Structural dynamics ($M\ddot{u}+C\dot{u}+Ku=f(t)$)

Structural dynamics studies the time-dependent motion of structures under applied forces and their response after a disturbance. Inertia, stiffness and damping jointly determine that motion. A common linear small-displacement model iswhere $u$ collects displacement coordinates, $M$ is the mass matrix, $C$ the damping matrix, $K$ the stiffness matrix and $f(t)$ the applied-force vector. Boundary conditions and the chosen coordinates are part of the model; a static force balance omits the velocity and acceleration terms.

A scalar version makes the consequences quantitative. Choose $m=1000\mathrm{kg}$, $k=40000\mathrm{N}{\mathrm{m}}^{-1}$ and $c=400\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. Its undamped natural frequency isFor a constant 100-newton force, static displacement is $100/k=2.5\mathrm{mm}$. For a sinusoidal force $100\cos ({\omega }_{0}t)$, the steady-state displacement amplitude of this linear forced harmonic oscillator is insteadThese are original model values, not measurements of a building. They show why the force amplitude alone does not determine displacement: forcing frequency and damping matter.

Real structures have multiple normal modes. A 2025 study of a viscous inerter damper uses a numerical 20-storey building example and shows that inertia and flexibility associated with discarded modes affect the reduced model's tuning. Matching one nominal mode while ignoring those contributions can therefore misrepresent the coupled device–structure behavior.^[ref]

Measured motion can test a chosen model, while its predictions remain limited by assumptions about linearity, boundary conditions and loading. A large predicted response is a reason to examine those assumptions, not proof that an actual structure will follow the linear response indefinitely.

A damping matrix $C$ maps generalized velocity to the velocity-dependent resisting-force term in a linear structural dynamics model:Here $M$ is the mass matrix and $K$ is the stiffness matrix. The sign convention places $C\dot{u}$ on the left, so the actual resisting force is $-C\dot{u}$. For pure translational coordinates, entries have units $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$; mixed rotational and translational coordinates require their corresponding generalized units.

Consider an original two-coordinate model with three viscous dampers: one from each coordinate to a fixed support, and a third between them. With positive coefficients $c_1,c_2,c_{12}$,For $\dot{u}=(v_1,v_2{)}^T$, the removed mechanical power isThis makes $C$ a positive semidefinite matrix. In particular, the connecting damper dissipates nothing when both coordinates move at the same speed. Choosing $c_1=c_2=10$ and $c_{12}=5$ in newton-seconds per metre gives $C=\left(\begin{array}{cc}15& -5\\ -5& 15\end{array}\right)$. Speeds $(1,0)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate15watts, while $(1,1)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate20watts.

For symmetric constant $M,K$, multiplying the equation by ${\dot{u}}^T$ gives an energy balance:The matrix therefore encodes where dissipation occurs and which relative motions activate it.

Rayleigh damping approximates $C$ by a weighted combination of the mass and stiffness matrices. OpenSees exposes separate coefficients for mass and different stiffness states, and cautions that this model can give incorrect results in some nonlinear concentrated-plasticity analyses.^[ref] A convenient matrix prescription must still represent the intended dissipation mechanism.

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.

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]

Modal damping coupling is the interaction between undamped modal coordinates produced by off-diagonal entries of the modal damping matrix. With mass-normalized modes and $D={\Phi }^{T}C\Phi$, the equations areThus $D_{ij}{\dot{q}}_j$ contributes to mode $i$'s force when $j\ne i$. Distinct undamped natural frequencies fix the modal directions up to scaling and ordering; repeated-frequency subspaces admit additional basis rotations and must be treated before interpreting an off-diagonal entry as unavoidable nonclassical damping.^[ref]

For an original example, takeIts eigenvalues are $0.5$ and $0.1$, so it is positive definite. At modal velocities $\dot{q}=(1,-2{)}^T$, total dissipated power isA diagonal-only approximation would give1.5 instead, despite leaving both diagonal coefficients unchanged.

The individual modal energies $E_i=\frac{1}{2}({\dot{q}}_i^2+{\omega }_i^2{q}_i^2)$ obey, with no external forcing,For these velocities, ${\dot{E}}_1=+0.1$ while ${\dot{E}}_2=-0.8$: one modal energy grows temporarily, even though the total falls at rate0.7. This modal energy transfer is compatible with a passive mechanical network; it is not evidence that an external energy source drives the growing coordinate.

