Scilab Textbook Companion for Mechanical Vibration by G. K. Grover 1 Created by Vishwas Anand B.E Mechanical Engineering B.M.S.COLLEGE OF ENGINEERING College Teacher None Cross-Checked by Reshma May 26, 2016
1 Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in
Book Description Title: Mechanical Vibration Author: G. K. Grover Publisher: Nem Chand And Bross, Roorkee . Edition: 8 Year: 2009 ISBN: 9788185240565
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Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular
Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.
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Contents List of Scilab Codes
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1 Fundamental of vibration
5
2 Undamped fre e vibrations of singl e degree of free dom system
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3 Damped free vibr ations of sin gle degree of freed om system
17
4 Forced vibra tions of single degree of freedo m system
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5 Two degrees of freedom systems
39
6 Multi degree of free dom systems exact an alysis
45
7 Multi degree of freedo m systems numerical methods
47
8 Critical speeds of shafts
52
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List of Scilab Codes Exa 1.4.2 Exa 1.5.1 Exa 1.6.1 Exa 1.6.2 Exa 1.7.1 Exa 1.8.2 Exa 2.3.4 Exa 2.4.1 Exa 2.5.1 Exa 2.5.2 Exa 3.3.2 Exa 3.3.3
sum of two harmonic motion . . . . . . . . . . . . . . 5 max and min amp litude of combined motion . . . . . 6 complex number in exponential form . . . . . . . . . . 7 complex number in rectangular form . . . . . . . . . . 8 work done by a force on displacement . . . . . . . . . 9 fourier components of periodic motion . . . . . . . . . 9 moment of inertia of flywheel . . . . . . . . . . . . . . 13 natural frequency of torsional pendulum . . . . . . . . 14 mass in a spring mass system . . . . . . . . . . . . . . 15 natural frequency of torsional oscillation . . . . . . . . 15 undamped and damped natural frequencies of system . 17 gun barrel recoilling . . . . . . . . . . . . . . . . . . . 18
Exa 3.4.1 disc of torsional pendulum . . . . . . . . . . . . . . . . 19 Exa 3.6.1 spring mass system with coulomb damping . . . . . . 20 Exa 3.6.2 vertical spring mass system . . . . . . . . . . . . . . . 21 Exa 3.7.1 time taken for complete damping . . . . . . . . . . . . 22 Exa 4.2.1 periodic torque on suspended flywheel . . . . . . . . . 24 Exa 4.2.2 damping factor and natural frequency of system . . . . 25 Exa 4.3.1 system of beams su pporting a motor . . . . . . . . . . 26 Exa 4.3.2 single cylinder vertical petrol engine . . . . . . . . . . 27 Exa 4.4.1 mass hung from end of vertical spring . . . . . . . . . 27 Exa 4.4.2 support of spring mass system . . . . . . . . . . . . . 28 Exa 4.4.3 spring of automobile trailer . . . . . . . . . . . . . . . 29 Exa 4.5.1 power required to vibrate spring mass dashpot . . . . 30 Exa 4.6.1 horizontal spring mass system in dry friction . . . . . 31 Exa 4.10.1 machine mounted on 4 identical springs . . . . . . . . 31 Exa 4.10.2 machine mounted on springs . . . . . . . . . . . . . . 33 Exa 4.10.3 radio set isolated from machine . . . . . . . . . . . . . 34 4
Exa 4.11.1 vibrotometer . . . . . . . . . . . . . . . . . . . . . . . 35 Exa 4.11.2 commercial type vibration pick up . . . . . . . . . . . 35 Exa 4.11.3 device used to measure torsional accerelartion . . . . . 37 Exa 4.11.4 Frahm tachometer . . . . . . . . . . . . . . . . . . . . 37 Exa 5.3.2 uniform rods pivoted at their upper en ds . . . . . . . . 39 Exa 5.3.3 shaft with 2 circular discs at the en d . . . . . . . . . . 40 Exa 5.7.1 Exa 5.7.2 Exa 6.8.2 Exa 7.2.1 Exa 7.3.1 Exa 7.4.1 Exa 7.7.2 Exa 8.2.1 Exa 8.3.1 Exa 8.4.1 Exa 8.6.1
torque applied to a tors ional system . . . . . . . . . . 41 section of pipe i n a machine . . . . . . . . . . . . . . . 42 system with rolling masses . . . . . . . . . . . . . . . 45 fundamental frequency by rayleigh method . . . . . . 47 Dunkerly method . . . . . . . . . . . . . . . . . . . . . 48 Stodola method . . . . . . . . . . . . . . . . . . . . . 49 four rotor system . . . . . . . . . . . . . . . . . . . . . 50 rotor mounted midway on shaft . . . . . . . . . . . . . 52 disc mounted midway between bearings . . . . . . . . 53 two critical speeds . . . . . . . . . . . . . . . . . . . . 54 right cantilever steel shaft with rotor at the end . . . . 55
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Chapter 1 Fundamental of vibration
Scilab code Exa 1.4.2 sum of two harmonic motion
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 1. 4. 2 \ n ’ ) 4 // given dat a 5 // case 1 6 //x1=(1/2) ∗ cos ( ( %p i /2) ∗ t ) . . . x1=a ∗ c o s (W1∗ t ) 7 // x2=s i n ( %p i ∗ t ) . . . x2=b ∗ s i n (W2∗ t ) 8 // calcu latio ns 9 W1=(%pi/2) 10 W2=%pi 11 t1=2*%pi/(W1) 12 t2=2*%pi/(W2) 13 p1=[t 1 t2] 14 T1 = lcm (p1) 15 // case 2 16 //x1=2 ∗ co s (( %pi ∗ t ) . . . x1=a ∗ c o s (W3∗ t ) 17 //x2=2 ∗ co s (2 ∗ t ) . . . x2=a ∗ c o s (W4∗ t ) 18 W3=%pi 19 W4=2 20 t3=2*%pi/(W3)
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\n
21 22 23 24 25
t4=2*%pi/(W4) p2=[t 3 t4] T2 = lcm (p2)
//output mprintf ( ’ Cas e ( i ) \ nTime period
s e c \ nTime period
of f i r s t wave is %f of f i r s t wave is %f sec \ nThe
ti me period of com bi ned wave is %f sec \ nCase ( i i ) \ nTime period of f i r s t wave is %f sec \ nTime peri od of f i r s t wave is %f sec \ nThe ti me period of co mb in ed wave is %f sec ’ ,t1,t2,T1,t3,t4,T2) 26 mprintf ( ’ \nNOTE: The time peri od of com bin ed mot ion in cas e ( i i ) ca nn ot be calculated \ n si n c e p i is a no n−terminating and non recu rrin g number . \ n But SCILAB take s th e val ue of pi to be 3.1 4159 3 and therefore \ n calcu lates th e LCM of pi and th e ti me period of f i r s t wave in case ( i i . ’ )
Scilab code Exa 1.5.1 max and min amplitude of combined motion
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 1. 5. 1 \ n ’ ) 4 // given dat a 5 //x1=a ∗ s i n (W1∗ t ) 6 //x2=b ∗ s i n (W2∗ t ) 7 // calcu latio ns 8 a=1.90 // amplit ude of f i r s t wave in cm 9 b=2.00 // amp lit ude of sec ond wave in cm 10 W1=9.5 // frequency of f i r s t wave in rad/ sec 11 W2=10.0 // frequenc y of sec ond wave in ra d/ sec 12 xmax=b+a // maximum amplitude of mot ion in cms 13 xmin= abs (a-b) //m inimum amplitude of mot ion in cms 14 f = abs (W1-W2)/(2*%pi) // bea t frequency in Hz 15 t=1/f //ti m e pe ri od of be at in se c
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16 //output 17 mprintf ( ’Th e maximum amplitude
of mot ion i s %4.4 f cm s \ nThe minimum amplitude of mot ion i s %4.4 f cms \ n The be at fre quen cy is %4.4 f Hz \n th e ti me per iod is %4.4 f sec ’ ,xmax,xmin,f,t)
Scilab code Exa 1.6.1
complex number in exponential form
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 1. 6. 1 \ n ’ ) 4 // given dat a 5 // case 1 6 //a co mp le x number is repr esen ted as Z=X+j ∗Y wh ere j 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
is ima gin ary //V=3 +j ∗ 7 x1=3 y1=7
// calcu latio ns r1 = sqrt (x1^2+y1^2) if (y1/x1)>0 then theta1= atan (y1/x1) else theta1=%pi- atan ( abs (y1/x1)) end theta1= atan (y1/x1)
// case 2 //V=−5 +j ∗ 4 x2=-5 y2=4
// calcu latio ns r2 = sqrt (x2^2+y2^2) if (y2/x2)>0 then theta1= atan (y2/x2) else theta2=%pi- atan ( abs (y2/x2)) end
25 //output
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26 mprintf ( ’ c a s e ( i ) V=3+ j ∗ 7 is
represented as V =%3.3 f ∗ e ˆ ( j ∗ (%3 . 3 f ) ) \ nc as e ( i i ) V= −5+j ∗4 is re pr es en te d a s V=%3 . 3 f ∗ e ˆ ( j ∗ (%3 . 3 f ) ) ’ ,r1,theta1 ,r2,theta2)
