I have a degree in chemistry and never took a course in mechanical engineering. Now that I am retired I have been reading this lessons of beam deflection in the Dr Structure and videos,even more, practicing the examples given. I am amazed of the clarity
Yes, m* would still be necessary. The only time that m* would not be necessary is if it equals 1. But, m* (bending moment equation in the beam) due to a unit load at the mid-span is not 1. So, we cannot just get rid of it. If the applied load (P) was at mid-span, and we wanted to calculate the deflection at mid-span, then we could write/say: M= P m*. Or, delta = P ( integral of m* m* /EI dx).
Question maam. Based on illustration at 5:28, the m* is at the vertex angle d(theta)? But when we use the virtual work formula, m* lies on x axis, m*(x). Im confused about this one.
M and m* both are bending moments in the beam segment. Note the small bent segment shown @5:28 where the radius of curvature is labeled as rho and the angle of bending is denoted as d(theta). The segment has been bent about the beam's neutral axis (shown using dashed lines). The cross section of that segment is also shown, just to the left of the diagram. Note the axis of bending of the cross-section. Both M and m* act about that axis, causing the segment to bend. They don't act at any vertex. The shown small segment can represent any part of the beam where M and m* are always present. But their numeric values depend on the location of the segment, which we denote as x. Put it differently, at x = 1, M and m* could have different values than at x = 2, or x = 3.5, ... Yet, regardless of the x value, when we cut the beam and isolate a tiny segment of it, we end up with the diagram shown @5:28 in which M and m* produce the bending of the segment. The difference between M and m* is that the former is caused by the real load and the latter by the virtual unit load.
No, the given moment equation is correct. If we had a unit vertical load acting at point A, m*(x) would have been (1, the unit load) times (x, the moment arm). But here, we don’t have a unit load. Rather, we have a unit moment. The moment of the unit moment at any point in the beam is: the unit moment. We don’t multiply the moment by a moment arm to get a moment. That would be incorrect.
hii Dr.structure , 1. In virtual work p* = 1 , it means we are assuming that 1N load won't make further deformation . What is the value of p*, if 1 N load is able to deform the structure ( it can happen in certain situations like if put load on my steel ruler of 30*8*4 cm3 etc.., ).. ? 2 . Generally in practical scenario , a load or force can be generate by applying mass ( F=mg), where as how can I generate moment at the end of section and at the middle of section ? 3.If a cycle wheel is located on spindle through bearing and wheel is rotating . force is acting on the wheel ( I.e tangential to diameter axis ) . To study the spindle , I want to transport that force to spindle because the wheel is on spindle and . I read the center will have moment and force . Can you explain this ?
The value of p* is not a factor. Before p* can cause any additional deformation, in needs to go through the existing deformation causing work to be done. It is that work/energy that we are considering in the formulation. If you wanted to actually cause a real bending moment at a given point in a beam, you can apply a load to the end of a lever rigidlyattached to that point. In real structures, a bending moment at the beam’s end is usually caused by (transferred from) an adjacent member. Generally speaking, the spindle is subjected to a torsion which can be calculated either from the tangential (friction) force between the wheel and the ground, or the tension in the chain caused by the force applied to the pedal. For example, if the friction force is F and the radius of the wheel is r, then the torque is (F)(r). There is also a vertical force at each end of the spindle cause by the weight of the person riding the bike which needs to be transferred to the ground through the wheel.
Dr. Structure I understood the points 2 and 3 but in 1 existing deformation means ... ? is it only at point of application of force or complete deformation of beam
To be sure, the entire beam deforms under the real load. But the deformation at a point other than the point of application of the virtual load does not play in role in the formulation. Work is done only under the (virtual) load. Since work = force x displacement, even if displacement is not zero but the force is zero, no work is done. So, beam deformation except under the virtual load does not enter into the formulation since those deformations do not result in work.
Can you make a video on explaining the analogy between strain energy method(Castiglano's theorem) and virtual work? Because the m* term in this video reminds me of the (dM/dP) term from the strain energy method.
Why is the deflection value at mid span is greater than at the point where "P" is applied? i.e: delta at mid span=22/3 (P/EI) delta at point "P"= 6 (P/EI) I thought the downward effect of force "P" at its point of application is more than that of mid span? Does that make sense?
