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7/25/2019 String Vibration
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String Vibration
Chapter 8
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Distributed Parameter Systems
Distributed mass and stiffness
Infinite DOF
Functional analysis
Exact solution for simple problems
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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
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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
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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
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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
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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
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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
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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,
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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)
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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
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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
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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
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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
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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)
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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
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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
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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
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Example
= 1 Kg/m T = 1 N L= 1 m
Initial Displacement :y0(x)
0.25
0.1
Initial Velocity :v0(x)
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1= rad/s
2= 2 rad/s
3= 3 rad/s
4= 4 rad/s
5= 5 rad/s
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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
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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
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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
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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
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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
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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
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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
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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
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Example
= 1 Kg/m T = 1 N L= 1 m F= 1 N
Yr(x) =2sinrx
L
a = 0.5; b = 0.5
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0 10 20 30-1
-0.5
0
0.5
1
y(b)
;
a = 0.5; b = 0.33333
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0 10 20 30-1
-0.5
0
0.5
1
y(b)
a = 0.1; b = 0.1
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0 10 20 30-1
-0.5
0
0.5
1
y(b)
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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
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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
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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)
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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)
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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
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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
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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
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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)
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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