Dr structure, thanks a lot for the video, I have a question , when I use the formulas in front cover of structural analysis book for Hibbeler , the delta B ( capital) is maximum at mid span equal 5wL^4/384 is it right? Because the final answer i got was 5wL/8 Please reply dr because I need it to teach my students in the college
Hi, my teacher is having me solve a problem with force method where there are no external forces on a cantilever beam with a roller support at the midpoint. It only gives a max deflection table. I think it is called the compatibility condition. Do you have a solution for a problem like this?
The problem description is rather vague. What is the maximum deflection table? The deflection of the beam is a always a function of the applied load and the material and section properties of the structure. The larger the applied load, or the smaller the stiffness of the member, the more the beam is going to deflect. That is, there is no theoretical upper limit to the amount of deflection.
Bending moment in a beam segment is considered positive if it causes the segment to bend concave up. So, when we are showing positive moment in a beam segment, we need to draw a clockwise moment at the left end of the segment, and a counterclockwise moment at the right end of the segment, causing the segment to bend downward (forming a concave up shape). Notice that bending moment, m*, is shown that way at the right end of the segments @6:28. The sign of the terms that appear in the moment equation is unrelated to the sign of the bending moment that we have drawn already. To write the moment equations, we have assumed counterclockwise direction to be positive. We could have assume the opposite, we could have assumed clockwise is positive. Please note that this has nothing to do with the sign convention I mentioned above. They are independent of each other. That is the sign convention for bending moment is not the same as the sign convention for writing the equilibrium equations. Using counterclockwise as positive, the first equation becomes: m(x) - (1/2)x = 0, or, m(x) = x/2 The second equation can be written as: m(x) - (1/2)x + 1(x-L) = 0. Note that here we have one additional term, the unit load. This term was not present for the first equation. After simplification, the second equation becomes: m(x) = L - x/2. As for EI being constant, that basically means the member does not change properties. But, we could have a composite member with different E values. Or, the shape of them member can change, like a tapered beam, making the moment of inertia (I) to change. When EI is not constant, then it cannot be treated as a constant when we are integrating. We need to have EI expressed in terms of x, then include it in our integration.
If there was an axial load on a beam with only one pin support (the rest of the supports were rollers), then the horizontal reaction at the pin would be equal to the magnitude of the horizontal force. The rest of vertical supports can be computed as usual using the force method. However, if there are two or more horizontal support reactions, and the beam is subjected to horizontal loads, then we need to consider one or more of those reactions (and possibly one or more of the vertical reactions) as redundant forces.
The Force Method involves writing and solving the displacement compatibility equations. These equations are defined using deflections in the beam under the applied load. We can calculate the deflection of a beam at a point under any loading scenario (i.e., distributed, concentrated,…) using the virtual work method. You can find a series of lectures covering these topics in our free online course. See the video description field for the link.
The lower case deltaB is deflection per unit load. When we multiply it by By, we get the deflection due to By. You may want to review the lecture on the virtual work method for the background knowledge.
Hello, I have 2 questions 1. What is the logic behind adding delta B + dell B = deflection of real beam 2. How did you consider unit load at B gives deflection= dell B. This is used to find deflection in Virtual work metho
The vertical deflection of a beam at a pin or roller support (say, at point B) is zero. When we treat the reaction force at a support as a redundant force and remove it from the beam, the beam is going to deflect downward by certain amount, say, Delta. That is, the support (its reaction force) prevents the beam from deflecting downward at point B. But now that we have remove the support, the beam is no longer restrained, hence, it is going to deflect downward by some amount which we referred to as Delta. This Delta we can compute, since by removing the support, we’ve made the beam statically determinate. Say, we calculate Delta to be 10 mm in the downward direction. To determine the redundant reaction force, call it R, we are going to place a unit upward force at B and calculate the upward deflection that results. We label this deflection as lowercase delta. Suppose this upward deflection is 0.5 mm. This means that if we increase the magnitude of the load to 2, the deflection becomes 1 mm, if the load magnitude is increased to 3, the deflection becomes 1.5 mm, and so on. So, if delta is the deflection due to a unit load, R times delta = the upward deflection due to the reaction force R. This upward deflection, which is the deflection due to the (redundant) reaction force at B must be equal to the downward deflection Delta, since we know the total deflection at the support must be zero. So, we write: R delta (upward) = Delta (downward) Using a more consistent sign convention where both deflections are defined in the upward direction, the above expression can be written as: R (delta) + Delta = 0 where the right hand side of the equation is the actual deflection of the beam at point B. Using the numbers given above, we can rewrite the equation as: R (0.5) = 10. Which yields R (the redundant force): R = 20.