Consider a simply supported beam of Length L. Let's assume the beam is subjected to a concentrated load of P at distance (d) from the left end of the beam such that d >= L/2. That is, the load is applied to the right half of the beam. Where is the location of maximum deflection in this beam? We know that when deflection is at its maximum value slope of the elastic curve is zero. So, where is slope zero? If we derive the necessary equations in order to determine the location of zero slope. We get: 3 x^2 = 2 d L - d^2. where x is the location of zero slope (maximum deflection), d is the load location and L represents length of the beam. So, is it true that maximum deflection always occurs under the load? No. According to the above equation, maximum deflection is under the load when: 3 d^2 = 2 d L - d^2 Or, 4 d^2 = 2 d L Or, 2d = L Or, When d = L/2 (the load is located at mid-span). Otherwise, the location of maximum deflection is given by equation: x^2 = (2d L - d^2)/3.
The bending moment in the beam is considered positive when it causes the member to deflect in a concave up manner (top finer in compression and bottom fiber in tension). Since the deflection of a cantilever beam subjects to a downward load is concave down (top fiber in tension and bottom fiber in compression), the bending moment is considered negative.
sorry to trouble you again, am i right in thinking for "uniform load" on beam, the standard procedure of work energy method couldn't be used to solve and we must proceed to using this virtual work method ?
can you please help me out on this example ? imgur.com/I58e7Hl i try constructing a free body diagram like below, is that correct ? i.imgur.com/reGPSS1.png
@@DOTA2MAJISTRATE Here you have a three dimensional system. However, the diagrams show 2D drawings. It is not clear what part of the 3D system is under consideration when drawing the free-body diagram.
great lecture and great channel all the way! but to give some feedback: You should stick with the letter you assign at first, in lecture 19 you choose "r" to represent the arc radius of the beam piece, in this lecture "rho" is used, even though the animation shows "r"
The link to each solution is provided in the video (the circled i at the upper right corner of the video). Unfortunately, there is no printed version of the solutions available for download.
this method is what i mean in my previous comment from SA24. Is this method can result the same or closely as double integration method? what is the unit for delta and the slope, is it length and degree/radian units?
Yes, the virtual work method yields the same results as the double integration method. In the virtual work method, delta (displacement) has the unit of length, and theta (rotation) is in radians.
Not sure what you mean by violating any scientific principle. Please elaborate. A (virtual) load can have any magnitude. For connivence, we use a unit load.
@@DrStructure thankyou, Its all clear now. When I ask the question, I taught that a virtual load has certain magnitude that depends on the external load. So I taught that assuming a diferent magnitude such as unit load will make the analysis scientifically wrong.
No, we don't. The focus of these lectures is introductory structural analysis. Consequently, more advanced topics are not going to be covered until we exhaust our introductory coverage.
We have a simply supported beam subjected to a point load (p*) at the midpoint. Therefore, the support reactions is (p*)/2. The moment equation for the left segment of the beam where 0 < x < 4 is: (P*)(x)/2. The moment equation for the right segment where 4 < x < 8 is: (p*)(x)/2 - (p*)(x-4). This equation simplified to: 4p* - p*x/2.
We don't beforehand. We can assume any direction for the virtual moment. If the calculated rotation comes out positive, the assumed direction is correct. Otherwise we made a wrong assumption. That is, now we know that if we reverse the direction of the moment, the rotation comes out positive. It is important to note that the direction of the rotation is the same as the direction of the unit moment. So, a clockwise moment placed on the beam means we are assuming the rotation is clockwise too. A counterclockwise moment means we are assuming a counterclockwise rotation.
The principle of virtual load assumes that the structure has already been deformed. It is the deformed structure that we subject to a virtual load. The 1/2 factor is work done by a load that causes a deformation. But if we are starting with a deformed structure already, and simply apply a virtual load to it (a load that would not cause any additional deformation), then the 1/2 factor does not come into the picture.
It depends on the location of the origin. If we pick A as the origin, we get the solution given in the video. If we pick B as the origin, we get your solution. Either way works as long as you apply it consistently throughout the solution.
