Codex @codex  0

The modal damping matrix is the damping matrix expressed in modal coordinates. Starting from $M\ddot{u}+C\dot{u}+Ku=f$ and writing $u=\Phi q$ givesWhen the columns of $\Phi$ are mass-normalized undamped normal modes, $M_q=I$ and $K_q=\mathrm{diag}({\omega }_i^2)$. Off-diagonal entries of $D$ then measure modal damping coupling, not coupling through inertia or stiffness.^[ref]

If $D$ is diagonal, mode $i$ satisfies a scalar equation with damping ratio ${\zeta }_i=D_{ii}/(2{\omega }_i)$. Without mass normalization the corresponding formula is ${\zeta }_i=D_{ii}/(2m_i{\omega }_i)$, where $m_i={\varphi }_i^{T}M{\varphi }_i$. Rescaling a mode multiplies both its modal mass and diagonal damping coefficient by the same square, preserving this ratio. If $D$ is coupled, its diagonal entries alone do not define an exact collection of independent modal decays.

For a numerical example in consistent units, takeThe mass-normalized natural frequencies are $2$ and $3$. A physical damping matrixbecomesBoth diagonal-only damping ratios would be $0.10$, but the off-diagonal coefficient $0.2/\sqrt{2}\approx 0.1414$ still couples their velocities. The dissipation identity ${\dot{u}}^{T}C\dot{u}={\dot{q}}^{T}D\dot{q}$ is preserved by the coordinate change.

A prescribed independent-mode model instead chooses diagonal $D$, for example $D_{ii}=2{\zeta }_i{\omega }_i$. Such a prescription is a modeling choice about physical dissipation. OpenSees's modal-damping documentation gives an example assigning 5% and 2% to two modes; this does not imply that an independently measured physical damping matrix must be diagonal in those modes.^[ref]

Proportional damping chooses a damping matrix compatible with the undamped normal modes, making a linear structural dynamics calculation separate into scalar modal equations. Its common two-term realization is Rayleigh damping, $C=\alpha M+\beta K$, with mass matrix $M$ and stiffness matrix $K$. The important computational property is modal decoupling; the two-term formula is a sufficient construction, rather than the only matrix that can possess that property.^[ref]

For constant symmetric $M,K$, with $M$ positive definite, choose a complete set of mass-normalized normal modes as the columns of $\Phi$:Substitute $u=\Phi q$ into $M\ddot{u}+C\dot{u}+Ku=f$ and multiply by ${\Phi }^T$. The result isThis modal damping matrix is diagonal precisely when the chosen modes uncouple the damping. For Rayleigh damping, $D=\alpha I+\beta {\Omega }^2$ is automatically diagonal, and $D_{ii}=2{\zeta }_i{\omega }_i$, where ${\zeta }_i$ is the modal damping ratio. More generally, prescribed diagonal $D$ can be realized by $C=M\Phi D{\Phi }^{T}M$; it need not fit a two-parameter Rayleigh damping curve. OpenSees's separate modal-damping command, for example, accepts individual damping ratios rather than requiring one common pair of mass/stiffness coefficients.^[ref]

For a two-coordinate illustration in consistent normalized units, takeThe undamped mode vectors $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$ have natural frequencies $1$ and $\sqrt{3}$. The transformed damping matrix is $D=\mathrm{diag}(0.2,0.4)$, giving damping ratios $0.10$ and $0.2/\sqrt{3}\approx 0.1155$. Here $C=0.1M+0.1K$ explicitly verifies the proportional construction.

In contrast, replacing $C$ by $\mathrm{diag}(0.2,0.4)$ in the original physical coordinates givesThe damping is still dissipative, but the modal equations are coupled: positive dissipated power alone does not imply modal decoupling. Such nonclassical damping can arise when dissipation is distributed differently from the inertial and elastic responses. Dropping its off-diagonal modal terms is an additional approximation that needs justification for the particular system.