String Vibration

7/25/2019 String Vibration

1/41

String Vibration

Chapter 8


2/41

Distributed Parameter Systems

Distributed mass and stiffness

Infinite DOF

Functional analysis

Exact solution for simple problems


3/41

f(x, t)(x),T(x)

y(x, t)

f(x, t)dx

T(x)

T(x) + T(x)x

dx

y(x,t)x

y(x,t)x +

2

y(x,t)x2 dx(x)dx

dx


4/41

T(x) +

T(x)

x dx

y(x, t)

x +

2y(x, t)

x2 dx+

f(x, t)dx T(x)y(x, t)x

=(x)dx2y(x, t)

t2

f(x, t)dx

T(x)

T(x) + T(x)x

dx

y(x,t)

x

y(x,t)x

+ 2y(x,t)x2

dx(x)dx

dx


5/41

T(x)

2y(x, t)

x2 +

T(x)

x

y(x, t)

x + f(x, t) =(x)

2y(x, t)

t2

x

T(x)y(x, t)

x

+ f(x, t) =(x)2y(x, t)

t2

y(0, t) = 0y(L, t) = 0


6/41

Initial (unforced) Response

x

T(x)

y(x, t)

x

=(x)

2y(x, t)

t2

y(0, t) = 0 y(L, t) = 0

y(x, t) =Y(x)F(t)

ddx

T(x) dY(x)

dx F(t)

=(x)Y(x) d

2F(t)dt2

Y(0) = 0 Y(L) = 0


7/41

1

(x)Y(x)

d

dxT(x)

dY(x)

dx= 1

F(t)

d2F(t)

dt2 =

2

d2F(t)

dt2 + 2F(t) = 0

F(t) =A sint + B cost=Ccos(t + )

ddx

T(x) dY(x)dx

= 2(x)Y(x)

Y(0) = 0 Y(L) = 0


8/41

d

dxT(x) dY(x)

dx= 2(x)Y(x)

Y(0) = 0 Y(L) = 0

For constant mass distribution and tension

d2Y(x)

dx2 +

2

T Y(x) = 0

d2Y(x)

dx2 + 2Y(x) =0 2 =2

TY(0) = 0 Y(L) = 0


9/41

d2Y(x)

dx2 + 2Y(x) =0

Y(0) = 0 Y(L) = 0

Solution : Y(x) =A sinx + B cosxBCs : Y(0) =B= 0 Y(L) =A sinL= 0

rL=r, r= 1, 2,

r =rs

TL2

, r = 1, 2,

Yr(x) =Arsinrx

L , r = 1, 2,


10/41

Free Vibration Solutions

Possible Solutions:y(x, t) =Y(x)F(t)

yr(x, t) =Yr(x)Fr(t), r= 1, 2,

=CrsinrxL cosrs T

L2t + r!

y(x, t) =

Xr=1 yr(x, t)

=

Xr=1 C

rYr(x)cos(rt + r)


11/41

Orthogonality

d

dxT(x) dYr(x)

dx=2r(x)Yr(x)

Ys(x)d

dx

T(x)dY

r(x)

dx

=2r(x)Ys(x)Yr(x)

Z L0

Ys(x) d

dx

T(x)

dYr(x)

dx

dx=2r

Z L0

(x)Ys(x)Yr(x)dx


12/41

Z L0 Y

s(x)

d

dx

T(x)

dYr(x)

dx

dx

= Ys(x)T(x)dYr(x)

dx

L

0 +Z L0 T(x)

dYs(x)

dx

dYr(x)

dx dx

=Z L0 T(x)

dYs(x)

dx

dYr(x)

dx dx

Z L

0

T(x)dYs(x)

dx

dYr(x)

dx dx=2r Z

L

0

(x)Ys(x)Yr(x)dx


13/41

Z L

0

T(x)dYs(x)

dx

dYr(x)

dx dx=2r Z

L

0

(x)Ys(x)Yr(x)dx

Z L

0

T(x)dYr(x)

dx

dYs(x)

dx

dx=2s Z L

0

(x)Yr(x)Ys(x)dx

0 = (

2

r 2

s)Z L0 (x)Yr(x)Ys(x)dx

0 =Z L0

(x)Yr(x)Ys(x)dx

0 =Z L0 T(x)

dYr(x)

dx

dYs(x)

dx dx


14/41

Normalization

Z L

0

T(x) dYr(x)

dx2

dx=2r Z L

0

(x) [Yr(x)]2 dx

Z L0

T(x)dYr(x)

dx2

dx=2r

Select Yr(x) such that :Z L0 (x) [Yr(x)]

