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intermediate beam analysis

intermediate beam analysis

intermediate beam analysis

(OP)
If you have two intermediate beams (cross beams) intersecting each other (imagine a square area cut into 4 equal areas by interior beams), with the same depth, material, length, loadings, tributary area, how would you analyze the intermediate beams and the assumption for support condition(without the use of FEM)?

RE: intermediate beam analysis

Load is distributed based on stiffness. Therefore, you would need to analyze the relative stiffness of each beam. If both beams have the same exact loading and have the same span, properties etc my first thought is that you could analyze them independently as I don't think their interaction would impact the results. If they aren't exactly identical you would need to solve by the stiffness method.

RE: intermediate beam analysis

Equate their deflections at the point of intersection.

Mike McCann
MMC Engineering

RE: intermediate beam analysis

Assumptions: Beams are both continuous through intersection, pin/roller at ends, and same size. Loading is by 4 plates, supported 4 sides each. No continuity over beams.

I would just use the diamond shaped loading area created by the lines between the support points. So triangular shaped loading on each of two beams.

Now if the loading is by one continuous plate, I would just multiply the results by 1.25.

RE: intermediate beam analysis

(OP)
Thanks all for your reply. Most of the replies above are correct but what if the other beam is 10% stiffer (all other parameters the same0? would this justify that the stiffer beam acts as a roller/pin support to the less stiff beam? My worry is that there is an over estimation of the support being provided by the stiffer beam. it is not really a rigid roller/pin support, but rather a spring support (due to the fact that it will deflect at the point of intersection).

Here is my understanding: In order to get the spring stiffness provided by the stiffer beam, one must get the deflection, but to get the deflection we must know how much load is the beam carrying, the amount of load the stiffer beam is carrying from the less stiff beam is dependent on the spring stiffness. hence it brings me into a loop

RE: intermediate beam analysis

You don't need to know the deflection for the actual loading. you need to derive a spring constant, which when in the elastic range, is the same no matter the load. Say you are loading a beam midspan by another. the deflection of a beam with a point load in the center is P*L^3/48EI = delta. A spring constant is P/delta, so rearrange and get 48EI/L^3 = P/delta. that is your spring at the end of the beam.

RE: intermediate beam analysis

if one beam is stiffer than the other, then it'll react more load whilst keeping displacement compatability at the common point.

this can be shown in two ways ...

1) let both beams react an equal amount of the applied load (ie 4 beams = 1/4 per) then allow the stiffer beam to react load from the weaker beam at their common point; or
2) let the stiffer beam react more of the applied load, something like (P/4)*x, and the weaker beam (P/4)*(1-x) and adjust the proportioning "x" untill displacements at teh common point at the same.

nit-picking the previous post, the loading can be considered uniform (for easy of calculation) but i suspect the "real" loading is non-uniform.

Quando Omni Flunkus Moritati

RE: intermediate beam analysis

Depending on the connection between the two beams it could be more complicated than adding a translational spring. You may also need to add a rotational spring (to account for torsion for instance) depending on the loading configuration and torsional stiffness of the beams. If you are neglecting any moment transfer between the beams, then yes you could do the virtual work method as described above and treat one of the beams as a translational spring.

RE: intermediate beam analysis

Let W1 be the load going onto first beam. W2 is load on second beam. Total W = W1+W2. The equate the deflections and the equations solve simultaneously.

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