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I am master student and doing an assignment of Finite element method. In the instruction I could not understand the derivation of the weak form, which should not be difficult. I'm sorry for posting this easy and most likely not helpful question for other people.

So the derivation is about the weak form of the integral formulation of 4th ODE. It is a simple beam deformation in the interval 0 and L. $$ \int_0^L\frac{d^2 w}{d x^2}EI\frac{d^2 \hat{u}}{d x^2}dx = ...$$ $$ \left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=0} \right. -\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=0} \right. -\left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=L} \right. +\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=L} \right. $$

I thought about partial integration $$\int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) \, u'(x) dx $$ then I ended up $$\frac{d^2u}{dx^2}\frac{dw}{dx} - \int \frac{dw}{dx}\frac{d^3x}{dx^3}d x$$ if I continue the partial integration to the 2nd term then derivative will be 4th order...

How can I get the following? $$ \left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=0} \right. -\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=0} \right. -\left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=L} \right. +\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=L} \right. $$

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  • $\begingroup$ Your integration is correct. Keep in mind that w is a test function which is zero on the boundary of the domain. Thus, the terms multiplied by w are still identically zero. $\endgroup$ – Paul Jan 10 '16 at 21:22
  • $\begingroup$ The expression $ \int_0^L\frac{d^2 w}{d x^2}EI\frac{d^2 \hat{u}}{d x^2}dx = ...$ is already in the weak form. Why are you integrating this equation again? Are you trying to acquire the strong form of the PDE? Please explain. $\endgroup$ – Paul Jan 11 '16 at 16:26
  • $\begingroup$ my question is how you can derive the right hand side from the left hand side in the equation, $$\int_0^L\frac{d^2 w}{d x^2}EI\frac{d^2 \hat{u}}{d x^2}dx = \left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=0} \right. -\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=0} \right. -\left( \left. w EI \frac{d^3 \hat{u}}{d x^3} \right|_{x=L} \right. +\left( \left. \frac{d w}{dx} EI \frac{d^2 \hat{u}}{d x^2} \right|_{x=L} \right.$$ $\endgroup$ – user26767 Jan 12 '16 at 14:36
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It's easier to understand this identity if you start with the partial differential equation for the Euler-bernoulli beam deflection equation

$$\frac{d^2}{dx^2}\left[ EI \frac{d^2u}{dx^2}\right] = 0$$

and work your way down to the weak form.

Multiply both sides of the equation by an arbitrary test function $w$. Then apply integration by parts (only once) over the domain [0,L]. You will obtain:

$$w\frac{d}{dx}\left[ EI \frac{d^2u}{dx^2}\right]\bigg|_0^L - \int_0^L \left[ EI \frac{d^2u}{dx^2}\right] \frac{dw}{dx}= 0.$$

Then, apply integration by parts again, we obtain:

$$w\frac{d}{dx}\left[ EI \frac{d^2u}{dx^2}\right]\bigg|_0^L - \frac{dw}{dx}EI \frac{d^2u}{dx^2}\bigg|_0^L + \int_0^L \frac{d^2w}{dx^2}EI \frac{d^2u}{dx^2}=0$$

Rearranging the terms evaluated at 0 and L to the right hand side, you will obtain the equation you state.

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