3-moment equation 3

[Kuhuh]

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/getfil...-9e97-4a69-9bcb-b041c6a7b5c6&file=problem.pdf

 

[MiketheEngineer]

Home work??

 

[Kuhuh]

no not homework... well its training so kinda like homework thus why I cant just go to my boss and say your wrong! haha that never flies to well.

 

[Kuhuh]

by the way he did hardi-cross method which I do not know how to do. If you have any recommendations,links, material I'd appreciate that as well.

 

[TLHS]

Hardy Cross is moment distribution, if you're familiar with that.

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.

 

[JoshPlumSE]

Hardy-Cross should work nicely for this problem. Unfortunately, you're both wrong. He's closer though. The right answer should end up being something like 17.4 k-ft.

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.

 

[hokie66]

TLHS is right, devise sanity checks. What are your reactions? With those, draw the shear and moment diagrams and you will find out if you have made an error in applying the three moment theorem.

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.

 

[dik]

I get a different number... haven't checked the numbers other than the input loads and offsets. Moments and Shears are visually OK.

Dik

 

[BAretired]

There are many ways of solving this problem. The three moment equation is one of them, but would not be my choice, although it works well if you apply it correctly (which you did not).

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

 

[rb1957]

looking at your calc, is your first line correct ? i think it applies when you have uniform loading over the whole span ?

as above many ways to skin the cat, without re-skining it the same way.

solve the single redundancy with unit force method

 

[MiketheEngineer]

Run it through RISA!!

 

[woodman88]

I ran it in my beam calculator and got -17.4 k-ft also.

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.

 

[spats]

Your boss must be from the stone age (and I'm no spring chicken). What's the point! With today's anlysis tools available to us, it makes no sense wasting your time with such an exercise. I suspect he's trying to show off, and rather badly at that (he's wrong, as others have pointed out). Getting correct results in a timely manner is what matters most. You can be a knowledgeable engineer without knowing how to do the mundane by hand. Next he'll be handing you a slide rule!

 

[fancypants]

I am guessing the boss done this by "Feel".

 

[rb1957]

sorry spats, but i disagree ... the more we surrender to the "black box" of canned s/ware the less we really know.

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.

 

[rb1957]

the first line in the derivation is wrong.

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)

 

[dhoward26]

I get 17.4 kip-ft for the max moment...but I just plugged your diagram into WoodWorks without the weight of the beam. Does your firm not have software that can calc reactions and moments of multiple spans? Both woodworks and enercalc do this in a matter of seconds as compared to a long hand written problem. But I suppose in your case since you are just learning the ropes, a hand calc is good practice

 

[IDS]

I also disagree with spats. Solving an exercise like this from first principles is a good way to lean the basics that will help you to spot when automated analyses are giving wrong answers. The blog post here:

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/