7.1 Solving the Structural Dynamics Equations for Periodic Applied Forces The dynamics of a linear structure subjected to periodic forces obeys the matrix differential equation ¨ + C X˙ + KX = F (t), MX with initial conditions ˙ X(0) = D0 , X(0) = V0 . The solution vector X(t) has dimension n and M , C, and K are real square matrices of order n. The mass matrix, M , the damping matrix, C, and the stiffness matrix, K, are all real. The forcing function F (t), assumed to be real and having period L, can be approximated by a Þnite trigonometric series as F (t) =

N

ck eıωk t where ωk = 2πk/L

k=−N

√

and ı = −1. The Fourier coefÞcients c k are vectors that can be computed using the FFT. The fact that F (t) is real also implies that c −k = conj(ck ) and, therefore, n ıωk t ck e . F (t) = c0 + 2 real k=1

The solution of the differential equation is naturally resolvable into two distinct parts. The Þrst is the so called particular or forced response which is periodic and has the same general mathematical form as the forcing function. Hence, we write n n Xp = Xk eıωk t = X0 + 2 real Xk eıωk t . k=−n

k=1

Substituting this series into the differential equation and matching coefÞcients of eıωk t on both sides yields Xk = (K − ωk2 M + ıωk C)−1 ck .

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The particular solution satisÞes initial conditions given by n n ck and X˙ p (0) = 2 real ıωk ck . Xp (0) = X0 + 2 real k=1

k=1

Since these conditions usually will not equal the desired values, the particular solution must be combined with what is called the homogeneous or transient solution Xh , where ¨ h + cX˙ h + KXh = 0, MX with Xh (0) = D0 − Xp (0) , X˙ h (0) = V0 − X˙ p (0). The homogeneous solution can be constructed by reducing the original differential equation to Þrst order form. Let Z be the vector of dimension 2n which is the concatenation of X and X˙ = V . Hence, Z = [X; V ] and the original equation of motion is dZ = AZ + P (t) dt where

0 I 0 A= and P = . −M −1 K −M −1 C m−1 F The homogeneous differential equation resulting when P = 0 can be solved in terms of the eigenvalues and eigenvectors of matrix A. If we know the eigenvalues λ j and eigenvectors U j satisfying AU = λ U , 1 ≤ ≤ 2n, then the homogeneous solution can be written as Z=

2n

z U eıω t .

=1

The weighting coefÞcients z are computed to satisfy the desired initial conditions which require z1

. X0 − Xp (0) U1 , U2 , · · · , U2n .. = . V0 − X˙ p (0) z2n We solve this system of equations for z 1 , · · · , z2n and replace each U by z U . Then the homogeneous solution is Xh =

n =1

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U (1 : n)eλ t

where U (1 : n) means we take only the Þrst n elements of column . In most practical situations, the matrix C is nonzero and the eigenvalues λ 1 , · · · , λ2n have negative real parts. Then the exponential terms e λ t all decay with increasing time, which is why X h is often known as the transient solution . In other cases, where the damping matrix C is zero, the eigenvalues λ are typically purely imaginary, and the homogeneous solution does not die out. In either instance, it is often customary in practical situations to ignore the homogeneous solution because it is usually small when compared to the contribution of the particular solution.

7.1.1 Application to Oscillations of a Vertically Suspended Cable Let us solve the problem of small transverse vibrations of a vertically suspended cable. This system illustrates how the natural frequencies and mode shapes of a linear system can be combined to satisfy general initial conditions on position and velocity. The cable in Figure 7.1 is idealized as a series of n rigid links connected at frictionless joints. Two vectors, consisting of link lengths [ 1 , 2 , · · · , n ] and masses [m1 , m2 , · · · , mn ] lumped at the joints, characterize the system properties. The accelerations in the vertical direction will be negligibly small compared to transverse accelerations, because the transverse displacements are small. Consequently, the tension in the chain will remain close to the static equilibrium value. This means the tension in link ı is n m . Tı = gbı where bı = =ı

