Maxwell's reciprocal theorem 4

According to Maxwell's reciprocal theorem the displacement at coordinate i due to a force at a coordinate j is equal to the displacement at coordinate j due to a force at a coordinate i.
What if the dispacement at coordinate i is a rotation and at j is a translation?
How can a translation and a rotation be equated?

 

[IRstuff]

Why would you think that they can be equated?

Isn't this why there are 6 degrees of freedom?

TTFN

 

[GregLocock]

IRstuff- because that IS what reciprocity is all about.

setee - I fully understand your confusion. I use reciprocity from time to time at work, or more accurately, I frequently use lack of reciprocity to find experimental errors. To give a concrete example, if I apply a torque TZ to station 1 and measure the displacement Y at station 2, and then reverse the two stations then reciprocity says that in a linear, passive, system Y2/TZ1=Y1/TZ2.

It is a corollary of the superposition theorem (according to Roark), and is closely associated with St Venants theorem.

The example you give is slightly incorrect "What if the dispacement at coordinate i is a rotation and at j is a translation?"

Reciprocity does not apply if you change the type, or the direction, of measurement. Cheers

Greg Locock

 

[austim]

Maxwell's theorem is entirely general, and there is no problem in applying it to relate loads and displacements even if one displacement is a translation and the other a rotation.

Just consider what we do when we use matrix analysis for any structure (whether 2D or 3D).

We start with [stiffness matrix]*{displacement vector}={force vector}

We quite happily include both translations and rotations in our displacement vector. Similarly we use both lineal loads and applied moments in our force vector, (with applied moments where the corresponding degree of freedom is a rotation).

The fact that the stiffness matrix is always symmetrical is evidence that Maxwell's theorem applies to both types of loads and displacement (linear and rotation).

All that we have to remember is that if a degree of freedom is a translation, then the unit force that is to be applied there is a lineal load. If the degree of freedom is a rotation, then the unit load to be applied is a unit moment.

eg, consider a simple propped cantilever of constant section.

With a vertical point load at midspan, the rotation at the free end is W*L^2/32EI. For a unit load, the end rotation = 1*L^2/32EI.

With a moment applied at the free end, the midspan vertical deflection = M*L^2/32EI, and for a unit moment it =1*L^2/32EI, and Maxwell's theorem is seen to apply.

 

[IRstuff]

Greg,

That was my point: a translation cannot be equated to a rotation.

TTFN

 

[satee]

That means rotation at the free end due to a unit load (vertical load) at the midspan is equal to the translation at at mid span due to a unit load(moment) at the free end?
Austim, this means that translation and rotation are equal numerically?

 

[GregLocock]

NO. That was the error I thought you were making.

The (vertical)deflection at the free end due to a load at midspan is equal to the (vertical)deflection at mid span due to a load at the free end.

Also


The rotation at the free end due to a load at midspan is equal to the rotation at midspan due to a load at the free end.

You can't muddle DOFs up, which is IRstuff's point (sorry, I thought you were confusing things, you were being cryptic)

Is that clearer?
Cheers

Greg Locock

 

[austim]

Satee,

I'm afraid that before we are done, this thread will have caused great confusion, but yes, your interpretation of my post is entirely correct. You do get the same numeric value whether you calculate the midspan vertical deflection due to "unit" moment at the propped end or the angular rotation at the propped end due to "unit" vertical load at midspan.

I urge you to do the same sort of calculation on any structure you choose, or try unit loads/moments on any pair of nodes in any structural model that have assembled and verify what I say for yourself.


Greg, regrettably I have to say that you are plain wrong.

What you have to do with Maxwell is to ensure that the loads that you apply are matched to the degree of freedom to which they are applied. So long as you do that, Maxwell does not care whether the degrees of freedom are two rotations or two linear displacements or one of each.

The whole point of my propped cantilever example was to show that the rotation at the propped end due to unit vertical load at midspan is numerically equal to the vertical deflection at midspan due to a unit moment applied at the propped end.

Note that I am not saying that a "translation is equivalent to a rotation", just that the computed values are numerically equal. That is all that Maxwell says.

As I have suggested to satee, why not try it out for yourself, on any structural model that you may have. Just try two load cases, each with a single unit load/moment and compare the resulting output displacements. You might be quite surprised by your findings.

 

[trilinga]

Austim, you are absolutely correct.

