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statically indeterminate vs rigidly indeterminate
4

statically indeterminate vs rigidly indeterminate

statically indeterminate vs rigidly indeterminate

(OP)
The difference between statically determinate and statically indeterminate is whether or not you can determine all the reaction forces and moments from a free bodying diagram without considering strain of the body (i.e. while considering the body rigid).

What does that have to do with static? To me static means not varying with time (opposite of dynamic).

Shouldn't it be called rigidly indeterminate?

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RE: statically indeterminate vs rigidly indeterminate

It has a nano-second of time involved (joke)

The "static" analysis assumes an instantaneous effect.

RE: statically indeterminate vs rigidly indeterminate

3
from the web http://www.personal.psu.edu/faculty/r/c/rce2/mcht111/111intro.html

The mechanics of rigid bodies is sub-divided into two areas, statics and dynamics, with dynamics being further subdivided into kinematics and kinetics.  Statics is the study of bodies in equilibrium.  This means there are no unbalanced forces on the body, thus the body is either at rest or moving at a uniform velocity.  Dynamics is the study of bodies which are not in equilibrium, thus there is acceleration.  Kinematics is the study of the motion of a body, without regard for how the motion is produced.  This is sometimes called the "geometry of motion".  Kinematic principles are often applied to the analysis of machine members to determine positions, velocities, or accelerations at various parts of the machines' operation.  Kinetics is the study of the forces which cause motion, or the forces which result from motion.

Mechanics of Deformable Bodies:  The mechanics of deformable bodies deals with how forces are distributed inside bodies, and with the deformations caused by these internal force distributions.  These internal force produce "stresses" in the body, which could ultimately result in the failure of the material itself.  Principles of rigid body mechanics often provide the beginning steps in analyzing these internal stresses, and resulting deformations.  These will be studied in courses called Strength of Materials or Mechanics of Materials.

RE: statically indeterminate vs rigidly indeterminate

(OP)
I don't quite understand. Are you guys saying the use of the term static here makes sense?

Aren't there many statically-indeterminate problems which can be solved by the methods of statics (where "statics" would include static deformation under static load) ?  

For example: find static reaction forces for various loadings of a beam with three supports.  This is classified as statically indeterminate but I can solve it using static methods.  But I can't solve it if I considered the beam to be rigid (non-bendable).  Hence to my mind it would be more logical to call it rigidly indeterminate?

I know none of this is new to you guys. Am I missing something or is this just another one of those ME terminology things that isn't supposed to make sense?

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RE: statically indeterminate vs rigidly indeterminate

I think the "static methods" you refer to are equating the forces and moments about the body as a whole so that it doesn't fly off into space.  Analyzing the internal forces (stresses) would be using "mechanics of deformable bodies" as a necessary addition to statics.

RE: statically indeterminate vs rigidly indeterminate

"dynamics" implies moving things (like velocity, acceleration, momentum, ...)

"statics" implies static equilibrium, that the thing isn't moving.

"statically determinate" means that you can solve for the external reactions using the equations of equilibrium (sum Forces and sum Moments).

"statically indeterminate" means that you have too many reactions (more reactions than there are equations of equilibrium).  This is solved by applying other methods (eg energy, unit force, ...) to assemble enough equations.

RE: statically indeterminate vs rigidly indeterminate

(OP)
Thanks Mike.  If I accept the framework of your first post, then I can see that the terminology makes sense:
1 – Rigid Body
1A – Statics
1B – Dynamics
2 – Mechanics of deformable bodies.
Statics would under this framework always imply rigid body.
Still I think the terminology is a little misleading. Once again it’s not the limitation of static (vs dynamic) analysis that determines whether or not we can solve this type of problem, but the limitation of rigid body (vs deformable) analysis.  

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RE: statically indeterminate vs rigidly indeterminate


Although I may be wrong, I believe that when speaking of structures, solutions to static indeterminacy, can sometimes be found by assuming displacements (related to stiffness), not necessarily body deformations from plasticity, rigidity or ductility considerations.

Thus it appears to me that the prevailing semantics still has a logical basis.

RE: statically indeterminate vs rigidly indeterminate

you're right 25362, in that statically indeterminate strcutures aren't solved by invoking plasticity.  rather energy concepts are applied.

the traditional terminology is still the best (99.99% of people involved understand it).  everything deflects, including statically determinate structures but this is still static.  as opposed to dynamic which would describe a moving structure, like something vibrating.

