SA21: Virtual Work Method (Trusses)
ฝัง
- เผยแพร่เมื่อ 15 ต.ค. 2024
- This lecture is a part of our online course on introductory structural analysis. Sign up using the following URL: courses.struct...
In addition to updated, expanded, and better organized video lectures,
the course contains quizzes and other learning content.
Solution for Exercise Problem 1: • SA21-1: Exercise 1
Solution for Exercise Problem 2: • SA21-2: Exercise 2
![[Gegagedigedagedago] NEW Help Gegagedigedagedago Nugget escape from Nikocado Avocado Challenge](http://i.ytimg.com/vi/hJWnwRC2Fz8/mqdefault.jpg)
You nailed it. After six years of my graduation, i am feeling this concept easy
Appreciate that you save my " structural Analysis " exam next week. Thanks Dr. Structure.
displacements=sum (f*.$), where $=(FL/EA), so;
dis.=sum (f*.F.L/EA), this is easier to solve.
f*: internal force due to the unit load(virtual)
F: real internal force
L: length of real member
$: Elongation
Thanks for the great work
I think you made a mistake at 10:00 when you calculate the internal virtual work. Shouldn't the virtual forces in members 2 (f2*) and 4 (f4*) be multiplied by 4.0*10^-6 instead of 3.5*10-6? I really liked the video btw.
Yes, you are right. Thanks for pointing the error.
Please let me know why virtual external work done is not 1/2*(P')*Del? Why didn't we use1/2 here?
of course, it will cancelled out from both sides as we calculate virtual work done in members as well, I am just curious why didn't we use it.
Suppose a beam has been displaced by D at some point (x) due to the applied loads. Imagine that we can remove the applied loads but keep the beam in its displaced form. Now, if we place a virtual load (f) at point x in the direction of D, the load going through the existing displacement D, we say, produces virtual work of fD. Since f is not causing D but simply “traveling” through it, the 1/2 factor does not come into play.
@@DrStructure Thank you so much. 🙏🏼
3:46 why is it p*∆v why not (1/2) p*∆v ? Also when should we take force×displacement and when (1/2) force × displacement for work please explain
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*.
This is good. How did you calculate f* in each case by placing unit load at desired point? Do you have a video which shows how to calculate that?
(f*) forces can be calculated using a truss analysis technique such as the method of joints or sections. Just analyze the statically determinate truss when it is subjected to a unit load and that gives the member forces (f*s).
We have a number of lecture on truss analysis on this channel. Alternatively, you can access them through our online course (link given in the project description field).
Good day
I have a question of a question that is externally statically indeterminate which is requiring me to find the deflection at a point with a give load. So my question is on getting the forces on the truss so that i can calculate the member elongations before i calculate the deflections.
So do i have to use SA26 first to calculate the forces of an indeterminate truss then i can use SA21 to calculate the deflections?
If there is a simpler way or short cut, please assist.
Yes, you need to analyze the truss first, for example using the force method (SA26), before you can determine the displacement using the virtual work method. Alternatively, you can use the matrix (displacement) method which gives the joint displacements as well as member forces for a statically indeterminate system, see SA48 for an explanation of its use for truss analysis.
@@DrStructure Thank you so much.
I thought there was a simpler way as the question was under multiple choice, so i thought less work was needed. But well let me take the long way and see if i win.
Can you elaborate on what exactly was done at 9:40 (Truss analysis) to get the values on the members. Thank you for this video also, really helpful.
This geometry of the truss and its supports were defined before. Starting @6:39, the video shows how to calculate the support reactions when the truss is subjected to two concentrated loads (500 N and 200 N). Then, starting @7:00 the procedure for calculating member forces is shown, again for the two applied loads.
Starting @9:40, we want to analyze the same truss but under a unit load only. So, remove the applied loads (500 N and 200 N), place a unit load at C and follow the same procedure as before to get the member forces.
@@DrStructure distances will remain same right? Just unit load will be applied right?
