SA23: Virtual Work Method (Frames)
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- Опубликовано: 8 сен 2024
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Solution (Exercise 1): • SA23-1: Virtual Work M...
Solution (Exercise 2): • Video
Thank you SO MUCH . Lifesavers out here
If I have your all lectures in pdf, it will be better for me to study everything from the beginning. Listening and then read and read again ; ) Anyway , many thanks to you.
Dear sir thank you for such a great help for the students ,teachers and all , I was really confused about SA but after watching your vidioes every thing is clear for me , I am proud of sir
Best explanation ever!
Great help Dr. Structure. But please clarify why it is integral of (1-x/4)dx NOT (1-x/8)dx [11:48].
Thank you for pointing out the typo. The expression should read (1 - x/8), not (1- x/4). The answer (1600/EI) is based on the correct expression.
I believed it was a typo. Dear Dr. Structure, could you please make tutorials on Stiffness methods. I and many students out there will really be grateful for your kindness.
Thank you Dr
I love it so much
It is very nice
I am looking for further videos in the future
I have learned...thanks
waooo ... very informative thanks
I just watched an example in a structural book and found out that it considered the effect of "shear" in frame analysis for each member. I wonder whether or not shear effect can be disregarded or included in the analysis?
It depends on the type of analysis we are performing. For displacement calculations, the effect of shear is negligible in most cases. For determining critical stresses for design purposes, however, shear is generally included in the analysis. Here in this video, since the discussion is about displacement in typical beams and frames, we can safely ignore shear.
Hi Dr.
at 6:46 why is there no internal bending moment on the RHS of the free body diagram. In other words at point C?
There was no hinge in the original question so in confused.
Correct, there is no hinge at C. But that does not mean there must be a non-zero bending moment at C.
Note that member CD carries an axial force only. This is due to the fact that we have a roller support at D. Therefore, no shear force develops in member CD. No shear force at the base of CD means no bending moment at the top of the member. That is why bending moment at C is zero.
thanks very much
+Dr. Structure very helpful videos.GREAT!!!One request can you provide us a video that applying all this lectures in multi-storey building or even simple 2 -storey building providing the actual loading conditions such as dead load, live load and other loads that can affect the building structure.thanks a lot
Could please elaborate that what are the rules to find member forces in frame member like direction and moment positions?
A frame consists of a series of connected beams and columns.
Generally, at each end of a frame member, we have an axial force, a shear force, and a bending moment. Although one of more of these forces can be zero for a given frame.
When the member end forces are unknown, we need to assume their directions. For example, we can use either a clockwise or a counterclockwise direction for a member-end moment.
For each member (shown in isolation) we show/draw the three unknown member forces at each end. This gives us a total of 6 unknown forces for each member.
We then transfer the forces at the ends of each member to their adjacent joints. So, if member AB is connected to joints A and B, we draw three forces at joint A and three forces at Joint B. When doing so, however, we need to change the direction of each force. For example, if we show the moment at end B of member AB in the clockwise direction, we need to draw the moment in the counterclockwise direction at joint B.
If the frame is statically determinate, assuming the support reactions have been calculated already, we can calculate the member end forces using the three static equilibrium equations applied to each isolated member.
@@DrStructuregot it now ! Thanks
Hi,
could you tell me why is the bending on a beam considered for horizontal displacement on a frame , despite it bending downward?
Very good questions.
Yes, the beam bends downward. But that does not meant it does not move laterally under the applied load. Although small, the beam does move laterally due to w.
Assume the lateral load (P) is zero. So, the only load applied to the frame is the uniformly distributed load (w). How does the beam deform under the load at D where there is roller? Common sense suggests that the roller moves to the right since CD is going to be compressed. (we can also prove this mathematically), Now if the lower end of CD moves, the upper end also has to move (though not by the same amount). That is, point C moves laterally even if P = 0. This is in part due to the stored bending energy in the beam. Of course this stored energy also makes the beam bend downward.
