1

Lecture 11: Resonance phenomena:

parametric resonance

P7.1 A rigid wire in a vertical plane is forced to oscillate up and down so that the y-coordinate of the symmetry point on the wire is given by

!

y0(t) = Lsin(t) . At

!

t = 0 the equation for the shape of the wire is given by

!

y = "x2

2+x4

4 .

Is it possible to balance a bead of mass

!

m close to the symmetry point in a stable way? Illustrate this motion. Adjust the gravitational field strength

!

g if necessary.

2

S7.1----------SOPHIA---------------- Balancing a bead on a double-well wire > read SophiaV6; Sophia for Maple R6 - 13 june 2000, based on version 2 December \ 1998 • Coordinates and frames > dependsTime(q1,u1): > rotList:=[[FN,FA,3,0]]: > chainSimpRot( rotList): Frame relation between FN and FA defined! > &kde 1; Generalized Velocity Form of Kinematic d.e. Saved as the GLOBAL Varable kde {q1t = u1} • Positions and velocities > rp:= (FN &ev [0,L*sin(nu*t),0]) &++ (FA &ev [q1,q1^4/4-q1^2/2,0]);

:= rp

!

"##

$

%&&,

!

"##

$

%&&, ,q1 + ' L ( )sin t

1

4q1

4 1

2q1

20 FA

> vp:=&simp subs(kde,(FN &fdt rp));

:= vp [ ],[ ], ,u1 + ! L ( )cos t q13u1 q1 u1 0 FA

3

• Momenta and forces > pp:= m&** vp: > Rp:=FN &ev [0,-m*g,0]: • K-vector velocity and tangent directions > vK:= &KM[vp]: > tauK:= KMtangents(vK,u,1);

:= tauK [ ][ ],[ ],[ ], ,1 ! q13

q1 0 FA 1 • K-vectors and projected equations of motion > pK:= &KM[pp]: > RK:= &KM[Rp]: > pKt:=&Ksimp (FN &Kfdt pK); > MGIF:=tauK &kane pKt; > GAF:= tauK &kane RK; MGIF m u1t ( ) ! q1

3q1 m L ( )sin t( ! [ :=

3 q12u1 q1t q1

3u1t q1t u1 q1 u1t ! ! + + )]

:= GAF [ ]!( ) ! q13

q1 m g > KaneEquations:= {MGIF[1]=GAF[1]}: • Maple set of equations > stateEquations:=KaneEquations union kde: > state:= solve(stateEquations,{q1t,u1t}): > stateFform:=simplify(subs(toTimeFunction,state));

4

• Linear approximation > MGIFapp:=simplify(mtaylor(MGIF[1],[q1,u1],2)); > GAFapp:=simplify(mtaylor(GAF[1],[q1,u1],2)); > AStateEqn:=solve(MGIFapp=GAFapp,{u1t}); := AStateEqn { } = u1t ! + q1 L ( )sin t g q1 •Compare with Mathieu equation in the course MMM

!

˙ ̇ y + " + #cos t( )( )y = 0 We may shift the time origin to identify our linearised equation with the Mathieu eq. Our parameters are

!

" = #g and

!

" = L. The Mathieu map shows the regions for

!

" and

!

" where motion is stable/unstable.

From this figure we may try some values in the full equations and see if it works.

5

• Numerical and graphical > with(plots,odeplot); > param:={m=1,g=0.1,L=0.4}; > statep:= subs(param,stateFform); > initc:={q1(0)=0.0001,u1(0)=0}; > deqns:=statep union initc: > st:=dsolve(deqns,{q1(t),u1(t)},type=numeric,output=procedurelist); := st proc( ) ... end procrkf45_x > st(0);st(10); > odeplot(st,[q1(t),u1(t)],0..100,numpoints=800);

•This is not a stable behaviour near the symmetry point q1=0.

6

g=0.1 and L=0.6

• This is a stable behaviour. -----------------

7

Externally forced resonance

P7.2 A three-body building are subject to a horizontal vibration of the basement given by x( t) = x0 sin!t , t ! 0 . The frictionless torque model is M = !k" , where k is the bending stiffness and ! is the relative angular deflection between neighbouring blocks of the building. We assume the same stiffness between the blocks as well as between the lowest block and the basement (which is performing the horizontal vibrations together with the earth). Each body segment is a rectangular surface of length L , width S and homogeneous mass m .

