1 vibration analysis using matlab

2

c1 c2 c3 c4 c5 c6 c7

k1 = 200 N/m k2 = 200 N/m k3 = 300 N/m

k4 = 300 N/m k5 = 300 N/m k6 = 400 N/m k7 = 400 N/m

m2 = 20kg

m3 =30kg

m4 =20kg

m5 = 40kg

m6 =60kg

m1 =10kg

x1 x2 x6x5x4x3

DYNAMICS OF VIBRATING SYSTEMS

SUMMARY

This report has successfully demonstrated the analytical derivation of the governing equations for

the vibration of an unforced 6DoF damped mechanism, using Lagrange’s equations. MATLAB® was

successfully implemented in several instances. The system characteristics were found, namely the six

natural frequencies and modal shapes. A comprehensive modal analysis was performed with two co-

ordinate transformations and proportional damping. A particular use for MATLAB was performing

the impulse response for two separate non-periodic excitation forces applied to the system. Other

types of excitation were discussed, and finally some of the fundamental assumptions of the system

were altered to observe the effect on damping.

1. INTRODUCTION

In this report, the system in figure 1 with six Degrees of Freedom (DoF) will be considered. The

governing equation will be derived using Lagrangian mechanics in section 2. Then it will be subject to

a number of different analysis procedures. The matrix method will determine the undamped natural

frequencies and modal shapes in section 3. A kinematic analysis will be performed in section 4, with

modal analysis that uses two co-ordinate transformations and proportional damping. This process

will find estimates for damping ratios, damping factors and damped frequencies. A kinetic analysis

will take place in section 5, where two separate pre-determined non-periodic excitation forces will

be applied to the system. Numerical and analytical procedures are thereby examined for their

effectiveness in predicting the impulse response. Finally, in sections 6 and 7, alternative excitation

functions, and methods for modelling damping will be discussed.

Figure 1 – 6DoF Mass-spring-damper model with parameters for mass and spring constants.

3

2. SYSTEM EQUATION

The analysis is presented for a purely translational system (where no rotation is involved). The

Lagrange equation for unforced oscillation with damping is given by:

ddt ( ∂T∂ q i

)−∂T∂qi

+∂V∂qi

+∂D∂ q i

= 0 (2.1)

Where T is the Kinetic energy (J), V is the Potential energy (J), D is the energy dissipation due to

damping (J), qi is the co-ordinate system, q i is the first derivative of the co-ordinate qi with respect to

time (ms-1).

The Kinetic energy of the system given in figure 1 can be expressed

T=12m1 x1

2+ 12m2 x2

2+ 12m3 x3

2+ 12m4 x4

2+12m5 x5

2+12m6 x6

2(2.2)

Where mi is the respective mass for each of the masses (kg).

The potential energy can also be expressed thus:

V=12k1 x1

2+ 12k2(x¿¿2−x1)

2+ 12k3 ¿¿ (2.3)

Where ki is the respective stiffness of each of the springs (Nm-1).

Energy dissipation due to damping is expressed:

D=12c1 x1

2+ 12c2 ¿ (2.4)

Where ci is the respective damping co-efficient for each damper.

By substituting into the Lagrange equation for i = 1,

∂T∂ q1

=m1 x1ddt ( ∂T∂ q1 )=m1 x1

∂V∂q1

=k1 x1−k2(x2−x1)∂ D∂ q1

=c1 x1−c2( x2− x1)

Using the Lagrange equation given in (2.1), the first equation (for i = 1) is therefore:

m1 x1+(k 1+k2 )x1−k2 x2+(c1+c2 ) x1−c2 x2=0 (2.5)

The same procedure may be continued for i = 2:

4

∂T∂ q2

=m2 x2ddt ( ∂T∂ q2 )=m2 x2

∂V∂q2

=k 2(x¿¿2−x1)−k3(x3−x2)¿

∂ D∂ q1

=c2( x2− x1)−c3( x3− x2)

And hence the equation for i = 2 is given:

m2 x2−k2 x1+(k 2+k3 )x2−k3 x3−c2 x1+ (c2+c3 )x2−c3 x3=0 (2.6)

Since the system has 6 DoF, it follows the same pattern through to i = 6, the set of Lagrange

equations can now be presented in matrix format.