Scilab code Exa 1.6.2
complex number in rectangular form
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 1. 6. 2 \ n ’ ) 4 // given dat a 5 //Z=r ∗ e ˆ ( i ∗ thet a ) is represente d as Z =r ∗ cos ( the ta ) + 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
i ∗ r ∗ si n ( the ta )= x +i ∗ y //wh er e r ∗ co s ( th et a )=x an d r ∗ si n ( th et a )=y // case 1 //V=5 ∗ e ˆ ( j ∗ 0.10) r1=5 theta1=0.1 x1=r1* cos (theta1) y1=r1* sin (theta1) v1=complex(x1,y1)
// case 2 //V=17 ∗ e ˆ( − j ∗ 3.74) r2=17 theta2=-3.74 x2=r2* cos (theta2) y2=r2* sin (theta2) v2=complex(x2,y2)
//output
mprintf ( ’ ca s e ( i ) :V= 5 ∗ e ˆ ( j ∗ 0.1 0) is rep res ent ed as ’ ) disp (v1) mprintf ( ’ \ n ca se ( i i ) :V=1 7 ∗ e ˆ( − j ∗ 3.7 4) is rep res ent ed as ’ ) 25 disp (v2) 26 mprintf ( ’ \nNOTE: compl ex nu mb er i s re pr es en te d as x+y
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∗ i in SCILAB ’ )
Scilab code Exa 1.7.1 work done by a force on displacement
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 1. 7. 1 \ n ’ ) 4 // given dat a 5 Po=25 // am pl it ud e of force in N 6 Xo=0.05 //amp li yu de of disp lace ment in m 7 W=20*%pi 8 // calcu latio ns 9 // case 1 10 t0=0 11 t1=1 12 v1 = integrate ( ’ s i n (W∗ t ) ∗ c o s (W∗ t−%p i /6) ’ , ’ t ’ ,t0,t1) 13 WD1=Po*Xo*W*v1 14 // case 2 15 t0=0 16 t1=1/40 17 v2 = integrate ( ’ s i n (W∗ t ) ∗ c o s (W∗ t−%p i /6) ’ , ’ t ’ ,t0,t1) 18 WD2=Po*Xo*W*v2 19 //output 20 mprintf ( ’ ( i )wo rk do ne durin g the f i r s t sec ond is %f
N−m\ n ( i i )wo rk do ne during se co nd is %f N −m ’ ,WD1,WD2)
the
f i r s t 1/ 40 th of
Scilab code Exa 1.8.2 fourier components of periodic motion
1 clc 2 clear
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3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 1. 8. 2 \ n ’ ) 4 // given dat a 5 T=0.1 //ti m e pe ri od of periodi c m o ti o n in se c 6 W=2*%pi/T 7 k=12/2 //n umber of el em en ts in half cycle 8 mprintf ( ’ \ tNo of ele men ts in one cycle degrees \n ’ ) 9 mprintf (’t(j) f(j) c o s(t(j))
\n
2k =12,t ( j ) in
f(j) ∗ cos ( t ( j ) ) sin ( t ( j ) ) f(j) ∗ sin ( t ( j ) co s (2 ∗t ( j )) f(j) ∗ co s (2 ∗ t ( j ) ) sin (2 ∗ t ( j ) ) f(j) ∗ sin ( 2 ∗t( j )) cos (3 ∗ t ( j ) ) f(j) ∗ cos (3 ∗ t ( j ) sin (3 ∗ t( j )) f ( j ) ∗ si n (3 ∗ t ( j ) \n ’ )
10 f(1)=10/6 11 for j=1:6 12 t(j)=j*(%pi/k) m(j)= cos (t(j)) 13 14 n(j)=f(j)*m(j) o(j)= sin (t(j)) 15 16 p(j)=f(j)*o(j) 17 q(j)= cos (2*t(j)) 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34
r(j)=f(j)*q(j) s(j)= sin (2*t(j)) u(j)=f(j)*s(j) v(j)= cos (3*t(j)) x(j)=f(j)*v(j) y(j)= sin (3*t(j)) z(j)=f(j)*y(j) f(j+1)=f(j)+f(1) mprintf ( ’% 3. 0 f \ t ’ ,t(j)*(180/%pi)) mprintf ( ’% 3. 4 f \ t \ t ’ ,f(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,m(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,n(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,o(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,p(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,q(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,r(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,s(j))
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35 36 37 38 39 40
mprintf mprintf mprintf mprintf mprintf end
( ’% 3. ( ’% 3. ( ’% 3. ( ’% 3. ( ’% 3.
4f 4f 4f 4f 4f
\ t \ t ’ ,u(j)) \ t \ t ’ ,v(j)) \ t \ t ’ ,x(j)) \ t \ t ’ ,y(j)) \ n ’ ,z(j))
41 f(7)=f(j)-f(1) 42 for j=7:12 43 t(j)=j*(%pi/k) 44 m(j)= cos (t(j)) 45 n(j)=f(j)*m(j) 46 o(j)= sin (t(j)) p(j)=f(j)*o(j) 47 48 q(j)= cos (2*t(j)) r(j)=f(j)*q(j) 49 50 s(j)= sin (2*t(j)) u(j)=f(j)*s(j) 51 52 v(j)= cos (3*t(j)) x(j)=f(j)*v(j) 53 54 y(j)= sin (3*t(j)) 55 z(j)=f(j)*y(j) 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
f(j+1)=f(j)-f(1) mprintf ( ’% 3. 0 f \ t ’ ,t(j)*(180/%pi)) mprintf ( ’% 3. 4 f \ t \ t ’ ,f(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,m(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,n(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,o(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,p(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,q(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,r(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,s(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,u(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,v(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,x(j)) mprintf ( ’% 3. 4 f \ t \ t ’ ,y(j)) mprintf ( ’% 3. 4 f \ n ’ ,z(j)) end sumf(j)=f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)
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73 74 75 76 77 78 79 80 81 82
83 84 85 86 87 88 89 90 91 92 93
+f(10)+f(11)+f(12) sumcos(t(j))=m(1)+m(2)+m(3)+m(4)+m(5)+m(6)+m(7)+m(8) +m(9)+m(10)+m(11)+m(12) sumfjcos(t(j))=n(1)+n(2)+n(3)+n(4)+n(5)+n(6)+n(7)+n (8)+n(9)+n(10)+n(11)+n(12) sumsin(t(j))=o(1)+o(2)+o(3)+o(4)+o(5)+o(6)+o(7)+o(8) +o(9)+o(10)+o(11)+o(12) sumfjsin(t(j))=p(1)+p(2)+p(3)+p(4)+p(5)+p(6)+p(7)+p (8)+p(9)+p(10)+p(11)+p(12) sumcos2(t(j))=q(1)+q(2)+q(3)+q(4)+q(5)+q(6)+q(7)+q (8)+q(9)+q(10)+q(11)+q(12) sumfjcos2(t(j))=r(1)+r(2)+r(3)+r(4)+r(5)+r(6)+r(7)+r (8)+r(9)+r(10)+r(11)+r(12) sumsin2(t(j))=s(1)+s(2)+s(3)+s(4)+s(5)+s(6)+s(7)+s (8)+s(9)+s(10)+s(11)+s(12) sumfjsin2(t(j))=u(1)+u(2)+u(3)+u(4)+u(5)+u(6)+u(7)+u (8)+u(9)+u(10)+u(11)+u(12) sumcos3(t(j))=v(1)+v(2)+v(3)+v(4)+v(5)+v(6)+v(7)+v (8)+v(9)+v(10)+v(11)+v(12) sumfjcos3(t(j))=x(1)+x(2)+x(3)+x(4)+x(5)+x(6)+x(7)+x (8)+x(9)+x(10)+x(11)+x(12) sumsin3(t(j))=y(1)+y(2)+y(3)+y(4)+y(5)+y(6)+y(7)+y (8)+y(9)+y(10)+y(11)+y(12) sumfjsin3(t(j))=z(1)+z(2)+z(3)+z(4)+z(5)+z(6)+z(7)+z (8)+z(9)+z(10)+z(11)+z(12) a0=sumf(j)/(2*k) a1=sumfjcos(t(j))/k b1=sumfjsin(t(j))/k a2=sumfjcos2(t(j))/k b2=sumfjsin2(t(j))/k a3=sumfjcos3(t(j))/k b3=sumfjsin3(t(j))/k disp (’T he fou rie r com pon en ts of period ic mo tio n shown in ex a m pl e 1.8.1 ar e as follows ’ ) mprintf ( ’ \ nao=%f \ na1=%f \ nb1=%f \ na2=%f \ nb2=%f \ na3=%f \ nb3=%f \n ’ ,a0,a1,b1,a2,b2,a3,b3)
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Chapter 2 Undamped free vibrations of single degree of freedom system
Scilab code Exa 2.3.4 moment of inertia of flywheel
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 2. 3. 4 \ n ’ ) 4 // given dat a 5 M=35 //m ass of flywhe el in Kgs 6 r=0.3/2 / / distanc e of cent re of mass fro m piv ot in m 7 T=1.22 //time peri od of o s c i l l a t i o n in sec 8 g=9.81 // accel arat ion due to gravity in m/( sec ˆ2 ) 9 // co nc ep t is as follows 10 //Jo =ma ss mo ment of i n er ti a abou t pivot , Wn=natu ral
freqency 11 // the tad d=theta dou ble do t ( dou ble di ff er en ti at io n ) 12 //Jo ∗ thetadd= −M∗ g ∗ r ∗ theta . . . . sum of moments is = to
zero 13 //Jo ∗ the tadd +(M∗ g ∗ r ∗ th et a )= 0 14 //Wn=s q r t ( (M ∗ g ∗ r ∗ )/Jo)=2 ∗ pi/T 15 // calcu latio ns 16 Jo=M*g*r/((2*%pi/T)^2)
14
17 Jg=Jo-M*r^2
//ma ss mo ment of in er ti a ab ou t geometri
c
axis 18 //output 19 mprintf ( ’M ass moment of
in ert ia ab ou t pivot is %4.4 f Kg−mˆ2\ n M ass m oment of in er ti a ab ou t geometric axis is %4.4 f Kg −mˆ2 ’ ,Jo,Jg)
Scilab code Exa 2.4.1 natural frequency of torsional pendulum
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 2. 4. 1 \ n ’ ) 4 // given dat a 5 l =1 // lenght in m 6 d=0.005 // dia of ro d im m 7 D=0.2 // di a of do to r in m 8 M =2 //ma ss of mo to r in Kg 9 G=0.83 *10 ^11 / /mo dulus of ri gi di ty in N /mˆ2 10 // calcu latio ns 11 J=M*((D/2)^2)/2 //ma ss mo ment of in er ti a in Kg −mˆ2 12 Ip=(%pi/32)*d^4 // se cti on mo du lu s in mˆ4 13 Kt=G*Ip/l // s t i f f n e s s in N −m/rad 14 Wn = sqrt (Kt/J) // natural freqe ncy in ra d/ sec 15 fn=Wn/(2*%pi) // natu ral freq in Hz 16 //output 17 mprintf (’ The na tur al fr eq enc y of vibr atio n of
