical location very much unless the stiffness of the supported structure allows it to change dimensions in plan by a significant amount. The plate curvature, under load, should probably not cause much dimensional change in plan. You are looking for compatibility of these two systems, the rod’s elongation at each node point, vs. the deflection of the supported structure due to these support loads. Are there temperature affects too?

 

[Denial]

Using the coordinate system in your second-posted diagram, the one with Z vertical, and assuming that the deflections are small, the deflection of the top surface of your "box" is entirely vertical and is given as a linear equation
Dz = a + b.X + c.Y
where (X,Y) is the location of the point of interest.

This equation has three unknowns.

You also have three equations of equilibrium:[ ] vertical forces to sum to zero;[ ] moments about the X axis to sum to zero;[ ] and moments about the Y axis to sum to zero.[ ] The only forces acting on the system are the weight of the box and the Z-direction forces in every hanger.[ ] The force in any hanger is proportional to the product of its axial stiffness and the Dz value at its lower end.

So, regardless of how many hangers in your problem you have a linear problem in only three unknowns.[ ] It is a fairly easy task to set up a matrix-based representation of the equilibrium equations, then solve it by inverting the appropriate 3x3 stiffness matrix.[ ] All easily done in VBA, possibly also able to be done without resort to VBA.

This is exactly the approach IDS was recommending, just worded a bit differently.

 

[rb1957]

i agree with denial ... the three variable are
1) the amount of translation (equal extension of all rods),
2) the amount of tilt due to Mx, and
3) the amount of tilt due to My.

to get the moment terms, you should strictly use the principal axes of the rods (ie a set of areas, calc Ixx, Iyy, and Ixy, calc the principal axes of inertia I11, I22). the moment due to the load is easy to determine, the load X the distance from the load to the centroid of the rods. then this moment will have components in the 11 and 22 directions.

Quando Omni Flunkus Moritati