3:22 in the previous video you said internal energy for beams is (1/2) integral m×d theta but here you are saying its integral m* × dtheta. where the (1/2) go ?
This is the same response that was offered for your previous question on SA21. For the sake of completeness, it is being posted here as well. Real work done by a force equals (P)(D)/2 where P is the force and D is the displacement in the direction of the force. We derived this expression in Lecture SA19. In that derivation, we started by placing a small force f on the beam and referred to the displacement caused by that tiny force as d. We then expressed work as (f)(d). We showed that by keep adding the small load increment f, we will eventually arrive at the expression (P)(D)/2. Basically, the virtual work method established the work-energy relationship based on f and d, indicating that the “virtual” work produced by the small load increment (f) must be equal to the internal energy caused by f. In that derivation the 1/2 factor does not come into the picture, yet the equality between work and energy holds true. Alternatively, you can think about it in the following way. A structure has real deformation under an applied load. Let’s refer to the load as P and to the resulting displacement under the load as D. After the deformation takes place, suppose somehow we can keep the structure in its deformed shape while removing the load. So, D remains present, yet P is no longer there. Now, if we place a virtual load (a very small load) in the position and direction of P, let’s call it p*, since p* goes through displacement D, it produced (p*)(D) work. The 1/2 factor is not going to be present here, since D was not caused by p*, the displacement was present by the time p* appeared on the scene. So p* simply travels through the existing distance (D) producing work: (P)(D). We refer to this product as virtual work, not real work. According to the law of conservation of energy, this virtual (external) work must be equal to the virtual internal energy/work caused by p*.
No, you can view the virtual load to be very small and applied in one chunk. After all, it is not a real load and it is not causing a real displacement.
Thanks for the note. The video shows the complete solution, it shows how to determine the deflection and the slope. It just goes blank afterward. But, we'll look into it.
Dr. Structure.... do you have other content like RCC design,,,geotechnical,,,,Transportation........ I am a student of Shahjalal university of science & technology.... your video is very helpful... it helps to understand the concept easily..... thank you
The flexibility methods of analysis invariably involve solving for unknown forces. The displacement methods of analysis use displacements/rotations as unknowns. The Force Method can be viewed as a flexibility method whereas the Slope-Deflection Method is a displacement method. The stiffness method is generalized version of the slope-deflection method.
Almost! According to our sign convention, the moment equation for the right half of the beam (@3:00) should be: px/2 - (x-4)p This slightly differs from the equation you gave above; One is the negative of the other. This equation can also be written as: px/2 - px + 4p. Or, 4p - px/2, which is the same as the one given in the video.
Same as SA21 ... Solution for Exercise Problem 1: ruclips.net/video/ZR9FcXIpK5k/видео.html Solution for Exercise Problem 2: ruclips.net/video/_Ihs8_OHjis/видео.html
No! where does that second P/2 come from? x is the distance from the left end of the beam. Going from left to right, we have (P/2) for the support reaction at the left end of the beam, and (P) for the applied load. The right support reaction (P/2) does not appear in that free-body diagram.
The left reaction is 1/2, its moment arm is x. And, the virtual load has a moment arm of (x - 4), assuming the beam has been cut at some distance to the right of the load. x is the distance from the left support to the cut point. (1/2)(x) - (1)(x - 4) = 4 - x/2
Because the beam has a concave down deflection. By convention, we take any bending moment that causes concave up deflection as positive, and a moment the cause concave down deflection as negative.
Set the origin of the coordinate system at the left end of the beam, x-axis being positive to the right. Now cut the beam at an arbitrary point along the x-axis, say at point O. Take the left segment, the segment of the beam between the origin and O, and draw the internal forces at O. There would be a shear force and a bending moment. Assume both are being positive. Label the bending moment M. A positive shear force, by convention, tends to rotate the segment in the clockwise direction. This means the shear force at O should be pointing downward for it to cause a clockwise rotation. A positive moment should cause a concave up deflection. This means moment at O should be drawn in the counterclockwise direction. If you are not sure why that is, experiment. Draw the beam segment and place these forces on it and visualize how the segment would rotate and deflect. Now, write the equilibrium equations, more specifically, the moment equation and solve for the unknown moment at point O, you would get: M = - w x x/2.
you people are better than most of phd doctors in my country
How concise but clear way you used to convey this complex stuff. Wonderful! Appreciated.