[m1 0 0 0 0 00 m2 0 0 0 00 0 m3 0 0 00 0 0 m4 0 00 0 0 0 m5 00 0 0 0 0 m6

]{x1x2x3x 4x5x6

}+[(c¿¿1+c2)¿−c2 0 0 0 0

−c2 (c¿¿2+c3)¿−c3 0 0 00 −c3 (c¿¿3+c4)¿−c4 0 00 0 −c4 (c¿¿ 4+c5)¿−c5 00 0 0 −c5 (c¿¿5+c6)¿−c60 0 0 0 ¿

(c¿¿6+c7)]{x1x2x3x4x5x6

}+[(k1+k2) −k 2 0 0 0 0−k2 (k¿¿2+k3)¿−k 3 0 0 ¿ ¿

−k3¿(k¿¿3+k4)¿−k4¿0¿0¿0¿0¿−k4¿(k¿¿4+k5)¿−k5¿0¿0¿0¿0¿−k5¿(k¿¿5+k 6)¿−k6¿0¿0¿0¿0¿−k6¿(k6+k 7)¿]{x1x2x3x4x5x6

}=0(2.7)

By substituting the appropriate values from figure 1, Lagrange’s equations can be found for this

particular system.

[10 0 0 0 0 00 20 0 0 0 00 0 30 0 0 00 0 0 20 0 00 0 0 0 40 00 0 0 0 0 60

]{x1x2x3x4x5x6

}+[(c¿¿1+c2)¿−c2 0 0 0 0

−c2 (c¿¿2+c3)¿−c3 0 0 00 −c3 (c¿¿3+c4)¿−c4 0 00 0 −c4 (c¿¿4+c5)¿−c5 00 0 0 −c5 (c¿¿5+c6)¿−c60 0 0 0 ¿

(c¿¿6+c7)]{x1x2x3x4x5x6

}+[400 −200 0 0 0 0

−200 500 −300 0 0 00 −300 600 −300 0 00 0 −300 600 −300 00 0 0 −300 700 −4000 0 0 0 −400 800

]{x1x2x3x4x5x6

}=0(2.8)

Equation (2.8) concludes the derivation of the governing equation for this particular system in matrix

form.

3. SYSTEM CHARACTERISTICS

In order to find the natural frequencies and modal shapes of the system, the Matrix method was used chosen over the Holzer method because it is more appropriate for computational analysis.

5

( [ A ]−ωn2 [ I ]) x=0 (3.1)

Where;

[ A ]=[m]−1[k ] (3.2)

Solving the characteristic equation for the eigenvalues of A, and then performing a square root on the resultant matrix, gives the natural frequencies for the system. The MATLAB code in figure 2 was used to do this:

K = [400,-200,0,0,0,0;-200,500,-300,0,0,0;0,-300,600,-300,0,0;0,0,-300,600,-300,0;0,0,0,-300,700,-400;0,0,0,0,-400,800]M = [10,0,0,0,0,0;0,20,0,0,0,0;0,0,30,0,0,0;0,0,0,20,0,0;0,0,0,0,40,0;0,0,0,0,0,60]% Mass and stiffness matrices are set up.A = inv(M)*K% A is defined and then the eigenvalues & eigenvectors are found in order% to solve according to the matrix method.[v,d] = eig(A)nat_freq = sqrt(d)% A plot of all the modal shapes is formedplot(v)pause;% Each time the user presses the return key, the modal shapes will appear% in ascending order of natural frequency.plot(v(:,5))pause;plot(v(:,6))pause;plot(v(:,4))pause;plot(v(:,3))pause;plot(v(:,2))pause;plot(v(:,1))

Figure 2 – MATLAB® code to solve for the natural frequencies and modal shapes

6

1.4514 rad/s

2.6591 rad/s

4.2858 rad/s

5.0976 rad/s

7.1167 rad/s

6.4541 rad/s

The code in figure 2 produced the following results for the natural frequencies.

ωn=[7.1167 0 0 0 0 00 6.4541 0 0 0 00 0 5.0976 0 0 00 0 0 4.2858 0 00 0 0 0 1.4514 00 0 0 0 0 2.6591

]The modal shapes may be plotted from the corresponding eigenvectors. They are given on figure 3 combined in one plot, and they are singled out in ascending order from the lowest to highest frequency in the subsequent figures 4 to 9, displayed on page 8.

