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CHAPTER 7: Continuous Systems. Axial and Torsional vibration of bars
In the previous chapters we have considered systems made up of a combination of rigid masses, springs and dampers. The number of degrees of freedom, was the sum of the degrees of freedom of each mass. If we remember, that the degrees of freedom equal the number of coordinates needed to specify the position of the mass, then for a continuous system we need an infinite number of degrees of freedom to specify the position. This is because we need to specify the position of each point on the continuous system. In this chapter, we will consider some continuous systems. This will allow us to make some general conclusions about continuous systems. We will consider the axial vibration of a uniform undamped continuous bar, then include damping and also consider tapered bars. It is then a simple matter to extend this to the torsional vibration of such bars. 7.1 Axial vibration of uniform bars Consider a uniform bar as shown in figure 7.1 and an element of the bar a distance x from one end. When vibrating the forces on the element and the deflection (u) of the element at some time t are as shown
Figure 7.1 Element of bar vibrating axially. For the element the stress/strain equation gives, ∂u ............................................................ (7.1) ∂x where A is the cross-section area and E is the elastic (Young's) modulus. F = AE Applying Newton II to the element, F + ∂F − F = m
∂2 u ∂t 2
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and if ρ is the density, the mass of the element m = ρA∂x so that, ∂2u ∂F = ρA∂x 2 ∂t so that ∂F ∂2 u = ρA 2 .................................................... (7.2) ∂x ∂t and substituting for F from (7.1) ∂ ∂u ∂2 u AE = ρA 2 ∂x ∂x ∂t and rearranging ∂2 u ρ ∂2u = ...................................................... (7.3) ∂x2 E ∂t 2 7.1.1 Steady-state sinusoidal motion Assume that u(x, t) = U(x)eiωt then substituting in (7.3) ρω2 U'' (x)e iωt = − U(x)e iωt E ρω 2 thus U'' (x) + U(x) = 0 E Let the solution for U(x) = Be β1 x + Ce β 2 x so that to find β1 and β2 we substitute U(x) = Be βx and find two values for β. Thus
β2 Beβx +
ρω Let λ2 = so that E
2
ρω2 βx Be = 0 E ρω2 and β2 = − E
β2 = −λ2 ∴β = ±iλ U(x) = Be iλx + Ce −iλx ............................................ (7.4)
hence
The constants B and C depend on the end conditions of the bar. If the bar is free/free then there is no force at either end. From equation (1) ∂u ∂u F = AE so that when there is no force =0 ∂x ∂x ∂u From (7.4) = iλBeiλx − iλCe −iλx = 0 ∂x so that when x=0 B − C = 0 ......... (a) iλL and when x=L Be − Ce −iλL = 0 ......... (b) From (a) C = B and substituting in (b) B e iλL − e −iλL = 0
(
)
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Thus either B=0 and there is no vibration or B may exist and there will be vibration without any continuing excitation (ie at a natural frequency) if e iλL − e −iλL = 0 ie when cosλL + isinλL − cosλL + isinλL = 2isinλL = 0 The natural frequencies are thus when sinλL = 0 which is when λL = nπ and n = 0 → ∞ . Substituting for ρ λ=ω E ρ nπ E ω L = nπ and hence ω n = n =0 →∞ E L ρ As a continuous bar has an infinite number of degrees-of-freedom we should not be surprised to find that it has an infinite number of natural frequencies. To find the mode shapes we return to equation (7.4) with C=B, U(x) = B e iλx + e −iλx = 2Bcosλx And substituting λL = nπ (ie the natural frequencies) x U(x) = Acos nπ L As the bar is free/free we have a zero frequency mode when n=0 and U(x) = A which is a solid body motion.
