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<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head>
<meta charset="utf-8">
<meta name="generator" content="quarto-1.3.294">
<meta name="viewport" content="width=device-width, initial-scale=1.0, user-scalable=yes">
<title>Structural Dynamics - 3 Forced Vibrations of Single Degree of Freedom Systems</title>
<style>
code{white-space: pre-wrap;}
span.smallcaps{font-variant: small-caps;}
div.columns{display: flex; gap: min(4vw, 1.5em);}
div.column{flex: auto; overflow-x: auto;}
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ul.task-list{list-style: none;}
ul.task-list li input[type="checkbox"] {
width: 0.8em;
margin: 0 0.8em 0.2em -1em; /* quarto-specific, see https://github.com/quarto-dev/quarto-cli/issues/4556 */
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}
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</head><body class="nav-sidebar floating">$$
% 05/16 change, define new env to convert multlined to multline* in HTML
\newenvironment{multlined}{\begin{multline*}}{\end{multline*}}
% define the command for indenting the first line of paragraphs after the first
\newcommand{\newpara}{ }
%\DeclareMathOperator{\expo}{e}
\def\expo{\mathrm{e}}
\def\expon#1{\expo^{#1}}
% \renewcommand\pi{\uppi} % date: 03/29 change: comment out this line since \uppi cannot be understood by qmd
\def\abs#1{\left| #1 \right|}
\def\matabs#1{\Bigl| #1 \Bigr|} % determinant
\def\RB#1{\mathcal{#1}}
% \def\rem#1{{\bfseries{#1}}} % regular emphasis
% \def\mem#1{{\bfseries{#1}}} % much emphasis
% date: 03/30 change \bfseries to \textbf
\def\rem#1{{\textit{#1}}} % regular emphasis
\def\mem#1{{\textit{#1}}} % much emphasis
% date: 03/30 change \emph to ** in qmd
% \def\lem#1{\emph{#1}} % less emphasis
\def\unit#1{\mathrm{\, #1}}
\def\punit#1{\mathrm{#1}}
\DeclareMathOperator{\dif}{d}
\def\diff{\dif \!}
\def\dt{\diff t} % dt
%%%%%%%%%%% VECTORS
\def\vctr#1{\underline{\mathrm{#1}}} % vectors with roman letters
\def\svctr#1{\underline{#1}} % vectors with symbols
\def\pvctr#1{\, \underline{\mathrm{#1}}} % vectors with roman letters and prior spacing
\def\vmag#1{\left| #1 \right|} % magnitude of a vector
\def\subvec#1#2{\vctr{#1}_{#2}} % vector (roman letter) with subscript
%%%%%%%%%%%
%%%%%%%%%%% MATRICES
\def\mtrx#1{[\mathrm{#1}]} % matrices with roman letters
\def\smtrx#1{[\mathnormal{#1}]} % matrices with symbols
\def\mmat{\mtrx{M}} % mass matrix
\def\cmat{\mtrx{C}} % damping matrix
\def\kmat{\mtrx{K}} % stiffness matrix
\def\nvec#1{\underline{{#1}}} % column matrix (vector) for symbols
\def\snvec#1{\underline{{#1}}} % column matrix (vector) for symbols
% \def\nvec#1{\underset{\sim}{\mathrm{#1}}} % column matrix (vector) for symbols
% \def\snvec#1{\underset{\sim}{\mathnormal{#1}}} % column matrix (vector) for symbols
\def\colmat#1{\left\{ \begin{array}{c} #1 \end{array} \right\}} % column matrix array
% \def\nvec#1{\undertilde{\mathrm{#1}}} % column matrix (vector) for symbols
% \def\snvec#1{\undertilde{#1}} % column matrix (vector) for symbols
% 03/31 \undertilde not found
\def\onecol{\nvec{1}}
\def\idmat{\mtrx{I}}
\def\zerocol{\nvec{0}}
\def\zeromat{\mtrx{0}}
\def\barmmat{\mtrx{{M'}}} % mass matrix in barred coordinates
\def\barcmat{\mtrx{{C'}}} % damping matrix in barred coordinates
\def\barkmat{\mtrx{{K'}}} % stiffness matrix in barred coordinates
%%%%%%%%%%%
\def\sint{\int \!}
\def\lsint#1#2{\int_{#1}^{#2} \hspace{-1ex}}
\def\divp#1#2{\frac{\diff #1}{\diff #2}}
\def\divt#1{\frac{\diff \, #1}{\diff t}}
\def\ddivt#1{\frac{\diff^2 #1}{\diff t^2}}
\def\pardiv#1#2{\frac{\partial #1}{\partial #2}}
\def\pdivt#1{\dot{#1}}
\def\pddivt#1{\ddot{#1}}
%\def{\ssum}{\mathsmaller{\sum}}
\DeclareMathOperator{\ssum}{\scaleobj{0.8}{\sum}}
\def\lowsum#1{\displaystyle{\ssum\limits_{#1}^{}}}
\def\allsum#1#2{\ssum\limits_{#1}^{#2}}
\def\volume{V}
\def\area{A}
\def\dV{\diff \volume} % differential volume
\def\dA{\diff \area} % differential area
\def\dm{\diff m} % differential mass
\def\totm{{m}} % total mass of a system of particles
\def\mden{\varrho} % mass density
\def\mlen{\widehat{m}} % mass per unit length
\def\gravity{\mathrm{g}}
\def\com{\mathrm{cm}} % center of mass
\def\cok{\mathrm{ck}} % center of stiffness
%%%%%%%%%%% GENERALIZED COORDINATE
\def\gc{q} % generalized coordinate
\def\dgc{\dot{q}} % dot-time derivative of gen. coord.
\def\ddgc{\ddot{q}} % second dot-time derivative of gen. coord.
\def\gct{\gc (t)}
\def\dgct{\dgc (t)}
\def\ddgct{\ddgc (t)}
\def\gcic{\gc_o} % \initial condition for gen. coord.
\def\dgcic{\dgc_o} % \initial condition for gen. vel.
%%%%%%%%%%%
%%%%%%%%%%% GENERALIZED SDOF SYSTEM
\def\gengc{q^{*}} % generalized coordinate in generalized SDOF sys
\def\dgengc{\dot{q}^{*}} % dot-time derivative of gen. coord. in generalized SDOF sys
\def\ddgengc{\ddot{q}^{*}} % second dot-time derivative of gen. coord. in generalized SDOF sys
\def\gengct{\gengc(t)}
\def\dgengct{\dgengc(t)}
\def\ddgengct{\ddgengc(t)}
\def\gengcic{\gc_o} % initial condition for gen. coord. in generalized SDOF sys
\def\gendgcic{\dgc_o} % initial condition for gen. vel. in generalized SDOF sys
\def\genm{m^{\ast}} % generalized mass
\def\genc{c^{\ast}} % generalized damping
\def\genk{k^{\ast}} % generalized stiffness
\def\genf{f^{\ast}} % generalized force
\def\gendamp{\damp^{\ast}} % generalized damping ratio
\def\shpf{\psi} % continuous shape function in generalized SDOF approach
% \def\vshpf{\snvec{\uppsi}} % shape function column matrix in generalized SDOF approach
%% \uppsi cannot be understood by qmd for html, replace \uppsi with \Psi
\def\vshpf{\snvec{\psi}} % shape function column matrix in generalized SDOF approach
\def\pvshpf#1{\psi_{#1}} % component of shape function column matrix in generalized SDOF approach
\def\genfreq{\freq^{\ast}} % undamped natural frequency in generalized SDOF sys.
%%%%%%%%%%%
% \def\u{u}
% \def\uvec{\nvec{\u}}
% \def\uvect{\nvec{\u}(t)}
% \def\du{\dot{\u}}
% \def\ddu{\ddot{\u}}
% \def\duvec{\dot{\nvec{\u}}}
% \def\duvect{\dot{\nvec{\u}}(t)}
% \def\dduvec{\ddot{\nvec{\u}}}
% \def\dduvect{\ddot{\nvec{{\u}}}(t)}
% \def\dmprat{\zeta}
% \def\ut{{\u}(t)}
% \def\dut{\dot{\u}(t)}
% \def\ddut{\ddot{\u}(t)}
% \def\gdis{\u_{g}}
% \def\gvel{\du_{g}}
% \def\gacc{\ddu_{g}}
% \def\gdist{\u_{g}(t)}
% \def\gvelt{\du_{g}(t)}
% \def\gacct{\ddu_{g}(t)}
% \def\maxgdis{D_{g}}
% \def\maxgvel{{V}_{g}}
% \def\maxgacc{{A}_{g}}
%\def\maxgdis{\u_{g,\mathrm{max}}}
%\def\maxgvel{\du_{g,\mathrm{max}}}
%\def\maxgacc{\ddu_{g,\mathrm{max}}}
%%%%%%%%%%% GROUND MOTION
\def\gdis{g} % ground displacement
\def\gvel{\dot{\gdis}} % ground velocity
\def\gacc{\ddot{\gdis}} % ground acceleration
\def\gdist{\gdis(t)} % ground displacement with time
\def\gvelt{\gvel(t)} % ground velocity with time
\def\gacct{\gacc(t)} % ground acceleration with time
\def\maxgdis{\mathsf{D}_{\! \gdis}} % absolute maximum ground displacement
\def\maxgvel{\mathsf{V}_{\!\gdis}} % absolute maximum ground velocity
\def\maxgacc{\mathsf{A}_{\!\gdis}} % absolute maximum ground acceleration
\def\adis{\alpha} % absolute displacement (earthquake analysis)
\def\avel{\dot{\adis}} % absolute velocity
\def\aacc{\ddot{\adis}} % absolute acceleration
\def\adisvec{\snvec{{\adis}}} % absolute acceleration
\def\aaccvec{\snvec{\ddot{\adis}}} % absolute acceleration
\def\gdisvec{\nvec{\gdis}} % multiple support motion displacement column
\def\gaccvec{\nvec{\gacc}} % multiple support motion acceleration column
\def\ininfmat{\mtrx{b}} % input influence matrix
\def\ininfvec{\nvec{b}} % input influence vector
%%%%%%%%%%%
%%%%%%%%%%% EARTHQUAKE RESPONSE
\def\baseshear{V_{b}} % base shear
\def\overturn{M_{b}} % overturning moment
\def\estat{f_{es}} % equivalent static force
\def\estati#1{f_{es,#1}} % equivalent static force
\def\estatvec{\nvec{f_{es}}} % equivalent static force
\def\modbaseshear#1{V_{b}^{(#1)}} % modal base shear
\def\modoverturn#1{M_{b}^{(#1)}} % modal overturning moment
\def\modestatvec#1{\nvec{f_{es}^{(#1)}}} % modal equivalent static force
\def\ccor#1{\varrho_{#1}}
\def\excifact{L} % earthquake excitation factor
\def\modpart{\Gamma} % modal participation factor
%%%%%%%%%%% SPECTRA