I think we should look at Maxwell's theorem from the fundamentals. The theorem is based on not just the displacements and forces but the equivalence of work done.

the theorem states that

' The work done by one set of forces in undergoing the corresponding displacements caused by the second set of forces is equal to the work done by the second set of forces in undergoing the corresponding displacements caused by the first set of forces.'

For simplicity, if you consider only one force (or moment) in eash set, the product of force-1 acting at point A and the displacement at A in the direction of force-1 caused by force-2 is equal to the product of force-2 acting at point B and the displacement at B in the direction of Force-2 due to force-1.

Hence, by making the values of the forces(or moments)numerically equal to unity, you get numerically equal displacements. The equivalence is only numerical irrespective of whether the displacement is linear or rotational.

 

[austim]

trilinga,

I am very grateful to you for your post.

One of the things that I have learnt over many years is that there is often no-one more wrong that someone who is quite sure that he is correct. With my certainty that my comments were valid came an uncomfortable feeling that I needed to verify them.

Accordingly, I have just been re-running my basic maths, deriving the deflected curve for a propped cantilever (by Macaulay's method of double integration of the BM diagram, much slower than when I used it in 1954 or so ).

I was greatly relieved to find that my result exactly matched Roark's formula for the end slope of a propped cantilever with varying location of point load (with the minor substitution of M for W).

My next step was going to be a couple of numeric trials with independent computer programs, and then I would have been ready to sleep. Thank for saving me those last steps .

 

[satee]

trilinga,
I since you mentioned about work done i would like to come out with a doubt which i have from years.
Now it is said, for a real work done we multiply the load and the corresponding displacement divided by 2.Right?
The reason for dividing by 2 is said to be that load being applied gradually.
I still don't follow that are we applying load gradually?For example consider the spring test we have done in U.G., we do not apply load gradually but just place the mass and as a result spring deflects.Only, thing is that we do not apply load under impact.
Next doubt,
As you mentioned,
"The work done by one set of forces in undergoing the corresponding displacements caused by the second set of forces is equal to the work done by the second set of forces in undergoing the corresponding displacements caused by the first set of forces."
Now both these works done you mention are actually not done by these forces, i.e. "work done by first set of first set of forces in undergoing the didplacement caused by second set of forces is actually done by second set of forces", right?Then why bring this work in expression for work done and give it names like virtual work???

 

[trilinga]

Satee

The answers to your questions

1. Gradually applied load does not mean that the load is applied gradually by you. It only means that the load gets transferred to the member from zero to maximum while the deformation increases simultaneously from zero to maximum. It is not the extent of time taken for the load transfer which makes it gradually applied load but the load transfer in proportion to the deformation which gives it the name. In other words, the load-displacement curve is not vertical but is inclined to the axes. The work done in this case is the area under the load-deflection curve which is a triangle (for linearly elastic behaviour)and hence the division by 2 for the work.

If you consider a body already loaded with a load P1 and an additional load P2 is applied to the body, the additional deformation under the load P1 will occur with P1 acting fully through the deformation. The additional work done by P1 in this case is the product of P1 and the additional deformation (no division by 2).

2. Regarding the virtual work and virtual displacements indicated in the reciprocal theorem, these are concepts developed to derive certain relationships among the structural parameters. Thay need not occur in reality.

 

[flamby]

The first part of your question is why the work done is load*deflection/2.

Consider a spring loaded by a weight W (gradually or suddenly). The spring will deflect by a distance say D. At any moment when the sping is deflecting from 0 to D, the deflection of spring will be Dt where 0<Dt<D. Now how much force the spring is exerting? It is K*Dt, where K is the spring constant.

So you see the spring reaction rising linearly from zero to KD. The area of this triangle is obviously (KD)*D/2. In other words, W*D/2, which is the potential energy of spring at the moment. Since this is a triangle, you see a factor of two there.

 

[GregLocock]

Well if I'm wrong I'm wrong, but I thought I was paraphrasing Roark:

Let a and b be any two points in a linear system. Let the displacement of b in any direction U due to a force P acting in any direction V at A be u, and let the displacement of a in direction V due to a force Q acting in direction U at B be v. Then Pv=Qu

So &quot;The (vertical)deflection at the free end due to a load at midspan is equal to the
(vertical)deflection at mid span due to a load at the free end.&quot;

is correct

&quot;The rotation at the free end due to a load at midspan is equal to the rotation at midspan due to a
load at the free end.&quot;

is wrong. aaagh. sorry.