The dynamic characteristics of the system are the properties during its active and changing states.

Therefore discussing these involves noting the order in which the masses and stiffness elements for

that constitute each degree of freedom hit their natural frequency. The cause of this is due to the

surrounding properties of the system. This information is ordered and summarised in table 1.

Natural Frequency (ωn – [rad/s]) Mass (m – [kg]) Surrounding stiffness elements – Left to right (k – [Nm-1])

1.4514 m5 40 k5 300; k6 4002.6591 m6 60 k6 400; k7 4004.2858 m4 20 k4 300; k5 3005.0976 m3 30 k3 300; k4 3006.4541 m2 20 k2 200; k3 300

FIGURE 3 - MODAL SHAPES. THE NATURAL FREQUENCIES ARE COLOUR CODED AS SHOWN ON THE LEFT OF THE FIGURE.

7

7.1167 m1 10 k1 200; k2 200TABLE 1 – NATURAL FREQUENCIES IN ASCENDING ORDER, ALLOCATED TO EACH MASS AND STIFFNESS ELEMENTS

The heavier mass and stiffer spring arrangements tend to have lower values for ω n. However, it may

be observed from table 1 that the lowest natural frequency actually occurs for m5, rather than m6.

Possible causes for this are the variance in surrounding stiffness elements for m5 whereas m6 has the

same k value on both sides; and also the fact that m5 has surrounding masses on both sides, whereas

m6 is against a wall on one side. Conversely, lighter masses with lower k values have higher natural

frequencies. This is a trend that loosely matches from m4 to m1. The exception occurs at m3, which is

located in the middle of the system, promoting a lower natural frequency. From the tabulation of

these results it is possible to see that the natural frequency for each of the six degrees of freedom is

highly dependent on every other element in the system.

8

FIGURE 4 – MODAL SHAPE FOR 1.4514 RAD/S FIGURE 5 – MODAL SHAPE FOR 2.6591 RAD/S



9

4. MODAL ANALYSIS

This is a kinematic analysis which approaches the 6DoF system assuming that free damped vibration

occurs. The damping will be proportional in order for the analysis to work, with α = 0.1 and β = 0.2.

The system equations of 2.8 were given thus:

[10 0 0 0 0 00 20 0 0 0 00 0 30 0 0 00 0 0 20 0 00 0 0 0 40 00 0 0 0 0 60

]{x1x2x3x4x5x6

}+[ [(c¿¿1+c2)¿−c2 0 0 0 0−c2 (c¿¿2+c3)¿−c3 0 0 00 −c3 (c¿¿3+c4)¿−c4 0 00 0 −c4 (c¿¿4+c5)¿−c5 00 0 0 −c5 (c¿¿5+c6)¿−c60 0 0 0 ¿

(c¿¿6+c7)]]{x1x2x3x4x5x6

}+[400 −200 0 0 0 0

−200 500 −300 0 0 00 −300 600 −300 0 00 0 −300 600 −300 00 0 0 −300 700 −4000 0 0 0 −400 800

]{x1x2x3x4x5x6

}=0(2.8 - repeated)

The problem was solved in MATLAB using the code given in figure 10.

10

% Stiffness and mass matrices defined, as beforeK = [400,-200,0,0,0,0;-200,500,-300,0,0,0;0,-300,600,-300,0,0;0,0,-300,600,-300,0;0,0,0,-300,700,-400;0,0,0,0,-400,800]M = [10,0,0,0,0,0;0,20,0,0,0,0;0,0,30,0,0,0;0,0,0,20,0,0;0,0,0,0,40,0;0,0,0,0,0,60]% Proportional damping applied to the system to form the damping matrixC = 0.1*M+0.2*K % Below - Used to check the order of undamped natural frequencies, will be needed later for the P matrix.A = inv(M)*K[v,d] = eig(A)nat_freq = sqrt(d) % First co-ordinate transformation...% Mass-Normalised Stiffness:K_tilda = (M^-0.5)*(K)*(M^-0.5) % Mass-Normalised Damping:C_tilda = (M^-0.5)*(C)*(M^-0.5) % Finding eigenvalues and eigenvectors for the normalised stiffness matrix[v1,d1] = eig(K_tilda) % Second co-ordinate transformation...% Arranging the eigenvectors into ascending order according to natural% frequency.P = [v1(:,5),v1(:,6),v1(:,4),v1(:,3),v1(:,2),v1(:,1)] % Diagonal matrices for use with the "r" co-ordinates after the second% transformationBig_Lambda_K = [P]'*[K_tilda]*[P]Big_Lambda_C = [P]'*[C_tilda]*[P] % Used for conversion from r(t) co-ordinates back to x(t) co-ordinates[S] = M^-0.5*[P]