RE: statically indeterminate vs rigidly indeterminate

(OP)
I don't see where the latest posts add anything to support the traditional terminology.

#1 - I don't see what difference it makes whether you call it deformation or displacement.  The key thing is that we cannot solve the indeterminate problem if the body is considered rigid (once again the reason for my suggestion rigidly indeterminate).

#2 - "as opposed to dynamic which would describe a moving structure, like something vibrating."  You're saying this supports the traditional terminology? I see it exactly the opposite.  There is no distinction static vs dynamic involved in this concept of statically-determinate vs statically indeterminate.  That was my original point.

One more comment, getting a little further into the realm of semantics.

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RE: statically indeterminate vs rigidly indeterminate

have it your own way, enjoy your journey (to the realm of semantics)

RE: statically indeterminate vs rigidly indeterminate

You most certainly can solve many indeterminate problems by assuming the body is rigid.  All "rigid" means is that it will have no deformation.  The reason this is never brought up is because it does not reflect real materials;  but mathematically speaking it would work in many cases.

RE: statically indeterminate vs rigidly indeterminate

(OP)
"You most certainly can solve many indeterminate problems by assuming the body is rigid."

Can you give an example to help clarify this?
Thx.

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RE: statically indeterminate vs rigidly indeterminate

Sure:  say you have a "rigid" beam with 3 supports underneath it.  The supports are oriented with one on each end and one in the middle.  Now add a distributed load across the beam.  The reactions at each support would equal 1/3 of the load applied.  Note that it wouldn't matter if the supports were pinned or moment resisting or even if they were rigid or not, as long as they were equal.  Another anomaly is that you get the same answer if you put a point load in the center.

Now, take the same setup and put a point load at one end.  If the supports are also perfectly rigid - the load goes straight into the support under the load and NO load goes to the other two.  If the two outer supports and equally not rigid and the center one is rigid, then P goes to the support under the load P, 2P goes to the center, and -P goes to the other end support.  Messed up, isn't it?

The assumptions made above are the same you make when doing a 4-bar linkage.  If those bars aren't rigid, the equations get goofy in a hurry.  In the 4-bar case, you assume the bars are rigid to simplify what's going on.  The difference is, this assumption actually approaches reality in most cases where my previous examples do not.

We all know this is hogwash in reality, but again, the mathematics work with the constraints given.

Looking back and reading, I have to apologize for this and the preceding post.  They're pretty far off topic and useless to boot...

RE: statically indeterminate vs rigidly indeterminate

(OP)
I would say rigid beam with three rigid supports and point load directly above one is still statically indeterminate.  A combination of those supports could take the load and we have no way of knowing which.  I could remove the support directly underneath the load and still the beam is in static equilibrium.

The case of center rigid support with two equally flexible supports is a little different than what I had considered but the same basic idea that to make the system become determinate you can no longer consider all parts rigid.  Before I focused on the "body"... didn't cover the support but the same concept applies.  

So I would still submit that to resolve static indeterminacy we need additional relationship which will describe actual movement of some part under the loading.

Still all this is nothing new to you guys I'm sure.

My argument wasn't intended to be mathematical, only semantic.  Semantically, one could interpret "statically indeterminate" to mean indeterminate (underspecified) when evaluated under static conditions and requires additional (nonstatic) information to resolve.

That is not the only way to semantically interpret the phrase and obviously not the way it was intended.

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RE: statically indeterminate vs rigidly indeterminate

(OP)
I think as a result of your comments I understand the sense of the phrase (why they chose those words).

Just to explain where I was coming from to begin with.  What if you substituted "underspecified" for "indeterminate".

Would the term "statically underspecified" make sense to you guys?

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RE: statically indeterminate vs rigidly indeterminate

(OP)
... to me the phrase "statically underspecified" better illustrates a phrase whose semantic interpretation would not very well match what we understand as statically indeterminate.

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RE: statically indeterminate vs rigidly indeterminate


It could also be statically overspecified.

RE: statically indeterminate vs rigidly indeterminate

(OP)
Yes I agree.  Overspecified in terms of number of variables (each reaction adds a variable).  Underspecified in terms of number of equations describing how the system will respond to the load (bending, deformation).

