SA38: Moment Distribution Method (Beam Analysis 1)

[@k283535]

The best structural engineering explanatory channel

[@nitin5777]

ohh thank u guys...my exams are coming...was really waiting for it...thanks a ton

[@sohamprajapati8784]

Clear.Concise.Crisp

[@huseyinarkan2047]

crystal clear. thanks :)

[@itachi0000]

thankyou very much...this is really helpful

[@JohnDiggle21]

I wish I found this before my exam. I'll probably have to retake anyway so I'll save this for later

[@antarajana9067]

Best explanation ever

[@samtorch6016]

Wonderfull explanation 🙂

[@jijithsudev3670]

You are the best

[@faisalamersadoon6711]

Thank you D.r

[@UpperXEgo]

thank you

[@moracacaca]

thanks a lot

[@thyathnarongny7967]

can you do a video demonstrat why the redistribution effect take place and why do need to consider them in the analysis ?

[@pankajmaurya5326]

thnx a lot

[@midnightson787]

Hello Dr Structure.. at 1:50 when calculating member stiffnes for both secitons, I was under the impression that when you have a pin or roller support then k = 3EI/L yet just EI/L was used...are you able to explain the difference please

[@DrStructure]

The general form of the stiffness factor is EI/L for the member divided by the sum of EI/L at the joint. When an end-support is a pin or a roller, the total sum of EI/L at that end is: EI/L, since there is only one member connected to the end. Hence, the factor becomes EI/L of the member divided by the total EI/L which in this case gives us 1. There is, however, a shortcut that can be used to speed-up the number of iterations of the moment distribution method. And that is using 3EI/L for end-pins or rollers, instead of using EI/L. Of course, this revised factor can be shown be true mathematically as well. Either way the method converges to the right value for the end moments, although the later factor results in a faster convergence.

[@agassimurray2186]

@@DrStructure So EI/L alone can be used to calculate the stiffness factor for a member with any support?

[@DrStructure]

@@agassimurray2186 Yes.

[@eng-mohmetyooz9732]

thanks Dr. structure

[@VocaFan4ever]

I just want to ask, is the table used like per category? For example if the beam had uniformly distributed load, a moment at point B or a vertical force going down somewhere else, do we just do a table for uniformly distributed load first then the rest, one by one, and then add them all up in the end to find out what the vertical forces would be?

[@DrStructure]

We can take the approach that you have suggested. But that results in going through the moment distribution process multiple times. To shorten the amount of computations, we can calculate the fixed-end moments due to each load, then add them up the values for each column before we start the moment distribution process. For example, let's say member BC is subjected to a uniformly distributed load of 3 kN/m, and there is a concentrated load of 8 kN applied at the midpoint of the member. If the length of the member is 4 m, then the fixed-end moments due to the distributed load are: left end (end B): (3)(4)(4)/12 = 4 right end (end C): -(3)(4)(4)/12 = -4 For the concentrated load the fixed-end moments are: left end (end B): (8)(4)/8 = 4 right end (end C) = -(8)(4)/8 = -4 Therefore, the total fixed-end moment B equals 8, and at end C equals -8. These would be the values that we enter in the moment distribution table, and perform computation on.

[@VocaFan4ever]

@@DrStructure Oh okay I get it, thank you so much for replying quickly and comprehensively!

[@bananian]

This is so much easier to understand than my instructor and it's free, wtf!

[@user-qy3zo6dz5v]

Can you please explain a worked example of solving a continuous beam with node moment and span loads.

[@engrshamsurrehman4599]

lucid explanation. thumps up

[@AjeetKumar-hy7cq]

good essay

[@chungken8496]

Excuse me, at 01:52, what is stiffness coefficient? What does that mean?

[@DrStructure]

Stiffness coefficient (EI/L) is a property of the beam that defines the relationship bending moment and rotation. We can compare it to the spring constant for an axially loaded spring. When subjecting the spring (beam) to a force (moment), the larger the stiffness coefficient (the stiffer the member), the smaller the displacement (rotation) would be.

[@tanooppersaud340]

how did you obtain the values 0.6 and 0.4 ? ( when carrying over )

[@DrStructure]

Please provide the video time for the question. Are you referring to the values at time 9:53? If so, 0.6 is half of 1.2 being carried over from joint B. And 0.4 is half of 0.8. The carry-over factor is always 1/2.