For harmonic forcing, the complete response solvesOff-diagonal effects therefore depend on forcing frequency and the other dynamic terms, not solely on the size of $D_{ij}$ relative to $D_{ii}$. Neglecting them is an approximation whose error can be enhanced near nearby resonances. Solving the coupled equations or the damped quadratic eigenvalue problem retains those interactions; assigning independent scalar damping ratios does not, by itself, justify dropping them.

Modal energy transfer describes redistribution of energy among the modal coordinates of a mechanical model. In mass-normalized undamped coordinates with ${\Omega }^2=\mathrm{diag}({\omega }_i^2)$, defineFor $\ddot{q}+D\dot{q}+{\Omega }^{2}q=g$, differentiation yieldsOff-diagonal terms of the modal damping matrix can therefore make an individual modal energy grow even when there is no external forcing. Their presence is modal damping coupling.^[ref]

For an original example, letThis damping matrix has positive eigenvalues0.7 and0.1. The energy rates are ${\dot{E}}_1=+0.2$ and ${\dot{E}}_2=-1.0$, whereasMode1 gains energy while the combined system loses energy. No negative total dissipation or external power source is needed.

The cross terms deserve care: $-D_{12}{\dot{q}}_1{\dot{q}}_2$ and $-D_{21}{\dot{q}}_2{\dot{q}}_1$ are not generally equal and opposite. They contribute to the total dissipation as well as to each coordinate's budget. Thus a damping-based redistribution should not automatically be interpreted as a unique lossless exchange between two modal reservoirs. A different permitted choice of basis, especially within a repeated-frequency subspace, can change individual modal energies while preserving total physical energy.

Summing the budgets gives the physically decisive balance $\dot{E}={\dot{q}}^{T}g-{\dot{q}}^{T}D\dot{q}$. With no forcing and positive-semidefinite $D$, total mechanical energy cannot increase, in agreement with the mechanical dissipation inequality. This supplies a stronger interpretation of an observed growing modal amplitude than assuming that every separately plotted mode must decay monotonically.

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.

Rayleigh damping approximates the damping matrix of a linear structural dynamics model by a combination of its mass matrix and stiffness matrix:Here $\alpha$ has units ${\mathrm{s}}^{-1}$ and $\beta$ has units $\mathrm{s}$. This is a convenient form of proportional damping: for constant symmetric $M,K$ and $M$ positive definite, the same normal modes that separate the undamped system also separate this damping matrix. The OpenSees implementation additionally permits coefficients multiplying current, initial and last-committed stiffness matrices; these are distinct choices in a nonlinear calculation.^[ref]

To see the modal consequence, take an undamped normal mode $K{\varphi }_i={\omega }_i^{2}M{\varphi }_i$ and its modal mass $m_i={\varphi }_i^{T}M{\varphi }_i$. ThenThus Rayleigh damping specifies a frequency-dependent damping ratio, rather than the same damping ratio for every normal mode. The mass-proportional term dominates at low nonzero angular frequency, and the stiffness-proportional term dominates at high angular frequency.

For example, fitting a damping ratio of $0.05$ at both ${\omega }_1=10rad/s$ and ${\omega }_2=50rad/s$ givesThe resulting damping ratio is $0.0375$ at $20rad/s$ and $0.0875$ at $100rad/s$. Agreement at two fitted frequencies therefore does not validate damping outside that band. In particular, a zero-frequency rigid motion has no modal damping ratio of this form, yet the term $\alpha M\dot{u}$ can still resist that motion.

For nonnegative $\alpha ,\beta$ and positive semidefinite $M,K$, the dissipated power ${\dot{u}}^{T}C\dot{u}$ is nonnegative. This supplies an immediate physical check on a fit. It does not establish that the chosen damping matrix represents a particular material or joint: an apparatus would need measured modal decay or another measured dissipative response to justify its coefficients. OpenSees specifically cautions that Rayleigh damping can give incorrect results in nonlinear time-history analysis with concentrated plasticity.^[ref]