to rs io na l pendulum is %4.4 f ra d/ sec \n or %4.4 f Hz ’ ,Wn,fn) 18 mprintf ( ’ \nNOTE: In bo ok the natu ral freqen cy of vibration of torsi onal pendulum \nis gi ve n a s 36 Hz wh ic h is wro ng . ’ )
15
Scilab code Exa 2.5.1 mass in a spring mass system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 2. 5. 1 \ n ’ ) 4 5 6 7 8 9 10 11 12
// given
\n
dat a
k1=2000 // s t i f f n e s s of sp ri ng 1 in N/m k2=1500 // s t i f f n e s s of sp ri ng 2 in N/m k3=3000 // s t i f f n e s s of sp ri ng 3 in N/m k4=500 // s t i f f n e s s of sp ri ng 4 in N/m k5=500 // s t i f f n e s s of sp ri ng 5 in N/m fn =1 0 / / natural freq uenc y of sy st em in Hz
// calcu latio ns
// e ff ec t i v e s t i f f n e s s t o p 3 spr in gs i n series i n N/m 13 Ke2=k4+k5 // ef fe ct iv e s t i f f n e s s of low er 2 springs in para llel in N/m 14 Ke=Ke1+Ke2 // total eff ect ive s t i f f n e s s of sring system 15 M=Ke/(2*%pi*fn)^2 // reqired mass su ch tha t th e Ke1=1/((1/k1)+(1/k2)+(1/k3))
of
natu ral fre que ncy of s ys t e m is 10 H z ( in K g) 16 //output 17 mprintf (’ The mass requ ired s uc h th at th e natu ral fr equ en cy of sy s te m is 10 H z \n is %4.4 f Kg ’ ,M )
Scilab code Exa 2.5.2
natural frequency of torsional oscillation
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 2. 5. 2 \ n ’ ) 4 // given dat a 5 G=0.83*10^11 // ri g id i ty mo du lu s in N/mˆ2 6 J=14.7
/ /m ass moment of in ert ia in kg −mˆ2 16
\n
7 8 9 10 11 12
l1=0.6 / / le ng ht of secti on 1 in m l2=1.8 / / le ng ht of secti on 2 in m l3=0.25 / / le ng ht of secti on 3 in m d1=0.05 / /di a of sec tio n 1 in m d2=0.08 / /di a of sec tio n 2 in m d3=0.03 / /di a of sec tio n 3 in m
13 // calcu latio ns 14 Kt1=(G/l1)*(%pi/32)*d1^4
//(%pi/32)
∗d ˆ4 is t h e
sec tio n mo du lu s 15 Kt2=(G/l2)*(%pi/32)*d2^4 16 Kt3=(G/l3)*(%pi/32)*d3^4 17 Kt=1/((1/Kt1)+(1/Kt2)+(1/Kt3)) 18 19 20 21
22
// tota l effective s t i f f n e s s of the to rs io na l sys tem Wn = sqrt (Kt/J) // natural freq in ra d/ sec fn=Wn/(2*%pi) // natu ral freq in Hz //output mprintf (’ The na tur al fr eq ue nc y of torsional o s c i l l a t i o n fo r the given syst em i s \ n %4.4 f rad/ sec or %4.4 f Hz. ’ ,Wn,fn) mprintf ( ’ \nNOTE: Since the value of Kt in the tex tboo k ha s be en ro un de d of \n to 3 decimal pl ac es , t he fina l a n s we r varies
17
slig htly . ’
)
Chapter 3 Damped free vibrations of single degree of freedom system
Scilab code Exa 3.3.2 undamped and damped natural frequencies of sys-
tem 1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns Ex am pl e 3. 3. 2 \ n ’ ) 4 // given dat a 5 m=10 / /m ass of solid in K g 6 Kr=3000 // s t i f f n e s s of natu ral 7 Kf=12000 // s t i f f n e s s of f e l t in 8 Cr=100 //d am pi ng c o e f f i c i e n t of
by G. K. Gr ov er
\n
rubbe r in N/m N/m natu ral rubbe r in N
−s e c /m //dampi ng c o e f f i c i e n t of f e l t in N −s e c /m // calcu latio ns Ke=1/((1/Kf)+(1/Kr)) // eq ui va le nt s t i f f n e s s in N/m Ce=1/((1/Cf)+(1/Cr)) // equ iv al en t da mp in g c o e f f i c i e n t N−s e c /m 13 Wn = sqrt (Ke/m) // undamped natural freq in ra d/ sec 14 fn=Wn/(2*%pi) // undamped natural freq in Hz 9 10 11 12
Cf=330
15 zeta=Ce/(2*
sqrt (Ke*m))
//da mp in g fa ct or 18
//d amped natural freuency in rad/ sec ( eq n 3. 3. 16 ) 17 fd=Wd/(2*%pi) // damped natural frequency in Hz 18 //output 19 mprintf ( ’ The undamped natural frequenc y is %4.4 f rad/ sec or %4.4 f Hz \ n Th e da mped natur al freuency 16 Wd = sqrt (1-zeta^2)*Wn
is %4.4 f rad/ sec
Scilab code Exa 3.3.3
or %4.4 f Hz ’
,Wn,fn,Wd,fd)
gun barrel recoilling
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 3. 3. 3 \ n ’ ) 4 // given dat a 5 m=600 //m ass of gun barrel in Kgs 6 k=294000 // s t i f f n e s s i n N/m 7 x=1.3 // rec oi l of gun in me te rs 8 // calcu latio ns 9 E=0.5*k*x^2 // en er gy sto re d at th e end of reco il 10 Vo = sqrt (2*E/m) // velo city of recoil 11 Cc=2* sqrt (k*m) // c r i t i c a l dampin g in N −s e c /m 12 Wn = sqrt (k/m) // natu ral frequency of undamped
vibration
in r ad/ sec
13 T=2*%pi/Wn //tim e period of undamped vibration in sec 14 Trecoil=(1/4)*T //ti m e pe ri od for recoi l or ou twa rd
st ro ke in se c 15 //x=(1.3+28.8 ∗ t ) ∗ e ˆ ( − 22.1 ∗t ) from eqn 3.3.2 4 16 mprintf ( ’a) t he i n i t i a l recoi l velocity of barrel
is %f m/s \ nb) c r i t i c a l damping co − e f f i c i e n t o f th e da sh po t whic h is en ga ge d at \ n t h e end of recoil st ro ke is %f N −s e c /m\ n\nsubstituting th e va lu e for t in eqn 3.3.24 , starting fro m t = 0.1 sec \ nwith an in cr em en t of 0.0 1 sec we ge t th e following observations
\ n ’ ,Vo,Cc) 19
17 t=0.1 18 for i=1:20 19 x=(1. 3 +28. 8*t)* exp (-22.1*t) mprintf ( ’ x=%f at t=%f \ n ’ ,x,t) 20 21 t=t+0.01 22 end 23 mprintf ( ’A s x app roa che s th e val ue of 0.05 m, th e value of t =0 .2 2 sec ’ ) 24 Trec=0.22 25 Tret=Trecoil+Trec 26 mprintf ( ’ \ nc) The ref ore t im e requ ired for barrel
ret urn to position 5cm fr o m \ n the posi tio n is %f se c ’ ,Tret)
to
initial
Scilab code Exa 3.4.1 disc of torsional pendulum
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 3. 4. 1 \ n ’ ) 4 // given dat a 5 J=0.06 //m oment of in ert ia of disc of pendulum in K g
−mˆ2 6 7 8 9 10 11 12 13
G=4.4*10^10 // ri g id i ty mo du lu s in N/mˆ2 l=0.4 / / le ng ht of shaf t in m d=0.1 // di am et re of shaft in m a1=9 // ampli tude of f i r s t o s c i l l a t i o n in de gr ees a2=6 // amplitude of second o s c i l l a t i o n in deg ree s a3=4 // amplitude of th ird o s c i l l a t i o n in deg rees
// calcu latio ns // logarithmic ex pl ain ed in se c 3.4
delta= log (a1/a2)
14 zeta=delta/
sqrt (4*%pi^2+delta^2)
fro m eqn 3.4.1 15 Kt=(G/l)*(%pi/32)*d^4
de cr em en t eqn 3.4 .1 // representing
zet a
in se c 3. 4 //(%pi/32) 20
∗ d ˆ4 is t h e sec tio n
modulus sqrt (Kt*J) // t o r s i o n a l da mp in g c o e f f i c i e n t wh ich is th e damping to rq ue at uni t velocity ( sim ila r t o eqn 3.3 .6 i n se c 3. 3) 17 Wn = sqrt (Kt/J) // undamped natural freq in ra d/ sec 18 T=2*%pi/( sqrt (1-zeta^2)*Wn) // periodic ti m e of 16 C=zeta*2*
vibration
19 fn=Wn/(2*%pi) // natural 20 //output 21 mprintf ( ’ a) logarithmic
freq
of undamped vibra tion
de cr em en t is %4.4 f \ n b ) damping to rq ue at uni t velocity is %4.4 f N −m/rad \ n c ) periodi c t i m e of vibr ation is %4. 5 f se c \n fr eq ue nc y of vibr ation i f t he disc is removed from viscou s flu id is %4.4 f Hz ’ ,delta,C,T,fn)
Scilab code Exa 3.6.1 spring mass system with coulomb damping
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 3. 6. 1 \ n ’ ) 4 // given dat a 5 m =5 //m ass in spring mass sy st em ) in kg) 6 k=980 // s t i f f n e s of sp ri ng in N/m 7 u=0.025 // c o e f f i c i e n t o f f r i c t i o n 8 g=9.81 // accelerati on due to gravi ty 9 // calcu latio ns 10 F=u*m*g // f r i c t i o n a l f o r c e in N 11 Wn = sqrt (k/m) // fre q of fr ee o s c i l l a t i o n s in rad/ sec 12 fn=Wn/(2*%pi) // fre q of fr ee o s c i l l a t i o n s in Hz 13 Ai=0.05 // i n i t i a l ampli tude in m 14 Ar=0.5*Ai // red uce d amp lit ude in m 15 totAreduc=Ai-Ar // total redu cti on in amp in m 16 Areducpercycl=4*F/k // reduction in amp lit ude/ cycle
ex pl ai ne d in sec tio n 3.6.2 21
in e qn 3.6.6
//n umber of cycl es in am pl it ud e //ti me ta ke n to ach iev e 50
17 n = round (totAreduc/Areducpercycl)
for
50% reduct
18 Treduc=n*(2*%pi/Wn)
