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2125 lines (1983 loc) · 205 KB
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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">
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<title>Structural Dynamics - 4 Numerical Solutions of the SDOF Equation of Motion</title>
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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="#preliminary-ideas" id="toc-preliminary-ideas" class="nav-link active" data-scroll-target="#preliminary-ideas"><span class="header-section-number">4.1</span> Preliminary Ideas</a></li>
<li><a href="#linear-interpolation-of-the-excitation" id="toc-linear-interpolation-of-the-excitation" class="nav-link" data-scroll-target="#linear-interpolation-of-the-excitation"><span class="header-section-number">4.2</span> Linear Interpolation of the Excitation</a></li>
<li><a href="#central-difference-method" id="toc-central-difference-method" class="nav-link" data-scroll-target="#central-difference-method"><span class="header-section-number">4.3</span> Central Difference Method</a></li>
<li><a href="#sec-numericalinterlude" id="toc-sec-numericalinterlude" class="nav-link" data-scroll-target="#sec-numericalinterlude"><span class="header-section-number">4.4</span> Interlude</a></li>
<li><a href="#houbolts-method" id="toc-houbolts-method" class="nav-link" data-scroll-target="#houbolts-method"><span class="header-section-number">4.5</span> Houbolt’s Method</a></li>
<li><a href="#sec-newmark" id="toc-sec-newmark" class="nav-link" data-scroll-target="#sec-newmark"><span class="header-section-number">4.6</span> Newmark’s Method</a>
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<h1 class="title"><span id="sec-Numerical_Solutions_of_the_SDOF_Equation_of_Motion" class="quarto-section-identifier"><span class="chapter-number">4</span> <span class="chapter-title">Numerical Solutions of the SDOF Equation of Motion</span></span></h1>
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<section id="preliminary-ideas" class="level2" data-number="4.1">
<h2 data-number="4.1" data-anchor-id="preliminary-ideas"><span class="header-section-number">4.1</span> Preliminary Ideas</h2>
<p>The equation of motion for a viscously damped linear SDOF system under the action of an external force was shown in previous sections to be <span id="eq-1"><span class="math display">\[
m \ddgct + c \dgct + k \gct = \extforcet
\qquad(4.1)\]</span></span> with the specialized equation for the case of a system excited by a ground motion given by <span id="eq-2"><span class="math display">\[
m \ddgct + c \dgct + k \gct = - m \gacct
\qquad(4.2)\]</span></span> Although we have discussed the analytical solutions of these equations for an important subset of excitations, it has to be said that most often analytical solutions are beyond our reach and numerical techniques must be employed to obtain values for the response at various instances. While discussing numerical techniques, we will employ the force excitation form of the equation only to minimize repetitive expositions; clearly the formulations may be specialized for ground motion excitation by substituting <span class="math inline">\(- m \gacc (t)\)</span> for <span class="math inline">\(\extforce(t)\)</span>.</p>
<p>The need for use of numerical techniques may arise from various reasons: it may be that the form of excitation is so complex that no known analytical solutions exist, it may be that the excitation is known only at discrete instances of time to begin with, or it may be that the nonlinearity of the system is prohibitive for any closed form solutions. In all cases, the method of attack for all numerical techniques is similar: a. time is discretized into a finite number of instances (using, if possible, small intervals in-between them); b. the behavior, in-between the time instances, of the force or the system or both is approximated via straight lines or simple curves; and c. the solution progresses from one step to the next through solution of algebraic equations.</p>
<p>In what follows, we will try to adhere to the following notation so as to provide some common language for all the methods to be discussed. We will assume that continuous time <span class="math inline">\(t\)</span> is discretized into a total of <span class="math inline">\(l+1\)</span> instances, each denoted by <span class="math inline">\(t_j\)</span> for <span class="math inline">\(j=0,1,2,\ldots,l\)</span>. The difference between any two consecutive instances is called a <em>time step</em>, to be denoted by <span class="math inline">\(\tstep = \dsct{t}{j+1} - \dsct{t}{j}\)</span>. The time step does not in general have to be constant, but when working on linear systems with relatively smooth inputs, it often is. To emphasize the discrete nature of the calculations, we’ll use brackets: for some time dependent variable <span class="math inline">\(\gc(t)\)</span>, <span class="math inline">\(\dsct{\gc}{j}\)</span> will denote the value of that variable at time <span class="math inline">\(t=\dsct{t}{j}\)</span> where, for constant time step, <span class="math inline">\(\dsct{t}{j} = j \tstep\)</span>. The construction of the numerical solution up to and including the final time, corresponding to step <span class="math inline">\(l+1\)</span> (counting time <span class="math inline">\(t=0\)</span> as the first time step) will require knowledge of the input <span class="math inline">\(\dsct{\extforce}{j}\)</span> for <span class="math inline">\(j=0,1,2,...,l\)</span> and the initial conditions <span class="math inline">\(\dsct{\gc}{0}\)</span>, <span class="math inline">\(\dsct{\dgc}{0}\)</span>.</p>
</section>
<section id="linear-interpolation-of-the-excitation" class="level2" data-number="4.2">
<h2 data-number="4.2" data-anchor-id="linear-interpolation-of-the-excitation"><span class="header-section-number">4.2</span> Linear Interpolation of the Excitation</h2>
<p>For linear SDOF systems, it is relatively straightforward to develop a numerical progression based on analytical solutions developed for simply varying forces. The fundamental idea is linear interpolation of the excitation in-between two consecutive time steps. Consider the sketch given in <a href="#fig-pem">Figure <span>4.1</span></a>. According to the this approach which we shall refer to as the <em>piecewise linear excitation</em><a href="#fn1" class="footnote-ref" id="fnref1" role="doc-noteref"><sup>1</sup></a> (PLE), the excitation <span class="math inline">\(\extforce(t)\)</span> is assumed to vary linearly between its values at <span class="math inline">\(\dsct{t}{j}\)</span> and <span class="math inline">\(\dsct{t}{j+1}\)</span> so that for <span class="math inline">\(\dsct{t}{j} \leq t \leq \dsct{t}{j+1}\)</span> the excitation is approximated by <span class="math display">\[