If there is a concentrated load on the beam, then the total displacement at a target point (at the point where the redundant force is located) would be due to the distributed load AND the concentrated load. We can calculate that displacement using the principle of superposition. Calculate the displacement due to the distributed load only, and calculate the displacement due to the concentrated load only, then add them together to get the total displacement.
Correct! However, there are a few typos in the youtube lectures. We have corrected them, and will continue to update the lectures on our (free) online course. The link is given in the video description field.
How did you make this video ( i mean which software used to make it ) if you dont mind please kindly tell i am a teacher too i cant make this kinda video ?
For this particular video, we created a set of powerpoint slides with all the text and diagrams, then traced over them using a stylus on an ipad, redrawing them. This resulted in a set of SVG files. we then used VideoScribe, an online app to animate the content of each SVG file yielding a video segment. And, used Camtasia studio to put the segments together in order to produce the entire lecture.
SIR great work. please give me answer of one question . Q.how we can determine that: that force is taken as redundent? please answer is very necessary for me .
@@kirolousyoussef8609 You can select any reaction force to be redundant so long as the resulting structure remains stable. How you choice such a redundant force is more or less arbitrary.
We can't choose Ax force as redundant its because due to this force the system become unstable it will move in x direction. Is i am right?? Yes or no please comment
Yes, and no. Yes, because we should not making a force that causes instability in the structure a redundant force. No, because in this case, Ax = 0. Since the sum of the forces in the x direction must be zero, and Ax is the only force in that direction, then Ax = 0. So, it serves no purpose to make Ax a redundant force.
In this case, it really does not matter. We have two identical free-body diagrams and moment equations. Both the real load and virtual load have a magnitude of 1, are placed at the same point, and act in the same direction.
@@joseh7612 If we want to determine the displacement at a point, say, A due to a load that is being applied at a different point (B), then the virtual and real loads result in different moment equations. In such a case, the virtual load needs to be placed at A while the real load is at B.
We are summing the moments about the cut point, assuming the counterclockwise direction as positive. There are two forces that create a moment about the point. The reaction force at the left end of the beam has a magnitude of 1/2 and a moment arm of x about the cut point. The force creates a clockwise moment of (1/2)(x) about the cut point. The unit virtual load has a moment arm of (x-L) about the cut point. Hence, it creates a moment of (1)(x-L) about the point. And there is a counterclockwise concentrated moment of m*(x) at the cut point. So, the total moment can be written as: m*(x) - (1/2)(x) + (1)(x-L) = 0. Solving this equation for m*(x), we get: m*(x) = (1/2)(x) - x + L. Or, m*(x) = L - x/2.
Is it possible or correct way to use it on a statically determined beam case? because I was taught like that in my class. I am a little confused right now, I need your wisdom. Thanks
The force method is for analyzing indeterminate beams. If the beam is determinate, there is no reason to use this method. We can simply use the three static equilibrium equations to calculate the support reactions.
Sorry, but that's not what I mean. I mean, is it possible to use the force method to calculate static beam deflection instead of the double integration method? is there any method similar to the force method to calculate the static beam deflection?
Also, please keep in mind that we are not using the force method to calculate deflections. The deflection computations that take place here are for determining the unknown reaction forces. Put it differently, the force method uses displacements of statically determinate beams in order to determine the unknown reactions in the statically indeterminate beam. The displacements themselves can be calculated using any technique of your choice. Generally speaking, the available techniques for displacement calculations require a statically determinate structure.
No, the force method cannot be used to calculate deflections. The force method is for calculating unknown reaction forces in indeterminate beams. It so happens that the method uses deflections for that purposes. The deflections themselves are calculated using techniques such as double integration.
Thanks for your wisdom, speaking of the calculation method for deflection, is there any method beside the double integration method? As you said that I can calculate the displacement using any technique I prefer.
No, in principle the method can be used to analyze beams with N degrees of indeterminacy. The method yields N equations in N unknowns which can be used to calculate N redundant reaction forces. See SA25 (ruclips.net/video/N6FDLx0MrOI/видео.html) for the use of the method for analyzing a beam with 2 degrees of indeterminacy.