ion
%reduction 19 //output 20 mprintf ( ’ a) The frequency
of fr ee
o s c i l l a t i o n s is %4
.4 f ra d/ sec or %4.4 f Hz \n b)nu mber of cycles ta ke n for 50 per ce nt redu cti on in am pl it ud e is %1 .0 f cycles \ n c ) ti me ta ke n to achiev e 50 perc ent red uct ion in am pl it ud e is %4.4 f sec ’ ,Wn,fn,n, Treduc)
Scilab code Exa 3.6.2
vertical spring mass system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 3. 6. 2 \ n ’ ) 4 // given dat a 5 k=9800 // s t i f f n e s of sp ri ng in N/m 6 m=40 //m ass in spring mass sy st em ) in kg) 7 g=9.81 // accelerati on due to gravi ty 8 F=49 // f r i c t i o n a l fo r c e in N 9 x=0.126 // total ex ten sio n of spr ing in m 10 xeq=m*g/k // ex te nsi on of spr ing a t equillib rium in m 11 xi=x-xeq // i n i t i a l exten sio n of spri ng fr om
equ illi briu m in m // reduction in amp lit ude/ cycle ex pl ai ne d in sec tio n 3.6.2 in e qn 3.6.6 13 n = int (xi/Alosspercycl) //n umber of com ple te cyc les that sy st em unde rgoe s 14 Af=xi-n*Alosspercycl // amp li tu de at th e end of n cycles 15 SF=k*Af // spri ng force acti ng on th e upward direction 12 Alosspercycl=4*F/k
for an ex ten sio n of Af 22
16 if F
will move u p since spri ng force gre ate r th an f r i c t i o n a l fo rce ’ ) Xa=Af // assigning Af to a new variable Xa Xb=0 //assume Xb=0 at f i r s t // solving th e quadrat ic equ atio n in Xb whose is
18 19 20 21 22 23 24 25 26 27 28 29 30
roo ts are Xb1 and Xb2
Xb1=(F+ sqrt ((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F *Xa))))/k Xb2=(F- sqrt ((-F)^2-(4*(0.5*k)*((-(1/2)*k*Xa^2)+F *Xa))))/k if int (Xb1-Xa)==0 then Xb=Xb2 else Xb=Xb1 end finalext=xeq+Xb mprintf (’ The final ex te nt io n of sp ri ng is %f m ’ ,finalext) els e dis p (’T he spri ng will n ot move up since
spring force ’
31 end
for ce is no t great er th an f r i c t i o n a l )
Scilab code Exa 3.7.1 time taken for complete damping
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 3. 7. 1 \ n ’ ) 4 // given dat a 5 fnA=12 / / fr eq ue nc y of free vibrat ions of s y s t e m A in
Hz 6 fnB=15
/ / fr eq ue nc y of free
vibrat
Hz 23
ions
of s y s t e m B in
//tim e tak en by sy st em A to damp ou t com ple tel y in sec 8 // calcu latio ns 9 TdB=fnA*TdA/fnB //tim e tak en by sy st em B to damp ou t com ple te ly in sec 10 //output 7 TdA=4.5
11 mprintf ( ’ The ti me ta ken by sy st em B to damp ou t comp let ely is %4.4 f sec ’ ,TdB)
24
Chapter 4 Forced vibrations of single degree of freedom system
Scilab code Exa 4.2.1 periodic torque on suspended flywheel
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 2. 1 \ n ’ ) 4 // given dat a 5 //T=To∗ s i n (W∗ t ) 6 To=0.588 //m aximum value of per iod ic torqu e in N −m 7 W =4 // fr eq en cy of app lie d force in r a d/ se c 8 J=0.12 //m oment of iner tia of wh e el in kg −mˆ2 9 Kt=1.176 // s t i f f n e s s of wire in N −m/rad 10 Ct=0.392/1 // damping c o e f f i c i e n t in N −m sec/rad 11 // calcu latio ns 12 theta=To/ sqrt ((Kt-J*W^2)^2+(Ct*W)^2) //Eq uat ion for
torsional vibr atio n am pl it ud e fro m F i g (4.2.2 ) and Eqn (4. 2. 5) 13 MaxDcoup=Ct*W*theta //m aximum da mp in g cou ple in N −m 14 if at an ((Ct*W)/(Kt-J*W^2))>0 then phiD=(180/%pi)* atan ((Ct*W)/(Kt-J*W^2)); 15
eqn 4.2 .6( in degree s ) 25
//from
16 17 18 19 20 21
else phiD=180+(180/%pi)*
atan ((Ct*W)/(Kt-J*W^2));
end
//output mprintf ( ’ a) The maximum angular
displacement
fr om
rest positio n is %4. 4 f ra dia ns \ n b ) The maximum co upl e appl ied to da sh po t is %4.4 f N −m\n c ) angl e by whi ch th e an gul ar dis pla ce me nt lags th e to rq ue is %4.4 f degree s ’ ,theta,MaxDcoup ,phiD)
Scilab code Exa 4.2.2 damping factor and natural frequency of system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 2. 2 \ n ’ ) 4 // given dat a 5 Wd=9.8*2*%pi // damped natural freqency in ra d/ sec 6 Wp=9.6*2*%pi // fre qen cy fro m force d vibration test 7 8 9
in
rad/ se c // calcu latio ns // (Wp/Wn)=s q r t (1 −2∗ zeta ˆ2) . . . ( 1 ) fr om Eqn 4. 2. 18 from S ec 4.2.1 //(Wd/Wn)=s q r t (1 − zeta ˆ2) . . . ( 2 ) fr om Eqn 4. 2. 19 fr om S ec 4.2.1 // dividing (1) by (2)
10 11 x=(Wp/Wd) 12 //x =[ s q rt (1 −2∗ zeta ˆ2) ] / [ sqr t (1 − ze ta ˆ2) ] 13 zeta= sqrt ((1-x)/(2-x)) //da mping fac tor obtaine
d on th e a b o ve eqn for ze ta in eqn 2 a b o ve 15 Wn=Wd/ sqrt (1-zeta^2) // natural freq uenc y of sy st em in rad/ se c simplifying 14 // substitut ing
16 fn=Wn/(2*%pi)
// natural
freq uenc y of sy st em in Hz 26
17 //output 18 mprintf (’T he damping facto r for
th e sy st em is %f an d \ n th e nat ural fre que ncy is %4.4 f r ad/ sec or % 4.2 f Hz ’ ,zeta,Wn,fn) 19 mprintf ( ’ \nNOTE: The da mp in g fa ct or zeta given in tex tboo k is 0.196 , wh ic h is wrong . ’ )
Scilab code Exa 4.3.1 system of beams supporting a motor
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns b G. K. Gr ov er \n Ex am pl e 4. 3. 1 \ n ’ ) 4 // given dat a 5 m=1200 //m ass of motor in kg 6 mo=1 // unba lanc ed ma ss on mo to r in kg 7 e=0.06 // locat ion of un ba la nc ed mass from motor in m 8 Wn=2210*(2*%pi/60) // reso nant freq in ra d/ sec 9 W=1440*(2*%pi/60) // operatin g freq 10 // calcu latio ns 11 // case 1 12 zeta=0.1 13 bet=(W/Wn) 14 y=(mo/m) //fr om eq n 4.3. 2 15 X1=(y*e)*(bet)^2/ sqrt ((1-bet^2)^2+(2*zeta*bet)^2) //
from eqn 4.3. 2 16 // case 2 17 zeta=0 18 X2=(y*e)*(bet)^2/
sqrt ((1-bet^2)^2+(2*zeta*bet)^2)
from eqn 4.3. 2 19 //output 20 mprintf (’ If th e damping is less t h a n 0. 1 t h e n t he ampl itu de of \n vibration will be b e t we e n %f m an d %f m’ ,X1,X2)
27
//
Scilab code Exa 4.3.2
single cylinder vertical petrol engine
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 3. 2 \ n ’ ) 4 // given dat a 5 m=320 //m ass of en gin e in kg 6 mo=24 // recip roca ting mass on motor in kg 7 r=0.15 // vertical st ro ke in m 8 e=r/2 9 delst=0.002 // stati de fln in m 10 C=490/(0.3) //d amping recis tanc e in N −s e c /m 11 g=9.81 // gravity in m/ sec ˆ2 12 N=480 // sp ee d in rpm in case b) 13 // calc ulat ion 14 Wn = sqrt (g/delst) // natural freqe ncy in ra d/ sec 15 Nr=Wn/(2*%pi)*60 // resonant spe ed in rpm 16 17 18 19 20
W=(2*%pi*N/60) bet=(W/Wn) zeta=(C/(2*m*Wn)) //da mp in g fa cto r y=(mo/m) //fr om eq n 4.3. 2 X=(y*e)*(bet)^2/ sqrt ((1-bet^2)^2+(2*zeta*bet)^2)
// from eqn 4.3. 2 21 //output 22 mprintf (’ a) sp ee d of driving shaft at wh ich es on an ce occu rs i s %4.4 f RPM \ n b)T he am pl it ud e of ste ady state for ced vibrat ions when t he dri ving shaf t \n of th e en gin e rotates at 4 80 RPM is %f m’ ,Nr,X)
Scilab code Exa 4.4.1 mass hung from end of vertical spring
28
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 4. 4. 1 \ n ’ ) 4 // given dat a 5 T=0.8 //t i m e pe ri od of free vib rat ion in se c 6 t=0.3 //ti m e for
wh ich t he vertica be calculated 7 //y=18 ∗ si n (2 ∗ p i ∗ t ) 8 Y=18 //m ax amplit ude in m m 9 // calcu latio ns 10 11 12 13
\n
l dis tanc e h a s t o
W=2*%pi Wn=(2*%pi/T) bet=(W/Wn) x=(Y/(1-bet^2))*(
sin (W*t)-bet* sin (Wn*t)) // fr om eq n 4.4.1 7 explain ed in th e same pr ob le m 14 //output 15 mprintf (’T he ver ti ca l distance moved by ma ss in th e ,x ) f i r s t 0.3 sec is %4.4 f mm’
Scilab code Exa 4.4.2 support of spring mass system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 4. 2 \ n ’ ) 4 // given dat a 5 m=0.9 //m ass in kg 6 K=1960 // s t i f f n e s s i n N/m 7 Y =5 //amp of vibration of su pp or t in m 8 N=1150 // freq uenc y in cycles pe r min 9 // calcu latio ns 10 Wn = sqrt (K/m) 11 W=N*2*%pi/60 // fr eq ue nc y of vibr ation of su pp or t 12 bet=(W/Wn)