(
)
The program shows the first three non-zero modes. For an excited end From the diagram above it should be noted that at the right hand end when x = L (the length of the bar) the force F is in the same positive direction as u. ∂u(L) F Thus for x = L from (7.1) = ∂x AE However when x = 0 the force F is in the negative u direction ∂u(0) F Thus for x =0 from (7.1) =− ∂x AE Response when excitation is at x = L For this case when x=L F = FL e iωt and when x=0 F=0. The response is u(x) = U(x)e iωt When x=0 from equation (7.4) so that when x=0 ∂u = iλBeiλx − iλCe −iλx = 0 ∂x B − C = 0 and hence B = C
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When x = L,
F = F Le
iωt
∂u(L) FL e iωt iωt and = U' (L)e = ∂x AE iλBe iλL − iλCe −iλL = FL AE FL AE
therefore
Put C = B so that Therefore,
B(iλcosλL − λsinλL − iλcosλL − λsinλL) = -F L 2AEλsinλL
B=C= returning to equation (7.4)
................................................. (7.5)
U(x) = Be iλx + Ce −iλx and substituting for B and C from (7.5) U(x) = e iλx + e −iλx
............................................. (7.4)
(
-F L ) 2AEλsinλL -FL 2AEλsinλL
∴ U(x) = (cosλx + isinλx + cosλx − isinλx ) so that U(x) = F Lcosλx AEλsinλL
and hence the response at a position x along the bar when excited at x=L is U(x) cosλx =− .............................................. (7.6) FL AEλsinλL for the case when x = L we obtain the response at the excitation position, U(L) cosλL =− .............................................. (7.7) FL AEλsinλL for the case when x = 0 we obtain the response at the free end, U(0) 1 =− ................................................ (7.8) FL AEλsinλL Response when excitation is at x = 0 Following similar maths to the above it can be shown that, U(0) cosλL =− ................................................... (7.9) Fo AEλsinλL and as expected from symmetry Also it is found that, Contents U(0) U(L) = . Compare equations (7.9) and (7.7). Fo FL
U(L) 1 =− .................................................. (7.10) Fo AEλsinλL U(0) U(L) and as expected from Maxwell's reciprocal theorem = . Compare equations (7.10) FL Fo and (7.8). Finally it may be shown that U(x) cosλ(L − x) =− ................................................. (7.11) Fo AEλsinλL The major point to note is that all these responses have sinλL in the denominator. Thus the response will tend to infinity and resonance will occur when sinλL = 0 which was the natural U(L) frequency equation for a free/free bar. The response is shown in figure 7.2 for a bar of FL length L = 2m diameter 0.2m, E = 2.0e11 N/m2 and ρ =7800.0 kg/m2 .
Figure 7.2 Tip response of a free/free bar
It should be noted that there are anti-resonances. If we consider the response at the end of the bar, u(L) cosλL =− FL AEλsinλL
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it is apparent that which is when λL = Substituting for
2n − 1 π and n = 1 → ∞ . 2 λ=ω ρ E n =1 → ∞
ω
ρ 2n − 1 (2n − 1)π E L= π and hence ω = E 2 2L ρ
From equation (7.6) we know the response at any position x for an exciting force at the end, U(x) cosλx =− FL AEλsinλL we may therefore show the animated deflected shape at any excitation by calculating the response at discrete positions along the bar.
Clamped/free bar
For sinusoidal excitation and response we have shown, U(x) = Be iλx + Ce −iλx Therefore when x = 0 U(0) = 0 = B + C so that C = -B as for the free/free case When x = L, F = F Le iωt and ∂u(L) F e iωt = U' (L)e iωt = L ∂x AE iλBe iλL − iλCe −iλL = iλBe iλL + iλBe −iλL = FL AE ............................................. (7.4)
therefore and substituting C = -B
FL AE
so that Biλ(cosλL + isinλL + cosλL − isinλL) =
FL AE
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Therefore, B = −C = FL .................................................... (7.12) 2GJiλ cosλL
returning to equation (7.4) and substituting for B and C from (7.12) e iλx − e −iλx FL FL sinλx U(x) = AEλ cosλL = AEλcosλL 2i so that F Lsinλx AEλcosλL and hence the response at a position x along the bar when excited at x=L is U(x) = U(x) sinλx = FL AEλ cosλL ......................................................... (7.13)
for the case when x = L we obtain the response at the excitation position, U(L) sinλL = .......................................................... (7.14) FL AEλ cos λL The major point to note is that the responses have cosλL in the denominator. Thus the response 2n − 1 will tend to infinity and resonance will occur when cosλL = 0 which is when λL = π 2 and n = 1 → ∞ . Substituting for ρ λ=ω E ρ 2n − 1 (2n − 1)π E ω L= π and hence ω = n =1 → ∞ E 2 2L ρ
The program shows the first three non-zero modes. It is of interest to note that these natural frequencies are the frequencies at which anti-resonances occurred for the response of a free/free bar excited at one end.
If the response at the free end (equation (7.14)) is plotted (figure 7.3) then it may be seen by comparing figure 7.3 with 7.2 that all the clamped/free resonances coincide with the antiresonances for the response of the free/free bar. Note the dimensions of the bar are the same in
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both cases. This is now a general conclusion for undamped systems. When the response at an end is zero then the frequency at which this occurs is the natural frequency the system would have if clamped at the excitation position.