Thank you so much!! can't believe you are still replying questions after all these years!!
Thank you, Jackson! It has been a real pleasure interacting with students from all corners of the world.
Now I m falling in love with virtual work method ... Thank you
I have learnt a lot of from you....better than college
I have a degree in chemistry and never took a course in mechanical engineering. Now that I am retired I have been
reading this lessons of beam deflection in the Dr Structure and videos,even more, practicing the examples given.
I am amazed of the clarity
@@joserios4159 Thank you for your feedback!
Can we expect this sort of lectures for other civil engineering courses
If the actual load is already at midspan is the m* still necessary? or just disregard that and have delta= integral of (M/EI)(dx)?
Yes, m* would still be necessary. The only time that m* would not be necessary is if it equals 1. But, m* (bending moment equation in the beam) due to a unit load at the mid-span is not 1. So, we cannot just get rid of it.
If the applied load (P) was at mid-span, and we wanted to calculate the deflection at mid-span, then we could write/say: M= P m*. Or,
delta = P ( integral of m* m* /EI dx).
Just summarized hours of lectures in minutes. Thanks!!
The way you express the bending moments (ex m = x/2 when 0
Yes!
@@DrStructure Thank you so much
you guys are the best if only if u guys have all other materials i would be an engineer already xD
Question maam. Based on illustration at 5:28, the m* is at the vertex angle d(theta)? But when we use the virtual work formula, m* lies on x axis, m*(x). Im confused about this one.
M and m* both are bending moments in the beam segment. Note the small bent segment shown @5:28 where the radius of curvature is labeled as rho and the angle of bending is denoted as d(theta). The segment has been bent about the beam's neutral axis (shown using dashed lines). The cross section of that segment is also shown, just to the left of the diagram. Note the axis of bending of the cross-section. Both M and m* act about that axis, causing the segment to bend. They don't act at any vertex.
The shown small segment can represent any part of the beam where M and m* are always present. But their numeric values depend on the location of the segment, which we denote as x. Put it differently, at x = 1, M and m* could have different values than at x = 2, or x = 3.5, ... Yet, regardless of the x value, when we cut the beam and isolate a tiny segment of it, we end up with the diagram shown @5:28 in which M and m* produce the bending of the segment.
The difference between M and m* is that the former is caused by the real load and the latter by the virtual unit load.
Thank god! I love you eventhough you are a robot
8:44 hi,is the bending moment of the unit load applied not supposed to be -1*x ?
No, the given moment equation is correct.
If we had a unit vertical load acting at point A, m*(x) would have been (1, the unit load) times (x, the moment arm).
But here, we don’t have a unit load. Rather, we have a unit moment. The moment of the unit moment at any point in the beam is: the unit moment. We don’t multiply the moment by a moment arm to get a moment. That would be incorrect.
@@DrStructure alright,I understand now,thank you
hii Dr.structure ,
1. In virtual work p* = 1 , it means we are assuming that 1N load won't make further deformation . What is the value of p*, if 1 N load is able to deform the structure ( it can happen in certain situations like if put load on my steel ruler of 30*8*4 cm3 etc.., ).. ?
2 . Generally in practical scenario , a load or force can be generate by applying mass ( F=mg), where as how can I generate moment at the end of section and at the middle of section ?
3.If a cycle wheel is located on spindle through bearing and wheel is rotating . force is acting on the wheel ( I.e tangential to diameter axis ) . To study the spindle , I want to transport that force to spindle because the wheel is on spindle and . I read the center will have moment and force . Can you explain this ?
The value of p* is not a factor. Before p* can cause any additional deformation, in needs to go through the existing deformation causing work to be done. It is that work/energy that we are considering in the formulation.
If you wanted to actually cause a real bending moment at a given point in a beam, you can apply a load to the end of a lever rigidlyattached to that point. In real structures, a bending moment at the beam’s end is usually caused by (transferred from) an adjacent member.