BTW thank you so much for these lectures I wish I would have known this channel 2 years before, my life would've been completely different then!!
And yes please upload other subjects also.... ❤❤
More power to you Doc! 🙌
@@DK-db4ot Yes, the distances remain the same. And thanks for the feedback. :)
completely threw me off do you add in the same pin and roller as well
Support from Hong Kong!!
Thanks Dr. Structure. But I have a question, how can you handle a case where the deflection is needed at somewhere in the middle of a member of a truss; not at the joint.
Since truss members don't bend, they just displace (translate and rotate), if we know the displacements at the ends of the truss member, we can use simple linear interpolation to determine the displacement at a specific point within the member.
@@DrStructure How I wish you sent a simple illustration on the same. Or I send you a sample quiz. I'll appreciate. farashanongeri@gmail.com
Hi, Dr. Structure. You mean I come up with a graph then an equation of the same which is in the form of y=mx+c
Yes, this is a pure geometry problem. We know the end positions of the truss member. These are two points in the xy-coordinate system. So, we can construct a line equation (i.e., y = mx + c) for the member. The equation enables us to determine the x and y position on every point on the member.
If the ends of the member have displaces by some amount, say the left end of the member is located at (x1, y1) and had displaced by (dx1, dy1). Then, the new (displaced) position of the left end of the member is point (x1+dx1,y1+dy1). The same is true for the right end of the member. Its new/displaced position can be expressed as (x2+dx2,y2+dy2).
Knowing these two new points, we can come up with a linear equation (of the form y = mx + c) for the displaced member. This equation can then be used to determine the displacement at any point within the length of the member.
@@DrStructure Wow!!! Superb. I wanna try this. Thanks for your continued help.
shouldnt the work done around 5:08 be (1/2)*P*(delta)..?
The virtual load is always placed on the deformed structure, meaning delta (due to the real load) has already taken place. So, the virtual load travels through an existing displacement causing work to be done. This work does not have the 1/2 factor associated with it. The 1/2 factor comes into the picture if the force is the cause of the displacement.
Dr. Structure thank you for your clarification sir
sorry but the link for the solution is private, where else can I find the solution
The solution to the exercise problems can be found in the (free) online course referenced in the video description field.
Where can I find the solution Dr. Structure? Thanks in advance. Great Explanation, I have done the two problems, I just would like to check if my answers match.
The links are given in the video description field.
@@DrStructure thanks so much
You're welcome.
At 4:08 I find it counter-intuitive that internal virtual work for member 1 is f1star*delta1 (virtual force times real displacement). Why isn`t there a contribution from f1star*delta1star (virtual force times virtual displacement)?
Right, the underlying concept could be a bit confusing. Imagine a rigid body (in this case it is a truss, but it could be any sort of 2D or 3D body). When we subject the body to some external loads, it is going to deform. Now imagine that somehow we can keep that body in that deformed shape/position, but remove the loads that have caused the deformation. Or, if you like just ignore the loads. The principle of virtual work simply states that given such a state of deformation, if we place an imaginary (virtual) load on the body, the external work done by that virtual load equals to the internal energy produced by the load. Whether or not the virtual load creates additional deformation is not significant. The principle holds true for the existing (real) deformation, without any additional deformation added to the system. That is, the virtual load (without causing any additional deformation) going through the real deformation produces virtual external work. That work equals to the internal energy caused by the same load. The internal energy being the virtual internal forces/stresses going thought the real internal deformations/strains.
Thanks for the video, could you show me the answers of exercise ? (When I click the link, it say videos are in private) Thanks
The solutions are provided in the (free) course referenced in the video description field.
For exercise 1, horizontal displacement at A = 19.14mm
Horizontal displacement at C = 2.5mm.
Vertical displacement at A calculation is a bit confusing....
Are the answers correct????