2

dx= 1


15/41

Modal Expansion

Eigenvectors:

Linearly independent Complete set

Any continuous displacement distributioncan be represented as a linear

combination of the modal shapes

y(x, t) =

Xr=1

Yr(x)r(t)


16/41

Initial (unforced) Response

x

T(x)

y(x, t)

x

=(x)

2y(x, t)

t2

y(0, t) = 0 y(L, t) = 0

y(x, 0) =y0(x) y(x, 0) =v0(x)

Xr=1d

dx

T(x)

dYr(x)

dx r(t) =

Xr=1(x)Yr(x)

d2r(t)

dt2

BCs Identically Satisfied :Y(0) 0 Y(L) 0

y(x, t) =

Xr=1Yr(x)r(t)

Yr(x) are mass

normalized

mode shapes


17/41

Uncoupled Modal EquationsZ L0

Ys(x)

Xr=1d

dx

T(x)

dYr(x)

dx

r(t)dx=

Z L0

Ys(x)

Xr=1(x)Yr(x)r(t)dx

Xr=1

Z L0

Ys(x) d

dx

T(x)

dYr(x)

dx

dxr(t) =

Xr=1

Z L0

(x)Ys(x)Yr(x)dxr(t)

2

ss(t) = s(t)

s(t) + 2

ss(t) = 0

s(t) =s(0) cosst+

s(0)

s sin st


18/41

Initial Conditions

y(x, 0) =y0(x) y(x, 0) =v0(x)

y(x, 0) =

Xr=1

Yr(x)r(0) =y0(x)(0)

Z L0 (x)Ys(x)

Xr=1

Yr(x)r(0)dx=Z L0 (x)Ys(x)y0(x)dx

s(0) =Z L0

(x)Ys(x)y0(x)dx

s(0) =Z L0

(x)Ys(x)v0(x)dx

Do not forget:

Yr(x) are mass

normalized

mode shapes


19/41

Example

= 1 Kg/m T = 1 N L= 1 m

Initial Displacement :y0(x)

0.25

0.1

Initial Velocity :v0(x)


20/41

1= rad/s

2= 2 rad/s

3= 3 rad/s

4= 4 rad/s

5= 5 rad/s


21/41

Normalized Mode Shapes

& Initial Condition

Yr(x) = 2sinrx

L

1(0)2(0)3(0)

4(0)5(0)

=

0.05400.01910.0060

0.00000.0022


22/41

0 0.2 0.4 0.6 0.8 1-0.05

0

0.05

0.1

0.15

x

y0

(x)

Initial Displacement

Modal ApproximationMode 1

Mode 2

Mode 3

Mode 4

Mode 5


23/41

0 0.2 0.4 0.6 0.8 1

-0.1

-0.05

0

0.05

0.1

0.15

x

y(x,

t)

Exact

t = 0.00 st = 0.25 s

t = 0.50 s

t = 0.75 s

t = 1.00 s

F d R


24/41

Forced Response

(x)2y(x, t)

t2

x

T(x)

y(x, t)x

=f(x, t)

y(0, t) = 0 y(L, t) = 0y(x, 0) =y0(x) y(x, 0) =v0(x)

y(x, t) =

Xr=1Yr(x)r(t)

Yr(x) are mass

normalized

mode shapes

Xr=1(x)Yr(x)

d2r(t)

dt2

Xr=1d

dx

T(x)

dYr(x)

dx r(t) =f(x, t)

BCs Identically Satisfied :Y(0) 0 Y(L) 0


25/41

Uncoupled Modal Equations

Multiply by Ys(x)

Integrate over 0 to L

Z L

0

Ys(x)

Xr=1(x)Yr(x)r(t)dx Z

L

0

Ys(x)

Xr=1d

dx

T(x)

dYr(x)

dx r(t)dx

=

Z L0

Ys(x)f(x, t)dx

s(t) + 2ss(t) =Ns(t)

Ns(t) =Z L0

Ys(x)f(x, t)dx

Do not forget:

Yr(x) are massnormalized

mode shapes

I iti l C diti


26/41

Initial Conditions

y(x, 0) =y0(x) y(x, 0) =v0(x)

y(x, 0) =

Xr=1

Yr(x)r(0) =y0(x)(0)

Z L0 (x)Ys(x)

Xr=1

Yr(x)r(0)dx=Z L0 (x)Ys(x)y0(x)dx

s(0) =Z L0

(x)Ys(x)y0(x)dx

s(0) =Z L0

(x)Ys(x)v0(x)dx

Do not forget:

