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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