The direction of an unknown force or moment has to be assumed. The sign of the force/moment magnitude, after it is calculated, tells us if the initial assumption was correct. If the magnitude comes out positive, we know the assumption was correct. Otherwise, the actual force/moment direction is opposite to the assumed direction. In our slope-deflection formulation, we assume the member-end moments to be acting in the counterclockwise direction. If a computed moment value turns out to be negative, then we know that moment is actually acting in the clockwise direction at the end of the member.
Member-end moments are the actual moments found at the ends of the member, once the structure assumes its deformed shape. Fixed-end moments are moments at the end of a member that is assumed to be fixed at its ends. Consider a continuous beam having 4 spans. Suppose we load the second and fourth span with distributed loads, there are no loads applied to the remaining spans. The beam is going to deform, its five joints are going to rotate, and moments develop at the ends of each member. These are the actual (member-end) moments which we can then use to design the member. So, even though spans 1 and 3 are not subjected to any loads, member-end moments do develop at their ends. Fixed-end moments are not the actual moments in a beam, unless the beam is fixed at both ends. These moments should be viewed as a representation of the applied loads. In the slope-deflection method, we simply replace an applied load with its equivalent moments at its ends (with its fixed-end moments.) That is, in the eyes of this method of analysis, a distributed load of W applied to a beam of length L is equivalent to a bending moment of WL^2/12 (the fixed end moments) applied at either end of the beam. So, to analyze the beam, we first replace each load with its equivalent fixed-end moments, then go through the analysis to determine joint rotation and then the actual (member-end) moments for each beam segment.
@@DrStructureThanks for explaining. But I have a doubt. We know actual bending moment at hinge or roller support is zero. And the hinge or roller position and the end of a beam represent same point. So how is it possible that the bending moment value at the same point (hinge/roller position or end of a beam) have two different values (one is zero from hinge/roller consideration and other is non zero from member end consideration)?
Yes, it is always the case that bending moment at a hinge or a roller support is zero, if the support is located at the end of the beam. If however the hinge or the roller is an interior support, then bending moments do develop to its left and right. Regardless, when using the slope deflection method, we always start by labeling the member-end moments for each member, even if we know the moment is zero. Say, we have a beam fixed at the left end with a roller support at the right end. Suppose we have labeled the ends of the member A and B. Even though we know bending moment at B is zero, we can still label that moment Mba. Then, when we write the joint equilibrium equation at B ( Sum of the moment at B must be zero), we set Mba = 0, we enforce the fact that the moment at the roller is zero. This enables us to calculate the slope of the beam at B while enforcing the constraint.
When dealing with beams (not beam-columns) the only forces that are present (according to the beam theory) are shear and moment. There is no axial force in a pure beam. The slope-deflection method is based on beam theory meaning axial forces do not show up in the formulation. The difference between a pin support and a roller support is a horizontal (axial) force. The pin has one, the roller doesn’t. But since our beam carries no axial force that horizontal reaction force at the pin would be zero. So, the pin acts like a roller as far as our analysis is concerned. In a way, they are interchangeable in our formulation.
Dr. Structure but doing that makes it determinate from indeterminate so we were able to apply virtual work. Had it been any indeterminate structure, how would we apply virtual work method then?
Replacing the pin with a roller does not make the beam determinate. That beam is already determinate with a pin and a roller support. We have picked such a beam deliberately. The specific sub problem that we are trying to solve here always involves a determinate beam, a beam that is going to rotate at its ends resulting in the end slopes theta_A and theta_B. Otherwise we would not have be able to come up with the equations that relate member end moments to the slops. Also keep in mind that here we are deriving the slope deflection equations, we are not analyzing an indeterminate structure. What we are doing is establishing the mathematical relationships between member end forces and rotations. If we had an indeterminate beam and wanted to calculate some slope or deflection, we would have started by analyzing the beam first, then we would have applied the virtual work method to calculate the desired displacement/rotation.
Before calculating them, we don't know the direction of the member-end shear forces. When drawing the free-body diagram for the member, we always assume a direction for the unknown shear forces. We then write and solve the equilibrium equations for the unknowns. If we end up with a positive value for the shear force, we assumed the correct direction. If we get a negative value for the shear magnitude, then we know its actual direction is the opposite of the assumed direction.