Figure 7.3 Tip response of a clamped/free bar
7.1.2 Damping In all the previous chapters the damping has been considered to be viscous when it has been added. Thus there has been a damping force proportional to the velocity. For steady state vibration this has given rise to terms of the form iωcX where X was the amplitude of vibration and c the viscous damping coefficient. This form of damping has an increasing effect with the frequency. In practice for many systems the damping force does not increase with frequency. Rather there is a constant energy loss each cycle depending on the amplitude but not on the frequency. Such a form of damping is called 'hysteretic' damping and gives rise to damping terms of the form ihX where h is called the hysteretic damping coefficient. When damping is included in the axial vibration of bars both viscous and hysteretic damping should be considered to determine which is the most appropriate. It is of interest to note that several authors have reported simply taking the equations (7.6) to (7.14) and replacing the elastic modulus by a complex quantity such as E (1+ iη)) with the expectation that this will include hysteretic damping. This does not give the correct response for a damped shaft but simply multiplies the undamped response by a constant complex number 1/(1 + iη). The response thus clearly still has an infinite value at each of the undamped natural
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frequencies. It is necessary to return to the basic equations and solve them with a complex modulus. This has been done [1] for torsional vibration and the receptances (ie responses) for axial vibration can be shown to be, for a free/free bar excited at the end x=L,
U(x) 2[(a cosh λSx cos λRx- bsinhλSxsin λRx)-i(a sinhλSxsin λRx+b cosh λSx cos λRx)] = FL a 2 + b2
(
)
....... (7.15)
where,
[(1+ η2 )1/2 − 1]1/2 and R = [(1 + η2 )1/2 + 1]1/2 S= −
2(1+ η2 )1/ 2 2(1+ η2 )1/ 2 a = 2AEλ[(S + ηR)cos(RλL)sinh(SλL) + (ηS − R)sin(RλL)cosh(SλL)] b = 2AEλ [(ηS − R)cos(RλL)sinh(SλL) − (S + ηR)sin(RλL) cosh(SλL)]
If viscous damping is to be modelled then we simply replace η by ωξ in the equations above. From equation (7.15) the responses at either end of the bar may be found by putting x = 0 and U(L) x = L. The response is shown in figure 7.4 for a bar of length l= 2m diameter 0.2m, E = FL 2.0e11 N/m2 and ρ =7800.0 kg/m2 and for various values of η.
Figure 7.4 Tip response of a damped free/free bar for various hysteretic damping values.
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7.1.3 Tapered bars Entwistle et al [2] have derived the responses of an undamped tapered bar and their results are given below.
Figure 7.5 Free/free tapered shaft - definition of terms r ρ r −r Defining z = λ x + o where λ = ω as before and B = L o Entwistle et al [*] B E L showed that, cosz L sin z L U(z) −1 .sin z + cosz L − sinz L .cosz ................ (7.16) = . + z Fo zL Eπr2 λΔz z 2 z2 L o L L where cosz o − sin zo cosz L + sinz L − cos zo + sinz o cosz L − sin z L ............ (7.17) 2 2 Δ= zo z 2 z L zL zo zo z L z2 o L Hence cosz L sin zL U(z o ) −1 .sinz o + cosz L − sin z L .cosz o .......... 2 + α oo = = . z Fo zL Eπr2 λΔz o z L z2 L o L (7.18) and α Lo = Eπr 2λΔz 2 o L −1 ....................................................... (7.19)
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Also Entwistle et al [2] showed that, cosz o sin z o U(z) −1 .sin z + cos zo − sinz o .cos z 2 + = . z FL zo Eπr2 λΔz zo z2 o L o Consequently; cosz U(z L ) −1 sin z o .sin zL + cosz o − sin z o .cosz L ......(7.20) α LL = = . 2 o + z FL zo Eπr2 λΔz L zo z2 o L o and α oL = Eπr 2 λΔz2 L o −1 ......................................................... (7.21)
Note: For a linear taper r 2 z 2 = r2 z2 . Hence α oL = α Lo ........................................ (7.22) L o o L Figure 7.6 shows α oo and α LL the responses at each end of a bar length L =2m and Do = 0.05m and = 0.2m. The resonances are the same for each response as the natural frequencies do not depend on the excitation position. However, the anti-resonances are at different frequencies as they correspond to the natural frequencies of a bar clamped at the excitation position.
Figure 7.6 Tip responses of a free/free tapered bar.
