3-moment equation
3-moment equation
(OP)
I need help. My boss gave me this beam to calc the supports reactions and moments. He insists he is right but wont go through my calcs to say why mine is wrong. I have follow 3 different text books that lead me to the same answer. His M2= -13,334 #ft and I got -6,984 #ft could someone please review what I did and see if I did something wrong.
http://files.engineering.com/getfile.aspx?folder=9cec9985-9e97-4a69-9bcb-b041c6a7b5c6&file=problem.pdf
http://files.engineering.com/getfile.aspx?folder=9cec9985-9e97-4a69-9bcb-b041c6a7b5c6&file=problem.pdf






RE: 3-moment equation
RE: 3-moment equation
RE: 3-moment equation
RE: 3-moment equation
I'm not really going to look at this, but your first internal sanity check should always be to draw a shear and moment diagram. Get your reactions, draw your diagram and then cross check the different bits to make sure they're internally consistent (forces add to zero, moments sum properly, things are zeroed where they're supposed to be, etc). If that's still not fixing things, try another method. If you don't know one, teach yourself one, or do a check on a computer. If you've got two different methods coming to the same answer it generally means you're right or it means that your assumptions are wrong somehow.
Learning to check and verify your own work is important, so it's a good exercise.
I'd probably check by doing the simple span analysis of the two exterior supports, checking the deflection at the middle support location, calculate the load you need to apply on the two support span to deflect the centre support back to 0 deflection and then add the two envelopes.
RE: 3-moment equation
As a reference, there is a good example of hardy-cross moment distribution on wikipedia.
http://en.wikipedia.org/wiki/Moment_distribution_method
Remember to start with the Fixed-End Moments and then re-distributed them based on the carry over factors.
RE: 3-moment equation
I did my own sanity check, distributed the entire load (17490#) uniformly on the two spans, and that gives a moment greater than your boss's moment. So you are both wrong...if that is any consolation.
RE: 3-moment equation
Dik
RE: 3-moment equation
Or, you could cut the beam at support 2 and calculate the rotation in each span, then determine the moment required to give each beam the same slope at that support.
Or, you could remove the central reaction and solve for the deflection under the given load, then determine the magnitude of R2 in order to bring the beam back to zero deflection.
The method favored by some engineers for many years was moment distribution. In this method, you calculate the fixed end moment of each beam and distribute the difference in accordance with beam stiffness.
BA
RE: 3-moment equation
as above many ways to skin the cat, without re-skining it the same way.
solve the single redundancy with unit force method
RE: 3-moment equation
RE: 3-moment equation
Garth Dreger PE - AZ Phoenix area
As EOR's we should take the responsibility to design our structures to support the components we allow in our design per that industry standards.
RE: 3-moment equation
RE: 3-moment equation
RE: 3-moment equation
how do you know FEA is giving you the right results (or at least a reasonable representation) ? hand calcs, sanity checks.
sure there are plenty of beam calculators around that'll solve this problem in a minute ... it is a simple problem (wonder why the OP didn't do this first, before asking us to check his calc?). the solution to the problem isn't really the point of the problem. i think it's an exercise in understanding how to apply calcs, verifying that you're right (there are many ways to solve this problem, to check your answer) something that hasn't beend done (someone did a calc, someone else said the answer was wrong, repeating the calc probably isn't the best way to see who's right). as stated there is one solution to the problem.
RE: 3-moment equation
it assumes (to get Ax = wL^3/3) that the loading is applied on the full span.
the A term is right (the area of the load is w*L^2/2)
but the centroid, from the LH side, x = l+L*2/3 (not L*2/3)
where l is the unloaded portion of the span (and L the loaded)
RE: 3-moment equation
RE: 3-moment equation
http://newtonexcelbach.wordpress.com/2011/11/19/continuous-beam-analysis-by-macaulays-method/
has a spreadsheet with open source code that solves continuous beam problems using Macaulays method that will quickly confirm that both you and your boss have the wrong answer (-17 425 #.ft is right), but more importantly, if you take the time, you can follow through the analysis and see where you went wrong.
Doug Jenkins
Interactive Design Services
http://newtonexcelbach.wordpress.com/