String Vibration

Chapter 8

Distributed Parameter Systems

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

f(x, t)ρ(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 dxρ(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

= ρ(x)dx∂2y(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 dxρ(x)dx

dx

T (x)∂2y(x, t)

∂x2+∂T (x)

∂x

∂y(x, t)

∂x+ f(x, t) = ρ(x)

∂2y(x, t)

∂t2

∂

∂x

∙T (x)

∂y(x, t)

∂x

¸+ f(x, t) = ρ(x)

∂2y(x, t)

∂t2

y(0, t) = 0

y(L, t) = 0

Initial (unforced) Response

∂

∂x

∙T (x)

∂y(x, t)

∂x

¸=ρ(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)

¸=ρ(x)Y (x)

d2F (t)

dt2

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

1

ρ(x)Y (x)

d

dx

∙T (x)

dY (x)

dx

¸=

1

F (t)

d2F (t)

dt2= −ω2

d2F (t)

dt2+ ω2F (t) = 0

⇒ F (t) = A sinωt+B cosωt = C cos(ωt+ φ)

d

dx

∙T (x)

dY (x)

dx

¸=− ω2ρ(x)Y (x)

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

d

dx

∙T (x)

dY (x)

dx

¸=− ω2ρ(x)Y (x)

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

• For constant mass distribution and tension

d2Y (x)

dx2+ω2ρ

TY (x) = 0

d2Y (x)

dx2+ β2Y (x) =0 β2 =

ω2ρ

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

d2Y (x)

dx2+ β2Y (x) =0

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

Solution : Y (x) = A sinβx+B cosβx

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

⇒ βrL = rπ, r = 1, 2, · · ·

ωr = rπ

sT

ρL2, r⇒ = 1, 2, · · ·

Yr(x) = Ar sinrπx

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 sin³rπxL

´cos

Ãrπ

sT

ρL2t+ φr

!

y(x, t) =∞Xr=1

yr(x, t)

=∞Xr=1

CrYr(x) cos (ωrt+ φr)

Orthogonality

− ddx

∙T (x)

dYr(x)

dx

¸= ω2rρ(x)Yr(x)

−Ys(x)d

dx

∙T (x)

dYr(x)

dx

¸= ω2rρ(x)Ys(x)Yr(x)

−Z L

0

Ys(x)d

dx

∙T (x)

dYr(x)

dx

¸dx = ω2r

Z L

0

ρ(x)Ys(x)Yr(x)dx

−Z L

0

Ys(x)d

dx

∙T (x)

dYr(x)

dx

¸dx

= −Ys(x)T (x)dYr(x)

dx

¯L0

+

Z L

0

T (x)dYs(x)

dx

dYr(x)

dxdx

=

Z L

0

T (x)dYs(x)

dx

dYr(x)

dxdx

Z L

0

T (x)dYs(x)

dx

dYr(x)

dxdx = ω2r

Z L

0

ρ(x)Ys(x)Yr(x)dx

Z L

0

T (x)dYs(x)

dx

dYr(x)

dxdx = ω2r

Z L

0

ρ(x)Ys(x)Yr(x)dxZ L

0

T (x)dYr(x)

dx

dYs(x)

dxdx = ω2s

Z L

0

ρ(x)Yr(x)Ys(x)dx

0 = (ω2r − ω2s)Z L

0

ρ(x)Yr(x)Ys(x)dx

0 =

Z L

0

ρ(x)Yr(x)Ys(x)dx

0 =

Z L

0

T (x)dYr(x)

dx

dYs(x)

dxdx

NormalizationZ L

0

T (x)

∙dYr(x)

dx

¸2dx = ω2r

Z L

0

ρ(x) [Yr(x)]2 dx

⇒Z L

0

T (x)

∙dYr(x)

dx

¸2dx = ω2r

SelectYr(x) such that :

Z L

0

ρ(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)ηr(t)

Initial (unforced) Response∂

∂x

∙T (x)

∂y(x, t)

∂x

¸= ρ(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

¸ηr(t) =

∞Xr=1

ρ(x)Yr(x)d2ηr(t)

dt2

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

y(x, t) =∞Xr=1

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

Uncoupled Modal EquationsZ L

0

Ys(x)

∞Xr=1

d

dx

∙T (x)

dYr(x)

dx

¸ηr(t)dx =

Z L

0

Ys(x)

∞Xr=1

ρ(x)Yr(x)ηr(t)dx

∞Xr=1

Z L

0

Ys(x)d

dx

∙T (x)

dYr(x)

dx

¸dxηr(t) =

∞Xr=1

Z L

0

ρ(x)Ys(x)Yr(x)dxηr(t)

−ω2sηs(t) = ηs(t)

ηs(t) + ω2sηs(t) = 0

ηs(t) = ηs(0) cosωst+ηs(0)

ωssinωst

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

y(x, 0) =∞Xr=1

Yr(x)ηr(0) = y0(x)(0)Z L

0

ρ(x)Ys(x)∞Xr=1

Yr(x)ηr(0)dx =

Z L

0

ρ(x)Ys(x)y0(x)dx

ηs(0) =

Z L

0

ρ(x)Ys(x)y0(x)dx

ηs(0) =

Z L

0

ρ(x)Ys(x)v0(x)dx

Do not forget:Yr(x) are mass normalizedmode shapes

Exampleρ = 1Kg/m T = 1N L = 1m

Initial Displacement : y0(x)