FIGURE 10 – MATLAB® CODE TO PERFORM KEY OPERATIONS REQUIRED FOR MODAL ANALYSIS.

The process of co-ordinate transformation and eventually finding the necessary parameters will be

detailed through each step.

Firstly, appropriate values should be substituted for C, for the damping matrix. The parameters α =

0.1 and β = 0.2 are given, and it is known that:

[C ]=α [ M ]+β [K ]

Therefore:

M {x }+(α [ M ]+β [K ] ) {x }+ [ K ] {x }=0

So, the equation of motion in terms of {x(t)} for 6DoF may be stated thus:

11

[10 0 0 0 0 00 20 0 0 0 00 0 30 0 0 00 0 0 20 0 00 0 0 0 40 00 0 0 0 0 60

]{x1x2x3x4x5x6

}+[81 −40 0 0 0 0

−40 102 −60 0 0 00 −60 123 −60 0 00 0 −60 126 −60 00 0 0 −60 144 −800 0 0 0 −80 166

]{x1x2x3x4x5x6

}+[400 −200 0 0 0 0

−200 500 −300 0 0 00 −300 600 −300 0 00 0 −300 600 −300 00 0 0 −300 700 −4000 0 0 0 −400 800

]{x1x2x3x 4x5x6

}=0(4.1)

In order to form a symmetric eigenvalue problem, the mass-normalised stiffness and damping, ~K

and ~C , must be found:

~K= [M ]−12 [K ][M ]

−12

~C=[M ]−12 [C ] [M ]

−12

These matrices were calculated through the MATLAB calculations shown in figure 10:

~K=[400 −141.4214 0 0 0 0

−282.8427 500 −244.9490 0 0 00 −367.4235 600 −367.4235 0 00 0 −244.9490 600 −212.1320 00 0 0 −424.2641 700 −326.59860 0 0 0 −489.8979 800

]~C=[

83 −29.6985 0 0 0 0−59.3970 105 −51.4393 0 0 0

0 −77.1589 125 −75.9342 0 00 0 −50.6228 126 −45.2548 00 0 0 −90.5097 150 −70.21870 0 0 0 −105.3281 166

]To form the first co-ordinate transformation, ~K and ~C can be expressed in terms of the new co-

ordinate system, {q(t)}.

[ I ] {q i }+(0.1 [ I ]+0.2 [~K ]) {qi }+[~K ] {q i }=0

And therefore:

[ I ] {q i }+[~C ] {q i }+[~K ] {qi }=0

12

Substituting the mass-normalised and stiffness and damping matrices, equation (4.2) is formed:

[1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

]{q1q2q3q4q5q6

}+[83 −29.6985 0 0 0 0

−59.3970 105 −51.4393 0 0 00 −77.1589 125 −75.9342 0 00 0 −50.6228 126 −45.2548 00 0 0 −90.5097 150 −70.21870 0 0 0 −105.3281 166

]{q1q2q3q4q5q6

}+[400 −141.4214 0 0 0 0

−282.8427 500 −244.9490 0 0 00 −367.4235 600 −367.4235 0 00 0 −244.9490 600 −212.1320 00 0 0 −424.2641 700 −326.59860 0 0 0 −489.8979 800

]{q1q2q3q4q5q6

}=0(4.2)

This completes the first co-ordinate transformation.

The eigenvalues and eigenvectors for ~K were found with MATLAB. [P] is defined as the normalised

matrix of stiffness eigenvectors, in ascending order of natural frequency.