[@_jojo11]

At 05:47 shouldn't moment at left end be negative because it is counterclockwise and vice-versa?

[@DrStructure]

Please elaborate. Which moment are you referring to? At the left end of AB? According to our sign convention, for member-end moments counterclockwise is considered positive.

[@_jojo11]

Dr. Structure thank you I've got it cleared up now

[@mohdfarhan1662]

Nice

[@ruchiawasthi3664]

So can we conclude that the use of distribution factor in the whole table appears only once?

[@DrStructure]

We write the distribution factores only one time in the table, but we use them repeatedly until the entire table is generated.

[@berkgumus4047]

Hi so what if there is a free end? Do we consider it like there is a joint there and nothing changes?

[@DrStructure]

If the beam has an overhang, we remove it before proceeding with the analysis. Suppose our beam rests on a pin at A, a roller at B, and a roller at C. And, there is an overhang to the right of C. Let's say the length of the overhand is 2 meters. If there is no load being applied to the overhang, we can simply get rid of it, resulting in a continuous beam that rests on a pin and two rollers with no overhang (no free end). However, if the overhang is subjected to a load, we need to include its effect on the continuous beam. Say a load of 40 kN has been placed at the free end of the overhang. This force creates a clockwise moment of (40 kN)(2 m) = 80 kN-m about point C. So, we can remove and replace the overhang with a clockwise moment of 80 kN-m placed at the roller support at C before proceeding with the analysis.

[@berkgumus4047]

@@DrStructure thank you so much for quick reply i also have a problem with having more than 3 supports. The table becomes so confusing and calculations are all over the place. Can we create seperate tables for each joint? Thats my last question.

[@DrStructure]

@@berkgumus4047 The table is our common way of representing the moment distribution data. But we can represent/tabulate the data differently. So it would be okay to create three separate tables for a beam with three supports. As long as we keep track of the carry-over moments, knowing how/where to transfer them from one table to another, we will get the same results at the end.

[@fanbingbing7390]

3:40 why did you put -10 you said counter-clockwise is positive in slope deflection medthod ?

[@DrStructure]

Yes, a member-end moment is considered positive if it is counterclockwise. When that moment is transferred to the adjacent joint, it becomes clockwise, since the equilibrium condition (sum of moments = 0) must be maintained. So, a positive moment at the end of the member becomes a clockwise moment at the joint. Here, since the joint load is counterclockwise, it suggests that the corresponding member-end moment would be clockwise (or negative), hence the negative sign.

[@fanbingbing7390]

Dr. Structure thanks for answering you are really great

[@fanbingbing7390]

Dr. Structure as I understood , the picture is at 1:32 right? we will have 2 clockwise moments at the end of member which support B is , and 2 of them must be included to be 10 . This 10 clockwise is negative.

[@DrStructure]

The member-end moments drawn in black are all considered positive. When we cut a member, say AB, closet to the joint, then we draw a counterclockwise (a positive) moment at the member end, and an accompanying clockwise moment at the other end (joint end) of the cut. So that the sum of the moments at the cut equals zero. So, if the applied load (shown in red) was acting in the clockwise direction, like the other two (black) moments acting at B, then we would have considered that a positive moment. But since the applied load is acting in the counterclockwise direction at the joint, it is considered a negative moment.

[@fanbingbing7390]

Dr. Structure Crystal clear thanks a lot

[@midnightson787]

Hello Dr Structure Please help... Case 1 : at 7:02 member end BA = +6, then at 7:38 it is put into the equilibrium equation as -6 Case 2 : at 10:11 member end BA = -2.4, then at 10:23 it is in the equilibrium equation as -2.4. So my question is, how can the SAME member end be summed up as positive in Case 1 and CHANGE to negative in the equilibrium equation for case 1, but in Case 2 it is negative when summed up and REMAINS negative in the equilibrium equation for Case 2?

