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Statics 3 equations 4 unkowns

Statics 3 equations 4 unkowns

Statics 3 equations 4 unkowns

(OP)
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?

RE: Statics 3 equations 4 unkowns

The platform is oververconstrained and statically indeterminate. No solution.

Yes, you need to include deformation, etc. in your solution. I've had luck supporting platform on stiff springs in FEA to get results.

RE: Statics 3 equations 4 unkowns

(OP)
Thanks alot TheTick, I thought it was indeterminate. Is there a any rules of thumb for this calculation I can through a huge safety factor at? I do not currently have FEA access and it took my IT department 6+ months to get me autocad. I am thinking over a year for anykind of FEA.lol

RE: Statics 3 equations 4 unkowns

Don't believe you will. Ever sat on a four legged stool - one is NEVER resting on the ground.

So either assume three legs take all the load (my choice) or all 4 legs take the same load - doubbtful

RE: Statics 3 equations 4 unkowns

This can be solved two ways, statically determinate or statically indeterminate.

In the first case, you need to add another equation to get four equations in four unknowns. Typically you use material properties and deformation, say, add the assumption that at the onset of plasticity, the system has failed. This is a very common approach taken by second year engineering students working with Statics.

The second case you need to look at the possiblilities in generating a sequence of linear dependent solution sets. This is an approach taken by Linear Algebra Calculus programs involving Markov's Chains. I think this is beyond the scope of this forum, but mention it only because I have used it in the past to get around theroetical assumptions that seemed unreasonable to me at the time.

I suggest the first approach, by need to add another equation by consideration in the mechanics of the materials. Look at a second year statics textbook, there are typically several good examples in those references.

Good luck with it.

Regards,
Cockroach

RE: Statics 3 equations 4 unkowns

Interesting !
We have platform all the time with numerous bolting. Your 4 bolts will have loads and dis-proportional to the distance from the loading point, that is the way we solve this kind of problem easily, very simple, no fancy deformation formula, energy formula or FEA. Unless you are doing a high tech project or school project, using engineering judgement is good enough to solve the problem.

RE: Statics 3 equations 4 unkowns

"dis-proportional" means inversely proportional. it is a simple resolution to the indeterminancy (ie assuming the table is rigid).
so P1/d1 = P2/d2 = P3/d3 = P4/d4
so P2 = P1*d2/d1, etc
and P1+P2+P3+P4 = P
ie P1*(d1+d2+d3+d4)/d1 = P

it will definative solve sum forces, not obvious that is solves sum moments (not saying it doesn't).

another approach would be sum moments about one reaction so (P2+P4)*d2 = P*d
an then use sum moments in the other plane to proportion between P2 and P4 (ie if P is in the middle then P2 = P4; if P is 40% from P2 then P2 is 60% of (P2+P4) ...

calear as mud ?

the point is there are simple ways to approximate a solution.

RE: Statics 3 equations 4 unkowns

oops ... should be P1d1 = P2d2 = P3d3 = P4d4 (the bigger d1, the smaller P1)
so that P1*(1/d1+1/d2+1/d3+1/d4)*d1 = P

RE: Statics 3 equations 4 unkowns

It is just two simple beams, in orthogonal directions. Solve the first beam, which gives you the sum of two reactions on each end. Rotate 90 degrees, solve the other beam, which tells you how each end reaction is distributed.

RE: Statics 3 equations 4 unkowns

i agree hokie, balance the moments about the two planes ... comes out the same as the inverse distance method (just clearer to see the blanace)

RE: Statics 3 equations 4 unkowns

"Do I need to consider deformation of the platform"

Absolutely.

Does the platform make continuous contact with whatever is supporting it? Perhaps a floor? Are there pads or baseplates at each corner? When it comes time to install it, If it's a big floppy platform it will just droop to set on the floor . If the platform is "stiff enough" then it is most likely 3 "legs" will touch due to variations in the floor, the platform manufacturing, or both. If this is just some mezzanine then shim the gap under the short leg. If the platform needs to be leveled then shimming at multiple legs will probably be required. If rotating machinery is involved then the mounting points need to be hard shimmed and the anchor bolts tightened securely after the grout has set. Then the calculated loads will be supported as predicted. If the support is not uniform, as the anchors are tightened the short leg will require extra downward loading (upward loading on the anchor) to twist the platform to contact the floor, and the two adjacent corners will press that much harder on the floor.

RE: Statics 3 equations 4 unkowns

I don't think you understand your own problem, Nech0604. This far from a trivial problem and not as simple as you believe it is.