Suppose if there is a beam joined with the steel cable as we see in Hibblers chapter 4 of mechanics of materials 9th edition in which we have to perform compatibility to calculate unknown in statically indeterminate beam Then how this method will help in calculating force in that cable Please clear this small thing
I don't know what specific problem you are solving. Regardless, you want to pick the reaction at the pin/roller where displacement is know to be zero, as the redundant force. That would turn the system into a determinate beam. Then you can use the equilibrium equations to find the remaining forces, including the one in the cable, calculate the elongation of the cable, then use basic geometry to determine displacement at the support where the redundant force has been removed...
All right Just Make one thing clear This method is applicable on all kinds of problems in mechanics of materials no matter which one i pick even those problems in which beam is fully supported by cables from starting to ending
This method requires that we know the displacement, often zero, at the point of redundancy. If a structure is supported such that none of the joint displacements are know, or can be assumed to be zero, then the method would not work.
OK then please assure one thing can we start with the pin which is connected at the starting or ending of the beam instead of roller just like you have done with the Roller.
The force method involves using various displacements for formulating the compatibility equations. We can use an energy method (e.g., virtual work method) or a non-energy method (e.g., conjugate beam method) for calculating the displacements. So, the force method may, or may not, be viewed as an energy method, depending on which method we use to calculate the displacements. But in a strict sense, it is not an energy method.
The work-energy principle which establishes mathematical relationships between external work and internal energy of systems is the basis of modern structural analysis techniques such as the displacement method, finite element methods (which is a generalized form of the displacement method), and boundary element methods.
Correct, the classical methods of structural analysis are time-consuming, especially when applied by hand without the use of computers. Although they are not practical for solving problems with more than a few degrees of indeterminacy, these classical techniques offer useful insights into the behavior of structural systems and a conceptual framework for thinking about their analysis.
@@DrStructure Yea I do know that,, in work we use computers to analyze beam and frame, truss structures however, I just wanna understand this in order to pass structural analysis 2 tho. Btw I have a question, can we use conjugate beam method for slope and deflection for the first case instead of virtual work? And then apply virtual work on the virtual beam
It is one of the best and cleanest lectures in youtube
First time i like animation type of explanation in SA. Thanks a lot to this channel. Hats of to Teacher who explained this very well.
This video, having a very tidy and lucid way of presentation proves that online learning is better than the physical one.
This helps me a lot. Thank you so much for your explanation!
Amazing job! Very clear explanation. Thank you very much.
Thank you for making mechanics of materials easy.
Such a clear representation and linup of data.........
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keep up the good work Doc !
Thank you Soo much.. It really helped me.. Great job... 👍👍👍👍
Very clear Explanation. Thank you very much
is the best method of explanation
you deserve way more subscribers.
Great video.thank you❤
love u Dr. structure, this video is awesome
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fantastic job. so many thanks.
Awesome demonstration
It is super helpful thank you so much
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tq.. its really helpful
thanks you helped me a lot.
Please upload a video on the theorem of least work for indeterminate beam
wow nice one.
thank you so much.
Great 😍😊
Thanks Rifat Hassan
thanks a lot for the explanation that was really good. Is there explanation of the question that u asked at the end of the vıdeo?
Here it is: ruclips.net/video/40MWuULaZ-Q/видео.html
thanks so much
wow amazing
Nice sir ji
Would you also post about the stiffness method??
Thank you.
Yes, the stiffness method will also be covered.
My dear your channel have utupe talk about anaylsis stress and strain
hey can you tell me which part of ur lecture talks about castigliano's theorem please
Bad news: We have not discussed Castigliano's Theorem in any of the lectures yet. Good news: It will be covered eventually.
Please upload video for Frame
Perfect👍🏻
Dr structure, thanks a lot for the video, I have a question , when I use the formulas in front cover of structural analysis book for Hibbeler , the delta B ( capital) is maximum at mid span equal 5wL^4/384 is it right? Because the final answer i got was 5wL/8
Please reply dr because I need it to teach my students in the college
Yes, the expression in the book is correct. The value is obtained by integrating:
M m* dx where
M = ( w L/2) x - w x^2 /2
m* = x/2 for 0
Dr. Structure thanks for replying, but why the final answer i obtained was half of that you obtained ?
My solution is
By=(5 wL^4/384)/(1*L^3/48)=5wL/8
I am not sure why your answer is different. I need to see your calculations in order to be able to detect the source of the problem.
Hi, my teacher is having me solve a problem with force method where there are no external forces on a cantilever beam with a roller support at the midpoint. It only gives a max deflection table. I think it is called the compatibility condition. Do you have a solution for a problem like this?
