Yo really appreciate the efforts to explain this method, I spend so much time reading textbook and understand nothing until i watch this video..Thankss
This the perfect channel. And I am a perfectionist that's y I understand what Dr. Structure(s) teach. For this era this is one of the ruling CZcams channel for structural analysis (SA). Different teachers, different voices, quality content in time, short video, awesome notations techniques and spacing, class lecture vibe in you tube. It helps a lot.
Hi Doctor Structure. Thank you for this very useful learning video that really improves my knowledge for structure, is it possible that you also explain modified slope deflection equation and its derivation based on far end being pinned or simply supported? Thank you!!!
See SAPS01 (czcams.com/video/E-qfuKL2rtY/video.html) for details. Yes, we can express rotation in the beam as a function of x by integrating M/EI. Here however, our objective is something else. We want to establish a relationship between member-end rotations and moments, and not have an expression in terms of x.
Hi ! I can't find the lecture SAP01 for the details of finding Theta ... Could you please write it on the description of the video or here at the comment ? Thank you :)
Hello Dr. Structure, Why 6:16 says that it is the equation for a beam not be subjected by any external loads. Isn't Mab and Mba external loads that cause the rotations at both ends of the beam? Or Maybe I'm confused by what is the definition of internal and external force... Thank you
Mab and Mba are member-end moments which are internal to the member. They are not forces that one physically places on the member. Yes, Mab and Mba cause the member-end rotations but that does not make them external (applied) loads. @7:00 the red moment applied at A, having a magnitude of 10 kN-m is considered an applied (external) load. Here, since equilibrium at A must be maintain, we can clearly see that Mba must be equal to 10 kN-m as well. Put it differently, Mba develops inside the member as a result the load applied at A. Or, the external load causes the internal loads to develop.
Right! The moment at the fixed support (Mb) is not zero; sum of the moments at B is zero. Note that there are two moments at B: Mb and Mba. Their sum must be zero. Perhaps the notation is a bit confusing since Mb is being used twice, once to indicate that the sum at B must be zero, and once to refer to the moment at the fixed support.
So that we can eliminate Mab from both equations and solve for Mba. Multiplying both sides of an equation by a constant is a standard technique for solving a system of linear equations.
In 2nd equation (θb) (at 5:30) is it correct? I think it should be "6 EI θb /L = 2 Mba - Mab" , since everything is same and denominator as well except only Mab and Mba are switched together...so how the result came i.e 12 EI θb/L = 4Mba - 2 Mab??
Yes, the equation for theta_b is correct. If we take that equation and multiply both sides of it by (6EI/L), then we get the first equation you wrote above. And if we multiple both side of that equation by 2, we get the second equation you wrote. But why multiple both sides of the equation by 2? Because now if we add the two equations given at 5:30, then the 2Mab in the first equation cancels the -2Mab in the second equation resulting in an equation in terms of Mba only.
Consider a continuous beam having two spans resting on three supports, a pin at the very left end (A), a roller at the very right end (C), and a roller in the middle (B). Now, suppose we placed a load on the right span (segment BC) only, causing the beam to deflect downward in that span. In that scenario, what happens to the shape of the left span (segment AB)? If BC deflects downward, then AB is going to move/deflect upward. Why? Because the downward deflection of BC means point B must turn clockwise, and C must turns counterclockwise. If B is turning clockwise, then it forces segment AB to deflect upward (and end A to turn counterclockwise), it causes the segment to deform/deflect. Therefore, while AB is not subjected to any direct applied load, it ends up deflecting, resulting in end rotations, and subsequently end-moments to develop in the segment.
The bending moment in the slope-deflection equations are internal to the beam. In structural members, internal forces and deflections are results of applied loads. Loads cause the structure to deform and internal forces (and stresses) to form. Generally speaking, deflection don't develop without loads. There are exceptions, but it most cases loads need to be present for the structure to deflect and the internal forces to develop.
@@DrStructure I was confused as to how did the beam deflect on it's own with any external load. That answer did it for me. Thank you so much for this explanation. You are doing a wonderful service to the Engineering community.
If the two shear forces were drawn in the same direction, say both were pointing up, then we would have written: Vab + Vba = 0. But when the forces are drawn in the opposite directions, the statement: "the sum of the forces in the y-direction must be zero" translates into Vab + (-Vba) = 0.
Dr structure I really love your video, I am confused; I want to learn from you and you use counterclockwise positive and my teacher, book and therefore all friend use clockwise positive. So if I go with them I am going to miss your video if I walk with you then I will be alone in class getting opposite joint moment and ultimately same diagram. Any suggestion will be very appreciated.