Yes, the numbers for Exercise 1 seem to be correct except for the unit. It should read m, no mm. Your numbers are not the same as those given in the video, even for the horizontal displacements. Why the discrepancy? What makes you think something is wrong with the given solution?
@@DrStructure but the solutions are not provided anywhere....& that's why I asked it to u....
You should see a circled i at the upper right corner of the video where the two exercise problems are presented. Click on the i. I believe the i should be visible in all platforms, desktop, mobile,...
Please how can i determined horizontal displacement at B
We place a horizontal virtual load at B and follow the procedure explained in the lecture.
Why create a virtual load P* in a place where a REAL load "Q" exists?
Conceptually, the virtual load is different from the real load. The principle of virtual work cannot be properly explained without having a virtual load.
Computationally, if there is only one real load applied to the truss, and the virtual load is placed at the same location and in the same direction as the real load, then we do not need to analyze the structure twice. We can adjust the analysis result from the real load to obtain the effect of the virtual load.
@@DrStructure i see..
How do you obtain the virtual forces in the member due to the applied virtual load?
Exactly the same way that we determine real member forces due to real applied loads, by analyzing the truss under the applied virtual load.
would u give me a proof that external virtual work equals internal virtual work even if it might be difficult?
The aim of these lectures is to help students develop an intuitive understanding of structural analysis concepts and their applications. You may want to review SA19 and SA20 for a discussion on the work-energy principle. But, for a mathematical proof of the virtual work principle, you need to consult advanced structural analysis and engineering mechanics texts.
Thank you guys for the great content. Could please do more 'Frame' examples using the virtual work method
9:42 how did you get internal forces at members f*1,f*2,f*3,f*4 and f*5? please help me solving this one... Thanks in advance
There is a pin at A and a roller at D for the truss which makes it stable and statically determinate. Analyze the truss using the method of joints, as if it was a typical truss subjected to a unit horizontal load at C. There is nothing special about this truss except that the unit load is not real, it is made up. But that does not prevent us from applying the method of joints to calculate the member forces (f*i, i = 1,2 ...,5) due to the made up unit load.
thanks a lot, it is helpful 🙏🏽
Great video, thanks!
Is there a working example to determine the slope of a displacement in a Truss and Frames. Because the examples I've gone through so far are only on displacements in Trusses and Frames. So how do you determine slope of deflection in a truss and Frame.
There is no single joint rotation in trusses. Each member attached to the joint rotates differently. You can determine such a rotation using simply geometry knowing the member end positions. We don't use the slope-deflection method to calculate slope in trusses,
There are a few examples of joint rotation calculations for beams, See:
th-cam.com/video/E-qfuKL2rtY/w-d-xo.html
th-cam.com/video/Fst1A4tI0YI/w-d-xo.html
and the other videos labeled SAPS.
Thank you for the help. I'll have a look at these examples.
Dr. Structure, what is the advantage of the virtual work method, in comparison with the double integration method?
The double integration method gives us the deflection equation for the entire beam. The virtual work method is for calculating the deflection at a specific point. The virtual work method can handle trusses as well as beams and frames. The double integration method is suitable for beams only.
@@DrStructure Great. Many thanks for your reply!
You're welcome.
I think... Min.10:06 it's wrong. The correct form should be:
Internal Virtual Work = (0.86)(6.6X10-6)+(0.5)(4.0X10-6)+(0)(3.5X10-6)+(0.5)(4.0X10-6)+(-0.86)(-12X10-6)
Internal Virtual Work =1041X10-6 N.m
(1N) (Δ)= 1041X10-6 N.m
Δ= (1041X10-6 N.m)/ 1N
Δ= 1041X10-6 m = 1.040 mm
Thanks for pointing out the error(s) in the equation. As you have correctly pointed out, (0.5)(3.5x10-6) should be: (0.5)(4.0x10-6). But, the final answer remains to be correct, the displacement is (20x10-6 m). The corrections are now added to the video using the TH-cam annotation tool.
@@DrStructure sir can we apply the virtual work method if the truss is indeterminate?
