The explanation is very clear, I understand better than the course of practical engineers. Examples could be more complex, such as triangular loads. But explanations are so detailed and interesting, you're really good.
Thank you very much! Best videos on SA on youtube! Keep up the good work!! P.S. Could you make a couple of videos on the method of finite elements, please?
@@DrStructure I know that B and C are taken as redundant in this beam I am talking about an other example like e.g I am taking C and D as redundant roller support at B will be remain same. In that case while I am calculating delta D virtual/unit load (1KN) will be applied at D then what will be the equation of m* ? Or Is it compulsory to Always take B and C force as redundant?
In that case, there would be two moment equations when a unit load is placed at D. We need one moment equation for segment AB (there would be a reaction force at A and one at B), and another moment equation for segment BD.
@9:00, we need to write only one set of moment equations, since M = m*. For the sake of completeness, however, we are writing both M and m* even though they are the same.
I am assuming you are referring to the calculations at 9:19. To determine deltaBC using the virtual work method, we need to write two moment equations: We need the moment equation M(x) for the beam when the load is at C (this is assumed to be the real load), and we need the moment equation m*(X) for the beam when a virtual unit load is placed at B (this is the point at which the displacement takes place). According to the virtual work method, to determine the displacement at point P, we need to place a virtual unit load at point P in order to get m*(x).
it's understandable. I appreciate for that but how did u calculated the bending moment for real structure. I,e while calculating Delta B we need to calculate bending moment for real structure as well so how did u calculated that , can you elaborate it I couldn't find answer for that. thank you
Why do you write displacements as By*deltaB ? is not that a unit of work? actual displacement is delta B, so it is unclear how you come to writing force*displacement = displacement ?
deltaB is displacement due to a unit load, or displacement per unit load. So, displacement due to 2 unit loads would be 2*deltaB. Displacement due to 3 unit load = 3 deltaB, and so on. Displacement due to By unit loads, therefore, equals By*deltaB.
dr structure if we suppose that the span distribution will not be equal like shown in the video 10m 10m 10m,what if its like 10m,10m,5m will still be the deflections be equal when unit load is applied on the constraints please reply dr strcuture
Hi doc. @ 9:19 , for delta bc, i get the first equation , however , the second term i.e. the limits 10-20 intergral, how do you get that ? because in the previous equations, there were only two sets of intergrals and i noted that you used virtual load at bb in this bc equation which i understand as it relates to the relationship between b and c. Im just really confused as to how the middle term at 9:19 (10-20 limits) becomes part of the delta bc equation. please explain. thank you.
For M(x) we have two equations, one for interval [0,20] and one for [20,30]. We also have two intervals for m*(x), [0,10] and [10,30]. However, these intervals are not a perfect match, they overlap. To integrate Mm*, we need to make sure the limits of the intervals are perfect match. So, we can divide the first interval for M(x) into two intervals, like this: M(x) = -x/3 [0,10] M(x) = -x/3 [10,20] This does not change the nature of the equation. But now the first interval of M(x) matches the first interval of m*(x). Therefore, M(x) can be written in terms of three equations: M(x) = -x/3 [0,10] M(x) = -x/3 [10,20] M(x) = 2x/3 - 20 [20,30] We also need to divide the second interval of m*(x) into two intervals: [10,20] and [20,30]. This gives us three equations for m*(x) as well: m*(x) = -2x/3 [0,10] m*(x) = x/3 - 10 [10,20] m*(x) = x/3 - 10 [20,30] Now we can integrate the three parts knowing that their limits match.
@@DrStructure thank you doc. That explains a whole bunch :). Im currently dealing with a static indeterminacy of 3, so having to formulate 3 different equations really complicates the method. I haven't seen anyone do a beam with that many unknowns. If possible doctor, could you do something like that? It would really help. Thanks again for the explanation. Subscribed and liked 👍
Make the reaction at the support with the settlement the redundant force. Assuming that is the only redundant force, then the compatibility equation becomes: D + (F d) = s where D is the vertical displacement at the location due to the applied load. If there is no applied load, then D = 0. F is the unknown redundant force, d is vertical displacement at the point due to a unit load. and s is support settlement. We'll do a lecture on this at a later time.
what happens if you have a beam with a varied value of EI (says the same beam in the problem above but has linear variation from EI at the left to 2EI at the right, how is that alter the force method solution ?)