The problem description is rather vague. What is the maximum deflection table? The deflection of the beam is a always a function of the applied load and the material and section properties of the structure. The larger the applied load, or the smaller the stiffness of the member, the more the beam is going to deflect. That is, there is no theoretical upper limit to the amount of deflection.
i have a small doubt regarding bending moment sign convention, i.e, you have written positive sign for reaction force while considering 0
Bending moment in a beam segment is considered positive if it causes the segment to bend concave up. So, when we are showing positive moment in a beam segment, we need to draw a clockwise moment at the left end of the segment, and a counterclockwise moment at the right end of the segment, causing the segment to bend downward (forming a concave up shape).
Notice that bending moment, m*, is shown that way at the right end of the segments @6:28.
The sign of the terms that appear in the moment equation is unrelated to the sign of the bending moment that we have drawn already. To write the moment equations, we have assumed counterclockwise direction to be positive. We could have assume the opposite, we could have assumed clockwise is positive. Please note that this has nothing to do with the sign convention I mentioned above. They are independent of each other. That is the sign convention for bending moment is not the same as the sign convention for writing the equilibrium equations.
Using counterclockwise as positive, the first equation becomes:
m(x) - (1/2)x = 0, or, m(x) = x/2
The second equation can be written as: m(x) - (1/2)x + 1(x-L) = 0. Note that here we have one additional term, the unit load. This term was not present for the first equation. After simplification, the second equation becomes:
m(x) = L - x/2.
As for EI being constant, that basically means the member does not change properties. But, we could have a composite member with different E values. Or, the shape of them member can change, like a tapered beam, making the moment of inertia (I) to change.
When EI is not constant, then it cannot be treated as a constant when we are integrating. We need to have EI expressed in terms of x, then include it in our integration.
Hi how can i find the solution of problem that is at the end
The solutions are available in the free online course referenced in the video description field.
If we had an axial load in the same beam (like in the case of a roof for example) would the calculations change?
If there was an axial load on a beam with only one pin support (the rest of the supports were rollers), then the horizontal reaction at the pin would be equal to the magnitude of the horizontal force. The rest of vertical supports can be computed as usual using the force method.
However, if there are two or more horizontal support reactions, and the beam is subjected to horizontal loads, then we need to consider one or more of those reactions (and possibly one or more of the vertical reactions) as redundant forces.
@@DrStructure Okey thanks for your clarification
Example! If it has a force at between point A and B, how I slove
it ? Can you show me, please?
The Force Method involves writing and solving the displacement compatibility equations. These equations are defined using deflections in the beam under the applied load.
We can calculate the deflection of a beam at a point under any loading scenario (i.e., distributed, concentrated,…) using the virtual work method.
You can find a series of lectures covering these topics in our free online course. See the video description field for the link.
hi ma'am. @4:37, where did By lower case deltaB=deltaB come from?
isn't the expression on the left a work? why did you equate it with deflection?
The lower case deltaB is deflection per unit load. When we multiply it by By, we get the deflection due to By.
You may want to review the lecture on the virtual work method for the background knowledge.
@@DrStructure thank you
Hello, I have 2 questions
1. What is the logic behind adding delta B + dell B = deflection of real beam
2. How did you consider unit load at B gives deflection= dell B. This is used to find deflection in Virtual work metho
The vertical deflection of a beam at a pin or roller support (say, at point B) is zero. When we treat the reaction force at a support as a redundant force and remove it from the beam, the beam is going to deflect downward by certain amount, say, Delta. That is, the support (its reaction force) prevents the beam from deflecting downward at point B. But now that we have remove the support, the beam is no longer restrained, hence, it is going to deflect downward by some amount which we referred to as Delta.
This Delta we can compute, since by removing the support, we’ve made the beam statically determinate. Say, we calculate Delta to be 10 mm in the downward direction.
To determine the redundant reaction force, call it R, we are going to place a unit upward force at B and calculate the upward deflection that results. We label this deflection as lowercase delta. Suppose this upward deflection is 0.5 mm. This means that if we increase the magnitude of the load to 2, the deflection becomes 1 mm, if the load magnitude is increased to 3, the deflection becomes 1.5 mm, and so on. So, if delta is the deflection due to a unit load, R times delta = the upward deflection due to the reaction force R. This upward deflection, which is the deflection due to the (redundant) reaction force at B must be equal to the downward deflection Delta, since we know the total deflection at the support must be zero. So, we write:
R delta (upward) = Delta (downward)
Using a more consistent sign convention where both deflections are defined in the upward direction, the above expression can be written as:
R (delta) + Delta = 0
where the right hand side of the equation is the actual deflection of the beam at point B.