29
13 // case 1 14 zeta=0 15 X1=Y*( sqrt (1+(2*zeta*bet)^2)/ sqrt ((1-bet^2)^2+(2* zeta*bet)^2)) //E qn (4 .4 .6 ) 16 // case 2 17 zeta =0. 2 18 X2=Y*( sqrt (1+(2*zeta*bet)^2)/ sqrt ((1-bet^2)^2+(2* zeta*bet)^2)) //E qn (4 .4 .6 ) 19 //output 20 mprintf ( ’T he amp lit ude of vibra tion when damping
fa ct or =0 is %4.4 f mm \n If damping fac tor =0 .2 , th en amp li tu de of vibration is %4.4 f mm’ ,X1,X2)
Scilab code Exa 4.4.3 spring of automobile trailer
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 4. 3 \ n ’ ) 4 // given dat a 5 delst=0.1 // st ea dy state defln in m 6 g=9.81 // accelerati on due to gravi ty 7 Y=0.08 //amp of vibration of au to mo bi le in m 8 lambda=14 // wa ve le ng ht of pr ofi le in m 9 // calcu latio ns 10 Wn = sqrt (g/delst) 11 fn=Wn/(2*%pi) // fre que ncy of vibration of au to mo bi le
in Hz 12 13 14 15 16 17
Vc=(3600/1000)*lambda*fn // c r i t i c a l spee d in km/hr V=60 // speed in km/hr W=V*(1000/3600)*(2*%pi/lambda) bet=(W/Wn) zeta=0 X=Y*( sqrt (1+(2*zeta*bet)^2)/ sqrt ((1-bet^2)^2+(2*zeta *bet)^2))
//E qn (4 .4 .6 ) 30
18 //output 19 mprintf ( ’ The c r i t i c a l speed
of automobile %4.4 f km/ h r \ n T he am pl it ud e of vibration at 60 Km/H r is %4 .4 f m’ ,Vc,X)
Scilab code Exa 4.5.1 power required to vibrate spring mass dashpot
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 4. 5. 1 \ n ’ ) 4 // given dat a 5 X=0.015 // am pl it ud e of vibration of spri ng mass
das hpo t sy st em in m of vibration sy st em in Hz
6 f=100 // fr que ncy 7 8 9 10 11 12 13
zeta=0.05 fnD=22 // damped natural m=0.5 //m ass in kg
frequenc
of sprin g mass da sh po t
y in Hz
// calcu latio ns W=2*%pi*fnD c=2*m*W*zeta // from E qn 3.3. 6 and E qn 3.3. 7 Epercycl=%pi*c*(2*%pi*f)*X^2 //Eqn 4 . 5 . 1 . . . en er gy
dissipat
ed p e r cyc le
14 Epersec=Epercycl*f // en er gy dissipated 15 //output 16 mprintf (’ The power requ ired to vibrat
pe r sec
e sprin g mass dash pot sys te m wit h \ n an am pl it ud e of 1.5 cm a nd at fre quen cy of 100 Hz is %4.4 f Watts ’ ,Epersec) 17 mprintf ( ’ \nNOTE: sl ig ht di ff er nc e in an sw er co mp ar ed to text book \ n is due ap pr ox im at io n of va lu e of pi ’)
31
Scilab code Exa 4.6.1 horizontal spring mass system in dry friction
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Ex am pl e 4. 6. 1 \ n ’ )
\n
4 // given dat a 5 mprintf ( ’NOTE: The ma ss given 6 7 8 9 10 11 12 13 14 15 16 17
in textbook should equal \ n to 3.7 kg s and no t 8.7 Kgs ’ ) m=3.7 //m ass in kg g=9.81 // gravit y K=7550 // // s t i f f n e s s of in N/m u=0.22 // c o e f f i c i e n t o f f r i c t i o n Fo=19.6 //amp of force in N f =5 // freq uenc y of force // calcu latio ns F=u*m*g // f r i c t i o n a l f o r c e W=2*%pi*f Wn = sqrt (K/m) bet=(W/Wn) X=(Fo/K)* sqrt (1-(4*F/(%pi*Fo))^2)/(1-bet^2)
be
//Eqn
4.6 .2 i n S e c 4. 6// equ iva lent 18 Ceq=4*F/(%pi*W*X)
vis cou s dam pi ng Eq n 4.6 .1 i n S e c 4. 6 19 //output 20 mprintf ( ’ \ nThe am pl it ud e of vibration of mass is %f m\n T he equivalent viscous damping is %f N −s ec / m’ ,X,Ceq) 21 mprintf ( ’ \nNOTE:
sl ig ht di ff er nc e in an sw er co mp ar ed to text book \ n is due ap pr ox im at io n of va lu e of pi in th e ta xt boo k ’ )
Scilab code Exa 4.10.1 machine mounted on 4 identical springs
1 clc
32
2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Examp le 4.1 0.1 \ n ’ ) 4 // given dat a 5 m=1000 //m ass of ma ch in e in kg 6 Fo=490 //amp of force in N 7 8 9 10 11 12 13 14 15 16
\n
f=180 // f r e q inRPM
// calcu latio ns // case a) K=1.96*10^6 // to ta l s t i f f n e s s of spr ing s in N/m Wn = sqrt (K/m) W=2*%pi*f/60 bet=(W/Wn) zeta=0 Xst1=Fo/K // am pl it ud e of ste ady state X1=Xst1*(1/( sqrt ((1-bet^2)^2+(2*zeta*bet)^2)))
//amp
of vib rat ion E qn 4.2. 15 in S e c 4.2.1 17 Ftr1=Fo* sqrt (1+(2*zeta*bet)^2)/ sqrt ((1-bet^2)^2+(2* zeta*bet)^2) // force tra nsm itt ed ,E qn 4.10. 2 in Se c
4.10.1 b)
18 // case 19 20 21 22 23 24 25
K=9.8*10^4 // to ta l s t i f f n e s s of spr ing s in N/m Wn = sqrt (K/m) W=2*%pi*f/60 bet=(W/Wn) zeta=0 Xst2=Fo/K // am pl it ud e of ste ady state X2=Xst2*(1/( sqrt ((1-bet^2)^2+(2*zeta*bet)^2)))
//amp
of vib rat ion E qn 4.2. 15 in S e c 4.2.1 sqrt ((1-bet^2)^2+(2* 26 Ftr2=Fo* sqrt (1+(2*zeta*bet)^2)/ zeta*bet)^2) // force tra nsm itt ed ,E qn 4.10. 2 in Se c
4.10.1 27 //output 28 mprintf (’ a)T he am pl it ud e of mo ti on of mac hi ne is %f
m a nd the maximum fo rce transm itted \n t o t h e foun dati on bec aus e of th e un ba la nc ed force when \ n K=1.96 ∗ 10ˆ 6 N/m is %4.4 f N \ n b) for same cas e as in a) i f K=9. 8 ∗ 10ˆ4 N/m then \ n th e am pl it ud e of 33
mo ti on of mac hi ne is %f m \ n an d the maximum fo rc e tra nsm itt ed to th e fou nda tio n be ca us e of \n the unb ala nce d for ce %4.4 f N’ ,X1,Ftr1,X2,Ftr2)
Scilab code Exa 4.10.2 machine mounted on springs
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Examp le 4.1 0.2 \ n ’ ) 4 // given dat a 5 m=75 //ma ss of ma ch in e in kg 6 K=11.76*10^5 // s t i f f n e s s of sp ri ng s in N/m 7 zeta=0.2 8 mo=2 //m ass of pist on in kg 9 stroke=0.08 // in m 10 e=stroke/2 // in m 11 N=3000 // spee in c . p .m 12 // calcu latio ns 13 Wn = sqrt (K/m) 14 W=2*%pi*N/60 15 bet=(W/Wn) 16 y=(mo/m) 17 Fo=mo*W^2*e // max for ce exerted 18 X=y*e*bet^2/( sqrt ((1-bet^2)^2+(2*zeta*bet)^2))
\n
//Eqn
4.3.2 19 Ftr=Fo* sqrt (1+(2*zeta*bet)^2)/ sqrt ((1-bet^2)^2+(2* zeta*bet)^2) // force tra nsm itt ed ,E qn 4.10. 2 in Se c
4.10.1 20 mprintf (’ a)T he am pl it ud e of vibration
of m ach i ne is %f m an d the \ n th e vib rat ory force F t r ,X,Ftr) tra nsm itt ed to th e fo und ati on is %5.4 f N’ 21 mprintf ( ’ \nNOTE: sl ig ht di ff er nc e in an sw er co mp ar ed to text book \ n is due ap pr ox im at io n of val ue s in textbook ’
)
34
Scilab code Exa 4.10.3 radio set isolated from machine
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Examp le 4.1 0.3 \ n ’ ) 4 // giv en da ta 5 m=20 //ma ss in kg s 6 k=125600 // ov er al l eqi val ent s t i f f n e s s i . e 4 7
\n
∗ 31400 in N/m c=1568 // o v e r a l l dam pin g c o e f f i c i e n t i . e 4 ∗ 3 9 2 in N − s e c /m n=500 // vibr atin g spe ed of ma ch in e in cpm // y=Ysi n (w ∗ t ) Y=0.00005 // vibrating amp li tu de of ma chi ne in m W=2*%pi*n/60 // vibrating freq uenc y in ra d/ sec Wn = sqrt (k/m) // natural freq uenc y in ra d/ sec bet=(W/Wn) // spe ed rat io zeta=c/(2* sqrt (k*m)) //da mp in g fa ct or // calcu latio ns
8 9 10 11 12 13 14 15 16 X=Y* sqrt ((1+(2*zeta*bet)^2)/((1-bet^2)^2+(2*zeta*bet )^2)) / / absol ute am pl it ud e of vibration of radi o
from eqn (4. 4.6 ) 17 Z=Y*((bet^2)/
sqrt (((1-bet^2)^2+(2*zeta*bet)^2)))
//
from eqn 4.4.1 1 18 19 20 21
FdynT=Z* sqrt ((c*W)^2+k^2) //d yna mic load to tal Fdyn=FdynT/4 //dy namic loa d on ea ch is ol at or FdynTmax=m*W^2*X //m ax dy namic loa d on th e is ol at or s Fdynmax=FdynTmax/4 // max dy na mi c load on each
isolator 22 //output 23 mprintf (’a)
The am pl it ud e of vibr atio n of ra dio is %f metre s \ n b) th e dynamic lo ad on ea c h iso lat or ,X,Fdyn) due to vibration is %3.3 f N’ 35
Scilab code Exa 4.11.1 vibrotometer
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Examp le 4.1 1.1 \ n ’ ) 4 // given dat a 5 T =2 // pe ri od of free vib rat ion in se c 6 f =1 // ver ti cal ha rm on ic freq uenc y of mac hi ne in in Hz 7 Z=2.5 // amp li tu de of vibr otome ter mass rel at ive to 8 9 10 11 12 13
vibrotometer // calcu latio ns
fr am e in m m
Wn=2*%pi/T W=2*%pi*f bet=(W/Wn) zeta=0 // for vibrotometers Y=Z*( sqrt ((1-bet^2)^2+(2*zeta*bet)^2))/bet^2
