Prob 23gdldd

7/29/2019 Prob 23gdldd

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4.8 EXAMPLES

PROBLEMA 1

Sever al ex amples of multiple- degree- of- frecdom systems, (heir schematics, and equa

tions of motion are presented in this section. The art" in vibration analysis and de

sign is often related to choosing an appropriate mathematical model to describe agiven structure or machine. The follow ing ex amples arc intended to provide additional 'practice in modeling and analysis.

F.xumplc 4.8.1

A drive shaft for a holt- driven machine such as a lathe is illustrated in Figure 4.21 (a). Thevibration model of this system is indicated in Figure 4.21(b), along with a frce- body

diagram of the machine. Write the equations of motion in matrix form and solve for the

= ^2~ ~ 10 kg m2/rad.A:i = k 2 m 1 0 ' N* m/rad. c = 2 N m s/rad for zeroinitial conditions and where the applied moment M ( t ) is a unit impulse function.

c(&2 - *,)

M03- 0j)

c(8j - 6 2 )

(h)

Figure 4.21 (a) Schematic of the moving pans of a lathe. The bcannp that support

the routing shaft axe modeled as providing viscous damping while tin. shaft* p

vide stiffness and the belt drive provides an applied unque


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S olution In Figure 4.21 (a) the bearings and shaft lubricant arc modeled as lumped vis

cous damping, and the s hafts are modeled as torsional springs .T he pulley and machine

disks are modeled as rotational inertias. T he motor is modeled s imply as supplying a

moment to the pulley. Figure 4.21 (b) illustrates a free- body diagram for each of the three

disks, where the damping is assumed to act in proportion to the relative motion of

the masses and of the same value at each coordinate (other damping models may bemore appropriate, but this choice yields an easy form to solve).

Ex amining the free- body diagram of Figure 4.21(b) and summing the moments on

each of the disks yields

= 9i) + 81)

J#i = * , ( , - e; ) + c(43- e ,) - * ,( e , - ,) - c(; - 6,)

3 " 9j) c(03 0;) + M ( l )

where 0 ,, 02, and 6, are the rotational coordinates as indicated in Figure 4.21. T he unitfor B is radians* Rearranging these equations yields

7j0i + c6| + /ct0, - c02 kxQ2 0

j 2ot + 2fe2 - cfl, - c03 + (*, + *2)o2 - AfjO, - *2, o

7,0, + C0J - c02 - k,0j + AtjBt = M(/)

In matnx form this becomes

y , o o

o j 2 o

o /, j

c - c 0I

1jt*

o

________________

0+ -c 2c -c 0 - f * , + * 2 - k2 0 0

0 c c- ^ ^2 _ - * ( ' L

where 0(/) - [ 0j(/) 02(f) )]r. Using the values for the coefficientsously, this becomes given previ-

1 0 / 0 + 2

1 - 1


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Note that one of the eigenvalues is /ero.T hiis the matrix k is singular. The physicalmeaning of this is interpreted in this ex ample.T he normalized eigenvectors of K yield

0.5774 0.7071 0.4082 0.5774 0.5774 0.5774 p = 0.5774 0 - 0.8165 P T = 0.7071 0 - 0.7071

_ 0.5774 - 0.7071 0.4082 _ _ 0.4082 - 0.8165 0.4082

Further computation yields

P TC P = diag[ 0 0.2 0.6]

P r K P * diag[ 0 100 300]

P tM 1/2F(/) =

0.1826

- 0.2236

0.12918(/)

T he dec oupled modal equations are

? ,(/) 0.18265(f)

r2(l) + 0.2r , (f ) + 100r2(f) - - 0.22368(0

rs(/) +0.6r3(f) +300r}(f) - 0.12916(f)

Obviously,u>j *O.w* * 10 rad/s. and = 17.3205 rad, *.Comparing coefficients of r, with 2C*>, yields the three modal damping ratios

0.2

2( 10) "

0.6

0.01

0.017322(17.3205)

so that the second two modes arc underdamped. Hence the two damped natural fre

quencies become

W(/2 = 2 V l ~ C ! * 9.9995 rad/s

= >v T ~ = 17.3179 rad/s


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Figure 4.23 Res ponse of each of the disk* o f Figure 4.21 to an impulse at 0, with

out the rigid- body mode, illustrating the v ibration that occurs in each disk.