Using the numbers given above, we can rewrite the equation as:
R (0.5) = 10.
Which yields R (the redundant force): R = 20.
@@DrStructure ok understood now, Thanks for explaining. Good work!! Keep going👍
What if there is a point load added to the question
If there is a concentrated load on the beam, then the total displacement at a target point (at the point where the redundant force is located) would be due to the distributed load AND the concentrated load. We can calculate that displacement using the principle of superposition. Calculate the displacement due to the distributed load only, and calculate the displacement due to the concentrated load only, then add them together to get the total displacement.
Won't the deflection due to udl be 5wl^4/384EI? At 6:36
For a beam of length L, yes. In our case, the length of the beam is 2L. So constant 384 in the denominator needs to be divided by 2^4.
perfect
at 6:25
Could you please let me know of the Moment equation is right? I am calculating it as - 1/2(L-x).
Please let me know. thanks
got it, x begins from far left end of the beam and now I got the answer (L-x/2). thank you :)
Correct!
However, there are a few typos in the youtube lectures. We have corrected them, and will continue to update the lectures on our (free) online course. The link is given in the video description field.
How did you make this video ( i mean which software used to make it ) if you dont mind please kindly tell i am a teacher too i cant make this kinda video ?
For this particular video, we created a set of powerpoint slides with all the text and diagrams, then traced over them using a stylus on an ipad, redrawing them. This resulted in a set of SVG files. we then used VideoScribe, an online app to animate the content of each SVG file yielding a video segment. And, used Camtasia studio to put the segments together in order to produce the entire lecture.
Dr. Structure Ok thanks i am grateful for your helpful videos. I will try to make a video like that
@ Dr structure can you plz provide the solution of excercise
The solutions for the exercise problems are provided in the course referenced in the video description field. Course registration is free.
Subscribed
can this also be known as the flexibility method?
Yes!
SIR great work. please give me answer of one question .
Q.how we can determine that: that force is taken as redundent?
please answer is very necessary for me .
Please elaborate a bit more. I am not sure what you are asking, what you are looking for.
@@DrStructure how do you decide a certain force to be the redundent force??
@@kirolousyoussef8609 You can select any reaction force to be redundant so long as the resulting structure remains stable. How you choice such a redundant force is more or less arbitrary.
We can't choose Ax force as redundant its because due to this force the system become unstable it will move in x direction.
Is i am right??
Yes or no please comment
Yes, and no.
Yes, because we should not making a force that causes instability in the structure a redundant force.
No, because in this case, Ax = 0. Since the sum of the forces in the x direction must be zero, and Ax is the only force in that direction, then Ax = 0. So, it serves no purpose to make Ax a redundant force.
are the two deltas in the steel handbook?
Not sure what you are asking. Please elaborate. Steel design handbooks do not usually deal with the analysis of structures.
@@DrStructure The equation for deflection is in the steel manual but not delta
Most structural analysis textbooks have a table that gives the slope and deflection for a limited number of lcases.
at 7:00 what is the difference between real and virtual load. And which one is which?
In this case, it really does not matter. We have two identical free-body diagrams and moment equations. Both the real load and virtual load have a magnitude of 1, are placed at the same point, and act in the same direction.
@@DrStructure when would they be different?
@@joseh7612 If we want to determine the displacement at a point, say, A due to a load that is being applied at a different point (B), then the virtual and real loads result in different moment equations. In such a case, the virtual load needs to be placed at A while the real load is at B.
why the m*(x)=L-x/2 (L
We are summing the moments about the cut point, assuming the counterclockwise direction as positive. There are two forces that create a moment about the point. The reaction force at the left end of the beam has a magnitude of 1/2 and a moment arm of x about the cut point. The force creates a clockwise moment of (1/2)(x) about the cut point. The unit virtual load has a moment arm of (x-L) about the cut point. Hence, it creates a moment of (1)(x-L) about the point. And there is a counterclockwise concentrated moment of m*(x) at the cut point. So, the total moment can be written as: m*(x) - (1/2)(x) + (1)(x-L) = 0. Solving this equation for m*(x), we get: m*(x) = (1/2)(x) - x + L. Or, m*(x) = L - x/2.
Can you share slides? It would be easier to revise these concepts.
We don’t have these lectures/videos in slide form.
What is the ans of last numerical ?