\def\dspec{\mathsf{D}} % disp spectral resp
\def\vspec{\mathsf{V}} % vel spectral resp
\def\aspec{\mathsf{A}} % acc spectral resp
\def\maxstifforce{\extforce_{el}} % max spring force (SDOF)
\def\maxlindisp{\gc_{el}} % maximum linear disp
\def\maxtotdisp{\gc_{max}} % maximum total disp
\def\yielddisp{\gc_{yl}} % yield displacement
\def\redfac{R} % Yield strength reduction factor
\def\duct{\mu} % ductility factor
\def\ydspec{\dspec_{yl}}
\def\yvspec{\vspec_{yl}}
\def\yaspec{\aspec_{yl}}
%%%%%%%%%%%
%%%%%%%%%%% FORCE
\def\extforce{f} % external force
\def\extforcet{f (t)} % external force with time (t)
\def\extf{\nvec{\extforce}} % external force vector
\def\extft{\extf (t)} % external force vector with time (t)
\def\fvec{\vctr{f}} % resultant force vector
\def\compf{f} % scalar components of the resultant force
\def\sforce{F} % scalar single external force, amplitude of external force
\def\force{\vctr{\mathrm{\sforce}}} % single external force vector
\def\stifforce{\extforce_{S}}
\def\intf{\nvec{\extforce_{I}}} % inertia force vector
\def\stff{\nvec{\extforce_{S}}} % stiffness force vector
\def\dmpf{\nvec{\extforce_{D}}} % damping force vector
\def\ldvec{\nvec{\extforce}} % load (force) vector in MDOF systems
\def\ldtvec{\ldvec (t)} % load (force) vector with time in MDOF systems
\def\barldvec{\nvec{\extforce'}} % load (force) vector in MDOF systems barred coord
\def\resistforce{\extforce_{R}} % resisting force (stiffness or stiffness+damping)
%%%%%%%%%%%
%%%%%%%%%%% GENERALIZED COORDINATE VECTORS
\def\gcvec{\nvec{\gc}} % generalized coordinate vector
\def\dgcvec{\dot{\gcvec}} % generalized velocity vector
\def\ddgcvec{\ddot{\gcvec}} % generalized acceleration vector
\def\gctvec{\gcvec (t)} % generalized coordinate vector with time
\def\dgctvec{\dgcvec (t)} % generalized velocity vector with time
\def\ddgctvec{\ddgcvec (t)} % generalized acceleration vector with time
\def\adisvec{\snvec{\alpha}} % absolute displacement column (earthquake analysis)
\def\avelvec{\dot{\adisvec}} % absolute velocity column
\def\aaccvec{\ddot{\adisvec}} % absolute acceleration column
\def\gcvecic{\nvec{\gc_o}} % initial displacement column
\def\dgcvecic{\nvec{\dot{\gc}_o}} % initial displacement column
\def\bargcvec{\nvec{{\gc}}'} % barred generalized coordinates
\def\bardgcvec{\nvec{{\dot{\gc}}}'} % barred generalized velocities
\def\barddgcvec{\nvec{{\ddot{\gc}}}'} % barred generalized accelerations
%%%%%%%%%%%
%%%%%%%%%%% MODAL COORDINATES
\def\modcor{z} % modal coordinates
\def\modcort{\modcor(t)} % modal coordinate with time
\def\dmodcor{\dot{\modcor}} % modal velocity
\def\ddmodcor{\ddot{\modcor}} % modal acceleration
\def\modcorvec{\nvec{\modcor}} % modal coordinate vector
\def\modcortvec{\nvec{\modcor}(t)} % modal coordinate vector with time
\def\dmodcorvec{\dot{\modcorvec}} % modal velocity vector
\def\ddmodcorvec{\ddot{\modcorvec}} % modal acceleration vector
\def\modm{\widehat{M}} % modal mass
\def\modc{\widehat{C}} % modal damping matrix coefficient
\def\modk{\widehat{K}} % modal stiffness
\def\modmmat{\mtrx{\widehat{M}}} % modal mass matrix
\def\modcmat{\mtrx{\widehat{C}}} % modal damping matrix
\def\modcmatm{\mtrx{\widehat{C}^{\sharp}}} % modal damping matrix modified
\def\modkmat{\mtrx{\widehat{K}}} % modal stiffness matrix
\def\modcoric#1{{\modcor_{#1 o}}} % initial ith modal coordinate
\def\dmodcoric#1{{\dmodcor_{#1 o}}} % initial ith modal velocity
\def\modcorvecic{\nvec{\modcor_o}} % initial modal coordinate vector
\def\dmodcorvecic{\nvec{\dot{\modcor}_{o}}} % initial modal velocity vector
\def\modcoramp{Z} % amplitude of free vibration in modal coordinates
\def\modextforce{\widehat{\extforce}} % load (force) component in modal coords
\def\modldvec{\nvec{\modextforce}} % load (force) vector in modal coords
\def\modgcvec#1{\nvec{\gc}^{(#1)}}
%%%%%%%%%%%
% absolute disp variable ua MDOF (currently same as gen coord.)
% \def\ua{q}
% \def\dua{\dot{\ua}}
% \def\ddua{\ddot{\ua}}
% \def\dmprat{\zeta}
% \def\uat{\ua(t)}
% \def\duat{\dot{\ua}(t)}
% \def\dduat{\ddot{\ua}(t)}
% \def\uavec#1{\nvec{\ua}_{#1}}
% \def\xuavec{\nvec{\ua}} % modified on 06/20 because $\uavec{}_{N\times1}$ does not work in html
% \def\duavec#1{\dot{\nvec{\ua}_{#1}}}
% %\def\dduavec#1{\nvec{\ddua}_{#1}}
% \def\xduavec{\dot{\nvec{\ua}}}
% \def\xdduavec{\ddot{\nvec{\ua}}}
% \def\uavect{\xuavec (t)}
% \def\duavect{\xduavec (t)}
% \def\dduavect{\xdduavec (t)}
%%%%% EIGENVALUES; EIGENVECTORS; RAYLEIGH-RITZ
\def\eigvecs{\phi} % mode shape symbol
\def\eigvec{\snvec{\eigvecs}} % mode shape vector
\def\eigveci#1{\snvec{\eigvecs_{#1}}} % mode shape vector for the ith mode
\def\meigveci#1{\snvec{\overline{\eigvecs_{#1}}}} % mass normalized mode shape
\def\teigveci#1{\snvec{\eigvecs_{#1}}^T} % mode shape vector transposed for the ith mode
\def\rayvecs{\psi} % rayleigh vector symbol
\def\rayvec{\snvec{\rayvecs}} % rayleigh vector
\def\trayvec{\snvec{\rayvecs}^T} % rayleigh vector
\def\ritzvec#1{\snvec{\rayvecs_{#1}}}
\def\tritzvec#1{\snvec{\rayvecs_{#1}}^T}
\def\ritzcandvec#1{\nvec{u_{#1}}}
\def\tritzcandvec#1{\snvec{u_{#1}}^T}
\def\ritzmat{\mtrx{U}}
\def\rayfreq{\freq^{\ast}} % Rayleigh's quotient
\def\ritzfreq#1{\freq^{\ast}_{#1}} % Ritz frequencies
\def\ritzmmat{\mtrx{M^{\ast}}}
\def\ritzkmat{\mtrx{K^{\ast}}}
\def\deigvec{\snvec{\widehat{\eigvecs}}}
\def\deigveci{\snvec{\widehat{\eigvecs_{i}}}}
\def\eigval{\lambda}
% \def\modmat{\smtrx{\Phiup}}
\def\modmat{\mtrx{\Phi}} % date: 05/01 change
\def\spectmat{\mtrx{\omega^2}} % date: 05/01 change
\def\dmodmat{\mtrx{\widehat{\Phi}}}
%%%%%%%%%%
%%%%%%%%%%% TIME STEPPING & NONLINEARITY
% \def\dsct#1#2{#1_{#2}}
% \def\edsct#1#2#3{#1^{(#3)}_{#2}}
\def\dsct#1#2{#1_{[#2]}}
\def\edsct#1#2#3{#1^{(#3)}_{[#2]}}
\def\yieldforce{\extforce_{yl}}
% \def\residual{\Delta \stifforce}
\def\residual{\Delta \resistforce}
%%%%%%%%%%%%
\def\puvec#1{\, \vctr{u}_{#1}}
% Cartesian rectangular unit vectors
\def\xuvec{\, \vctr{i}}
\def\yuvec{\, \vctr{j}}
\def\zuvec{\, \vctr{k}}
\def\pos{r} % position variable
\def\vel{v} % velocity
\def\acc{a} % acceleration
\def\posrelcom{R} % position relative to the center of mass
\def\pvec{\vctr{\pos}} % position vector
\def\vvec{\vctr{\vel}} % velocity vector
\def\avec{\vctr{\acc}} % acceleration vector
\def\zerovec{\vctr{0}} % zero vector
%\def\pdpvec{\vctr{\pdivt{\pos}}} % time (dot) derivative of the position vector
%\def\pdvvec{\vctr{\pdivt{\vel}}} % time (dot) derivative of the velocity vector
\def\pdpvec{\pdivt{\pvec}} % time (dot) derivative of the position vector
\def\pdvvec{\pdivt{\vvec}} % time (dot) derivative of the velocity vector
\def\prelcom{\vctr{\posrelcom}} % position vector relative to the center of mass
\def\vrelcom{\pdivt{\prelcom}} % velocity vector relative to the center of mass
\def\arelcom{\pddivt{\prelcom}} % velocity vector relative to the center of mass
\def\dpvec{\diff \pvec} % differential of position vector
\def\dvvec{\diff \vvec} % differential of velocity vector
\def\angvel{\omega} % scalar angular velocity
\def\dtangvel{\pdivt{\angvel}} % dot-time derivative of scalar angular velocity
%\def\vecangvel{\svctr{\angvel}} % angular velocity vector
\def\angvelvec{\svctr{\omega}} % angular velocity vector
\def\dtangvelvec{\pdivt{\svctr{\angvel}}} % dot-time derivative of angular velocity vector
\def\angacc{\alpha} % scalar angular acceleration
\def\angaccvec{\vctr{\angacc}} % angular acceleration vector
%\def\om{\Omega}
%\def\omvec{\vctr{\om}}
\def\vprod{\times} % vector product
\def\sprod{\cdot} % scalar product
\def\spvprod{\!\! \times} % vector product with spacing adjusted
\def\spsprod{\!\! \cdot} % scalar product with spacing adjusted
\def\linmom{\vctr{\mathrm{L}}} % linear momentum vector
\def\angmom#1{\vctr{\mathrm{H}}_{#1}} % angular momentum vector with respect to point #1
\def\dlinmom{\pdivt{\vctr{\mathrm{L}}}} % dot-time derivative of the linear momentum vector
\def\dangmom#1{\pdivt{\vctr{\mathrm{H}}}_{#1}} % dot-time derivative of the angular momentum vector with respect to point #1
\def\smoment{{M}}
\def\moment#1{\vctr{\mathrm{\smoment}}_{#1}}
\def\pmoment#1#2{\vctr{\mathrm{\smoment}}_{#1}^{(#2)}}
%%%%%%%%%%%%%%% FREQUENCIES etc.
