SA27: Slope-Deflection Method (Overview)

[@anamijatovic4824]

Why did you assumed negative internal bending moments (counterclockwise) (2:38), and later (3:08) for the middle beam you assumed it positive (clockwise)?

[@DrStructure]

Thank you for pointing out that inconsistency in notation. Mbc and Mcb at 3:08 should have been drawn in the counterclockwise direction. When dealing with unknown moments, we always should draw them in the assumed positive direction.

[@VocaFan4ever]

Could you guys have smaller/thinner pens? It makes it easier to see when I have to pause and trying to understand the points being made, thanks!

[@zumbatan550]

Sir, the 3-span beam is symmetrical except the supports at A & B are different. That means they were almost opposite along the line of symmetry. On your free BOD on 2:24, you drew them - the rotational forces are all in one direction. Shouldn't they be drawn in the opposite to be understood and undisputable ? And also shouldn't M (AB) & M (BA) for example be in the opposite direction?

[@DrStructure]

The beam segments are subjected to generalized distributed loads with various lengths and moments of inertia. Why assume the symmetry? Here, we are developing a general formulation in which the beam segments could be subjected to any loads and could have any length and moment of inertia. As a matter or principle, the direction of an unknown force or moment has to be assumed; the assumed direction is arbitrary. We are not trying to decide whether the internal moment in the beam is positive or negative, since at this point we don’t have enough information for that determination. We are simply stating that there is a moment (with the understanding that its magnitude could be zero, positive, or negative) at each end of the member, and we choose to use the commonly used beam sign convention to show/draw it. The M_ab, M_ba … notation is associated with the slope-deflection method. This method uses a different sign convention for representing member end moments; we show the (unknown) member end moments in the counterclockwise direction. Please see the previous lecture on the slope-deflection method on this.

[@ARON_1100]

How can someone teach so easily and clearly.... 😢

[@vivek5380]

Thank you so much for all these lovely lectures💐

[@anamijatovic4824]

Hi Dr. Structure, Can a slope-deflection method be used on simply supported (or any other statically determinate) beams? If not, why?

[@DrStructure]

Interesting question! Yes, we can apply the slope-deflection method to determinate structures. For example, if we apply it to a simply supported beam, it would give us the beams's end joint rotations. Of course, the method has not utility for calculating the support reactions since those can be easily determined using the equilibrium equations.

[@norodomiiamil7961]

@3:39 how did you get equation Ma. how did you get wL cube over 24EI? i would much appreciate if can you elaborate for me. tia

[@DrStructure]

Mab = 0 (2EI/L) (2 theta_a + theta_b) + wL^2/12 = 0 Multiply both sides of the above equation by L/(2EI), you get: 2theta_a + theta_b + wL^3 / 24EI = 0

[@norodomiiamil7961]

Dr. Structure thanks so much. very much appreciated

[@DrStructure]

You're welcome.

[@eminashindahouse]

Dr plz help me with the sign convention for beam AB @ 1:50. From my understanding, the actual direction for Mab has to be clockwise, while the Mba is anti clockwise. But why do u wrote anti clockwise for both Mab and Mba? Is it has to do with initial assumption since we are talking about an unknown bending moment?

[@DrStructure]

Exactly! We are making the initial assumption that all the member-end moments are acting in the counter-clockwise direction. That enables us to formulate the problem in a consistent manner without trying to determine the correct direction for each moment beforehand. After doing all the necessary calculations, if a member-end moment ends up having a negative value, then we know that it actually acts in the clockwise direction.

[@eminashindahouse]

Dr. Structure thank you very much Dr. Appreciate your explanation

[@derrickmaina3202]

Hey doctor, can I get more examples dealing with slope deflection methods?

[@umdbest001]

awesome as always

[@jcoc8129]

excelentes videos, tengo una duda se esta usando el metodo de las secciones para analizar los segmentos y puntos de la viga por separado cierto? por ejemplo, realizo 2 cortes a ambos lados del punto B para poder calcular M_B=0 y sum(FyB=0)

[@michelbaya4293]

Thank you for this great lecture. Thumbs 👍

[@manishgautam1318]

Can you please explain difference between member - end shear forces and support reactions.. I mean you are actually calculating support reactions then why are u calling them "member-end shear forces" (at 5: 15)... how and where are that came from? And I also want to know that why are u always taking clockwise moment as -ve and anti clockwise moment as +ve (4:35).. however while calculating support reactions, when you taking moment about one end support in moment equation (i.e sigma Ma or sigma Mb...whatever) then you are doing just reverse i.e. clockwise moment as +ve and anti clockwise as -ve (at 5:25 in the moment equation "-LVba", at 5:45 "-LVdc")... I mean How???

