maxwells reciprocal theorem

huangyhg

maxwell's reciprocal theorem
why would you think that they can be equated?
isn't this why there are 6 degrees of freedom?
ttfn
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
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).
greg,
that was my point: a translation cannot be equated to a rotation.
ttfn
satee (visitor)16 jul 02 0:30
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?
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
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.   
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.
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 (visitor)16 jul 02 9:00
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???

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   
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.
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 "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."
is correct
"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."
is wrong. aaagh. sorry.
cheers
greg locock