SA20: Work-Energy Principle (Part 2)

[@kashf8694]

These mini lectures are far more understandable than any I've seen given by any lecturers. It seems like teachers are more involved with writing their names on the slides more than the content. I don't know who to be thankful for these lectures but uncovering knowledge to many who watch these videos advances the human race. Being the roof cause of this, no acknowledgement is needed. You are the foundation of a better future.

[@DrStructure]

Thank you for your thoughtful and kind words.

[@appdroide100]

Bro this videos are so good , you've got a talent , I really mean it . Thanks a lot

[@supertv2783]

Thank you Dr. Structure it is so great videos...

[@mohammadammouri2387]

Thank you so much Dr , Your lectures incredibly useful . Please upload much more

[@accessuploads7834]

This lecture is a brain food such as fish

[@rahul1268]

dr please upload more lectures ,they are really useful and to the point

[@DrStructure]

akash poddar Thanks! Will start uploading new lectures in September.

[@amitb3262]

Sir,Upload more examples,so we can make practice & perfect it

[@NinjaMartin]

Such top content! I really enjoy the emphasis on learning, and that you take your time proving your points with your animations. One question: I didn’t understand the steps for M r = EI. Isn’t E the slope in the elastic linear part on a «work diagram» (sorry my broken English, in Norway we call it a work diagram, with tension on the y-axis and epsilon on the x-axis) and I is the second moment of inertia, which is a geometric property of the cross section of the beam? Have a great day.

[@DrStructure]

Thank you for your feedback. The derivation of the relationship M r = EI was intentionally left out of the presentation to reduce derivation overload. We will put together a short video to explain why M r = E I shortly. Look for a link here in couple of days.

[@DrStructure]

You can find the derivation of equation M r = E I here: kzbin.info/www/bejne/qJ-Tc6V-jbx-n80

[@yishakabrham5002]

fabulous channel tnx

[@SANDEEPKUMARsandeepky1012]

appreciable work...

[@patrisagar6548]

1:53 can you explain a bit on how we can use integral in the place of summation? Thanks

[@DrStructure]

This comes from the definition of integral. If you have a continuous function f, the integral of the function gives us the total area under the curve. That area can also be approximately computed by dividing it into an infinitesimal number of rectangular elements each having a base of (dx) and a height of (f_i). The total area in summation form can be written as: SUM (f_i)(dx) for i 1 to N.

[@chortejie1halliru774]

Hey thanks so much for the lectures sir. They're really helpful 🙏🙏 Please are you also having a page for mathematics? And also please 🥺 how do you make your videos i mean with Which software? 🙏🙏

[@DrStructure]

You can find the written lecture notes for this lecture in the online course referenced in the video description field. This particular video was compiled using VideoScribe and Camtasia Studio.

[@Sudhakarsen10]

But we ignore axial deformation in frame and beams. Member are inextensible. So deflection due to moment only is to be considered

[@DrStructure]

We can ignore them only if they are insignificant.

[@gentpolloshka6716]

When you took the right side of the beam (from L/2 to L), why is the moment (P/2)*x - P(X-L)/2. Can you explain that? Why do we take it like this?

[@DrStructure]

For a simply supported beam of length L subjected to a concentrated load of P at its midpoint, the left support reaction is P/2. Therefore, the bending moment equation for the right segment can be written in terms of the moment of the left support reaction about an arbitrary point (x), and the moment of the applied load about the same point. If we choose the position of the left support as the origin of the coordinate system, the moment arm for the support reaction is x. Therefore, the moment of the support reaction about the arbitrary point equals (P/2)(x). The moment of the applied load about the same arbitrary point is (-P)(x - L/2), where x - L/2 is the moment arm for the applied load. That is, the distance between the applied load and the arbitrary point in the right segment of the beam is x - L/2. So, the moment equation becomes: (P/2)(x) - (P)(x - L/2). This expression simplifies to PL/2 - Px/2, as shown at time 5:44 in the lecture.

[@sep_shrk]

Is this the end of videos? Your videos are great Thanks

[@DrStructure]

sepehr sharaki Thanks! No, we will resume adding more videos in couple of months.

[@sep_shrk]

Dr. Structure Great, thanks :)

[@fathereshark8421]

At 2:23 where does the x come from, and how did you get the equations M(x)=PX/2 and M(x)=PL/2-PX/2. I tried taking moments about point A (left side) but I didn't get the equations you got. Also at 6:02 how did you get p^2l^3/96EI? At 8:40 why doesn't joint CD have a moment force at the end?

