SA30: Slope-Deflection Method (Fixed-End Moments)
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- เผยแพร่เมื่อ 25 ก.ค. 2024
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.structure.education/
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Solution for Exercise Problem 1: • Fixed-End Moments: Exe...
Solution for Exercise Problem 2: • Fixed-End Moments: Exe...
SASP03: • SAPS03: Example of Joi...
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thank u very much Dr. Structure... will never forget your help....
well done guys. you are genius.
I'm a master degree student still whatching your videos! Thank you very much !!
Hello...don't know how to thank you for your awesome lectures..mind blowing and damn good these lectures are...so proud to be a subscriber of your channel...
Thanks for the feedback.
Thanks any of my teachers can't express like this.
Awesome
I looked for sources to prepare my lectures for easy understand ...and booom. I found my rest
could you please explain forced method on symmetry and anti-symmetry structures ?
Are you asking for a lecture on this topic? If so, we plan to cover it in the future, hopefully sooner rather than later.
But if you have a specific question, feel free to ask. Depending on the nature of the question, we may be able to answer it here using text only.
what about if there is a roller in the middle or somewhere in between the beam how do you solve that ?
By definition, the fixed-end moments (not to be confused with member-end moments) are computed for a beam segment that is assumed to be fixed at both ends. So, if we have a two-span beam (with a roller at the middle), consisting of two segments (say, AB, and BC), we need to calculate two sets of fixed-end moments. We need to calculate one set for segment AB and another set of fixed-end moments for BC, assuming fixed ends for each segment.
but in some books in slope deflection they take conterclockwise fem to be negative and clockwise to be positive but u have taken the opposite ??
Yes, the sign convention is just that, a convention. It is a frame of reference that we pick rather arbitrarily. We can choose as positive whichever direction we wish as long as we consistently use it throughout.
where are the links of the solutions?? can you plz provide me
Here they are:
Solution for Exercise Problem 1: th-cam.com/video/Iv5mp8MiQ7U/w-d-xo.html
Solution for Exercise Problem 2: th-cam.com/video/Jvkk2pJ41BA/w-d-xo.html
Can i do this FEM from another method?
Fixed end moments are support reactions associated with an indeterminate beam that is fixed at both ends. You can use any appropriate analysis techniques to calculate these reactions, like the force method.
hey, I have a question. I'm dying to find the answer to this I don't know whether this is stupid or not, but anyway I'm gonna ask.
So when there's a settlement at the supports there's a moment generated right? to resist it? when both ends are fixed the moment is
6EIdelta/L^2 right? my question is what's the moment when only one end is fixed and the other pinned?
Any technical question is relevant and welcomed, and its of use to at least one person.
You are right, if a beam is fixed at both ends, and has a support settlement, the end moments are equal to 6EI Delta/L^2. Note that in this case, both end rotations are known; they are zero.
If one of the supports is a pin, say the right one, then the rotation at that support is not zero. Let's refer to that rotation as t_b. We can write the two slope-deflection equations as:
M_ab = (2 EI/L)( t_b - 3 Delta/L)
M_ba = (2 EI/L) (2 t_b - 3 Delta/L)
Since there is a pin at B, M_ba = 0. So, setting the second equation above equal to zero, we can determine t_b in terms of Delta.
t_b = 3 Delta / 2L
Then, from the first equation, we get:
M_ab = (2 EI/L) (3 Delta/2L - 3 Delta/L) = - 3 EI Delta/ L^2
@@DrStructure thnx guys
Why is clockwise FEM negative?
This is purely a matter of convention. We could have assumed clockwise direction as positive.
Can slope deflection address load case as if the load applied at the support?
Yes. When there are no member loads, the fixed-end moments would be zero. But we can still follow the process and write the joint moment equilibrium equations in terms of joint rotations. When the beam is subjected to only concentrated moment(s) at the joints, these applied moments would appear in the joint equilibrium equations (when summing the moments at each joint, we need to include the applied moment at the joint in the equation). However, if the applied loads are vertical forces only and they are located at the supports, then there is no need to analyze the beam. The reaction force at each support would be equal to the vertical load applied to the support, and the beam segments themselves would not carry any load (no internal forces develop in the segments and no joint rotation takes place).
+Dr. Structure
thank you Fri the good clarification
however how can we follow up on the issue if the joint load is only vertical load
+Dr. Structure
good ti be in touch again with you sir
just for clarification
assume we have two span beam abc
assume that there are two loads
at b
vertical load and moment
how to tack this problem using slope deflection equation
thank you sir
@@munirasad273 The beam can be analyzed, the support reactions can be calculated, using the principle of superposition. Placing the vertical load at B by itself, the support reaction at B would be equal to the load, and at A and C the reactions would be zero. Then remove the vertical load and place the bending moment at B. Now analyze the beam using the slope-deflection method. This will give us three support reactions, one at A, one at B and one at C. Finally, add the reaction force you got from the first loading case to B to get the reactions due to both loads.
can i know the solutions for problem 1 and 2?? because the videos aren't available anymore.