[@DrStructure]

We need to distinguish between two different sign conventions: the sign convention that we use for member-end moments in the moment distribution method (which is derived from the slope-deflection method), and the sign convention that we use when writing static equilibrium equations. The two sign conventions are not the same. For the moment distribution method, we take counterclockwise as the positive direction. So, @7:02 we get a positive value for the bending moment at the right end of member AB. Since the moment is positive, we’ve placed it in the counterclockwise direction at the right end of AB @7:38. For Case 2, the computed moment at the right end of AB came out negative. So, we’ve placed that moment in the clockwise direction at the right end of the member @10:23. Note that in both cases, since we are placing the member-end moment in the correct direction on the free-body diagram, we specify the moment magnitude as a positive value. As for the other sign convention, the one that we use for writing the moment equilibrium equation, we can assume either direction as positive. When we are summing up the moments about point A (@7:38 and 10:23), we can take either clockwise or counterclockwise as positive. In Case 1, at 7:38, clockwise direction is taken as positive. That is why 4Vba has a positive sign (since it is creating a clockwise moment about A) and 6 has a negative sign. In case 2, at 10:23, counterclockwise is taken as the positive direction. Hence, 4Vba has a positive sign and 2.4 has a negative sign. Note that even if we had done it differently, we would have gotten the exact same results. For example, for Case 1, when writing the moment equation, if we took counterclockwise as the positive direction, the first equation would have been written as: 6 - 4Vba = 0 which is the exact equation as 4Vba - 6 = 0.

[@midnightson787]

@@DrStructure thank you,

[@mangeshtajne3061]

For member BC far end is 'C' and it is roller support there and for roller support stiffness factor is 3/4.EI/L as far as I know so please tell me how do u take stiffness factor of EI/L there?

[@DrStructure]

How did you come up with 3/4? More broadly, how do we calculate the distribution factors? The process described here is a systemic approach for calculating distribution factors. The use of 3/4 is a shortcut that works when certain condition is met. Either way we end up with the correct solution.

[@roushanrounak8648]

Hello sir...good morning...sir i have a doubt...suppose a continuous beam consisting three span AB BC CD...ALL are pinned support...there is no load in any span on the beam...it is given that support B settles by an amount 15mm...Now i know that A FEM will be induced in span AB and BC...FEM(AB),FEM(BA),FEM(BC),FEM(CB)..ALL will be equal to 6EI(delta)/L^2...but sir what will be the sign convention if i consider anticlockwise rotation as positive..please help sir...really urgent...thank you

[@roushanrounak8648]

I m solving it by moment distribution method...

[@DrStructure]

In this case, we end up having positive values for FEM for member AB, and negative values for FEM for member BC. This is the case because AB is rotating clockwise which means Delta is negative, while BC is turning counterclockwise implying a positive Delta.

[@amitkumarmeena3701]

@Dr. Structure Q .1 Can you please tell me why carrying over member-end moment always taken Half of the previous end moment ? Q.2 What will be the procedure to solve it if the Joint B having 3 or more members passing through it ( example : Frame Analysis using Moment Distribution Method ) ? Q. 3 How can we solve a problem having Degree of indeterminacy 2 or more ?

[@DrStructure]

On Question 1: The reason can be found in the underlying slope-deflection equations. Take a look at SA37 where I show the math behind the process. More specifically, go to 8:20 and examine the two sets of equations shown. When we have a rotation at B (theta B), then moment at the right end of AB is calculated using a factor of 4 (rotation multiplied by 4EI/L) where as the moment at the left end of the segment is calculated using a factor of 2 (rotation multiplied by 2EI/L). So, if we have the moment at B, then the moment at A would be half of that. Questions 2 & 3: The procedure would be the same, it is just that the table would get a bit bigger, and a bit irregular in shape (in the case of frames). Stay tuned for the next few lectures for further discussion on this topic.

[@amitkumarmeena3701]

Thank You very much , You are awesome

[@nserekomuhammad9580]

Can you pls explain what is done for beam with internal hinge for that method

[@DrStructure]

The moment distribution method works best when the beam nodes correspond to the pin/roller/fixed supports. Meaning, the beam does not have any vertical displacement at the nodes. An internal hinge introduced in the beam causes a vertical displacement at the hinge node. This makes the use of the moment distribution method rather cumbersome. Such beams are best analyzed using the matrix displacement method.

