Statics 3 equations 4 unkowns

[Liftingengineer]

This problem seems so simple but I am having trouble. I more or less have a rectangular platform with bolts on all for corners. There is a weight causing a force downward on some point on the platform. I was looking for the force at the bolts. Summations of the forces and moments only gave me 3 unqiue equations. How do I find a fourth? Do I need to consider deformation of the platform. Is their a simple trick for another equaiton?

 

[Cockroach]

Seems to be plenty of lip service on this problem, only a few ideas on the mathematical solution to the problem at hand. Attached is a reference that will serve as a guide to that solution.

Good luck with it.

Regards,
Cockroach

 

[jtseng123]

There are academic people and engineering people here for this simple question. So the answer is so divided.
Inverse proportion (or my original term dis-proportion) is the way to go. This is not a 3-leg problem. Period. Just look at every table at your home, and you step on top of any place, you will know what I mean. Or, you can build a very flimsy 4 leg frame and hanging a dummy weight at any place, as long as it will not crash, regardless how it deforms, all legs will carry loads, not 3 legs only. You try to pull up each leg to feel it, and you will know what I mean.
If you are doing a academic project or working in aerospace or NASA, go use FEA, or looking for the 4th equation by the differential equation of stress-strain energy approach (if I remember the term correctly) Otherwise, inverse proportion is good enough.
Best Regards,

 

[rb1957]

ok, except for "period". there are many simple approaches that arrive at the same (or Very similar) results.

my original thought was to solve it as a beam in one plane, so you've get sum(1&3) and sum(2&4), then solve moments in the orthogonal plane to get the proportions of 1:3 and 2:4, and so get the reactions.

i did look at the inverse proportion method, saw i got the same reactions, and saw how the geometry drove the answer. so i'd be happy to use that in the future.

the idea behind the three legs is that it's statically determinate. solve all four "triples", average them, you'll probably get the same answer. i don't think the suggestion was meant as "ignore one leg ('cause it makes life hard)".

 

[Cockroach]

Well, haven't heard from Nech0604 for quite some time.

So I guess we're saying if 100 lbf is sitting on your kitchen table, then 25 lbf is on each of the four (4) legs. If you prefer the three (3) legged approach, then go with 33.3 lbf per leg.

Forget the academic approach, don't worry about the real world and design it with a factor of safety of five (5) to allow for your uncertainty with the mathematics. Done.

Regards,
Cockroach

 

[rb1957]

actually the loads aren't equal (in either 3 or 4 legged cases) unless the load is applied at the centroid of the legs.

and i'm pretty sure Nech hs turned off "accept post notice emails" !?