REM—-VIBRATIONS PROGRAM CONFORCE
REM (PIECEWISE-CONSTANT FORCING FUNCTION)
NI = NUMBER OF TIME STEPS
$T=$ TIME; $D=$ TIME INTERVAL
$a=$ MAGNITUDE OF FORCING FUNCTION
$x$ = DISPLACEMENT; $v=$ VELOCITY
$K$ = SPRING CONSTANT
T1 = NATURAL PERIOD; $P=$ ANGULAR FRFQUENCY
$X O$, VO INITIAL VALUES OF $X$ AND $V$

DIM $T(100), D(100), Q(100), X(100), V(100)$
READ K, T1,XO,VO,NI
$P=2 * P 1 / T 1$
MAT READ T(N1), Q(N1)
PRINT ‘UNDAMPED RESPONSE TO PIECEWISE-CONSTANT FORCING FUNCTION’
PRINT
PRINT
PRINT ‘ $T(1) \quad D(1) \quad Q(1) \quad x(1) \quad V(1)$ ‘
PRINT
PRINT 0 – 0 – ‘;X0, VO
R(1) T(1) AND PRINT RESPONSE
$V_{1}=V_{0}$
$1=1$ TO $\mathrm{N} 1$
I THEN GO TO 30
$x_{1}=x(1-1)$
$v 1=v(1-1)$
$C 1=\cos (P * D(1))$
$C 2=\operatorname{SiN}(P * D(1))$
$x(1)=X 1 * C 1+V 1 * C 2 / P+Q(1) *(1,-C 1) / K$
$V(1)=-X_{1} * C 2 * P+V 1 * C 1+Q(1) * C 2 * P / K$
PRINT T(1);D(1);Q(1);X(1),V(1)

NEXT 1
PRINT
STOP
DATA $1,10,0,0,5$
DATA $1,2,3,4,5$
DATA $1,1,1,1,1$
END
COMMAND ? GO

UNDAMPED RESPONSE TO PIECEWISE-CONSTANT FORCING FUMCTION
\begin{tabular}{lccll}
T(1) & $D(1)$ & $Q(1)$ & $x(1)$ & $V(1)$ \\
0 & – & – & 0 & 0 \\
1 & 1 & 1 & .190983 & .3693163 \\
2 & 1 & 1 & .6909829 & .597564 \\
3 & 1 & 1 & 1.309016 & .5975664 \\
4 & 1 & 1 & 1.809015 & .3693166 \\
5 & 1 & 1 & 1.999998 & $1.430511 E-0.6$ \\
LINE & 37 & (4nO0) STOP EXECUTED \\
COHMAND ?
\end{tabular}

COMPUTER PROGRAM

REM—-VIBRATIONS PROGRAM–AVAC3A
REM (AVERAGE-ACCELERATION METHOD)
REM—NOTATION
REM $T=$ TIME; $D$ a TIME INTERVAI.
REM $X, Y, Z=$ DISPLACEMENT, VELOCITY, AND ACCELERATION
REM $M, C, K=$ MASS, DAMPING, AND SPRING CONSTANTS
REM $S=$ SCALAR MULTIPLYING NONLINEAR TERM
REM $X O, Y O=I N I T I A L$ VALUES OF $X$ AND $Y A T$ TIME $T=0$
REM TI = TIME RANGE OF INTEREST
REM NI = NUMBER OF TIME INTERVALS
REM EI = ALLOWABLE ERROR RATIO
REM 01 = MAGNITUDE OF STEP FUNCTION
DIM T(100), X(100),Y(100),Z(100)
READ M, C, K,S, XO, YO, T1,N1, E1, Q1
$D=$ TI/NI
PRINT
PRINT
PRINT
PRINT
ITERS.
APPROX. $x$
APPROX. Y’