Thanks for the note. Eventually you want to be able to rise above the sign convention and warp your mind around the concepts and the process/method. Sign conventions could become confusing, but you need to be able to operate (solve problems) effectively regardless of the notation/convention. The impetus behind these lecture is to help you grasp the method. If you can follow the lectures here, follow the derivation , and can duplicate it, then perhaps you should do the derivation yourself using the reverse sign convention. Once you master the derivation, you may find it relatively easy to go back and forth between the two notations. Obviously if your professors are using the reverse convention, it makes sense for you to do the same. That minimizes the potential for misunderstanding/miscommunication.
@@DrStructure , Thank you for your reply and thank you for your videos they helped me to grap concept. and I am going to look your every video in future also. I am with you.
We end up with a statically determinate beam, if we replace the fixed support with a pin or roller. In that case, we can analyze the beam using the static equilibrium equations; there is no need to use the slope-deflection method to analyze the beam.
I just watched structurefree's video on this subject and he says that you must ALWAYS write the moments as clockwise because that is what the equations require. Now I am confused :/
It all depends on the sign convention we adopt. There is no such thing as ALWAYS … with regard to the positive/negative direction. Our adopted frame of reference dictates which side/direction is positive and which is negative.
These are the assumed directions, not the actual ones. At this point, we don't know the direction of any of these forces, we are just making an assumption. When setting up formulations and solving problems, we always go with a sign convention, and place forces/moments on diagrams according to that convention. So, here, we are placing all the moments in the counter-clockwise direction. After the problem is solved and the moment values are determined, the sign of the computed values tell us if we made the correct assumption at the beginning. For example, if we assumed a moment to act in the counterclockwise direction at the beginning, and its value comes out to be negative at the end, then we know that the actual direction of the moment is opposite to what was assumed.
I am not sure if I understand the essence of your question correctly, feel free to elaborate if my response does not address your question. One can view a fixed support as having a rotational stiffness (EI/L) that approaches infinity. The larger the stiffness of the joint (not of the member) the less rotation takes place at the joint. A pin support, on the other hand, has a rotational stiffness of zero. A semi-rigid connection is one that has a finite stiffness (something between zero and infinity) associated with it. The basic slope-deflection method is not the most suitable technique for modeling such support conditions. The displacement/stiffness method is a more suitable technique to use in which we can estimate the stiffness of the joints/supports and incorporate them into the system stiffness matrix.
Dr. Structure I am taking this course currently,and I don't know those methods yet.I don't know is there any case can exist in slope deflection,I saw semi rigid(partially fixed) supports in portal method, and i wondered. Anyway thank you for quick reply sir!
Hi Dr. Struture I really like your explanation, but I don't understand because I don't understand English, I'm from Indonesia, please subtitle indonesia
Correct, there is a typo in the expression written on the diagram. Although the equation below the diagram is written correctly. For updated version of this (and other lectures), please see our Udemy course. You can find the link to the course in the video description field.
I feel privileged to find this channel...
Feel Gratitude that I found this video!
awesome .....so excited to watch next videos....the sooner the better
thank god for this channel... you guys are awesome
Yo really appreciate the efforts to explain this method, I spend so much time reading textbook and understand nothing until i watch this video..Thankss
Is of great help in understanding the concept..thanks
This the perfect channel. And I am a perfectionist that's y I understand what Dr. Structure(s) teach. For this era this is one of the ruling CZcams channel for structural analysis (SA). Different teachers, different voices, quality content in time, short video, awesome notations techniques and spacing, class lecture vibe in you tube. It helps a lot.
Thank you for your feedback.
ur my professor Dr structure tnx alot!! ...anyone from AAIT give it thumbs up
Hi Doctor Structure. Thank you for this very useful learning video that really improves my knowledge for structure, is it possible that you also explain modified slope deflection equation and its derivation based on far end being pinned or simply supported? Thank you!!!
Keep up the good work
thank you simply understood
@04:54, should the rotation of support B should be Mba * L /(3EI) - Mab * L/(6EI) ?
Thanks for the correction. On the diagram, yes! For Theta_B, M_AB and M_BA are in the wrong place. They should switch places.
awsome
how you find the thetas at 3:23, whats the formula for it because i thought it would be integration of mM/EI
See SAPS01 (czcams.com/video/E-qfuKL2rtY/video.html) for details. Yes, we can express rotation in the beam as a function of x by integrating M/EI. Here however, our objective is something else. We want to establish a relationship between member-end rotations and moments, and not have an expression in terms of x.