[@davexkaranja]

Why do you keep changing the sign convention for moments? First its -ve for anticlockwise at 3:29 then +ve for the same at 8:38

[@DrStructure]

We are using the sign convention adopted for the slope-deflection method: an anticlockwise moment at the ends of the member is considered positive. Note that an anticlockwise moment at the end of the member (connected to a joint) translates into a clockwise moment at the joint since the sum of the two moments must be zero. So, at 3:28, the moment shown is at the joint (not at the end of the member). That anticlockwise moment at the joint results in a clockwise moment at the end of the member; therefore, we treat it as a negative moment. At 8:38, however, the referenced moment is at the end of the member. The counterclockwise moment at the left end of the member is positive, and the clockwise moment at the right end is negative.

[@Dr.DineshaK]

Thank you very much for your effort. Arrows at 7.47 min at BC member for shear is incorrect. Is should be opposite.

[@DrStructure]

The shear force (Vbc) at the left end of segment BC needs to be upward (since the bending moment at that end of the segment is counterclockwise). Otherwise the sum of the moments at the right end of the segment would not come out to be zero.

[@karlcorrz]

What if the beam has 4 rollers; 2 being on the inside?

[@DrStructure]

The process remains the same, except that the table would end up having 4 columns instead of 3. See: kzbin.info/www/bejne/baCynaaElMR7d9U

[@tutstorial8474]

but why does moment distribution factor for a fixed support equals zero?

[@DrStructure]

See Lecture SA39 (kzbin.info/www/bejne/bHTYn5VjmJ5ne9k) for the explanation/derivation.

[@jenni2896]

mam, please tell the sign convention used here again

[@DrStructure]

The member-end moments are assumed to be acting in the counterclockwise direction. That is, when we place the moments at the ends of the member, counterclockwise is considered positive. These moments become clockwise when placed at the neighboring joints. This means, joint moments are positive when clockwise. Consequently, a joint load (moment) acting in a clockwise direction, is considered positive. If the moment is counterclockwise at the joint, we treat it as a negative moment.

[@isla2974]

2:58 - Left should be right and right should be left?

[@DrStructure]

At A, the total bending moment at the joint is going to be distributed between the ends of the members connected to A. Since there is only one member connected to the joint, we could say, the moment at A is going be distributed to member AB only. More specifically, to the left end of the member where joint A resides. Similarly, at C the entire moment is going to be distributed to the C-end (the right end) of member BC. Please keep in mind that here we are talking about the distribution of the moment at the joint, and not the carry-over moment.

[@SheekoSom]

thanks that is usefull but in 7:44 where 5/6 in B came from

[@DrStructure]

The reaction force at B is the algebraic sum of Vba and Vbc: Vba - Vbc = (3/2) - (2/3) = 5/6

[@roushanrounak8648]

Hello sir....again your patient is suffering from a disease of structure.. this time moment distribution method...first of all i wana ask as we know that moment distribution method is an approximate analysis so can answer vary in range...like one has obtained a value of moment at some point as 20 According to book through other method like stifness factor modification and hence distribution factors gets change accordingly and i obtained as 24 with normal process...is the difference is acceptable or am i doing any error...for making it more clear to you i have mailed you today a screenshot of my solution...please take a look and tell me if i m doing any blunder.please do take a look sir.waiting for your response.thank you.i do need your suggestion....

[@DrStructure]

The moment distribution method should provide reasonably accurate results, when enough number of iterations are performed. Any deviation from the actual results by more than 1% is problematic and points to potential calculation error(s).

[@roushanrounak8648]

Dr. Structure got you sir...I will keep it in mind... Thank you

[@yishakabrham5002]

plz can we get final exercise answer plz

[@DrStructure]

Solution (Exercise Problem 1): kzbin.info/www/bejne/onS1m4ShbNmNg6s Solution (Exercise Problem 2): kzbin.info/www/bejne/hmPRooKrdtKhaMk

[@yishakabrham5002]

tnx by the way u make every thing clear

[@dawidmencel6739]

ei/4 +ei/6 =5ei/12 almost understood that...

[@merensow200]

mam, how did we get 0, -2.4, 2.4 and 0 in 10:15

[@DrStructure]

By adding up the numbers in each column, starting from the numbers shown in red (fixed-end moments).

[@merensow200]

@@DrStructure but, mam, how? I can't understand how it is done. Please show me an example of how we got -2.4 and 2.4 in the bottom.

