# Moment Distribution Method



## ipswitch (Jun 25, 2012)

Been trying to review moment distribution method: http://en.wikipedia.org/wiki/Moment_distribution_method

For the life of me I can't figure out why there's a 3, a 4, and a 4 added to the numerator of the flexural stiffnes/distribution factors:

The flexural stiffness of members AB, BC and CD are




,



and



, respectively. Therefore, expressing the results in repeating decimal notation:

Can anyone help me with this? thanks.


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## Ble_PE (Jun 25, 2012)

It's been a while since I've worked a moment distribution problem, so I had to pull out my old structural analysis notes. The 4EI/L is the absolute flexural stiffness of a member between two supports. This is true for all prismatic members. The 3EI/L on the other hand is a little more complicated. In a moment distribution problem, whenever you have a simply supported end span with or without a cantilever, you can multiply the stiffness by 3/4. That is, your flexural stiffness becomes 3/4(4EI/L) = 3EI/L. The only catch is you have to balance that end and not carry over any moment back to it. The example problem in the article you linked to shows this:



> Joint A Joint B Joint C Joint D Distrib. factors 0 1 0.2727 0.7273 0.6667 0.3333 0 0 Fixed-end moments -14.700 +6.300 -8.333 +8.333 -12.500 +12.500 Step 1 +14.700 → +7.350 Step 2 -1.450 -3.867 → -1.934 Step 3 +2.034 ← +4.067 +2.034 → +1.017 Step 4 -0.555 -1.479 → -0.739 Step 5 +0.246 ← +0.493 +0.246 → +0.123 Step 6 -0.067 -0.179 → -0.090 Step 7 +0.030 ← +0.060 +0.030 → +0.015 Step 8 -0.008 -0.022 → -0.011 Step 9  +0.004 ← +0.007 +0.004 → +0.002 Step 10 0.001 0.003 Sum of moments 0 +11.569 -11.569 +10.186 -10.186 +13.657


Does this help any?

Edit: That table didn't quote very good so you'll need to look at it on the Wikipedia page.


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## kevo_55 (Jun 25, 2012)

In the MDM, it doesn't really matter if you really use a 3 or 4 for your stiffness. Basically, if you follow Ble's post you'll find that your problem will solve itself faster than simply assuming that everything is just 4EI/L.

If I ever have to do this, I prefer the long way (assume 4EI/L). It's just less things that I need to remember.


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## McEngr (Jun 25, 2012)

kevo_55 said:


> If I ever have to do this, I prefer the long way (assume 4EI/L). It's just less things that I need to remember.


bingo!


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## civilized_naah (Jun 25, 2012)

If the far end of the member is moment free (pinned), the stiffness is taken as 3EI/L while if the far end is rigid (either fixed support, or continuous), then the stiffness is 4EI/L. Doing it this way should be consistent with the way you calculate fixed end moments, i.e. for a span with one end pinned, the fixed end moment (at the rigid end) would be different than the fixed end moment that would exist there if the far end were rigid.

With the approach of taking 3EI/L as the stiffness for a member with a pinned far end, the moment at the pin is always zero, so there is no moment unbalance to take care of. The iteration converges/progresses faster.


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## ipswitch (Jun 25, 2012)

Thanks so much for the input. I was able to find it in Chaper 13, Section 5 of the Ken Leet and Chia-Ming Uang book "Fundamentals od Structural Analysis." A prismatic element in double curvature has a stiffness modification of 6EI/L, btw.


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## civilized_naah (Jun 26, 2012)

ipswitch said:


> A prismatic element in double curvature has a stiffness modification of 6EI/L, btw.


Not quite sure what you mean. How would you know - a priori - that the member is in double curvature?


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## ipswitch (Jun 26, 2012)

civilized_naah said:


> ipswitch said:
> 
> 
> > A prismatic element in double curvature has a stiffness modification of 6EI/L, btw.
> ...


Opposite acting moments on each end of the beam create a double curvature. They counteract rotation on the ends so-to-speak and increase stiffness of the element.


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