The solutions are still available online through the course link provided in the description field.
@@DrStructure Thank you so much!! Your explanations help a lot in my exams!
Glad to be of help. :)
Please tell me how to draw FEM in case of temperature variation or settlement?!
Please see Lecture SA32 (th-cam.com/video/dwXudm5IPDE/w-d-xo.html) for the slope-deflection formulation for support settlement. This method however is not suitable for handling change in temperature in the material. More advanced (finite element based) techniques are needed for that.
@@DrStructure thank you, Do you have any source (video or PDF) for this technique (finite element based) ?!
@@abdelsattar6007 We have a series of lectures on the displacement method which is a precursor to the finite element analysis (see Lectures SA45 thru SA50). However we don't discuss the effect of temperature change, since our focus here is mostly on the basics of structural analysis.
You should be able to find the coverage of temperature change in some advanced structural analysis textbooks/reports.
@@DrStructure thank you 💙
I'm sorry, I can't find SASP04. Please help. Thank you.
Here is the link: th-cam.com/video/paVihWpaUQU/w-d-xo.html
@@DrStructure oh, it is SAPS04. Thanks a lot.
@03:17, should the FEM ba should be counterclockwise, since your equation of FEM ba = 2EI(theta A+2theta B)/L ?
The direction of the arrows are placed to correspond to the actual deformation of the beam. In this case, theta_B is going to have a negative value resulting in a negative value for FEM_ba.
There are errors at 4:40 and 6:32 the Mba direction is wrongly set twice in two different examples
The member-end moments are drawn in the positive (counterclockwise) direction at both places. If they were to be actually calculated, the sign of the moment would correctly indicate the direction.
Thank you so much for the explanation.
1:22 can anyone please explain what is the difference between fixed end moments and Mab, Mba etc...
When analyzing continuous beams, we seek to calculate the internal forces at the two ends of beam segments. If we know the bending moment and shear force at the ends of the member, we can easily calculate these forces within the member using the static equilibrium equations.
The forces at the ends of the member are called member-end forces. So, for beams we have member-end shear forces and member-end bending moments. When using the slope-deflection method, we label the member-end (bending) moments as Mab and Mba, assuming the ends of the beam segment are labeled A and B. These are the actual moments that develop in the member when the structure is subjected to loads.
Consider a beam with length L that is fixed at both ends. Suppose there is a uniformly distributed load (w) acting on the entire beam. The moments that develop at the ends of the member are wL^2/12. These indeed are the member-end moments. But in the special case that the beam is fixed at both ends, we can call them fixed-end moments as well. So, the term fixed-end moment applies only when the ends of the member are actually fixed or assumed to be fixed.
Suppose we replace one of the fixed supports in the beam with a roller support. Now the beam is fixed at one end and rests on the roller at the other end. If we are asked what are the moments at the ends of the beam. We could say, there is an unknown moment at the fixed end, and a zero moment at the roller end. That is, we know one of the member-end moments (it is zero), but we need to compute the other one (at the fixed end). In this case, we cannot call these fixed-end moments since the beam is no longer fixed at both ends, but we can call them member-end moments.
In the slope-deflection formulation, the member-end moments are formulated using the method of superposition. That is, we say, the member-end moments are a function of the fixed-end moments and the moments that develop due to the rotations at the ends of the member. The equation:
Mab = (2EI/L)(2 t_a + t_b) + FEM_ab
Where t_a is the rotation of the beam at end A, t_b is the rotation at end B, and FEM_ab is the fixed-end moment at end A, can be written as:
M_ab = (2EI/L)(2t_a) + (2EI/L)(t_b) + FEM_ab
The first term is the moment at end A of the beam segment due to t_a,
the second term is the moment at end A due to t_b,
and the last term is the moment at end A assuming that the beam is fixed at both ends.
That is, when using the slope-deflection method, we start by assuming the ends of the member are fixed and calculate the (fixed-end) moment that results. We also calculate the moments at the end of the member due to rotation t_a and t_b. Then we say the actual (member-end) moment is the sum of the three terms.
@@DrStructuresir, can i say that fixed end moment is associated only with a particular span with a certain type of loading while the member end moment is the total moment developed in the beam due to effect of all the span loadings.
Thank you so much for this wonderful explaination..I got my all doubts clear.👍👍
Yes, that is a correct way of thinking about the two moments.
@@keertisingh894 was it doc's or mine. 😜😁
virtual work SAPS03
in 7:37 isn't the arrow a counterclockwise? why is it clockwise?
Are you referring to the applied moment in the second exercise problem? That is the direction in which the moment force is being applied to the beam. Such a force can be placed on the beam in either direction.
@@DrStructure sorry the time stamp is wrong, I mean the time at 5:00. the direction is counter but the label is clockwise
@5:00, two types of entities are shown on the diagram: moments (FEMab and FEMba) and rotations (ThetaA and Theta B).