// am pl it ud e of vibration of m a chi n e E qn 4.4.11 in S ec 4.4.2 14 //output 15 mprintf (’ The am pl it ud e of vibration of su pp or t of mac hin e i s %4.4 f mm’ ,Y )
Scilab code Exa 4.11.2 commercial type vibration pick up
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Examp le 4.1 1.2 \ n ’ ) 4 // given dat a 5 fn=5.75 // natural
freq uenc y in Hz 36
\n
6 7 8 9
zeta=0.65 ZbyY=1.01
// case 1 // su bs ti tu ti ng fo r (Z/ Y) =1. 01 an d (W/Wn)=r ˆ2 in Eqn 4.4.11 we ge t t he qu ad ra tic eqn a s follows 10 // 0. 02 ∗ r ˆ 4 −0.31 ∗ r ˆ2+ 1=0 11 // solving
r2
for
r in a b o v e eqn whose rootes
ar e r1 an d
12 r1 = sqrt (((0.31)+ sqrt (((-0.31)^2) -4*0.0 2*1))/(2*0.02) ) 13 r2 = sqrt (((0.31)- sqrt (((-0.31)^2) -4*0.0 2*1))/(2*0.02) ) 14 if r1>r2 then 15 r=r1 else r=r2 16 17 end 18 bet=r // b e t =(W/Wn) 19 f1=bet*fn 20 // case 2 21 ZbyY=0.98 22 // su bs ti tu ti ng fo r (Z/ Y) =0. 98 an d (W/Wn)=r ˆ2 in Eqn
we ge qu ad ra tic eqn a s follows 23 // 0.4.4.11 04 ∗ r ˆ4+0 .31t ∗ trhe ˆ2−1=0 24 // solving for r in a b o v e eqn whose rootes ar e r3 an d r4 25 26 27 28 29 30 31 32 33 34 35
r3 = sqrt ((-0.31+ sqrt (((0.31)^2)-4*0. 04*-1)) /(2*0.04)) r4 = sqrt ((-0.31- sqrt (((0.31)^2)-4*0. 04*-1)) /(2*0.04)) t1 = real (r3) t2 = real (r4) if t1>t2 then r=r3 else r=r4 end bet=r // b e t =(W/Wn) f2=bet*fn mprintf (’T he lowest frequency be yon d wh ic h the
ampl itu de can be me as ur ed within \ n ( i ) on e per ce nt error is %4.4 f Hz \n ( i i )t wo perc ent 37
error
is %4.4 f Hz ’ ,f1,f2)
Scilab code Exa 4.11.3 device used to measure torsional accerelartion
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Examp le 4.1 1.3 \ n ’ ) 4 // given dat a 5 J=0.049 //m oment of in ert ia in kg −mˆ2 6 Kt=0.98 // s t i f f n e s s i n N −m/rad 7 Ct=0.11 // damping c o e f f i c i e n t i n N −m sec/rad 8 N=15 //R.P.M 9 thetaRD=2 // rel at iv e am pl it ud e be twe en ring and shaft
in degr ees ns // fr eq ue nc y of vibr atin g shaf t in r a d/ sec 12 Wn = sqrt (Kt/J) // natural freqe ncy in ra d/ sec 10 // calcu latio 11 W=N*2*%pi/60
13 zeta=(Ct/(2* sqrt (Kt*J))) 14 thetaRR=(thetaRD/(57.3))
//da mp in g fa ct or // rela tiv e am pl it ud e in
radians 15 bet=(W/Wn) 16 thetamax=thetaRR*(( sqrt ((1-bet^2)^2+(2*zeta*bet)^2)/ bet^2)) 17 maxacc=(W^2)*thetamax 18 //output 19 mprintf (’T he m aximum acce lera tio n of th e shaft is %4 .4 f rad /( se c ˆ2) ’ ,maxacc)
Scilab code Exa 4.11.4 Frahm tachometer
1 clc
38
2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er Examp le 4.1 1.4 \ n ’ ) 4 // given dat a 5 RF=1800 // resonant frequency in rpm 6 L=0.050 // le ng ht of steel re ed in m et re s 7 8 9 10 11 12 13 14 15
\n
B=0.006 //w i d t h of steel re ed in me tr es t=0.00075 // thic kne ss of steel re ed in me t re s E=19.6*10^10 //yo un g ’ s mod ulu s in N/( mˆ2)
// calcu latio ns Wn=2*%pi*RF/60 // natural fre quen cy in radians I=(B*t^3)/12 //m oment of in er ti a in (mˆ4) m=3*E*I/((Wn^2)*L^3) // requ ired ma ss
//output mprintf (’T he required
of th e reeds
mass M t o be plac ed at th e end of Frahm ta ch om et er is %f Kgs ’ ,m )
39
Chapter 5 Two degrees of freedom systems
Scilab code Exa 5.3.2 uniform rods pivoted at their upper ends
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 5. 3. 2 \ n ’ ) 4 // given dat a 5 m1=5*0.75 //m ass of r o d 1 in k gs 6 m2=5*1.00 //m ass of r o d 2 in k gs 7 l1=0.75 // le ng ht of r o d 1 in m 8 l2=1.00 // le ng ht of r o d 2 in m 9 K=2940 // s t i f f n e s s of sp ri ng in N/m 10 // calcu latio ns 11 Wn = sqrt (3*(m1+m2)*K/(m1*m2)) // natural fre quen cy in
rad/ se c 12 fn=Wn/(2*%pi)
the textbook
// natu ral itself
fre que ncy
in Hz as solv ed in
13 b1=(K*l2) 14 b2=(K*l1-m1*l1*Wn^2/3) 15 x=(b2/b1) 16 Fmax=K*(l1*1-l2*x)/57.3
// to con ve rt into 40
radi ans
17 //output 18 mprintf (’T he fr eq ue nc y of t he resulting
if t h e efec t of gr av it y \ n is neg lec rad/ sec or %4.4 f Hz. \ n The angul ar is %3.3 f degrees ( ou t of ph ase ) \n mo ve me nt of AB. \ n Th e m aximum for ce
vibrat ions ted is %4. 4 f movement of CD wi t h th e in the spring
is %4.4 f N’ ,Wn,fn,x,Fmax)
Scilab code Exa 5.3.3 shaft with 2 circular discs at the end
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 5. 3. 3 \ n ’ ) 4 // given dat a 5 m1=500 //m ass of disc 1 in Kgs 6 m2=1000 //m ass of disc 2 in Kgs 7 D1=1.25 // ou te r di a of dis c 1 in m 8 D2=1.9 // ou te r di a of dis c 2 in m 9 l=3.0 // le ng ht of shaf t in m 10 d=0.10 // di a of shaf t in m 11 G=0.83*10^11 // ri g id i ty mo du lu s in N/mˆ2 12 // calcu latio ns 13 J1=m1*(D1/2)^2/2 //m ass m oment of in ert ia in kg −mˆ2 14 J2=m2*(D2/2)^2/2 //m ass m oment of in ert ia in kg −mˆ2 15 Ip=(%pi/32)*d^4 // sectio n mod ulus of shaft in m ˆ4 16 Kt=G*Ip/l // s t i f f n e s s i n N −m/rad 17 Wn = sqrt (Kt*(J1+J2)/(J1*J2)) //fr om Eqn 5. 3. 28 , Sec
5.3.3 18 19 20 21 22
fn=Wn/(2*%pi) Kt1=2*Kt Kt2=2*Kt*2^4 Kte=1/((1/Kt1)+(1/Kt2)) x = sqrt (Kte/Kt) // ratio
of mo di fi ed nat ur al freq
srcin al nat ura l fr eq ue nc y 41
to
23 //output 24 mprintf (’T he nat ur al fr eq ue nc y of t he torsional
vib rat ion is \ n %4.4 f rad/ sec or %3.3 f Hz. \ n The ratio of mo di fi ed nat ur al fr eq ue nc y to srcina l natural freq uenc y \ n i s %3. 3 f . Which me an s stiffeni ng a sy s t e m incr ease s its na tu ra l frequency ’
,Wn,fn,x)
Scilab code Exa 5.7.1 torque applied to a torsional system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 5. 7. 1 \ n ’ ) 4 // given dat a 5 J1=0.735 //m oment of in ert ia of main sy st em in Kg −mˆ2 6 Kt1=7.35*10^5 // t o r s i o n a l s t i f f n e s s 7 To=294 // am pl it ud e of applied tor que 8 W=10^3 // freq uenc y of applied tor que 9 // u=rat io of absorber ma ss to main ma ss i . e M 2/M1 10 //Wn is exita tion freq uenc y 11 // calcu latio ns 12 W1 = sqrt (Kt1/J1) 13 // cas e1 14 x1=0.8 // wh er e x=(W/W2) 15 u1=[x1^2-1]^2/x1^2 //fr om Eqn 5.7. 9 , Se c 5. 7. 1. 16 // case 2 17 x2=1.2 // wh er e x=(W/W2) 18 u2=[x2^2-1]^2/x2^2 //fr om Eqn 5.7. 9 , Se c 5. 7. 1. 19 if u1>u2 then 20 u=u1 21 else 22 u=u2 23 end 24 J2=u*J1 //m oment of
iner tia
of abs orb er in K g −mˆ2 42
25 26 27 28 29
Kt2=u*Kt1 // tota l tor sio nal s t i f f n e s s of abso rber K=Kt2/(4*0.1^2) // s t i f f n e s s of each sp rin g in N/m b2=-(To/Kt2) // am pl it ud e of vibration in r ad
//output mprintf ( ’T he maximum moment of
in er ti a of absorber ( J2) is %4.4 f Kg −mˆ2 and \n %f is the s t i f f n e s s of e a c h of th e fou r ab so rb er spri ngs s u c h th at \n the re so na nt freque ncies ar e at least 20 pe rc en t from exita tion freq uenc y . \ n T he ampl itu de of vibr atio n of this ab so rb er (b 2) at exitation frequency \ n is %f ra dia ns ’ ,J2,K,b2)
Scilab code Exa 5.7.2 section of pipe in a machine
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 5. 7. 2 \ n ’ ) 4 // given dat a 5 W1=220*2*%pi/60 // vib ra tin g frequency at 22 0 RPM ( in
rad / se c ) 6 W2=W1 // freq uenc y to whic h th e spring 7 8 9
mass sy st em is tun ed to . M2=1 //m ass in spring mass sy st em in kg s N1=188 // f i r s t res onant freq of spring mass sy st em in cpm N2=258 // se co nd reso nant freq of spring mass sy st em in cpm // u=rat io of absorber ma ss to main ma ss i . e M 2/M1 // calcu latio ns
10 11 12 K2=M2*W2^2 13 Wn1=N1*2*%pi/60
// f i r s t reso nant freq of spring mass sy st em in ra d/ sec 14 Wn2=N2*2*%pi/60 // se co nd res onant freq of spring mass sy st em in ra d/ sec 43
15 16 17 18 19 20
// case
1
21 22 23 24 25 26 27 28
x2=(W/W2) u2=[x2^2-1]^2/x2^2
W=Wn1 x1=(W/W2) u1=[x1^2-1]^2/x1^2
// case
//fr om Eqn 5.7. 9 , Se c 5. 7. 1.