P=[0.1436 −0.1880 0.2826 0.4092 −0.3943 −0.73570.3848 −0.4379 0.4323 0.4055 0.0461 0.55390.5534 −0.4238 −0.0922 −0.5051 0.3925 −0.31040.4238 −0.0096 −0.4446 −0.1587 −0.7402 0.22280.4738 0.4687 −0.3811 0.5232 0.3601 −0.07540.3457 0.6111 0.6180 −0.3377 −0.1038 0.0165

]Λ~K and Λ~C will be used in the equation in terms of {r(t)}, the result of the second co-ordinate

transformation. They produce ordered values that relate to natural frequency and damping, as

shown below.

Λ~K=[P]T [~K ] [P ]=[ωn1

2 0 0 0 0 0

0 ωn2

2 0 0 0 0

0 0 ωn3

2 0 0 0

0 0 0 ωn4

2 0 0

0 0 0 0 ωn5

2 0

0 0 0 0 0 ωn6

2]

13

Λ~C=[P]T [~C ] [ P ]=[2ζ ω1 0 0 0 0 00 2 ζ ω2 0 0 0 00 0 2 ζ ω3 0 0 00 0 0 2 ζ ω4 0 00 0 0 0 2ζ ω5 00 0 0 0 0 2 ζ ω6

]The MATLAB calculations produced the following results for Λ~K and Λ~C :

Λ~K=[2.1067 0 0 0 0 00 7.0710 0 0 0 00 0 18.3677 0 0 00 0 0 25.9854 0 00 0 0 0 41.6552 00 0 0 0 0 50.6474

]Λ~C=[

0.5213 0 0 0 0 00 1.5142 0 0 0 00 0 3.7735 0 0 00 0 0 5.2971 0 00 0 0 0 8.4310 00 0 0 0 0 10.2295

]To complete the second co-ordinate transformation, Λ~K and Λ~C are substituted into the system

equation expressed in terms of {r(t)} . This provides the format for decoupled equations to be

formed. Rather than expressing the system as a single 6DoF equation as before, it may now be

stated in terms of 6 separate single DoF (sDoF) equations.

[ I ] {ri }+[ ΛC ] {ri }+[ ΛK ] {ri }=0

14

[1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

]{r1r2r3r 4r5r6

}+[2 ζ ω1 0 0 0 0 00 2 ζ ω2 0 0 0 00 0 2 ζ ω3 0 0 00 0 0 2 ζ ω4 0 00 0 0 0 2 ζ ω5 00 0 0 0 0 2 ζ ω6

]{r1r2r3r4r5r6

}+[ωn1

2 0 0 0 0 0

0 ωn2

2 0 0 0 0

0 0 ωn3

2 0 0 0

0 0 0 ωn4

2 0 0

0 0 0 0 ωn5

2 0

0 0 0 0 0 ωn6

2]{r1r2r3r4r5r6

}=0Substituting in values for Λ~K and Λ~C gives (4.3):

[1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

]{r1r2r3r 4r5r6

}+[0.5213 0 0 0 0 00 1.5142 0 0 0 00 0 3.7735 0 0 00 0 0 5.2971 0 00 0 0 0 8.4310 00 0 0 0 0 10.2295

]{r1r2r3r 4r5r6

}+[2.1067 0 0 0 0 00 7.0710 0 0 0 00 0 18.3677 0 0 00 0 0 25.9854 0 00 0 0 0 41.6552 00 0 0 0 0 50.6474

]{r1r 2r3r4r5r6

}=0(4.3)

The equivalent 6 sDoF equations are:

r1+0.5213 r1+2.1067 r1=0

r2+1.5142 r2+7.0710 r2=0

r3+3.7735 r3+18.3677 r3=0

r 4+5.2971 r 4+25.9854 r 4=0

r5+8.4310 r5+41.6552r 5=0

r6+10.2295 r6+50.6474 r6=0

To obtain the final solution for x{t}, simply:

{x ( t ) }=[M ]−12 [P ]{r ( t )}=[S ]{r ( t )} (4.4)

Where:

[S ]=[M ]−12 [ P ]

15

In order to tabulate the results for each DoF, each of the sDoF equations found in terms of {r(t)}

must be addressed (equation (4.3)). Therefore the natural frequencies are calculated from the

values of Λ~K , where each term is equal to ωni

2 :

