This video and surrounding playlist (Structural Dynamics) is infinitely more informative than all the structural dynamics courses I took in university combined. Thank you so much for your clear, comprehensive explanations.
I wish there was a good sharing of this video so that every student who has struggles in this field could overcome. This video is so helpful specially for me as a phd student working on the fracture mechanics. Thanks 🙏🏼🙏🏼🙏🏼
Kindly make a video on phase field model for cracking in brittle material and how we do the discretization of weak form. I am doing research on it and have a lot of difficulties in it.
Most textbooks on Variation Calculus will discuss this. Initially, the derivation of the Euler-Langrane equation assumes geometric boundary conditions and so the boundary terms drop away (meaning that just the integral part will need to be zero on its own). Then after this, most textbooks will discuss the case of natural boundary conditions. It is in this context that it is determined that each of the conditions must be zero independently. My go-to book for Variational Calculus tends to be Dym & Shames, "Solid Mechanics: A Variational Approach", page 100.
@@Freeball99 Thank you for making some of the best lectures on the subject matter. I haven't found much better material on UA-cam. Like the water that knows the lesser potential surface to flow naturally, you have the ability to present things in absolute clarity. Please preserve your videos forever. Regarding the question posed by @jorgeluismedina1548, there is an excellent reference in the lecture notes by H Kardestuncer - "Discrete Mechanics - A Unified Approach " Chapter 2 Calculus of Variations, Where the author expanded variational $\delta \Pi $using an approach similar to Taylor series for the function and its variable. Then integrating the terms by parts he showed that the parts integrated are the natural boundary conditions. He presented a second-order Euler-Lagrange differential equation.
Governing equations in mechanics, one way or another, ultimately come out of Newton's Law. So it's an equilibrium equation between inertial and dynamic loads.
Very elegant. However the thing that has always kept me at unease is the intuition behind the expression of total potential energy. The contribution of strain energy term is understandable. The addition of load potential (potential due to external loads) seems kinda unintuitive. All the souces I have seen so far just define the total potential energy to be so. Hope you see this comment and respond to this query. Thank you
The idea is that if I place an external load on a system then the amount of energy stored by the system is exactly equal to the work performed on the system by the external load. Take the example of a spring which is fixed on one end. If I push on the other other end then the spring deflects (ie it stores potential energy). We understand intuitively that, due to the stored potential energy, the spring now has the ability to do work (ie if we replaced the force with a mass, the spring will extend and will perform work on the mass on the process. So the amount of potential energy stored by the spring is equal to the external work performed ON THE SYSTEM. When work is performed ON THE SYTEM by the external loads, we find that the sign of the external load is the opposite from the sign of the displacement - this is defined as NEGATIVE WORK - so we need to flip the sign - i.e. the increase in the potential of the system is the NEGATIVE of the external work performed on the system. Conversely, if the external work and the displacement are in the same direction, then the systems is actually performing work on the external load (instead of vice versa) and as a result the system potential decreases. So, the increase (or decrease) of the system potential can simply be written as the negative of the external work performed on the system. RULE OF THUMB: If external work is being performed ON THE SYSTEM, then the external work is negative and the potential of the system (to do work) increases. Conversely if external work is being performed BY THE SYSTEM, then the external work is positive and the potential of the system (to do work) decreases. Hope this covers your confusion.
@@Freeball99 Thankyou for responding to the query. I agree that the workdone on the system is stored as potential energy of the system. But that stored energy is what accounts for the elastic strain energy of the system. Thats what I referred to as 'understandable' in the previous comment. By defining the work potential as the work done by the external force(s) on the system, am I somehow not taking into account the same effect twice. Because the effect of load in terms of raising the potential energy of the system is already accounted for by the strain energy term. That's where my confusion lies.
@@m_tahseen No you aren't taking it twice into account, it's more like a balance. Because there is actually a difference of sign. Let's say U is the strain energy of the considered system and W is the potential of the generalized external forces. Then the total potential Energy of this system is V= U+W. However, the work imposed by the external forces is NOT equal to W, rather it's equal to -W. Hope that helps.
Probably because you brought the differential under the integral when you calculated de variation. But I wonder why the strong form gives exact solutions and the weak one appropriate solutions.
The fundamental Lemma of The Calculus of Variations states that because the variation of the displacement 𝛿w is arbitrary, therefore everything multiplying it must be zero FOR EVERY POINT IN THE DOMAIN. This can only be applied to the strong form of the equation. It is from here that the exact solution emerges.
This video and surrounding playlist (Structural Dynamics) is infinitely more informative than all the structural dynamics courses I took in university combined. Thank you so much for your clear, comprehensive explanations.
Happy to see you're still making videos!
I wish there was a good sharing of this video so that every student who has struggles in this field could overcome. This video is so helpful specially for me as a phd student working on the fracture mechanics. Thanks 🙏🏼🙏🏼🙏🏼
Could you please make videos about Castigliano's theorems and Virtual work?
Excellent video...
Very informative...
I cannot express my gratitude to you....