%T he three solutions 6 t(f ). e 2(/),a nd tt3(r) arc plotted in Figure 4.22. Figure 4.23 plots the

three solutions without the rigid- body term. T his represents the vibrations ex perienced

by each disk as it rotatct


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PROBLEMA 2


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since the springkxexperiences a displacementx - /,0 and k2experiences a displacementjf + /jO. S imilaily. the velocity ex perienced by the damper c, isx - /t0 and that of c2 isx 4- /20. Taking moments about the ccnter of gravity yields

ye - c ,/,(i - /,8) - c ,/j( i + /;0 ) + - (, ) - k, lx + /;8) (4.153)

where J mr. Here r is the radius of gyration of the vehicle (recall Example 1.4.6).

liquations (4.152) and (4.153) can be rewritten as

mi + (c, + c2)jr + (/jCj - lxci)8 + (fc, + k3)x + (l2k2- Z,*,) * 0

/ wr e + (c2/2 c, / t )x + (f| c2 + / f cj e + (ft2/2 - Ac,/,X-v + (/f/c, + o (4. 154)

In matrix form, these two coupled equations become

/mT 1 ] * r c* + Cj ^ c2 ~ * -t i * * + ki ^ \ 01.0 L/zCj - t\C{ /f o + f f a j L M l - M l *1*1 + /!* ? _]

(4.155)

where the vector x is defined by

r * ( o

Led)]Reasonable values for a truck arc

r 0.64 irr m - 4000 kg c, = c 2 - 2000 N s/m

A, * k2* 20,000 N/m /, = 0.9 m l2 - 1.4 ra

With these values, equation (4.155) becomes

f 4000 0 1 _ . f-WXH) 1000~1 . f 40,000 10,000*] r o l

I. 0 2560J * + [_1000 5540 J ' + | _10,000 55,400 J * | _0J (*

Nole that C =*(O.t )A',so that the damping is proportional. If a momentM{t)is appliedto the angular coordinate 0(f) the equations of motion become

I 4000 O ' ] , [*4000 10001 . ^ f 40,000 10,000 I = f 0 1 ,

L 0 2560 J X ^ |_1000 5540J * * [_ 10,000 55.400J X L m o Jv , , _

Following the usual procedures of modal analysis, calculation ofM 1/1yields

M l/2

Thus

T 0.0158 0

0 0.0198 J

' (* 1.0000 0.31251 r = r 10000 3.1250*]

' L- 3125 2.1641 J am 13.1250 21.6406J


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Solving the eigenvalue problem for K yields

0.9698 0.24391 / '= ( "09698 "- 2439l0.2439 0.9698 J an L0:24390.9698J

with e igenvalues = 9.2141 and \ 2 = 22.4265, so that the natural frequencies arc

u>i 3.0355 rad/s and w2 = 4.7357 rad/s

Thus

P r /e/> =diag[9,2141 22.4265] and PrCI> * diag[ 0.9214 2.2426]

Comparing the elements ofF rC P to u>i and w2yields the modal damping ratios

- S r 0 1 5 1 8 -

Usingthe formula from Window 4.7 for damped natural frequencies yields

(*,, 3.0003 rad/s and 4.6009 rad/s. T he modal forces are calculated from

rw i/r 0 1 r 0.0153 - 4 .S8 (0 ]r 0 ] r -4 .8 6( ,n

L()J L0.0039 19.2 8(1) JLM')J L*9.2 8(f) J

The decoupled modal equations become

r ,(/) + (0.9 21 4)r ,(f) + (9.2141 )r ,(f) = - 0.00488 (f)

r 2(/) + (2.2436 )/2(/) + (22.4265)r2(/) * 0.01 92 8(0

From equation (3.6) these have solutions

r , ) = " i A e - w S in = 7T T7, (3 .0 00 3/)/(ujr, (1 >( 3.0003)

= - 1.6066 e- * 7' sin (3.0003/)

r2(f) = sin (4.6(X>y,)

= 4 .1 6 5 9 ^'12* sin (4.6009/)

The solution in physical coordinates is obtained from

[ * ( 0

Led)

which yields

* (/) = - 2.41 X 10"V


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7

6

5

4

3

2

1

0

1

2

3

lime (s)

Figure 4.26 Plot of the bounce and pitch vibrations of the vchicle of Figure 4.24 os

the result of the engine being shut off (in meter* versus seconds).

fhese coordinates arc plotted in Figure 4.26.


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PROBLEMA 3

Fxamplc 4.8.3

T he punch press of Figure 4.27 can be modeled for vibration analysis in the x direction

as indicated by the three- degrce- of- frecdom system of Figure 4.28. Discuss the solutionfor the response due to an impact at w, using modal analysis.