PRINT – 0 – 1, X0,YO
REM—-INITIALIZE ZO AND ITERATE $Y(1), X(i), Z(1)$
$Z O=(01-C * Y 0-K *(X O+S * X O \bullet X O \bullet X O)) / M$
FOR $I=1$ TO NI
$T(1)=1 * D$
$|F|=1$ THEN GO TO 111
IF $1=2$ THEN GO TO 115
$A=Y(1-1)+Z(1-1) \div D / 2$
$B=X(1-1)+Y(1-1) * D / 2$
$Y 1=Y(1-2)+2 * Z(1-1) * 0$
G0 TO 118
$A=Y O+Z O+D / 2$
$\mathrm{B}=\mathrm{XO}+\mathrm{YO} 0 \mathrm{D} / 2$
$Y 1=Y 0+Z O * D$
GO TO 11 \%
$A=Y(1)+Z(1) * D / 2$
$B=X(1)+Y(1) * D / 2$
$Y_{1}=Y_{0}+2 * Z(1) *$ ?
$J=1$
GO To 123
$J=J+1$
$x(1)=X 1$
$Y_{1}=A+Z I * D / 2$
$X_{1}=B+Y 1 * D / 2$
$Z 1=\left(01-C * Y 1-K *\left(X_{1}+s * X_{1} * X_{1} * X_{1}\right)\right) / M$
IF $J=10$ THEN GO TO 127
IF ABS (X1- $X(1))>=1 \pm A R S\left(X_{1}\right)$ THEN GO TO 120
$X(1)=X 1$
$Y(1)=Y 1$
$Z(1)=Z 1$
PRINT T(1), J,X(1),Y(1)

NEXT I
PRINT
STOP
DATA $100,0,400,2,0,10,0,5,20,0,0001, n$
END
COMMAND?:

COMPUTER PROGRAM

REM—-VIBRATIONS PROGRAM–AVAC2A
REM (AVERAGE-ACCELERATION METHOD)
REM—- NOTATION
REM T = TIME; $D=$ TIME INTERVAL
REM $X, Y, Z=$ DISPLACEMENT, VELOCITY, AND ACCELERATION
REM KI $\quad P * 2=G / L$ FOR A SIMPLE PÉNOULUM
RFM $X 0, Y O=$ INITIAL VALUES DF $X$ AND Y AT TIME $T=0$
REM T1 = TIME RANGE OF INTEREST
REM NI $=$ NLMBER OF TIMF INTERVALS
REM EL = ALLOWABLE ERROR RATIO
DIM T $(100), X(100), Y(100), Z(100)$
READ K1, XO, YO, T1,N1, E1
$D=T I / N I$
PRINT \”AVAC2A—-TIME INTERVAI = ‘, D
PRINT
PRINT
PRINT ‘TIME ITERS. APPROX. $X$ APPROX, Y’
PRINT
APPROX. $X$
, $\times 0, Y 0$
$Z O=-K 1 * \operatorname{SIN}\left(X_{0}\right)$
FOR, $T$ (i) $1=1
eq D$
$|F|=1$ THEN. GO TO 111
IF $I=2$ THEN GO TO 115
$A=Y(1-1)+Z(1-1) \div D / Z$
$B=X(1-1)+Y(1-1)=D / 2$
$Y 1=Y(1-2)+2 * Z(1-1) * D$
GO TO 118
$A=Y O+Z O * D / 2$
$B=X O+Y O * 0 / 2$
$Y 1=Y 0+Z 0 * 0$
Go ro 118
$A=Y(1)+Z(1)+D / 2$
$B=X(1)+Y(1) * D / 2$
$Y 1=Y 0+2 * Z(1) * D$
$J=1$
GO TO 123
$J=J+1$
$X(1)=X 1$
$Y 1=A+Z 1 * D / 2$
$X 1=B+Y 1 * D / 2$
$Z 1=-K 1 * \operatorname{SIN}\left(X_{1}\right)$
IF $J=10$ THEN GO TO 127
IF $A B S(x)-x(1))=E 1 * A B S(x))$ THFN GO TO 120
$X(1)=X 1$
$Y(1)=Y 1$
$Z(1)=Z 1$
PRINT T(1), J, X(1),Y(1)

NEXT 1
PRINT
STOP
DATA $3.437687,1,570796,0,2,20,0.0001$
END
COMMAND ?