\def\freq{\omega} % undamped natural frequency
\def\cfreq{f} % undamped cyclic frequency
\def\period{T} % period
\def\dfreq{\overline{\freq}} % damped frequency
\def\dperiod{\overline{\period}} % damped period
\def\phs{\theta} % phase angle
\def\damp{\zeta} % damping ratio
\def\extfreq{\Omega} % frequency of external harmonic excitation
\def\extphs{\varphi} % phase of external harmonic excitation
\def\ratfreq{\rho}
\def\ratdur{\beta}
%%%%%%%%%%%%% INERTIA
%\DeclareMathOperator{\inrt}{\mathds{I}}
\def\inrt{{I}} % symbol for inertia
\def\inertia#1{\mkern1mu \inrt_{#1}} % inertia with subscript
\def\dtinertia#1{\mkern1mu \pdivt{\inrt}_{#1}} % dot-time derivative of inertia with subscript
\def\ke{\mathscr{T}} % kinetic energy
\def\pe{\mathscr{V}} % potential energy
\def\te{\mathscr{E}} % total energy
\def\work{\mathscr{W}} % work
\def\virt{\delta} % virtual (variation, work, etc.)
% \def\virt{\updelta} % virtual (variation, work, etc.) date: 03/29 change: comment out this line
\def\vgc{\virt q} % (virtual) variation in gen. coord.
\def\vgengc{\virt \gengc} % (virtual) variation in gen. coord. in generalized SDOF sys
\def\vwork{\virt \mathscr{W}} % virtual work
\def\lagrangian{\mathscr{L}}
\def\vforce{\mathscr{F}} % Lagrangian force; coefficient of the virtual disp. in virtual work.
\def\wnc{{\mathscr{W}}^{nc}} % work of non-conservative forces
\def\vwnc{{\mathscr{W}}^{nc}} % work of non-conservative forces
\def\pwork#1{\mathscr{W}_{#1}} % work on some particle called #1
\def\pwnc#1{{\mathscr{W}}^{nc}_{#1}} % work of non-conservative forces on some particle called #1
\def\dbar{{\mathchar'26\mkern-12mu \mathrm{d}}} % not total derivative
\def\grad{\vctr{\nabla}} % gradient operator
\def\imag{\mathrm{j}}
\def\curv{\kappa} % curvature (as in moment-curvature)
\def\ratio#1#2{\displaystyle{\frac{#1}{#2}}}
\def\dfrm{\Delta} % deformation
\def\maxdfrm{\Delta_{\mathrm{max}}} % deformation
% \def\impresp{\mathscr{h}} %impulse response function
% \def\frf{\mathscr{H}} %frequency response function
\def\impresp{\mathsf{h}} %impulse response function
\def\frf{\mathsf{H}} %frequency response function
\def\tshift{t_{\star}}
\def\cosp#1{\cos \left( #1 \right)}
\def\sinp#1{\sin \left( #1 \right)}
\def\dynamp{\mathbb{D}}
\def\dynampr{\dynamp (\ratfreq,\damp)}
\def\tstep{t_{\Delta}}
\def\ststep{{t^2_{\Delta}}}
%%% State Space Models
% \def\dscs#1{\nvec{{x}_{#1}}}
%\def\dscs#1{\nvec{x}_{#1}}
%\def\dscf#1{\nvec{{f}}_{#1}}
\def\dscs#1{\nvec{x}_{[#1]}}
\def\dscf#1{\nvec{{f}}_{[#1]}}
\def\dscA{\mtrx{A}}
\def\dscB{\mtrx{B}}
\def\eigval{\lambda}
\def\eigvecmat{\mtrx{V}}
% \def\eigvalmat{\mtrx{\uplambda}}
\def\eigvalmat{\mtrx{\lambda}} %05/01 change
$$
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<li><a href="#general-methodology" id="toc-general-methodology" class="nav-link active" data-scroll-target="#general-methodology"><span class="header-section-number">3.1</span> General Methodology</a>
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<li><a href="#constant-force-step-input" id="toc-constant-force-step-input" class="nav-link" data-scroll-target="#constant-force-step-input"><span class="header-section-number">3.1.1</span> Constant Force: Step Input</a></li>
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<li><a href="#constant-load-applied-in-finite-time" id="toc-constant-load-applied-in-finite-time" class="nav-link" data-scroll-target="#constant-load-applied-in-finite-time"><span class="header-section-number">3.2.2</span> Constant Load Applied in Finite Time</a></li>
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<li><a href="#sec-harmonicforce" id="toc-sec-harmonicforce" class="nav-link" data-scroll-target="#sec-harmonicforce"><span class="header-section-number">3.3</span> Harmonic Force Excitations</a>
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<li><a href="#sec-dynamp" id="toc-sec-dynamp" class="nav-link" data-scroll-target="#sec-dynamp"><span class="header-section-number">3.3.1</span> Dynamic Amplification</a></li>
<li><a href="#sec-harmforceudmpd" id="toc-sec-harmforceudmpd" class="nav-link" data-scroll-target="#sec-harmforceudmpd"><span class="header-section-number">3.3.2</span> Response of Undamped Systems</a></li>
<li><a href="#sec-harmforcedmpd" id="toc-sec-harmforcedmpd" class="nav-link" data-scroll-target="#sec-harmforcedmpd"><span class="header-section-number">3.3.3</span> Response of Damped Systems</a></li>
<li><a href="#beat-phenomenon" id="toc-beat-phenomenon" class="nav-link" data-scroll-target="#beat-phenomenon"><span class="header-section-number">3.3.4</span> Beat Phenomenon</a></li>
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<li><a href="#sec-rectangularpulse" id="toc-sec-rectangularpulse" class="nav-link" data-scroll-target="#sec-rectangularpulse"><span class="header-section-number">3.4.1</span> Rectangular Pulse</a></li>
<li><a href="#sec-sinepulse" id="toc-sec-sinepulse" class="nav-link" data-scroll-target="#sec-sinepulse"><span class="header-section-number">3.4.2</span> Half-Sine Pulse</a></li>
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<li><a href="#sec-harmonicGM" id="toc-sec-harmonicGM" class="nav-link" data-scroll-target="#sec-harmonicGM"><span class="header-section-number">3.7.2</span> Response to Harmonic Base Excitations</a></li>
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<h1 class="title"><span id="sec-sdofforcedvib" class="quarto-section-identifier"><span class="chapter-number">3</span> <span class="chapter-title">Forced Vibrations of Single Degree of Freedom Systems</span></span></h1>
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<p>While investigation of free vibrations is a necessary starting point in discussions, critical performance issues are encountered almost always during forced vibrations while the system is acted upon by external load effects, either in the form of external forces or support motions such as those occurring during earthquakes. We’ll see that, while only the excitation amplitudes matter in static analyses, how the excitation varies with time and in relation to the time constants of the system matters as much, if not even more, in dynamic analyses. So let us begin by introducing some typical excitation patterns that are frequently encountered, hoping to lay the groundwork for the tools we will have to introduce for solution of various problems.</p>
<p>A static force is essentially a force of constant magnitude and direction. In principle, it is impossible to apply any load instantaneously, but when the duration it takes for the load to reach its peak value is much smaller in comparison with the period of the system, it may be feasible to model the load as what is called a step function, an example of which is sketched in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (a). In this model the load is assumed to be applied instantaneously and it is relatively simple to handle analytically. The response of a damped SDOF system will eventually settle to the response one would obtain by static analysis, given by the ratio of the amplitude of applied force to the stiffness of the system. A larger peak response may however occur before this steady value is attained, which may be an important issue in certain applications.</p>
<div id="fig-sampleforcetime" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/sampleforcetime.png" class="quarto-discovered-preview-image img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.1: Various types of forces frequently encountered in dynamic analysis. (a) constant (step), (b) impulsive, (c) harmonic, (d) periodic, (e) numerically defined, (f) random.</figcaption><p></p>
</figure>
</div>
<p>On the other end of the spectrum is a force of very small duration relative to the period of the system, an example of which is sketched in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (b). If the force duration is very small compared to the reaction time of the system, the effect is similar to imposing some initial velocity to the system. Such forces are called <em>impulsive</em>; as the force duration decreases to what one would call to be instantaneous, the force is mathematically modeled as an <em>impulse</em>. It is possible to obtain an analytical solution to the response of an SDOF system to an impulse and, moreover, this solution may be used to construct the response of a system to some arbitrary excitation.</p>
<p>Dynamic effects start to be more pronounced when one considers a harmonic excitation like the one shown in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (c). When the duration of the force is long relative to the period of the system, the system will eventually start to oscillate with the frequency of the force during what is called the <em>steady-state response</em>. We will investigate this type of input in great detail since it will be shown that a very critical condition called <em>resonance</em> occurs when the frequency of excitation approaches that of the system. Resonance is critical because the maximum response of the system under resonance may reach multiple times of the static response it would show under the same amplitude of force, with the amplification factor being some function of damping. As this phenomenon is the cause of many failures in real life, understanding it is of paramount importance.</p>
<p>It may be that the force is not just a single harmonic wave but it comprises repetitions of some particular pattern as is the case shown in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (d). When the duration of the force is long compared with the period of the system, such a repetitive force is called <em>periodic</em>. Under a periodic force the response will also eventually reach a repetitive cycle called the steady-state response; in that case the analysis may be conducted by modeling the periodic force as the superposition of a number of harmonics, and calculating the response as the superposition of the response to each harmonic, using the tools developed for the analysis under a single harmonic force. Periodic forces are important because the presence of multiple harmonics may lead to various risks regarding resonance.</p>
<p>Most often, however, we will have to deal with forces that are somewhat arbitrary in that their time variation can not be characterized analytically, as sketched in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (e). In such cases, if the values of the excitation are known at points in time, the response may be evaluated via numerical techniques. We will introduce some numerical techniques that may be used in a wide variety of systems. Earthquake response calculations are generally based on such numerical methods since ground motions, which act as inputs to structures, are of such arbitrary nature, as numerous recordings of earthquake induced ground vibrations have shown.</p>
