Nonclassical damping

Nonclassical damping occurs when a linear mechanical system's damping matrix cannot be diagonalized together with its mass matrix and stiffness matrix by an undamped normal mode transformation. Consequently the damping terms couple the modal equations even though their inertia and stiffness terms are separated. This can matter when a localized dissipative element is attached to a structure whose elastic and inertial responses are distributed differently.^[ref]

For $M\ddot{u}+C\dot{u}+Ku=0$, take a complete mass-normalized undamped mode matrix $\Phi$ with ${\Phi }^{T}M\Phi =I$ and ${\Phi }^{T}K\Phi ={\Omega }^2$. Then $u=\Phi q$ givesWith distinct undamped natural frequencies, a nonzero off-diagonal $D_{ij}$ shows modal damping coupling. If a frequency is repeated, first allow a rotation of the modes within its repeated-frequency subspace: an off-diagonal entry in one arbitrary basis is not by itself proof of nonclassical damping. For symmetric $M,K,C$ with $M$ positive definite, the basis-independent condition for classical decoupling isFailure of this condition establishes nonclassical damping under those assumptions.^[ref]

For an original example in consistent normalized units, letThe damping acts only on the first physical coordinate. The undamped mode vectors $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$ giveThus an initial modal state $q=(1,0{)}^T$, $\dot{q}=(1,0{)}^T$ has ${\ddot{q}}_2=-0.2$. Dropping the off-diagonal damping terms would incorrectly predict zero initial acceleration in the second modal coordinate. Nevertheless ${\dot{u}}^{T}C\dot{u}=0.4{\dot{u}}_1^2\ge 0$: passive dissipation does not require modal decoupling.

The damped quadratic eigenvalue problem follows from $u(t)=r{e}^{\lambda t}$:In oscillatory cases its complex vectors can encode different phases at different physical coordinates. One can solve the coupled equations directly or use an appropriate damped-mode method; assigning separate scalar decay equations to the undamped modes is an additional approximation, not a consequence of positive dissipation.

The damped quadratic eigenvalue problem finds exponential motions of a linear mechanical model with mass matrix $M$, damping matrix $C$ and stiffness matrix $K$. Substituting $u(t)=r{e}^{\lambda t}$ into $M\ddot{u}+C\dot{u}+Ku=0$ givesThe dependence on $\lambda$ is quadratic, so solving the undamped problem $Kr={\omega }^{2}Mr$ alone does not find the damped decay rates and mode shapes.^[ref]

A useful linearization sets $z=(r,\lambda r{)}^T$ and rewrites the problem asThis generalized eigenvalue problem has twice the displacement dimension. With invertible $M$, it becomes an ordinary first-order state eigenproblem by multiplication by the block mass inverse. Singular $M$ requires more care and cannot be handled by silently using $M^{-1}$.

For an original one-coordinate example in consistent units, take $M=1$, $C=2$ and $K=5$. Then ${\lambda }^2+2\lambda +5=0$, giving $\lambda =-1\pm 2i$. Real free motions have the formThe decay rate is $1$ and damped oscillation frequency is $2$, while the undamped natural frequency is $\sqrt{5}$. Increasing $C$ to6 instead gives real roots $-1,-5$, showing that the quadratic problem also describes nonoscillatory decay.

For real symmetric positive-definite $M,K$ and positive-semidefinite $C$, an oscillatory eigenpair satisfies a useful energy check. Multiply by the conjugate transpose $r^{\ast }$ and write $m=r^{\ast }Mr>0$, $c=r^{\ast }Cr\ge 0$, $k=r^{\ast }Kr>0$. The imaginary part of $m{\lambda }^2+c\lambda +k=0$ gives $\mathrm{Re}\lambda =-c/(2m)\le 0$ when $\mathrm{Im}\lambda \ne 0$. In a multi-coordinate system, complex eigenvectors can encode relative phase differences between coordinates; nonclassical damping generally prevents replacing them by the undamped real mode shapes.