(Structural Analysis) Moment Distribution Method: Help?


by Khamul
Tags: analysis, distribution, method, moment, structural
Khamul
Khamul is offline
#1
Nov13-12, 03:46 PM
P: 24
Hello everyone, as the thread title implies, I'm in a bit of a bind when it comes to understanding the Moment Distribution Method for displacement methods of analysis.

In particular, I am having a lot of trouble in determining Stiffness Factors. I have attempted to browse my book and online for clarification, but I feel like every source I look at contradicts the other.

For example...page two and page three of this pdf:
http://www.sut.ac.th/engineering/civ...stribution.pdf

It states that if the far end member is fixed, then the stiffness factor, K, is K=4EI/L. Okay.
But page 4 of this pdf then goes on to state that K(ab) = 3EI/L, even though the far end D on this is fixed.

Could someone please help me make sense of this? :(
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afreiden
afreiden is offline
#2
Nov14-12, 01:30 AM
P: 104
1) You can always use 4EI/L for all members. If you choose to do this, then the moment distribution process will be a bit longer and you will have to repeatedly "carry-over" the moments back and forth until they become small enough ("small enough" is based on your own judgement).

2) Alternatively, you can use the "modified stiffness" moment distribution method. In this case, one way to do it for your beam is to observe that there is a pin at the end of one of the spans. This means that you can choose to use 3EI/L for that span. Leave the other spans as 4EI/L. The advantage, in general, of using a "modified stiffness" is that you only "carry-over" the moments one-time.

It is unfortunate that the beam in question is not solved by either method 1) or 2) in your attached file, even though they mention both of these methods on pg 2, pg 3, as you mentioned (or did I just miss it?). Nevertheless, it appears that they solve a variety of problems later on...

Note the beam on pg 7, for example -- Here, they use a modified stiffness factor for all 3 spans. They get the 2EI/L by observing "symmetry" in the first span. This problem is solved in one step, but it is a simpler problem. Later on, in other examples, 6EI/L pops up, which is the modified stiffness factor for "anti-symmetry."


I've attached my own notes on the subject. It's only 8 pages, and would probably be worth your time if you are still "in a bind." It is adapted from "Elementary Theory of Structures" by Hsieh and Mau. Back when I learned this stuff that was the best text on the subject. I think Hibbeler has since gotten in on the action.
Attached Files
File Type: pdf afreiden_moment_dist_no_joint_trans.pdf (117.9 KB, 20 views)


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