Kindly make a video on phase field model for cracking in brittle material and how we do the discretization of weak form. I am doing research on it and have a lot of difficulties in it.
Good to see you back !
Amazing sir,you are back.
By the way can you tell the aap you use to write?
The app is called "Paper" by WeTransfer. Running on an iPad Pro 13 inch and using an Apple pencil.
Thanks, a wonderful presentation Regards Major Edmund Keen, India
Hey freeball!
Thanks. You are amazing!
Thank you for this video. where can i find the proof for wich the boundary conditions go to zero independently? Is there a reference book for this?
Most textbooks on Variation Calculus will discuss this. Initially, the derivation of the Euler-Langrane equation assumes geometric boundary conditions and so the boundary terms drop away (meaning that just the integral part will need to be zero on its own). Then after this, most textbooks will discuss the case of natural boundary conditions. It is in this context that it is determined that each of the conditions must be zero independently. My go-to book for Variational Calculus tends to be Dym & Shames, "Solid Mechanics: A Variational Approach", page 100.
@@Freeball99 Thank you for making some of the best lectures on the subject matter. I haven't found much better material on UA-cam. Like the water that knows the lesser potential surface to flow naturally, you have the ability to present things in absolute clarity. Please preserve your videos forever. Regarding the question posed by @jorgeluismedina1548, there is an excellent reference in the lecture notes by H Kardestuncer - "Discrete Mechanics - A Unified Approach " Chapter 2 Calculus of Variations, Where the author expanded variational $\delta \Pi $using an approach similar to Taylor series for the function and its variable. Then integrating the terms by parts he showed that the parts integrated are the natural boundary conditions. He presented a second-order Euler-Lagrange differential equation.
Love your videos. Can you do some videos about PDEs?
That would be great.
Oh my god, he's back!
Yay New Video!
divσ+ ρ=0 is the governing equation of my variational method… what is the nature and origin of a governing equation. What is it saying to us?
Governing equations in mechanics, one way or another, ultimately come out of Newton's Law. So it's an equilibrium equation between inertial and dynamic loads.
great!!! Thank you
Thank you, please kindly note that the units of "w(0)" and "w_,x(0)" are different and it is more approperiate to set them equal to zero seperately.
Yip. You're right about this. Was just using some shorthand to save space.
Very elegant. However the thing that has always kept me at unease is the intuition behind the expression of total potential energy. The contribution of strain energy term is understandable. The addition of load potential (potential due to external loads) seems kinda unintuitive. All the souces I have seen so far just define the total potential energy to be so. Hope you see this comment and respond to this query. Thank you
The idea is that if I place an external load on a system then the amount of energy stored by the system is exactly equal to the work performed on the system by the external load.
Take the example of a spring which is fixed on one end. If I push on the other other end then the spring deflects (ie it stores potential energy). We understand intuitively that, due to the stored potential energy, the spring now has the ability to do work (ie if we replaced the force with a mass, the spring will extend and will perform work on the mass on the process.
So the amount of potential energy stored by the spring is equal to the external work performed ON THE SYSTEM. When work is performed ON THE SYTEM by the external loads, we find that the sign of the external load is the opposite from the sign of the displacement - this is defined as NEGATIVE WORK - so we need to flip the sign - i.e. the increase in the potential of the system is the NEGATIVE of the external work performed on the system.
Conversely, if the external work and the displacement are in the same direction, then the systems is actually performing work on the external load (instead of vice versa) and as a result the system potential decreases.
So, the increase (or decrease) of the system potential can simply be written as the negative of the external work performed on the system.
RULE OF THUMB: If external work is being performed ON THE SYSTEM, then the external work is negative and the potential of the system (to do work) increases. Conversely if external work is being performed BY THE SYSTEM, then the external work is positive and the potential of the system (to do work) decreases.
Hope this covers your confusion.
@@Freeball99 Thankyou for responding to the query. I agree that the workdone on the system is stored as potential energy of the system. But that stored energy is what accounts for the elastic strain energy of the system. Thats what I referred to as 'understandable' in the previous comment. By defining the work potential as the work done by the external force(s) on the system, am I somehow not taking into account the same effect twice. Because the effect of load in terms of raising the potential energy of the system is already accounted for by the strain energy term. That's where my confusion lies.
@@m_tahseen No you aren't taking it twice into account, it's more like a balance. Because there is actually a difference of sign. Let's say U is the strain energy of the considered system and W is the potential of the generalized external forces. Then the total potential Energy of this system is V= U+W.
However, the work imposed by the external forces is NOT equal to W, rather it's equal to -W. Hope that helps.
Probably because you brought the differential under the integral when you calculated de variation. But I wonder why the strong form gives exact solutions and the weak one appropriate solutions.
The fundamental Lemma of The Calculus of Variations states that because the variation of the displacement 𝛿w is arbitrary, therefore everything multiplying it must be zero FOR EVERY POINT IN THE DOMAIN. This can only be applied to the strong form of the equation. It is from here that the exact solution emerges.
Great!
Let's go finite elements
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