Cheers

Greg Locock

 

[austim]

Greg,

please don't forget Austim's Nth rule of engineering - Don't believe anything you are told unless you can verify it somehow. If you haven't already run your own numeric verifications Please do so.

That would be a much more satisfactory conclusion than simply being bludgeoned into changing your belief by some anonymous Australian engineering fundamentalist that you have never even met.

 

[GregLocock]

I did this at uni, we even did a lab on it to try and get the idea imprinted properly. At a conservative estimate that was 5000 bottles of red ago, not that that is any excuse. Anyway, now I've looked the proper definition up I'll have to think of a use for it! Cheers

Greg Locock

 

[satee]

Thank you Trilinga!!
Trilinga you mentioned-----&quot;In other words, the load-displacement curve is not vertical but is inclined to the axes----.&quot;Should'nt it be &quot;horizontal&quot;(usually with load on x-axes and displacement on y axes)?Hope i am not wrong!!!!!

Flame,

The spring deflects because it is first in a state of unstable eqilbrium and then gradully when it reaches a state of stable equilbrium it stops deflecting.As the load is getting transferred, simultaneouly spring is aquiring equilbriting force , the reaction in the spring is at any instant the amount of load transfered.
So the multiplication of the spring constant with the displacement should be done only when the spring stops deflecting not at any moment when the sping is deflecting from 0 to D.What i am trying to tell is , that , is the theorem of potential energy related to the reason for division by 2.Is'nt the theorem of minimum potential energy only an explanation for stable/unstable equilbrium?
Trilinga please comment

 

[trilinga]

Satee

I have considered the load - deflection curve with deflection on the x-axis. This is the normal practice since deflection is taken as the independent variable since it is observable and independently measurable.

Flame's explanation is right. Read his post once more. Flame is not referring to the displacement of an oscillating spring which comes to static equillibrium after some time. that situation corresponds to a dynamic or impact loading whose equation of equillibrium is different. What is referred here is a static loading on the spring which deflects gradually and reach the maximum. The condition 0<Dt<D mentioned in his post means that. He has considered the load transfer proportionate to the deflection under static loading.

I am not clear about the last part of your question regarding the potential energy. Can you elaborate?

 

[satee]

Sorry, sorry i meant load on y axis and deflection on x axis should it be read as &quot;horizontal&quot; instead of &quot;vertical&quot;?
Sorry, Flame if i got you wrong.Appologies!!!
Trilinga, according to the theorem of minimum potential energy th structure is in equilbrium if no change occurs in total potential energy of the system when its displacement is changed by a small arbritrary amount.In case of the spring above , is the spring in a state of stable equilbrium throughout its course of deflection from 0 to D.??

 

[trilinga]

Thanks for kindling my thought process and also make me do some homework.

I would like to modify one of the statements in my previous post. In reality, there is no load transfer that takes place completely static. Any load that is applied to a system causes a disturbance to the system and makes it respond to the change. The system undergoes a time varying deformation before it 'settles down'. During this transient phase, the system is governed by the equation of dynamics. The governing equation reduces to the equation of static equilibrium once the dynamic components die down.

So, the spring is in equilibrium even during the moving phase but is governed by a different equation.

However, we reckon the response as static or dynamic depending on the time taken by the system to get rid of the 'ripples' and come back to static equilibrium. For a system recovering fast the response will appear more static.

I can state a very common example. You observe a sequence of ripples on the surface when water is poured into a bowl. But the surface looks practically undisturbed while pouring a very viscous liquid like lubricating oil. The dynamic disturbance is absorbed immediately due to the viscosity of the liquid and the dynamic phase is unseen.

The minimum energy principle is common for both static and dynamic problems. It defines the unique displacement configuration, which keeps the system in equilibrium.

For a static system, it is stated as :

‘Of all the displacement configurations which satisfy the boundary conditions, the one which minimizes the potential energy keeps the system in stable equilibrium.’

In case of your spring example this yields the equation

stiffness *deflection = applied force

or k*v = P where v is the deflection.

For a dynamic system the same principle is stated as :

‘Of all the time histories which satisfy the boundary conditions and also the initial and final conditions, that which makes the term (Kinetic energy – Potential energy) a minimum, will keep the system in equilibrium.’

the equation in this case becomes

m*d^v/dt^2+C*dv/dt +k*v =P(t),where v is the time varying displacement

You can refer to any standard text book which will show how the minimum energy principle results in the equilibrium equation.

Hope I have not deviated very much in answering your query.