Essentially you are saying that a "platform" is subjected to a point load placed anywhere upon its surface. The "platform" which assumes a rectangular shape, is constrained at the four corners using a rigid system, say fasteners. You are interested with the load through the fastener at these four points.

First of all, deflection of a plate given placement of a point load anywhere on the surface is not a simple problem. You are dealing with a manifold which is not rigid. To simplify the problem, you make the manifold rigid so that it acts as a plate. I suggest you are now adding weight or at least load intensity to the problem, in addition to your point load.

Take a look at Roarks textbook. Plates subjected to boundary constrains typically have maximum moments of bending along the edges. This necessitates considerably less bending moment at the centre of the plate, but due to non rigidity, there is deflection. Since you are rigidly pinned at the corners of the plate, you must therefore be stretching the plate and so, you have stain.

So making the point load variable across the surface of the plate greatly complicates the problem. Finding the reactive loads at the four corners of this manifold will necessitate an understanding in manifold deformation, suggested by several responses to your question.

This is not a trivial question. I suggest trying to get a recipe solution set out of Roark's and moving on. Otherwise you're talking boundry valued differential equations, which are far from "kind of easy".

Regards,
Cockroach

RE: Statics 3 equations 4 unkowns

""dis-proportional" means inversely proportional. it is a simple resolution to the indeterminancy (ie assuming the table is rigid)."

Where's the proof?


"...The second case you need to look at the possiblilities in generating a sequence of linear dependent solution sets. This is an approach taken by Linear Algebra Calculus programs involving Markov's Chains. I think this is beyond the scope of this forum, but mention it only because I have used it in the past to get around theroetical assumptions that seemed unreasonable to me at the time."

Why is this "approach beyond the scope of this forum and furthermore In any case, I don't believe that a fancy mathematical manipulation can help solve a fundamentally unsolvable equation.

The 4th independent equation has to be based on a physical reality, not smoke and mirrors.

We are engineers, not snake salesmen. The idea that one's company has some mickey mouse method, not grounded in physics is not good engineering at best and dangerous at worst.

RE: Statics 3 equations 4 unkowns

So, maybe the simple approach of admitting that only 3 points will sustain the load is the only logical conclusion, but just how would require looking at the 4 triplet planes that are possible.

I would take the arbitrary point load and find all the triplets that enclose that point.

For each triplet, the 3 equations will yield the support forces. Now you take the maximum force for each answer, i.e., worst case.

Not the best, but logical, I think.

If one uses deformation, then the characteristics that define that deformation must be understood before writing the equations.

RE: Statics 3 equations 4 unkowns

@zeke, """dis-proportional" means inversely proportional. it is a simple resolution to the indeterminancy (ie assuming the table is rigid)."

Where's the proof?" ... i was trying to clarify the previous post. "disproportional" is IMO an unusual mathematical term, "inversely proportional" is more commonly used and correct (as intended, i believe, and as i've derived for myself).

@zeke, yes, i agess you could solve the 4 statically determinate "triples" and avarage them ... i'm willing to bet that that'll give you the same result (as assuming inversely proportiona, or assuming a rigid frame).

@greg, if this was an exam question, and if (as in this case) data isn't provided, then you make (and state) assumptions. you could assume that you know the bending stiffness of the frame, you could assume it's rigid, you could assume it is some finite value. you could assume that the supports line on a plane, or you could assume that they don't. you could assume that all four supports are in contact with the "rest of the world" or you could assume that one isn't. you could assume that the frame, part of a structure, is as dynamic as the "springs for a car" or you could assume that it is a static structure.

geez, can we bury this horse, pls ? the poor thing has been flogged enough, IMHO.

RE: Statics 3 equations 4 unkowns

"Where's the proof?" ... i was trying to clarify the previous post. "disproportional" is IMO an unusual mathematical term, "inversely proportional" is more commonly used and correct (as intended, i believe, and as i've derived for myself)."

Then why don't you share that derivation with the rest of the world?

RE: Statics 3 equations 4 unkowns

sorry, i'll distribute it once i've copyrighted it ... 1/2 pg of scribble showed me that reactions distributed this way (F1d1 = F2d2 = F3d3 = F4d4) satisfy static equilibrium (assuming a rigid frame). it gives you the same solution as balancing the moments about the two planes, but is easier to remember (i guess).

RE: Statics 3 equations 4 unkowns

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,

RE: Statics 3 equations 4 unkowns

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)".

RE: Statics 3 equations 4 unkowns

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

RE: Statics 3 equations 4 unkowns

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" !?

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