Solution The mass and stiffness of the various components can be easily approximated

using the static methods suggested in Chapter 1. However, it is very difficult to estimate

values for the damping coefficients. Hence an educated guess is made for the modaldamping ratios. Such guesses arc often made based on experience or from measure

ments such as the logarithmic decrement. In this case the values of various masses and

stiffness coefficients arc [ in mks units and /( f ) = 10006(f)]

mj 400 kg m2 * 2000 kg = 8000 kg

Arj = 300,000 N /m k} = 80,000 N/m * , = 800,000 N/m

X

Figure 4.27 Schematic of a purvch-press

machine.

bas*Isolation mounting pa


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Figure 4.28 Vibration model of the punch

press of Figure 4.27.

fr om frec- body diagrams of each mass, the s umming of forces in thex direction yieldsthe three coupled equations

- - * ,( * ! - * 2) + / ( ' )

= ^ 3(^1 * 2) k x 2 * * j)

m3*3 - - * 3x* + k ^X i - at,)

Rewriting this set of coupled equations in matrix form yields


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where x * ( * i ( f ) x2(t) Aj ( f )] r. Substituting the numerical values for m, and k, yields

'0.4 0 0

(10) 0 2 0 x + (104)

0 0 8_

30 - 301

10006(f)

30 38 - 8 * = 0

0 - 8 88 1 0

Following the modal analysis procedure for an undamped system yields

"20 0 0 0.0500 0 0

Mxn - 0 44.7214 0 M'K = 0 0.0224 0

_ 0 0 89.4427 _ L o 0 0.0112

and

K

750

- 335.4102

0

- 335.4102

190

- 20

Solving the eigenvalue problem for K yields

- 0.4116 - 0.1021 0.9056" '- 0 .41 16 - 0.8848 - 0.2185'

P - - 0.8848 0.1935 - 0.4239 PT - - 0.1021 - 0.1935 0.9758

.- 0.2185 0.9758 0.0106 _ 0.9056 - 0.4239 0.0106 _

and

X, = 29.0223

\2 = 113.9665

X3 = 907.0112

The modal force vector becomes

, - 5.3872

j = 10.6755

u3 - 30.1166

~ 10005(f)" "- 20.5805"

0 SB - 5.1026

0 45.2814 _

5(f)

Hcncc, ihc undamped modal equations arc

r,(f ) + 29.0223r,(f) = - 20.58058(f)F2(f) + 113.9665r2(/) - - 5.10265(f)

r ,(f ) 4- 907.0112r*(f) = 45.28145(f)


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To model the damping, note that each mode shape is dominated by one element. From

examining the first column of the matrix P, the second element is larger than the othertwo elements Hence if the system were vibrating only in the first mode, the motion of

jr?(0 would dominate. T his element corresponds to the platform mass, which receives

high damping from the rubber support. Hence it is given a large damping ratio ofi, - 0.1(rubber provides a lot of damping). Similarly, the second mode is dominated by its third

element, corresponding to the motion of s a predominantly metal part, so itis given a low damping ratio of fc - 0.01. The third mode shape is dominated by the

first element, which corresponds to the mounting pad. Hence it is given a medium damp

ing ratio of = 0.05. Recalling that the velocity coefficient in modal coordinates hasthe form 2& w,, the damped modal coordinates bccome 2:to, = 2(0.1 )(5.387 2),

2 ^ 2 = 2(0.01)(10.6755), and 2fcw3 2(0.05)(30.116 6).T here fore, the damped modal

equations become

r,(/) + 1.0774?,(I) + 29.0223r,(r) = - 20.58055(f)

r2(f ) + 0.2135r->(f) + U3.9665/- 2(r) - - 5.102 68 (f)

r3(/) + 3.0117r,(/) + 907.0112r(/) = 45.28148(f)

These have solutions given by equation (3.6) as

r,(f) = - 3X3956- *** sin (5.3602/)

r2(f) = - 0.4780e'alt**' sin ( 10.6750r)

r ,( f) = 1.5054e-1sos' sin (30.0789:)

Using the transformation x (/) = M' - zPr(i) yields

A,( f) * 0.0790e~05587' sin (53602/) + 0.0024e'aM* s' sin ( I0.6750f) + 0.0682


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*l(')

0.10

0.05

0.00

-0.05

*}(')

Figure 4.29 Numerical s imulation of the vibration of the punch press of figures 4 2 7

and 4.28 as the result of the machine tool impacting the tool base.

I his ex ample illustrates a method of ass igning modal damping to an analytical

mode l.T his is a somewhat arbitrary proc edure that falls in the category of an educated

guess. A more s ophisticated method is to measure the modal damping.T his is discussedin Chapter 7. Note that the floor, .t3(/), vibrates much longer than the machine parts do.

T his is something to consider in des igning how and where the machine is mounted to thefloor of a buildinc.

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