Kinetic and Potential Energy of a multi degree of freedom (MDOF) system

Question

Consider the following MDOF system:

$M\ddot x+Kx=F$

where $M$ and $K$ are the mass and stiffness matrix respectively, and $x$ and $F$ are the displacement and force vectors.

How can one determine the potential and kinetic energy?

Answer 1

$\def \b {\mathbf}$

$$\b M\,\ddot{\b{x}}+\b K\,\b x=\b F$$

multiply from the left with $~\frac{d}{dt}{\b{x}}$

$$\dot{\b{x}}\,\cdot \left(\b M\,\ddot{\b{x}}\right)+\frac{d}{dt}{\b{x}}\,\cdot\,\left(\b K\,\b x\right)=\frac{d}{dt}{\b{x}}\cdot\b F\tag 1$$

thus equation (1)

$$\frac{d}{dt}\left(\frac 12\dot{\b{x}}^T\b M\,\dot{\b{x}}\right)+\frac{d}{dt}{\b{x}}\,\cdot\,\left(\b K\,\b x\right)=\frac{d}{dt}{\b{x}}\cdot\b F$$

multiply with $~dt~$ and integrate (assume $~\b F~$ is constant)

$$ \underbrace{\frac 12\dot{\b{x}}^T\b M\,\dot{\b{x}}}_{\text{kinetic energy}} + \underbrace{\frac 12\b x^T\,\b K\,\b x-\b F\cdot x}_{\text{potential energy}}=0$$

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assume you have this scalar equation

$$ m\ddot x+k\,x=F$$

multiply with $~\dot x~$ $$\dot x\,m\,\ddot x+\dot x\,k\,x=\dot x\,F$$

with $$~\dot x\,m\,\ddot x~=\frac m2\frac {d}{dt}\,\dot x^2$$ you obtain

$$\frac m2\frac {d}{dt}\,\dot x^2+\frac{dx}{dt}\,k\,x= \frac{dx}{dt}\,F\\ \frac m2 {d}\,(\dot x^2)+{dx}\,k\,x= {dx}\,F\\ \int \frac m2 {d}\,(\dot x^2)+\int{dx}\,k\,x= \int{dx}\,F\\$$

$$\frac m2 \,(\dot x^2)+\frac{k}{2}\,x^2=F\,x$$

If you deals with vectors and matrices you obtain the above result