Advanced Engineering Dynamics 2010 Part 9 potx

1 54 Impact and one-dimensional wave propagation Fig. 6.29 Wave velocities for positive curvature dispersion curve upwards. It is conventional to plot the curves with k as the independent variable because o is univalued for a given k, but not vice versa. Figure 6.30 shows four symmetrical pulse shapes and their respective Fourier transforms. The ordinate is the square of the amplitude as this gives a measure of the energy and it is Fig. 630 Fourier transform pairs Waves in a uniform beam 155 seen that most of the energy is accounted for before the abscissa value reaches x. This gives credence to the approximations r/T, = 1 and L/h, = 1. The practical difficulty is the determination of the width of the peak and hence the appro- priate value of k,,,. Measured values tend to be towards the phase velocity but are far from the group velocity. Although the pulse-peak velocity is as precisely defined as the other two it does emphasize the fact that the group velocity is valid only for a narrow frequency band. In a highly dispersive situation the group velocity gives the arrival time of specific wave- lengths which were generated by an impact. Dispersion indicates that short pulses will spread out but the total energy remains con- stant. The term is not to be confused with dissipation in which some of the mechanical energy is converted to thermal energy. 6.12 Waves in a uniform beam In this section we shall be examining lateral waves in a long uniform beam, shown in Fig. 6.3 1, with a cross-sectional area A and a second moment of area Z about the z axis through the centroid. The xy plane is a plane of symmetry. The material has a Young’s modulus E, a shear modulus G and the density is p. We are going to use Hamilton’s principle to obtain the equations of motion because it is easier to modify the model. The exact equations are very involved and therefore approximations are required. A first approximation is to consider only kinetic energy due to lateral motion and strain energy due to bending strains. Later we shall include rotary inertia and shear strain energy. The simple case can be obtained by free-body diagrams and Newton’s laws but we shall use the vari- ational method and subsequently modify the Lagrangian to take into account the extra terms. Refemng to Fig. 6.32 we can write an expression for the kinetic energy (note that in this section v is the deflection in they direction) TI =soy (4 dr (6.101) Lp~ av ’ 156 Impact and one-dimensional wave propagation Fig. 632 and for the strain energy L~~ a0 z V,=S,1(3+ (6.102) If we assume that there are no external forces Hamilton’s principle states that t2 6 J (TI - Vl)dt = 0 tl or 2 t2 L pA aV 2 EI 80 61 s[ -T(z 1-1 (dx) I drdt=o (6.103) 0 tl Carrying out the variation first L t2 /,It [PA g 0 (g ) - EI; 0 (z) dxdt = 0 (6.104) I For the first term we reverse the order of integration and integrate by parts to obtain L t2 t2 t2 / [ It pA z 6 ($) dt ] dr =IL[ pA$ 6v It -Itl$ 6v dt] dx (6.105) 01 0 I Because GVvanishes, by definition, at t, and t2 the first term in the square brackets is zero. It, I [ EI E 6 ( z)] dx dt =I [ EIg 60 lo - 1; 2 60 dx] dt (6.106) For the second term we integrate by parts with respect to x to obtain t2 LL 12 L 0 0 Waves in a unifonn beam 157 We have already assumed for this exercise that there are no external active forces. Therefore the end constraints must be workless and this implies that either 60 or a0lax must be zero at the ends. Hence the first term is zero. Combining the two results, equations (6.105) and (6.106), gives 1; I,' ( EZ$ 60 - pA v and B are not independent but are related by geometry dV -0 ax so that av 60 = 6 ( dx) (6.107) (6.108) (6.109) Integrating the first term once more by parts and noting that the end forces are workless leads to and finally since 0 = avBx (6.110) Because 6v is arbitrary (except at t, and t2 where it is zero) the expression in the large paren- theses must be zero. Thus a4v d2V ax at EI- + PA 7 = 0 (6.1 11) This equation is known as Euler 's equation for beam vibration and is widely used. This form can readily be deduced from free-body diagrams in a similar method to that used for longi- tudinal waves in a bar. The reason for using Hamilton's principle here is to expose the details of the method and to form the basis for development of a more refined model. We now add an extra kinetic energy term to take into account rotary inertia. For a thin ele- ment of beam the moment of inertia about a z axis through the centroid is pZ dx, where Z is the second moment of area. Therefore the kinetic energy is (6.112) Thus 1 58 Impact and one-dimensional wave propagation t2 (6.1 13) The rate at which the lateral deflection changes with x is now augmented by y, the shear strain, giving av ax _- -0+y and the additional strain energy is (6.1 14) (6.115) The constant K (kappa), which is greater than unity, corrects for the fact that the shear stress is not uniformly distributed across the cross-sectional area. For a rectangular cross-section K = 6/5; this is based on a parabolic shear stress distribution. Thus This time we shall not make the first term zero as we are now admitting external forces. vention is used) From Fig. 6.3 1 the virtual work done by the external forces is (note that a vector sign con- (6.1 17) 6W = MI601 + MI602 + SI6vl + S26~2 Therefore Hamilton's principle for the modified model including external forces is t2 tl tl 6 Ifi(q + T2 - VI - V2) dt + GWdt = 0 (6.1 18) Using equations (6.105), (6.106), (6.1 13), (6.1 16) and (6.1 17), equation (6.118) may be written J: Ji { [ -PAY,, + KGA(v, - 0.~1 dv . . . + [ -pI0+,, + EI0, + KGA(v, - 0)] 60 } dxdt (6.1 19) Here the notation a2vldx' = v, etc. is used. double integrals must each equate to zero. Thus the two equations of motion are Because 6v and 60 are arbitrary between I, and t2 the factors of 6v and of 60 under the Waves in a uniform beam 159 and 2 2 (6.120) (6.121) Summing each of the coefficients of 60,, 60,, 6v, and 6v2 to zero gives the boundary conditions (EI g ), = -MI (EZ )2 = M2 au KGA ( dx - 0) = -VI I au KGA ( dx - 0) = -V2 2 (6.122) (6.123) (6.124) (6.125) It is now possible to eliminate 0 between equations (6.120) and (6.12 1). Equation (6.120) can be written as 2 2 V.lf - KC, V-cr KC, 0, = 0 Therefore 1 0, - v.u - - 2 VJI KC, - Equation (6.12 1) is written as and differentiating partially with respect to x gives (6.126) (6.127) or This is known as the Emoshenko beam equation. 160 Impact and one-dimensional wave propagation For a running wave solution v = ;d(ar - W so substituting into equation (6.128) and dividing through by the common factor gives 1 2 o4 = 0 (6.129) which is the dispersion equation for bending waves in a uniform beam. This equation is a quadratic in o2 and therefore yields two values of o for any value of k. The lower of the two, the first mode, approximates to Euler's equation for small values of k (k c 0.2, i.e. wave- lengths longer than about five times the beam depth). Plots of a, cp and cg are shown in Fig. 6.33. The phase velocity tends to a maximum value which is close to the velocity of pure shear waves (see section 7.5). The group velocity also tends to the same value but passes through a maximum for wavelengths of the order of the depth of the beam. It follows that after a short-duration impact these wavelengths are the first to arrive at a distant point but most of the energy will follow at longer wavelengths. The model becomes invalid when the wavelengths are very short compared with the depth of the beam in which case the wave speed will tend to that of surface waves which have a speed a little less than the shear wave speed. Figure 6.34 is a similar plot but for the higher, or second, mode. It is seen that there is a minimum frequency; below this no travelling wave is possible in this mode. The consequence of this is that the phase velocity tends to infinity at very low wavenum- bers but the group velocity remains finite and less than c,. The