Generally speaking, the spindle is subjected to a torsion which can be calculated either from the tangential (friction) force between the wheel and the ground, or the tension in the chain caused by the force applied to the pedal. For example, if the friction force is F and the radius of the wheel is r, then the torque is (F)(r). There is also a vertical force at each end of the spindle cause by the weight of the person riding the bike which needs to be transferred to the ground through the wheel.
Dr. Structure I understood the points 2 and 3 but in 1
existing deformation means ... ? is it only at point of application of force or complete deformation of beam
To be sure, the entire beam deforms under the real load. But the deformation at a point other than the point of application of the virtual load does not play in role in the formulation.
Work is done only under the (virtual) load. Since work = force x displacement, even if displacement is not zero but the force is zero, no work is done. So, beam deformation except under the virtual load does not enter into the formulation since those deformations do not result in work.
Dr. Structure thank you :)
Can you make a video on explaining the analogy between strain energy method(Castiglano's theorem) and virtual work? Because the m* term in this video reminds me of the (dM/dP) term from the strain energy method.
Why is the deflection value at mid span is greater than at the point where "P" is applied?
i.e: delta at mid span=22/3 (P/EI)
delta at point "P"= 6 (P/EI)
I thought the downward effect of force "P" at its point of application is more than that of mid span? Does that make sense?
Consider a simply supported beam of Length L. Let's assume the beam is subjected to a concentrated load of P at distance (d) from the left end of the beam such that d >= L/2. That is, the load is applied to the right half of the beam.
Where is the location of maximum deflection in this beam?
We know that when deflection is at its maximum value slope of the elastic curve is zero. So, where is slope zero?
If we derive the necessary equations in order to determine the location of zero slope. We get:
3 x^2 = 2 d L - d^2.
where x is the location of zero slope (maximum deflection), d is the load location and L represents length of the beam.
So, is it true that maximum deflection always occurs under the load? No.
According to the above equation, maximum deflection is under the load when:
3 d^2 = 2 d L - d^2
Or,
4 d^2 = 2 d L
Or,
2d = L
Or,
When d = L/2 (the load is located at mid-span).
Otherwise, the location of maximum deflection is given by equation:
x^2 = (2d L - d^2)/3.
Can you please explain why M(x)=-wx^2/2.. why is there a negative sign here.. isn't it suppose to be positive? 7:25
The bending moment in the beam is considered positive when it causes the member to deflect in a concave up manner (top finer in compression and bottom fiber in tension).
Since the deflection of a cantilever beam subjects to a downward load is concave down (top fiber in tension and bottom fiber in compression), the bending moment is considered negative.
sorry to trouble you again, am i right in thinking for "uniform load" on beam, the standard procedure of work energy method couldn't be used to solve and we must proceed to using this virtual work method ?
can you please help me out on this example ?
imgur.com/I58e7Hl
i try constructing a free body diagram like below, is that correct ?
i.imgur.com/reGPSS1.png
correct, the virtual work method is a more general approach for calculating slope and deflection in beams.
@@DOTA2MAJISTRATE Here you have a three dimensional system. However, the diagrams show 2D drawings. It is not clear what part of the 3D system is under consideration when drawing the free-body diagram.
great lecture and great channel all the way!
but to give some feedback: You should stick with the letter you assign at first, in lecture 19 you choose "r" to represent the arc radius of the beam piece, in this lecture "rho" is used, even though the animation shows "r"
Thanks for the feedback.
very great lecture but i have one question where can I get the manual solved problem of the exercise that u gave us at the end of the videos
The link to each solution is provided in the video (the circled i at the upper right corner of the video). Unfortunately, there is no printed version of the solutions available for download.
Is this the Betti's theorem ? At least, in my country (Italy) is known as Betti's theorem
Yes, it is Maxwell-Betti’s theorem turned into a method.
this method is what i mean in my previous comment from SA24. Is this method can result the same or closely as double integration method? what is the unit for delta and the slope, is it length and degree/radian units?
Yes, the virtual work method yields the same results as the double integration method. In the virtual work method, delta (displacement) has the unit of length, and theta (rotation) is in radians.
Does shear force have no role in deflection of beam ? Why we ignore it ?
Correct, deformation due to shear in most beams is negligible.
5:37 why we can assume that the virtual load is a unit load? is it not voilating any science principles?
Not sure what you mean by violating any scientific principle. Please elaborate.