8

S7.2-Reading SOPHIA > #restart; > read SophiaV6;

Frames and kinematic differential equations > set2:={L2=L,L1=L,L3=L}: > &rot[FN,FA,3,q1]: &rot[FN,FB,3,q2]: &rot[FN,FC,3,q3]: > &kde 3;

Inertias and Force Description >IA:=EinertiaDyad(IA1,IA2,JA,0,0,0,FA); IB:=EinertiaDyad(IBC1,IBC2,JBC,0,0,0,FB);IC:=EinertiaDyad(IBC1,IBC2,JBC,0,0,0,FC):inertias:={IA1=m1*s^2/12,IA2=m1*(L1^2)/12,IBC1=(m2)*s^2/12,IBC2=(m2)*((L2)^2)/12}:Jnertias:=subs(set2,subs(inertias,{JA=IA1+IA2,JBC=IBC1+IBC2}));

Jnertias :=

{ }, = JBC + 1

12m2 s

2 1

12m2 L

2 = JA +

1

12m1 s

2 1

12m1 L

2

9

> Ta:=FN &ev [0,0,-k*q1+k*(q2-q1)]; Tb:=FN &ev [0,0,-k*(q2-q1)+k*(q3-q2)]; Tc:=FN &ev [0,0,-k*(q3-q2)]; Ra:=FN &ev [0,-m1*g,0]; Rb:=FN &ev [0,-m2*g,0]; Rc:=FN &ev [0,-m3*g,0]; A substitution that removes the fourth segment. > set3:={m1=m,m2=m,m3=m}:

Positions and velocities. The basis point is subject to a horizontal vibration, with a displacement amplitude x0. > r0:=FN &ev [x0*sin(omega*t),0,0]: rhinge1:=r0 &++ (FA &ev [0,L1,0]): rhinge2:=rhinge1 &++ (FB &ev [0,L2,0]): rha:=FA &ev [0,-L1/2,0]: rhb:=FB &ev [0,L2/2,0]: rhc:=FC &ev [0,L3/2,0]: r1:=rhinge1 &++ rha: r2:=rhinge1 &++ rhb: r3:=rhinge2 &++ rhc: v1:=FN &to subs(kde,FN &fdt r1): v2:=FN &to subs(kde,FN &fdt r2): v3:=FN &to subs(kde,FN &fdt r3): w1:=subs(kde,FN &aV FA): w2:=subs(kde,FN &aV FB):w3:=subs(kde,FN &aV FC):

10

Kvelocity and Tangent direction > vK:= subs(set2,&KM [v1,w1,v2,w2,v3,w3]): > tauK:=KMtangents(vK,u,3);

tauK!"##

$%&&,

!"##

$%&&, ,'

1

2( )cos q1 L '

1

2( )sin q1 L 0 FN [ ],[ ], ,0 0 1 FA, ,

!"##!"## :=

[ ],[ ], ,' ( )cos q1 L ' ( )sin q1 L 0 FN [ ],[ ], ,0 0 0 FB, ,

[ ],[ ], ,' ( )cos q1 L ' ( )sin q1 L 0 FN [ ],[ ], ,0 0 0 FC 6, ,$%&& [ ],[ ], ,0 0 0 FN ,!"##,

[ ],[ ], ,0 0 0 FA!"##

$%&&,

!"##

$%&&, ,'

1

2( )cos q2 L '

1

2( )sin q2 L 0 FN [ ],[ ], ,0 0 1 FB, , ,

[ ],[ ], ,' ( )cos q2 L ' ( )sin q2 L 0 FN [ ],[ ], ,0 0 0 FC 6, ,$

%&& [ ],[ ], ,0 0 0 FN ,!

"##,

[ ],[ ], ,0 0 0 FA [ ],[ ], ,0 0 0 FN [ ],[ ], ,0 0 0 FB, , ,

!

"##

$

%&&,

!

"##

$

%&&, ,'

1

2( )cos q3 L '

1

2( )sin q3 L 0 FN [ ],[ ], ,0 0 1 FC 6, ,

$

%&&$

%&&

Equations of motion > p1:=m1 &** v1: p2:=(m2) &** v2: p3:=(m3) &** v3; h1:=IA &o w1: h2:=IB &o w2: h3:=IC &o w3;

p3 m31

2( )cos q3 u3 L3 ( )cos q2 u2 L2 ( )cos q1 u1 L1! ! !

"#$$

%&''%&'' :=

x0 ( )cos ( t ( + )*++,

m3"

#$$

)

*++! ! !

1

2( )sin q3 u3 L3 ( )sin q2 u2 L2 ( )sin q1 u1 L1 0,

,

-.. FN,

,

-..