2
W=Wn2
//fr om Eqn 5.7. 9 , Se c 5. 7. 1.
// the ref ore u=(u1+u2)/2 //wh ic h is equal to M 2/M1 M1=M2/u // mass of main sy st em in kg s K1=K2/u // s t i f f n e s s of mai n system in N/m
//now Wn21=150*2*%pi/60
//n ew f i r s t resonant
frequenc
y in
rad/ se c 29 Wn22=310*2*%pi/60
//n ew se co nd reso nant
fre quen cy in
rad/ se c 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
W=Wn21 x1=(W/W2) u1=[x1^2-1]^2/x1^2
// case
//fr om Eqn 5.7. 9 , Se c 5. 7. 1.
2
W=Wn22 x2=(W/W2) u2=[x2^2-1]^2/x2^2
// choo sing
//fr om Eqn 5.7. 9 , Se c 5. 7. 1. th e highe r valu e
if u1>u2 then u=u1 else u=u2 end M3=M1*u // mass of main sy st em in kg s K3=K1*u // s t i f f n e s s of mai n system in N/m
//output mprintf ( ’ The mass of main sy st em required
is %4.4 f k g s \n s t i f f n e s s of main sys tem req ire d is %5.5 f N /m\ n If th e re so na nt freque ncies l i e out sid e th e ra ng e of 15 0 to 31 0 rpm th en \n mass of main sy st em is %4.4 f kg s \ n s t i f f n e s s of mai n syste m i s 44
%5. 5 f N/m’ ,M1,K1,M3,K3)
45
Chapter 6 Multi degree of freedom systems exact analysis
Scilab code Exa 6.8.2 system with rolling masses
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 6. 8. 2 \ n ’ ) 4 // given dat a 5 m1=250;m2=100 //m ass of two blocks in Kgs 6 c1=80;c2=60,c=20 // damping c o e f f i c i e n t s i n N −s e c /m 7 F1=1000;F2=1500 // am pl it ud e of force acti ng on bl oc k
1 and 2 rspt ly 8 9 10 11 12 13 14 15 16
k=250000 // s t i f f n e s s of sp ri ng in N/m W=60 // fr equ enc y of appl ied force in r ad/ sec
// calcu latio ns M=[m1,0;0,m2]; K=[k,-k;-k,k]; C=[c+c1,-c;-c,c+c2]; R=[F1;F2;0;0]; X=K-(W^2)*M Y=W*C
17 G=[X,-Y;Y,X]
46
18 19 20 21 22
AB = inv (G ) *R //fr om Eqn6 .8 .4 in Se c 6.8 X1 = sqrt (AB(1,1 )^2 +AB(3,1 )^2) X2 = sqrt (AB(2,1 )^2 +AB(4,1 )^2)
//output mprintf (’T he am pl it ud e of vibrat ions are % fm for ma ss 1 and %fm fo r ma ss 2 ’ ,X1,X2)
47
Chapter 7 Multi degree of freedom systems numerical methods
Scilab code Exa 7.2.1 fundamental frequency by rayleigh method
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 7. 2. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 I=4*10^-7 //m oment of area in m ˆ4 7 M1=100;M2=50 //m ass of discs 1 and 2 in Kgs 8 c=0.18 // dis tanc e of disc 1 from su pp or t in m 9 l=0.3 // dist anc e of disc 2 fro m su pp or t in m 10 g=9.81 // acel erat ion due to gravity in m/ sec ˆ2 11 // calcu latio ns 12 a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I) *(3*l-c)),(l^3/(3*E*I))] // from SOM 13 y1=g*(M1*a(1,1)+M2*a(1,2)) 14 y2=g*(M1*a(2,1)+M2*a(2,2)) 15 Wn = sqrt (g*(M1*y1+M2*y2)/(M1*y1^2+M2*y2^2)) 16 // now to find o ut lo we r natu ral fre que ncy 17 F1=M1*y1*Wn^2
48
18 19 20 21
F2=M2*y2*Wn^2 y1new=F1*a(1,1)+F2*a(1,2) y2new=F1*a(2,1)+F2*a(2,2) Wnnew= sqrt ((F1*y1new+F2*y2new)/(M1*y1new^2+M2*y2new ^2)) // actual natural freq uenc y in ra d/ sec
22 //output 23 mprintf ( ’ The pract
ical natu ral fre que ncy W n is %4.4 f ra d/sec , bu t the lowe r \ n natural frequency W n‘ is %4. 4 f r a d/ se c whi ch is closer to t he act ual \n natural freq uenc y ’ ,Wn,Wnnew)
Scilab code Exa 7.3.1
Dunkerly method
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 7. 3. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 I=4*10^-7 //m oment of area in m ˆ4 7 M1=100;M2=50 //m ass of discs 1 and 2 in Kgs 8 c=0.18 // dis tanc e of disc 1 from su pp or t in m 9 l=0.3 // dist anc e of disc 2 fro m su pp or t in m 10 g=9.81 // acel erat ion due to gravity in m/ sec ˆ2 11 // calcu latio ns 12 a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I) *(3*l-c)),(l^3/(3*E*I))] // from SOM 13 y1=g*M1*a(1,1) // considering onl y M1 to be acting 14 y2=g*M2*a(2,2) // considering onl y M2 to be acting 15 W1 = sqrt (g/y1) 16 W2 = sqrt (g/y2) 17 Wn = sqrt (1/((1/W1^2)+(1/W2^2))) // applying Eqn 7. 3. 7 ,
Sec7 .3 18 //output 19 mprintf (’ The nat ural
fre que ncy 49
of transverse
vibration obt aine d from \ n Dun ke rly metho d is %4 .4 f r a d/ se c whi ch is slig htly lo we r \n tha n th e correct va lu e ’ ,Wn)
Scilab code Exa 7.4.1 Stodola method
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 7. 4. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 I=4*10^-7 //m oment of area in m ˆ4 7 M1=100;M2=50 //m ass of discs 1 and 2 in Kgs 8 c=0.18 // dis tanc e of disc 1 from su pp or t in m 9 l=0.3 // dist anc e of disc 2 fro m su pp or t in m 10 g=9.81 // acel erat ion due to gravity in m/ sec ˆ2 11 // calcu latio ns 12 a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I) *(3*l-c)),(l^3/(3*E*I))] // from SOM 13 x1(1)=1;x2(1)=1 14 for i=1:10 //u p t o 10t h iteration for more perfect
answer 15 16 17 18 19 20 21 22 23 24 25
F1(i)=100*x1(i) // ’ i ’ repr esen ts F2(i)=50*x2(i) x1(i)=F1(i)*a(1,1)+F2(i)*a(1,2) x2(i)=F1(i)*a(2,1)+F2(i)*a(2,2) r=(x2(i)/x1(i)) x2(i+1)=r x1(i+1)=1 end x1dd=1 W1=(x1dd/x1(10)) W2=(r/x2(10))
26 Wn = sqrt ((W1+W2)/2)
// natural
the da sh ( ’)
freq uenc y in ra d/ sec 50
27 mprintf (’T he natural
fre quen cy of sy st em in i i l u s t r a t i v e examp le 7. 2 . 1 obt ain ed by \ nStodala met hod i s Wn=%f rad / s ec ’ ,Wn) 28 mprintf ( ’ \nNOTE: The obtain ed answ er i s mo re near to th e perfect a ns we r \ since 10 it er at io ns / t r i a l s \ nh as be en carried ou t . In te xt boo k onl y up to 3rd iteration
h a s b ee n carrie
d out ’
)
Scilab code Exa 7.7.2 four rotor system
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 7. 7. 2 \ n ’ ) 4 // given dat a 5 J(1)=100 //m oment of in ert ia of f i r s t rotor in Kg −mˆ2 6 J(2)=50 //m oment of iner tia of se co nd rotor in K g −mˆ2 7 J(3)=10 //m oment of iner tia of third rotor in K g −mˆ2 8 J(4)=50 //m oment of iner tia of four th rotor in K g −mˆ2 9 Kt(1)=10^4 // s t i f f n e s s of sha ft be tw ee n 1 and 2 in N −
m/rad 10 Kt(2)=10^4
// s t i f f n e s s of sha ft be tw ee n 2 and 3 in N
m/rad // s t i f f n e s s of sha ft be tw ee n 3 and 4 in N−m/rad To=10000 // am pl it ud e of applied tor que in N −m W =5 // freq uenc y of applied tor que in ra d/ sec // calcu latio ns b(1)=-(0.789*To)/3825 // tw is t of sha ft 1 in r a d
11 Kt(3)=2*10^4 12 13 14 15 16 17 18 19 20
P(1)=J(1)*W^2 Q(1)=P(1)*b(1) R(1)=Q(1) S(1)=R(1)/Kt(1) b(2)=b(1)-S(1)
