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I am just looking at it as a free body diagram so no member strength/stiffness has been added at this stage

What if we assume that both the plate and the beams used in both methods are infinitely stiff?

I think that treating the solution as a plate/box means the loads can propagate through the plate to each support evenly. Where as within the actual structure of the product, when represented by a beam model, the loads have discrete paths that they need to take and as such will load up...

We are also a bit surprised by the answer which is where this discussion has come from. I'm not sure what you mean by "why do the models have different couples in the reactions ??" there are no applied moments added to either model.. only the loads and supports shown.. the supports are what...

Sorry.. no rat here... just a little clueless hahaha So my actual structure is represented by the beam model shown above, then these are the results of my reaction loads I should be using? And not simplifying it to a "rigid black box"

So if I treat it as 5 beams then I get this scenario A = 7.66kN B = 14.29kN C = 17.04kN D = 10.01kN Much the same as my calcs above.. If I treat it as a solid box A = 11.51kN B = 11.08kN C = 13.99kN D = 13.00kN which are much the same as using the centroid..

The structure is as per this diagram The load at the centroid in the attached to the individual load points via a wire rope As I said.. It has been simplified at this stage to remove an horizontal loads, and just concentrate on the verticals. So 49kN at the centroid will equal 24.5kN on...

So the load at the centroid position is attached to the individual load points via a steel wire rope. Obviously this brings horizontal loads into the formula but for simplicity at this stage I was just working on the verticals as the horizontals should all cancel out. So I guess my question...

So that's two way's of working it out. with both giving the same answers. But others in the office believe that using the centroid load should give the same results for the reactions. i.e as I calculated earlier. AC=(49*186)/369 = 24.699kN BD=(49*183)/369 = 24.301kN A=AC*(262.5/586) =...

haha sorry at this stage just using the extra figures to make sure they match exactly will pull them back in final calcs

Do you mean like this?

The single load of 49kN is at the centroid of the three other load points So for the single load. Taking moments around an axis BD AC=(49*186)/369 = 24.699kN Taking moments around an axis AC BD=(49*183)/369 = 24.301kN A=AC*(262.5/586) = 11.064kN B=BD*(262.5/586) = 10.886kN C=AC*(323.5/586) =...

Structure would look something like this

Dimensions are in mm

Hi all! Apologies if this is in the wrong spot as I am new here. This also may seem pretty basic problem, but there is some argument about it going on in my office so I wanted to see what others thought. Below are two scenarios. Basically just looking for the reactions at A, B, C and D All...