[@DrStructure]

x is the horizontal axis in the Cartesian coordinate system. The origin of the system is assumed to be at the left support. Please see our shear and moment equations lectures to better understand how to write shear and moment equations for beams. In this case, the left support reactions is P/2. Therefore, bending moment for the left half of the beam where x is between 0 and L/2 is (P/2)(x). And for the right half of the beam where x is between L/2 and L, we get (P/2)(x) - (P)(x-L/2) which simplifies to PL/2 - Px/2. The internal energy was computed via integration (see 6:00 for the integral expression). Member BD carries an axial force only, there is no bending moment in the member. Therefore, bending moment at its upper end (at joint D) is zero.

[@fathereshark8421]

@@DrStructure Thank you

[@zhiwang6885]

Can you tell me which software you use to tech us?

[@DrStructure]

Zhi Wang Primarily VideoScribe and Camtasia Studio.

[@rizalpurnawan3796]

Can you provide the proof that the internal energy is equal to 1/2 M d\thetha? Or may be you could provide a reference. As far as I can guess, it is analogous to the work which is equal to 1/2 P \delta. However I would like to be more comprehensive. Thank you.

[@DrStructure]

Please see Lecture SA19. kzbin.info/www/bejne/r5jUn6x-aN1qeac You can go through that derivation using moment/rotation instead of force/displacement to obtain the results/proof you are seeking.

[@rizalpurnawan3796]

@@DrStructure I have seen that video before I asked the question. I got that as P is directly proportional to delta (increment), so that we can express P as a continuous function with respect to delta whose derivative is constant. Let say that there exist some real number eta such that P = eta delta. Then the work increment can be given as the multiplication of P and the delta increment. And the total work is given by the integral of P with respect to delta, which will end up equal to 0.5 P Delta. Then you mean, we can just apply the same method by replacing P with M and delta with theta, right? Okay, I think my interpretation on 'work' wasn't correct. But now I think I get it, as I have also read on wikipedia about work on other physical systems. Thank you for your reply.

[@rizalpurnawan3796]

@@DrStructure By the way, that's a great lecture! Keep it up!

[@DrStructure]

Correct, in thinking about work or deriving its equation, we can replace force with moment and displacement with rotation. Put it differently, moment is a type of force and rotation is a type of displacement. Work is the product of force and the displacement caused by the force. So, a moment (think of it as a rotational force) causes rotation, therefore, the product of the moment and its corresponding rotation is work done by the moment.

[@rajsah8487]

In the frame example,for the member AC,why shouldn't we take M(x)=Px - 2PL as we have taken in the member CD????

[@DrStructure]

In the example, we set the origin of the coordinate system for AC at the lower end (joint A) of the member. Since there is no bending moment at A, then M(x) = P x only. We could have picked C as the origin with positive x pointing down toward A. Then, the term 2PL would have been appeared in the moment equation since there is such a moment at C.

[@marcrogue5268]

@Dr.Structures why does the work energy method neglects the direction of the force being applied?

[@DrStructure]

You need to elaborate more. Work is force times displacement in the direction of the force, and energy is the product of the internal force times the deformation caused by the force. So, the direction of the force is not neglected in determining either external work or internal energy.

[@marcrogue5268]

Dr. Structure I should have been more specific, what I was trying to say was the deformation of some member may be in the vertical, horizontal direction or both. So my question is if the force is applied let’s say horizontally to the right as in your example it seems reasonable to me to only take the horizontal component of the deformation in the members to calculate the displacement in such direction. So my question was more directed towards calculating displacement using the work energy method, I’m sure there must be a reason for this, the only thing I could think of is that the displacements are soo small that it doesn’t really matter if we don’t considered direction but I’m not totally sure if this is why. Hope I was more clear this time, thank you for the quick response!!! Love the channel ❤️❤️❤️

[@DrStructure]