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7.2 Torsional vibration of uniform bars For torsional vibration very similar maths applies to that developed for axial vibration. For uniform bars there are only four parameter differences, E is replaced throughout by G; A by J; F by T and u by θ. The receptances are then the same as for the axial case but with λ2 = ρω2 / G thus for 7.2.1 Free/free Θ(x) cosλx =− ......................................................... TL GJλsinλL Θ(L) Θ(0) cosλL = =− .................................................... TL To GJλsinλL Θ(0) Θ(L) 1 = =− .......................................................... TL To GJλsinλL Θ(x) cosλ(L − x) =− ....................................................... To GJλsinλL (7.23) (7.24) (7.25) (7.26)
The natural frequencies are thus when sinλL = 0 which is when λL = nπ and n = 0 → ∞ . Substituting for ρ λ=ω G ρ nπ G ω L = nπ and hence ω n = n =0 →∞ G L ρ
The program shows the first three non-zero modes. 7.2.2 Clamped/free Θ(x) sinλx = ......................................................... (7.27) TL GJλ cosλL Θ(L) sinλL = .......................................................... (7.28) TL AEλ cosλL The natural frequencies are thus when cosλL = 0 which is when n =1 → ∞. Substituting for λ=ω ω ρ G n =1 → ∞ λL = 2n − 1 π and 2
ρ 2n −1 (2n − 1)π E L= π and hence ω = G 2 2L ρ
The program shows the first three modes.
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7.2.3 Damping For a free/free bar excited at the end x=L, again E is replaced throughout by G; A by J; F by T and u by θ. Θ(x) 2[( a coshλSxcosλRx - bsinh λSxsin λRx ) - i(a sinh λSxsinλRx + bcosh λSxcosλRx )] = TL a 2 + b2
(
)
....... (7.29) where,
[(1+ η2 )1/2 − 1]1/2 and R = [(1 + η2 )1/2 + 1]1/2 S= −
2(1+ η2 )1/ 2 2(1+ η2 )1/ 2 a = 2GJλ[(S + ηR)cos(RλL)sinh(SλL) + (ηS − R)sin(RλL)cosh(SλL)] b = 2GJλ [(ηS − R)cos(RλL)sinh(SλL) − (S + ηR)sin(RλL)cosh(SλL)]
If viscous damping is to be modelled then we simply replace η by ωξ in the equations above. From equation (7.29) the responses at either end of the bar may be found by putting x = 0 and x = L. 7.2.4 Tapered bars For the torsional vibration of tapered bars it not just a simple matter of changing parameters [2]. The solutions are significantly different. r ρ r −r Defining z = λ x + o where λ = ω and B = L o Entwistle et al [2] showed that, B G L Θ(z) 2z 2 o . − z 2 cosz − 3z sinz − 3cosz sinz − cos z L = L L L L z To ΔGπr 4λ o cosz + z2 sinz L + 3z L cos z L − 3sinz L sin z + .................. (7.30) L z Θ(z) 2z2 L . − z2 cosz − 3z sin z − 3cos z sin z − cosz o α xL = = o o o o z TL ΔGπr 4 λ L cosz + z2 sin z o + 3z o cosz o − 3sinz o sinz + .................. (7.31) o z α xo =
(
)
(
)
(
)
(
)
where Δ = z2 sinz o + 3z o cosz o − 3sin zo o
}{z2L cos zL − 3z L sin zL − 3coszL } − {z2 cosz o − 3z o sinz o − 3cosz o }{z2 sinz L + 3z L cos zL − 3sinz L } ................... o L { ( ) sinozo − coszo z
(7.32)
The tip receptances may then be shown to be, Θ(z o ) 2z2 o . − z2 cosz − 3z sin z − 3cosz L α oo = = L L L L To ΔGπr 4 λ o
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+ z2 sinz L + 3z L cos z L − 3sinz L L
(
)sin zo + coszo zo
.................. (7.33)
α LL =
Θ(z L ) 2z2 L . − z2 cosz − 3z sin z − 3cos z sin z L − cosz o = o o o o L TL ΔGπr 4 λ zL L cos zL .................. (7.34) + z2 sin z o + 3z o cosz o − 3sinz o sinz L + o zL
(
)
(
)
2z 4 2z 4 L o α oL = α Lo = 4 = ΔGπr L λ ΔGπr 4K o
............................................. (7.35)
7.3 Conclusions In this chapter we have seen that continuous systems have an infinite number of natural frequencies and associated mode shapes. 7.4 References 1. Derry, S., Drew, S. J. and Stone, B. J., 'The torsional vibration of a damped continuous bar.’ Abstract Proceedings of the VIII International Congress on Experimental Mechanics, Nashville Tennessee, 1996. 72-73. Entwistle R. D., Hesterman D. C. and Stone, B. J., 'Axial and Torsional Receptances for Tapered Circular Shafts', Modal Analysis: The International Journal of Analytical and Experimental Modal Analysis. Vol 11 (1997) 178-193. Next chapter
2.
.
