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# Fixed end moments

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 odendaal (Structural) (OP) 20 Dec 03 9:28
 Does anyone perhaps have formulas for the fixed end forces and moments for beams fixed at both ends with uniformly distributed loads and variably distributed loads where the loads do not start and end at the supports.Thanks
 EnglishMuffin (Mechanical) 20 Dec 03 10:00
 odendaal (Structural) (OP) 20 Dec 03 10:18
 EnglishMuffinThank you for your detailed reply which I wish to pursue further.I have Roark and noticed that his loading goes up to the support, but I could not find the solution for stopping short of the support. Maybe you can indicate that to me.What is TK and how do I get to that infomation ?My prime interest in this is that I have written a program in VB6 which solves space and other frames, but I need to include a routine whereby a user can insert any type of load and the routine would then break it down into fixed end actions which I can then handle at the nodes.I have tried to use the McCaulay theorem to deduce a standard formula, but the results I get from this does not correspond with known values for special cases. Maybe I do not remember the McCaulay theorem so well.Thanks P.K. Odendaalpk@odendaal.com
 Qshake (Structural) 20 Dec 03 10:28
 Any good text on energy theorems will help out here.  Virtual work is the staple for this type of derivation, since you're interested in a response to a unit displacement or rotation.  Take a look back in your old structures text, like McCormac.
 EnglishMuffin (Mechanical) 20 Dec 03 10:37
 In my 5th edition, there is an example in section 7.3 of a case where the distributed load does not fully extend to either end. By studying this, which I confess I have not in detail, you should be able to solve your problem. In all these sort of situations, you can combine the basic cases by matching the slopes and deflections at the junction between them, thus enabling combinations of any degree of complexity to be solved, although it can be tedious. Regarding Roark & Young on TK, see http://www.roarksformulas.com/ryupgrade.htmlThere is something called "Superposition capability" and "Superposition Wizard", which may do the trick - but you will have to check. From memory, I think this program used to cost between $500 and$1000 dollars.
 dtex (Civil/Environmental) 30 Dec 03 3:29
 You can derive such formulas "relatively easily" yourself ( as long as your beams cross section is prismatic ).  I've done several myself assuming that varying loads vary in a linear fashion.  You can also do this for non-linearly varying loads.  The algebra is just a bit stickier.You already know the fixed end moments for a single point load located at some location on a beam.  It's a simple formula and is shown in most analysis text.  A uniform load is nothing more than a "whole lot" of these single point loads all placed next to each other along a beams length.You can use basic integral calculus to sum up the results of a bunch of these point loads ( size of P*dx each ) along the length of your beam in question.For standard prismatic beams,  I think you'll like the formulas you come up with better than the ones you'll find in Roark's.   And... you get the satisfaction of having developed them yourself.Dan   www.dtware.com
 odendaal (Structural) (OP) 30 Dec 03 5:15
 Hi DanThanks for the response.I have tried using integration to get to a formula, but instead of starting from first principles (which I have long forgotten)I thought I might get by using McCaulays Theorem which is in effect in an already partially integrated from, but I just cannot get the correct answer. The theorem looks like :EId2y/dx2=Mx=Ma+Ra.x-w[x-a]^2/2I have searched the web for his theorem, but could not find anything - although I have used his theorem many times in my career.I agree Roarks formula is elaborate and involves deduction and adding various loads just to arrive at one loadcase. P.K. Odendaalpk@odendaal.com
 dtex (Civil/Environmental) 30 Dec 03 14:15

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