It may be useful to think of external work and internal energy separately, as two related but different phenomena. When we think about external work, we should think of the system as a whole, not in terms of its individual components (members). The system as a whole is going to deform in some manner under the applied load. This deformation can be described/measured at any point on the system as a x-displacement, y-displacement, and z-rotation (assuming a 2D system). That is, every point of the system is going to displace by some amount under the applied load. If the load (P) is applied at point A, then by definition, work done by P is a function of P and the displacement at A in the direction of P only. None of the other displacements (at points other than A) matter, they are inconsequential when we want to determine the external work being done by P. Moreover, the individual members of the system (e.g., beams and columns), how much they displace and in which direction don't play any role in determining the external work, as they are considered internal to the system. So, if we have a frame that has a horizontal displacement under an applied horizontal load, the forces and displacements of the individual members do not play any role when we are calculating the external work, which is solely a function of the applied load and the displacement of the whole system in the direction of the load. As far as the external work is concerned, the only thing that matters is the amount of the overall displacement caused by the force. Here, we have made no assumption or simplifications in defining work in terms of force and displacement. On the other hand, to determine the internal energy of the system caused by an external load, we need to dissect the system and examine individual components/members. Under the applied load, each member is going to be stressed (internal forces develop in the member). For example, in a frame structure, an internal shear force, bending moment, and axial force could develop in each member. And, the member is going to deform (we refer to these as axial deformation, shear deformation, and bending deformation). These deformations, however, are distinct from the external deformation of the system. If 3 internal forces (i.e., axial, shear, and bending) and 3 internal displacements (axial, shear, and bending) develop in a beam or column, then the internal energy stored in the member can be calculated as the product of each internal force and its associated internal displacement. Again, here the internal force and displacement are aligned with each other. That is, axial force in the member must be multiplied by the axial displacement of the member in order to determine the (axial) internal energy. Here too we are not making any simplifications. If internal forces develop, causing internal deformation, then internal energy emerges and can be calculated as the product of the internal force and displacement. At no time, whether we are talking about external work or internal energy, the direction of the displacement (of the whole system or its individual members) can be ignored. Work (energy) is always the product of the external/internal force and its related external/internal deformation. However, when we are performing the actual calculations, some components of internal energy can be ignored. For example, in most cases, for beams and columns, we can ignore internal energy due to shear force and shear deformation since the latter is generally very small and can be assumed to be zero. For calculating external work, since by definition work equals force time displacement in the direction of the force, we simply ignore displacements in other directions. For example, if point A displaces in both x and y directions, but the applied force (P) is horizontal, that y-displacement plays no role in work done by P. If, on the other hand, P is applied at an angle at A, both x and y displacements should be used to calculate external work. In that case, the x component of P times the x-displacement plus the y component of P times the y-displacement gives us the total external work done by P.

[@marcrogue5268]

Dr. Structure thank soo much for the elaborate response. If you could focus on internal energy for instance,I agree we need not to worry about with any other deformation but the one I’m the direction of the external force being applied. My question is going back to the horizontal force P being applied causing an external displacement delta in the direction of the force, But lets say every truss member that will develop an axial force will deformed based on the orientation of the member under such axial load and since the orientation of the axial load is determined by the physical direction of the member also the deformation will occur in such direction. There is the source of my questioning since the internal deformation in all the component contribute the the external displacement how come we are using the full magnitude of the internal axial force to calculate external displacement? I wish I could explain myself better. Thanks for the response anyways

[@wonghueihann2548]

I don't understand why it is PL/2 - Px/2 @2:23 ...why not PL/4-Px/2 ?

[@DrStructure]

+Wong Huei Hann To determine the moment equation for L/2 < x < L, cut the beam at some arbitrary point to the right of the applied load. Let's refer to the point as O, and refer to the distance from the left support to this point as x. Now, draw the free-body-diagram for the left beam segment, the segment between the left support and point O. The diagram should consist of an upward force of P/2 located at the left end of the beam. The distance from this load to point O is x. There is also a downward force of P located at L/2 from the left end of the beam. At point O, the diagram should show a bending moment M(x) and a shear force V(x). Now write sum of the moments about point O and set it equal to zero: (P/2)(x) - (P)(x - L/2) - M(x) = 0. This gives: M(x) = PL/2 - Px/2.

[@wonghueihann2548]

Dr. Structure thanks for replying me ! and it is helpful ! much thanks :)

[@clintonchijioke330]

Internal Energy due to bending_ You did not multiply with the 1/2 outside the integral. confirm?

[@DrStructure]

Are you referring to the equation at time 6:00? Yes, 1/2 has been taken into account in arriving at the result. The same is true for the equation at time 12:00, and for the equation at 13:10.