My underspecified analogy isn't perfect.

All in all if it were up to me I'd still call it rigidly indeterminate.  I guess it's a good thing it's not up to me!

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RE: statically indeterminate vs rigidly indeterminate

"Static" (the word) and "Dynamic" (the word) are nearly opposites.  However "Statics" (the discipline) and "Dynamics" (the discipline) are not opposites.  "Statically indeterminate" refers to the discipline, and means that the system cannot be solved with the equations of Statics alone.  Neither Statics nor Dynamics include any consideration of deformable bodies, as noted by IvyMike.  Your assertion that this is confusing, inaccurate, or misleading is equivalent to the following:

Why is it called "Statically Indeterminate" when the problem has nothing to do with electricity?

RE: statically indeterminate vs rigidly indeterminate

Maybe I'm way off topic now, but I'll throw this in anyway.

One thing that separate statics from dynamics is time.  In statics (actually getting into strength of materials), we can have deflection.  I can apply a point load to the end of a beam that is "rigidly" anchored at the opposite end (the classic cantilever beam) and observe a deflection.  This is still a statics situation becuase I'm only interested in the beam in it's deflected state (i.e. what the stresses are, etc.)  I don't care how long it took to get there and, in this case, the applied load doesn't change with time.  Therefore, the situation is one of equilibrium --- static, that is.

For Dynamics, there's going to be a time element involved.  i.e., the force applied at the end of the beam goes from 0 lbs to 1000 lbs back to 0 lbs every second.  Or, my beam is lying on the ground and I apply a net lateral force of 1000 lbs and the beam starts sliding faster and faster (F = ma, or in this case, a = F/m).

Does that help this conversation, or have I just been babblign to myself this morning?

Edward L. Klein
Pipe Stress Engineer
Houston, Texas

"All the world is a Spring"

All opinions expressed here are my own and not my company's.

RE: statically indeterminate vs rigidly indeterminate

(OP)
StressGuy - your suggestion that a beam can deflect in statics is in conflict with IvyMike's link which identifies statics as a subset of rigid-body mechanics.

handleman - You will forgive me if I don't agree that  electrically-indeterminate  describes the situation just as well as rigidly-indeterminate.  Bearing in mind that indeterminate is not too far from underspecified.  We can add additional non-rigid equations to solve the problem.   

Another analogy might be to identify a type of mathematical problem as "analytically unsolveable".  That means we need to try another approach such as numerical.  The word analytically here describes the bounds within which the problem is not unsolveable...but the problem can be solved by other means outside of those bounds.  From semantics alone, one might also expect that the word statically in statically indeterminate describes the bounds within which the problem is indeterminate.

However I acknowledge that is not how it is intended.

I am happy to call this matter closed.

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RE: statically indeterminate vs rigidly indeterminate

(OP)
No, not quite closed.  I'd like to ask if anyone on this forum agrees there is a semantically-possible analogy between "analytically unsolveable" (or "algebraically unsolveable") and "statically indeterminate".  

And further does anyone see that under the semantic interpretation implied  through this analogy , rigidly indeterminate would be more descriptive of the condition than statically indeterminate?

Thx.

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RE: statically indeterminate vs rigidly indeterminate

So does "analytically unsolveable" mean that the problem does not yield to any type of analysis?  Analysis would include numerical analysis.  If the problem is unsolveable with any type of analysis then the "analytically" tacked on the front is self-redundantationalifying, semantically speaking. winky smile

The difference between "Statics" and "static" is similar to the difference between "analytically" used in the mathematical sense and "analysis" in the more general sense.  

RE: statically indeterminate vs rigidly indeterminate

(OP)
The following might help clear up the confusion:

http://mitpress.mit.edu/books/FLAOH/cbnhtml/glossary-A.html#analytical_solution

"Analytical Solution     An exact solution to a problem that can be calculated symbolically by manipulating equations (unlike a numerical solution). "

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RE: statically indeterminate vs rigidly indeterminate

sorry, that link has nothing to do with the topic, since we numerically solve an expanded set of equations (to solve statically indeterminate structures).

regarding some other posts, if the structure is determiate then it's stiffness doesn't matter (that will only affect how much the structure deflects) ... the loads will be reacted as required by equilibrium.

if a force is applied at a conventionally pinned support, then the reaction will be at that support.  the only way the load will work its way out to the other (redundant) supports is if the support is elastic (rather than rigid), then effectly the applied load is reacted by the stiffness of the support and the stiffness of the structure.