:= h3 [ ],[ ], ,0 0 JBC u3 FC

11

> pK:= subs(set2,subs(0=0,&KM [p1,h1,p2,h2,p3,h3])): pKt:=simplify( subs(kde, FN &Kfdt pK)): RK:=simplify(subs(set2,&KM [Ra,Ta,Rb,Tb,Rc,Tc])):

Two projected equations (Kane's equations) > MGIF:=subs(Jnertias,tauK &kane pKt): GAF:=tauK &kane RK;

GAF1

2( )sin q1 L m1 g 2 k q1 k q2 ( )sin q1 L m2 g ! + +

"#$$ :=

( )sin q1 L m3 g + ,

! + + + 1

2( )sin q2 L m2 g 2 k q2 k q1 k q3 ( )sin q2 L m3 g,

+ 1

2( )sin q3 L m3 g k ( )! + q3 q2

%

&''

Implict equation > KaneEqn:= {seq(simplify(MGIF[j])=simplify(GAF[j]),j=1..3)}: Explicit equation >dynStateEqn3:=solve(subs(set3,KaneEqn),{u1t,u2t,u3t}): >CDS:=collect(combine(subs(dynStateEqn3,u3t),trig),[cos,sin]):

12

Possible simplifying approximation >dSE:=[seq(collect(u||i||t=subs(dynStateEqn3,u||i||t),q||i),i=1..3)]: > for i from 1 to 3 do AS||i:=lhs(dSE[i])=simplify(mtaylor(rhs(dSE[i]),[q1,q2,q3,u1,u2,u3],2)) od: >ASeqns:=[seq(collect(AS||i,[sin(omega*t),q||i]),i=1..3)]: > ASeqns[1]; >AdynStateEqn3:=collect(convert(ASeqns,set),[q1,q2,q3]);

>param:={L=2.0,omega=1,x0=0,s=2,k=1000,g=10,m=2}: sase:=subs(param,AdynStateEqn3): lase:=subs(sase,[u1t,u2t,u3t]); Q1:={q2=0,q3=0}:Q2:={q1=0,q3=0}:Q3:={q2=0,q1=0}:

lase ! + ! 454.9484536 q1 494.1030928 q2 185.1958763 q3,[ :=

! + 654.4329894 q1 1025.134021 q2 537.1546392 q3,

! + ! 239.3814433 q1 937.2371136 q2 716.3505156 q3]

13

> for i from 1 to 3 do M||i:=[seq(subs(Q||j,lase[i])/q||j,j=1..3)] od; KM:=simplify(matrix(3,3,[-M1,-M2,-M3]));

:= KM

!

"

#######

$

%

&&&&&&&

454.9484536 -494.1030928 185.1958763

-654.4329894 1025.134021 -537.1546392

239.3814433 -937.2371136 716.3505156 Finding eigenvalues of coupled oscillators. The square root of them are the frequences of the eigen-modes. > eig3:=map(sqrt,evalf(Eigenvals(KM)));

:= eig3 [ ], ,42.91948266 18.61086513 2.826074453 Note: If s=1

:= eig3 [ ], ,52.20524613 20.32794770 2.861460428 To hit a resonance, the forcing should be near these frequences. IF JUST AT LEAST ONE EIGENVALUE IS IMAGINARY the solutions will be unstable anyway!!!.!

Numerical and graphical illustrations > StateForm3:=subs(toTimeFunction,dynStateEqn3 union kde): > with(plots,odeplot):

14

Parameters >param:={L=2,omega=(2.82389),x0=0.01,s=2,k=1000,g=10,m=2}; statep3:= subs(param,StateForm3): initc:={q1(0)=0.0,q2(0)=0.0,q3(0)=0.0,u1(0)=0,u2(0)=0,u3(0)=0};

> deqns3:=statep3 union initc: st3:=dsolve(deqns3,{q1(t),q2(t),q3(t),u1(t),u2(t),u3(t)},type=numeric,output=procedurelist); > odeplot(st3,[t,q1(t)],0..10);

15

16

Lecture 12: Analysis of inertia

P7.3 Use SOPHIA to calculate the centre of mass inertia dyad of a rigid body composed of two massive rods forming an angle ! . Introduce the combined centre of mass reference frame and display the combined inertia dyad for the composed rigid body. Use SOPHIA to calculate the centre of mass inertia dyad of a rigid body composed of two massive rods forming an angle ! . Introduce the combined centre of mass reference frame and display the combined inertia dyad for the composed rigid body.