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Continuous
CHAPTER 7: Continuous Systems. Axial and Torsional vibration of bars
In the previous chapters we have considered systems made up of a combination of rigid masses, springs and dampers. The number of degrees of freedom, was the sum of the degrees of freedom of each mass. If we remember, that the degrees of freedom equal the number of coordinates needed to specify the position of the mass, then for a continuous system we need an infinite number of degrees of freedom to specify the position. This is because we need to specify the position of each point on the continuous system. In this chapter, we will consider some continuous systems. This will allow us to make some general conclusions about continuous systems. We will consider the axial vibration of a uniform undamped continuous bar, then include damping and also consider tapered bars. It is then a simple matter to extend this to the torsional vibration of such bars. 7.1 Axial vibration of uniform bars Consider a uniform bar as shown in figure 7.1 and an element of the bar a distance x from one end. When vibrating the forces on the element and the deflection (u) of the element at some time t are as shown
Figure 7.1 Element of bar vibrating axially. For the element the stress/strain equation gives, ∂u ............................................................ (7.1) ∂x where A is the cross-section area and E is the elastic (Young's) modulus. F = AE Applying Newton II to the element, F + ∂F − F = m
∂2 u ∂t 2
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and if ρ is the density, the mass of the element m = ρA∂x so that, ∂2u ∂F = ρA∂x 2 ∂t so that ∂F ∂2 u = ρA 2 .................................................... (7.2) ∂x ∂t and substituting for F from (7.1) ∂ ∂u ∂2 u AE = ρA 2 ∂x ∂x ∂t and rearranging ∂2 u ρ ∂2u = ...................................................... (7.3) ∂x2 E ∂t 2 7.1.1 Steady-state sinusoidal motion Assume that u(x, t) = U(x)eiωt then substituting in (7.3) ρω2 U'' (x)e iωt = − U(x)e iωt E ρω 2 thus U'' (x) + U(x) = 0 E Let the solution for U(x) = Be β1 x + Ce β 2 x so that to find β1 and β2 we substitute U(x) = Be βx and find two values for β. Thus
β2 Beβx +
ρω Let λ2 = so that E
2
ρω2 βx Be = 0 E ρω2 and β2 = − E
β2 = −λ2 ∴β = ±iλ U(x) = Be iλx + Ce −iλx ............................................ (7.4)
hence
The constants B and C depend on the end conditions of the bar. If the bar is free/free then there is no force at either end. From equation (1) ∂u ∂u F = AE so that when there is no force =0 ∂x ∂x ∂u From (7.4) = iλBeiλx − iλCe −iλx = 0 ∂x so that when x=0 B − C = 0 ......... (a) iλL and when x=L Be − Ce −iλL = 0 ......... (b) From (a) C = B and substituting in (b) B e iλL − e −iλL = 0
(
)
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Thus either B=0 and there is no vibration or B may exist and there will be vibration without any continuing excitation (ie at a natural frequency) if e iλL − e −iλL = 0 ie when cosλL + isinλL − cosλL + isinλL = 2isinλL = 0 The natural frequencies are thus when sinλL = 0 which is when λL = nπ and n = 0 → ∞ . Substituting for ρ λ=ω E ρ nπ E ω L = nπ and hence ω n = n =0 →∞ E L ρ As a continuous bar has an infinite number of degrees-of-freedom we should not be surprised to find that it has an infinite number of natural frequencies. To find the mode shapes we return to equation (7.4) with C=B, U(x) = B e iλx + e −iλx = 2Bcosλx And substituting λL = nπ (ie the natural frequencies) x U(x) = Acos nπ L As the bar is free/free we have a zero frequency mode when n=0 and U(x) = A which is a solid body motion.