In this formulation, counterclockwise is taken as the positive direction. That makes, theta_a1, theta_b2 and theta_b3 positive, and the other three joint rotations (theta_b1,theta_a2, theta_a3) negative in the two theta equations
This derivation seeks to establish the mathematical relationship between member-end moments, rotations and the applied load. Case 3 by itself is a simply supported beam that is free to rotate at its ends (with zero end-moments). That is not going to help us come up with a general formulation for the slope-deflection method as in a typical beam segment (in a continuous beam) the ends of the segment are not completely free to rotate; the bending moments (Mab and Mba) at the ends of the segment are not zero. So, we need a formulation that does take these end moments into consideration. The simply supported beam by itself does not do it. Now, if we treat these end moments as loads, and apply it to the beam (as in Loading Cases 1 and 2), then use the principle of superposition to add the resulting rotations to those obtain from Loading Case 3, then we can establish the necessary relationship between the non-zero end-moments, rotations and the applied load.
@@DrStructure I Understand that due to superposition cases and 1 and 2 are algebraically added. But I still don't relate how the algebraic sum that includes all cases relates them. In the force method I understand why you add the deflections of the unit value and the loads acting on the primary structure, somehow I miss it here. Please elaborate.
@@dear_dennis Let me give this another try. Suppose we have a simply supported beam subjected to a uniformly distributed load. What would be the rotations at the ends of the beam? This corresponds to Loading Case 3. We can determine the end rotations using the virtual work method, for example. Here, we have basically related the end rotations to the applied load. If we know W (the load intensity) then we can determine the total end rotations in terms of E and I. This scenario, however, assumes that the moments at the ends of the beam are zero. But, suppose we are told that the moment at the left end of the member is not actually zero. We don't know what that moment is, we just know it is not zero. How does that effect the end rotations? Obviously, the end rotations are not going to be that of Loading Case 3 since now there is a moment at the left end that somewhat restrains the rotations. So how do we adjust the end rotations from Loading Case 3 in the presence of Mab? That is where the principle of superposition comes into play. We treat Mab as an applied load/moment, calculate the end rotations in terms of Mab, then add them to the end rotation due to Loading Case 3. That gives us the mathematical expression(s) that relate the beam's end rotations to W (the applied load) and the unknown end moment (Mab). If we apply the same line of reasoning to Mba, then we get the three loading cases that we have here.
@@DrStructure Thank you! That was very explicit! I guess it was easier for me to understand first load case 3 since the effect of the load is applied directly to the beam, while load cases 1 and 2 relate the adjacent sides. Now I understand much better the simplification when the beam's end is an overhang or a simple support.
why does moment ab is positive when you apply it at the joints ? it should change to a clockwise direction and we picked that moment rotation as negative. Thanks :)
True, we show the moment at A is the clockwise direction since at the end of the member, the moment is shown in the counterclockwise direction. That is also true for the other joints. So, member-end moments shown at joints all are acting in the clockwise direction. When writing the joint equilibrium equations, we are summing the moments at each joint. Since they are all acting in the sane direction (clockwise) then we simply add them together, as positive terms, or as negative terms, it really does not matter. That is, -Mab = 0 is the same as Mab = 0. Or, -Mba - Mbc = 0 is the same as Mba + Mbc = 0. The sign convention that we use for adding moments at the joints is independent of the sign convention that we use for writing the slope-deflection equations for each member.
Fixed end moments are the reaction forces at the supports due to the applied load; they are the resisting moments. A downward load has a tendency to rotate the beam in the clockwise direction at the left support, and in the counterclockwise direction at the right support. So, the resisting moment (FEM) at the left support is counterclockwise, and the direction of the FEM at the right support is clockwise.
Since the shear force is unknown, you can assume it to be upward, or downward. Make a choice and draw the force in the assumed direction on the free-body diagram. Then write the equilibrium equations accordingly and solve them for the unknown shear forces. If your assumption is correct, the computed value would come out positive. Otherwise, you get a negative value for the shear force. Either way, once the equilibrium equations are solved, you know the actual direction and magnitude of the shear force.
I think there’s a slight mistake . I mean the part of vbc and vcb I don’t get it why 160*6*3 -450- 6vcb ? I think it’s 150*6*3+405 +6vcb right ?😅 I’m confused
Correct. There is a typo in that equation. The magnitude of the distributed load is 150 not 160. So, the correct moment equations is (taking clockwise direction as positive) : (150)(6)(3) - 405 - 6 Vcb = 0 Or, Vcb = 382.5 N
Correct! The analysis of indeterminate structures is rather involved and time consuming. The good news is that once we understood the theory behind these methods, we can resort to computers to automate the process.