0.25

0.1

Initial Velocity : v0(x)

ω1 = π rad/s

ω2 = 2 π rad/s

ω3 = 3 π rad/s

ω4 = 4 π rad/s

ω5 = 5 π rad/s

Normalized Mode Shapes & Initial Condition

Yr(x) =√2 sin

rπx

L⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩η1(0)η2(0)η3(0)η4(0)η5(0)

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0.05400.01910.00600.0000−0.0022

⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭

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

ρ(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)ηr(t)Yr(x) are mass normalizedmode shapes

∞Xr=1

ρ(x)Yr(x)d2ηr(t)

dt2−∞Xr=1

d

dx

∙T (x)

dYr(x)

dx

¸ηr(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

ρ(x)Yr(x)ηr(t)dx−Z L

0

Ys(x)

∞Xr=1

d

dx

∙T (x)

dYr(x)

dx

¸ηr(t)dx

=

Z L

0

Ys(x)f(x, t)dx

ηs(t) + ω2sηs(t) = Ns(t)

Ns(t) =

Z L

0

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)ηr(0) = y0(x)(0)Z L

0

ρ(x)Ys(x)∞Xr=1

Yr(x)ηr(0)dx =

Z L

0

ρ(x)Ys(x)y0(x)dx

ηs(0) =

Z L

0

ρ(x)Ys(x)y0(x)dx

ηs(0) =

Z L

0

ρ(x)Ys(x)v0(x)dx

Do not forget:Yr(x) are mass normalizedmode shapes

Steady-State Harmonic Response

f(x, t) = f(x) cosωt

ηs(t) =Ns

ω2s − ω2cosωt

Ns(t) = Ns cosωt where Ns =

Z L

0

Ys(x)f(x)dx

y(x, t) = y(x) cosωt =∞Xr=1

Yr(x)ηr(t)

y(x) =∞Xr=1

Yr(x)Nr

ω2r − ω2=∞Xr=1

Yr(x)

ω2r − ω2Z L

0

Yr(x)f(x)dx

Point Force • Force at a point a

• Displacement at a point b

• Reciprocal Theorem

F=F cos(ωt)ab

Ns =

Z L

0

Ys(x)Fδ(a)dx = FYs(a)

y(b) = F∞Xr=1

Yr(b)Yr(a)

ω2r − ω2

Exampleρ = 1Kg/m T = 1N L = 1m F = 1N

Yr(x) =√2 sin

rπx

L

0 10 20 30-1

-0.5

0

0.5

1

ω

y(b)a = 0.5; b = 0.5

0 10 20 30-1

-0.5

0

0.5

1

ω

y(b)

a = 0.5; b = 0.33333

0 10 20 30-1

-0.5

0

0.5

1

ω

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 dx

m(x)dx

P (x, t) = σA = EA² = EA∂u(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= −ω2

d2F (t)

dt2+ ω2F (t) = 0

⇒ F (t) = A sinωt+B cosωt = C cos(ωt+ φ)

d

dx

∙EA(x)

dU(x)

dx

¸=− ω2m(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

¸=− ω2m(x)U(x)

U(0) = 0dU(x)

dx

¯x=L

= 0

d2U(x)

dx2+ω2m

EAU(x) = 0

d2U(x)

dx2+ β2U(x) =0 β2 =

ω2m

EA

U(0) = 0dU(x)

dx

¯x=L

= 0

Fixed-Free

Solution : U(x) = A sinβx+B cosβx

⇒ βrL =

µr − 1

2

¶π, r = 1, 2, · · ·

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

dx

¯x=L

= Aβ cos βL = 0¯

⇒ Ur(x) = Ar sin

¡r − 1

2

¢πx

L

ωr =

µr − 1

2

¶π

rEA

mL2⇒ , r = 1, 2, · · ·

, r = 1, 2, · · ·

Fixed-Fixed

Solution : U(x) = A sinβx+B cosβx

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

⇒ βrL = rπ, r = 1, 2, · · ·

ωr = rπ

rEA

mL2⇒ , r = 1, 2, · · ·

Ur(x) = Ar sinrπx

L, r⇒ = 1, 2, · · ·

Similar Problem: Torsional Vibrations

θ(x, t)I(x), GJ(x)

τ(x, t)

τ(x, t)dx

T (x, t) + ∂T (x,t)∂x dxT (x, t)

I(x)dx

T (x, t) = GJ(x)∂θ(x,t)∂x

∙T (x) +

∂T (x)

∂xdx

¸+ τ (x, t)dx − T (x) = I(x)dx∂

2θ(x, t)

∂t2

∂T (x)

∂x+ τ(x, t) = I(x)

∂2θ(x, t)

∂t2

∂

∂x

∙GJ(x)

∂θ(x)

∂x

¸+ τ(x, t) = I(x)

∂2θ(x, t)

∂t2

θ(0, t) = 0 T (L, t) = GJ(x)∂θ(x, t)

∂x

¯x=L

= 0