(
)
The program shows the first three non-zero modes. For an excited end From the diagram above it should be noted that at the right hand end when x = L (the length of the bar) the force F is in the same positive direction as u. ∂u(L) F Thus for x = L from (7.1) = ∂x AE However when x = 0 the force F is in the negative u direction ∂u(0) F Thus for x =0 from (7.1) =− ∂x AE Response when excitation is at x = L For this case when x=L F = FL e iωt and when x=0 F=0. The response is u(x) = U(x)e iωt When x=0 from equation (7.4) so that when x=0 ∂u = iλBeiλx − iλCe −iλx = 0 ∂x B − C = 0 and hence B = C
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When x = L,
F = F Le
iωt
∂u(L) FL e iωt iωt and = U' (L)e = ∂x AE iλBe iλL − iλCe −iλL = FL AE FL AE
therefore
Put C = B so that Therefore,
B(iλcosλL − λsinλL − iλcosλL − λsinλL) = -F L 2AEλsinλL
B=C= returning to equation (7.4)
................................................. (7.5)
U(x) = Be iλx + Ce −iλx and substituting for B and C from (7.5) U(x) = e iλx + e −iλx
............................................. (7.4)
(
-F L ) 2AEλsinλL -FL 2AEλsinλL
∴ U(x) = (cosλx + isinλx + cosλx − isinλx ) so that U(x) = F Lcosλx AEλsinλL
and hence the response at a position x along the bar when excited at x=L is U(x) cosλx =− .............................................. (7.6) FL AEλsinλL for the case when x = L we obtain the response at the excitation position, U(L) cosλL =− .............................................. (7.7) FL AEλsinλL for the case when x = 0 we obtain the response at the free end, U(0) 1 =− ................................................ (7.8) FL AEλsinλL Response when excitation is at x = 0 Following similar maths to the above it can be shown that, U(0) cosλL =− ................................................... (7.9) Fo AEλsinλL and as expected from symmetry Also it is found that, Contents U(0) U(L) = . Compare equations (7.9) and (7.7). Fo FL
U(L) 1 =− .................................................. (7.10) Fo AEλsinλL U(0) U(L) and as expected from Maxwell's reciprocal theorem = . Compare equations (7.10) FL Fo and (7.8). Finally it may be shown that U(x) cosλ(L − x) =− ................................................. (7.11) Fo AEλsinλL The major point to note is that all these responses have sinλL in the denominator. Thus the response will tend to infinity and resonance will occur when sinλL = 0 which was the natural U(L) frequency equation for a free/free bar. The response is shown in figure 7.2 for a bar of FL length L = 2m diameter 0.2m, E = 2.0e11 N/m2 and ρ =7800.0 kg/m2 .
Figure 7.2 Tip response of a free/free bar
It should be noted that there are anti-resonances. If we consider the response at the end of the bar, u(L) cosλL =− FL AEλsinλL
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it is apparent that which is when λL = Substituting for
2n − 1 π and n = 1 → ∞ . 2 λ=ω ρ E n =1 → ∞
ω
ρ 2n − 1 (2n − 1)π E L= π and hence ω = E 2 2L ρ
From equation (7.6) we know the response at any position x for an exciting force at the end, U(x) cosλx =− FL AEλsinλL we may therefore show the animated deflected shape at any excitation by calculating the response at discrete positions along the bar.
Clamped/free bar
For sinusoidal excitation and response we have shown, U(x) = Be iλx + Ce −iλx Therefore when x = 0 U(0) = 0 = B + C so that C = -B as for the free/free case When x = L, F = F Le iωt and ∂u(L) F e iωt = U' (L)e iωt = L ∂x AE iλBe iλL − iλCe −iλL = iλBe iλL + iλBe −iλL = FL AE ............................................. (7.4)
therefore and substituting C = -B
FL AE
so that Biλ(cosλL + isinλL + cosλL − isinλL) =
FL AE
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Therefore, B = −C = FL .................................................... (7.12) 2GJiλ cosλL
returning to equation (7.4) and substituting for B and C from (7.12) e iλx − e −iλx FL FL sinλx U(x) = AEλ cosλL = AEλcosλL 2i so that F Lsinλx AEλcosλL and hence the response at a position x along the bar when excited at x=L is U(x) = U(x) sinλx = FL AEλ cosλL ......................................................... (7.13)
for the case when x = L we obtain the response at the excitation position, U(L) sinλL = .......................................................... (7.14) FL AEλ cos λL The major point to note is that the responses have cosλL in the denominator. Thus the response 2n − 1 will tend to infinity and resonance will occur when cosλL = 0 which is when λL = π 2 and n = 1 → ∞ . Substituting for ρ λ=ω E ρ 2n − 1 (2n − 1)π E ω L= π and hence ω = n =1 → ∞ E 2 2L ρ
The program shows the first three non-zero modes. It is of interest to note that these natural frequencies are the frequencies at which anti-resonances occurred for the response of a free/free bar excited at one end.
If the response at the free end (equation (7.14)) is plotted (figure 7.3) then it may be seen by comparing figure 7.3 with 7.2 that all the clamped/free resonances coincide with the antiresonances for the response of the free/free bar. Note the dimensions of the bar are the same in
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both cases. This is now a general conclusion for undamped systems. When the response at an end is zero then the frequency at which this occurs is the natural frequency the system would have if clamped at the excitation position.
