Thanks a lot, Dr. Structure. Your videos are highly informative and easy to understand. I have a question. If the horizontal member in a frame contains an internal hinge, then how do we proceed with the method? What would be the distribution factors?
Generally speaking, the moment distribution method should not be a method of choice for analyzing indeterminate beams with internal hinges. In dealing with such a beam, the vertical displacement of the beam at the hinge needs to be brought into the formulation, similar to what we do when analyzing frames with sidesway. This adds layer of complexity makes the use of the moment distribution method significantly less efficient (more time consuming).
A better classical technique for dealing with internal hinges in beams is the slope-deflection method. The presence of an internal hinge in a continuous beam introduces three (3) additional variables: the slope at the left side of the hinge, the slope at the right side of the hinge, and the deflection at the hinge. To analyze the beam using the slope-deflection method, we would need to introduce three more equations. These equations are: 1) Moment at the left side of the hinge = 0 2) Moment at the right side of the hinge = 0 3) Shear at the left side + shear at the right side of the beam = 0 These equations added to the joint equilibrium equations at the supports give us sufficient number of equations to determine all the joint slopes (including the ones at the hinge) and the vertical displacement at the hinge. We can then substitute these slopes and deflection back into the slope-deflection equations in order to determine the member-end moments.
dr u r really providing quality education to us ,thanx for u r lecture, are u gonaa also cover stiffness matrix of structure analysis and what are the other topic u gonna cover
Thank you for the note. Yes, the stiffness method will be covered. We plan to cover topics in both directions: the more basic (statics and mechanics of solids) as well as the more advanced (structural design) topics. We will also continue adding lectures and case studies on the analysis topics that have been covered already in order to make the content richer and more comprehensive.
We would add the couple/moment to the table @6:42 under column DB, DC, or DE (does not matter which one). Then, we go through the moment distribution process as usual. Note that now the total moment at that joint would be the sum of (-6) and the added moment. If the added moment is counter-clockwise, we enter it in the table as a negative value. If it is clockwise, we enter it as a positive value.
A sign convention is rather arbitrary. For the slope-deflection and the related methods we have assumed counterclockwise rotations/moments at the ends of beam segments to be positive. We could have make a different assumption, but given the arbitrariness of it all, our assumption is as good as the alternative.
If the frame is subjected to horizontal or unsymmetrical loading, if it is unsymmetrical geometrically, or if member section or and/or material properties is not constant throughout the structure, sidesway could develop. When in doubt, assume it is there. If it turns out to be insignificant, it can then be ignored.
In the moment distribution method, the distribution factors are ratios: The ratio of the rotational stiffness of the member to the total rotational stiffness at the joint. Therefore, any constant that appears in the numerator and denominator of that ratio cancel each other out. This reduction/simplification not only applies to 4, but it also applies to E, if the modulus of elasticity of the material is constant throughout the structure.
Sir if member end moment and fixed end moment are showing opposite sign still why are u taking both the moment in the same sense(clockwise and counterclockwise)?
We have determined the end moments for segment BD. This makes the segment statically determinate, with the shear forces at the ends of the segment being unknown. We can determine these forces using static equilibrium equations. Writing the moment equation about point B, we get: (2 kN/m) (6 m)(3 m)+ 5.17 - 3.19 + 6Vd= 0 By solving the above equation for the shear force at D, we get Vd= 6.33 kN. Next, summing the forces in the y-direction, we get: Vb + 6.33 = (2 kN/m)(6 m) which yields Vb = 5.67 kN.
In the first example, the columns are said to have 2I for the moment of inertia. In the second example, "assume a constant EI" means the beams and columns have the same moment of inertia.
How the moment is distributed between the ends of a joint using the distribution factors was explained in the previous moment distribution lectures. This lecture series needs to be watched in the right sequence; please review the prerequisite lectures first.