[@DrStructure]

Imagine a beam having a pin support at its left end. Say, the reaction at the support is 10 kN. This is the force that the beam transmits to the substructure (through the support). If we cut the beam, say 0.0001 mm, away from the support, then we end up with a shear force. This is a force internal to the beam. We are not going to see it on the free-body diagram unless we cut the beam. In this case, the shear force has to be equal to the reaction force in magnitude but opposite in direction since the sum of the forces in the y direction in that tiny segment must be zero. So, when considering the left end of a beam segment, we talk about shear force at that end. If that end is next to a support, then the reaction at the support is going to have the same magnitude as the shear force. In a sense, we can refer to it as either the shear (at the left end of the beam segment) or the reaction force (at the support). As for your second question. There are two distinct sign conventions here: One for the slope-deflection equations and one for writing the equilibrium equations. When writing the slope-deflection equations, we have assumed counterclockwise to be positive, both for end rotations and end moments. The derivation of the slope-deflection equations is based on that assumption. The other sign convention is for summing the moments about a point. This is a more general sign convention that we use when writing the equilibrium equations. Here, we can assume either direction to be positive. The choice is really yours.

[@salimmohammed3945]

Overview has been to enough for me to subscribe your channel and press the damn 🔔. nice lec

[@gospelnews123]

How do I solve the equation for the unknowns....3:49

[@DrStructure]

We need to use our knowledge of algebra to solve the equations. In this case, we have four equations in four unknowns. A practical technique for solving such a system of equations is Gaussian Elimination Method.

[@gospelnews123]

@@DrStructure okay thanks

[@niteshmeena9961]

Can you provide lectures link other than structure and statics.

[@DrStructure]

You probably can find some other content online on this, try google search...

[@KjCav]

You lost me at 3:52. How did you get the "Equilibrium Equations in terms of joint rotations"? Also, how did you get those values once you solved for the unknowns?

[@DrStructure]

@3:30 the slope-deflection equations define Mab, Mba, ... in terms of joint rotations. So we just have to substitute the slope-deflection equations in the equilibrium equations to arrive at the set of equations in terms of the joint rotations. As for the solution of the system of equations, it is a linear systems, except that instead of having numbers we have parameters. Treat the joint rotations as unknowns, then solve the equations for the unknowns. Think of the system as something like this: a1 X + b1 Y + c1 Z + d1 W = e1 a2 X + b2 Y + c2 Z + d2 W = e2 a3 X + b3 Y + c3 Z + d3 W = e3 a4 X + b4 Y + c4 Z + d4 W = e4 where X, Y, Z, and W are the four rotations, and a1, b1, c1,... are the constants/parameters. Solving such a system by hand is a tedious task. We can use an app like Mathlab or Mathematica to facilitate the process. There is an online version of Mathematica that can be used freely. It should suffice for solving this size equations. Here is the link, in case you want to give it a try: www.wolframalpha.com/

[@siguerhakim4723]

very nice method,

[@anamijatovic4824]

In slope-deflection equations, when the part of the equation (wL^2/12) is positive and when it is negative? Does that depend of the bending moment and applied load directions?

[@DrStructure]

Our sign convention for bending moment in the slope-deflection method is one that assumes counterclockwise direction to be positive. For fixed-end moments, assuming w is acting downward, bending moment at the left end of the beam is going to be counterclockwise, hence positive wL^2/12, and bending moment at the right end of the beam would be clockwise, or negative wL^2/12. Therefore, Mij, bending moment at the left end of the beam has the positive wL^/12, and Mji, bending moment at the right end of the beam has the negative wL^2/12. Obviously, if the direction of w is reversed, if w would be acting upward, then the wL^2/12 term in each equation changes sign.

[@oluwatomirhodes795]

Hello Dr. Sturcture please how did you solve for thera a to be 30EI. Thanks

[@DrStructure]

There we have a system of linear equations; we have four equations in four unknowns. Theta_a is one of the unknowns. To find all the unknowns, we need to solve the equations simultaneously. Linear algebra provides various techniques for solving such systems of equations.

[@oluwatomirhodes795]

@@DrStructure thank you sir/ma

[@이현민-k7h]

when will you post next lecture???I'm very looking forward to watching next lecture. Thank you for always good lectures!