RE: statically indeterminate vs rigidly indeterminate

I realize that in mathematical terminology analytical solutions exclude numerical solutions.  However, in the broader definition of analysis, numerical solutions would be included.  You are restricting the definition of "Analytically" to the discipline out of which the term "analytically unsolvable" is being taken.

In the same way, in Engineering terminology the "Statically" part of "Statically indeterminate" refers to the discipline of Statics, not to the fact that the system is not in motion.  "Rigidly indeterminate" would make no sense in this application because the term "Statically indeterminate" is from the discipline of statics, which excludes deformation of bodies!  You cannot remove "statically indeterminate" from its context of Statics an attempt to make it stand on its own semantically.  Otherwise, you must include all definitions of "static", including fields such as electricity, biology, computer science, etc.

Statically indeterminate: Can't solve with Statics alone.
Algebraically indeterminate: Can't solve with Algebra alone.

How are these different?

RE: statically indeterminate vs rigidly indeterminate

(OP)
[quote]
Statically indeterminate: Can't solve with Statics alone.
Algebraically indeterminate: Can't solve with Algebra alone.

How are these different?[/quoute]

I think they are very similar.  And I think my argument is supported to the extent that we believe they are similar and we accept your words:
"Statically indeterminate: Can't solve with Statics alone."

The other implied half of "Can't solve with Statics alone" is "Can solve by going beyond statics".

Well I can go beyond statics to dynamics and it does nothing to help me solve the problem.  

What I need to do is go beyond rigid body mechanics to solve the problem.

So again from this train of logic one would prefer to call it "rigidly indeterminate"  (cannot solve by rigid-body analysis).

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RE: statically indeterminate vs rigidly indeterminate

I can't help myself;

"A Rose by any other name is still a Rose."

RE: statically indeterminate vs rigidly indeterminate

(OP)
rb1957 - the link was a response to the post which immediately preceded it. Sorry if I wan't clear.  It sounds to me like you are saying that the scenario of rigid beam with three rigid supports and point load above one is determinate (similar to what another poster suggested).  I have a hard time making that leap.  As I said we can remove the support directly underneath the load and nothing will change... so how do I know which supports were carrying the load before I removed that support? Seems indeterminate to me.  As the previous poster mentioned this is a somewhat idealized scenario and I am happy with his conclusion that it's not that important.  (Happy to agree to disagree on that point)

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RE: statically indeterminate vs rigidly indeterminate

Quote:

... I can go beyond statics to dynamics ...

The implied statement here is that dynamics is beyond/above statics.  Dynamics is not necessarily "beyond" Statics.  It's just different.  In fact, it's not even the next logical direction to go once you determine that a system is Statically Indeterminate.  Keeping the discipline and logic of Engineering in mind, one would not arrive at the conclusion that a system is Statically Indeterminate (as defined by the discipline of Statics) without first concluding/assuming that the system is not in motion.  This rules out Dynamics to begin with.  Statically Indeterminate systems are usually (always?) overconstrained rather than underconstrained.  If you had a

Also, since "Statically Indeterminate" originates in the discipline of Statics, all that can be said is that Statics cannot solve for the reactions at supports.  Statics is not "qualified" to tell the engineer which direction to take.  It is up to the experience and knowledge of the engineer to know that deformable body mechanics are needed after that.  

RE: statically indeterminate vs rigidly indeterminate

if you have a beam (rigid or flexible) on three rigid (not elastic springs) supports, and you have a load directly above any one of the supports, then load will be reacted by that support and the other two support reactions will be zero.

if you have an idealised rigid beam on three rigid supports and load it then ...
1) you can't apply unit load methods (as you've got no deflection to cancel out) so
2) i suspect that energy methods will indicate that the load is reacted by the two adjacent supports; as i think this minimises the internal energy in the beam, by minimising the area of the bending moment diagram.

if you apply the load to a real flexible beam between the supports, the problem is indeterminate (statics won't be able to tell you how much load is reacted by any of the supports).  i would typically use the unit load method to the load at the middle support and then the others fall out

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