COMPUTER PROGRAM 462

COMPUTER PROGRAM

REN:–VIBRATIONS PROGRAM=-EIGIT3
REM (EIGENVALUES AND VECTORS BY ITERATION)
AEM- – NOTATION
REM $\quad$ = COEFFICIENT MATRIX (OF ORDER N)
REM $\quad M=$ MASS VECTOR
REM $X, Y, Z=$ EIGENVECTORS
REM EM EIGENVALUE
REM S = SWEEPING MATRIX
REM E1 = ALLOWABLE ERROR
$\operatorname{DIM} A(10,10), M(10), X(10), Y(10), Z(10,2), S(10,10)$
READ N, E1
MAT READ A(N,N),M(N)
PRINT ‘EIGIT3—THREE EIGENVALUES AND VECTORS BY ITERATION’
PRINT
$1=0$
REM—-ITERATE AND PRINT EIGENVALUE AND EIGENVECTOR
$1=1+1$
MAT $X=\operatorname{CON}(N)$
FOR $K=1$ TO 20
MAT $Y=A * X$
$F=Y(N) / X(N)$
$J 1=0$
FOR $J=1$ TO N
$Y(J)=Y(J) / Y(N)$
IF ABS(Y(J) – $x(J)$ ) < EI THEN J1 = J1 – 1 .
$X(J)=Y(J)$
NEXT.J
If JI = N THEN GO TO 30

NEXT $K$
PRINT
PRINT ‘MODE ‘;1;’Eigenvalue = ‘;E;’NO. OF ItERs. = ‘;K
PRIITT ‘EIGENVÉCT́OR’
MAT PRINT $X$
IF 1 = 3 THEN GO TO 57
REM—SET UP AND APPLY SWEEPING MATRIX
MAT $S=\operatorname{IDN}(N, N)$
$|F|=2$ THEN CO TO 46
$\$(1,1)=0$
$Z(1,1)=x(1)$
$C=M(1) * x(1)$
FOR $J=2$ TO $N$
$S(1, j)=-M(J) * x(J) / C$
$Z(J, 1)=x(J)$

NEXT $J$
GO TO 54
$S(2,2)=0$
$z(1,2)=x(1)$
$z(2,2)=x(2)$
$C=M(2) *(Z(1,1) * Z(2,2)-Z(1,2) * Z(2,1))$
FOR $J=3$ TO $\mathrm{N}$
$z(J, 2)=x(J)$
$S(2, J)=-M(J) *(Z(1,1) \cdot Z(J, 2)-Z(1,2) \cdot Z(J, 1)) / C$

NEXT $J$
MAT $B=A \cdot S$
MAT $A=B$
GO TO 17
PRINT
STOP
DATA 3, 0.0001
DATA 1,1,1, 1,2,2, 1,2,3
DATA 1,1,1
COMMAND ?

COMPUTER PROGRAM

REM—-VIBRATIONS PROGRAM–DYNACON3
REM (DYNAMIC RESPONSE OF MULTI-DEGREE SYSTEM
REM TO PIECEWISE-CONSTANT FORCING FUNCTION)
REM—-NOTATION
REM $N=$ NUMBER OF DEGREES OF FREEDOM
REM N1 = NUMBER OF TIME STEPS
REM EI = ALLOWABLE ERROR FOR EIGENVECTORS
REM G1 = MODAL DAMPING RATIO (GAMMA)
REM II = TYPE INDICATOR (O-STIFFNESS; 1-FLEXIBILITY)
REM $\quad A=$ STIFFNESS OR FLEXIBILITY MATRIX
REM $M=$ MASS VECTOR
REM B,C = INITIAL DISPLACEMENTS AND VELOCITIES
REM $P=$ VECTOR OF LOAD FACTORS
REM T = TIME; $D=$ TIME INTERVAL
REM $\quad F=$ FORCING FUNCTION (PIECEWISE-CONSTANT)
REM E = EIGENVALUES; $X, Y, Z=$ EIGENVECTORS
REM $S=$ SWEEPING MATRIX
REM G,H = INITIAL DISPLS. AND VELS. IN NORMAL COORDS.
REM $Q=$ NORAAL-MODE LOADS
REM P1,P2 = UNDAMPED AND DAMPED ANGULAR FREQUENCIES
REM U,V NORMAL-MODE DISPLACEMENTS AND VELOCITIES
REM R RESPONSE IN ORIGINAL COOROINATES
REM W, $L$ WORKING STOPAGE
$0 \operatorname{IM} A(10,10), M(10), B(10), C(10), P(10), T(100), D(100), F(100), E(3), X(10)$
$D I M Y(10), Z(10,3), Q(3,100), U(3,100), V(3,100), R(1,10), W(10,10), i(10)$
READ N, N1, E1, G1, I 1
MAT READ A(N,N),M(N), B(N),C(N),P(N), T(N1),F(N1)
IF II = 0 THEN MAT $A=I N V(A)$
FOR $K=1$ TO $N$
FOR $J=1$ TO $N$
$A(J, K)=A(J, K) * M(K)$
NEXT $J$

