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### Transcript of String Vibration - dept.aoe.vt. mpatil/courses/ Expansion • Eigenvectors: – Linearly independent...

• String Vibration

Chapter 8

• Distributed Parameter Systems

Distributed mass and stiffness Infinite DOF Functional analysis Exact solution for simple problems

• 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 +

2y(x,t)x2 dx

(x)dx

dx

• T (x) +

T (x)

xdx

y(x, t)

x+2y(x, t)

x2dx

+

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

• 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) = 0

y(L, t) = 0

• 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)

d

dx

T (x)

dY (x)

dxF (t)

=(x)Y (x)

d2F (t)

dt2

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

• 1

(x)Y (x)

d

dx

T (x)

dY (x)

dx

=

1

F (t)

d2F (t)

dt2= 2

d2F (t)

dt2+ 2F (t) = 0

F (t) = A sint+B cost = C cos(t+ )

d

dx

T (x)

dY (x)

dx

= 2(x)Y (x)

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

• d

dx

T (x)

dY (x)

dx

= 2(x)Y (x)

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

For constant mass distribution and tension

d2Y (x)

dx2+2

TY (x) = 0

d2Y (x)

dx2+ 2Y (x) =0 2 =

2

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

• d2Y (x)

dx2+ 2Y (x) =0

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

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

BCs : Y (0) = B = 0 Y (L) = A sinL = 0

rL = r, r = 1, 2,

r = r

sT

L2, r = 1, 2,

Yr(x) = Ar sinrx

L, r = 1, 2,

• Free Vibration Solutions Possible Solutions:

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

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

= Cr sinrxL

cos

r

sT

L2t+ r

!

y(x, t) =Xr=1

yr(x, t)

=Xr=1

CrYr(x) cos (rt+ r)

• Orthogonality

ddx

T (x)

dYr(x)

dx

= 2r(x)Yr(x)

Ys(x)d

dx

T (x)

dYr(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

• Z L0

Ys(x)d

dx

T (x)

dYr(x)

dx

dx

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

dx

L0

+

Z L0

T (x)dYs(x)

dx

dYr(x)

dxdx

=

Z L0

T (x)dYs(x)

dx

dYr(x)

dxdx

Z L0

T (x)dYs(x)

dx

dYr(x)

dxdx = 2r

Z L0

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

• Z L0

T (x)dYs(x)

dx

dYr(x)

dxdx = 2r

Z L0

(x)Ys(x)Yr(x)dxZ L0

T (x)dYr(x)

dx

dYs(x)

dxdx = 2s

Z L0

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

0 = (2r 2s)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)

dxdx

• NormalizationZ L0

T (x)

dYr(x)

dx

2dx = 2r

Z L0

(x) [Yr(x)]2 dx

Z L0

T (x)

dYr(x)

dx

2dx = 2r

SelectYr(x) such that :

Z L0

(x) [Yr(x)]2 dx = 1

• Modal Expansion

Eigenvectors: Linearly independent Complete set

Any continuous displacement distribution can be represented as a linear combination of the modal shapes

y(x, t) =Xr=1

Yr(x)r(t)

• 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=1

d

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=1

Yr(x)r(t)Yr(x) are mass normalizedmode shapes

• Uncoupled Modal EquationsZ L0

Ys(x)

Xr=1

d

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)

2ss(t) = s(t)

s(t) + 2ss(t) = 0

s(t) = s(0) cosst+s(0)

ssinst

• Initial Conditionsy(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 normalizedmode shapes

• Example = 1Kg/m T = 1N L = 1m

Initial Displacement : y0(x)

0.25

0.1

Initial Velocity : v0(x)

• 1 = rad/s

2 = 2 rad/s

3 = 3 rad/s

4 = 4 rad/s

5 = 5 rad/s

• Normalized Mode Shapes & Initial Condition

Yr(x) =2 sin

rx

L1(0)2(0)3(0)4(0)5(0)

=0.05400.01910.00600.00000.0022

• 0 0.2 0.4 0.6 0.8 1-0.05

0

0.05

0.1

0.15

x

y 0 (x

)

Initial DisplacementModal ApproximationMode 1Mode 2Mode 3Mode 4Mode 5

• 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)Exactt = 0.00 st = 0.25 st = 0.50 st = 0.75 st = 1.00 s

• Forced Response

(x)2y(x, t)

t2 x

T (x)

y(x, t)

x

= f(x, t)

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

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

y(x, t) =Xr=1

Yr(x)r(t)Yr(x) are mass normalizedmode shapes

Xr=1

(x)Yr(x)d2r(t)

dt2Xr=1

d

dx

T (x)

dYr(x)

dx

r(t) = f(x, t)

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

• Uncoupled Modal Equations

Multiply by Ys(x) Integrate over 0 to LZ L

0

Ys(x)

Xr=1

(x)Yr(x)r(t)dxZ L0

Ys(x)

Xr=1

d

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 mass normalizedmode shapes

• Initial Conditionsy(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 normalizedmode shapes

• Steady-State Harmonic Response

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

s(t) =Ns

2s 2cost

Ns(t) = Ns cost 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

Yr(x)Nr

2r 2=Xr=1

Yr(x)

2r 2Z L0

Yr(x)f(x)dx

• Point Force Force at a point a

Displacement at a point b

Reciprocal Theorem

F=F cos(t)ab

Ns =

Z L0

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

y(b) = FXr=1

Yr(b)Yr(a)

2r 2

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

Yr(x) =2 sin

rx

L

• 0 10 20 30-1

-0.5

0

0.5

1

y(b)a = 0.5; b = 0.5

• 0 10 20 30-1

-0.5

0

0.5

1

y(b)

a = 0.5; b = 0.33333

• 0 10 20 30-1

-0.5

0

0.5

1

y(b)a = 0.1; b = 0.1

• Similar Problem: Axial Vibrationsf(x, t)

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

f(x, t)dx

P (x, t) P (x, t) + P (x,t)x dxm(x)dx

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

• P (x, t) +

P (x, t)

xdx

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

2u(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

• 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)

dxF (t)

=m(x)U(x)

d2F (t)

dt2

U(0) = 0dU(x)

dx

x=L

= 0

• 1

m(x)U(x)

d

dx

EA(x)

dU(x)

dx

=

1

F (t)

d2F (t)

dt2= 2

d2F (t)

dt2+ 2F (t) = 0

F (t) = A sint+B cost = C cos(t+ )

d

dx

EA(x)

dU(x)

dx

= 2m(x)U(x)

U(0) = 0dU(x)

dx

x=L

= 0

• For constant mass distribution and cross-sectional stiffness

d

dx

EA(x)

dU(x)

dx

= 2m(x)U(x)

U(0) = 0dU(x)

dx

x=L

= 0

d2U(x)

dx2+2m

EAU(x) = 0

d2U(x)

dx2+ 2U(x) =0 2 =

2m

EA

U(0) = 0dU(x)

dx

x=L

= 0

• Fixed-Free

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

rL =r 1

2

, r = 1, 2,

BCs : U (0) = B = 0dU (x)

dx

x=L

= A cos L = 0

Ur(x) = Ar sinr 12

x

L

r =

r 1

2

rEA

mL2 , r = 1, 2,

, r = 1, 2,

• 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) = Ar sinrx

L, r = 1, 2,

• Similar Problem: Torsional Vibrations

(x, t)I(x), GJ(x)

(x, t)

(x, t)dx

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