A (virtual) load can have any magnitude. For connivence, we use a unit load.
@@DrStructure thankyou, Its all clear now. When I ask the question, I taught that a virtual load has certain magnitude that depends on the external load. So I taught that assuming a diferent magnitude such as unit load will make the analysis scientifically wrong.
Excuse me ! Do you have any video on geometric stiffness matrix (36 3L 36 3L) ,?
No, we don't. The focus of these lectures is introductory structural analysis. Consequently, more advanced topics are not going to be covered until we exhaust our introductory coverage.
can you explain how you got the virtual and real moment equations at 2:55
We have a simply supported beam subjected to a point load (p*) at the midpoint. Therefore, the support reactions is (p*)/2.
The moment equation for the left segment of the beam where 0 < x < 4 is:
(P*)(x)/2.
The moment equation for the right segment where 4 < x < 8 is:
(p*)(x)/2 - (p*)(x-4). This equation simplified to:
4p* - p*x/2.
Where can I find the solutions for the last two problems?
Click on the circled i at the upper right corner of the video.
Thanks!
You're welcome.
how do you know whether the applied virtual moment at A 8:35 is clockwise or counterclockwise?
We don't beforehand. We can assume any direction for the virtual moment. If the calculated rotation comes out positive, the assumed direction is correct. Otherwise we made a wrong assumption. That is, now we know that if we reverse the direction of the moment, the rotation comes out positive. It is important to note that the direction of the rotation is the same as the direction of the unit moment. So, a clockwise moment placed on the beam means we are assuming the rotation is clockwise too. A counterclockwise moment means we are assuming a counterclockwise rotation.
@@DrStructure Thanks! I really appreciate the content & response.
@@edwardplumb8071 You're welcome!
In the above lecture no need to use half before work like avg work
Perfect👍🏻
Why is the formula for the external virtual work, not 1/2*P*delta?
The principle of virtual load assumes that the structure has already been deformed. It is the deformed structure that we subject to a virtual load.
The 1/2 factor is work done by a load that causes a deformation.
But if we are starting with a deformed structure already, and simply apply a virtual load to it (a load that would not cause any additional deformation), then the 1/2 factor does not come into the picture.
Dr. Structure woowwwww a very good explanation Dr. Thank youuuu
Hi , for cantilever beam example my solution for virtual moment =x-L why it s different from yours
It depends on the location of the origin. If we pick A as the origin, we get the solution given in the video. If we pick B as the origin, we get your solution.
Either way works as long as you apply it consistently throughout the solution.
3:22 in the previous video you said internal energy for beams is (1/2) integral m×d theta but here you are saying its integral m* × dtheta. where the (1/2) go ?
This is the same response that was offered for your previous question on SA21. For the sake of completeness, it is being posted here as well.
Real work done by a force equals (P)(D)/2 where P is the force and D is the displacement in the direction of the force. We derived this expression in Lecture SA19. In that derivation, we started by placing a small force f on the beam and referred to the displacement caused by that tiny force as d. We then expressed work as (f)(d). We showed that by keep adding the small load increment f, we will eventually arrive at the expression (P)(D)/2.
Basically, the virtual work method established the work-energy relationship based on f and d, indicating that the “virtual” work produced by the small load increment (f) must be equal to the internal energy caused by f. In that derivation the 1/2 factor does not come into the picture, yet the equality between work and energy holds true.
Alternatively, you can think about it in the following way. A structure has real deformation under an applied load. Let’s refer to the load as P and to the resulting displacement under the load as D. After the deformation takes place, suppose somehow we can keep the structure in its deformed shape while removing the load. So, D remains present, yet P is no longer there. Now, if we place a virtual load (a very small load) in the position and direction of P, let’s call it p*, since p* goes through displacement D, it produced (p*)(D) work. The 1/2 factor is not going to be present here, since D was not caused by p*, the displacement was present by the time p* appeared on the scene. So p* simply travels through the existing distance (D) producing work: (P)(D). We refer to this product as virtual work, not real work. According to the law of conservation of energy, this virtual (external) work must be equal to the virtual internal energy/work caused by p*.