NEXT K
$D(1)=T(1)$
FOR $J=2$ TO NI
$D(J)=T(J)-T(J-1)$

NEXT J
PRINT ‘DYNACON3–DYNAMIC RESPONSE OF FIRST THREE MODES OF DAMPED’
PRINT ‘MULTI-DEGEE SYSTEM TO PIECEWISE-CONSTANT FORCING FUNCTION’
PRINT
PRINT
PRINT ‘THREE EIGENVALUES AND VECTORS BY ITERATION’
MAT $Z=Z E R(N, 3)$
$1=0$
REM—-ITERATE AND PRINT EIGENVALUE AND ANGULAR FREQUENCY
$1=1+1$
MAT $X=\operatorname{CON}(N)$
FOR $K=1$ TO 20
MAT $Y=A * X$
$E(I)=Y(N) / X(N)$
$\mathrm{Jl}=0$
FOR $J=1$ TO $N$
$Y(J)=Y(J) / Y(N)$
IF ABS (Y(J)-X(J))<E1 THEN J1-J1+1
$x(J)=Y(J)$
NEXT $J$
IF Jl = N THEN GO TO 59

NEXT $K$
FOR $J=1$ TO $N$
$Z(J, 1)=X(J)$

NEXT J
$P 1=1 / S Q R(E(1))$

PRINT
PRINT ‘MODE ‘ ; ‘;’E-VAL. = ‘; E(1);’ANG. FREQ. = ‘;P1;’ITERS. = ‘;K
$|F|=3$ THEN GO TO 84
REM—SET UP AND APPLY SWEEPING MATRIX
MAT $S=1 D N(N, N)$
IF I – 2 THEN GO TO 75
$\mathrm{S}(1,1)=0$
$C 1=M(1) * x(1)$
FOR $J=2$ TO $\mathrm{N}$
$S(1, J)=-M(J) \star x(J) / C .1$

NEXT $J$
GO To 81
IF $N=2$ THEN GO TO 84
$\mathrm{S}(2,2)=0$
$C 2=M(2) *(Z(1,1) * Z(2,2)-Z(1,2) * Z(2,1))$
FOR $J=3$ TO $\mathrm{N}$
$S(2, J)=-M(J) *(Z(1,1) * Z(J, 2)-Z(1,2) * Z(J, 1)) / C 2$

NEXT $J$
MAT $W=A * S$
MAT $A=W$
GO TO 46
PRINT
PRINT ‘MODAL MATRIX’
PRINT
MAT PRINT $Z$
REM—NORMALIZE MODAL MATRIX WITH RESPECT TO M
FOR $1=1$ TO 3
$\mathrm{Cl}=0$
FOR $J=1$ TO $N$
$C 1=C 1+M(J) * Z(J, 1) * Z(J, 1)$
NEXT $J$
IF $C 1=0$ THEN GO TO 100
$C 1=\operatorname{SQR}(\mathrm{C} 1)$
FOR $J=1$ TO $N$
$Z(J, 1)=Z(J, 1) / C 1$
NEXT $J$

NEXT 1
REM—-TRANSFORM INFORMATION TO NORMAL COORDINATES
FOR $1=1$ TO 3
$G(1)=H(1)=L(1)=0$
FOR $K=1$ TO $N$
$G(1)=G(1)+Z(K, 1) * M(K) * B(K)$
$H(1)=H(1)+Z(K, 1) * M(K) * C(K)$
$L(1)=L(1)+z(k, 1) * P(k)$
NEXT $K$
FOR $J=1$ TO N1
$Q(1, J)=L(1) * F(J)$
NEXT J