L

b1

m

M

l

c1

S7.3 #We start with the two triads indicated in the figure. > restart; readlib(trigsubs): with (linalg): read `SophiaV6`; > &rot [FB,FC,3,alpha]:

Frame relation between FB and FC defined!

#Each single centre of mass position > rOB:=FB &ev [L/2,0,0]:

17

rOC:=FC &ev [l/2,0,0]: #Simplification > l:=L; M:=m;

:= l L

:= M m

#The combined centre of mass position uses a SOPHIA command taking the original centres of mass as arguments > R0D:= FB &cm [[M, rOB],[m,rOC]];

:= R0D

!"##

$%&&,

!"##

$%&&+

1

4L

1

4( )cos ' L

1

4( )sin ' L 0 FB

#The body-fixed inertia dyads, Symmetric of course. > IB11:=0; IB22:=M*L^2/12;IB33:=M*L^2/12; IB12:=0: IB13:=0: IB23:=0: IDB:= EinertiaDyad(IB11,IB22,IB33,IB12,IB13,IB23,FB):

:= IB11 0

:= IB22

1

12m L2

:= IB33

1

12m L2

> IC11:=0; IC22:=m*l^2/12;IC33:=m*l^2/12; IC12:=0: IC13:=0: IC23:=0: IDC:= EinertiaDyad(IC11,IC22,IC33,IC12,IC13,IC23,FC):

:= IC11 0

:= IC22

1

12m L2

18

:= IC33

1

12m L2

#Displacement of the new combined centre of mass >dBD:= R0D &-- rOB; dCD:= R0D &-- rOC;

:= dBD

!"##

$%&&,

!"##

$%&&' +

1

4L

1

4( )cos ( L

1

4( )sin ( L 0 FB

dCD + + !

1

4( )cos " L

1

4( )cos " 2 L

1

4( )sin " 2 L

1

2l

#$%%#$%% :=

!1

4( )sin " L 0

&'(( FC,

&'((

#Move the reference points of each rod to the combined centre. Each inertia dyad will change according to the formula:

!

IO=

ICM

+m rCMB 2

U ! rCMBrCMB"

# $

% & '

to the new origin. > IDBD:=([M,IDB] &->dBD): IDCD:=([m,IDC] &-> dCD): #Here we add the dyads and obtain the total inertia dyad > IDD:= &simp(IDBD &++ IDCD):

19

#Lets look at it in the particular case

!

" = # . This is a new rod, twice as long and with twice the mass. > IDD1:= subs({alpha=Pi},IDD); IDD1

!"## :=

$524

m L2 524

m ( )cos % 2 L2!"## ,

$ +18m ( )sin % L2 5

24( )cos % m ( )sin % L2 0,

&'((

$ +18m ( )sin % L2 5

24( )cos % m ( )sin % L2!

"## ,

+ $524

m L2 524

m ( )cos % 2 L2 14m ( )cos % L2 0,

&'((

!"## , ,

&'((0 0 $

512

m L2 14m ( )cos % L2 FC,

&'((

#We need simplifications here, extracting the matrix itself >MIDD1:=simplify(vPart IDD1);

:= MIDD1

!

"

########

$

%

&&&&&&&&

0 0 0

02

3m L2 0

0 02

3m L2

20

#Lets look at a less trivial system with ! = " / 2 . > IDD2:= subs({alpha=Pi/2},IDD): MIDD2:=simplify( &vPart IDD2);

:= MIDD2

!

"

#########

$

%

&&&&&&&&&

5

24m L2 '

1

8m L2 0

'1

8m L2 5

24m L2 0

0 05

12m L2

#Among the eigenvalues we find the maximal and minimal moments of inertia. > Egen1:=[eigenvals(MIDD1)]; Egen2:=[eigenvals(MIDD2)];

:= Egen1

!"##

$%&&, , 0

2

3m L2 2

3m L2

:= Egen2

!"##

$%&&, ,

5

12m L2 1

12m L2 1

3m L2

#The eigenvectors tell us the directions of pricipal axes > Vekt1:= eigenvects(MIDD1); Vekt2:=eigenvects(MIDD2);

21

Vekt1!"##

$%&&, ,

2

3m L2 2 { }, [ ]0 0 1 [ ]0 1 0 , :=

[ ], , 0 1 { }[ ]1 0 0

Vekt2!"##

$%&&, ,

512

m L2 1 { }[ ]0 0 1 , :=

!"##

$%&&, ,

13m L2 1 { }[ ]-1 1 0 ,

!"##

$%&&, ,

112

m L2 1 { }[ ]1 1 0

#Selecting the pure vectors > Vekt2[1][3][1]; Vekt2[2][3][1]; Vekt2[3][3][1];

[ ]0 0 1

[ ]-1 1 0

[ ]1 1 0

#Selecting a new triad, with the first direction d1 along the minimal axis. The #relevant angle is calculated > beta:=angle(Vekt2[3][3][],vector([1, 0 ,0])) ;

:= !