Ok thanks! That makes sense. But what I still can't figure out is this: If I was to draw the BMD for the beam, I would get a diagram in which there is bending downward and/or upward , no matter what the loads are. Why is that vertical bending considered for horizontal dispalcement?
Is it because the direction of the load is horizontal on the virtual frame , which would mean that all displacement resulting would be horizontal?
If I was to evaluate the bending with another method, would the bending in the beam still be considered for horizontal displacement?
You need to shift slightly your analysis of the problem here.
Here is what I suggest: Ask yourself two questions:
1. Does the beam bend under the (real) applied load? Does a bending moment develop in the beam? In this case the answer is YES.
2. Does the same beam bend under the virtual load? Again, the answer is YES.
If the answer to both questions is YES, then the beam does indeed store internal bending energy that contributes to the horizontal displacement.
If the answer to either question is NO, then no internal energy is stored in the beam, hence no impact on the displacement.
So, if the placement of the virtual unit load would not have caused a bending moment to develop in beam BC, then the bending of the beam due to applied load would not have contributed to the horizontal displacement at C.
In response to your statement in the second paragraph, we intentionally place the virtual unit load at C, because we want to calculate horizontal displacement at C. If we were interested in calculating, say vertical displacement at the mid-point of the beam, then we would have place a vertical unit load at the mid-point.
So, using this method we can calculate only one displacement at a time, not all displacements.
Concerning the third paragraph of your post, if the method that you are going to use requires calculating bending moment equations in the beam, then yes, it plays a role in horizontal displacement computations.
Ok. Thanks a lot!
@@DrStructure I just want to say thank you for responding to comments like this, i learned a lot and this wasn't even a question I had, but now i understand more deeply whats happening. Very awesome, i love it
Sorry Dr. May i ask a question regarding my university example exercise?
It is regarding the force method (unit load) for frame. The question’s diagram look exactly like @ 8:21..except that both support at A and B are pinned (indeterminate frame). The only actual load is UDL 5kN/m from B to C. With height (A to B) is 5m and length (B to C) is 10m. Now the question is to determine the reaction at each support.
In the example (referring to my university text book example), the redundancy is 1. So they replaced the support A with a roller support, and a horizontal unit load is applied at that particular support. To make it short, they have come out with a vertical reaction at A = 29.17kN (upward), and at B = 20.83kN (upward). And horizontal reaction at A = 8.33kN (right direction) and at B = 8.33kN (left direction.
However, my own working method results in different direction for the horizontal reaction in both supports, and also the vertical support at A and B is swapping position. Then i try to figure out what is the difference in my calculation and the book example, the only differences are the unit load location (i placed at C direction to the left, meanwhile in the textbook placed at A to the right), and another one is the way i cut the section BC to find the moment equation, i made a cut section origin from B. In my text book, they made the cut origin from point C.
As u know, in force method, we have to separate the calculation in two cases, which is when the beam is subjected to the actual load (actual load system case), and when the beam is only subjected only to a unit load (unit load system case). The problem arose during the calculation in the unit load system, (kindly note that the Ra = 0.5 upward and Rc = 0.5 downward for both my working calculation and my textbook calculation). BUT since the reaction at A and C is not in the same direction, when i cut the span BC and made B as the origin, my moment equation (lower case m star) due to unit load at B is positive. My text book made the cut at span BC with C as the origin, this results in their moment equation to be in negative.
The above is the main reason why my vertical reaction magnitude at A and B is swapped position when compared to my textbook answer, and it also has flipped the horizontal reaction direction at both support as compared to my text book.
Do u have any rule of thumb for formulating the moment equation in frame? What should i do to avoid mistakes (especially in assuming direction)?
I am your online student from Malaysia, i do a part time studies here while working full time. I hope u can really help me in understanding the mechanic of this structure. Should u need my email, u can contact me at affiq1990@gmail.com . I really appreciate your work Dr. Thank you so much for all the educations taught to us.
