castiglianos theorem111

huangyhg

castigliano's theorem!!!
the answer to your question lies in the understanding of the physical meaning of the theorem.
every stable structural system responds to the external loads satisfying two important conditions viz. equillibrium and compatibility. in a statically determinate structure,there is only one set of reactions which satisfy the equillibrium conditions and this set of reactions also ensures the compatibility. however, in a statically indeterminate system, due to the redundancy, it is possible to obtain a number of sets of reactions which satisfy the equillibrium conditions. but the realistic set of reactions is that which also satisfies the compatibility conditions that introduce the redunduncy.
castigliono's theorem states that of all the sets of reactions which satisfy the equillibrium condition,that set of reactions which makes the rate of change of strain energy with respect to the respective reactions zero (for  non-sinking supports)is the realistic one satisfying the compatibility also.
any system (a staically indeterminate system in this case) under a given set of external loading creates its reactive forces in such a way to minimise the work done by the external forces. in a system with non-sinking supports, the work done by the reactive forces is zero and so the work done by the external loads is equal to the strain enegy of the system . hence, minimising the external work means minimising the strain energy which in turn means the rate of change of strain energy with respect to any reaction is zero.
hope this clarifies your doubt (partly at least?).
gendoubt!!! (visitor)30 jun 02 14:26
thanks very much for the reply.can you give me one example with respect to a propped cantilever beam taking the reaction at the prop as the redundant about a set of reaction that satisfies equilbrium but not compatibility?
gendoubt,
your second question is much easier than your first .
if we assume that your propped cantilever has a uniformly distributed load over its full length, then two sets of reactions which satisfy equibrium are :
(a) wl/2 vertical reaction at each end, but no fixing moment.  equilibrium is satisfied, and so is the requirement for zero deflection at both ends, but the reuirment for zero rotation at the fixed end is not met.
(b) wl vertical reaction at the fixed end, plus a moment of wl^2/2.  ok for zero deflection and rotation at the fixed end, but does not give zero deflection at the propped end.
well stated austim.
gendoubt
in fact these are not the only two sets of reactions possible. mathematically, you have four unknowns (three reactions at fixed end and one at the propped end) but you will have only three equations of equillibrium if you do not consider compatibility condition. so you have three linear equations with four unknowns. solving the three equations for any three of the unknowns will yield expressions for the three reactions in terms of the fourth reaction. for each arbitrary value of the fourth reaction, you will get one set of the other three reactions. all these four reaction values will satisfy the equillibrium equations. you can hence generate infinite no. of reaction sets which satisfy equillibrium. but, there is only one set which is realistic which is fixed by the compatibility condition.
gendoubt!! (visitor)1 jul 02 20:48
thanks very much for the reply!!!much obliged!!lat question,
i.e. in case of a pinned support deflection in the vertical direction and horizontal direction are zero, fixed support
deflection in the vertical direction and horizontal direction , slope are zero.why are these conditions called compatibility conditions?what is the literal meaning of the word compatible here?  
compatibility in this case refers to compatibility of displacements. physically, it means that at the location of the support, the member and the support have consistent displacements. in other words, there is neither a gap or overlapping between the   
gendoubt,
my 2-cent contribution to this thread:
the steel manual shows on the beam diagrams and formulas a propped cantilever under uniform load, concentrated load at the center and concentrated load at any point.    the reactions for the uniform load are 3/8*w*l at the pinned end and 5/8*w*l at the fixed end.
one system to obtain these reactions (not using the castigliano's theorem, but superimposing loads) is to consider a cantilever beam under any loading condition, and find what force at the pinned end would bring back the deflection of that support to zero.
i know this does not answer your question about the castigliano's theorem, but is another easier approach to the propped cantilever problem.
gendoubt!!! (visitor)2 jul 02 12:14
trilinga,austim,dlew thanks very much for your interests.a very special thanks to trilinga for you made my thoughts a step clearer in the subject.  much grateful and obliged.
thanks again!!!!!!
thank you gendoubt.
i visit this forum almost regularly. the forum is very useful and interesting and the response to any type of question is prompt and helps to broaden one's knowledge base. you can come across varied experiences of the practicing engineers and their valuable suggestions.i find that there are many experts  who participate in this forum and who can clarify your doubts much better than me.
however, if you need any help just let me know.
trilinga
you mentioned:
"in fact these are not the only two sets of reactions possible. mathematically, you have four unknowns (three reactions at fixed end and one at the propped end) but you will have only three equations of equillibrium if you do not consider compatibility condition. so you have three linear equations with four unknowns. solving the three equations for any three of the unknowns will yield expressions for the three reactions in terms of the fourth reaction. for each arbitrary value of the fourth reaction, you will get one set of the other three reactions. all these four reaction values will satisfy the equillibrium equations. you can hence generate infinite no. of reaction sets which satisfy equillibrium. but, there is only one set which is realistic which is fixed by the compatibility condition. "
is that the case in displacement method of analysis too?, that is it so in displacement method of analysis, there could be a number of possible joint displacements(rotation or translation ) possible but only one satisfies equilbrium??
it is true.
both force and displacement methods of analyses aim at identifying the unique set of realistic reactive forces and displacement field by satisfying both equillibrium and the compatibility conditions.
unlike the force method where the forces are treated as unknowns,in displacement method, we assume the displacement field as unknown, express the various nodal/reactive forces in terms of these displacements and apply the conditions of nodal equillibrium along with the geometric boundary conditions to solve for the unknown displacements.
slope deflection method and stiffness method of matrix analysis are examplels of this approach.
ok, but as you mentioned in case of force method of analysis for indeterminate structures the no. of unknowns exceed the no. of available equilbrium equations.as mentioned by austim and yourself there could be a no. of possible solutions for the set of forces which satisfy equilbrium but only one set of forces satisfies compatibility which is got through compatibility equation.a wonderful example with reference to a propped cantilever beam was given by austim and was fantastically elaborated by you.can u give one such example for displacement method too(that is where a no. of sets of forces satisy compatibility but only one realistic se satisfies equilbrium)
rogy (visitor)15 aug 02 14:23
when you say compatibility it is meant compatibility of displacements not forces so your question "can u give one such example for displacement method too(that is where a no. of sets of forces satisy compatibility but only one realistic se satisfies equilbrium" is wrong instead it shoould be
"can u give one such example for displacement method too (that is where a no. of sets of ""displacements"" satisy compatibility but only one realistic set satisfies equilbrium)"
now just think a bit and let me know if you get the answer for yourself else i shall let you know.
rogy (visitor)16 aug 02 8:52
since you have'nt replied i presume you have'nt got the answer:
consider a two span continuous beam of spans l1, l2 respectively, this has three supports all pinned.say both spans have a udl of x kn/m. now assume rotation of support 1(left extreme) is  say "y", rotation to the left of intermediate support is say "z", to the right of  intermediate support "z",support at right extreme say rotation is "z1", then clearly all such dispalcements are compatible set of displacements, infact there can infinite such sets of compatible displacements, but the realistic one is that which satisfies eqilbrium too.
followed?
is rogy right?i somewhat, am not able to follow what rogy tries to say!!!
rceng (visitor)26 aug 02 22:43
yes, rogy is right. i think that in this context, compatibility means that the member displacements are compatible with the boundary conditions. in the case of the two span continuous beam the boundary conditions are that at each of the three supports the translation of the   
great insight in to catligano, compatibility and equilibrium, full marks to all
raj