\extforce(t) \approx \extforce(\dsct{t}{j}) + \frac{\extforce(\dsct{t}{j+1}) - \extforce(\dsct{t}{j})}{\dsct{t}{j+1}-\dsct{t}{j}} (t - \dsct{t}{j}) = \dsct{\extforce}{j} + \frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{\tstep} (t - \dsct{t}{j}) \quad
\]</span> It is easier to express this equation using a shifted time variable <span class="math inline">\(\tau=t-\dsct{t}{j}\)</span> such that (replacing the approximation with an equality) <span class="math display">\[
\extforce(\tau+\dsct{t}{j})=\dsct{\extforce}{j} + \frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{\tstep} \tau \quad \text{for} \quad 0 \leq \tau \leq \tstep
\]</span> For a linear system, it is possible to employ superposition to develop the ‘exact’ response of the system under such an excitation: The response <span class="math inline">\(\gc (t)\)</span> for <span class="math inline">\(\dsct{t}{j} \leq t \leq \dsct{t}{j+1}\)</span> may be considered to be the superposition of the following three responses, all of which were previously developed:</p>
<ul>
<li>The response due to nonzero initial conditions <span class="math inline">\(\dsct{\gc}{j}\)</span> and <span class="math inline">\(\dsct{\dgc}{j}\)</span> at time <span class="math inline">\(t=\dsct{t}{j}\)</span>. This response is given (see <a href="singledof.html"><span>Chapter 2</span></a>) by <span class="math display">\[
\gc (\tau + \dsct{t}{j}) = \expon{- \damp \freq \tau} \left(\dsct{\gc}{j} \cos \dfreq \tau + \frac{\dsct{\dgc}{j} + \damp \freq \dsct{\gc}{j}}{\dfreq} \sin \dfreq \tau \right)
\]</span></li>
<li>The response due to the constant force <span class="math inline">\(\dsct{\extforce}{j}\)</span> acting on a system initially at rest. This response is given (see <a href="singledofforced.html"><span>Chapter 3</span></a>) by <span class="math display">\[
\gc (\tau + \dsct{t}{j}) = \frac{\dsct{\extforce}{j}}{k} \left[1-\expon{- \damp \freq \tau} \left(\cos \dfreq \tau + \frac{\damp}{\sqrt{1 - \damp^2}} \sin \dfreq \tau \right) \right]
\]</span></li>
<li>The response due to the linearly varying (ramp) force <span class="math inline">\((\dsct{\extforce}{j+1} - \dsct{\extforce}{j}) \tau / \tstep\)</span> acting on a system initially at rest. This response is given (see <a href="singledofforced.html"><span>Chapter 3</span></a>) by <span class="math display">\[
\gc (\tau + \dsct{t}{j}) = \frac{1}{k} \frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{\tstep} \expon{- \damp \freq \tau} \left(\frac{2 \damp}{\freq} \cos \dfreq \tau + \left[\frac{2 \damp^2}{\dfreq} - \frac{1}{\dfreq}\right] \sin \dfreq \tau \right)
+ \frac{1}{k}\frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{\tstep} \left[\tau -\frac{2 \damp}{\freq} \right]
\]</span> By taking the derivatives of these expressions we may obtain expressions for velocity, which will also be the superposition of the contributions from the three cases considered above.</li>
</ul>
<div id="fig-pem" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/pem.png" class="quarto-discovered-preview-image img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.1: Approximating the force as a series of linear segments.</figcaption><p></p>
</figure>
</div>
<p>The displacement at <span class="math inline">\(\tau = \tstep\)</span> (corresponding to <span class="math inline">\(t=\dsct{t}{j+1}\)</span>) is therefore given by <span class="math display">\[\begin{align*}
\gc(\dsct{t}{j+1}) = \dsct{\gc}{j+1} & = \expon{- \damp \freq \tstep} \left(\dsct{\gc}{j} \cos \dfreq \tstep + \frac{\dsct{\dgc}{j} + \damp \freq \dsct{\gc}{j}}{\dfreq} \sin \dfreq \tstep \right) \\
& \quad + \frac{\dsct{\extforce}{j}}{k} \left[1-\expon{- \damp \freq \tstep} \left(\cos \dfreq \tstep + \frac{\damp}{\sqrt{1 - \damp^2}} \sin \dfreq \tstep \right) \right]\\
& \quad \quad + \frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{k \tstep} \expon{- \damp \freq \tstep} \left(\frac{2 \damp}{\freq} \cos \dfreq \tstep + \left[\frac{2 \damp^2}{\dfreq} - \frac{1}{ \dfreq}\right] \sin \dfreq \tstep \right) \\
& \quad \quad \quad + \frac{\dsct{\extforce}{j+1} - \dsct{\extforce}{j}}{k \tstep} \left[\tstep -\frac{2 \damp}{\freq} \right]
\end{align*}\]</span> Despite the somewhat complicated structure of this result, it is actually composed of a number of coefficients such as <span class="math inline">\(\cos \dfreq \tstep\)</span>, <span class="math inline">\(\sin \dfreq \tstep\)</span>, etc, multiplying the time-varying variables <span class="math inline">\(\gc\)</span>, <span class="math inline">\(\dgc\)</span> and <span class="math inline">\(\extforce\)</span>. Rearranging the terms, it is possible to express the result above as <span id="eq-pemgc"><span class="math display">\[
\dsct{\gc}{j+1} = A_1 \dsct{\gc}{j+1} + A_2 \dsct{\dgc}{j} + A_3 \dsct{\extforce}{j} + A_4 \dsct{\extforce}{j+1}
\qquad(4.3)\]</span></span> where <span id="eq-pemai"><span class="math display">\[
\begin{array}{l}
A_1 = \expon{- \damp \freq \tstep} \left(\cos \dfreq \tstep + \frac{\damp}{\sqrt{1 - \damp^2}} \sin \dfreq \tstep \right) \\
A_2 = \expon{- \damp \freq \tstep} \left(\frac{1}{\dfreq} \sin \dfreq \tstep \right) \\
A_3 = \frac{1}{k} \left[\frac{2 \damp}{\freq \tstep} + \expon{- \damp \freq \tstep} \left( - \left[1 + \frac{2 \damp}{\freq \tstep} \right]\cos \dfreq \tstep + \left[\frac{1 - 2 \damp^2}{\dfreq \tstep} - \frac{\damp}{\sqrt{1-\damp^2}} \right] \sin \dfreq \tstep \right) \right] \\
A_4 =\frac{1}{k} \left[1 - \frac{2 \damp}{\freq \tstep} + \expon{- \damp \freq \tstep} \left( \frac{2 \damp}{\freq \tstep} \cos \dfreq \tstep + \frac{2 \damp^2-1}{\dfreq \tstep} \sin \dfreq \tstep \right) \right]
\end{array}
\qquad(4.4)\]</span></span> It is noteworthy that these coefficients are time invariant if a constant time step is used, in which case they would have to be calculated only once. Similarly, if we obtain the velocity by differentiating <span class="math inline">\(\gc (\tau + t_{j})\)</span> and evaluate its value at <span class="math inline">\(\tau = \tstep\)</span>, we find that this value may also be expressed in the form <span id="eq-pemdgc"><span class="math display">\[
\dsct{\dgc}{j+1} = B_1 \dsct{\gc}{j} + B_2 \dsct{\dgc}{j} + B_3 \dsct{\extforce}{j} + B_4 \dsct{\extforce}{j}