The force method uses displacements for formulating and solving the compatibility equations for the redundant forces. The EI for the beam effects the displacements. When solving for displacements (Delta), we integrate the product of M and m*, where EI appears as a constant to the left of the integral symbol. If however, EI becomes a variable, in terms of x, then we need to write it to the right of the integral symbol. This means that now instead of just integrating M m*, we need to integrate M m*/ f where f is the function that describe EI for the beam. For example, if EI changes linearly as you have indicated above, then f can be written as: EI(1+x). This means that now we need to integrate M m*/(1+x) in order to calculate Delta...
Yes, the force method can analyze any indeterminate beam or frame. However, as the degree of indeterminacy of the system increases, the use of the method becomes more laborious.
+Dr. Structure sorry let me try again. delta_bc and delta_cb both = 10500/27EI and I was wondering if delta_bc always equals delta_cb or if it was only this problem
Good day Dr Structure, I am have emailed you are small request of assistance in formulating the equations for the whole beam when the beam has only the reactions at A and G
In such a case we need to pick three of the reactions as redundant forces. For example, we could pick Fx, Fy, and M at either end as redundant forces, thereby turning the beam into a cantilever one. Or, we could turn the beam into a simply supported one by making the end moments and one of the horizontal reactions forces redundant.
Dr. Structure ya..from your tutorial i know how to solve when my redundant force is in vertical direction... but how to do it if it is in horizontal or moments
Please see: ruclips.net/video/Ho_yrY8efy4/видео.html You can also access the example directly from SA25, just click on the circled i at the upper right corner of the video, or see the video description field for the link.
The force method involves removing redundant forces from the structure and calculating the displacements that result. The shape of the structure is inconsequential. As long as we can calculate the displacement due to the removed the redundant force, we can apply the method to any analyze the system.
No, we often use the principle of superposition when we wish to determine the net response (say, the displacement, or support reactions) of a structure subjected to multiple loads by adding up all such responses due to the individual loads. Although that principle is being used in the force method, they are not one and the same. Superposition is a more general concept and works for solving all sorts of problems in engineering and physics.
@@nikadekasrirahayu6197 We have our updated video lectures, and exercise problem solutions available through the (course) link given in the video description field. The registration is free. Register and log in there to find the exercise problem solutions for The Force Method (Frame Analysis) in PDF format.
The team responsible for creating and maintaining these lecture are from around the globe. You can learn about the person who oversees the broader project at lab101.space
I would not have ever passed this exam without your explanations! Thank you!
Happy to help!
You are The best
in explaining
best lecture ever tnx Dr. structure.... any one from 5kilo AAIT give me thumbs up !!
Best explanation so far!
The explanation is very clear, I understand better than the course of practical engineers.
Examples could be more complex, such as triangular loads.
But explanations are so detailed and interesting, you're really good.
Thanks for the feedback.
Thank you very much! Best videos on SA on youtube! Keep up the good work!!
P.S. Could you make a couple of videos on the method of finite elements, please?
how to apply maxwell reciprocal theorm to solve quickly please make video on that dr structure
very good channel
If you could upload some videos of "Moment distribution" and "Stiffness matrix" method, it would be very helpful for us.Thank you.
Plz post video for the end support are A and C and intermediate support B and support A is fixed support and B&C supports are roller.
Best channel for structural analysis!
Where can I find the solution for the exercise problems?
Click on the i at the upper right corner of the video screen for the links. They are also listed in the video description field.
e.g we are talking force at C and D as redudants then while calculating delta D, will m* will have only 1 equ? as force at D will have 0 distance.
The reaction forces at B and C (not C and D) are taken to be redundant. Please rephrase your question.
@@DrStructure I know that B and C are taken as redundant in this beam I am talking about an other example like e.g I am taking C and D as redundant roller support at B will be remain same. In that case while I am calculating delta D virtual/unit load (1KN) will be applied at D then what will be the equation of m* ? Or Is it compulsory to Always take B and C force as redundant?
In that case, there would be two moment equations when a unit load is placed at D. We need one moment equation for segment AB (there would be a reaction force at A and one at B), and another moment equation for segment BD.
@@DrStructure ok thank you
At 9:00 can you explain why you you make the bending moment twice for the virtual load?
@9:00, we need to write only one set of moment equations, since M = m*. For the sake of completeness, however, we are writing both M and m* even though they are the same.
Love you mam the way you explain is wonderful
Why are we applying the unit load at b when finding m*? for deltaBC, isn't the load applied at C?
I am assuming you are referring to the calculations at 9:19.
To determine deltaBC using the virtual work method, we need to write two moment equations: We need the moment equation M(x) for the beam when the load is at C (this is assumed to be the real load), and we need the moment equation m*(X) for the beam when a virtual unit load is placed at B (this is the point at which the displacement takes place).