[@DrStructure]

@@merensow200 First column (adding the moment values from top to bottom): 4 + (-4) + 1.2 + (-1.2) + 0.6 + (-0.6) + 0.3 + (-0.3) + 0.1 + (-0.1) = 0 Second column: (-4) + 2.4 + (-2) + 1.2 + (-0.6) + 0.6 + (-0.3) + 0.3 + (-0.1) + 0.1 = -2.4 Third column: 1.6 + 0.8 + (-0.4) + 0.4 + (-0.2) + 0.2 + (-0.1) + 0.1 = 2.4 Fourth column: 0.8 + (-0.8) + 0.4 + (-0.4) + 0.2 + (-0.2) + 0.1 + (-0.1) = 0

[@merensow200]

@@DrStructure Sir/mam, I am really thankful for your reply and help. God bless your kindness.

[@DrStructure]

@@merensow200 Thank you for you well wishes.

[@biniambelayneh4774]

how 0.1 came since our carry over factor is 0.5?

[@DrStructure]

Rounding down from 0.15 to 0.1. Needless to say, we can do the calculations using more significant digits, producing more accurate numbers.

[@AyaanKhan-ei3gk]

At 6:09 , half of 6 goes to support a , but shouldn't there be a 3 already existing there , ? Why are we considering 6 , when it gets halved as the other half is carried over to support A , so why are we considering it in full when we sum the moments ??

[@DrStructure]

You wrote “shouldn’t there be a (moment) of 3 at A already?” At the starting point the applied moment of -10 is distributed to joints A, B, and C. The moment at A is 3 which is half of 6 (the moment at the right end of segment AB). Why should there be any other moment at A already? Where does that additional moment come from?

[@AyaanKhan-ei3gk]

@@DrStructure sorry I meant the moment at joint b , when the 10 moment gets distributed and 6 comes towards segment ab and 4 goes to segemt BC , then we say the half of 6 goes to support a due to carry over ,but my question is that shouldnt there be moment 3 left at b because 3 goes to a , then when we sum moments at BA then why do we consider the whole 6 and not 3 ?

[@DrStructure]

You are assuming that we cut the moment in half, keep half of it where it is, and carry the other half to the other end of the member. That is not the right way of thinking about/visualizing it. We take the total moment acting at a joint (say, the moment of 10 kN.m acting at B), and we distribute it to the ends of the members based on the distribution factors. In this case, we need to distribute the 10 kN.m moment to four member ends: the left end of member AB, the right end of member AB, the left end of member BC and the right end of member BC. Since the 10 kN.m is applied at B (where the right end of AB and the left end of BC meet), the moment is distributed using factors 3/5 and 2/5 (obtained from member stiffness coefficients). The 10 kN-m moment also impacts (distributes to) the far end of each member. We say the 10 kN-m moment distribution to the far end of each member is equal to half of the moment distributed to the near end of the member. So, since the distribution of the 10 kN-m moment to the near end of AB is 6, then the distribution of the moment to the far end of the member is half of 6. See Lecture SA37 for an explanation of why this is the case.

[@AyaanKhan-ei3gk]

@@DrStructure if half of 6 goes to other end due to carry over and it comes out as 3 then that means the 4 that goes to the other segment will also be halved to 2 when it carries over to the far end , but if we sum of all of them up if comes as 6+3+4+2=15 , but you said the 10 will be distributed to all the four parts , what exactly is carry over ? Because some books it says it's a sort of reaction that a fixed end makes when a moment is aplied on the other end , so in this case when 6 is applied to the segment the 3 or the carry over is basically the reaction made due to 6 moment in the other end or fixed support, is this correct

[@DrStructure]

You can think of the moment distribution method as a derivative of the slope-deflection method. Please review Lecture SA37 ( kzbin.info/www/bejne/r6XdnZ9to5x_j9k) to see the connection between the two methods. In Lecture SA37 at 4:31, you can see how the member-end moments can be expressed as mathematical series; each end moment can be written as a sum of many terms. The moment distribution method uses that formulation to find the member end moments, one series term at a time. So, in each iteration, one term of the moment (series) is determined. Naturally, since the end moments are determined one term at a time, only approximate values are obtained at the early stages of the process. That is why the numbers don’t add up in the first iteration. But as the process continues and more terms of the series are considered, the numbers converge toward the actual values.