[@momenabdallh5258]

thank you dr where can i found example for this lesson

[@DrStructure]

Please see Lectures SA28, SA29, and SA31 for examples and exercise problems on the slope-deflection method.

[@dillonhaire9195]

Hi, How did you seperate theta and determine the values of: theta a, theta b etc. @4.10?

[@DrStructure]

There we have 4 linear equations in terms of 4 unknowns. The unknowns are thetaA, thetaB, thetaC and thetaD. w, L, E and I are considered to be known. Generally speaking, we need to solve such equations simultaneously. A technique like Gaussian Elimination is a good choice here.

[@dillonhaire9195]

Thankyou.

[@sagantisharathsai8216]

grt and simple explanation. can we get lectures for other engineering​ subjects

[@willrichards1454]

Are these slope deflection moment equations the same for every problem?

[@DrStructure]

Yes, these derived equations work for analyzing any continuous beam.

[@sherlingyuan5618]

+Dr. Structure Very impressive lecture, Doctor. But my question is, how many more topics or videos would be covered and introduced for Structural Analysis, after this latest one?

[@DrStructure]

On Slope-Deflection Method, three or four more. Overall, about 30 more.

[@intririthlim6892]

Dr, where can I find the answers to the problems in each lecture?

[@DrStructure]

They should be listed in the video description field. Alternatively, you can find them here: lab101.space/Library-Exercise.asp

[@alierdaadam4863]

Thank you a lot.

[@muralidhar9511]

Dr structure you calculated Moments for simply supported beam ( left end is pin joint M AB= 2EI /L ( 2* theta A + theta B )and M BA =2EI /L ( theta A + 2*theta B ) which is a roller joint ) I.e … but your are using the same formulae for the beam section rested on two roller joints i. e in B C section.Is this assumption is valid for all ?

[@DrStructure]

Yes. The moments that we are calculating here are not the support reactions, or at the supports. They are the moments at the ends of the member as if you cut the member close to whatever support is at the left end and at the right end of it. So, the supports themselves are not relevant here.

[@anamijatovic4824]

What if we know that EI is not constant, can we use and how do we adjust the equilibrium equations in term of joint rotations?

[@DrStructure]

If two or more beam segments have different E or I, that would impact the slope-deflection equations. The two slope-deflections, Mij and Mij are defined in terms of (EI/L). So, members with different E or I would have different (EI/L) terms which in turn would be reflected in the joint equilibrium equations.

[@meshalbataineh4472]

thank you doctor ... excuse me I have a question ..in (5:22) in segment AB ..you wrote MA =WL^2/2 ..i think there is a mistake doctor

[@DrStructure]

Thanks for the note. Why do you think there is a mistake? What is the mistake? Please elaborate.

[@vivek5380]

meshal bataineh thats how UDL is dealt with W*L*L/2

[@Onlinedistancelearning]

Can any one tell me which software is being used to solve problem in such way

[@DrStructure]

Mathematica: www.wolfram.com/mathematica/ A limited online version of it can be found here: www.wolframalpha.com/

[@Onlinedistancelearning]

@@DrStructure Thanks sir..your courtesy

[@accessuploads7834]

thanks

[@zumbatan550]

Can you explain how u determine the directions of moment at the beam MAB & MBA - both CCW, why MBA is not CW?

[@DrStructure]

When drawing free-body diagrams, we have to assume a direction for the unknown forces. In this case, we need to assume a direction for the member-end moments. In the slope-deflection method, we assume the member-end moments are acting in the counterclockwise direction (but we could also assume them to be acting in the clockwise direction, and then adjust the formulation accordingly). This is very similar to how we show the support reactions in a beam. When the reaction forces are unknown, we generally assume them to be in the upward direction. If the computed member-end moments turn out be be positive, then our assumption is correct. However, a computed negative magnitude for a member-end moment tells us that the moment actually acts in the opposite direction to what was assumed initially. That is, the moment acts the clockwise direction. So, we are not actually determining the direction of the moments at the onset of the process, when drawing the free-body diagram. What determines the actual direction of the moment is the sign of the magnitude of the moment, after it is computed. If the sign is positive, then the moment is acting in the assumed (counterclockwise) direction. If the sign is negative, the moment is acting in the clockwise direction.