I got M(a)=-75 N-m, M(ab)=75 N-m, M(ba)=150 N-m, M(bc)=150 N-m, M(cb)=-300N-m. Vertical reaction at A=56.25 N; vertical reaction at B=181.25 N; vertical reaction at C=175 N. Are these the answers?
No, your solution is not correct. The correct solutions for the exercise problems are provided as a part of the online course referenced in the video description field. See: courses.structure.education/courses/take/structural-analysis-i/lessons/19946240-slope-deflection-equations-part-2
At 8:00 , why is Mbc positive and Mcb negative when they are both going counter-clockwise, which I thought would be positive?
Why do you think Mcb is negative? More significantly, why do you think Mbc and Mcb are having different signs?
Perfect lessons..thans for explanations
So all moments at the joints act clockwise everytime ??
The direction of an unknown force or moment has to be assumed. The sign of the force/moment magnitude, after it is calculated, tells us if the initial assumption was correct. If the magnitude comes out positive, we know the assumption was correct. Otherwise, the actual force/moment direction is opposite to the assumed direction.
In our slope-deflection formulation, we assume the member-end moments to be acting in the counterclockwise direction. If a computed moment value turns out to be negative, then we know that moment is actually acting in the clockwise direction at the end of the member.
@@DrStructure thanks Dr i understand now
What is the difference between member end moment and fixed end moment practically? They both are acting at the end of the member. Please explain.
Member-end moments are the actual moments found at the ends of the member, once the structure assumes its deformed shape. Fixed-end moments are moments at the end of a member that is assumed to be fixed at its ends.
Consider a continuous beam having 4 spans. Suppose we load the second and fourth span with distributed loads, there are no loads applied to the remaining spans. The beam is going to deform, its five joints are going to rotate, and moments develop at the ends of each member. These are the actual (member-end) moments which we can then use to design the member. So, even though spans 1 and 3 are not subjected to any loads, member-end moments do develop at their ends.
Fixed-end moments are not the actual moments in a beam, unless the beam is fixed at both ends. These moments should be viewed as a representation of the applied loads. In the slope-deflection method, we simply replace an applied load with its equivalent moments at its ends (with its fixed-end moments.) That is, in the eyes of this method of analysis, a distributed load of W applied to a beam of length L is equivalent to a bending moment of WL^2/12 (the fixed end moments) applied at either end of the beam. So, to analyze the beam, we first replace each load with its equivalent fixed-end moments, then go through the analysis to determine joint rotation and then the actual (member-end) moments for each beam segment.
@@DrStructureThanks for explaining.
But I have a doubt. We know actual bending moment at hinge or roller support is zero. And the hinge or roller position and the end of a beam represent same point. So how is it possible that the bending moment value at the same point (hinge/roller position or end of a beam) have two different values (one is zero from hinge/roller consideration and other is non zero from member end consideration)?
Yes, it is always the case that bending moment at a hinge or a roller support is zero, if the support is located at the end of the beam. If however the hinge or the roller is an interior support, then bending moments do develop to its left and right. Regardless, when using the slope deflection method, we always start by labeling the member-end moments for each member, even if we know the moment is zero.
Say, we have a beam fixed at the left end with a roller support at the right end. Suppose we have labeled the ends of the member A and B. Even though we know bending moment at B is zero, we can still label that moment Mba. Then, when we write the joint equilibrium equation at B ( Sum of the moment at B must be zero), we set Mba = 0, we enforce the fact that the moment at the roller is zero. This enables us to calculate the slope of the beam at B while enforcing the constraint.
@@DrStructure Okay. Thank you very much.
@0:36 the beam had two rollers at the end, then how come while superimposing the beam is having one roller and other pinned?
When dealing with beams (not beam-columns) the only forces that are present (according to the beam theory) are shear and moment. There is no axial force in a pure beam. The slope-deflection method is based on beam theory meaning axial forces do not show up in the formulation.
The difference between a pin support and a roller support is a horizontal (axial) force. The pin has one, the roller doesn’t. But since our beam carries no axial force that horizontal reaction force at the pin would be zero. So, the pin acts like a roller as far as our analysis is concerned. In a way, they are interchangeable in our formulation.
Dr. Structure but doing that makes it determinate from indeterminate so we were able to apply virtual work. Had it been any indeterminate structure, how would we apply virtual work method then?