Hi !
I can't find the lecture SAP01 for the details of finding Theta ...
Could you please write it on the description of the video or here at the comment ?
Thank you :)
Here is the link: czcams.com/video/E-qfuKL2rtY/video.html
Hello Dr. Structure,
Why 6:16 says that it is the equation for a beam not be subjected by any external loads.
Isn't Mab and Mba external loads that cause the rotations at both ends of the beam?
Or
Maybe I'm confused by what is the definition of internal and external force...
Thank you
Mab and Mba are member-end moments which are internal to the member. They are not forces that one physically places on the member. Yes, Mab and Mba cause the member-end rotations but that does not make them external (applied) loads.
@7:00 the red moment applied at A, having a magnitude of 10 kN-m is considered an applied (external) load. Here, since equilibrium at A must be maintain, we can clearly see that Mba must be equal to 10 kN-m as well. Put it differently, Mba develops inside the member as a result the load applied at A. Or, the external load causes the internal loads to develop.
Where do I get the solutions to the questions left at the end of the lecture
The solution links are listed in the video description field.
7：25 why the MB(sum of moment at B) is 0, I remember that fixed joint has moment, right?
Right! The moment at the fixed support (Mb) is not zero; sum of the moments at B is zero. Note that there are two moments at B: Mb and Mba. Their sum must be zero.
Perhaps the notation is a bit confusing since Mb is being used twice, once to indicate that the sum at B must be zero, and once to refer to the moment at the fixed support.
Thx for this usefull video... I think there is a missing at 4.18 -----> teta A2 = MBA.L%6EI ------> L is forgotten
Thanks for pointing out the missing L.
@ 5:30 I am wondering why theta B is rearranged to have 2 times each value, as compared to theta A.
So that we can eliminate Mab from both equations and solve for Mba. Multiplying both sides of an equation by a constant is a standard technique for solving a system of linear equations.
In solved example for propped cantilever beam why rotation takes place at support B even though it is fixed support ?
The rotation at B is zero, as indicated @8:07
In 2nd equation (θb) (at 5:30) is it correct? I think it should be "6 EI θb /L = 2 Mba - Mab" , since everything is same and denominator as well except only Mab and Mba are switched together...so how the result came i.e 12 EI θb/L = 4Mba - 2 Mab??
Yes, the equation for theta_b is correct. If we take that equation and multiply both sides of it by (6EI/L), then we get the first equation you wrote above. And if we multiple both side of that equation by 2, we get the second equation you wrote. But why multiple both sides of the equation by 2? Because now if we add the two equations given at 5:30, then the 2Mab in the first equation cancels the -2Mab in the second equation resulting in an equation in terms of Mba only.
HI Dr. structure am new to this course i was wondering if u could explain to me how a beam segment got END MOMENT (rotation ) with out loading thanks
Consider a continuous beam having two spans resting on three supports, a pin at the very left end (A), a roller at the very right end (C), and a roller in the middle (B). Now, suppose we placed a load on the right span (segment BC) only, causing the beam to deflect downward in that span. In that scenario, what happens to the shape of the left span (segment AB)? If BC deflects downward, then AB is going to move/deflect upward. Why? Because the downward deflection of BC means point B must turn clockwise, and C must turns counterclockwise. If B is turning clockwise, then it forces segment AB to deflect upward (and end A to turn counterclockwise), it causes the segment to deform/deflect. Therefore, while AB is not subjected to any direct applied load, it ends up deflecting, resulting in end rotations, and subsequently end-moments to develop in the segment.
@@DrStructure wow thanks for the deep explanation
@@abukarhassan943 You're welcome!
Can someone tell me why both of the rotations are equal at 4:56?
Oh it's an error
The moments that are produced are external or internal? Are the moments due to the deflection or load? I'm new to this course.
The bending moment in the slope-deflection equations are internal to the beam.
In structural members, internal forces and deflections are results of applied loads. Loads cause the structure to deform and internal forces (and stresses) to form. Generally speaking, deflection don't develop without loads. There are exceptions, but it most cases loads need to be present for the structure to deflect and the internal forces to develop.
@@DrStructure I was confused as to how did the beam deflect on it's own with any external load. That answer did it for me. Thank you so much for this explanation. You are doing a wonderful service to the Engineering community.
@@sumfun103 Glad to be of help.