<p>In some cases it may not be possible to even measure the input. Some of such forces, an example of which is sketched in <a href="#fig-sampleforcetime">Figure <span>3.1</span></a> (f), may be modeled using tools and techniques from analysis of random processes, and the response of a system subject to such inputs is generally referred to as <em>random vibrations</em>. Random vibrations are generally analyzed in the <em>frequency domain</em> since random inputs are more easily characterized in that domain though their spectra. Although random vibrations are important in some applications, our focus in this chapter will be introducing basic analysis tools and understanding of forced vibrations.</p>
<section id="general-methodology" class="level2" data-number="3.1">
<h2 data-number="3.1" data-anchor-id="general-methodology"><span class="header-section-number">3.1</span> General Methodology</h2>
<p>To develop the general method for developing analytical solutions to forced vibrations, let us once again consider the prototypical SDOF model, shown for ease of reference again in <a href="#fig-sdofmckfbdrepeat">Figure <span>3.2</span></a>. The equation of motion for the system, obtained by summing the forces shown in the free body diagram and considering initial conditions, is given by <span id="eq-eomdsdofgen"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \extforce(t) \, ; \quad \left\{\gc(0) = \gcic, \dgc(0) = \dgcic\right\}
\qquad(3.1)\]</span></span></p>
<div id="fig-sdofmckfbdrepeat" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/sdofmckfbd.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.2: (a) Viscously damped single degree of freedom system and (b) its free body diagram including the d’Alembert force.</figcaption><p></p>
</figure>
</div>
<p>Studies in differential equations tell us that the solution to <a href="#eq-eomdsdofgen">Equation <span>3.1</span></a> may be constructed via two components as <span class="math display">\[
\gct = \gc_{c} (t) + \gc_{p}(t)
\]</span> where <span class="math inline">\(\gc_{c}(t)\)</span> is referred to as the complementary solution, given by the solution to the homogeneous equation so that <span class="math display">\[
m \ddgc_{c}(t) + c \dgc_{c}(t) + k \gc_{c}(t) = 0
\]</span> and <span class="math inline">\(\gc_{p}(t)\)</span> is referred to as the particular solution satisfying <span class="math display">\[
m \ddgc_{p}(t) + c \dgc_{p}(t) + k \gc_{p}(t) = \extforce (t)
\]</span> The particular solution will depend on the forcing function and so it must be determined anew for each different type of force, whereas the complementary solution for an <em>underdamped system</em> will always be given by some form of <span id="eq-compsol"><span class="math display">\[
\gct = \expon{-\damp \freq t}\left(C_1 \cos \dfreq t + C_2 \sin \dfreq t \right)
\qquad(3.2)\]</span></span> where <span class="math inline">\(C_1\)</span> and <span class="math inline">\(C_2\)</span> are two coefficients to be determined. The existence of these two coefficients is what allows us to solve the problem since that freedom is needed to incorporate the contribution of initial conditions, something that the particular solution alone cannot do. That the superposition of the complementary and the particular solutions still satisfy the force equilibrium equation is trivial due to the linearity of the differential equation since <span class="math display">\[\begin{align*}
m \left[\ddgc_{c} + \ddgc_{p}\right] & + c \left[\dgc_{c} + \dgc_{p} \right] + k \left[\gc_{c} + \gc_{p} \right] \\
& = \left[m \ddgc_{c} + c \dgc_{c} + k \gc_{c} \right] + \left[m \ddgc_{p} + c \dgc_{p} + k \gc_{p} \right] \\
& = [0] + \left[m \ddgc_{p} + c \dgc_{p} + k \gc_{p} \right] = \extforce(t)
\end{align*}\]</span></p>
<p>The linearity of the governing equation has even more significant consequences which has proven extremely useful in analysis: since <em>superposition</em> holds, the simultaneous action of a number of disturbances may be analyzed by evaluating the response of the system to each individual disturbance separately, and then superpose all the solutions thus obtained to evaluate the cumulative response; the principle is schematically explained in <a href="#fig-superposition">Figure <span>3.3</span></a>. To express this statement symbolically, if <span class="math inline">\(\gc^{(1)}\)</span> is the response of the system to some input <span class="math inline">\(\text{inp}^{(1)}\)</span> and if <span class="math inline">\(\gc^{(2)}\)</span> is its response to some input <span class="math inline">\(\text{inp}^{(2)}\)</span>, then the response of the system will be given by <span class="math inline">\(\gc^{(1)} + \gc^{(1)}\)</span> when the system is acted upon by both inputs simultaneously, i.e. by <span class="math inline">\(\text{inp}^{(1)} + \text{inp}^{(2)}\)</span>. If the inputs are amplified so that the system is acted upon by <span class="math inline">\(a_1 \text{inp}^{(1)} + a_2 \text{inp}^{(2)}\)</span>, then the responses to individual inputs will likewise be amplified so that the system response will be given by <span class="math inline">\(a_1 \gc^{(1)} + a_2 \gc^{(1)}\)</span>. Note that we have used the more generic term <em>input</em> to denote the effect that sets the system in motion for it can refer to a set of initial conditions and/or external forces and/or base motion; all of these may be investigated separately and the results thus obtained may then be combined to calculate the response that would occur under the combined action of all. We may therefore finally state the following: Let <span class="math inline">\(\gc^{(1)}\)</span> be the solution to the initial condition problem <span id="eq-eomdsdofgenic"><span class="math display">\[
m \ddgc^{(1)} (t) + c \dgc^{(1)} (t) + k \gc^{(1)} (t) = 0 \, ; \quad \left\{\gc^{(1)}(0) = \gcic, \dgc^{(1)} (0) = \dgcic \right\}
\qquad(3.3)\]</span></span> so that it is given for a viscously underdamped system by <span class="math display">\[
\gc^{(1)}(t) = \expon{-\damp \freq t}\left(\gcic \cos \dfreq t + \frac{\dgcic + \damp \freq \gcic}{\dfreq} \sin \dfreq t \right)
\]</span></p>
<div id="fig-superposition" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/superposition.png" class="img-fluid figure-img" style="width:60.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.3: Superposition in linear systems.</figcaption><p></p>
</figure>
</div>
<p>as we had shown while discussing free vibrations. If the same system were subjected only to some external force with zero initial conditions, the response <span class="math inline">\(\gc^{(2)}(t)\)</span> would have to satisfy<a href="#fn1" class="footnote-ref" id="fnref1" role="doc-noteref"><sup>1</sup></a> <span id="eq-eomdsdofgenzero"><span class="math display">\[
m \ddgc^{(2)} (t) + c \dgc^{(2)} (t) + k \gc^{(2)} (t) = \extforce (t) \, ; \quad \left\{\gc^{(2)} (0) = 0, \dgc^{(2)} (0) = 0 \right\}
\qquad(3.4)\]</span></span> When the system is subjected to both the external excitation <span class="math inline">\(\extforce(t)\)</span> and the nonzero initial conditions <span class="math inline">\(\gc(0) = \gcic, \dgc(0) = \dgcic\)</span>, the response <span class="math inline">\(\gct\)</span> would have to satisfy <span id="eq-eomdsdofgen2"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \extforce(t) \, ; \quad \left\{\gc(0) = \gcic, \dgc(0) = \dgcic\right\}
\qquad(3.5)\]</span></span> and the linearity of the system allows superposition so that <span class="math inline">\(\gct\)</span> is simply given by <span id="eq-soldsdofgen"><span class="math display">\[
\gc (t) = \gc^{(2)} (t) + \gc^{(2)} (t) = \expon{-\damp \freq t}\left(\gcic \cos \dfreq t + \frac{\dgcic + \damp \freq \gcic}{\dfreq} \sin \dfreq t \right) + \gc^{(2)} (t)
\qquad(3.6)\]</span></span> This result will allow us to investigate all forced vibration problems with zero initial conditions as the additional contribution of any nonzero initial condition may simply be superposed afterwards.</p>
<!-- [^1]: That the initial conditions are zero DOES NOT mean the complementary solution has to be zero; $\gc_2$ will still need a particular and a complementary component. -->
<section id="constant-force-step-input" class="level3" data-number="3.1.1">
<h3 data-number="3.1.1" data-anchor-id="constant-force-step-input"><span class="header-section-number">3.1.1</span> Constant Force: Step Input</h3>
<p>Let us start with the case of a constant force shown in <a href="#fig-forcestep">Figure <span>3.4</span></a>, defined mathematically as <span class="math display">\[
\extforce (t) = \begin{cases} \sforce & t \geq 0 \\ 0 & t < 0 \end{cases}
\]</span></p>
<!-- change from margin figure to figure, since no margin figure command is found in Quarto -->
<div id="fig-forcestep" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/forcestep.png" class="img-fluid figure-img" style="width:60.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.4: Constant force (step function)</figcaption><p></p>
</figure>
</div>
<p>For <span class="math inline">\(t \geq 0\)</span>, the underdamped system is governed by <span id="eq-xforcestep"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \sforce \, ; \quad \left\{\gc(0) = 0, \dgc(0) = 0\right\}
\qquad(3.7)\]</span></span> and the complementary solution is given by <span id="eq-compsolstep"><span class="math display">\[
\gc_{c} (t) = \expon{-\damp \freq t}\left(C_1 \cos \dfreq t + C_2 \sin \freq t \right)
\qquad(3.8)\]</span></span> We have to find the particular solution before evaluating the unknown coefficients since the initial conditions are to be imposed on the actual response, not just the complementary part. As the forcing function is a constant, we may initially try a constant particular solution of the form <span id="eq-partsolstep"><span class="math display">\[
\gc_{p} (t) = C
\qquad(3.9)\]</span></span> Substituting the trial solution of <a href="#eq-partsolstep">Equation <span>3.9</span></a> into the equation of motion in <a href="#eq-xforcestep">Equation <span>3.7</span></a> leads to<a href="#fn2" class="footnote-ref" id="fnref2" role="doc-noteref"><sup>2</sup></a> <span class="math display">\[
0 + 0 + k C = \sforce
\]</span> so that we conclude the particular solution to be of the trial form with <span class="math display">\[
C = \frac{\sforce}{k}
\]</span> which, it should be noted, is equal the static displacement <span class="math display">\[
\Delta_{st}=\frac{\sforce}{k}
\]</span> if this were a static problem. Now the solution becomes <span id="eq-static"><span class="math display">\[
\gct = \expon{-\damp \freq t}\left(C_1 \cos \dfreq t + C_2 \sin \dfreq t \right) + \frac{\sforce}{k}
\qquad(3.10)\]</span></span> and applying the initial conditions lead to <span class="math display">\[\begin{align*}
\gc (0) = C_1 + \frac{\sforce}{k} = & 0 \quad \rightarrow \quad C_1 = - \frac{\sforce}{k} \\
\dgc (0) = - \damp \freq C_1 + \dfreq C_2 = & 0 \quad \rightarrow \quad C_2 = - \frac{\sforce}{k} \frac{\damp}{\sqrt{1-\damp^2}}