ωn1=√2.1067=1.4514 rad /s

ωn2=√7.071=2.6591 rad /s

ωn3=√18.3677=4.2858 rad /s

ωn4=√25.9854=5.0976 rad / s

ωn5=√41.6552=6.4541rad /s

ωn6=√50.6474=7.1167rad /s

The damping factors are found directly from each value of Λ~C . The damping ratios are found from

the each value of Λ~C where each term is equal to 2 ζ ωi:

ζ 1=0.5213

(2∗√2.1067)=0.1796Ns/m

ζ 2=1.5142

(2∗√7.071)=0.2847Ns/m

ζ 3=3.7735

(2∗√18.3677)=0.4402Ns/m

ζ 4=5.2971

(2∗√25.9854)=0.5196Ns/m

ζ 5=8.4310

(2∗√41.6552)=0.6532Ns/m

ζ 6=10.2295

(2∗√50.6474 )=0.7187Ns /m

Finally, damped frequencies (ωd) are found from equation (4.5).

ωd=ωn√1−ζ i2 (4.5)

16

Natural frequencies, 𝜔n

(rad/s)Damping ratio, ζ

(Ns/m)Damping Factor

(Ns/m)Damped frequencies, 𝜔d

(rad/s)1.4514 0.1796 0.5213 1.42792.6591 0.2847 1.5142 2.54914.2858 0.4402 3.7735 3.84815.0976 0.5196 5.2971 4.35556.4541 0.6532 8.4310 4.88727.1167 0.7187 10.2295 4.9484

TABLE 2: COMPARISON BETWEEN PARAMETERS OBTAINED THROUGH MODAL ANALYSIS.

Table 2 shows the trend between parameters obtained through

From table 2 it may be observed that damping ratio increases with the natural frequency. This is intuitive because with a higher oscillating frequency it is likely that more damping occurs. The damped frequencies increase with natural frequency, but not at a linear rate, as shown by graph 1. The relationship between them is actually convergent because damping factor increases to a limit of 1. Therefore at higher values the damped frequencies are significantly lower than the natural frequencies.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

1

2

3

4

5

6

7

8

Natural FrequenciesDamped Frequencies

Damping ratio (Nm/s)

Freq

uenc

y (r

ad/s

)

GRAPH 1: COMPARISON BETWEEN NATURAL FREQUENCIES AND DAMPED FREQUENCIES ACROSS DAMPING RATIOS.

17

5. NON-PERIODIC EXCITATION

5.1 IMPULSE FORCE (FORCE_1.M)This section performs a kinetic analysis on the system shown in figure 1 for an impulse force acting specifically on m1. The impulse force is defined as F.

F=∫F ( t )dt (5.1)

When considering a unit impulse:

F=f ( ξ )∆ξ (5.2)

The convolution integral can be formed:

x (t )=∫0

t

f (ξ )h ( t−ξ )dξ (5.3)

The rectangular pulse can be considered analytically as the sum of two step functions, where F(t) = F0

for a damped system.

x1 ( t )=F0k

(1−cosωnt) t<t1 (5.4)

x2 (t )=F0k

[1−cosωn (t−t 1) ] t=t 1 (5.5)

The sum of which is:

x (t )=F0k

[−cosωn t+cosωn (t−t 1 )] t>t1 (5.6)

For this system, the values are:

x (t )= 1200

¿ (5.7)

The original excitation files for use with MATLAB are given in Appendix 2. To apply these forces to the particular system shown in figure 1, additional sections of code were added to perform the transfer function, impulse response and convolution integral. In figures 11A and 12A, the original sections are shaded out in grey, and the added sections are in black.