You may have made a mistake in calculating the support reactions due to the unit load placed at A. If you are placing the unit load pointing to the left, the support reaction at A is going to be downward with a magnitude of 0.5, and the reaction at C should be upward. In another word, the direction of your reactions should be opposite to the direction given in the textbook.
As long as you are consistent in your calculations, the assumed direction of the load (displacement) does not matter. At the end, you should get the same support reactions.
As for the bending moment equation for BC, it does not matter whether you set the origin at B or at C. As long as you take all the forces into consideration, the final results would be the same.
To determine the sign of the moment equation, you need to define the beam/column bottom fiber. For horizontal beams, we know where the bottom fiber is located. For columns, you need to assume the bottom fiber. The internal bending moment in the beam should be drawn in the positive direction with respect to the bottom fiber. This means, if we cut the beam, at the left end of the cut, the moment needs to be drawn in the counter-clockwise direction, while the moment at the right end of the cut should be drawn in the clockwise direction.
Dr. Structure i have solved for the reaction at both supports, and it did give the same answer as in my textbook. THANKS A LOT DR...i really love the way you explain. May god bless you always. Thanksss
Glad to be of help :)
Dear Dr. Structure, why when you divide the various rods, the horizontal force P, which is present in the node is not reported on the rod BC?
Thanks.
In some cases the horizontal (axial) force is shown and in some cases the force is omitted from the free-body diagram. We can omit showing a force when we know it is zero. So, when you see a free-body diagram of a member for which the axial force is not drawn, it means the axial force in that particular member is zero.
Thnx for this amazing video .... Excuse me I wanna the answer of this problems where I can find its ?
hi.
Can you please explain why you didn't consider the moment of 2PL in member AB in your analysis. Thanks
Bending moment is part of the formulation for member AB. The solution given involves writing the bending moment equation where point A is taken as the origin. So, x is measured from A upward. That means 2PL does not appear in the free-body diagram of the lower part of the member. The only force present in that diagram is P. So, M(x) = Px.
If we would have taken x from the top, then we would have gotten M(x) = 2PL - Px.
Either way, we would end up with the same results.
11:47 - shouldn't the integration be done on (1- x/8) ......... instead of (1 - x/4) ?
Because there is just one equation of m*
+Saif Chowdhury It is true that we only need one equation for m*, but two equations are needed for M. This means we need to break the interval into two segments so that both m* and M remain continuous within each interval.
Does anyone know if i can get hold of the pdf for these notes that they make? I would find it a lot more helpful if I could just listen and watch and not have to take notes. Thanks
+Peter Manthorpe Unfortunately, a written version of these lectures is not available at this time.
Hellooo, i have a question about the negative sign using infront of Fab in member AB. Where did it come from? ( - (wl/2 - 2P))
The negative sign is placed there since the member is in compression, which means the member shortens due to the axial force. When a member elongates (due to a tensile force), we consider delta to be positive. When the member shortens (due to a compressive force), delta is expressed as a negative value.
I feel embarresed to ask but where did the +2p and -2p come from in the reactions stated at 3:30 ?
There are no embarrassing questions around here. :)
There are two horizontal forces of magnitude P separated by distance 2L. They form a couple resulting in a clockwise moment of 2PL. This has to be counteracted by a counterclockwise moment of 2PL. The 2P and -2P (separated by L) form such a moment.
Alternatively, just write the three equilibrium equations and solve them for the unknown reaction forces, since the frame is statically determinate.
We can also approach this by analyzing the frame twice, once due to the distributed load w and then due to the horizontal force P. When the frame is subjected to w only, the vertical reactions at the base would be (wL/2), both upward.
When the frame is subjected to P (at joint B) only, if we sum the moments about A, we get a vertical reaction of 2P at D. When we sum the forces in the y direction, we get -2P at A, and we sum the forces in the x direction, we get P at A.
Adding the results of the two analyses, we get (wL/2) -2P at A and (wL/2)+2P at D.
Ah, Okay I would never have thought of that . Thank you very much for your quick reply. You are very good at this
If the moment diagrams are to be geometric shapes, and since we're assuming EI is constant, wouldn't the method of "Values of Product Integrals" be a simpler way to solve this? Just saying...