\end{align*}\]</span> so that the solution is finally obtained as <span id="eq-xsolstep"><span class="math display">\[
\gct = \frac{\sforce}{k} \left[ 1 - \expon{-\damp \freq t}\left(\cos \dfreq t + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq t \right) \right] \quad \text{for } t \geq 0
\qquad(3.11)\]</span></span></p>
<p>How the system responds to the force and the effects of damping on the response may be most easily visualized via the plots sketched in <a href="#fig-responsestep">Figure <span>3.5</span></a>. There are a few important things to be noted. First of all, in the presence of damping, the response eventually converges to the static displacement; how fast this convergence occurs depends on the amount of damping with faster rates occurring for larger damping values. Results obtained via static analyses are therefore the asymptotic values the dynamic responses eventually settle to whenever the duration of the load application is long enough. Another point worth making is that the maximum response that occurs may exceed the static response and the peak value depends again on the amount of damping. In fact, the velocity as a function of the normalized time <span class="math inline">\(\tau = t/\dperiod\)</span> may be shown to be given by <span id="eq-xsolstepvel"><span class="math display">\[
\dgc (t) = \frac{\sforce}{k} \expon{- \damp \freq t}\frac{\freq}{\sqrt{1-\damp^2}} \sin \dfreq t
\qquad(3.12)\]</span></span></p>
<div id="fig-responsestep" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/responsestep.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.5: Response to a constant force (step function) for various values of damping ratio (noted in percentage) as a function of normalized time <span class="math inline">\(\tau = t / \dperiod\)</span>.</figcaption><p></p>
</figure>
</div>
<p>which will take on a value of zero for <span class="math inline">\(t > 0\)</span> whenever <span class="math inline">\(\dfreq t\)</span> is a positive integer multiple of <span class="math inline">\(\pi\)</span>, or in other words, whenever <span class="math display">\[
t = \ratio{i \pi}{\dfreq} = i \ratio{\dperiod}{2} \quad \text{for } i=1,2,\ldots
\]</span> which corresponds to <span class="math display">\[
\tau = \ratio{t}{\dperiod} = \ratio{1}{2} i \quad \text{for } i=1,2,\ldots
\]</span> i.e. whenever <span class="math inline">\(\tau\)</span> is a positive integer multiple of <span class="math inline">\(1/2\)</span>. Therefore, the first peak is reached, if at all, when <span class="math inline">\(\tau = 1/2\)</span>, i.e. halfway through the first cycle when <span class="math inline">\(t = \dperiod / 2\)</span>, and the value of the peak displacement is given by <span class="math display">\[
\gc_{max} = \Delta_{st}\left[1+\expon{-\pi \damp / \sqrt{1-\damp^2}}\right]
\]</span> so that for an undamped system we have <span class="math display">\[
\frac{\gc_{max}}{\Delta_{st}}= 2
\]</span> and for damping values of <span class="math inline">\(\zeta=5\%\)</span>, <span class="math inline">\(\zeta=20\%\)</span> and <span class="math inline">\(\zeta=50\%\)</span> the amplification above will be calculated as <span class="math inline">\(1.85\)</span>, <span class="math inline">\(1.53\)</span> and <span class="math inline">\(1.16\)</span>, respectively.</p>
</section>
<section id="footnotes" class="footnotes footnotes-end-of-section" role="doc-endnotes">
<hr>
<ol>
<li id="fn1"><p> That the initial conditions are zero DOES NOT mean the complementary solution has to be zero; <span class="math inline">\(\gc^{(2)}\)</span> will still need a particular and a complementary component.<a href="#fnref1" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
<li id="fn2"><p>This approach is commonly referred to as the method of undetermined coefficients.<a href="#fnref2" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
</ol>
</section>
</section>
<section id="linearly-increasing-force-ramp-input" class="level2" data-number="3.2">
<h2 data-number="3.2" data-anchor-id="linearly-increasing-force-ramp-input"><span class="header-section-number">3.2</span> Linearly Increasing Force: Ramp Input</h2>
<p>Consider a linearly increasing external force as shown in <a href="#fig-forcestep">Figure <span>3.4</span></a>. This force, having a rate of increase of <span class="math inline">\(1\)</span> (unit of force/unit of time), is defined mathematically as <span class="math display">\[
\extforce (t) = \begin{cases} t & t \geq 0 \\ 0 & t < 0 \end{cases}
\]</span> so that for <span class="math inline">\(t \geq 0\)</span>, the underdamped system is governed by <span id="eq-xforceramp"><span class="math display">\[
m \ddgct + c \dgct + k \gct = t \, ; \quad \left\{\gc(0) = 0, \dgc(0) = 0 \right\}
\qquad(3.13)\]</span></span></p>
<div id="fig-forceramp" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/forceramp.png" class="img-fluid figure-img" style="width:40.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.6: Linearly increasing force (ramp function).</figcaption><p></p>
</figure>
</div>
<p>The complementary solution is still given by <a href="#eq-compsolstep">Equation <span>3.8</span></a>. Since the right hand side of the equation is a linear function of <span class="math inline">\(t\)</span>, for the particular solution, we may try <span id="eq-partsolramp"><span class="math display">\[
\gc_p (t) = At + B
\qquad(3.14)\]</span></span> which leads, when substituted into the equation of motion of <a href="#eq-xforceramp">Equation <span>3.13</span></a>, to <span class="math display">\[
0 + cA + Akt + Bk = t
\]</span> In this case, the coefficients of the left and right hand sides of the equation may be matched to conclude <span class="math display">\[
A = \frac{1}{k}, \quad B = - \frac{c}{k^2} = - \frac{2 \damp \freq m}{k^2} = - \frac{1}{k}\frac{2 \damp}{\freq}
\]</span> so that the solution <span class="math inline">\(\gc = \gc_{c} + \gc_{p}\)</span> is of the form <span id="eq-sol"><span class="math display">\[
\gct = \expon{-\damp \freq t}\left(C_1 \cos \dfreq t + C_2 \sin \dfreq t \right) + \frac{1}{k} \left[t - \frac{2 \damp}{\freq} \right]
\qquad(3.15)\]</span></span> Applying the initial conditions leads to <span class="math display">\[\begin{align*}
\gc (0) = C_1 - \frac{2 \damp}{k \freq} = & 0 \quad \rightarrow \quad C_1 = \frac{2 \damp}{k \freq} \\
\dgc (0) = - \damp \freq C_1 + \dfreq C_2 + \frac{1}{k}= & 0 \quad \rightarrow \quad C_2 = \frac{2 \damp^2 - 1}{k\dfreq}
\end{align*}\]</span> with the final solution now given for <span class="math inline">\(t \geq 0\)</span> by <span id="eq-finalsol"><span class="math display">\[
\gct = \frac{1}{k\freq} \left[\expon{-\damp \freq t}\left(2 \damp \cos \dfreq t + \frac{2 \damp^2 - 1}{\sqrt{1-\damp^2}} \sin \dfreq t \right) + \freq t - 2 \damp \right]
\qquad(3.16)\]</span></span></p>
<p>Having determined the solution for a unit rate, we may easily extend this solution to the case when the rate of increase of the force is not unity but some <span class="math inline">\(\lambda\)</span> so that the force is defined as <span class="math display">\[
\extforce (t) = \begin{cases} \lambda t & t \geq 0 \\ 0 & t < 0 \end{cases}
\]</span> and the system is governed by <span id="eq-xforceramp2"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \lambda t \, ; \quad \left\{\gc(0) = 0, \dgc(0) = 0\right\}
\qquad(3.17)\]</span></span> Due to the linearity of the system, the principle of superposition summarized in <a href="#fig-superposition">Figure <span>3.3</span></a> allows us to conclude that when the input is scaled by some value <span class="math inline">\(\lambda\)</span>, the output will also have to be scaled by the same value so that the response will simply be given for <span class="math inline">\(t \geq 0\)</span> by <span id="eq-solforceramp"><span class="math display">\[
\gct = \frac{\lambda}{k \freq} \left[\expon{-\damp \freq t}\left(2 \damp \cos \dfreq t + \frac{2 \damp^2 - 1}{\sqrt{1-\damp^2}} \sin \dfreq t \right) + \freq t - 2 \damp \right]
\qquad(3.18)\]</span></span> which of course includes the initial problem as a special case with <span class="math inline">\(\lambda = 1\)</span>. If the system is undamped, then using <span class="math inline">\(\damp = 0\)</span> in <a href="#eq-solforceramp">Equation <span>3.18</span></a> leads to <span id="eq-solforcerampud"><span class="math display">\[
\gct = \frac{\lambda}{k\freq} \left[ \freq t - \sin \freq t \right] \quad \text{for } t\geq 0
\qquad(3.19)\]</span></span></p>
<div id="fig-responseramp" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/responseramp.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.7: Response to a linearly increasing force (step function) for various values of damping ratio (noted in percentage) as a function of normalized time <span class="math inline">\(\tau = t / \period\)</span>.</figcaption><p></p>
</figure>
</div>
<p>A graphical comparison of the variation of response with damping in this case is not as straightforward as it was in the response to a step function. To help discuss general trends, <a href="#fig-responseramp">Figure <span>3.7</span></a> presents the responses that would be observed for a particular system for various values of the damping ratio. The normalized time used in these plots is <span class="math inline">\(\tau = t / \period\)</span> which differs from that used for previous investigations in that here we use the undamped period as opposed to the damped one. To help facilitate a visual reference, the plots also include what may be referred to as instantaneous static response <span class="math inline">\(\extforce (\tau) / k\)</span>. The tendency of the response is to converge to some percentage of the instantaneous static response. Since <span class="math inline">\(\freq t = 2 \pi \tau\)</span>, for an undamped system we have <span class="math display">\[
\frac{\extforce(\tau)}{k} - \gc (\tau) = \frac{\lambda}{\freq k} 2 \pi \tau - \frac{\lambda}{\freq k} \left[ 2 \pi \tau - \sin 2 \pi \tau \right] = \frac{\lambda}{\freq k}\sin 2 \pi \tau
\]</span> so that the response oscillates around the instantaneous static response. On the other hand, whenever there is damping in the system we have, from <a href="#eq-solforceramp">Equation <span>3.18</span></a>, <span class="math display">\[
\underset{\tau \rightarrow \infty}{\lim}\left[\frac{\extforce(\tau)}{k} - \gc (\tau)\right] = \frac{\lambda}{\freq k} 2 \pi \tau - \frac{\lambda}{\freq k} \left[ 2 \pi \tau - 2 \damp\right] = 2 \damp \frac{\lambda}{\freq k}
\]</span> so that eventually the transient vibrations die out and the steady state response tracks the force with a certain lag that depends on the damping ratio for a given system and force.</p>
<section id="sec-xmpl:timeshift" class="level3" data-number="3.2.1">
<h3 data-number="3.2.1" data-anchor-id="sec-xmpl:timeshift"><span class="header-section-number">3.2.1</span> Input Shifted in Time</h3>