The parameters for the transfer function were formed thus:

G' ( s )=X (s)F (s)

=

1m

s2+2 ζ ωn s+ωn2

18

For m1: 1m

=0.1, 2 ζ ωn6=10.2295 and ωn6

2=50.6474

The impulse solution is given on the next page, using a linear simulation and convolution integral. The plots are given in order of the corresponding solutions noted by the code.

dt=0.1;t=0:dt:50;f=[zeros(50,1);ones(100,1);zeros(length(t)-150,1)];subplot(311);plot(t,f);title('Square input');axis([0 50 0 1.5]);pause; %Numerical solution – impulse responsesys=tf(0.1,[1 10.2295 50.6474]);% using m1 = 10kg and values from sect 4 % in the stiffness and damping matrices in terms of r(t) for nat freq #6 h=impulse(sys,t);sol_1=lsim(sys,f,t); %linear simulationsol_2=conv(f,h)*dt; %convolutionsubplot(312);plot(t,sol_1(1:length(t)),t,sol_2(1:length(t)));title('Impulse response of system');xlabel('Time');ylabel('Output');pause; %Attempted analytical validation – step responsey=0.005.*((-cos(7.1167*t))+(cos(7.1167.*(t-(t-0.1))))) %Using Nat. Freq #6.subplot(313);plot(t,y)title('System Response to a step input');xlabel('Time');ylabel('x(t)');FIGURE 11A – MATLAB® CODE FOR THE NON-PERIODIC EXCITATION FORCE_1.M APPLIED TO m1 IN THE 6DOF SYSTEM.

19

The impulse response showed a resemblance to the rectangular input used to excite m 1. The system

reacts dynamically for the duration of the impulse, and then the force dies away over time due to

the system damping. The analytical result for step input did not produce a result as expected as it is

a continuous function rather than a solution to the step input. This is due to an unknown flaw in the

equation presented in the code.

5.2 HALF-SINE PULSE EXCITATION FORCE (FORCE_2.M)In this case a half-sine pulse was applied to m1 on the system given in figure 1.

The analytical solution in this case is given by:

x (t )=F0k [1− e−ζ ωn t

√1−ζ 2 ]sin (ωd t−ϕ) (5.8)

Where:

ϕ=tan−1 ζ

√1−ζ 2

Stating (5.8) with values relevant to this problem:

FIGURE 11B – IMPULSE RESPONSE PLOT FOR FORCE_1.M APPLIED TO m1 IN THE 6DOF SYSTEM

20

x (t )= 1200 [1− e−5.1148 t

√1−0.71872 ]sin (4.9484 t−0.802) (5.9)

The solution using MATLAB code for a linear simulation and convolution integral is given on the following page.

The plots are given in order of the corresponding solutions noted by the code.

21

DT=0.1;t=0:DT:50; %Time (high resolution)sys=tf(0.1,[1 0.5213 2.1067]); f=(sin(t).*t+0.1).*(t<8.1);subplot(311);plot(t,f,'b');xlabel('Time');ylabel('Input');title('Input vs Time');pause; % Impulse response (below)h=impulse(sys,t);sol_1=lsim(sys,f,t); %linear simulationsol_2=conv(f,h)*dt; %convolutionsubplot(312);plot(t,sol_1(1:length(t)),t,sol_2(1:length(t)))title('Impulse response of system');xlabel('Time');ylabel('Output');pause; % Attempted Analytical validation - step response (below)y = 0.2.*(1-(exp(-5.1148*t)/sqrt(1-0.7187^2).*(sin(4.9484*t- (atan(0.7187/sqrt(1-0.7187^2)))))))subplot(313);plot(t,y)title('System Response to a step input');xlabel('Time');ylabel('x(t)');axis([0 5 0 0.5]);FIGURE 12A – MATLAB® CODE FOR THE NON-PERIODIC EXCITATION FORCE_2.M APPLIED TO m1 IN THE 6DOF SYSTEM

FIGURE 12B – RESPONSE PLOT FOR FORCE_2.M APPLIED TO m1 IN THE 6DOF SYSTEM

22

The system response appears to be valid for the convolution integral approach (impulse response,

middle plot). Here, the vibrations follow the trend of the input before dying away due to damping in

the system. This shows that a numerical approach can provide accurate results with appropriate

conditions for linear simulation. Unfortunately the validation was not very accurate for the half-sine

pulse (analytical approach, lower plot). This could be due to an error in the equation or a mistake

with the input for MATLAB. However it still shows an initial transient period followed by stability.

6. OTHER TYPE OF EXCITATION

An excitation is considered of the following form in (6.1):

F (t )=e−t sinωt (6.1)

To illustrate the implications of this on the system, the function is plotted for two different values of

ω, the oscillating frequency, in figures 13 and 14.