Figure 7.3 Tip response of a clamped/free bar
7.1.2 Damping In all the previous chapters the damping has been considered to be viscous when it has been added. Thus there has been a damping force proportional to the velocity. For steady state vibration this has given rise to terms of the form iωcX where X was the amplitude of vibration and c the viscous damping coefficient. This form of damping has an increasing effect with the frequency. In practice for many systems the damping force does not increase with frequency. Rather there is a constant energy loss each cycle depending on the amplitude but not on the frequency. Such a form of damping is called 'hysteretic' damping and gives rise to damping terms of the form ihX where h is called the hysteretic damping coefficient. When damping is included in the axial vibration of bars both viscous and hysteretic damping should be considered to determine which is the most appropriate. It is of interest to note that several authors have reported simply taking the equations (7.6) to (7.14) and replacing the elastic modulus by a complex quantity such as E (1+ iη)) with the expectation that this will include hysteretic damping. This does not give the correct response for a damped shaft but simply multiplies the undamped response by a constant complex number 1/(1 + iη). The response thus clearly still has an infinite value at each of the undamped natural
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frequencies. It is necessary to return to the basic equations and solve them with a complex modulus. This has been done [1] for torsional vibration and the receptances (ie responses) for axial vibration can be shown to be, for a free/free bar excited at the end x=L,
U(x) 2[(a cosh λSx cos λRx- bsinhλSxsin λRx)-i(a sinhλSxsin λRx+b cosh λSx cos λRx)] = FL a 2 + b2
(
)
....... (7.15)
where,
[(1+ η2 )1/2 − 1]1/2 and R = [(1 + η2 )1/2 + 1]1/2 S= −
2(1+ η2 )1/ 2 2(1+ η2 )1/ 2 a = 2AEλ[(S + ηR)cos(RλL)sinh(SλL) + (ηS − R)sin(RλL)cosh(SλL)] b = 2AEλ [(ηS − R)cos(RλL)sinh(SλL) − (S + ηR)sin(RλL) cosh(SλL)]
If viscous damping is to be modelled then we simply replace η by ωξ in the equations above. From equation (7.15) the responses at either end of the bar may be found by putting x = 0 and U(L) x = L. The response is shown in figure 7.4 for a bar of length l= 2m diameter 0.2m, E = FL 2.0e11 N/m2 and ρ =7800.0 kg/m2 and for various values of η.
Figure 7.4 Tip response of a damped free/free bar for various hysteretic damping values.
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7.1.3 Tapered bars Entwistle et al [2] have derived the responses of an undamped tapered bar and their results are given below.
Figure 7.5 Free/free tapered shaft - definition of terms r ρ r −r Defining z = λ x + o where λ = ω as before and B = L o Entwistle et al [*] B E L showed that, cosz L sin z L U(z) −1 .sin z + cosz L − sinz L .cosz ................ (7.16) = . + z Fo zL Eπr2 λΔz z 2 z2 L o L L where cosz o − sin zo cosz L + sinz L − cos zo + sinz o cosz L − sin z L ............ (7.17) 2 2 Δ= zo z 2 z L zL zo zo z L z2 o L Hence cosz L sin zL U(z o ) −1 .sinz o + cosz L − sin z L .cosz o .......... 2 + α oo = = . z Fo zL Eπr2 λΔz o z L z2 L o L (7.18) and α Lo = Eπr 2λΔz 2 o L −1 ....................................................... (7.19)
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Also Entwistle et al [2] showed that, cosz o sin z o U(z) −1 .sin z + cos zo − sinz o .cos z 2 + = . z FL zo Eπr2 λΔz zo z2 o L o Consequently; cosz U(z L ) −1 sin z o .sin zL + cosz o − sin z o .cosz L ......(7.20) α LL = = . 2 o + z FL zo Eπr2 λΔz L zo z2 o L o and α oL = Eπr 2 λΔz2 L o −1 ......................................................... (7.21)
Note: For a linear taper r 2 z 2 = r2 z2 . Hence α oL = α Lo ........................................ (7.22) L o o L Figure 7.6 shows α oo and α LL the responses at each end of a bar length L =2m and Do = 0.05m and = 0.2m. The resonances are the same for each response as the natural frequencies do not depend on the excitation position. However, the anti-resonances are at different frequencies as they correspond to the natural frequencies of a bar clamped at the excitation position.
Figure 7.6 Tip responses of a free/free tapered bar.
