SA39: Moment Distribution Method (Beam Analysis 2)

[@OsakpolorOgieriakhi]

To be honest, I never knew this method would be so easy to grasp. These videos really made the structural design of continuous beams easy to understand. Thank you

[@fireblade211]

Brilliant. I wish I had this video when I was at college.

[@civilmix]

great content it was really helpful

[@QuantumDestra]

Hi, in my Complex Structures course we are taught to use 4EI/L for a far fixed end and 3EI/L for a far pinned or roller supported end. And to use 4EI/L for all internal, continuous spans. However using this, I get distribution factors of 0.66 and 0.33 for members AB and BC respectively for the second example (fixed - UDL - roller - unloaded - roller). Member AB: 4EI/L = 4EI/4 = EI Member BC: 3EI/L = 3EI/6 = EI/2 = 0.5EI Joint Stiffness = sum(AB+BC)= 3/2EI = 1.5EI DFab = (EI)/(1.5EI)= 0.666 DFbc = (0.5)/(1.5EI) = 0.333 Could you please explain why you use EI/L to compute your member stiffnesses and why our answers are different, I'm rather confused :/ edit: using these distribution factors I get: Mab = 5.44, Mba = -1.16, Mbc = 1.16 and Mcb = 0 These internal moments are similar to your results, but I gather are inherently different because the distribution factors are not the same, but are reasonably close together. Thanks! :) And I'm sorry about the long comment!

[@DrStructure]

Yes, what you are describing is correct, it is based of a revised version of the slope-deflection equations. Basically, you are using a shortcut to solve the problem; It gets you to the answer with fewer number of iterations.

[@QuantumDestra]

Ah okay, thank you for the clarification! I appreciate the content, it's a great alternate source for my studies.

[@D3vilunheard]

I love these videos. They help so much, thank you!

[@DrStructure]

You're welcome!

[@dougwoolery9721]

in the example, when balancing out. where does the 0.16,0.24 and 0.02 come from. in previous example the number would have been cancled out by subtracting the number from it self and transferring over half. but this time i am a bit confused. did i miss understand ?

[@DrStructure]

The moment of -0.4 at joint B has to be balanced. To balance the joint, we need to add a positive moment of 0.4. That way, the total moment at the joint becomes zero (i.e., the joint is balanced). This +0.4 moment is distributed between the two columns in the table using the moment distribution factors 3/5 and 2/5. That is where 0.24 and 0.16 come from: +0.4(3/5) = 0.24 +0.4(2/5) = 0.16

[@amitkumarmeena3701]

@Dr. Structure Q.1 ) According to video at 3.45 / 8.02 , DF for pin joint 1 & DF of fixed end is 0 but what is the value of DF if the joint is Hinge ( in between the span of beam ) & just a free end ? Q.2 ) In case of Free End does it contribute anything in carry over moments ? Q.3 ) In the example ( of video ) let's say joint B undergoes downward displacement of 20 mm due to a certain load P acting downward at joint B along with other given loadings . Then does the method remains same or do I have to make some changes ?

[@DrStructure]

1) If an internal hinge is present in the beam, the analysis is not going to be straightforward when using the moment distribution method, it becomes a two-step process. It is not just a matter of assigning a distribution factor to the joint. 2) When a free end is present, we do this: kzbin.info/www/bejne/hmPRooKrdtKhaMk 3) When there is a know support settlement, we can determine a fixed-end moment for it, then insert the value in the moment distribution table and proceed as usual. Bending moment (fixed-end moment) due to the settlement can be calculated form the slope-deflection equations.

[@amitkumarmeena3701]

Thank you for explaining me these & as always you are Awesome

[@dayalgodara]

At the start of the moment distribution method you followed the sign convention that clockwise moment is considered as positive but in this video you considered clockwise moment as negative. Why so?

[@DrStructure]

The moment distribution method is based on the slope-deflection method in which we have taken the counterclockwise direction as positive for rotations and moments at the *ends* of the member. The counterclockwise (positive) moment at the end of a member, when transferred to the adjacent joint must assume the opposite (clockwise) direction if the static equilibrium is to be maintained. That clockwise joint moment is still considered positive. So, depending on whether our reference point is an end of the member or the joint attached to the member, counterclockwise or clockwise direction is considered positive.

[@dayalgodara]

@@DrStructure Thanks for the speedy reply, tomorrow I have examination of structural analysis, wish me luck 🙂

[@DrStructure]

@@dayalgodara Wishing you success!

[@theenendravasudevan5183]

Where did you get the 0.16 from ?

[@DrStructure]

You mean the 0.16 in the moment distribution table? The balancing moment for the right side of joint B? It comes from the product of the joint moment to be balanced (-0.4) and the distribution factor (2/5). -(-0.40)(2/5) = 0.16

[@ZiHeQin]

if it happens support settlement ， can I still use the moment disttibution method？ How to use？😢

[@DrStructure]

The moment distribution method is not suited for solving support settlement problems.

[@ZiHeQin]

ok，thank you❤

[@yishakabrham5002]

column 3 line 6 ....0.24 where did that come from????? i think it is 0.6 since 1/2 *1.2

[@DrStructure]

0.24 is the product of (0.4) and (3/5). It is a balancing moment, not a carry-over moment.

[@yishakabrham5002]

ok tnx i got it

[@yishakabrham5002]

can u do the same video for frame

[@DrStructure]

Yes, it will be available soon.

[@ruchiawasthi3664]

Is there a method of calculating FEM other than through slope deflection method?

[@DrStructure]

Any technique for analyzing indeterminate beams can be used for calculating fixed end moments (e.g. force method). Fixed end moments are the support reactions of a beam that is fixed at both ends. So, any technique that can analyze such a beam can handle the job.

[@vedprakashsingh7152]

how we will distribute 10 kn-m moment at joint B , either with +ve sign or -ve sign (1:41)

[@DrStructure]

See SA38.

[@samtorch6016]

Please provide. Solution of these two exercise problem

[@DrStructure]

The links to the solution videos are given in the video description field.

[@samtorch6016]

@@DrStructure thanks 😊 And I have read this topic in entirety different way, Specially,when we are dealing with Distribution factor.

[@samtorch6016]

You are only one,who has provided this method in this way. I'm so confusing,

[@vedprakashsingh7152]

Pease provide B.M diagram for mdm method