+Vinny Lemae Generally, a given problem can be solved using multiple methods. The point of this lecture however is not to find the easiest or the most convenient way to solve a specific example problem. Rather, it is to explain how a particular method works. The example is there for illustration purposes only.
In practice, frame structures are generally analyzed on a computer using the displacement method (or more broadly, the finite element method) where only a few seconds is needed to perform the necessary computations. But, as engineers we still need to have a fundamental understanding of how structures behave and why. The classical analysis techniques provide a basis for developing such an understanding.
+Dr. Structure U ROCK
Dear Dr Structure @5:44 , Should M(x) = Px-2P?
No, M(x) = Px is correct.
The origin of the coordinate system is at the base of the column with x going up. If we cut the column at some distance x from the origin, and examine the segment from the origin to the cut point, the only force that causes a moment about the cut point would be the horizontal force P at x = 0. The forces that you see on top of the column do not appear in that free-body diagram since the diagram is that of the lower part of the member. So, a force of P placed at distance x from the cut point creates a bending moment of Px at that point.
Hi mate, what is the software you have been using for the women voice?
+Abdelrahim Elfarra It is real human voice, not a software.
+Dr. Structure ;)
why do we not consider deformation due to shear here
?
+Funi Kwon Similar to axial deformation, shear deformation in beams and frames is generally very small and can be ignored. Since internal (virtual) work is a function of internal deformation of the member, then internal virtual work due to shear is generally negligible-- it is less than 1% of the virtual work due to bending.
If however we are dealing with a deep beam, one with a large cross-sectional height (like a bridge girder), then shear stress and deformation may become significant and may have to be considered in design calculations.
+Dr. Structure crystal clear
Why is M(x) in member BC with limits 4
There is a shear force of 200 at the left end of the segment, having a moment arm of x.
There is an applied load of 400 having a moment arm of (x-4) about the cut point.
So, M(x) = 200 x - 400(x-4). Or, M(x) = 1600 -200x.
@@DrStructure Thank you, Doctor!
You're welcome.
Hi, at the example: where is the moment in member BC due to 400N load?for example at the FBD of joint B? am I confused?!!
+Omid Hassani The free-body diagram of member BC is shown @9:00.
The diagram shows no bending moment at B or C. Basically, the FBD is that of a simply supported beam. So, we are writing moment equation for a simply supported beam subjected to a concentrated load of 400N at mid span.
This means bending moment at B is zero (moment at the ends of a simply supported beam is always zero). Then moment increases as we move toward the middle of the beam. Where does this moment come from? From the shear force at B. That is, at distance x from B, bending moment equals: x times shear force (or support reaction, if we saw the beam as simply supported) at B, or 200x. This equation is valid for the left half of the beam where 0 < x < 4.
Past the middle of the beam, the bending moment equation changes due to the presence of the 400N concentrated load. Now, the moment equation becomes M(x) = 1600 - 200x. This is valid for the right segment of the beam where x is between 4 and 8 as indicated in the video.
Just like a simply supported beam subjected to a concentrated load, the moment diagram for member BC takes the shape of a triangle defined using the two equations above, if we were to graph them.
+Dr. Structure
What if member BC( horizontal one) makes a 120 degree angle with AB and just one load of P (vertical) at B (down word)?
+Dr. Structure thx
+Omid Hassani In that case, member BC carries no load (no internal shear, no bending moment). The entire vertical load at B (assuming it is the only load applied to the frame) is carried by AB. No horizontal displacement at B or A, only a vertical displacement at B.
Even when the A is a roller, still no horizontal displacement for A?? ( AB is not vertically, 60 degree with ground and 120 degree with BC)
💙💚
In the numerical example, shouldn't the axial effort due to the virtual force in the member BC equal to 1 instead of -1 (f*bc = +1)?
The axial force in BC due to the virtual unit load placed at A is compressive, that translates into -1 per our sign convention.
Got it, thanks!