@@rikdevghosh7324 No, the truss needs to be determinate, or it needs to be analyzed before we can apply the virtual work method to calculate its displacements.
I believe the value for Ay @6:47 is wrong. Please I'd like an explanation if I'm wrong
Why do you believe Ay is wrong? What value do you get for it? and how do you arrive at that value?
Great video, so if the structure was statically indeterminate would you have to use a finite method to calculate the member forces first?, after which we can apply the unit load and use the virtual work method with the member displacements?
+Liam O'Donnell Yes, if the structure is statically indeterminate an appropriate analysis technique (such as the Force Method) needs to be used to determine member forces. However, it would become unnecessary to use the method of virtual work to determine joint displacements if you want to resort to the finite element method (FEM) for calculating forces. Why? Because the widely used FEM can determine joint displacements as well as member forces.
Great answer, thank you!
can"t understand how we are getting virtual members force , Dr please intervene
The same way that we get member forces due to the real loads, we need to analyze the truss under the virtual load. We remove all the real loads, place the virtual load on the truss, then use the method of joint to analyze the truss.
It's Good lecture thanks.
Please i cant understand why in the internal virtual work we use the real elongation why we did not use the virtual elongation as we do in the external virtual work when we use in it the virtual displacment ?
You wrote: "... why we did not use the virtual elongation as we do in the external virtual work when we use it in the virtual displacement?"
That is not correct. We do not use virtual displacement in any case. External virtual work equals the virtual unit load (P*) times the real displacement (Delta_V).
Think of it this way:
The structure deforms under the applied loads. In this case, the truss deforms, its joints displace in the x and y directions, and its members elongate and shorten.
Suppose we can magically hold the truss in its deformed shape yet remove the real loads that have caused the deformation.
Now. if we apply a virtual load to this deformed structure (when the real loads are not present), without considering the additional effect/displacements that virtual load could cause, then the principle goes like this:
The virtual load times the real displacement at the point of application of the load (external virtual work) EQUALS the sum of the virtual member forces (caused by the virtual load) times the real elongation/shortening of the members (internal virtual work)
@@DrStructure Now i understand great explanation ✌️🙏
Thank you very much for the fast and detailed reply 🙏😄
You're welcome!
At 8:18 how did you get 8.6 as l1
The height of the truss is 7 m. So the height of the right triangle in which our member forms the hypotenuses is 7. The base of the right triangle is 5 m. That makes the length of hypotenuse the square root of the sum of the squares of the two sides.
Square root of 7^2 + 5^2 is 8.6
9:45 what are the calculations for f1,2,3,4,5*?
We have a statically determinate truss subjected to a horizontal unit load at C. Apply the method of joints to analyze the truss and determine the member forces f1 thru f5.
Start with Joint C. There are two three forces acting at the joint: the applied unit load, the force in member AC (f1) and the force in member CD (f5). The two members make a 54.46 degrees angle with the horizontal (x-)axis. By setting the sum of the forces in the x-direction to zero, and setting the sum of the forces in the y-direction to zero, we can calculate f1 and f5.
The remaining member forces can be determined in a similar manner.
If you are not familiar with the method of joints, you can review our lectures on the topic.
@@DrStructure When I do method of joints, is the unit load 1N going to replace 500N and 200N
The real loads (200 N and 500 N) need to be removed from the truss. They play no role in the analysis of the truss subjected to the virtual unit load. As for the unit of the virtual load, it does not matter what is used, you can consider it to be 1 N.
I remember virtual work method is used to calculate the support reactions too!?
Virtual work method is for calculating displacement. However, it can be used as a part of other techniques (e.g., the force method) for analyzing indeterminate systems.
at 10:17 why is .5 multiplied by 3.5 and not 4.0?
Yes, the should have been 4, not 3.5Thanks for pointing this out.
Thank you so much for great job you did...
are you sure that f*2 and f*4 = 0.5? no compression?