After the member-end moments are calculated using the moment distribution method, the beam segments become statically determinate. From that point on we can use the static equilibrium equations to determine the member-end shear forces. The reactions at a support are the algebraic sum of the member-end forces near the support.
Suppose there is a fixed support at the left end of a beam. To determine the two distribution factors associated with the support (one factor to the left of the support and one to its right), we can view the support itself as a very (infinitely) rigid beam segment that does not rotate. To model this fictitious rigid beam, we say its modulus of Elasticity is infinity (say, B represent infinity). And let's denote the modulus of elasticity of the beam segment attached to the fixed support as b. Compared to B, b has a very small value. For the sake of brevity, let's assume that both beam segments have the same length (L). The two DFs can be determined using the following equations (see SA39): The DF to the left of the fixed support = B /( B + b). Since b is negligible compared to B. The equation becomes B/B = 1. The DF to the right of the fixed support = b/(B + b). Since the numerator is negligible compared to the denominator, the expression becomes 0/B = 0. So, the DF to the left of a fixed support is 1, and the DF to the right of it is 0, as indicated in the video.
Once the moments are calculated, the frame becomes a statically determinate structure. We can then use the static equilibrium equations to determine the rest of the unknowns (the shear and axial force in each member, and then the support reactions).
Dr. Structure but if I'll start at member AB, I don't get the vertical reactions at A.. what should I do? Sorry for my weak understanding ma'am, hope you don't get mad at me..
@@RaymartGDamot You can view it as a fill-in-the-blank or a kind of puzzle. We have all the pieces, we just have to put them together. From member AB, we calculate the shear force in AB which gives us the horizontal reaction force at A. From member CD, we calculate the shear force in CD which gives us the horizontal reaction at D. From member BC, we calculate the shear in BC, and since the member is not directly attached to a support, we transfer the load from BC, through joints B and C to the AB and CD, and then to the support.
Knowing the member-end moments, we can apply the equilibrium equations to the individual members in order to calculate member-end shear forces. For example @3:54, member AB is subjected to a clockwise moment of 5.13 kN-m at the top and a clockwise moment of 2.58 kN-m at the bottom. Assuming the shear force at end A of the member to be V, then since the sum of the moments about the top end of the member must be zero, we get: (5.13 kN-m + 2.58 kN-m) = (4 m) V. Solving for V, we get: V = 1.93 kN.
Your Way of teaching is so beautiful and helpful
the only best place to clear doubts online...thnx a lot
Your videos are the best, very thorough and concepts are clearly explained at a good pace.
Thank you for the feedback.
This is incredibly helpful. Thanks for posting this.
You are the best Dr.
Thanks a lot, Dr. Structure. Your videos are highly informative and easy to understand.
I have a question. If the horizontal member in a frame contains an internal hinge, then how do we proceed with the method? What would be the distribution factors?
Generally speaking, the moment distribution method should not be a method of choice for analyzing indeterminate beams with internal hinges. In dealing with such a beam, the vertical displacement of the beam at the hinge needs to be brought into the formulation, similar to what we do when analyzing frames with sidesway. This adds layer of complexity makes the use of the moment distribution method significantly less efficient (more time consuming).
A better classical technique for dealing with internal hinges in beams is the slope-deflection method. The presence of an internal hinge in a continuous beam introduces three (3) additional variables: the slope at the left side of the hinge, the slope at the right side of the hinge, and the deflection at the hinge. To analyze the beam using the slope-deflection method, we would need to introduce three more equations. These equations are:
1) Moment at the left side of the hinge = 0
2) Moment at the right side of the hinge = 0
3) Shear at the left side + shear at the right side of the beam = 0
These equations added to the joint equilibrium equations at the supports give us sufficient number of equations to determine all the joint slopes (including the ones at the hinge) and the vertical displacement at the hinge.
We can then substitute these slopes and deflection back into the slope-deflection equations in order to determine the member-end moments.
these videos are the best!