\qquad(4.5)\]</span></span> where <span id="eq-pembi"><span class="math display">\[
\begin{array}{l}
B_1 = - \expon{- \damp \freq \tstep} \left(\frac{\freq}{\sqrt{1 - \damp^2}} \sin \dfreq \tstep \right) \\
B_2 = \expon{- \damp \freq \tstep} \left(\cos \dfreq \tstep - \frac{\damp}{\sqrt{1 - \damp^2}} \sin \dfreq \tstep \right) \\
B_3 = \frac{1}{k \tstep} \left[-{1} + \expon{- \damp \freq \tstep} \left( \cos \dfreq \tstep + \frac{\freq \tstep + \damp}{\sqrt{1-\damp^2}} \sin \dfreq \tstep \right) \right] \\
B_4 =\frac{1}{k \tstep} \left[1 - \expon{- \damp \freq \tstep} \left( \cos \dfreq \tstep + \frac{\damp}{\sqrt{1 - \damp^2}} \sin \dfreq \tstep \right) \right]
\end{array}
\qquad(4.6)\]</span></span> and once again all coefficients are independent of time if a constant time step is used. Once the displacement and the velocity at an instant are evaluated, the acceleration at that instant may be calculated via the the equation of motion as: <span id="eq-pemddgc"><span class="math display">\[
\dsct{\ddgc}{j} = \frac{1}{m}(\dsct{\extforce}{j} - c \dsct{\dgc}{j} - k \dsct{\gc}{j}) = \frac{\dsct{\extforce}{j}}{m} - 2 \damp \freq \dsct{\dgc}{j} - \freq^2 \dsct{\gc}{j}
\qquad(4.7)\]</span></span> Having developed the analytical results based on the initial assumptions, the flow of the PLE implemented with a constant time step may be summarized as in <a href="#fig-algpem">Figure <span>4.2</span></a>.</p>
<div class="callout callout-style-default callout-note no-icon callout-titled">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon no-icon"></i>
</div>
<div class="callout-title-container flex-fill">
<strong>Pseudocode for the Piecewise Linear Excitation method.</strong>
</div>
</div>
<div class="callout-body-container callout-body">
<p><em>viscously damped SDOF system, for constant time step</em></p>
<p><strong>Input</strong>: <span class="math inline">\(m\)</span>, <span class="math inline">\(\damp\)</span> (or <span class="math inline">\(c\)</span>), <span class="math inline">\(k\)</span>, <span class="math inline">\(\tstep\)</span>, <span class="math inline">\(\dsct{\gc}{0}\)</span>, <span class="math inline">\(\dsct{\dgc}{0}\)</span>, <span class="math inline">\(\left\{\dsct{\extforce}{0}, \ldots, \dsct{\extforce}{l} \right\}\)</span></p>
<p><strong>Initialize</strong>: calculate <span class="math inline">\(A_1,A_2,A_3,A_4,B_1,B_2,B_3,B_4\)</span> using <a href="#eq-pemai">Equation <span>4.4</span></a> and <a href="#eq-pembi">Equation <span>4.6</span></a></p>
<p><strong>for</strong> <span class="math inline">\(j=0,\ldots,l-1\)</span>:</p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\gc}{j+1}=A_1 \dsct{\gc}{j} + A_2 \dsct{\dgc}{j} + A_3 \dsct{\extforce}{j} + A_4 \dsct{\extforce}{j+1}\)</span></p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\dgc}{j+1}=B_1 \dsct{\gc}{j} + B_2 \dsct{\dgc}{j} + B_3 \dsct{\extforce}{j} + B_4 \dsct{\extforce}{j+1}\)</span></p>
<p><strong>for</strong> <span class="math inline">\(j=0,\ldots,l\)</span>:</p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\ddgc}{j} = \frac{1}{m}(\dsct{\extforce}{j} - c \dsct{\dgc}{j} - k \dsct{\gc}{j})\)</span></p>
</div>
</div>
<div id="fig-algpem" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/blank.png" style="width:0.0%" height="0" class="figure-img"></p>
<p></p><figcaption class="figure-caption">Figure 4.2: Pseudocode for PLE.</figcaption><p></p>
</figure>
</div>
<section id="sec-sdofnum" class="level4 unnumbered">
<h4 class="unnumbered" data-anchor-id="sec-sdofnum">EXAMPLE 4 .1</h4>
<p>A single story structure, initially at rest, is idealized as an SDOF system for studying its lateral vibrations when subjected to unidirectional, pulse type lateral forces. The mass<a href="#fn2" class="footnote-ref" id="fnref2" role="doc-noteref"><sup>2</sup></a> of the structure is assumed to be condensed at a certain level and estimated as <span class="math inline">\(125 \unit{tons}\)</span>, its lateral stiffness is estimated as <span class="math inline">\(2\times 10^5 \unit{kN/m}\)</span>, and it is assumed to be viscously damped with a damping ratio of <span class="math inline">\(2\%\)</span>. The equivalent model of the system and the forces it is subjected to are shown in <a href="#fig-numericalex1">Figure <span>4.3</span></a>. Consider a constant time step of <span class="math inline">\(\tstep = 0.01 \unit{seconds}\)</span>, which corresponds to less than one-tenth of the system’s period, and let us say we want to evaluate the response of the system for a duration of <span class="math inline">\(1 \unit{second}\)</span>, so that a total of <span class="math inline">\(101\)</span> time instances are to be taken into consideration including the initial time <span class="math inline">\(t=0\)</span>.</p>
<div id="fig-numericalex1" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalex1.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.3: (a) SDOF model of a single story structure, (b) triangular pulse with duration <span class="math inline">\(0.2 \unit{seconds}\)</span>, (c) half-sine pulse with duration <span class="math inline">\(0.4 \unit{seconds}\)</span>.</figcaption><p></p>
</figure>
</div>
<p>Let us start with the analysis of the triangular pulse with an amplitude of <span class="math inline">\(100 \unit{kN}\)</span> and duration <span class="math inline">\(t_d = 0.2 \unit{seconds}\)</span>. The force increases linearly until it reaches its peak value of <span class="math inline">\(100 \unit{kN}\)</span> at <span class="math inline">\(t = t_d / 2 = 0.1 \unit{seconds}\)</span> corresponding to <span class="math inline">\(j = 10\)</span>, after which it declines linearly until it disappears at <span class="math inline">\(t = t_d / 2 = 0.1 \unit{seconds}\)</span> corresponding to <span class="math inline">\(j = 20\)</span>. The value of the force may therefore be defined at instances <span class="math inline">\(j\)</span> as <span class="math display">\[
\dsct{\extforce}{j} = \begin{cases} 100 \frac{\dsct{t}{j}}{t_d/2} = 10 j \unit{kN} & \text{for } j = 0,1,\ldots,10 \\
200 - 200 \frac{\dsct{t}{j}}{t_d} = 200 - 10 j \unit{kN} & \text{for } j = 11,12,\ldots,20 \\
0 & \text{for } j = 21, 22, \ldots, 100
\end{cases}
\]</span> The first order of business is to calculate the coefficients <span class="math inline">\(A\)</span>’s and <span class="math inline">\(B\)</span>’s that appear in the recursive formulas of <a href="#eq-pemgc">Equation <span>4.3</span></a> and <a href="#eq-pemdgc">Equation <span>4.5</span></a>. These coefficients are evaluated using the expressions given in <a href="#eq-pemai">Equation <span>4.4</span></a> and <a href="#eq-pembi">Equation <span>4.6</span></a> to obtain: <span class="math display">\[