Yr(x) are mass

normalized

mode shapes

St d St t H i R


27/41

Steady-State Harmonic Response

f(x, t) =f(x)cost

s(t) =

Ns

2s 2 cost

Ns(t) =Nscost where Ns=Z L0 Ys(x)f(x)dx

y(x, t) =y(x)cost=

Xr=1

Yr(x)r(t)

y(x) =

Xr=1 Y

r(x)

Nr

2r 2 =

Xr=1

Yr(x)

2r 2Z L0 Yr(x)f(x)dx

P i t F


28/41

Point Force

Force at a point a

Displacement at a point b

Reciprocal Theorem

F=F cos(t)

a b

Ns= Z L

0

Ys(x)F(a)dx=F Ys(a)

y(b) =FXr=1

Yr(b)Yr(a)2r 2


29/41

Example

= 1 Kg/m T = 1 N L= 1 m F= 1 N

Yr(x) =2sinrx

L

a = 0.5; b = 0.5


30/41

0 10 20 30-1

-0.5

0

0.5

1

y(b)

;

a = 0.5; b = 0.33333


31/41

0 10 20 30-1

-0.5

0

0.5

1

y(b)

a = 0.1; b = 0.1


32/41

0 10 20 30-1

-0.5

0

0.5

1

y(b)


33/41

Similar Problem: Axial Vibrationsf(x, t)

m(x),EA(x) u(x, t)

f(x, t)dxP(x, t) P(x, t) + P(x,t)x

dxm(x)dx

P(x, t) =A=EA=EAu(x,t)x


34/41

P(x, t) +P(x, t)

x

dx+ f(x, t)dx P(x, t) =m(x)dx2u(x, t)

t2

P(x, t)

x + f(x, t) =m(x)

2u(x, t)

t2

u(0, t) = 0 P(L, t) = EA(x)u(x, t)x

x=L

= 0

x

EA(x)u(x, t)

x

+ f(x, t) =m(x)

2u(x, t)t2


35/41

Free Vibration

u(0, t) = 0 P(L, t) = EA(x)u(x, t)

x

x=L

x

EA(x)

u(x, t)

x =m(x)

2u(x, t)

t2

u(x, t) =U(x)F(t)

d

dx

EA(x)dU(x)

dx F(t)

=m(x)U(x)d2F(t)

dt2

U(0) = 0

dU(x)

dxx=L = 0

1 d dU( ) 1 d2F (t)


36/41

1

m(x)U(x)

d

dxEA(x)

dU(x)

dx= 1

F(t)

d2F(t)

dt2

=2

d2F(t)

dt2 + 2

F(t) = 0 F(t) =A sint + B cost=Ccos(t + )

d

dxEA(x) dU(x)

dx= 2m(x)U(x)

U(0) = 0 dU(x)

dxx=L

= 0

d

EA(x)dU(x)

2m(x)U(x)


37/41

For constant mass distribution and cross-sectional stiffness

dx

EA(x)

dx = 2m(x)U(x)

U(0) = 0 dU(x)

dx

x=L

= 0

d

2

U(x)dx2 + 2

mEAU(x) = 0

d2

U(x)dx2

+ 2U(x) =0 2 = 2

mEA

U(0) = 0

dU(x)

dxx=L = 0

Fixed-Free


38/41

Fixed-Free

Solution : U(x) =A sinx + B cosx

rL= r 1

2, r= 1, 2,

BCs : U(0) =B = 0

dU(x)

dxx=L

=Acos L= 0

Ur(x) =Arsin r 1

2x

L

r =

r 1

2rEA

mL2 , r= 1, 2,

, r= 1, 2,

Fixed-Fixed


39/41

Fixed-Fixed

Solution : U(x) =A sinx + B cosx

BCs : U(0) =B = 0 U(L) =A sin L= 0

rL=r, r= 1, 2,

r =r

rEA

mL2 , r= 1, 2,

Ur(x) =Arsinrx

L , r = 1, 2,

S


40/41

Similar Problem: Torsional Vibrations

(x, t)

I(x), GJ(x)

(x, t)

(x, t)dxT(x, t) + T(x,t)

x dxT(x, t)

I(x)dx

T(x, t) =GJ(x)(x,t)x

T ( ) 2( t)


41/41

T(x) +T(x)

x

dx +(x, t)dx T(x) =I(x)dx2(x, t)

t2

T(x)

x + (x, t) =I(x)

2(x, t)

t2

x

GJ(x)(x)

x

+ (x, t) =I(x)

2(x, t)t2

(0, t) = 0 T(L, t) = GJ(x)(x, t)x

x=L

= 0

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