According to the virtual work method, to determine the displacement at point P, we need to place a virtual unit load at point P in order to get m*(x).
@@DrStructure Thank you so much! Now I understand why. I had thought that was the reason but couldn't tell why.
Thank you..great work!
In the case that we have more than 2 degree of indeterminacy , for example 3 pins, what we will do?
The same principle applies except that we would end up with three equations and three unknowns (redundant forces).
Thank you for your answer! So there is not any other shorter way?
Not really. The number of equations to be solved equals to the degree of indeterminacy of the system. N unknown forces means N equations.
it's understandable. I appreciate for that but how did u calculated the bending moment for real structure. I,e while calculating Delta B we need to calculate bending moment for real structure as well so how did u calculated that , can you elaborate it I couldn't find answer for that. thank you
The beam is statically determinate. We can easily analyze it and determine its bending moment equation(s). See Lectures SA06 & SA07, for example.
Where can I find the solutions of the exercise problems?
The solutions are provided in the free online course referenced in the video description field.
Why do you write displacements as By*deltaB ? is not that a unit of work? actual displacement is delta B, so it is unclear how you come to writing force*displacement = displacement ?
deltaB is displacement due to a unit load, or displacement per unit load.
So, displacement due to 2 unit loads would be 2*deltaB. Displacement due to 3 unit load = 3 deltaB, and so on.
Displacement due to By unit loads, therefore, equals By*deltaB.
Can you please tell me,which software are you using to create this explainer video?Thanks.
A combination of software tools: Camtasia Studio, VideoScribe, CrazyTalk, Autodesk Graphic.
Thanks for your help.
congratulations is very didactic videos excuse me ask you a consultation with that program make the equations thank you very much
The equations are hand-written. More accurately, they are generated by hand-tracing over software (MathType) generated equations.
When the title tells me that support settlement occurs at some point, how do I use this force method to solve it?
Instead of setting the displacement at the support to zero, you can set it equal to the amount of settlement.
thank you❤
very helpful , thanks doc
dr structure if we suppose that the span distribution will not be equal like shown in the video 10m 10m 10m,what if its like 10m,10m,5m will still be the deflections be equal when unit load is applied on the constraints please reply dr strcuture
Yes, Delta_B and Delta_C are going to be different if points B and C are not placed at the 1/3 and 2/3 points along the length of the beam.
@@DrStructure kindly make a video in how to apply maxwell reciprocal theorm dr structure with examples,and many thanks for the reply
Hi doc. @ 9:19 , for delta bc, i get the first equation , however , the second term i.e. the limits 10-20 intergral, how do you get that ? because in the previous equations, there were only two sets of intergrals and i noted that you used virtual load at bb in this bc equation which i understand as it relates to the relationship between b and c. Im just really confused as to how the middle term at 9:19 (10-20 limits) becomes part of the delta bc equation. please explain. thank you.
For M(x) we have two equations, one for interval [0,20] and one for [20,30]. We also have two intervals for m*(x), [0,10] and [10,30]. However, these intervals are not a perfect match, they overlap. To integrate Mm*, we need to make sure the limits of the intervals are perfect match.
So, we can divide the first interval for M(x) into two intervals, like this:
M(x) = -x/3 [0,10]
M(x) = -x/3 [10,20]
This does not change the nature of the equation. But now the first interval of M(x) matches the first interval of m*(x). Therefore, M(x) can be written in terms of three equations:
M(x) = -x/3 [0,10]
M(x) = -x/3 [10,20]
M(x) = 2x/3 - 20 [20,30]
We also need to divide the second interval of m*(x) into two intervals: [10,20] and [20,30]. This gives us three equations for m*(x) as well:
m*(x) = -2x/3 [0,10]
m*(x) = x/3 - 10 [10,20]
m*(x) = x/3 - 10 [20,30]
Now we can integrate the three parts knowing that their limits match.
@@DrStructure thank you doc. That explains a whole bunch :). Im currently dealing with a static indeterminacy of 3, so having to formulate 3 different equations really complicates the method. I haven't seen anyone do a beam with that many unknowns. If possible doctor, could you do something like that? It would really help. Thanks again for the explanation. Subscribed and liked 👍
@@AK--if1su Here is an analysis example of a beam with three unknowns, in PDF format:
lab101.space/pdf/examples/SA25-Example1.pdf
@@DrStructure thanks a lot Doc. I will have a look. Really appreciate the help. 👍
Excellent, well explained! Thank you very much! :)
Thank you
thanx a lot
Dr structures, will u please explain how to solve the support settlement problems using the FORCE METHOD?