\begin{array}{rr}
A_{1} = 9.2148 \times 10^{-1}, & A_{2} = 9.6580\times 10^{-3} \\
A_{3} = 2.6086\times 10^{-7}, & A_{4} = 1.3174\times 10^{-7} \\
B_1 = -1.5453\times 10^{1}, & B_2 = 9.0603 \times 10^{-1} \\
B_3 = 3.8004\times 10^{-5}, & B_4 = 3.9260\times 10^{-5}
\end{array}
\]</span> Note that these coefficients are independent of the force insofar as the time step does not change. The next step is to calculate the displacement and the velocity through <a href="#eq-pemgc">Equation <span>4.3</span></a> and <a href="#eq-pemdgc">Equation <span>4.5</span></a>, and the acceleration at each time step through <a href="#eq-pemddgc">Equation <span>4.7</span></a>. At the initial instance corresponding to <span class="math inline">\(j=0\)</span> the system is at rest, and so <span class="math inline">\(\dsct{\gc}{0} = 0\)</span>, <span class="math inline">\(\dsct{\dgc}{0} = 0\)</span>. Since <span class="math inline">\(\dsct{\extforce}{0} = 0\)</span>, we find <span class="math inline">\(\dsct{\ddgc}{0} = 0\)</span> as well. <span class="math inline">\(\dsct{\gc}{1}\)</span> and <span class="math inline">\(\dsct{\dgc}{1}\)</span> are to be evaluated from <span class="math display">\[\begin{align*}
\dsct{\gc}{1} & = 9.2148 \times 10^{-1} \dsct{\gc}{0} + 9.6580\times 10^{-3} \dsct{\dgc}{0} + 2.6086\times 10^{-7} \dsct{\extforce}{0} \\
& \phantom{9.2148} + 1.3174\times 10^{-7} \dsct{\extforce}{1} = (1.3174\times 10^{-7})(10) = 1.3174\times 10^{-6} \unit{m} \\
\dsct{\dgc}{1} & = -1.5453\times 10^{1} \dsct{\gc}{0} + 9.0603 \times 10^{-1} \dsct{\dgc}{0} + 3.8004\times 10^{-5} \dsct{\extforce}{0} \\
& \phantom{-1.5} + 3.9260\times 10^{-5} \dsct{\extforce}{1} = (3.9260\times 10^{-5} )(10) = 3.9260\times 10^{-4} \unit{m/s} \\
\end{align*}\]</span> and the acceleration at <span class="math inline">\(j=1\)</span> is consequently given by <span class="math display">\[
\dsct{\ddgc}{1} = \frac{1}{m}(\dsct{\extforce}{1} - c \dsct{\dgc}{1} - k \dsct{\gc}{1}) = 7.7264 \times 10^{-2} \unit{m/s}
\]</span> The calculations proceed in this manner, and the instances until <span class="math inline">\(t=0.1 \unit{seconds}\)</span> are obtained as: <span class="math display">\[
\small{
\begin{array}{rrr}
\dsct{\gc}{0} = 0.0000 \times 10^{-3} \unit{m}, & \dsct{\dgc}{0} = 0.0000 \times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{0} = 0.0000 \unit{m/s^2}\\
\dsct{\gc}{1} = 0.0013\times 10^{-3} \unit{m}, & \dsct{\dgc}{1} = 0.3926\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{1} = 0.0773 \unit{m/s^2}\\
\dsct{\gc}{2} = 0.0102\times 10^{-3} \unit{m}, & \dsct{\dgc}{2} = 1.5006\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{2} = 0.1412 \unit{m/s^2}\\
\dsct{\gc}{3} = 0.0331\times 10^{-3} \unit{m}, & \dsct{\dgc}{3} = 3.1391\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{3} = 0.1820 \unit{m/s^2}\\
\dsct{\gc}{4} = 0.0739\times 10^{-3} \unit{m}, & \dsct{\dgc}{4} = 5.0430\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{4} = 0.1937 \unit{m/s^2}\\
\dsct{\gc}{5} = 0.1338 \times 10^{-3} \unit{m}, & \dsct{\dgc}{5} = 6.9100 \times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{5} = 0.1748 \unit{m/s^2}\\
\dsct{\gc}{6} = 0.2110\times 10^{-3} \unit{m}, & \dsct{\dgc}{6} = 8.4482\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{6} = 0.1289 \unit{m/s^2}\\
\dsct{\gc}{7} = 0.3009\times 10^{-3} \unit{m}, & \dsct{\dgc}{7} = 9.4219\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{7} = 0.0635 \unit{m/s^2}\\
\dsct{\gc}{8} = 0.3971\times 10^{-3} \unit{m}, & \dsct{\dgc}{8} = 9.6876\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{8} = -0.0108 \unit{m/s^2}\\
\dsct{\gc}{9} = 0.4922\times 10^{-3} \unit{m}, & \dsct{\dgc}{9} = 9.2149\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{9} = -0.0823 \unit{m/s^2}\\
\dsct{\gc}{10} = 0.5792\times 10^{-3} \unit{m}, & \dsct{\dgc}{10} = 8.0896\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{10} = -0.1397 \unit{m/s^2}
\end{array}
}
\]</span> These values are provided as reference for readers wishing to check the answers they obtain through their calculations and thereby verify their codes. The response obtained for the whole <span class="math inline">\(1 \unit{second}\)</span> duration is plotted in <a href="#fig-numericalex1pem">Figure <span>4.4</span></a>.</p>
<div id="fig-numericalex1pem" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalex1pem.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.4: Displacement, velocity and acceleration response calculated for the system of <a href="#fig-numericalex1">Figure <span>4.3</span></a> subjected to the triangular pulse. The response quantities calculated by the PLE are marked with solid circles while the dashed lines are interpolations included for visualisation.</figcaption><p></p>
</figure>
</div>
<p>Let us repeat the analysis for the case of the half-sine pulse, of amplitude <span class="math inline">\(100 \unit{kN}\)</span> and duration <span class="math inline">\(0.4 \unit{seconds}\)</span>, shown in <a href="#fig-numericalex1">Figure <span>4.3</span></a>.c. This time the force is defined by <span class="math display">\[
\dsct{\extforce}{j} = \begin{cases} 100 \sinp{\frac{\pi}{0.4}\dsct{t}{j}} = 100 \sinp{\frac{\pi}{40}j}; & j = 0,1,\ldots,40 \\
0; & j = 41, 42, \ldots, 100
\end{cases}
\]</span></p>
<div id="fig-numericalex1pem2" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalex1pem2.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.5: Displacement, velocity and acceleration response calculated for the system of <a href="#fig-numericalex1">Figure <span>4.3</span></a> subjected to the half-sine pulse. The response quantities calculated by the PLE are marked with solid circles while the dashed lines are interpolations included for visualisation.</figcaption><p></p>
</figure>
</div>
<p>Since the coefficients <span class="math inline">\(A\)</span>’s and <span class="math inline">\(B\)</span>’s of the PLE do not depend on the input but only on the system parameters and the time step, the values calculated previously for the triangular pulse case are still valid. Proceeding with the recursive calculations, the response quantities until <span class="math inline">\(t=0.1 \unit{seconds}\)</span> are obtained as follows, presented for reference: <span class="math display">\[
\small{
\begin{array}{rrr}
