# Hamilton's principle

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1. Mar 22, 2015

### Last-cloud

I want to obtain equation using Hamilton principle but I just couldn't figure it out;
i have The kinetic energy :

E_{k}=\dfrac{1}{2}m_{z} \displaystyle\int\limits_{0}^{L}\ \left[ \left( \dfrac{\partial w(x,t)}{\partial t}\right)^{2}+\left( \dfrac{\partial v(x,t)}{\partial t}\right)^{2}\right] dx

and The potential energy $E_{p}$

\begin{split}
E_{p} &= \dfrac{1}{2}EI \displaystyle\int\limits_{0}^{L}\ \left[\dfrac{\partial^{2} w(x,t)}{\partial x^{2}} \right]^{2} dx + \dfrac{1}{2}T \displaystyle\int\limits_{0}^{L}\ \left[\dfrac{\partial w(x,t)}{\partial x} \right]^{2} dx \; + \\
& \dfrac{1}{2}EA \displaystyle\int\limits_{0}^{L}\ \left\lbrace \dfrac{\partial v(x,t)}{\partial x} \; + \; \dfrac{1}{2} \left[\dfrac{\partial w(x,t)}{\partial x} \right]^{2} \right\rbrace^{2} dx
\end{split}

The work is given by :

\begin{split}
W &=W_{F}+W_{d}+W_{m} \\
&=\displaystyle\int\limits_{0}^{L}\ \left\lbrace \left[ f(x,t)-c_{1} \dfrac{\partial w(x,t)}{\partial t}\right] w(x,t) \;-\; c_{2}\left[ \dfrac{\partial v(x,t)}{\partial t}\right] v(x,t) \right\rbrace \;dx \\
&+ u_{T}w(x,t)+u_{L}v(x,t)
\end{split}

and i should use The extended Hamilton’s principle to obtain the equation

\displaystyle\int\limits_{t_{1}}^{t_{2}}\ \delta\left( E_{k}-E_{p}+W\right) dt = 0

shuch that :

\delta \displaystyle\int\limits_{t_{1}}^{t_{2}}\ L dt = \displaystyle\int\limits_{t_{1}}^{t_{2}}\ \delta L dt = \displaystyle\int\limits_{t_{1}}^{t_{2}}\ \left( \dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right) \right) \delta q dt

the variation for the Kinetic energy i think it's :

\displaystyle\int\limits_{t_{1}}^{t_{2}}\ \delta E_{k_{1}} dt = -\displaystyle\int\limits_{0}^{L}\ \displaystyle\int\limits_{t_{1}}^{t_{2}}\ m_{z}\ddot{w} \; \delta w \; dt \; dx

and

\displaystyle\int\limits_{t_{1}}^{t_{2}}\ \delta E_{k_{2}} dt = -\displaystyle\int\limits_{0}^{L}\ \displaystyle\int\limits_{t_{1}}^{t_{2}}\ m_{z}\ddot{v} \; \delta v \; dt \; dx

but the variation for potential energy i couldn't do it , because what i think is

\dfrac{\partial E_{p}}{\partial w} = 0

because

\dfrac{\partial w''}{\partial w} = 0

i know there is something wrong with my reasoning but I don't know what it is.
what should I do?

2. Mar 22, 2015

### Dr.D

What system are you applying Hamilton's Principle to?
What you have called the "extended Hamilton's Principle" looks a bit strange to me, but I hesitate to say too much until I know what problem you are working.

3. Mar 25, 2015

### Last-cloud

the system that i'm applying Hamilton's Principle to , is the Lagrangian "L" such that

\begin{split}
L&=E_{k}-E_{p}+W=\\
&\dfrac{1}{2}m_{z} \displaystyle\int\limits_{0}^{L}\ \left[ \left( \dfrac{\partial w(x,t)}{\partial t}\right)^{2}+\left( \dfrac{\partial v(x,t)}{\partial t}\right)^{2}\right] dx-\\
&\dfrac{1}{2}EI \displaystyle\int\limits_{0}^{L}\ \left[\dfrac{\partial^{2} w(x,t)}{\partial x^{2}} \right]^{2} dx + \dfrac{1}{2}T \displaystyle\int\limits_{0}^{L}\ \left[\dfrac{\partial w(x,t)}{\partial x} \right]^{2} dx \; +
\dfrac{1}{2}EA \displaystyle\int\limits_{0}^{L}\ \left\lbrace \dfrac{\partial v(x,t)}{\partial x} \; + \; \dfrac{1}{2} \left[\dfrac{\partial w(x,t)}{\partial x} \right]^{2} \right\rbrace^{2} dx+\\
&\displaystyle\int\limits_{0}^{L}\ \left\lbrace \left[ f(x,t)-c_{1} \dfrac{\partial w(x,t)}{\partial t}\right] w(x,t) \;-\; c_{2}\left[ \dfrac{\partial v(x,t)}{\partial t}\right] v(x,t) \right\rbrace \;dx \\
&+ u_{T}w(x,t)+u_{L}v(x,t)
\end{split}

where

m_{z},EI,EA,T,c_{1},c_{2} \;\;\; are \;\;\; constants

but i thought i should use Hamilton's Principle like this :

\displaystyle\int\limits_{t_{1}}^{t_{2}}\ \delta\left( E_{k}-E_{p}+W\right) dt = 0 \Longrightarrow
\displaystyle\int\limits_{t_{1}}^{t_{2}}\ \left( \delta E_{k}-\delta E_{p}+\delta W\right) dt = 0

and i am calling it "extended Hamilton's Principle" because the Lagrangian is usualy kinetic energy - potential energy but when now there is work in the equation