[@farrasdaffa1418]

hi sorry for late comment, what about solve some indeterminate beam? What is the best method to use? Are there any example about it? @Dr.Structure

[@DrStructure]

Do you mean techniques for calculating displacements in indeterminate beams? One can use the slope-deflection method, or the force method, to analyze an indeterminate beam turning it into a determinate one. Then, use the virtual work principle (or other similar methods) to determine the beam's deflections. You can check out our lectures on the slope-deflection and the force methods for their details.

[@farrasdaffa1418]

Dr. Structure i have tried using force method, but what happens if both of the support are fixed support? How to change it to be a determinate one? And also what about a beam with fixed support(left side) and a roller(right side)? I don’t know which one i have to use to solve a problems.. so much methods ya..

[@DrStructure]

You can use either method (slope-deflection or force) for analyzing indeterminate beams. If a beam is fixed at both ends, the slope-deflection method is a better choice. Why? because the method tells you that the moments at the fixed ends are equal to the fixed-end moments which can be easily calculated using the tables given in most structural analysis textbooks. But, if you want to solve the problem using the force method, you can make the reaction forces at one of the ends redundant, which would turn the beam into a cantilever one. In the case of a beam with a fixed and a roller support, the reaction at the roller can be made the redundant force. Alternatively, you can make the bending moment at the fixed support the redundant force.

[@farrasdaffa1418]

@@DrStructure wow thanks !! hm what about Castigliano Theoreme that makes Work-Energy Method can solve Distribution Load and more than 1 load?(using partial derivative) is it right? do you have the lesson here?

[@DrStructure]

Yes, Castigliano's theorem can also be used to solve displacement problems. We don't have a video on that. But all these methods can handle distributed and multiple loads. Also, one can use the principle of superposition when multiple loads are present. The total displacement at a point due to multiple loads = the sum of the displacements due to each load.

[@laughinggood]

dr, at 8:39 u are telling about the various forces at different section...my doubt is how come at the support A there is no moment,what i meant was at section AC at A there is only forces but at C there is a moment can u please explain???

[@DrStructure]

+Ajay Balan There need not be a moment at A in order to have a moment at C. The moment at C is caused by the horizontal support reaction at A. This reaction force causes a bending moment at C equals to the magnitude of the force (P) times the length of the column (2L). Hence, we have a bending moment of (P)(2L) at C. Think about a cantilever beam having a length of 2L subjected to concentrated load (P) at the free end. The load causes a bending moment of (P)(2L) at the fixed end of the beam. That is what we have here.

[@laughinggood]

+Dr. Structure could u please specify the reason behind the direction of moment...isnt the moment caused by P clockwise???could u pls also tell why there is no moment of mag 2PL at D...could we provide a counter clockwise moment at P???

[@DrStructure]

+Ajay Balan The free-body diagram (FBD) for each member of the frame is shown @9:00. Notice the FBD for member AC. The horizontal force (P) at the bottom of the member is pointing to the left. This means shear force at the top of member must have a magnitude of P and must act in the opposite direction to the force at the bottom so that the force equilibrium equation (sum of the forces in the x direction = 0) is satisfied. Similarly, to satisfy the moment equilibrium equation (sum of moments = 0), bending moment at the top of the member must act counter-clockwise. If it acted in the clockwise direction, then the moment equilibrium equation (say, taken about A) would have given us 4PL instead of zero. The FBD for member CD tells us that there is a downward shear force of 2P and a clockwise moment of 2PL at the left end of the member. This means sum of the moments about the other end of the member is already zero. The shear force causes a counter-clockwise moment of 2PL, added to the clockwise 2PL moment, we get zero. If we place a non-zero moment at the right end of the beam (at D), the moment equilibrium equation cannot be satisfied, it is not going to remain zero.

[@laughinggood]

+Dr. Structure thank u very much.....means a lot :D

[@ssattor]

for member cd, the moment equation is M(x)=2PL-2Px Shouldn't it be M(x)=2Px-2PL instead?

[@DrStructure]

+Sattar Sattarov Depends on your sign convention. Here, the member was assumed to have its tension fiber on top and compression fiber on the bottom.

[@zhiwang6885]

Dr. I don't understand why the enegy by shear had done in frame is negetive.

[@DrStructure]

Not sure what you are referring to, please elaborate. Energy due to shear is assumed to be zero, it is not calculated here.

[@zhiwang6885]

@@DrStructure Dr. at 11.22 ,I saw the shear is not zero but P , I don't understand why the energy due to sheer should be neglected .