\dsct{\gc}{0} = 0.0000\times 10^{-4} \unit{m}, & \dsct{\dgc}{0} = 0.0000 \times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{0}= 0.0000 \unit{m/s^2} \\
\dsct{\gc}{1}= 0.0103\times 10^{-4} \unit{m}, & \dsct{\dgc}{1}= 0.3080\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{1}= 0.0606 \unit{m/s^2} \\
\dsct{\gc}{2}= 0.0804\times 10^{-4} \unit{m}, & \dsct{\dgc}{2}= 1.1755\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{2}= 0.1104 \unit{m/s^2} \\
\dsct{\gc}{3}= 0.2591\times 10^{-4} \unit{m}, & \dsct{\dgc}{3}= 2.4518\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{3}= 0.1414 \unit{m/s^2} \\
\dsct{\gc}{4}= 0.5772\times 10^{-4} \unit{m}, & \dsct{\dgc}{4}= 3.9214\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{4}= 0.1486 \unit{m/s^2} \\
\dsct{\gc}{5}= 1.0416\times 10^{-4} \unit{m}, & \dsct{\dgc}{5}= 5.3378\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{5}= 0.1310 \unit{m/s^2} \\
\dsct{\gc}{6}= 1.6350\times 10^{-4} \unit{m}, & \dsct{\dgc}{6}= 6.4633\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{6}= 0.0913 \unit{m/s^2} \\
\dsct{\gc}{7}= 2.3181\times 10^{-4} \unit{m}, & \dsct{\dgc}{7}= 7.1061\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{7}= 0.0357 \unit{m/s^2} \\
\dsct{\gc}{8}= 3.0361\times 10^{-4} \unit{m}, & \dsct{\dgc}{8}= 7.1495\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{8}= -0.0270 \unit{m/s^2} \\
\dsct{\gc}{9}= 3.7271\times 10^{-4} \unit{m}, & \dsct{\dgc}{9}= 6.5696\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{9}= -0.0873 \unit{m/s^2} \\
\dsct{\gc}{10}= 4.3315\times 10^{-4} \unit{m}, & \dsct{\dgc}{10}= 5.4370\times 10^{-3} \unit{m/s}, & \dsct{\ddgc}{10}= -0.1361 \unit{m/s^2}
\end{array}
}
\]</span> The displacement, velocity and acceleration responses calculated for the first <span class="math inline">\(1 \unit{second}\)</span> are plotted in <a href="#fig-numericalex1pem2">Figure <span>4.5</span></a>.</p>
</section>
<section id="footnotes" class="footnotes footnotes-end-of-section" role="doc-endnotes">
<hr>
<ol>
<li id="fn1"><p>This approach is referred in many references as the <em>piecewise exact method</em> which we think is somewhat misleading in that the ‘exactness’ alluded to with this name pertains only the analytical expressions used and not to the accurateness of the method itself.<a href="#fnref1" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
<li id="fn2"><p>Many structures involve relatively large masses such that in practice metric ton (<span class="math inline">\(\punit{ton}\)</span> or <span class="math inline">\(\punit{t}\)</span>) is often used, with <span class="math inline">\(1 \unit{ton} = 1000 \unit{kg}\)</span>.<a href="#fnref2" class="footnote-back" role="doc-backlink">↩︎</a></p></li>
</ol>
</section>
<section id="sec-sdofnum2" class="level4 unnumbered">
<h4 class="unnumbered" data-anchor-id="sec-sdofnum2">EXAMPLE 4 .2</h4>
<p>To illustrate how the calculations would proceed for an SDOF system subjected to ground motion, let us consider the system of <a href="#sec-sdofnum">Ex. 4 .1</a>, subjected to a ground motion defined by <span class="math display">\[
\gdist = \begin{cases} \ratio{\maxgvel}{2}t - \ratio{\maxgvel t_{d}}{2 \pi} \sinp{\ratio{2\pi}{t_{d}}t}; 0 \leq t \leq t_{d} \\
\ratio{\maxgvel t_{d}}{2}; t > t_{d}\end{cases}
\]</span> with <span class="math inline">\(\maxgvel = 1 \unit{m/s}\)</span> and <span class="math inline">\(t_{d} = 0.4 \unit{s}\)</span>. The ground acceleration is therefore given for <span class="math inline">\(0 \leq t \leq t_{d}\)</span> by <span class="math display">\[
\gacct = \ratio{\pi \maxgvel}{t_{d}} \sinp{\ratio{2\pi}{t_{d}}t} = 2.5 \pi \sinp{{5 \pi}t}
\]</span> and it is zero at all other times. Recall that for this system, <span class="math inline">\(\damp = 2 \%\)</span>.</p>
<div id="fig-numericalex2" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalex2.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.6: (a) SDOF model of a single story structure subjected to ground motion, (b) A velocity-pulse type ground motion with duration <span class="math inline">\(0.5 \unit{s}\)</span>.</figcaption><p></p>
</figure>
</div>
<p>The model and the ground acceleration are shown in <a href="#fig-numericalex2">Figure <span>4.6</span></a>, and the governing equation of motion for <span class="math inline">\(0 \leq t \leq 0.4 \unit{s}\)</span> is given by <span class="math display">\[
m \ddgct + c \dgct + k \gct = - m \gacct \rightarrow 125 \ddgct + c \gct + 2 \times 10^5 \gct = - 312.5 \pi \sinp{5 \pi t}
\]</span> subjected to initial conditions <span class="math inline">\(\gc (0) = 0, \dgc (0) = 0\)</span>; or equivalently by <span class="math display">\[
\ddgct + 2 \zeta \freq \dgct + \freq^2 \gct = - \gacct \rightarrow \ddgct + 1.6 \gct + 1600 \gct = - 2.5 \pi \sinp{5 \pi t}
\]</span> where we have taken into consideration that <span class="math inline">\(\damp = 0.02\)</span> and <span class="math inline">\(\freq = \sqrt{k/m} = 40 \unit{rad/s}\)</span> as per the properties stated for the system. Free vibrations ensue after <span class="math inline">\(t = 0.4 \unit{s}\)</span>. Either form of the equation of motion may be used while calculating the response, but note that for the second equation the numerical values for the mass, damping and stiffness coefficients would have to be defined as <span class="math inline">\(m = 1\)</span>, <span class="math inline">\(c = 1.6\)</span> and <span class="math inline">\(k = 1600\)</span>, respectively. Let us use this second form in our calculations with a constant time step of <span class="math inline">\(\tstep=0.01 \unit{s}\)</span>. The input is defined at each step by <span class="math display">\[
\dsct{\extforce}{j} = - m\dsct{\gacc}{j} = \begin{cases} - 7.854 \sinp{0.1571 j} & \text{for } j = 0,1,\ldots,40 \\
0 & \text{for } j = 41, \ldots, 100
\end{cases}
\]</span></p>
<div id="fig-numericalex2pem" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalex2pem.png" class="img-fluid figure-img" style="width:100.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.7: Relative displacement, relative velocity and relative acceleration response calculated for the system of <a href="#fig-numericalex2">Figure <span>4.6</span></a> subjected to the pulse-type ground motion. The response quantities calculated by the PLE are marked with solid circles while the dashed lines are interpolations included for visualisation.</figcaption><p></p>
</figure>
</div>
<p>For <span class="math inline">\(m=1\)</span>, <span class="math inline">\(\damp=0.02\)</span> and <span class="math inline">\(k=1600\)</span>, the coefficients for the PLE are calculated via <a href="#eq-pemai">Equation <span>4.4</span></a> and <a href="#eq-pembi">Equation <span>4.6</span></a> to obtain <span class="math display">\[