Make the reaction at the support with the settlement the redundant force. Assuming that is the only redundant force, then the compatibility equation becomes:
D + (F d) = s
where D is the vertical displacement at the location due to the applied load. If there is no applied load, then D = 0.
F is the unknown redundant force, d is vertical displacement at the point due to a unit load. and s is support settlement.
We'll do a lecture on this at a later time.
@@DrStructure Thank u so much. May God bless u more with the light of knowledge!
what happens if you have a beam with a varied value of EI (says the same beam in the problem above but has linear variation from EI at the left to 2EI at the right, how is that alter the force method solution ?)
The force method uses displacements for formulating and solving the compatibility equations for the redundant forces. The EI for the beam effects the displacements. When solving for displacements (Delta), we integrate the product of M and m*, where EI appears as a constant to the left of the integral symbol. If however, EI becomes a variable, in terms of x, then we need to write it to the right of the integral symbol. This means that now instead of just integrating M m*, we need to integrate M m*/ f where f is the function that describe EI for the beam. For example, if EI changes linearly as you have indicated above, then f can be written as: EI(1+x). This means that now we need to integrate M m*/(1+x) in order to calculate Delta...
@@DrStructure It makes great sense. Thanks
You're welcome.
pls Is this applicable in all case of indeterminate beams
Yes, the force method can analyze any indeterminate beam or frame. However, as the degree of indeterminacy of the system increases, the use of the method becomes more laborious.
Wonderful. Thanks.
Would you folks be kind enough and upload more often ? That would be great
Thanks. Always a fan : )
where i can get ur all videos these are very helpful
They are available only on youtube.
Excuse me, does force method could be solve, when moment is applied?
Yes, the force method can be used if there are moments applied to the beam.
Does f_BD always equal f_DB when considering a system of Degree of Indeterminacy=2?
Not sure what you mean, please rephrase.
+Dr. Structure sorry let me try again. delta_bc and delta_cb both = 10500/27EI and I was wondering if delta_bc always equals delta_cb or if it was only this problem
They would always be equal. delta_ij always equals to delta_ji where i and j are points on the beam.
well done
Good day Dr Structure,
I am have emailed you are small request of assistance in formulating the equations for the whole beam when the beam has only the reactions at A and G
Yes, we got your email and responded accordingly.
@@DrStructure Thank you soo much, finally i have cracked it. My problem was of the segments, but now the answer is out thank you. 😊
may i know how to solve when both ends and fixed(Degree of indeterminacy=3)
In such a case we need to pick three of the reactions as redundant forces. For example, we could pick Fx, Fy, and M at either end as redundant forces, thereby turning the beam into a cantilever one. Or, we could turn the beam into a simply supported one by making the end moments and one of the horizontal reactions forces redundant.
Dr. Structure ya..from your tutorial i know how to solve when my redundant force is in vertical direction... but how to do it if it is in horizontal or moments
I'll do an example video for you in couple of days.
thank you so much
Please see: ruclips.net/video/Ho_yrY8efy4/видео.html
You can also access the example directly from SA25, just click on the circled i at the upper right corner of the video, or see the video description field for the link.
How do you use the force method on frames?
The force method involves removing redundant forces from the structure and calculating the displacements that result. The shape of the structure is inconsequential. As long as we can calculate the displacement due to the removed the redundant force, we can apply the method to any analyze the system.
thanx
ıs there a possıbılıty that force method mıght also be named as superposition?
No, we often use the principle of superposition when we wish to determine the net response (say, the displacement, or support reactions) of a structure subjected to multiple loads by adding up all such responses due to the individual loads. Although that principle is being used in the force method, they are not one and the same. Superposition is a more general concept and works for solving all sorts of problems in engineering and physics.
when solving for delta_CC , your integration calculation I think should be 0 to 20 for the first term not 0 to 10.
Thanks for pointing out the typo.
No problem! You're amazing keep it up
Can I know the answer of exercise problems?
@@nikadekasrirahayu6197 We have our updated video lectures, and exercise problem solutions available through the (course) link given in the video description field. The registration is free. Register and log in there to find the exercise problem solutions for The Force Method (Frame Analysis) in PDF format.
Hi everybody. Does anyone has the Matlab code of 'Force Method'?
👍🏻
Where are you from
The team responsible for creating and maintaining these lecture are from around the globe. You can learn about the person who oversees the broader project at lab101.space
That last integration for dcc should have had the first bound be 0-20
I guess it's correct, because M(x) is different in regions 0
dr. who? thank you.
It's too advance but I like it