At this point, the moments are unknown, therefore, we assume them to be acting in the counterclockwise (positive) direction. In the slope-deflection equations, counterclockwise is considered the positive direction. Note that both moment arrows are shown in the counterclockwise direction.
Since the rotations have been calculated at this point, we know their actual direction and magnitude. ThetaA is counterclockwise and ThetaB is clockwise, which is the result of an upward deflection of the beam. These rotations are shown in the correct direction on the diagram. The magnitude of the rotations are also given. However, since we are going to plug in these values in the slope-deflection equations at the bottom of the screen, and in those equations rotations are assumed to be in the counterclockwise direction, we have to enter a negate the value for ThetaB. That is, a positive value of PL^2/16EI for ThetaB in the slope-deflection equations, implies a counterclockwise rotation at B. But since the rotation is clockwise, we need to negate that value.
I want a favour from your channel...plz make a video on to calculate fixed end moment due to three concentrated point load...I really need your help....waiting for your response...
Thanks for the note. These videos take some time to develop, by the time we are done making such a video you have figured out the process already. Perhaps you can solve this problem on your own, with a bit of hint:
Place once concentrated load on the beam at a time and determine the fixed-end moments due to that single load. So, for concentrated load (P1) you get mLeft1 and mRight1. For P2, you get mLeft2 and mRight2, and for P3 gives mLeft3 and mRight3. Then, you can find the fixed-end moments due to all three loads placed on the beam simultaneously by adding the three moments on either side of the beam. That is,
mLeft = mLeft1 + mLeft2 + mLeft3
mRight = mRight1 + mRight2 + mRight3
Dr. Structure dear sir can't it be solved by the same process you have solved some problems and specific results too..considering the beam to be simply supported first and then calculate the end rotation by unit load method considering four sections..and finally putting the obtained value of slope in slope deflection equation to get the desired FEM...
Dr. Structure Got you...I will try to solve it as per your precious guidance...thank you once again for your support...
Hello sir...as per your hint I tried to solve the problem but couldn't get the desired result..I obtained the m1 left value for the first load at(L/4) distance from left as (9PL/64)...m2 i obtained for load at (L/2) from left as (PL/8)..m3 left due to load at(3L/4) from left as (3PL/16)..adding( m1+m2+m3) left,total comes out equals(29PL/64)...m1,m2,m3 are calculated as per specific result by formula (Pb^2a/L^2)...answer is (15PL/48)
Yes, it can be solved that way. But that is a lengthy process for solving a problem that can be solved quickly using the principle of superposition.
What are advantages and disadvantages of slop deflection???
A method can be examined in that light (i.e., advantages and disadvantages) in a comparative way only. The slope-deflection is a classical analysis technique, similar to the force method. The techniques was developed about a century ago, when numerical and computer-based techniques were not available. The slope-deflection method works best for problems that can be solved by hands, without using the computer. If the degree of indeterminacy of the structure is more than 4 or 5, hand-calculations become laborious and error-prone, rendering the use of the classical methods impractical.
A comparison of the slope-deflection method with the force method may be of use when trying to decide which technique to use for solving a problem. In the slope-deflection method the unknowns are joint rotations and relative displacements. In the force method, the unknowns are reaction forces. Consider a beam that is fixed at both ends and rests on a roller at its midpoint. The degree of indeterminacy of this system is 4, meaning if we were to analyze the structure using the force method, we would need to set up and solve 4 equations in 4 unknowns (if the horizontal force at the supports can be ignored, we can solve the problem using 3 equations only). The slope-deflection method, however, requires only one equation in one unknown (the unknown being the rotation of the beam at the roller support, since the end rotations are zero) to analyze the beam. Now, consider the case in which the end supports are pins/rollers instead of being fixed. So, we have a continuous beam with three pin/roller supports. In this case, the degree of indeterminacy of the system is 1, meaning that we can solve the problem using the force method by setting up one equation in one unknown. Here, the slope-deflection method requires using 3 equations in three unknowns (the three joint rotations) to solve the problem.
We can also examine and compare the two methods in terms of the actual steps that each require and ascribe time requirement to each step in order to compare and contrast the time efficiency of the techniques in various scenarios.
Thank you very much❤❤
The need to use statically indeterminate structures???
@@ahmedbarakat1384 Indeterminacy naturally arises in some type of structures. Consider a building (frame) having three or more columns at its base. Each column is sitting/resting on a foundation which can be viewed, depending on the details of the connection, as either a pin connection or a fixed connection. Either way, there would be too many support reactions to calculate using the three static equilibrium equations only; here we have an indeterminate system to be analyzed using techniques such as slope-deflection or matrix displacement method.
Can I ask you the last question, please?
I found this application in youtube. th-cam.com/video/Ld4V6GfS1TY/w-d-xo.html. Is this app ok to use?