Best series of videos, can you please add column analogy method ?
very clear
dr u r really providing quality education to us ,thanx for u r lecture, are u gonaa also cover stiffness matrix of structure analysis and what are the other topic u gonna cover
Thank you for the note.
Yes, the stiffness method will be covered. We plan to cover topics in both directions: the more basic (statics and mechanics of solids) as well as the more advanced (structural design) topics. We will also continue adding lectures and case studies on the analysis topics that have been covered already in order to make the content richer and more comprehensive.
very good
Greetings thank you very much for the lesson.
TÜRKİYE den selamlar
Dr structure. What if there was a couple applied at joint D of the second frame? How would we distribute it ?
We would add the couple/moment to the table @6:42 under column DB, DC, or DE (does not matter which one). Then, we go through the moment distribution process as usual.
Note that now the total moment at that joint would be the sum of (-6) and the added moment. If the added moment is counter-clockwise, we enter it in the table as a negative value. If it is clockwise, we enter it as a positive value.
@@DrStructure ☺You're the best Dr Structure. I clearly understand you,thank you very much for such quality help.
You're welcome!
2:56 why the bending moment distributed from B to A is -2.30 instead of -2.25?
It should've been written as -2.25.
Both -2.25 and -2.30 is right
very thanks
Why are you considering clockwise moment negative?
A sign convention is rather arbitrary. For the slope-deflection and the related methods we have assumed counterclockwise rotations/moments at the ends of beam segments to be positive. We could have make a different assumption, but given the arbitrariness of it all, our assumption is as good as the alternative.
Hello. How come we didnt consider stiffnes of column to be twice of the beam in the 2nd question?
In the second problem, EI is assumed to be constant. So, the members have the same stiffness.
thank,s alot
Ly
@07:20, the rightest column 0.82 should be negative.
Thanks for the correction.
Dr Structure how do we know if a frame is with sidesway or not?
If the frame is subjected to horizontal or unsymmetrical loading, if it is unsymmetrical geometrically, or if member section or and/or material properties is not constant throughout the structure, sidesway could develop. When in doubt, assume it is there. If it turns out to be insignificant, it can then be ignored.
Others use 4EI/L for member stiffness factor , but you use only EI/L. Can u pls explain why?
In the moment distribution method, the distribution factors are ratios: The ratio of the rotational stiffness of the member to the total rotational stiffness at the joint. Therefore, any constant that appears in the numerator and denominator of that ratio cancel each other out. This reduction/simplification not only applies to 4, but it also applies to E, if the modulus of elasticity of the material is constant throughout the structure.
@@DrStructure Thanks for the very clear explanation; I appreciate it.
You're welcome.
Sir if member end moment and fixed end moment are showing opposite sign still why are u taking both the moment in the same sense(clockwise and counterclockwise)?
I am not sure what you are referring to. Please point to the video segment that is the source of your question and elaborate.
Dr. Structure I think I got the answer sir..thank you
How did you get 5.67 and 6.33 KN in 8:15?
We have determined the end moments for segment BD. This makes the segment statically determinate, with the shear forces at the ends of the segment being unknown. We can determine these forces using static equilibrium equations. Writing the moment equation about point B, we get:
(2 kN/m) (6 m)(3 m)+ 5.17 - 3.19 + 6Vd= 0
By solving the above equation for the shear force at D, we get Vd= 6.33 kN.
Next, summing the forces in the y-direction, we get:
Vb + 6.33 = (2 kN/m)(6 m)
which yields Vb = 5.67 kN.
Dr structure you use 2EI/L for columns in 1st example and in 2nd only EI/L why is it so ?
Because the moment of inertia for the columns is defined as 2I while for the beam it is I.
@@DrStructure means In Exm U use 2EI/L for Columns and In 2nd Example ..U use EI/L for Columns .
I m Talking About This Point
In the first example, the columns are said to have 2I for the moment of inertia. In the second example, "assume a constant EI" means the beams and columns have the same moment of inertia.