validity of this mode is not as good as the first mode and is probably only witnessed in I-section beams where the end load is camed mainly by the flanges and the shear is carried mainly by the web. c',k '4 - - 12 0 -( 1 +$)02k2+o (I/A 1 Fig. 633 Uniform beam mode 1 Waves in periodic structures 16 1 Fig. 6.34 Uniform beam mode 2 6.13 Waves in periodic structures The type of structure envisaged here is the continuous mass-spring system shown in Fig. 6.35. Away from any boundary we can use the same form of expression for displacement as used in the earlier sections but in place of the continuous location x we have a discrete num- ber of locations n. The mass of each body in the system is m and the stiffness of each spring is s. For the nth body the equation of motion is S(U,+~ - u,) - s(u, - n,-,) = mu, (6.130) Let us assume (6.13 1) Substituting equation (6.13 1) into equation (6.130) and dividing through by the common factor Ue'" we obtain u, = (,r&"' - kn) = ue'" e-Jh - e-Jkn) - s (e-Jh - e-Jhn-l)) = -mI1zoze-Jh s (e-JMn+1) (e-Jk - 1) - (1 - e'') = - mw'/s Dividing further by KJh gives and as e*Jk = cosk fj sink Fig. 6.35 162 Impact and one-dimensional wave propagation we get 2 2 COS (k) - 2 = - - mw S or 2 2( -2 sin2 (W2)) = - = S giving o = 21 (s/m) sin(W2) (6.132) Figure 6.36 is a plot of the dispersion diagram. From this we see that there is a cut-off fie- quency, a,, = 2J (slm), above which no continuous wave will propagate. At this point k = X SO 1( = vi@ e-jnn = vi" (-1)n that is, each body is in phase opposition with its neighbours. The phase velocity is 0 sin (k/2) k k/2 cp - - J (s/m) (6.133) and the group velocity is do gdk c =-= J (s/m) cos(W2) (6.134) If we impose a vibration above the cut-off frequency then one solution is to assume that each body has the opposite phase to its neighbour and that the amplitude decays exponen- tially, (6.135) U" = U(- 1)" e" e-Kn Substituting equation (6.135) into equation (6.130) leads to o = 2J (dm) cosh (k'/2) (6.136) for o > oca. Here the disturbance remains local to the point of initial excitation and does not propagate; such a mode is said to be evanescent. Fig. 636 Dispersion diagram for mass-spring system Waves in a helical spring 163 6.14 Waves in a helical spring The helical spring will be treated as a thin wire, that is plane cross-sections remain plane. This assumption has been shown to be acceptable by mechanical testing. An element of the wire has six degrees of freedom, three displacements and three rotations, so the possibility exists for six modes of propagation. If the helix angle is small these modes separate into two groups each having three degrees of freedom, one set consisting of the in-plane motion and the other the out-of-plane motion. The effect of helix angle will be discussed later but here we shall develop the theory for out-of-plane motion for a spring with zero helix angle. This, as we shall see, is associated with axial motion of the spring, that is with the spring being in its compression or tensile mode. Figure 6.37 defines the co-ordinate system to be used. The unit vector i is tangent to the axis of the wire,j is along the radius of curvature directed towards the centre and k com- pletes the right-handed triad. For zero helix angle k is parallel to the axis of the spring. The radius of curvature is R and s is the distance measured along the wire. 8 is the angle through which the radius turns. Thus ds=Rde (6.137) First we shall consider the differentiation of an arbitrary vector V with respect to s (6.138) where the prime signifies differentiation with respect to non-rotating axes and SZ is the rate of rotation of the axes with distance s. Thus - d’V+ R x V dV dsds de de dsds R=-=-k and using equation (6.137) 1 R Q=-k (6.139) Fig. 637 [...]... 6 39 Dispersion diagram for helical spring =o (6.1 89) Waves in a helical spring 1 69 Fig 6.40 For low values of wavenumber equation (6.1 89) reduces to K2q (6. 