Therefore it is a function which dies away in amplitude at a fixed rate, regardless of the natural

frequency. It is clear to see that this is unlike the other excitation functions in section 5 which were

largely solved through numerical analysis, as it is dissimilar to simple rectangular, half-sine pulse or

constant with rise time functions.

The best way to tackle this problem analytically is through the forced modal analysis method.

[ M ] {x }+[C ] { x }+ [K ] {x }=[B ] {F (t ) }=[B ]{e−t sinωt } (6.2)

FIGURE 13 – FORCING FUNCTION (6.1) APPLIED WHERE ω = 5 RAD/S

FIGURE 14 – FORCING FUNCTION (6.1) APPLIED WHERE ω = 50 RAD/S

23

Proportional damping would occur in the same way as in section 4, and the co-ordinate

transformations would also be applied to the right hand side of the equation. [B] is defined here as

an identity matrix.

24

7. EFFECTS OF DAMPING

7.1 SELECTING PROPORTIONAL DAMPING CONSTANTS

In section 4 it was assumed that the values for α and β we given. But this raises the question as to

how they were found in the first place. When the system is in terms of r co-ordinates, its equations

follow this format for unforced vibration:

{r (t ) }+(α [ I ]+β [ Λ ]) {r (t ) }+ [ Λ ] {r (t ) }=0 (7.1)

The damping constant, c, can be found through empirical measurements or chosen for desired

values.

ζ i=c

2√km(7.2)

Then, through the relationship given in (7.2), The damping ratios, ζi , can thus be found. Proportional

values of α and β can be chosen such that these values of ζi are matched.

7.2 SYSTEM RESPONSE WITHOUT HORIZONTAL ROLLERS

If the masses were resting on a flat surface this would introduce additional losses through friction.

Due to the conservation of energy, the governing equation would follow this form:

[ M ] {x }+[C ] { x }+ [K ] {x }+∑ FrictionalLosses=0

The precise values for frictional losses could either be measured empirically or from the time taken

for the vibrations of the system to diminish. The latter method assumes that all other damping

parameters in the system are known.

7.3 NON-RIGID WALLS

Rigid walls essentially act as a place for the energy to “bounce” back through the system. This goes

by the assumption that no energy is lost at the walls and that it is all transferred back into the

system. If the walls lost their rigidity, there would have to be a proportional constant for how much

of the energy was lost each time a vibration hit a wall and travelled back into the system.

25

8. CONCLUSIONS

The 6DoF system was treated in a number of ways, both computationally and analytically. The

system equations and characteristics were successfully found. Modal analysis provided a useful way

of tackling vibration problem in conjunction with MATLAB, which computed the co-ordinate

transformations. All of the natural frequencies, damping ratios, damping factors and damped

frequencies were found. Forced, non-periodic excitation was analysed numerically with the

convolution integral, for a rectangular pulse and a half-sine pulse. The results were accurate using

the impulse response in MATLAB. However, the analytical validation proved to be a challenge and

the results were likely to be incorrect. Forced modal analysis was considered to model an excitation

function of F ( t )=e−t sinωt . Overall, the analysis of this 6DoF system proved to be successful and a

good understanding has been formed of its fundamental properties in free and forced vibration.

26

9. REFERENCES

[1] TONGUE, B., 1996, Principles of Vibration. Oxford University Press

[2] THOMSON W. T., DAHLEH M. D., 1998, Theory of Vibration with Applications, 5th ed. Prentice Hall.

[3] ES386 Student Resources. Available online: http://www2.warwick.ac.uk/fac/sci/eng/euo/modules/year3/es386a/resources/

10. APPENDICES

APPENDIX 1ES386 Assignment handout: Dr. Xian Ping Liu, School of Engineering, Warwick University. March

2010.

APPENDIX 2Excitation files for sections 5.1 and 5.2

Force_1.m

dt=0.1;t=0:dt:50;f=[zeros(50,1);ones(100,1);zeros(length(t)-150,1)];subplot(211)plot(t,f)title('Square input')axis([0 50 0 1.5])%pause

Force_2.m

DT=0.1;t=0:DT:50; %Time (high resolution)f=(sin(t).*t+0.1).*(t<8.1);plot(t,f,'b');xlabel('Time');ylabel('Input');title('Input vs Time');pause;

1 vibration analysis using matlab

Documents