<p>Consider what the response would be if the step and the ramp forces of the discussions above were applied not at time <span class="math inline">\(t=0\)</span> but they started to act at some time <span class="math inline">\(t=\tshift\)</span> as shown in <a href="#fig-shiftinput">Figure <span>3.8</span></a>.</p>
<p>Since the coefficients in the differential equation are time invariant, or in other words since the mass, damping and stiffness properties of the system remain unaltered as time progresses, the solution to these problems is relatively straightforward. Let us take the step input to begin with. The problem statement for zero initial conditions is given by <span class="math display">\[
m \ddgct + c \dgct + k \gct = \sforce \, ; \quad \left\{\gc(\tshift) = 0, \dgc(\tshift) = 0\right\} \quad \text{for } t \geq \tshift
\]</span> If we define a new time variable <span class="math inline">\(\tau = t - \tshift\)</span>, then <span class="math display">\[
\frac{\diff q}{\diff \tau} = \frac{\diff q}{\diff t} \frac{\diff t}{\diff \tau} = \frac{\diff q}{\diff t}, \quad \frac{\diff^2 q}{\diff \tau^2} = \frac{\diff^2 q}{\diff t^2}
\]</span> so that the problem may be stated in terms of this new time variable as <span class="math display">\[
m \ddgc (\tau) + c \dgc (\tau) + k \gc (\tau) = \sforce \, ; \quad \left\{\gc(0) = 0, \dgc(0) = 0\right\} \quad \text{for } \tau \geq 0
\]</span> Note that this is exactly the same problem as that in <a href="#eq-xforcestep">Equation <span>3.7</span></a>, with <span class="math inline">\(\tau\)</span> replacing <span class="math inline">\(t\)</span>. The solution may therefore immediately be written following <a href="#eq-xsolstep">Equation <span>3.11</span></a> <span class="math display">\[%\begin{xequation}\label{eq:xsolstepshift}
\gc (\tau) = \frac{\sforce}{k} \left[ 1 - \expon{-\damp \freq \tau}\left(\cos \dfreq \tau + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq \tau \right) \right] \quad \text{for } \tau \geq 0
\]</span> and substituting <span class="math inline">\(\tau = t - \tshift\)</span> leads to <span class="math display">\[%\begin{xequation}\label{eq:xsolstepshift2}
\gct = \frac{\sforce}{k} \left[ 1 - \expon{-\damp \freq (t-\tshift)}\left(\cos \dfreq (t-\tshift) + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq (t-\tshift) \right) \right] \quad \text{for } t \geq \tshift
\]</span></p>
<div id="fig-shiftinput" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/shiftinput.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.8: Step and ramp inputs shifted in time.</figcaption><p></p>
</figure>
</div>
<p>As the methodology described above is general, the conclusion is obvious: since the coefficients of the linear equation of motion are time invariant, a time shift in input results simply results in a shift in the output. If, for example, the initial conditions for the system were nonzero at time <span class="math inline">\(t=\tshift\)</span> so that the system were to be governed by <span class="math display">\[
m \ddgct + c \dgct + k \gct = \sforce \, ; \quad \left\{\gc(\tshift) = \gc_{*}, \dgc(\tshift) = \dgc_{*}\right\} \quad \text{for } t \geq \tshift
\]</span> then, based on equations <a href="#eq-soldsdofgen">Equation <span>3.6</span></a> and <a href="#eq-xsolstep">Equation <span>3.11</span></a> wherein we replace <span class="math inline">\(t\)</span> by <span class="math inline">\(t-\tshift\)</span>, and <span class="math inline">\(\gcic\)</span> and <span class="math inline">\(\dgcic\)</span> by <span class="math inline">\(\gc_{*}\)</span> and <span class="math inline">\(\dgc_{*}\)</span>, respectively, the response would be given by: <span class="math display">\[\begin{align*}
\gc (t) & = \expon{-\damp \freq (t-\tshift)}\left[\gc_{*} \cos \dfreq (t-\tshift) + \frac{\dgc_{*} + \damp \freq \gc_{*} }{\dfreq} \sin \dfreq (t-\tshift) \right] + \\
& \; \frac{\sforce}{k} \left[ 1 - \expon{-\damp \freq (t-\tshift)}\left(\cos \dfreq (t-\tshift) + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq (t-\tshift) \right) \right] \quad \text{for } t \geq \tshift
\end{align*}\]</span></p>
<p>Similarly, if an underdamped SDOF system, initially at rest, is subjected to the ramp function shown in <a href="#fig-shiftinput">Figure <span>3.8</span></a> so that it is governed by <span class="math display">\[
m \ddgct + c \dgct + k \gct = \lambda (t-\tshift) \, ; \quad \left\{\gc(\tshift) = 0, \dgc(\tshift) = 0\right\} \quad \text{for } t \geq \tshift
\]</span> then its response is obtained simply by shifting the solution in <a href="#eq-solforceramp">Equation <span>3.18</span></a> as: <span class="math display">\[
\gct = \frac{\lambda}{k \freq} \left[\expon{-\damp \freq (t-\tshift)}\left(2 \damp \cos \dfreq (t-\tshift) + \frac{2 \damp^2 - 1}{\sqrt{1-\damp^2}} \sin \dfreq (t-\tshift) \right) + \freq (t-\tshift) - 2 \damp \right]
\]</span></p>
</section>
<section id="constant-load-applied-in-finite-time" class="level3" data-number="3.2.2">
<h3 data-number="3.2.2" data-anchor-id="constant-load-applied-in-finite-time"><span class="header-section-number">3.2.2</span> Constant Load Applied in Finite Time</h3>
<p>To model the fact that it is not physically possible to instantaneously apply a load, let us consider a forcing function which starts from 0 and increases linearly to some value <span class="math inline">\(\sforce\)</span> in duration <span class="math inline">\(t_{r}\)</span>, called rising time, after which it remains at the constant value of <span class="math inline">\(\sforce\)</span>. Such a loading, shown in <a href="#fig-forcerampstep">Figure <span>3.9</span></a>, is described mathematically by <span class="math display">\[
\extforce (t) = \begin{cases} \sforce {\ratio{t}{t_{r}}} & 0 \leq t < t_{r} \\ F & t \geq t_{r} \end{cases}
\]</span></p>
<div id="fig-forcerampstep" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/forcerampstep.png" class="img-fluid figure-img" style="width:50.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.9: Force linearly increasing to a constant value.</figcaption><p></p>
</figure>
</div>
<p>There is more than one way to approach this problem, as will probably be the case in many of the problems that will be considered throughout this work. One approach is to make use of the results previously derived and construct a solution via superposition. Assume the system is underdamped and initially (at time <span class="math inline">\(t=0\)</span>) at rest. The system is governed by two different equations, one corresponding to each force segment, as follows: <span class="math display">\[\begin{align*}
m \ddgct + c \dgct + k \gct & = \sforce {\ratio{t}{t_{r}}} \, ; \quad \left\{\gc(0) = 0, \dgc(0) = 0\right\} \quad \text{for } 0 \leq t < t_{r} \\
m \ddgct + c \dgct + k \gct & = \sforce \, ; \quad \left\{\gc(t_{r}) = \gc_{*}, \dgc(t_{r}) = \dgc_{*} \right\} \quad \text{for } t \geq t_{r}
\end{align*}\]</span> The solution to the first part for <span class="math inline">\(0 \leq t < t_{r}\)</span> is simple in that it is given by <a href="#eq-solforceramp">Equation <span>3.18</span></a> with <span class="math inline">\(\lambda = \sforce / t_{r}\)</span>: <span id="eq-solforcerampstep1"><span class="math display">\[
\gct = \frac{\sforce}{k \freq t_{r}} \left[\expon{-\damp \freq t}\left(2 \damp \cos \dfreq t + \frac{2 \damp^2 - 1}{\sqrt{1-\damp^2}} \sin \dfreq t \right) + \freq t - 2 \damp \right]
\qquad(3.20)\]</span></span> The second part is essentially a problem we have investigated in <a href="#sec-xmpl:timeshift"><span>Section 3.2.1</span></a>, with a shifted step input and non-zero initial conditions with <span class="math inline">\(t_{r}=\tshift\)</span>. The solution was shown to be given for <span class="math inline">\(t \geq t_{r}\)</span> by <span class="math display">\[\begin{align*}
\gc (t) & = \expon{-\damp \freq (t-t_{r})}\left[\gc_{*} \cos \dfreq (t-t_{r}) + \frac{\dgc_{*} + \damp \freq \gc_{*} }{\dfreq} \sin \dfreq (t-t_{r}) \right] + \\
& \; \frac{\sforce}{k} \left[ 1 - \expon{-\damp \freq (t-t_{r})}\left(\cos \dfreq (t-t_{r}) + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq (t-t_{r}) \right) \right]
\end{align*}\]</span> wherein the initial conditions <span class="math inline">\(\gc_{*}\)</span> and <span class="math inline">\(\dgc_{*}\)</span> will have to be determined as the values from the last instant governed by the solution to the first part. Substituting <span class="math inline">\(t=t_{r}\)</span> in <a href="#eq-solforcerampstep1">Equation <span>3.20</span></a> yields <span class="math display">\[%\begin{xequation}\label{eq:solforcerampstep1}
\gc_{*} \equiv \gc ({t_{r}}) = \frac{\sforce}{k \freq t_{r}} \left[\expon{-\damp \freq t_{r}}\left(2 \damp \cos \dfreq t_{r} + \frac{2 \damp^2 - 1}{\sqrt{1-\damp^2}} \sin \dfreq t_{r} \right) + \freq t_{r} - 2 \damp \right]
\]</span> The time derivative of <span class="math inline">\(\gc\)</span> may be obtained from <a href="#eq-solforcerampstep1">Equation <span>3.20</span></a> as <span id="eq-solforcerampstepvel"><span class="math display">\[
\dgct = \frac{\sforce}{k t_{r}} \left[1 - \expon{-\damp \freq t}\left(\cos \dfreq t + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq t \right) \right]
\qquad(3.21)\]</span></span> so that <span class="math display">\[
\dgc_{*}\equiv \dgc ({t_{r}})=\frac{\sforce}{k t_{r}} \left[1 - \expon{-\damp \freq t_{r}}\left(\cos \dfreq t_{r} + \frac{\damp}{\sqrt{1-\damp^2}} \sin \dfreq t_{r} \right)\right]
\]</span> Although we have obtained the full solution, the presence of damping somewhat makes the algebra complicated and the presentation symbolically overloaded; let us therefore first simplify the results by considering the undamped case with <span class="math inline">\(\damp = 0\)</span>. When there is no damping, the solution reduces to <span class="math display">\[
\gct = \begin{cases}\ratio{F}{k}\frac{1}{\freq t_{r}}\left[\freq t - \sin \freq t \right] \; \text{for } 0 \leq t < t_{r} \\
% \vspace{-1em} \\
\gc_{*} \cos \freq (t-t_{r}) + \ratio{\dgc_{*}}{\freq} \sin \freq (t-t_{r}) + \frac{F}{k}\left[ 1 - \cos \freq (t-t_{r}) \right] \; \text{for } t \geq t_{r}
\end{cases}
\]</span> The initial conditions for the second part will be given by <span id="eq-rampstepudic"><span class="math display">\[
\gc_{*}= \ratio{\sforce}{k \freq t_{r}} \left[\freq t_{r} - \sin \freq t_{r}\right], \quad \dgc_{*}=\ratio{\sforce}{k t_{r}} \left[1 - \cos \freq t_{r} \right]
\qquad(3.22)\]</span></span> and substituting these into the solution leads, after some algebra and trigonometric combinations, to <span id="eq-solrampstepud2"><span class="math display">\[
\gct = \begin{cases} {\ratio{\sforce/k}{\freq t_{r}}}\left[\freq t - \sin \freq t \right] & \text{for } 0 \leq t < t_{r} \\
% \vspace{-1em}\\
{\ratio{\sforce/k}{\freq t_{r}}} \left[ \freq t_{r} - \sin \freq t + \sin \freq (t-t_{r}) \right] & \text{for } t \geq t_{r}
\end{cases}