\begin{array}{rrrr}
A_{1} = 9.2148\times 10^{-1}, & A_{2} = 9.6580\times 10^{-3} \\
A_{3} = 3.2607\times 10^{-5}, & A_{4} = 1.6468\times 10^{-5} \\
B_1 = -1.5453\times 10^{1}, & B_2 = 9.0603 \times 10^{-1} \\
B_3 = 4.7504\times 10^{-3}, & B_4 = 4.9076\times 10^{-3}
\end{array}
\]</span> The recursive calculations via <a href="#eq-pemgc">Equation <span>4.3</span></a>, <a href="#eq-pemdgc">Equation <span>4.5</span></a> and <a href="#eq-pemddgc">Equation <span>4.7</span></a> yield, for the first few steps, the following results:</p>
<p><span class="math display">\[
\small{
\begin{array}{rrr}
\dsct{\gc}{ 0 } = 0.0000\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 0 } = 0.0000\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 0 } = -0.0000 \unit{m/s^2} \\
\dsct{\gc}{ 1 } = -0.0202\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 1 } = -0.0603\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 1 } = -1.1866 \unit{m/s^2} \\
\dsct{\gc}{ 2 } = -0.1569\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 2 } = -0.2290\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 2 } = -2.1393 \unit{m/s^2} \\
\dsct{\gc}{ 3 } = -0.5036\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 3 } = -0.4735\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 3 } = -2.6841 \unit{m/s^2} \\
\dsct{\gc}{ 4 } = -1.1136\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 4 } = -0.7471\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 4 } = -2.7151 \unit{m/s^2} \\
\dsct{\gc}{ 5 } = -1.9897\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 5 } = -0.9967\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 5 } = -2.2106 \unit{m/s^2} \\
\dsct{\gc}{ 6 } = -3.0818\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 6 } = -1.1712\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 6 } = -1.2357 \unit{m/s^2} \\
\dsct{\gc}{ 7 } = -4.2934\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 7 } = -1.2302\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 7 } = 0.0683 \unit{m/s^2} \\
\dsct{\gc}{ 8 } = -5.4955\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 8 } = -1.1501\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 8 } = 1.5073 \unit{m/s^2} \\
\dsct{\gc}{ 9 } = -6.5461\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 9 } = -0.9284\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 9 } = 2.8651 \unit{m/s^2} \\
\dsct{\gc}{ 10 } = -7.3110\times 10^{-3} \unit{m}, & \dsct{\dgc}{ 10 } = -0.5835\times 10^{-1} \unit{m/s}, & \dsct{\ddgc}{ 10 } = 3.9370 \unit{m/s^2}
\end{array}
}
\]</span> The relative displacement, relative velocity and relative acceleration responses calculated for the first <span class="math inline">\(1 \unit{second}\)</span> are plotted in <a href="#fig-numericalex2pem">Figure <span>4.7</span></a>.</p>
</section>
</section>
<section id="central-difference-method" class="level2" data-number="4.3">
<h2 data-number="4.3" data-anchor-id="central-difference-method"><span class="header-section-number">4.3</span> Central Difference Method</h2>
<p>A family of numerical methods commonly referred to as <em>finite difference methods</em> makes use of the Taylor series expansion of response and assumes that the series may be truncated after some terms with negligible error if the time step is relatively small. What is known as the Central Difference Method makes use of a one step forward and a one step backward expansion to reduce the order of error. Recall that for some sufficiently smooth continuous function <span class="math inline">\(\gc = \gc (t)\)</span>, Taylor series expansion expresses the value of the function at <span class="math inline">\(t=t_2\)</span> as a function of the value of itself and its derivatives at <span class="math inline">\(t=t_1\)</span> through <span class="math display">\[\begin{align*}
\gc(t_2) & = \gc(t_1) + (t_2 - t_1) \frac{\diff \gc }{\dt}\biggr|_{t=t_1} \!\!\!\!\! + \frac{(t_2 - t_1)^2}{2}\frac{\diff^2 \gc}{\dt^2}\biggr|_{t=t_1} \!\!\!\!\! + \frac{(t_2 - t_1)^3}{6} \frac{\diff^3 \gc}{\dt^3}\biggr|_{t=t_1} \!\!\!\!\! + \cdots \\
& = \gc(t_1) + \sum_{j=1}^{\infty} \frac{(t_2 - t_1)^j}{j!}\frac{\diff^j \gc}{\dt^j}\biggr|_{t=t_1}
\end{align*}\]</span> Consider now the Taylor series expansions for displacement <span class="math inline">\(\gct\)</span> at <span class="math inline">\(t+\tstep\)</span> and <span class="math inline">\(t - \tstep\)</span>, given by <span class="math display">\[\begin{align*}
\gc (t + \tstep) & = \gc (t) + \dgc (t) \tstep + \frac{1}{2}\ddgc (t) \tstep^2 + o(\tstep^3) \\
\gc (t - \tstep) & = \gc (t) - \dgc (t) \tstep + \frac{1}{2}\ddgc (t) \tstep^2 - o(\tstep^3)
\end{align*}\]</span> where <span class="math inline">\(o(\tstep^p)\)</span> is the remainder of the expansion with leading term of order <span class="math inline">\(\tstep^p\)</span>. The sum of the two equations above leads to <span class="math display">\[
\gc (t + \tstep) + \gc (t - \tstep) = 2 \gc (t) + \ddgc (t) \tstep^2 + o(\tstep^4)
\]</span> and if the fourth-order remainder were to be neglected, the acceleration at time <span class="math inline">\(t\)</span> could be approximated by <span id="eq-cemacc"><span class="math display">\[
\ddgc (t) \approx \frac{\gc (t + \tstep) - 2 \gc (t) + \gc (t - \tstep)}{\tstep^2}
\qquad(4.8)\]</span></span> The error so incurred is expected to decrease in an absolute sense as <span class="math inline">\(\tstep\)</span> decreases. On the other hand, the difference of the two expansions leads to <span class="math display">\[
\gc (t + \tstep) - \gc (t - \tstep) = 2 \dgc (t) \tstep + + o(\tstep^3)
\]</span> so that neglecting the third order remainder yields the following approximation for the velocity: <span id="eq-cemvel"><span class="math display">\[
\dgc (t) \approx \frac{\gc (t + \tstep) - \gc (t - \tstep)}{2 \tstep}
\qquad(4.9)\]</span></span> <a href="#eq-cemacc">Equation <span>4.8</span></a> and <a href="#eq-cemvel">Equation <span>4.9</span></a> are called the central difference approximations to the acceleration and the velocity, respectively. These approximations may be used in the equation of motion to express the equation solely in terms of the generalized displacement. Substituting the central difference approximations into the equation of motion and considering that time is discretized with constant time step <span class="math inline">\(\tstep\)</span> so that <span class="math inline">\(t = j \tstep\)</span> and <span class="math inline">\(t \pm \tstep = (j \pm 1) \tstep\)</span>, we get <span class="math display">\[\begin{align*}
