For any body not clear on how 1/2 is introduced at 8:46 We have a formula ∆(F^n) =n.[F^(n-1) ].∆F replace F with du/dx in above form Here ∆ is only for first du/dx, so to exchange ∆ with integral it has to be applied to second du/dx
At 10:22, there is an error. EA*(du/dx)=force, not stress. (1/2)*force*displacement is strain energy. Hence, (1/2)*force*(du/dx) is strain energy per unit length. The integration over whole domain length will give total strain energy.
You are correct that it is force. But integration of the dot product of Force and du (i.e., strain times dx) will give the strain energy. There is no point having '1/2' here!
If you are talking about the 1/2 at 8:50, then it is required. Let me explain. Since the delta has come out of the integral, to match with the above step, we need 1/2. δ[〖1/2 (u^' )〗^2 ] = u' δu
It might be difficult for you to understand that equation because TH-cam comments have formatted the equation. I will try to spell it. Variation of (half * (u') squared) is equal to u' * variation of u
@@srikant519 I'm not dumb, I can read what you've written. So, give me a logical reason or just leave it be... And if you listen to the professor carefully you'll find that in the second 8:51 he says that variation and differentiation/intergation can be interchanged. He applies this logic to the next term. So, do your research before answering.
@@ashish24294 Bro calm down. I have not said anything offensive or I don't mean you are dumb. It's FEM and not so easy a subject and I understand that. I too got many doubts while going through the course. Anyway, all I am trying to say is: When the delta comes out of d(u) but not the integral, it becomes delta of [1/2 (du/dx) squared]. This is because delta follows the same rules as of differential 'd'. Here the delta is still inside the integral. This step is skipped by the prof. Then, in the next step, he takes the delta out, and like the prof said, the integral and variation can be interchanged. I hope this clears the confusion.
It comes from the fact that B(u,u) is bilinear and symmetric as one of the other comments inquires. I.e.: initially the weak form is recast into the variational problem => B(w,u) = L(w) then w = delta(u) and from the above property B(delta(u), u) = 1/2 * delta*B(u,u)
Hi, It is very simple. the delta operator is only for one u' inside the integral. not for other u'. So, to bring the delta operator on complete integration the following operation is done: B(delta(u), u) = 1/2 * delta*B(u,u). Hope u can understand
For any body not clear on how 1/2 is introduced at 8:46
We have a formula
∆(F^n) =n.[F^(n-1) ].∆F
replace F with du/dx in above form
Here ∆ is only for first du/dx,
so to exchange ∆ with integral it has to be applied to second du/dx
This the previous lecture I was grasping things this is a complete bouncer
At 10:22, there is an error. EA*(du/dx)=force, not stress. (1/2)*force*displacement is strain energy. Hence, (1/2)*force*(du/dx) is strain energy per unit length. The integration over whole domain length will give total strain energy.
Well said.
You are correct that it is force. But integration of the dot product of Force and du (i.e., strain times dx) will give the strain energy. There is no point having '1/2' here!
Why there is the need to introduce 1/2 in term 1, the integration itself should do that upon solving/calculation?
If you are talking about the 1/2 at 8:50, then it is required. Let me explain. Since the delta has come out of the integral, to match with the above step, we need 1/2.
δ[〖1/2 (u^' )〗^2 ] = u' δu
It might be difficult for you to understand that equation because TH-cam comments have formatted the equation. I will try to spell it.
Variation of (half * (u') squared) is equal to u' * variation of u
@@srikant519 But previously we have established that δ and d/dx are interchangeable...
@@srikant519 I'm not dumb, I can read what you've written. So, give me a logical reason or just leave it be... And if you listen to the professor carefully you'll find that in the second 8:51 he says that variation and differentiation/intergation can be interchanged. He applies this logic to the next term. So, do your research before answering.
@@ashish24294 Bro calm down. I have not said anything offensive or I don't mean you are dumb. It's FEM and not so easy a subject and I understand that. I too got many doubts while going through the course. Anyway, all I am trying to say is:
When the delta comes out of d(u) but not the integral, it becomes delta of [1/2 (du/dx) squared]. This is because delta follows the same rules as of differential 'd'. Here the delta is still inside the integral. This step is skipped by the prof. Then, in the next step, he takes the delta out, and like the prof said, the integral and variation can be interchanged. I hope this clears the confusion.
I have still not understood how the half came. Please someone explain in detail
It comes from the fact that B(u,u) is bilinear and symmetric as one of the other comments inquires. I.e.:
initially the weak form is recast into the variational problem => B(w,u) = L(w)
then w = delta(u)
and from the above property B(delta(u), u) = 1/2 * delta*B(u,u)
Hi, It is very simple. the delta operator is only for one u' inside the integral. not for other u'. So, to bring the delta operator on complete integration the following operation is done: B(delta(u), u) = 1/2 * delta*B(u,u).
Hope u can understand
What you teaching completely bouncers, jumping steps. When you teaching for nptel please keep in mind an avg student can understand
Clear to karle bhai video