psyclones Posted April 14, 2015 Share Posted April 14, 2015 (edited) Please excuse [deflection] error. Could you help me with the following.Using continuous beam theory, constructing BM diagram from points b to c, to calculate the max deflection. I only found a have a single solution, though the BM digram show two points of zero bending. I can provide the solution.your thoughts. [edit: Rb = 685 N] Please find revised attachment. attach1(revised).pdf Edited April 14, 2015 by psyclones Link to comment Share on other sites More sharing options...
studiot Posted April 14, 2015 Share Posted April 14, 2015 What do you mean by 'continuous beam theory' ? I don't see any sign of the three moment equation and the resulting simultaneous equations in your working pdf. It would also be nice to see the reactions identified and labelled. Link to comment Share on other sites More sharing options...
psyclones Posted April 14, 2015 Author Share Posted April 14, 2015 (edited) Thanks for your post. I can assure you I've used 'three moment equation(s)' (solving three equations simultaneously) to arrive at the given moments at b and c, and reaction at b (reaction b is under the support b). My question is, given the manner in which I've formulated the M(x) equation. Is it correct- given I have fixed unequal (in this case) BM's at points b and c, with a free moment due to dist load? [edit] Using the integration method to solve for angle = 0, to solve for deflection. [edit] Edited April 14, 2015 by psyclones Link to comment Share on other sites More sharing options...
studiot Posted April 15, 2015 Share Posted April 15, 2015 (edited) Here is a good pdf on the 3-moment method for continuous beams like yours. http://people.duke.edu/~hpgavin/cee201/three-moment.pdf and here is a simpler worked hand calculated example. http://www.roymech.co.uk/Useful_Tables/Beams/Continuous_Beams.html Edited April 15, 2015 by studiot Link to comment Share on other sites More sharing options...
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