We assume that the transverse displacement y ı for mass mı is small compared to the total length of the cable. A free body diagram for mass ı is shown in Figure 7.2. The small deßection angles are related to the transverse deßections by θı+1 = (yı+1 − yı ) ı+1 and θı = (yı − yı−1 ) /ı . Summation of forces shows that the horizontal acceleration is governed by mı y¨ı = g(bı+1 /ı+1 ) (yı+1 − yı ) − g(bı /ı ) (yı − yı−1 ) = g(bı /ı )yı−1 − g(bı /ı + bı+1 /ı+1 )yı + g(bı+1 /ı+1 )yı+1 . In matrix form this equation is M Y¨ + KY = 0 where M is a diagonal matrix of mass coefÞcients and K is a symmetric tridiagonal matrix. The natural modes of free vibration are dynamical states where each element of the system simultaneously moves with harmonic motion of the same frequency. This means we seek motions of the form Y = U cos(ωt), or equivalently Y = U sin(ωt), which implies KU = λ M U where λ = ω2 for 1 ≤ ≤ n.

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1 m1 2 m2

3 n m3 mn

Figure 7.1: Transverse Cable Vibration

Solving the eigenvalue problem (M −1 K)U = λU gives the natural frequencies ω1 , · · · , ωn and the modal vectors U 1 , · · · , Un . The response to general initial conditions is then obtained by superposition of the component modes. We write Y =

n

cos(ω t)U c + sin(ω t)U d /ω

=1

where the coefÞcients c 1 , · · · , cn and d1 , · · · , dn (not to be confused with Fourier coefÞcients) are determined from the initial conditions as c1 . U1 , · · · , Un .. = Y (0) , c = U −1 Y (0), cn d1 U1 , · · · , Un ... dn

= Y˙ (0) , d = U −1 Y˙ (0).

The following program determines the cable response for general initial conditions. The natural frequencies and mode shapes are computed along with an animation of the motion. The cable motion produced when an initially vertical system is given the same initial transverse velocity for all masses was studied. Graphical results of the analysis appear in Figures 7.3 through 7.6. The surface plot in Figure 7.3 shows the cable deßection pattern in terms of longitudinal position and time. Figure 7.4 shows the deßection pattern at two times. Figure 7.5 traces the motion of the middle and the

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Tı = gbı θı

mı θı+1

Tı+1 = gbı+1

mı g Figure 7.2: Forces on ı’th Mass

free end. At t = 1, the wave propagating downward from the support point is about halfway down the cable. By t = 2, the wave has reached the free end and the cable is about to swing back. Finally, traces of cable positions during successive stages of motion appear in Figure 7.6.

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Surface Showing Cable Deflection

2

transverse deflection

1.5 1 0.5 0 −0.5 −1 10

−1.5 8 −2 0

6 0.2

0.4

4 2

0.6

0.8

1

0

time

y axis

Figure 7.3: Surface Showing Cable Deßection

Cable Transverse Deflection at t = 1 and t = 2 1 t=1 t=2

The cable was initially vertical and was given a uniform transverse velocity. A 30 link model was used.

0.9 0.8

distance from bottom

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.2

0.4

0.6 0.8 1 transverse displacement

1.2

1.4

Figure 7.4: Cable Transverse Deßection at t = 1 and t = 2

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1.6

Position versus Time for the Cable Middle and End 2 Midpoint Lower end 1.5

transverse displacement

1

0.5

0

−0.5

−1

−1.5

−2

0

1

2

3

4 5 6 dimensionless time

7

8

9

10

Figure 7.5: Position Versus Time for the Cable Middle and End

Trace of Linearized Cable Motion

Figure 7.6: Trace of Cable Motion

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MATLAB Example Program cablinea

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function cablinea % Example: cablinea % ~~~~~~~~~~~~~~~~~ % This program uses modal superposition to % compute the dynamic response of a cable % suspended at one end and free at the other. % The cable is given a uniform initial % velocity. Time history plots and animation % of the motion are provided. % % User m functions required: % cablemk, udfrevib, canimate

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% Initialize graphics hold off; axis(’normal’); close;

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% Set physical parameters n=30; gravty=1.; masses=ones(n,1)/n; lengths=ones(n,1)/n;

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% Obtain mass and stiffness matrices [m,k]=cablemk(masses,lengths,gravty);

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% Assign initial conditions & time limit % for solution dsp=zeros(n,1); vel=ones(n,1); tmin=0; tmax=10; ntim=30;

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% Compute the solution by modal superposition [t,u,modvc,natfrq]=... udfrevib(m,k,dsp,vel,tmin,tmax,ntim);

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% Interpret results graphically nt1=sum(t