Cross / Moment distribution method

[kingnero]

I'm checking an old storage building, which was all drawn and calc'ed by hand.
Method for the hyperstatically determinate beams = Hardy Cross method.

Apparently back then (dated 1970), they used different distribution coefficients then what I learned at school (15 yrs ago).
Anyone cares to share his notes?

for a continuous beam on three supports (A - B - C),
k(AB) = 3 EI / L or 4 EI / L (factor 4 is used for fixed opposite supports, 3 for pinned on the opposite end).
DF (Distribution Factor) : DF(AB) : k(AB) / ( k(AB) + k(BA) )
The method the architect/engineer used then was just k(AB) = I/L (E always cancels out), so no distinguishing for pinned or fixed supports.
Also, he used other sign conventions, but not consequently.

one of us is wrong (but the fact the building still stands pushes the odds towards his side ) as it gives a 18% difference in solutions.

Google was not very helpful, apparently Cross is no longer tought at schools (found very few detailed material on-line).

all suggestions are very welcome...

 

[desertfox]

Never heard that they used different coefficients, have a look at this link it might help

http://www.engr.mun.ca/~swamidas/ENGI6705-ClassNotesHandout7.ppt

 

[kingnero]

Thanks for the reference, that's about the same as what I was tought in school.

One thing though, on the slide 7.4 "Solutions of Problems", point 7.4.1.2 "Stiffness factors", there's a factor I do not agree with: K(CD) is 4/8 EI, and I would say that needs to be 3/8 EI. (other side (D) is a hinge! as explained on the previous slide)

Am I understanding this incorrectly?

 

[rb1957]

"hyperstatically determinate" = statically indeterminate ... the names people come up with ... yes, i realise it's directing the attention of the reader to how to solve these problems ("indeterminate" can be read as "unsolvable", as i've been told in interviews).

"no distinguishing for pinned or fixed supports" ... of course there's a difference (a fixed joint is much stiffer than a pinned one). my reference for MDM (Bruhn, chapter A11) shows that the difference shows up in the distribution factors ... since the method starts with assuming each joint is fixed, then balances each joint (by the relative stiffness of the adjacent spans), then distributes moment to adjacent joints "releasing the joint". a fixed support develops the assumed fixed end moment, doesn't redistribute any moment to the adjacement joints, and accepts moment from adjacent pinned joints.

i'd suggest that truly fixed joints provide limits for moment distribution ... divide the problem into sub-problems based on the fixed joints, solve each sub-problem separately, sum the results.

Quando Omni Flunkus Moritati

 

[zekeman]

"Am I understanding this incorrectly? "

Yes, incorectly

The method takes each section as built in and iteratively releases the moments, including the socalled carryover moments

I think it is brilliant.

Why don't you test the solution given to see how accurately it was done?