+JAnosik099 Why do you suggest there might be a compression force in the bottom chord of the truss?
Are there answers to the exercises at the end of the videos ?
You should see a small i at the upper right corner of the screen, click on it for the links.
Solution for Exercise Problem 1: th-cam.com/video/x54AtTdhnrk/w-d-xo.html
Solution for Exercise Problem 2: th-cam.com/video/vcG_M7SfZj8/w-d-xo.html
This is awesome... It helped me a lo :) Could you make a video about Castiglano's Theorem??
great job
Why is the angle 54.4 and not 60.5?
really appreciate your work mate.
The side of the right triangle facing the angle is 7 m. The side adjacent to the angle is 5 m. Therefore,
angle = inverse tangent (7/5) = 54.46 degrees.
Hi, I don`t understand that how can i calculate the f* force at 9:40? i am very confused. Could u clearly explain that?
IF there is unit load in point C, and remove 500N and 200N. I calculated reaction on A and B first, got Ay=-1.43, By=1.43, Ax=-1. And then, I used method of joints to calculate the f1*, the answer is 1.75, which comes from (f1* X sin(54.46)-1.43)=0, I followed before steps. So, why i cant get ur ans?
Oh, I got the answer. I made a mistake in calculation. Thank you for your videos, help me a lot in studying structure.
I don't get it mate, can you explain it a bit? Thank you in advance.
oh i get it mate.
thank u very much Dr
Is it okay if value of f*2 and f*4 is -0.36 at 11:00?
The negative sign implies a compressive force is present in the member. But in that particular loading case, members 2 and 4 are not in compression, they are in tension. So it should read +0.36, not -0.36.
In Exercise two >>
Are the dimensions given enough ?
Yes, all the necessary dimensions are given.
Is (BC)=2m
Yes, similar triangles. It has to be 2 meters.
at 1:20 how we can find deltaB hence there is no external load acting at B?
Using the virtual work method, yes.
Dr. Structure
But there is zero load then there will be zero displacement!?
No, there needs not to be a load at the point in order for it to displace. The displacement could be the result of a load applied at a different location. Generally, a single load applied to a structure at point A causes displacement at points A, B, C, D, ...
isnt this method is exactly like castigliano second theorem ?
in principle, yes.
how do you calculate the internal member forces
The truss is statically determinate; the method of joints can be used to analyze the truss, to determine the member forces.
@@DrStructure ouh thanks
IS THERE ANY THING RELATED TO THAT CALLED " ENERGY TABLE ' ?
Please elaborate. I am not sure what you mean by "...anything related to..."
there is something called "energy table" one of the steps i will use to do the integration and find the slope and the displacement.
writing the energy table is important to take the values from it. So, from where i can know how this table look like and how can i put the values on it or the concept of this table?
thanks
i hope you understant what i mean by the "energy table".
Show me a sample "energy table." You can email an image of it to:
Dr.Structure@EducativeTechnologies.net
how to find the values of f1 to f5 i dont know
+Tharun Thaarun You can use the method of joints for that. See SA04 (th-cam.com/video/Evsjp0zKeGw/w-d-xo.html) for details.
Great video! Thanks!
how to calculate the length of the members
They are given.
Sir please add castigaliono theorem based problem of frame ...
You are great...I would like to study PhD...could you guide me to get scholarship?
Your best option is to identify the university at which you want to pursue your PhD studies, apply for admission and inquire about scholarship at that time.
Thankyou !!!
Thank You So So Much!!!
what happens if B was pinned ?
Then we would have an indeterminate truss with no displacement at B. But still the principle of virtual work would be applicable.
and then , there is no deflection at AB,BD results from load p ,right?
Correct, AB is not going to have any axial deformation since the member would be pin-connected at both ends. But BD is going to elongate since there is roller at D.
ok ,thank you very much
Thank you so much! Help me a lot haha.
Temperature effect on truss ... needed do it
We need more practice
*+1 subscriber