@@DrStructure thanks a lot ..
6 can be divided 4plus 2 or 3plus3 or 4.2 plus 1.8, why it is 3.5 plus 1.5, explain these simple things
How the moment is distributed between the ends of a joint using the distribution factors was explained in the previous moment distribution lectures. This lecture series needs to be watched in the right sequence; please review the prerequisite lectures first.
Sir will you please deliver us the solution of the excersize questions?
The links are given in the project description field.
Thanx a lot for Ur quick cooperation,sir.. respect from core of my heart
How did you calculate the support reactions
After the member-end moments are calculated using the moment distribution method, the beam segments become statically determinate. From that point on we can use the static equilibrium equations to determine the member-end shear forces. The reactions at a support are the algebraic sum of the member-end forces near the support.
Could you please do a video, showing how you done the calculation for the reactions. It would really help
Hello, I went over lecture SA39 and still do not know why the DF of the fixed support it 1?
Suppose there is a fixed support at the left end of a beam. To determine the two distribution factors associated with the support (one factor to the left of the support and one to its right), we can view the support itself as a very (infinitely) rigid beam segment that does not rotate. To model this fictitious rigid beam, we say its modulus of Elasticity is infinity (say, B represent infinity). And let's denote the modulus of elasticity of the beam segment attached to the fixed support as b. Compared to B, b has a very small value.
For the sake of brevity, let's assume that both beam segments have the same length (L). The two DFs can be determined using the following equations (see SA39):
The DF to the left of the fixed support = B /( B + b). Since b is negligible compared to B. The equation becomes B/B = 1.
The DF to the right of the fixed support = b/(B + b). Since the numerator is negligible compared to the denominator, the expression becomes 0/B = 0.
So, the DF to the left of a fixed support is 1, and the DF to the right of it is 0, as indicated in the video.
Thanks sir but the first carry over at point A and D were supposed to be 2.25. Why it is 2.30?
Thanks for the comment. Yes, the value should be 2.25. However, since this is an approximate technique, reasonable rounding of numbers is permitted.
@@DrStructure OK Sir thanks a lot for your amazing videos!💗💗
i need to know calculation system of multi stories
Where are the answer? So that i can check of im wrong or not
The links are given in the video description field.
Hi ma'am, may I know how did you get the reactions? Thanks.
Once the moments are calculated, the frame becomes a statically determinate structure. We can then use the static equilibrium equations to determine the rest of the unknowns (the shear and axial force in each member, and then the support reactions).
Dr. Structure I just don't understand ma'am why did you start at segment BC instead of segment AB.
@@RaymartGDamot The order is arbitrary. You can start the process from any joint or member.
Dr. Structure but if I'll start at member AB, I don't get the vertical reactions at A.. what should I do? Sorry for my weak understanding ma'am, hope you don't get mad at me..
@@RaymartGDamot You can view it as a fill-in-the-blank or a kind of puzzle. We have all the pieces, we just have to put them together. From member AB, we calculate the shear force in AB which gives us the horizontal reaction force at A. From member CD, we calculate the shear force in CD which gives us the horizontal reaction at D. From member BC, we calculate the shear in BC, and since the member is not directly attached to a support, we transfer the load from BC, through joints B and C to the AB and CD, and then to the support.
How to get 1.93 Kn?
Knowing the member-end moments, we can apply the equilibrium equations to the individual members in order to calculate member-end shear forces. For example @3:54, member AB is subjected to a clockwise moment of 5.13 kN-m at the top and a clockwise moment of 2.58 kN-m at the bottom. Assuming the shear force at end A of the member to be V, then since the sum of the moments about the top end of the member must be zero, we get: (5.13 kN-m + 2.58 kN-m) = (4 m) V. Solving for V, we get: V = 1.93 kN.
@@DrStructure Thank You so much! :)