\qquad(3.23)\]</span></span> The response is very sensitive to the value of the rise time <span class="math inline">\(t_{r}\)</span> in relation to the system’s period <span class="math inline">\(\period\)</span> as signified by the ratio <span class="math inline">\(t_{r}/T\)</span>. Some trends may be observed better by writing the solution in terms of a normalized time <span class="math inline">\(\tau = t / \period\)</span> and replacing <span class="math inline">\(\freq\)</span> with <span class="math inline">\(2 \pi / \period\)</span> so that <span id="eq-solrampstepud3"><span class="math display">\[
\ratio{\gc (\tau)}{\sforce/k} = \begin{cases} {\ratio{\tau}{ (t_{r}/\period)}}- {\ratio{\sin 2 \pi \tau}{2 \pi (t_{r}/\period)}} & \text{for } 0 \leq \tau < (t_{r}/\period) \\
% \vspace{-1em} \\
1 - {\ratio{\sin 2 \pi \tau - \sin 2 \pi (\tau-(t_{r}/\period))}{2\pi(t_{r}/\period)}} & \text{for } \tau \geq (t_{r}/\period)
\end{cases}
\qquad(3.24)\]</span></span> A somewhat interesting result is observed when it is noted that whenever <span class="math inline">\((t_{r}/\period)\)</span> is some integer, <span class="math inline">\(\sin 2 \pi \tau - \sin 2 \pi (\tau-(t_{r}/\period)) = 0\)</span> since the sine of an angle is the same as the sine of the same angle plus any integer multiple of <span class="math inline">\(2\pi\)</span>. In such cases, the response after the rise time becomes equal to <span class="math inline">\(\sforce/k\)</span> and no oscillations occur. Incidentally this corresponds to those cases in which the velocity of the system at the end of the ramp loading is zero. For non-integer values of <span class="math inline">\((t_{r}/\period)\)</span>, the behavior is determined by the magnitude of the ratio: if <span class="math inline">\((t_{r}/\period) \gg 1\)</span> (load is applied very slowly relative to the period of the system), the maximum response is close to <span class="math inline">\(\sforce/k\)</span>, which again could be considered as the static response of the system. At the other end, if <span class="math inline">\((t_{r}/\period) \ll 1\)</span> (load is applied very quickly compared to the period of the system), the maximum response is close to <span class="math inline">\(2 (\sforce/k)\)</span>; recall that in the limiting case of an instantaneously applied step load we had previously shown the maximum response to be <span class="math inline">\(2 (\sforce/k)\)</span><a href="#fn3" class="footnote-ref" id="fnref3" role="doc-noteref"><sup>3</sup></a>. The variation of the ratio <span class="math inline">\(q(\tau)/(\sforce/k)\)</span> with the relative rise time <span class="math inline">\((t_{r}/\period)\)</span> is examplified for various cases in <a href="#fig-responserampstepundmpd">Figure <span>3.10</span></a>.</p>
<div id="fig-responserampstepundmpd" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/responserampstepundmpd.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.10: Normalized response of an undamped system to step input with a finite rise time for various values of <span class="math inline">\((t_{r}/\period)\)</span>. The dashed lines corresponding to <span class="math inline">\((\extforce (\tau)/k)/(\sforce/k)\)</span> are presented to indicate the behavior of the force.</figcaption><p></p>
</figure>
</div>
</section>
<section id="footnotes" class="footnotes footnotes-end-of-section" role="doc-endnotes">
<hr>
<ol start="3">
<li id="fn3"><p>As <span class="math inline">\((t_{r}/\period) \rightarrow 0\)</span>, <a href="#eq-solrampstepud3">Equation <span>3.24</span></a> may be shown to converge to <span class="math inline">\(1 - \cos 2 \pi \tau\)</span>, which is the same result as that obtained for the step input applied instantaneously.<a href="#fnref3" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
</ol>
</section>
</section>
<section id="sec-harmonicforce" class="level2" data-number="3.3">
<h2 data-number="3.3" data-anchor-id="sec-harmonicforce"><span class="header-section-number">3.3</span> Harmonic Force Excitations</h2>
<p>We have already seen that the nature of vibrations occurring in an SDOF system depend significantly on the time variation of the input relative to the period of the system. This dependence may lead to catastrophic outcomes in the case of repeated loads under which the amplifications in the response may become excessively large so as to induce failure. The most significant parameter in the response to repeated loads turns out to be the ratio of the frequency of excitation to the frequency of the system. Therefore we start the discussions of this phenomenon with the analysis of response to a single frequency input in order identify critical issues. The results developed for this case may then be used to investigate the responses to a broader set of repeated loads with the help of a well-known expansion that has been used in many branches of engineering sciences.</p>
<p>A <em>harmonic excitation</em> is essentially a single frequency sinusoidal wave. A harmonic force may be expressed as <span id="eq-harm"><span class="math display">\[
\extforce (t) = \sforce \sinp{\extfreq t - \extphs}
\qquad(3.25)\]</span></span> where <span class="math inline">\(\sforce\)</span> is the amplitude, <span class="math inline">\(\extfreq\)</span> is the <em>excitation frequency</em>, and <span class="math inline">\(\extphs\)</span> is the phase angle of the force. When analyzing long term behavior the phase generally does not have a significant bearing on design critical issues so that most often it is neglected (i.e. <span class="math inline">\(\extphs\)</span> is assumed to be <span class="math inline">\(0\)</span>); we will however consider the possibility of a nonzero phase to promote a general discussion.</p>
<p>We will assume that such a force acts on our system for a long duration<a href="#fn4" class="footnote-ref" id="fnref4" role="doc-noteref"><sup>4</sup></a>, long enough so that the transient vibrations have died out completely and the system is in what is called to be <em>steady state</em>. Recall that the forced vibration response of a viscously underdamped system is given by <span class="math display">\[
\gct = \gc_{c} (t) + \gc_{p}(t) = \expon{-\damp \freq t}\left(C_1 \cos \dfreq t + C_2 \sin \dfreq t \right) + \gc_{p}(t)
\]</span> so that as <span class="math inline">\(t\)</span> progresses, the exponential term tends to die out; hence the name <em>transient vibrations</em>. As the transient vibrations die out, the response is defined more and more solely by the particular solution, and this state of things is referred to as <em>steady state vibrations</em>. Since a harmonic excitation by assumption acts for a long duration, we shall initially neglect the transients and focus solely on the steady state. This does not mean that the critical response is observed always during steady state vibrations; it may be that for some cases the maximum deformation in the system occurs before the transients die out. It turns out, however, that the worst of the worst occurs for particular ranges of the excitation frequency and in those cases the transients play an insignificant role.</p>
<section id="sec-dynamp" class="level3" data-number="3.3.1">
<h3 data-number="3.3.1" data-anchor-id="sec-dynamp"><span class="header-section-number">3.3.1</span> Dynamic Amplification</h3>
<p>The steady state vibrations under the action of a harmonic force are governed by <span id="eq-harmsdofeom"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \sforce \sinp{\extfreq t - \extphs}
\qquad(3.26)\]</span></span> or equivalently, after dividing through by <span class="math inline">\(m\)</span>, <span id="eq-harmsseom"><span class="math display">\[
\ddgct + 2 \damp \freq \dgct + \freq^2 \gct = \frac{\sforce}{m} \sinp{\extfreq t - \extphs}
\qquad(3.27)\]</span></span> where it should be mentioned that no initial condition information is provided simply because the effects of initial conditions are irrelevant for steady state vibrations. When the forcing function is harmonic the response may be expected to be harmonic as well since, at the end of the day, the left hand side of equation ought to match the right hand side at all times, and therefore appearance of sines and cosines on the left hand side should not come as a surprise. A trial solution for steady state vibrations may therefore be formulated as <span id="eq-harmsssol_1"><span class="math display">\[
\gct = A_1 \cosp{\extfreq t - \extphs} + A_2 \sinp{\extfreq t - \extphs}
\qquad(3.28)\]</span></span> which, when substituted into <a href="#eq-harmsseom">Equation <span>3.27</span></a>, leads to <span class="math display">\[\begin{align*}
& \left[- \extfreq^2 A_1 + 2 \damp \freq \extfreq A_2 + \freq^2 A_1 \right ] \cosp{\extfreq t- \extphs} + \\
& \phantom{XXXXXXX} \left[- \extfreq^2 A_2 - 2 \damp \freq \extfreq A_1 + \freq^2 A_2 \right ] \sinp{\extfreq t- \extphs} = \frac{\sforce}{m} \sinp{\extfreq t - \extphs}
\end{align*}\]</span> and equating the coefficients of the sines and cosines on both sides leads to <span id="eq-130a"><span class="math display">\[
\left(\freq^2 - \extfreq^2 \right) A_1 + 2 \damp \freq \extfreq A_2 = 0
\qquad(3.29)\]</span></span> <span id="eq-130b"><span class="math display">\[
- 2 \damp \freq \extfreq A_1 + \left(\freq^2 - \extfreq^2 \right) A_2 = \frac{\sforce}{m}
\qquad(3.30)\]</span></span> Solving for the coefficients <span class="math inline">\(A_i\)</span>, one obtains, <span class="math display">\[
A_1 = \ratio{\sforce}{m} \ratio{\left(- 2 \damp \freq \extfreq \right)}{\left( \freq^2 - \extfreq^2 \right)^2 + \left(2 \damp \freq \extfreq \right)^2}, \quad A_2 = \ratio{\sforce}{m} \ratio{\left( \freq^2 - \extfreq^2 \right)}{\left(\freq^2 - \extfreq^2 \right)^2 + \left(2 \damp \freq \extfreq \right)^2}
\]</span> and after dividing the nominators and denominators by <span class="math inline">\(\freq^4\)</span> we get <span class="math display">\[
A_1 = \ratio{\sforce}{k} \ratio{ - 2 \damp \ratfreq }{\left( 1 - \ratfreq^2 \right)^2 + \left(2 \damp \ratfreq \right)^2}, \quad A_2 = \ratio{\sforce}{k} \ratio{ 1 - \ratfreq^2}{\left(1 - \ratfreq^2 \right)^2 + \left(2 \damp \ratfreq \right)^2}
\]</span> where <span class="math inline">\(\ratfreq\)</span> is the ratio of the excitation frequency to the frequency of the system, i.e. <span class="math display">\[
\ratfreq = \ratio{\extfreq}{\freq}
\]</span> Since the response is given by <span class="math display">\[
\gct = A_1 \cosp{\extfreq t - \extphs} + A_2 \sinp{\extfreq t - \extphs}
\]</span> these two harmonics may, via the expansion <span class="math inline">\(\sinp{a-b} = \sin a \cos b - \cos a \sin b\)</span>, be combined into a single wave, as was done on numerous previous instances, to obtain <span id="eq-harmsssol_2"><span class="math display">\[
\begin{array}{rcl}
\gct & \!\!\! = & \!\!\! Q \sinp{\extfreq t - \extphs - \phs} = Q\left[\sinp{\extfreq t - \extphs}\cos{\phs} - \cosp{\extfreq t - \extphs}\sin{\phs}\right] \\
& \!\!\! = & \!\!\! (-Q \sin \phs) \cosp{\extfreq t - \extphs} + (Q \cos \phs ) \sinp{\extfreq t - \extphs}