The link to the solution video is given in the description field. Here it is: ruclips.net/video/40MWuULaZ-Q/видео.html
Is it possible or correct way to use it on a statically determined beam case? because I was taught like that in my class. I am a little confused right now, I need your wisdom. Thanks
The force method is for analyzing indeterminate beams. If the beam is determinate, there is no reason to use this method. We can simply use the three static equilibrium equations to calculate the support reactions.
Sorry, but that's not what I mean. I mean, is it possible to use the force method to calculate static beam deflection instead of the double integration method? is there any method similar to the force method to calculate the static beam deflection?
Also, please keep in mind that we are not using the force method to calculate deflections. The deflection computations that take place here are for determining the unknown reaction forces. Put it differently, the force method uses displacements of statically determinate beams in order to determine the unknown reactions in the statically indeterminate beam. The displacements themselves can be calculated using any technique of your choice. Generally speaking, the available techniques for displacement calculations require a statically determinate structure.
No, the force method cannot be used to calculate deflections. The force method is for calculating unknown reaction forces in indeterminate beams. It so happens that the method uses deflections for that purposes. The deflections themselves are calculated using techniques such as double integration.
Thanks for your wisdom, speaking of the calculation method for deflection, is there any method beside the double integration method? As you said that I can calculate the displacement using any technique I prefer.
Is the force method applicable ONLY for having one degree of indeterminacy?
No, in principle the method can be used to analyze beams with N degrees of indeterminacy. The method yields N equations in N unknowns which can be used to calculate N redundant reaction forces. See SA25 (ruclips.net/video/N6FDLx0MrOI/видео.html) for the use of the method for analyzing a beam with 2 degrees of indeterminacy.
@@DrStructure Thanks.
Suppose if there is a beam joined with the steel cable as we see in Hibblers chapter 4 of mechanics of materials 9th edition in which we have to perform compatibility to calculate unknown in statically indeterminate beam
Then how this method will help in calculating force in that cable
Please clear this small thing
I don't know what specific problem you are solving. Regardless, you want to pick the reaction at the pin/roller where displacement is know to be zero, as the redundant force. That would turn the system into a determinate beam. Then you can use the equilibrium equations to find the remaining forces, including the one in the cable, calculate the elongation of the cable, then use basic geometry to determine displacement at the support where the redundant force has been removed...
All right
Just Make one thing clear This method is applicable on all kinds of problems in mechanics of materials
no matter which one i pick
even those problems in which beam is fully supported by cables from starting to ending
This method requires that we know the displacement, often zero, at the point of redundancy. If a structure is supported such that none of the joint displacements are know, or can be assumed to be zero, then the method would not work.
OK then please assure one thing can we start with the pin which is connected at the starting or ending of the beam instead of roller just like you have done with the Roller.
Yes, any support reaction (at a roller, a pin or a fixed support) can be taken as the redundant force.
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is force method based on strain energy
The force method involves using various displacements for formulating the compatibility equations. We can use an energy method (e.g., virtual work method) or a non-energy method (e.g., conjugate beam method) for calculating the displacements. So, the force method may, or may not, be viewed as an energy method, depending on which method we use to calculate the displacements. But in a strict sense, it is not an energy method.
@@DrStructure thank you 😎
@@DrStructure then what are the different energy methods to analyse a structure?
The work-energy principle which establishes mathematical relationships between external work and internal energy of systems is the basis of modern structural analysis techniques such as the displacement method, finite element methods (which is a generalized form of the displacement method), and boundary element methods.
@@DrStructure does castigliano theorem come under energy method?
This is a good method! But it takes about 30 minutes for a man who is utterly professional lol
Correct, the classical methods of structural analysis are time-consuming, especially when applied by hand without the use of computers. Although they are not practical for solving problems with more than a few degrees of indeterminacy, these classical techniques offer useful insights into the behavior of structural systems and a conceptual framework for thinking about their analysis.
@@DrStructure Yea I do know that,, in work we use computers to analyze beam and frame, truss structures however, I just wanna understand this in order to pass structural analysis 2 tho. Btw I have a question, can we use conjugate beam method for slope and deflection for the first case instead of virtual work? And then apply virtual work on the virtual beam
In principle, yes. We can use any of the available techniques to determine slope and/or deflection in beams.
Please Dr I need help iam from Iraq 😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭
We are here to help as much as humanly and ethically possible. Please feel free to tell us about the conceptual or procedural challenges you face.
@@DrStructure please can slove example 😭😭😭