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7.2 Torsional vibration of uniform bars For torsional vibration very similar maths applies to that developed for axial vibration. For uniform bars there are only four parameter differences, E is replaced throughout by G; A by J; F by T and u by θ. The receptances are then the same as for the axial case but with λ2 = ρω2 / G thus for 7.2.1 Free/free Θ(x) cosλx =− ......................................................... TL GJλsinλL Θ(L) Θ(0) cosλL = =− .................................................... TL To GJλsinλL Θ(0) Θ(L) 1 = =− .......................................................... TL To GJλsinλL Θ(x) cosλ(L − x) =− ....................................................... To GJλsinλL (7.23) (7.24) (7.25) (7.26)
The natural frequencies are thus when sinλL = 0 which is when λL = nπ and n = 0 → ∞ . Substituting for ρ λ=ω G ρ nπ G ω L = nπ and hence ω n = n =0 →∞ G L ρ
The program shows the first three non-zero modes. 7.2.2 Clamped/free Θ(x) sinλx = ......................................................... (7.27) TL GJλ cosλL Θ(L) sinλL = .......................................................... (7.28) TL AEλ cosλL The natural frequencies are thus when cosλL = 0 which is when n =1 → ∞. Substituting for λ=ω ω ρ G n =1 → ∞ λL = 2n − 1 π and 2
ρ 2n −1 (2n − 1)π E L= π and hence ω = G 2 2L ρ
The program shows the first three modes.
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7.2.3 Damping For a free/free bar excited at the end x=L, again E is replaced throughout by G; A by J; F by T and u by θ. Θ(x) 2[( a coshλSxcosλRx - bsinh λSxsin λRx ) - i(a sinh λSxsinλRx + bcosh λSxcosλRx )] = TL a 2 + b2
(
)
....... (7.29) where,
[(1+ η2 )1/2 − 1]1/2 and R = [(1 + η2 )1/2 + 1]1/2 S= −
2(1+ η2 )1/ 2 2(1+ η2 )1/ 2 a = 2GJλ[(S + ηR)cos(RλL)sinh(SλL) + (ηS − R)sin(RλL)cosh(SλL)] b = 2GJλ [(ηS − R)cos(RλL)sinh(SλL) − (S + ηR)sin(RλL)cosh(SλL)]
If viscous damping is to be modelled then we simply replace η by ωξ in the equations above. From equation (7.29) the responses at either end of the bar may be found by putting x = 0 and x = L. 7.2.4 Tapered bars For the torsional vibration of tapered bars it not just a simple matter of changing parameters [2]. The solutions are significantly different. r ρ r −r Defining z = λ x + o where λ = ω and B = L o Entwistle et al [2] showed that, B G L Θ(z) 2z 2 o . − z 2 cosz − 3z sinz − 3cosz sinz − cos z L = L L L L z To ΔGπr 4λ o cosz + z2 sinz L + 3z L cos z L − 3sinz L sin z + .................. (7.30) L z Θ(z) 2z2 L . − z2 cosz − 3z sin z − 3cos z sin z − cosz o α xL = = o o o o z TL ΔGπr 4 λ L cosz + z2 sin z o + 3z o cosz o − 3sinz o sinz + .................. (7.31) o z α xo =
(
)
(
)
(
)
(
)
where Δ = z2 sinz o + 3z o cosz o − 3sin zo o
}{z2L cos zL − 3z L sin zL − 3coszL } − {z2 cosz o − 3z o sinz o − 3cosz o }{z2 sinz L + 3z L cos zL − 3sinz L } ................... o L { ( ) sinozo − coszo z
(7.32)
The tip receptances may then be shown to be, Θ(z o ) 2z2 o . − z2 cosz − 3z sin z − 3cosz L α oo = = L L L L To ΔGπr 4 λ o
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+ z2 sinz L + 3z L cos z L − 3sinz L L
(
)sin zo + coszo zo
.................. (7.33)
α LL =
Θ(z L ) 2z2 L . − z2 cosz − 3z sin z − 3cos z sin z L − cosz o = o o o o L TL ΔGπr 4 λ zL L cos zL .................. (7.34) + z2 sin z o + 3z o cosz o − 3sinz o sinz L + o zL
(
)
(
)
2z 4 2z 4 L o α oL = α Lo = 4 = ΔGπr L λ ΔGπr 4K o
............................................. (7.35)
7.3 Conclusions In this chapter we have seen that continuous systems have an infinite number of natural frequencies and associated mode shapes. 7.4 References 1. Derry, S., Drew, S. J. and Stone, B. J., 'The torsional vibration of a damped continuous bar.’ Abstract Proceedings of the VIII International Congress on Experimental Mechanics, Nashville Tennessee, 1996. 72-73. Entwistle R. D., Hesterman D. C. and Stone, B. J., 'Axial and Torsional Receptances for Tapered Circular Shafts', Modal Analysis: The International Journal of Analytical and Experimental Modal Analysis. Vol 11 (1997) 178-193. Next chapter
2.
.
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