1

4"

#Defining the new center of mass frame > &rot[FC,FD,3,beta]:

Frame relation between FC and FD defined!

#In the new rotated frame we find the combined inertia dyad: > IDD2:= FD &to subs({alpha=Pi/2},IDD);

:= IDD2

!

"

#########

$

%

&&&&&&&&&

,

!

"

#########

$

%

&&&&&&&&&

1

12m L2 0 0

01

3m L2 0

0 05

12m L2

FD

22

Statics - equilibria P7.2 (compare ML: Illustration p235) A force screw is defined as the equipollent force system consisting of the combined applied torque with respect to an arbitrary point and the total force, i.e.

TA ,R( ) , where

A is an arbitrary point. Apply SOPHIA's screw commands to calculate the reaction forces on the tetrahedron truss of side length L loaded vertically at the point F by R . The floor is smooth.

23

S7.2 Let us look at the geometry. Floor Geometry: Face orientation:

Floor Geometry: Consider two similar triangles with the same top angle ! / 6 .

We get :

L / 2

L=

x

y together with

x + y = 3L / 2 : ! y =

L

3.

Let the origin be the midpoint between A and B with a1 in the C-direction and a3 pointing vertically upwards. Face orientation: The AB face is seen to be rotated an

angle arcsin 1

3

!

" # $

% & from the vertical direction. Let b3 be the

leaning direction of this face.

24

-------------------SOPHIA------------------------------ > restart; readlib(trigsubs): read `KET-PM:Program & tillbehör:MapleV R3:sophia21_3`; > &rot [A,B,2,,arcsin(1/3)]: #Important points of application > rOA:=A &ev [0,L/2,0]: rOB:=A &ev [0,-L/2,0]: rOC:=A &ev [sqrt(3)*L/2,0,0]: rOF:=B &ev [0,0,S]: rzero:= A &ev [0,0,0]: #Reaction forces at smooth surface contact > RA:= A &ev [0,0,RA3]: RB:= A &ev [0,0,RB3]: RC:= A &ev [0,0,RC3]: #Load at point F > Rw:= A &ev [0,0,-w]: #The equipollent screw of 'applied forces' load+ reaction. At a point A the reduced screw system is SA=(MA,R), where MA is the measured torque and R is the total applied force. We measure the equipollent screw from the point A, where RA is applied. > rAB:= rOB &-- rOA; rAC:= rOC &-- rOA; rAF:= rOF &-- rOA;

:= rAB [ ], [ ]0 !L 0 A

:= rAC

!"##

$%&&,

!"##

$%&&

1

23 L '

1

2L 0 A

:= rAF

!"##

$%&&,

!"##

$%&&

1

3S '

1

2L

2

32 S A

25

#To obtain the screw system we feed in the relative vectors (to point A) to respective forces. > TAscrew:= A &screw [[rAB,RB],[rAC,RC],[rAF,Rw],[rzero,RA]];

TAscrew ! !1

2L w

1

2L RC3 L RB3

"#$$"#$$"#$$ :=

!1

3S w

1

23 L RC3 0

%&'' A,

%&'',

[ ], [ ]0 0 ! + +RA3 w RC3 RB3 A%&''

#Equations are formed and solved: > Tex:= TAscrew[1]; Rex:= TAscrew[2];

Tex ! !1

2L w

1

2L RC3 L RB3

"#$$"#$$ :=

!1

3S w

1

23 L RC3 0

%&'' A,

%&''

:= Rex [ ], [ ]0 0 ! + +RA3 w RC3 RB3 A

> reactions:= solve({ Tex &c 2, Tex &c 1, Rex &c 3}, { RC3, RB3, RA3});

reactions =RA3 !1

18

w ( )! +9 L 2 3 S

L,

"

#$%% :=

=RC32

9

S w 3

L,

=RB3 !1

18

w ( )! +9 L 2 3 S

L

&

'(%%

#Simplification > expand(");

=RC32

9

S w 3

L=RB3 !

1

2w

1

9

S w 3

L, ,

"

#$%%

=RA3 !1

2w

1

9

S w 3

L

&

'(%%

#Testing the reaction forces as the load is shifted to the symmetric point at the top. > subs(S=sqrt(3)*L/2,reactions):