I thought since the virtual load is gradually applied then work becomes 1/2 int m* dtheta
No, you can view the virtual load to be very small and applied in one chunk. After all, it is not a real load and it is not causing a real displacement.
Solution of Exercise 2 has some problem...Last part of the video doesn't appears....please solve this problem
Thanks for the note. The video shows the complete solution, it shows how to determine the deflection and the slope. It just goes blank afterward. But, we'll look into it.
Dr. Structure.... thank's a lot....waiting for the video...
The revised video for Exercise 2 can be accessed by clicking on the upper right corner of the main video (on the circled i).
Dr. Structure.... do you have other content like RCC design,,,geotechnical,,,,Transportation........ I am a student of Shahjalal university of science & technology.... your video is very helpful... it helps to understand the concept easily..... thank you
Not at the moment, but we will eventually get to them. Although by then, you will already have graduated and have started your professional career. :)
cool
What difference between flexibility method and stiffness method?
The flexibility methods of analysis invariably involve solving for unknown forces. The displacement methods of analysis use displacements/rotations as unknowns. The Force Method can be viewed as a flexibility method whereas the Slope-Deflection Method is a displacement method.
The stiffness method is generalized version of the slope-deflection method.
Shouldn't the moment equation be (x-4)*p* - x*p*/2
Almost!
According to our sign convention, the moment equation for the right half of the beam (@3:00) should be:
px/2 - (x-4)p
This slightly differs from the equation you gave above; One is the negative of the other.
This equation can also be written as:
px/2 - px + 4p.
Or,
4p - px/2, which is the same as the one given in the video.
@@DrStructure Thanks for the answer. Helped clear my doubt
Are there answers to the exercises ?
Same as SA21 ...
Solution for Exercise Problem 1: ruclips.net/video/ZR9FcXIpK5k/видео.html
Solution for Exercise Problem 2: ruclips.net/video/_Ihs8_OHjis/видео.html
Can you post the links for example solution here
Exercise 1 Solution: ruclips.net/video/ZR9FcXIpK5k/видео.html
Exercise 2 Solution: ruclips.net/video/y9984N1lKDM/видео.html
Dr. Structure thank you
at 2:54 for m*(x) how 4P*-(P*x)/2 comes for 4
(P/2)(x) - P(x-4) = 4P - (P/2)(x)
it should be (P/2)(x)-(P/2)(x-4).....right?
No! where does that second P/2 come from?
x is the distance from the left end of the beam. Going from left to right, we have (P/2) for the support reaction at the left end of the beam, and (P) for the applied load. The right support reaction (P/2) does not appear in that free-body diagram.
am sorry but it should be (P/2)(x)-(P/2)(x-4)
because the applied p load is (x-4) distance from the cut isnt it?
is this Maxwll's Theorem ?
No, Maxwell's Reciprocal Theorem is not the same as the Virtual Work Principle.
How is m*(X)= 4 - X/2 4
The left reaction is 1/2, its moment arm is x. And, the virtual load has a moment arm of (x - 4), assuming the beam has been cut at some distance to the right of the load. x is the distance from the left support to the cut point.
(1/2)(x) - (1)(x - 4) = 4 - x/2
@@DrStructure thnx..I took the moment arm as (X+4)
how come M= - wx^2/2 why negative
Because the beam has a concave down deflection. By convention, we take any bending moment that causes concave up deflection as positive, and a moment the cause concave down deflection as negative.
How can we identify if it has a concave down or up deflection
Set the origin of the coordinate system at the left end of the beam, x-axis being positive to the right. Now cut the beam at an arbitrary point along the x-axis, say at point O. Take the left segment, the segment of the beam between the origin and O, and draw the internal forces at O. There would be a shear force and a bending moment. Assume both are being positive. Label the bending moment M.
A positive shear force, by convention, tends to rotate the segment in the clockwise direction. This means the shear force at O should be pointing downward for it to cause a clockwise rotation. A positive moment should cause a concave up deflection. This means moment at O should be drawn in the counterclockwise direction. If you are not sure why that is, experiment. Draw the beam segment and place these forces on it and visualize how the segment would rotate and deflect.
Now, write the equilibrium equations, more specifically, the moment equation and solve for the unknown moment at point O, you would get: M = - w x x/2.
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