// twisting
moment of shaft
1 in N
// twi st of shaf t 1 in ra dia ns // tw is t of sha ft 2 in r a d
21 P(2)=J(2)*W^2
51
−m
−
22 23 24 25 26 27
Q(2)=P(2)*b(2) R(2)=Q(1)+Q(2)+To // twisting moment of shaft 2 in N S(2)=R(2)/Kt(2) // twi st of shaf t 2 in ra dia ns b(3)=b(2)-S(2) // tw is t of sha ft 3 in r a d P(3)=J(3)*W^2 Q(3)=P(3)*b(3)
28 29 30 31 32 33 34
R(3)=Q(2)+Q(3) // twisting moment of shaft 3 in N S(3)=R(3)/Kt(3) // twi st of shaf t 3 in ra dia ns b(4)=b(3)-S(3) // tw is t of sha ft 4 in r a d P(4)=J(4)*W^2 Q(4)=P(4)*b(4) R(4)=Q(3)+Q(4) // twisting moment of shaft 4 in N mprintf (’T he am pl it ud es of discs ar e as follows
−m
−m
−m \n di sc 1=%4. 4 f rad \ n di sc2= %4.4 f rad \n di sc 3=%4. 4 f r a d \n di sc4= %4.4 f rad ’ ,b(1),b(2),b(3),b(4)) 35 mprintf ( ’ \ nThe twi sts of sha ft ar e a s foll ows \ nf irs t sh af t=%5. 5 f rad \ nsecond sh af t=%5.5 f rad \ nthird sh af t=%5. 5 f rad ’ ,S(1),S(2),S(3)) 36 mprintf ( ’ \ nThe twisting moments of shaft s are as follows \ n f i r s t sh af t=%5. 5 f N −m\ nsecond sh af t=%5.5 f N−m\ nthird sha ft= %5.5 f N −m ’ ,R(1),R(2),R(3)) 37 mprintf ( ’ \nNOTE: Theaccur sl ig ate ht due to th e more SCILAB ’ )
di ffe ren \ce in values values ncalcul ated
52
are by
Chapter 8 Critical speeds of shafts
Scilab code Exa 8.2.1 rotor mounted midway on shaft
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 8. 2. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 m =5 //m ass of rot or in kg 7 d=0.01 // di a of shaf t in m 8 I=(%pi/64)*d^4 ///m oment of area in m ˆ4 9 l=0.4 // bearing sp an in m 10 e=0.02 // distance of CG a way from geom etr ic centre
of
rotor in m m 11 N=3000 // spe ed of shaft in R PM 12 // calcu latio ns 13 k=48*E*I/l^3 // s t i f f n e s s of sh af t in N/m 14 15 16 17 18
Wn = sqrt (k/m) W=2*%pi*N/60 bet=(W/Wn) r=(bet^2*e/(1-bet^2)) rabs= abs ( r ) // absolute
19 Rd=k*rabs/1000
// total
//f ro m E qn 8.2.2 in S e c 8.2 valu e of disp lace ment dynamic lo ad in bear ings 53
in N (
div ide by 1 0 0 0 since
r is in mm)
20 F=Rd/2 //dy namic loa d on ea ch bearings in N 21 //output 22 mprintf (’ The am pl it ud e of st ea dy state vibr ation
of
shaft is %f mm \nNOTE: neg eti ve si gn sho ws that di sp la ce me nt is o u t of p h a s e w i t h centrifugal force \ nThe dy na mi c for ce tran smtt ed to the bear ings is %4.4 f N \ n Th e dy na mi c load on eac h beari ng is %4.4 f N’ ,r,Rd,F)
Scilab code Exa 8.3.1 disc mounted midway between bearings
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 8. 3. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 m =4 //m ass of rot or in kg 7 g=9.81 // acc due to gravity in m/ sec ˆ2 8 d=0.009 // di a of shaf t in m 9 I=(%pi/64)*d^4 ///m oment of area in m ˆ4 10 l=0.48 // bearing sp an in m 11 e=0.003 // dista nce of CG a way fr om geometr ic centre
of rotor 12 13 14 15 16 17 18 19 20
in mm
N=760 // spe ed of shaf t in R PM c=49 // equivalent viscous damping in N −s e c /m
// calcu latio ns K=48*E*I/l^3 // s t i f f n e s s of sh af t in N/m Wn = sqrt (K/m) W=2*%pi*N/60 bet=(W/Wn) zeta=c/(2* sqrt (K*m)) r=e*(bet^2/ sqrt (((1-bet^2)^2+(2*zeta*bet)^2)))
Eqn 8.3.4
, S e c 8.3 54
//from
//d yna mi c load on bearing in N 22 Fs=m*g // static lo ad in N 23 Fmax=Fd+Fs // maximum st at ic load on the shaft un de r dynamic condition in N 24 smax=(Fmax*l/4)*(d/2)/I //m aximum s t r e s s under 21 Fd = sqrt ((K*r)^2+(c*W*r)^2)
dy na mi c con diti on in N/mˆ2 // maximum st re s s und er de ad load con diti on in N /mˆ2 Fdamp=(c*W*r) //d amping force in N Tdamp=Fdamp*r //d amping torqu e in N −m P=2*%pi*N*Tdamp/60 //pow er in Wa tt s //output mprintf (’ The m amximum st re ss in the sha ft und er dy na mi c con diti on is %.3 f N/( mˆ2) \n The de ad load str ess is %.3 f N/( mˆ2 ) \n T he power required to ,smax, drive the shaft at 76 0 RPM is %4.4 f Watts ’
25 ss=(Fs*l/4)*(d/2)/I 26 27 28 29 30
ss,P)
Scilab code Exa 8.4.1 two critical speeds
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 8. 4. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 I=4*10^-7 //m oment of area in m ˆ4 7 M1=100;M2=50 //m ass of discs 1 and 2 in Kgs 8 c=0.18 // dis tanc e of disc 1 from su pp or t in m 9 l=0.3 // dist anc e of disc 2 fro m su pp or t in m 10 g=9.81 // acel erat ion due to gravity in m/ sec ˆ2 11 // calcu latio ns 12 a=[(c^3/(3*E*I)),(c^2/(6*E*I)*(3*l-c));(c^2/(6*E*I)
*(3*l-c)),(l^3/(3*E*I))]
// from SOM 55
13 p=M1*a(1,1)+M2*a(2,2) //fr om E qn 8.4 .6 , Se c 8.4 14 q=M1*M2*(a(1,1)*a(2,2)-(a(1,2)^2)) //f ro m Eqn 8.4.6
,
Se c 8.4 15 Wn1= sqrt ((p- sqrt (p^2-4*q))/(2*q))
//f ro m E qn 8.4.6
,
//f ro m E qn 8.4.6
,
Se c 8.4 16 Wn2= sqrt ((p+ sqrt (p^2-4*q))/(2*q)) 17 18 19 20
Se c 8.4
Nc1=Wn1*60/(2*%pi) Nc2=Wn2*60/(2*%pi)
// c r i t i c a l sp ee d in RPM // c r i t i c a l sp ee d in RPM
//output mprintf ( ’ The c r i t i c a l speeds
for the sys te m shown in fi g 7. 2. 1 are %4.4 f RPM an d %4.4 f RPM’ ,Nc1,Nc2
)
Scilab code Exa 8.6.1 right cantilever steel shaft with rotor at the end
1 clc 2 clear 3 mprintf ( ’ Mec hanic al vi br at io ns by G. K. Gr ov er \ n Ex am pl e 8. 6. 1 \ n ’ ) 4 // given dat a 5 E=1.96*10^11 //youngs mod ulu s in N/mˆ2 6 M=10 //m ass of rot or in kg 7 g=9.81 // acc due to gravity in m/ sec ˆ2 8 ra=0.12 // radi us of gyr atio n in m 9 l=0.3 // le ng ht of steel sha ft in m 10 b=0.06 // thic knes s of rot or in m 11 I=10*10^-8 //m oment of ine rtia of section in mˆ4 12 // calcu latio ns 13 r = sqrt ((ra^2/2)+(b^2/12)) 14 h=3*(r^2)/l^2 //fr om E qn 8.6. 4 , Se c 8.6 15 g1 = sqrt ((2/h)*((h+1)sqrt ((h+1)^2-h))) // nat ura l
fre quen cy , fr om Eqn 8.6 .4 16 g2 = sqrt ((2/h)*((h+1)+
, Se c 8.6
sqrt ((h+1)^2-h)))
fre quen cy , fr om Eqn 8.6 .4
, Se c 8.6 56
// nat ura l
17 18 19 20 21 22
W1=g1* sqrt (3*E*I/(M*l^3)) //fr om Eqn W2=g2* sqrt (3*E*I/(M*l^3)) //fr om Eqn Nc1=W1*60/(2*%pi) // c r i t i c a l sp ee d i n Nc2=W2*60/(2*%pi) // c r i t i c a l sp ee d i n
8.6. 4 , Se c 8.6 8.6. 4 , Se c 8.6 RPM RPM
//output mprintf ( ’ The operatin
g sp ee d of 10 00 0 RPM is no t
n e a r t o eith er of \n the c r i t i c a l speeds i . e %4.4 f RPM or %4 . 4 f RPM. \ n The re for e th e oper atin g sp e e d is safe . ’ ,Nc1,Nc2)
57