Replacing the pin with a roller does not make the beam determinate. That beam is already determinate with a pin and a roller support.
We have picked such a beam deliberately. The specific sub problem that we are trying to solve here always involves a determinate beam, a beam that is going to rotate at its ends resulting in the end slopes theta_A and theta_B. Otherwise we would not have be able to come up with the equations that relate member end moments to the slops.
Also keep in mind that here we are deriving the slope deflection equations, we are not analyzing an indeterminate structure. What we are doing is establishing the mathematical relationships between member end forces and rotations.
If we had an indeterminate beam and wanted to calculate some slope or deflection, we would have started by analyzing the beam first, then we would have applied the virtual work method to calculate the desired displacement/rotation.
@7:57 is it the same theta B on left and right of the joint B ? .. thanks in advance
Yes, the rotation immediately to the left and to the right of the joint indeed are the same as the rotation at the joint.
How do you do to know the directions on shear forces ? Cause I’m confused why VAB going downward and the rest upward?
Before calculating them, we don't know the direction of the member-end shear forces. When drawing the free-body diagram for the member, we always assume a direction for the unknown shear forces. We then write and solve the equilibrium equations for the unknowns. If we end up with a positive value for the shear force, we assumed the correct direction. If we get a negative value for the shear magnitude, then we know its actual direction is the opposite of the assumed direction.
@@DrStructure thanks!! I appreciate it
Please make a playlist for the solutions to problems.Can't seem to find them. Thank you
Done!
Thank you so much!
why is theta B3 positive unlike all the other thetas at 2:14
In this formulation, counterclockwise is taken as the positive direction. That makes, theta_a1, theta_b2 and theta_b3 positive, and the other three joint rotations (theta_b1,theta_a2, theta_a3) negative in the two theta equations
Load case 3 happens due to the load applied on the members, BUT why include Load Case 1 and Load Case 2 if the beam is already loaded?
This derivation seeks to establish the mathematical relationship between member-end moments, rotations and the applied load. Case 3 by itself is a simply supported beam that is free to rotate at its ends (with zero end-moments). That is not going to help us come up with a general formulation for the slope-deflection method as in a typical beam segment (in a continuous beam) the ends of the segment are not completely free to rotate; the bending moments (Mab and Mba) at the ends of the segment are not zero. So, we need a formulation that does take these end moments into consideration. The simply supported beam by itself does not do it.
Now, if we treat these end moments as loads, and apply it to the beam (as in Loading Cases 1 and 2), then use the principle of superposition to add the resulting rotations to those obtain from Loading Case 3, then we can establish the necessary relationship between the non-zero end-moments, rotations and the applied load.
@@DrStructure I Understand that due to superposition cases and 1 and 2 are algebraically added. But I still don't relate how the algebraic sum that includes all cases relates them. In the force method I understand why you add the deflections of the unit value and the loads acting on the primary structure, somehow I miss it here. Please elaborate.
@@dear_dennis Let me give this another try. Suppose we have a simply supported beam subjected to a uniformly distributed load. What would be the rotations at the ends of the beam? This corresponds to Loading Case 3. We can determine the end rotations using the virtual work method, for example. Here, we have basically related the end rotations to the applied load. If we know W (the load intensity) then we can determine the total end rotations in terms of E and I. This scenario, however, assumes that the moments at the ends of the beam are zero.
But, suppose we are told that the moment at the left end of the member is not actually zero. We don't know what that moment is, we just know it is not zero. How does that effect the end rotations? Obviously, the end rotations are not going to be that of Loading Case 3 since now there is a moment at the left end that somewhat restrains the rotations. So how do we adjust the end rotations from Loading Case 3 in the presence of Mab? That is where the principle of superposition comes into play. We treat Mab as an applied load/moment, calculate the end rotations in terms of Mab, then add them to the end rotation due to Loading Case 3. That gives us the mathematical expression(s) that relate the beam's end rotations to W (the applied load) and the unknown end moment (Mab).
If we apply the same line of reasoning to Mba, then we get the three loading cases that we have here.
@@DrStructure Thank you! That was very explicit! I guess it was easier for me to understand first load case 3 since the effect of the load is applied directly to the beam, while load cases 1 and 2 relate the adjacent sides. Now I understand much better the simplification when the beam's end is an overhang or a simple support.
why does moment ab is positive when you apply it at the joints ? it should change to a clockwise direction and we picked that moment rotation as negative. Thanks :)
Please be more specific. You mean in the example at the end of the lecture?