\end{array}
\qquad(3.31)\]</span></span> so that we have <span class="math inline">\(A_1 = - Q \sin \phs\)</span> and <span class="math inline">\(A_2 = Q \cos \phs\)</span>, leading to: <span class="math display">\[
Q = \sqrt{A_1^2 + A_2^2}, \quad \tan \phs = \ratio{-A_1/Q}{A_2/Q}
\]</span> The amplitude <span class="math inline">\(Q\)</span> of the response is therefore given by <span id="eq-harmQ"><span class="math display">\[
Q = \sqrt{A_1^2 + A_2^2} = \ratio{\sforce}{k}\ratio{1}{\sqrt{\left(1 - \ratfreq^2 \right)^2 + \left(2 \damp \ratfreq \right)^2}} = \ratio{\sforce}{k} \dynampr
\qquad(3.32)\]</span></span> where <span class="math inline">\(F/k\)</span> would be the response that would be observed if the force of amplitude <span class="math inline">\(\sforce\)</span> were to be applied statically, and <span id="eq-harmdynamp"><span class="math display">\[
\dynamp = \dynampr = \ratio{1}{\sqrt{\left(1 - \ratfreq^2 \right)^2 + \left(2 \damp \ratfreq \right)^2}} = \ratio{Q}{F/k}
\qquad(3.33)\]</span></span> is called the <em>dynamic amplification factor</em>.<a href="#fn5" class="footnote-ref" id="fnref5" role="doc-noteref"><sup>5</sup></a> With this definition, the phase angle <span class="math inline">\(\phs\)</span> may now be shown to be defined through <span id="eq-harmP"><span class="math display">\[
\tan \phs = \ratio{ ({2 \damp \ratfreq})/{\dynamp}}{({1-\ratfreq^2})/{\dynamp}}
\qquad(3.34)\]</span></span> and it is implied by definition that <span class="math inline">\(0 \leq \phs \leq \pi\)</span>.</p>
<p>The dynamic amplification factor is a measure of increase in maximum response due to the harmonic application of the force, depending on ratio of the frequencies and available damping. How this amplification factor varies as a function of the ratio of frequencies is naturally of greatest importance, and this variation is shown graphically in <a href="#fig-harmdynamp">Figure <span>3.11</span></a>, along with variations of the phase angle, for various levels of viscous damping. A few characteristics of these curves deserve special mention:</p>
<div id="fig-harmdynamp" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/harmdynamp.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 3.11: Variation of the dynamic amplification factor <span class="math inline">\(\dynamp\)</span> and the phase angle <span class="math inline">\(\phs\)</span> with ratio of frequencies, plotted for various levels of viscous damping.</figcaption><p></p>
</figure>
</div>
<ol type="i">
<li><p>For all levels of damping, <span class="math inline">\(\dynamp \rightarrow 1\)</span> as <span class="math inline">\(\ratfreq \rightarrow 0\)</span>. This is mathematically obvious as the limit of the expression in <a href="#eq-harmdynamp">Equation <span>3.33</span></a>, and it may physically be interpreted in a few different ways:</p>
<ol type="a">
<li>For a given excitation with some finite, non-zero excitation frequency <span class="math inline">\(\extfreq\)</span>, <span class="math inline">\(\ratfreq \rightarrow 0\)</span> implies <span class="math inline">\(\freq \rightarrow \infty\)</span>, which in turn requires <span class="math inline">\(m \rightarrow 0\)</span> for a system with finite stiffness. If so, then the inertial forces would be very small compared to the deformational forces so that <span class="math inline">\(m \ddgct\)</span> could be neglected in comparison with <span class="math inline">\(k \gct\)</span> and the governing differential equation would effectively simplify to <span class="math inline">\(k \gct = \sforce \sinp{\extfreq t - \extphs}\)</span> with the response given by <span class="math display">\[
\gct = \ratio{\sforce}{k} \sinp{\extfreq t - \extphs}
\]</span> and the system would deform in phase with the force (<span class="math inline">\(\phs \rightarrow 0\)</span>). A massless spring represents a limiting case as the spring simply deforms in phase with the force (<span class="math inline">\(\phs = 0\)</span>), with the maximum deformation be given by <span class="math inline">\(Q = \sforce / k\)</span>.</li>
<li>For a given excitation with some finite, non-zero excitation frequency <span class="math inline">\(\extfreq\)</span>, <span class="math inline">\(\ratfreq \rightarrow 0\)</span> implies <span class="math inline">\(\freq \rightarrow \infty\)</span>, which in turn requires <span class="math inline">\(k \rightarrow \infty\)</span> for a system with finite mass. Also in this case inertial forces could be neglected beside deformational forces so that <span class="math inline">\(m \ddgct\)</span> could be neglected compared to <span class="math inline">\(k \gct\)</span> and the governing differential equation would again effectively simplify to <span class="math inline">\(k \gct = \sforce \sinp{\extfreq t - \extphs}\)</span>. The ratio of maximum deformation to <span class="math inline">\(\sforce / k\)</span> is again given by <span class="math inline">\(Q / (\sforce / k) = 1\)</span>, no matter how small <span class="math inline">\(Q\)</span> and <span class="math inline">\(F / k\)</span> are due to the very high value of <span class="math inline">\(k\)</span>.</li>
<li>For a given system with some finite, non-zero frequency <span class="math inline">\(\extfreq\)</span>, <span class="math inline">\(\ratfreq \rightarrow 0\)</span> implies <span class="math inline">\(\extfreq \rightarrow 0\)</span>, which would mean that the force is applied ever so slowly and no significant accelerations develop, again allowing us to neglect <span class="math inline">\(m \ddgct\)</span> compared to <span class="math inline">\(k \gct\)</span>, with the governing differential equation simplifying once again to <span class="math inline">\(k \gct = \sforce \sinp{\extfreq t - \extphs}\)</span>, and the previous conclusion follow.</li>
</ol></li>
<li><p>For all levels of damping, <span class="math inline">\(\dynamp \rightarrow 0\)</span> as <span class="math inline">\(\ratfreq \rightarrow \infty\)</span>. This result is also obvious mathematically as the limit of the expression in <a href="#eq-harmdynamp">Equation <span>3.33</span></a>, and the physical system may correspond to one of the following:</p>
<ol type="a">
<li>For a given excitation with some finite, non-zero excitation frequency <span class="math inline">\(\extfreq\)</span>, <span class="math inline">\(\ratfreq \rightarrow \infty\)</span> implies <span class="math inline">\(\freq \rightarrow 0\)</span>, which in turn requires <span class="math inline">\(m \rightarrow \infty\)</span> for a system with finite stiffness. If so, then the inertial forces would be very large compared to the deformational forces so that <span class="math inline">\(k \gct\)</span> could be neglected in comparison with <span class="math inline">\(m \ddgct\)</span> and the governing differential equation would effectively simplify to <span class="math inline">\(m \ddgct = \sforce \sinp{\extfreq t - \extphs}\)</span>, the integration of which leads to: <span class="math display">\[
\gct = - \ratio{\sforce}{m\extfreq^2} \sinp{\extfreq t - \extphs}
\]</span> The minus sign implies that the response is out of phase with the force (<span class="math inline">\(\phs \rightarrow \pi\)</span>) so that whenever the force reaches a maximum, the displacement reaches a minimum and vice versa. Clearly the maximum deformation tends to zero as <span class="math inline">\(m\)</span> gets larger; in the limit, the inertia of the system is so great that no force can get it to start moving.</li>
<li>For a given excitation with some finite, non-zero excitation frequency <span class="math inline">\(\extfreq\)</span>, <span class="math inline">\(\ratfreq \rightarrow \infty\)</span> implies <span class="math inline">\(\freq \rightarrow 0\)</span>, which in turn requires <span class="math inline">\(k \rightarrow 0\)</span> for a system with finite mass. Also in this case deformational forces could be neglected beside inertial forces so that <span class="math inline">\(k \gct\)</span> could be neglected compared to <span class="math inline">\(m \ddgct\)</span> and the governing differential equation would again effectively simplify to <span class="math inline">\(m \ddgct = \sforce \sinp{\extfreq t - \extphs}\)</span>, leading to the same conclusions as in (ii.a).</li>
<li>For a given system with some finite, non-zero frequency <span class="math inline">\(\freq\)</span>, <span class="math inline">\(\ratfreq \rightarrow \infty\)</span> implies <span class="math inline">\(\extfreq \rightarrow \infty\)</span>, which would mean that the time variation of the force is extremely fast and significant accelerations develop as the mass tries to respond, again allowing us neglect <span class="math inline">\(k \gct\)</span> compared to <span class="math inline">\(m \ddgct\)</span>. The governing differential equation would again effectively simplify to <span class="math inline">\(m \ddgct = \sforce \sinp{\extfreq t - \extphs}\)</span> and the response would be given by <span class="math display">\[
\gct = - \ratio{\sforce}{m\extfreq^2} \sinp{\extfreq t - \extphs}
\]</span> which would tend to zero as <span class="math inline">\(\extfreq\)</span> gets larger and larger.</li>
</ol></li>
<li><p>The dynamic amplification factor reaches a peak value somewhere in the vicinity of <span class="math inline">\(\ratfreq = 1\)</span> when damping levels are low. We can investigate the derivative of the dynamic amplification factor with respect to <span class="math inline">\(\ratfreq\)</span> to locate the extremum points: <span class="math inline">\(\diff \dynamp / \diff \ratfreq\)</span> becomes zero at <span class="math inline">\(\ratfreq = 0\)</span> and <span class="math inline">\(\ratfreq = \sqrt{1-2 \damp^2}\)</span> for <span class="math inline">\(\damp \leq 1/\sqrt{2}\)</span>, while for <span class="math inline">\(\damp > 1 / \sqrt{2}\)</span> the derivative is zero only for <span class="math inline">\(\ratfreq = 0\)</span>. Therefore whenever <span class="math inline">\(\damp \leq 1/\sqrt{2} \approx 71\%\)</span>, the maximum value of the dynamic amplification factor is given by <span id="eq-last"><span class="math display">\[
\dynamp_{\max} = \dynamp \bigr|_{\ratfreq = \sqrt{1-2\damp^2}} = \ratio{1}{2 \damp} \ratio{1}{\sqrt{1-\damp^2}} \approx \ratio{1}{2 \damp}
\qquad(3.35)\]</span></span> where the last approximation is valid for small values of the damping ratio. To provide some numerical justification, for a damping ratio of <span class="math inline">\(\damp = 10 \%\)</span>, which is not so small in terms of damping ratios frequently encountered in structural dynamics, the exact value for the amplification factor is <span class="math inline">\(5.05\)</span> whereas the approximate value is <span class="math inline">\(5.00\)</span>, with the error of approximation about <span class="math inline">\(1\%\)</span> or, in other words, practically completely negligible.</p></li>