m \ddgc (t) + c \dgc (t) + k \gc (t) & = m \dsct{\ddgc}{j} + c \dsct{\dgc}{j} + k \dsct{\gc}{j} \\
& = m \frac{\dsct{\gc}{j+1} - 2 \dsct{\gc}{j} + \dsct{\gc}{j-1}}{\tstep^2} + c \frac{\dsct{\gc}{j+1} - \dsct{\gc}{j-1}}{2 \tstep} + k \dsct{\gc}{j} = \extforce(t)= \dsct{\extforce}{j}
\end{align*}\]</span> It is straightforward to recast this expression into a regression type formula so that given the response up to and including time <span class="math inline">\(t\)</span> (time step <span class="math inline">\(p\)</span>), the response at <span class="math inline">\(t+\tstep\)</span> (time step <span class="math inline">\(p+1\)</span>) may then be estimated using <span id="eq-cdmq"><span class="math display">\[
A_1 \dsct{\gc}{j+1} = \left(\dsct{\extforce}{j} - A_2 \dsct{\gc}{j} - A_3 \dsct{\gc}{j-1} \right)
\qquad(4.10)\]</span></span> where <span id="eq-cdmai"><span class="math display">\[
\begin{array}{l}
A_1 = {\ratio{m}{\tstep^2} + \ratio{c}{2 \tstep}} = m \left[\ratio{1}{\tstep^2} + \ratio{\damp \freq}{\tstep} \right] \\
A_2 = k - \ratio{2m}{\tstep^2} = m \left[\freq^2 - \ratio{2}{\tstep^2} \right] \\
A_3 = \ratio{m}{\tstep^2} - \frac{c}{2 \tstep} = m \left[\ratio{1}{\tstep^2} - \ratio{\damp \freq}{\tstep} \right] \\
A_4 = \ratio{1}{2\tstep}\\
A_5 = \ratio{1}{\tstep^2}
\end{array}
\qquad(4.11)\]</span></span> Once again, for a constant time step <span class="math inline">\(\tstep\)</span>, the coefficients <span class="math inline">\(A_i\)</span> are to be calculated only once. At each time step, the velocity is to be calculated from the central difference approximation in <a href="#eq-cemvel">Equation <span>4.9</span></a>, which may be written in discrete form as <span id="eq-cdmdq"><span class="math display">\[
\dsct{\dgc}{j} = \frac{1}{2\tstep}(\dsct{\gc}{j+1} - \dsct{\gc}{j-1}) = A_4 (\dsct{\gc}{j+1} - \dsct{\gc}{j-1})
\qquad(4.12)\]</span></span> and in the case of SDOF systems the acceleration may be calculated using the equilibrium equation for that time step via the discretized form: <span id="eq-cdmddq"><span class="math display">\[
m \dsct{\ddgc}{j} = \left(\dsct{\extforce}{j} - c \dsct{\dgc}{j} - k \dsct{\gc}{j} \right)
\qquad(4.13)\]</span></span> In the case of systems with numerous degrees of freedom the matrix computations involved above may be prohibitive and the accelerations may instead be calculated via the central difference approximations: <span id="eq-cdmddq2"><span class="math display">\[
\dsct{\ddgc}{j} = \frac{\dsct{\gc}{j+1}- 2 \dsct{\gc}{j} + \dsct{\gc}{j-1}}{\tstep^2} = A_5 (\dsct{\gc}{j+1}- 2 \dsct{\gc}{j} + \dsct{\gc}{j-1})
\qquad(4.14)\]</span></span></p>
<p>For initial conditions <span class="math inline">\(\dsct{\gc}{0}\)</span> and <span class="math inline">\(\dsct{\dgc}{0}\)</span>, the first iteration would be given by <span class="math display">\[
A_1 \dsct{\gc}{1} = \left(\dsct{\extforce}{0} - A_2 \dsct{\gc}{0} - A_3 \dsct{\gc}{-1} \right)
\]</span> where <span class="math inline">\(\dsct{\gc}{-1}\)</span> is needed. One practice for a system at rest may be to assume <span class="math inline">\(\dsct{\gc}{-1} = 0\)</span>. An alternative is to use a value consistent with the central difference approximation which may be derived as follows: The central difference approximations for the acceleration and the velocity at time <span class="math inline">\(t = 0\)</span> are <span class="math display">\[
\dsct{\ddgc}{0} = \frac{\dsct{\gc}{1} - 2 \dsct{\gc}{0} + \dsct{\gc}{-1}}{\tstep^2}, \quad \dsct{\dgc}{0} = \frac{\dsct{\gc}{1} - \dsct{\gc}{-1}}{2 \tstep}
\]</span> so that using these two equations to eliminate the <span class="math inline">\(\dsct{\gc}{1}\)</span> term, <span class="math inline">\(\dsct{\gc}{-1}\)</span> is given by <span class="math display">\[
\dsct{\gc}{-1} = \dsct{\gc}{0} - \tstep \dsct{\dgc}{0} - \frac{\tstep^2}{2} \dsct{\ddgc}{0}
\]</span> where the initial acceleration term <span class="math inline">\(\dsct{\ddgc}{0}\)</span> is to be calculated from the equilibrium equation at time <span class="math inline">\(t=0\)</span>: <span class="math display">\[
\dsct{\ddgc}{0} = \frac{1}{m} \left(\dsct{\extforce}{0} - c \dsct{\dgc}{0} - k \dsct{\gc}{0} \right)
\]</span> The calculations involved are summarized in <a href="#fig-algcdm">Figure <span>4.8</span></a>.</p>
<div id="label-name" class="callout callout-style-default callout-note no-icon callout-titled">
<div class="callout-header d-flex align-content-center">
<div class="callout-icon-container">
<i class="callout-icon no-icon"></i>
</div>
<div class="callout-title-container flex-fill">
<strong>Pseudocode for the Central Difference Method</strong>
</div>
</div>
<div class="callout-body-container callout-body">
<p><em>viscously damped SDOF system, for constant time step</em></p>
<p><strong>Input</strong>: <span class="math inline">\(m\)</span>, <span class="math inline">\(\damp\)</span> (or <span class="math inline">\(c\)</span>), <span class="math inline">\(k\)</span>, <span class="math inline">\(\tstep\)</span>, <span class="math inline">\(\dsct{\gc}{0}\)</span>, <span class="math inline">\(\dsct{\dgc}{0}\)</span>, <span class="math inline">\(\left\{\dsct{\extforce}{0}, \ldots, \dsct{\extforce}{l} \right\}\)</span></p>
<p><strong>Initialize</strong>: calculate</p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(A_1,A_2,A_3\)</span>, <span class="math inline">\(A_4\)</span> and <span class="math inline">\(A_5\)</span> using <a href="#eq-cdmai">Equation <span>4.11</span></a>, <span class="math inline">\(A_1^{-1}\)</span></p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\ddgc}{0} = \frac{1}{m}(\dsct{\extforce}{0} - c \dsct{\dgc}{0} - k\dsct{\gc}{0})\)</span></p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\gc}{-1} = \dsct{\gc}{0} - \tstep \dsct{\dgc}{0} - \frac{\tstep^2}{2} \dsct{\ddgc}{0}\)</span></p>
<p><strong>for</strong> <span class="math inline">\(j=0,\ldots,l-1\)</span>:</p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\gc}{j+1} = A_1^{-1} \left(\dsct{\extforce}{j} - A_2 \dsct{\gc}{j} - A_3 \dsct{\gc}{j-1} \right)\)</span></p>
<p><strong>for</strong> <span class="math inline">\(j=1,\ldots,l\)</span>:</p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\dgc}{j} = A_4(\dsct{\gc}{j+1} - \dsct{\gc}{j-1})\)</span></p>
<p><span class="math inline">\(\phantom{xxxxxxxxxx}\)</span> <span class="math inline">\(\dsct{\ddgc}{j} = A_5 (\dsct{\gc}{j+1}- 2 \dsct{\gc}{j} + \dsct{\gc}{j-1})\)</span></p>