8:57 why is the sum of shear forces Vab and Vba not equal to zero? Vab - Vba = 0?
If the two shear forces were drawn in the same direction, say both were pointing up, then we would have written: Vab + Vba = 0. But when the forces are drawn in the opposite directions, the statement: "the sum of the forces in the y-direction must be zero" translates into Vab + (-Vba) = 0.
Dr structure I really love your video, I am confused; I want to learn from you and you use counterclockwise positive and my teacher, book and therefore all friend use clockwise positive. So if I go with them I am going to miss your video if I walk with you then I will be alone in class getting opposite joint moment and ultimately same diagram. Any suggestion will be very appreciated.
Thanks for the note. Eventually you want to be able to rise above the sign convention and warp your mind around the concepts and the process/method. Sign conventions could become confusing, but you need to be able to operate (solve problems) effectively regardless of the notation/convention.
The impetus behind these lecture is to help you grasp the method. If you can follow the lectures here, follow the derivation , and can duplicate it, then perhaps you should do the derivation yourself using the reverse sign convention. Once you master the derivation, you may find it relatively easy to go back and forth between the two notations.
Obviously if your professors are using the reverse convention, it makes sense for you to do the same. That minimizes the potential for misunderstanding/miscommunication.
@@DrStructure , Thank you for your reply and thank you for your videos they helped me to grap concept. and I am going to look your every video in future also. I am with you.
Why you are not inculcated the displacement effect ok equation negative of 3delta/lenth
That is covered in Part 3 of the lecture series on the slope-deflection method.
See: czcams.com/video/HbqbIIzu0Bo/video.html
tnx guys
What if there's no fixed support. How do you eliminate one unknown?
We end up with a statically determinate beam, if we replace the fixed support with a pin or roller. In that case, we can analyze the beam using the static equilibrium equations; there is no need to use the slope-deflection method to analyze the beam.
@@DrStructure But that's the assignment you gave. Is there anywhere I can see the answers to your assignments?
You can find the solution to the exercise problems in the free online course referenced in the video description field.
@@DrStructure thanks a lot
I just watched structurefree's video on this subject and he says that you must ALWAYS write the moments as clockwise because that is what the equations require. Now I am confused :/
It all depends on the sign convention we adopt. There is no such thing as ALWAYS … with regard to the positive/negative direction. Our adopted frame of reference dictates which side/direction is positive and which is negative.
@@DrStructure I appreciate the timely response. Thank you
You’re welcome!
how Mbc and Mcb are on same direction at 01:18
These are the assumed directions, not the actual ones. At this point, we don't know the direction of any of these forces, we are just making an assumption. When setting up formulations and solving problems, we always go with a sign convention, and place forces/moments on diagrams according to that convention. So, here, we are placing all the moments in the counter-clockwise direction. After the problem is solved and the moment values are determined, the sign of the computed values tell us if we made the correct assumption at the beginning. For example, if we assumed a moment to act in the counterclockwise direction at the beginning, and its value comes out to be negative at the end, then we know that the actual direction of the moment is opposite to what was assumed.
@@DrStructure Thanks for your explanation
Sir what is the condition of slope and deflection right at the partially fixed support?
I am not sure if I understand the essence of your question correctly, feel free to elaborate if my response does not address your question.
One can view a fixed support as having a rotational stiffness (EI/L) that approaches infinity. The larger the stiffness of the joint (not of the member) the less rotation takes place at the joint. A pin support, on the other hand, has a rotational stiffness of zero. A semi-rigid connection is one that has a finite stiffness (something between zero and infinity) associated with it.
The basic slope-deflection method is not the most suitable technique for modeling such support conditions. The displacement/stiffness method is a more suitable technique to use in which we can estimate the stiffness of the joints/supports and incorporate them into the system stiffness matrix.
Dr. Structure I am taking this course currently,and I don't know those methods yet.I don't know is there any case can exist in slope deflection,I saw semi rigid(partially fixed) supports in portal method, and i wondered.
Anyway thank you for quick reply sir!
Hi Dr. Struture I really like your explanation, but I don't understand because I don't understand English, I'm from Indonesia, please subtitle indonesia
We'll do our best to accommodate your request.
@@DrStructure thank you
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if you just wouldn't skip stuff it would be better than refering back to old vid
Mistake in theta b at 4:58, should be 1/3 MBA - 1/6 MAB
Correct, there is a typo in the expression written on the diagram. Although the equation below the diagram is written correctly. For updated version of this (and other lectures), please see our Udemy course. You can find the link to the course in the video description field.