True, we show the moment at A is the clockwise direction since at the end of the member, the moment is shown in the counterclockwise direction. That is also true for the other joints. So, member-end moments shown at joints all are acting in the clockwise direction.
When writing the joint equilibrium equations, we are summing the moments at each joint. Since they are all acting in the sane direction (clockwise) then we simply add them together, as positive terms, or as negative terms, it really does not matter. That is, -Mab = 0 is the same as Mab = 0. Or, -Mba - Mbc = 0 is the same as Mba + Mbc = 0. The sign convention that we use for adding moments at the joints is independent of the sign convention that we use for writing the slope-deflection equations for each member.
You are a godsend.
coming from an athiest.
I'm still confused about signs for FEM. Does the sign represent the moment generated by the load or the moment resisting the load?
Fixed end moments are the reaction forces at the supports due to the applied load; they are the resisting moments. A downward load has a tendency to rotate the beam in the clockwise direction at the left support, and in the counterclockwise direction at the right support. So, the resisting moment (FEM) at the left support is counterclockwise, and the direction of the FEM at the right support is clockwise.
@@DrStructure
Many thanks!
How do I know what direction to assume the shear forces in?
Since the shear force is unknown, you can assume it to be upward, or downward. Make a choice and draw the force in the assumed direction on the free-body diagram. Then write the equilibrium equations accordingly and solve them for the unknown shear forces. If your assumption is correct, the computed value would come out positive. Otherwise, you get a negative value for the shear force. Either way, once the equilibrium equations are solved, you know the actual direction and magnitude of the shear force.
Can I know the links of each solution? The solution buttons don't work.
Problem 1: th-cam.com/video/Dvo7rJQjAZY/w-d-xo.html
Problem 2: th-cam.com/video/oMwbhRhhtI4/w-d-xo.html
Thank you Dr. Structure :)
I think there’s a slight mistake . I mean the part of vbc and vcb I don’t get it why 160*6*3 -450- 6vcb ? I think it’s 150*6*3+405 +6vcb right ?😅 I’m confused
Correct. There is a typo in that equation. The magnitude of the distributed load is 150 not 160. So, the correct moment equations is (taking clockwise direction as positive) :
(150)(6)(3) - 405 - 6 Vcb = 0
Or, Vcb = 382.5 N
@@DrStructure thanks 🙏🏾 you are the best
WHERE is SAPS01 idk how to find the thetas ??? ik its virtual method but how?????
th-cam.com/video/E-qfuKL2rtY/w-d-xo.html
Ms Do you have books about slope deflection and example i need so much
I not strong in this subject .Thx so much.
Most structural analysis textbooks cover the topic. See Structural Analysis by Hibbeler or Kassimali.
Do you have link can download?
As far as I know, you cannot download such textbooks.
where can i find the solution on exercises?
Solution for Exercise Problem 1: th-cam.com/video/Dvo7rJQjAZY/w-d-xo.html
Solution for Exercise Problem 2: th-cam.com/video/oMwbhRhhtI4/w-d-xo.html
the solution links are missing
The solution files are included in the course referenced in the description field.
@8.36 MCB you should put -450 not -540
Thanks for the correction.
how do you get this result?
Please elaborate. Which results?
theta A and theta B
To find the unknowns, the equations need to be solved simultaneously using techniques such as Gaussian Elimination.
nice video, but the calculation process is tedious.
Correct! The analysis of indeterminate structures is rather involved and time consuming. The good news is that once we understood the theory behind these methods, we can resort to computers to automate the process.
Mb=150*6*3-405-6Vcb=0
Thanks for pointing out the typo. Yes, @8:53 the moment equation should read:
(150)(6)(3) - 405 - 6Vcb = 0.
You are missing the solution (a video) for the first problem.
The solutions are provided in the free online course referenced in the video description field.
I got M(a)=-75 N-m, M(ab)=75 N-m, M(ba)=150 N-m, M(bc)=150 N-m, M(cb)=-300N-m. Vertical reaction at A=56.25 N; vertical reaction at B=181.25 N; vertical reaction at C=175 N. Are these the answers?
No, your solution is not correct. The correct solutions for the exercise problems are provided as a part of the online course referenced in the video description field. See:
courses.structure.education/courses/take/structural-analysis-i/lessons/19946240-slope-deflection-equations-part-2