NEXT I
REM—-COMPUTE RESPONSE IN NORMAL COOROINATES
FOR $1=1$ TO 3
$U_{1}=G(1)$
$V 1=H(1)$
IF $E(1)=0$ THEN GO TO 134
$P 1=1 / \operatorname{SQR}(E(1))$
$P 2=P 1 * S Q R(1-G 1 * G 1)$
FOR – J $=1$ TO N1
IF $J=1$ THEN GO TO 123
$v 1=u(1, J-1)$
$v_{1}=V(1, v-1)$
$C_{1}=E X P(-P 1 * G 1 * D(J))$
$C 2=\cos (P 2 * D(J))$
464

$C 3=\sin (P 2 * D(J))$
$C_{4}=(V 1+P 1 * G 1 * U 1) / P 2$
$C .5=P 1 * G 1 / P 2$
$C 6=Q(1, J) /(P 1 * P 1)$
$U(1, J)=C 1 *(U 1 * C 2+C 4 * C 3)+C 6 *(1-C 1 *(C 2+C 5 * C 3))$
$v(1, J)=C_{1} *\left(-\mathrm{U} 1 * \mathrm{C}_{3}+\mathrm{C}_{4} * \mathrm{C}_{2}-\mathrm{C}_{5} *\left(\mathrm{U}_{1} * \mathrm{C}_{2}+\mathrm{C}_{4} * \mathrm{C}_{3}\right)\right) * \mathrm{P}_{2}$
$v(1, J)=v(1, J)+c 6 * c 1 *(1+C 5 * C 5) * c 3 * p 2$
NEXT $J$
NEXT I
REM—-TRANSFORM AND PRINT RESPONSE
:PRINT
PRINT ‘RESPONSE IN ORIGINAL COORDINATES (PRINTED COLUMN-WISE)’
PRINT
FOR $J=1$ TO N1
MAT $R=\operatorname{ZER}(1, N)$
FOR $K=1$ TO $\mathrm{N}$
FOR $1=1$ TO 3
$R(1, K)=R(1, K)+Z(K, 1) * U(1, J)$
NEXT 1
NEXT $K$
MAT PRINT ‘R

NEXT $J$
PRINT
STOP
DATA $3,10,0,00001,0.05,0$
DATA $2,-1,0,-1,2,-1,0,-1,1$
DATA $1,1,1,0,0,0,0,0,0,0,0,1$
DATA $1,2,3,4,5,6,7,8,9,10$
DATA $1,1,1,1,1,1,1,1,1,1$
END
COMMAND ? go
DYNACONS–DYNAMIC RESPONSE OF FIRST THREE MODES OF DAMPED MULTI-DEGEE SYSTEM TO PIECEWISE-CONSTANT FORCING FUNCTION

THREE EIGENVALUES AND VECTORS BY ITERATION
MODE 1 E-VAL. = 5.048918 ANG. FREQ. = . 445042 .ITERS. $=7$
MODE 2 E-VAL. = . 6431057 ANG. FREQ. = 1.246973 ITERS, = 17
MODE 3 E-VAL. = . 3079775 ANG. FREQ. = 1.801941 ITERS, = 2
MODAL MATRIX
\[
\begin{array}{llr}
.445042 & -1.246984 & 1.801909 \\
i 8019375 & -.5549535 & -2.246983
\end{array}
\]

RESPONSE IN ORIGINAL COORDINATES (PRINTED COLUMN-WISE)
\begin{tabular}{lll}
$3.058553 E-03$ & $4.285055 E-02$ & .4488853 \\
$7.557404 E-02$ & .4513053 & 1.422269 \\
1499555 & 1.361509 & 2.416564 \\
1.195432 & 2.370045 & 3.339921 \\
1.856796 & 3.163614 & 4.245371 \\
2.014587 & 3.692737 & 5.019314 \\
1.878725 & 3.859205 & 5.430093 \\
1.811241 & 3.647371 & 5.271109 \\
1.74055 & 3.225843 & 4.54407 \\
1.438758 & 2.635145 & 3.558956
\end{tabular}