</div>
</div>
<div id="fig-algcdm" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/blank.png" class="img-fluid figure-img" style="width:0.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.8: Pseudocode for the central difference method.</figcaption><p></p>
</figure>
</div>
<p>It is extremely important to note that for the response to stay stable, meaning not to eventually get excessively and unphysically large so as to render all results useless, the time step used must be such that <span class="math display">\[
\tstep \leq \frac{T}{\pi}
\]</span> where <span class="math inline">\(T = 2 \pi \sqrt{m}/\sqrt{k}\)</span> is the natural period of the system. Because of this constraint, the CDM is said to be a <em>conditionally stable</em> method, as opposed to those that are stable for any time step (which are called <em>unconditionally stable</em>). Note that this condition does not necessarily guarantee accuracy, but only stability. In general, the smaller the time step, the more accurate the results would be expected to be; a commonly suggested value is <span class="math inline">\(\tstep < T/10\)</span>.</p>
<p>Geometrically, the central difference method is equivalent to fitting a second order curve passing through three data points <span class="math inline">\(\dsct{\gc}{j-1}\)</span>, <span class="math inline">\(\dsct{\gc}{j}\)</span> and <span class="math inline">\(\dsct{\gc}{j+1}\)</span>, and evaluating the derivatives (velocity and acceleration in this case) at the midpoint. Assume a discretization with constant time step <span class="math inline">\(\tstep\)</span> and consider a time variable <span class="math inline">\(\tau = t - (j-1) \tstep\)</span>, such that <span class="math display">\[
\dgc (t=\tau + (j-1) \tstep) = \frac{\diff \gc(\tau)}{\diff \tau}, \quad \ddgc (t=\tau + (j-1) \tstep) = \frac{\diff^2 \gc (\tau)}{\diff \tau^2}
\]</span> for <span class="math inline">\(j\tstep \leq t \leq (j+1)\tstep\)</span> and <span class="math inline">\(0 \leq \tau \leq 2\tstep\)</span>. Given three data points <span class="math inline">\(\dsct{\gc}{j-1}\)</span>, <span class="math inline">\(\dsct{\gc}{j}\)</span> and <span class="math inline">\(\dsct{\gc}{j+1}\)</span>, there is a unique second order curve <span class="math inline">\(\gc(\tau)\)</span> that passes through all the three points as shown in <a href="#fig-numericalparabolic">Figure <span>4.9</span></a>. Such a curve is defined by the equation <span class="math display">\[
\gc (\tau) = C_1 \tau^2 + C_2 \tau + C_3
\]</span> where <span class="math inline">\(C_i\)</span> are constants that may be determined through the available information. In fact, we have, <span class="math display">\[\begin{align*}
q(\tau = 0) & = C_3 = \dsct{\gc}{j-1} \\
q(\tau = \tstep) & = \tstep^2 C_1 + \tstep C_2 + C_3 = \dsct{\gc}{j} \\
q(\tau = 2\tstep) & = 4 \tstep^2 C_1 + 2 \tstep C_2 + C_3 = \dsct{\gc}{j+1}
\end{align*}\]</span></p>
<!-- margin figure to normal figure -->
<div id="fig-numericalparabolic" class="quarto-figure quarto-figure-center">
<figure class="figure">
<p><img src="images/numericalparabolic.png" class="img-fluid figure-img" style="width:40.0%"></p>
<p></p><figcaption class="figure-caption">Figure 4.9: Second order curve passing through data three points.</figcaption><p></p>
</figure>
</div>
<p>Solving for the coefficients leads to <span class="math display">\[
C_1 = \ratio{1}{2 \tstep^2}(\dsct{\gc}{j+1} - 2 \dsct{\gc}{j} + \dsct{\gc}{j-1}), \quad C_2 = \ratio{1}{2 \tstep}(-\dsct{\gc}{j+1} + 4 \dsct{\gc}{j} - 3 \dsct{\gc}{j-1}), \quad C_3 = \dsct{\gc}{j-1}
\]</span> so that the velocity and acceleration at the midpoint, i.e. at <span class="math inline">\(t=i \tstep\)</span> or <span class="math inline">\(\tau = \tstep\)</span>, are given by <span class="math display">\[
\dsct{\dgc}{j}=\dgc (t=j \tstep) = \frac{\diff \gc(\tau)}{\diff \tau}\biggr|_{\tau = \tstep} = (2 C_1 \tau + C_2)\bigr|_{\tau=\tstep} = \ratio{\dsct{\gc}{j+1}-\dsct{\gc}{j-1}}{2\tstep}
\]</span> <span class="math display">\[
\dsct{\ddgc}{j}=\ddgc (t=j \tstep) = \frac{\diff^2 \gc(\tau)}{\diff \tau^2}\biggr|_{\tau = \tstep} = (2 C_1)\bigr|_{\tau=\tstep} = \ratio{\dsct{\gc}{j+1}-2\dsct{\gc}{j}+\dsct{\gc}{j-1}}{\tstep^2}
\]</span> which are in fact the finite difference approximations used for these two quantities in the central difference method.</p>
<section id="sec-sdofnum3" class="level4 unnumbered">
<h4 class="unnumbered" data-anchor-id="sec-sdofnum3">EXAMPLE 4 .3</h4>
<p>To illustrate how the central difference calculations proceed, let us consider the system discussed previously in <a href="#sec-sdofnum">Ex. 4 .1</a>, subjected to the half-sine pulse shown in <a href="#fig-numericalex1">Figure <span>4.3</span></a>.c. The system is defined by parameters <span class="math inline">\(m=125 \unit{tons}\)</span>, <span class="math inline">\(k=200 \, 000 \unit{kN/m}\)</span> and <span class="math inline">\(\damp = 0.02\)</span> (or, equivalently, <span class="math inline">\(c = 2 \damp \freq m = 200 \unit{kN \cdot s / m}\)</span>). With a time step of <span class="math inline">\(\tstep = 0.01 \unit{seconds} \approx 0.06 \period\)</span>, the force is again given by <span class="math display">\[
\dsct{\extforce}{j} = \begin{cases} 100 \sinp{\frac{\pi}{40}j} & \text{for } j = 0,1,\ldots,40 \\
0 & \text{for } j = 41, 42, \ldots, 100
\end{cases}
\]</span> and the coefficients to be used in the central difference method are calculated via <a href="#eq-cdmai">Equation <span>4.11</span></a> as: <span class="math display">\[
A_{1} ^{-1}= 7.9365\times 10^{-7}, \; A_{2} = -2300 \times 10^{3}, \; A_{3} = 1240 \times 10^{3}, \; A_{4} = 50, \; A_{5} = 10^{4}
\]</span> With zero initial conditions so that <span class="math inline">\(\dsct{\gc}{0} = 0\)</span> and <span class="math inline">\(\dsct{\dgc}{0} = 0\)</span>, <span class="math inline">\(\dsct{\ddgc}{0}\)</span> and <span class="math inline">\(\dsct{\gc}{-1}\)</span> are calculated to obtain <span class="math display">\[
\dsct{\ddgc}{0} = \frac{1}{m} (\dsct{\extforce}{0} - c \dsct{\dgc}{0} - k \dsct{\gc}{0}) = 0, \quad \dsct{\gc}{-1} = \dsct{\gc}{0} - \tstep \dsct{\dgc}{0} - \frac{\tstep^2}{2} \dsct{\ddgc}{0} = 0
\]</span> after which, through the recursive formulas given by <a href="#eq-cdmq">Equation <span>4.29</span></a>, <a href="#eq-cdmdq">Equation <span>4.12</span></a> and <a href="#eq-cdmddq2">Equation <span>4.14</span></a>, the following values are calculated for the first few time steps: <span class="math display">\[
\small{