[@zumbatan550]

@@DrStructure Thanks. In other words, we can assume the other direction so long as we are consistent with the way we assumed. I am struggle with this: All the times, I read, solving problems, support types to the beams were given. In real life, how do we determine the support types. When we design a continuous beam, say, a reinforced 3 or 4 spans concrete beam. Is the end span taken as fixed support? Is the intermediate span, fixed, pinned or roller supports? Also assuming I have a 10 feet steel channel beam (supporting solar panels). The beam sit and bolted on top of 3 larger channels as two span continuous beam, what kind of support should each of the support be ? Can you do a video, with real life examples the type of continuous beam that their support should be analyzed as pinned, roller and fixed supports.

[@DrStructure]

@@zumbatan550 Regarding support types, we design real structures with specific support types in mind. So, it is not like we build the bridge and then ask ourselves what kind of support it uses. Rather, we decide beforehand what type(s) of support we wish the bridge to utilize. That is why we need to know the support types before analyzing the structural system. Put differently, as structural designers, we must decide on the support types before we can analyze the system and eventually construct it. Next time you come across a bridge, if its supports are visible, try to see if you can determine its support types. Generally speaking, the connection is considered rigid if you don’t see a roller or a friction pad between the beam and the column that supports it. When analyzing such a system, the continuous beam and the columns that support it must be analyzed as a frame structure. If the concrete beam is attached to the abutment in a monolithic fashion, we can consider it a fixed connection. In this discussion, it is important to distinguish between a fixed joint and a rigid joint. A fixed joint cannot rotate or displace. A rigid joint can displace and rotate, but all the members connected to the joint rotate together the same amount. A good example to keep in mind is a frame structure. The base of the frame could be fixed at the foundation, but the upper joints of the frame (the connections between the beams and the columns) are not considered fixed; but they can be rigid. The connections can be viewed as pins if a beam is supported by three channels using bolts. If the bolts prevent the lateral movement of the beam at the connection, it is not a roller. Although, in principle, we can build a fixed connection using bolts or a combination of bolts and welds. But in this case, the chances are we are dealing with pin connections.

[@zumbatan550]

@@DrStructure Thank you Dr. for educate me. Appreciate your response.

[@TheGeorge3011]

Where does the WL^2 / 20 come from ?

[@DrStructure]

Around @4:30? member-end moments Mba, Mbc, Mcb? By substituting the calculated slopes (shown to the left of the equations) in the slope-deflection equations we get W L^2/20 for the three member-end moments mentioned above.

[@TheGeorge3011]

Thank you so much I got it now ! and omg you actually replied and so fast!!!! faster than my own professor would! thank you

[@deependrashah793]

great

[@civl112]

Why can't we just keep integrating the load w(x) to get M and deflection (and all reactions)?

[@DrStructure]

We can determine shear, moment, and deflection by integrating the load function while considering the beam’s boundary conditions. This integration works when the load function is continuous. However, when point loads or support reactions are present (which generally cannot be expressed as continuous functions), the beam must be divided into segments where the load is continuous. This results in piecewise equations for shear, moment, and deflection.

[@civl112]

@@DrStructure Thank you!

[@jaffarkhan4487]

why the moments are same on the bothe sides of joint?bit cinfused

[@DrStructure]

The member-end moments are not the same (one is Mij, the other is Mji), but they are assumed to have the same direction (counterclockwise). When deriving the slope-deflection equations we need to have a sign convention for joint rotations and member-end moments. We can either use counterclockwise, or clockwise, as positive. This simply means when we are drawing our diagrams, we draw rotations and moments in the positive direction.

[@happyhappy-po6fm]

What if the EI is not constant

[@DrStructure]

If each member has its own constant EI, the method still works. Each set (for each member) of slope-deflection equations has its own EI which needs to be used in the equilibrium equations. So, a constant EI cannot be factored out in the moment equilibrium equations. But that is not a problem. We still can write the equation and solve them. If however, EI varies within a member, say, a beam segment has a variable moment of inertia, the slope-deflection equations as we have written them (as we derived them based on the assumption that EI is constant) become inapplicable. A revised and more complex set of equation would be needed to handle beams with variable moment of inertia.

[@niteshmeena9961]

In this lecture how you directly written the moment in sections

[@DrStructure]

If you are asking a question, please rephrase. I don't know how to interpret your statement.

[@civilideas1925]

👍🏻

[@sssy785]

you are star

[@dimitric3607]

I came to this video to understand where the equations came from and you completely skip over it and just state the equations... What help is that when I dont even know where the equations are coming from

[@DrStructure]

This video (SA27) provides an overview of the method without any derivation. See SA28 through SA33 for the derivations you are looking for and more.