190 ) 1 + qa2 Now a2 =2R2/r2for solid circular cross-section wire The spring index (Rlr) is unlikely to ’ be less than 3 so the minimum value of a is about 18; a more typical index of 5 gives a2 = 50 Since q is of the order unity equation (6. 190 )... P, = @0,/2 (6.168) (6.1 69) (6.170) Substituting the six equations of state ((6.155) to (6.160))into the equations of motion ((6.162-6.164)and (6.168-6.170))will yield six equations in the six co-ordinates and these Waves in a helical spring 167 will separate into two groups of three So, substituting equations (6.157), (6.158) and (6.1 59) into equations (6.164), (6.168) and (6.1 69) leads to G J ( p 2... typical value for circular cross-sections is 0 .9 The relationship between the elastic constants, Young's modulus E, shear modulus G and Poisson's ratio u is E = 2G(1 + u) ,so let (6.154) Combining equations (6.148) to (6.154) Pi = GA (2mpui - 2mujlR) (6.155) Pi = GA (VU, + p i I R - 90 3 (6.156) pk = GA (qPuk + qOj) (6.157) For bending we use the usual engineering relationships for bending and torsion... index of 5 gives a2 = 50 Since q is of the order unity equation (6. 190 ) is, to a close approximation, w 2 = W’ = (Kla)’ (6. 191 ) Returning to the dimensional form ~ = c2 k = rc2 k a R,2 from which the phase velocity and the group velocity are given by cp = cg = c2/a (6. 192 ) (6. 193 ) This approximation is quite reasonable for wavelengths longer than five turns Also shown on Fig 6.40 are the amplitude ratios... strain (4= [TI1 (u) - [@IX ( W I d s (6.146) = [TI1 (4+ [d9/WX0 ) ( Now (ds) = (ds 0 0) and therefore M=l 0 0 0 -1 0 ds 0 0 1 (6.147) = [T21 0 (see appendix 1) The components of strain are, therefore, axial strain = pui - ujlR shear strain = puj + u,lR - = puk + shear strain &k (6.148) 0k 0j These are related to the elastic constants by (6.1 49) (6.150) Waves in a helical spring E ' Pi EA =- 165 (6.151)... 1 - a2q ) 0 -qa2j ][ ik] = E'[ t] (6.177) 4p" 9j2 (Note that this matrix equation can be written in symmetrical form; however, we can discuss the manner of wave propagation just as well in the current form.) In a manner similar to previous cases we shall assume a wave travelling along the axis of the wire Thus 0, = 0, e J(0f - b) GkeJ("'-k') (6.178) (6.1 79) (6.180) = R (-Jk)0, (6.181) - 0' e ~ ( w f... Coilradius R Wmradius r Mlterial, steel I k (fi) o s" 11.5 50.0 mm 12.5 mm EN 498 4 Galcdatedjvm jidl theory U e appmximatethemy Meanrredm.wmantjkpency, mode s h q e idmt@d M a r e ~ o n a n t f h p e n c.mode shape not ichtij%d enrd y Fig 6.41 Dispersion curve for helical spring (data from Ph.D thesis, H.R Hamson 197 1) In section 6.12 the dispersion diagram for a periodic massspring system was developed... cross-section has point symmetry then with J being the polar second moment of area 4 = Ik = Ii/2 = J12 Also E = G2m so that E4 = EIk = mGJ, and therefore d0; Mi = GI; - = G J @ 0 ; - B~IR) ds (6.158) (6.1 59) (6.160) The equations of motion can be derived with reference to Fig 6.38 Resolving forces acting on the element, neglecting any external forces, a or, letting D = at 1 66 Impact and one-dimensional... direction of wave propagation In the case of the helical spring the path of propagation is curvilinear, namely that along the wire axis We now consider a homogeneous, isotropic, linearly elastic solid The dynamics of such a solid are completely defined by three constants: the density and two elastic moduli There are six elastic constants in general use: Young’s modulus E, the shear modulus or modulus of . qKZ) Fig. 6 39 Dispersion diagram for helical spring (6.1 89) =o Waves in a helical spring 1 69 Fig. 6.40 For low values of wavenumber equation (6.1 89) reduces to (6. 190 ) Now a2 =2R2/r2. is of the order unity equation (6. 190 ) is, to a close approximation, W’ = (Kla)’ (6. 191 ) w2 = K2q 1 + qa2 Returning to the dimensional form k (6. 192 ) c2 rc2 ~=-k=- a R,2 from. groups of three. So, substituting equations (6.157), (6.158) and (6.1 59) into equations (6.164), (6.168) and (6.1 69) leads to m R GJ(p2g - p0,IR) - GJ- + 0JR) = pJD20, (6.171)