[@DrStructure]

In most beams, shear deformation is not significant, compared with rotation and axial deformations, hence we can often ignore it. Further, the proper treatment of shear deformation requires a basic understanding of mechanics of solids which we have not yet covered in these lectures. Eventually, we will get into solid mechanics and discuss the topic then.

[@zhiwang6885]

@@DrStructure Thank you so much Dr. 非常感谢

[@JelenaW]

Dr Structure, please tell me, at 8:03, how do we get that the internal moment at C is 2PL?

[@DrStructure]

Assume the moment at C is unknown. Let's refer to it as Mc. Say, it is acting in the counterclockwise direction at the upper end of member AC. Now write the moment equilibrium equation (for member AC) about A. That is, the sum of the moments about A for member AC must be zero. The equation is: (P) (2L) - Mc = 0. Or, Mc = 2PL.

[@montecarlo5345]

Dear Dr.Structure, why did you not take (2P)(L) at C when you calculated Moment at A? Should it not be (P)(2L) - (2P)(L) - Mc = 0 ?

[@DrStructure]

I am not quite sure what you are referring to, so feel free to elaborate if the answer below is unsatisfactory. @9:05, considering vertical member AB, taking the sum of the moments about A, there are only two forces that contribute to the moment: the shear force at B having a magnitude of P, and the bending moment at B having the magnitude 2PL. The moment at C does not come into the picture here, since joint C is not a part of the free-body diagram for AB. So, the equation becomes: (bending moment at B) - (shear force at B times L) or, (2PL) - (P)(2L) = 0

[@montecarlo5345]

Thanks for the quick response. When we are considering the whole structure at @7:22, there are 4 forces, P and Ay go through C and can be neglected, then we have Ax and By left. So moment in C is (Ax)(2L) - (By)(L) = Mc. Since Ax = P, By = 2P, so we get Mc = 0 ?

[@DrStructure]

Not quite. Mc does not (should not) appear in any of the equations since it is not shown on the free-body diagram we are using to write the equations. Mc is an internal moment and unless we cut the frame at C, the moment is not going to show itself, it is not going to be drawn on the diagram. So, considering the entire frame, the only forces that appear on its free-body diagram are the applied loads and the support reactions. And these are the ony forces that we use to write the equilibrium equations. The moment equilibrium equation about Point C should read: the sum of the moments about C equals zero, not equal Mc. So, we write: (Ax)(2L) - (By)(L) = 0 In this problem, Mc is not actually zero. To determine it, after calculating the support reactions, we can cut the left column very close to C, and draw the free-body diagram for member AC. The diagram shows a horizontal foce of P and a vertical force of 2P at A, and a shear force of Cx, axial force of Cy and bending moment Mc at the top. Now, taking the sum of the moments about C based on the forces on that diagram, we get: Mc = 2PL.

[@DM-ce6su]

Where is the constant of integration?

[@DrStructure]

Say the function that we need to integrate is f(x). Let's refer to its integral as g(x) + C where C is the integration constant. Our integral here is bounded, we are evaluating it between a and b. That means, we get: [ g(b) + C ] - [ g(a) + C] = g(b) - g(a). The constant of integration vanishes.

[@DM-ce6su]

ah,so my calculus skills failed me.Thanks for the reply and keep up the great work!

[@patrisagar6548]

9:21 how is it negative 2P ? It should be 2P isn't it?

[@DrStructure]

Since the member is in compression, we take the force as being negative. A tensile force, on the other hand, is considered to be positive.

[@patrisagar6548]

@@DrStructure thanks

[@Islam_Hanafi]

can you tell me what books do you use ??

[@DrStructure]

+ Islam Hanafi This content does not come from any specific textbook. But if you are looking for a textbook recommendation, I suggest Structural Analysis by Russell Hibbeler.

[@Islam_Hanafi]

+Dr. Structure ok , thanks i have another question please , in the virtual work method can we develop a simple way to integrate ?

[@DrStructure]

+ Islam Hanafi No, there are no shortcuts for the integration. If the beam is subjected to several concentrated and distributed loads, the moment equation may becomes lengthy making the integration time consuming. But you can use a software tool, like Mathematica, for performing symbolic integration.

[@Islam_Hanafi]

in our class we sue another method i cant recognize we draw the bending moment due to loads as Mo and draw bending moment due to unit force and integrate the two areas Mo and M1 got it ?

[@kanduridinesh2608]

@@Islam_Hanafi is that moment area method?

[@civilideas1925]

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