Advanced Engineering Dynamics 2010 Part 8 docx

134 Impact and one-dimensional wave propagation Fig. 6.10 At x = -L the strain is always zero and hence g:-, (c,t + (-L) - (n - 1)2L) - fn (c,t - (-L) - n2L) = 0 or f"= gA-1 (6.29) At x = 0 there must be continuity of velocity. For the short bar the particle velocity is super- imposed on the pre-impact speed of J! Thus (6.30) V + cJ*, + c,gL = c2F; and the contact force is CIZlfn - c,z,g:, = -C2Z2Fn (6.3 1) (note that Z = force/velocity). From equations (6.29), (6.30) and (6.3 1) we obtain 2v/c, - (1 - z,/Z,)g:-, (1 + ZJZ2) (6.32) g'n = and (6.33) ) clzl c2z2 ( (1 + Z,/Z,) v/c, + 2gz-1 Fn = - Since the first waves are go and Fo it follows thatf, = 0 , g-, = 0 and F-, = 0. Let us first examine the waves immediately after the impact, that is for n = 0 - v/c, (1 + Z,/Z,) czz2 (1 + ZI/Z2) (6.34) Po = - ClZ, Vlc, ('6.35) gb = and for n = 1 Impact of two bars 135 v z,/z, (ZJZ, - 1) c2 (1 + Z,/Z,) (Z,/Z, + 1) F', = - (6.36) -2v/c, (1 + Z,/Z2) (6.37) If Z, is less than or equal to Z2 then F', is zero or negative. This means that the strain is zero or positive, that is tensile. Because a tensile strain is not possible at the interface the contact is terminated, the contact time being 2L/c,. fi=gb= The velocity at the interface is v = V + c,gb = c,Fb (6.38) ZJZZ (1 + z,/z2> =v In the special case when Z, = Z,, v = V/2. Figure 6.1 1 shows the progress of the wave. mine the wave functions. With a little algebra it can be shown that z, - z2 2c z, + z2 If Z, is greater than Z2 then equations (6.32) and (6.33) can be used repeatedly to deter- g:, = v [ 1 - ( )n+i] (6.39) and T: Fig. 6.1 1 I 36 Impact and one-dimensional wave propagation ZIYlC, z, - z2 (6.40) ( z, + z2 r F, = ZI + z2 so that the force transmitted is (6.41) from which we see that force decays exponentially. Also the velocities decay to zero so the coefficient of restitution, defined in the usual rigid body way, is zero. In the case where the two bars have the same properties and the second bar is the same length as the first the coef- ficient of restitution is unity, showing that this quantity can range from 0 to 1 even though the process is elastic. z, vz, z, - z2 Il ( ZI + z2 1 (EA)2F L = ZI + z2 6.7 Constant force applied to a long bar We shall now consider a long bar under the action of a constant force X applied to the face at x = 0 as shown in Fig. 6.12. If we assume that a wave travels into the bar with a speed c then we may use force = rate of change of momentum d dt X = - (~Av (et)) = ~AVC so -E = X/(AE) = pVdE By definition -E = vt I (et) = vlc Equating the two expressions for E gives pvclE = v/c or c2 = Elp as before. Fig. 6.12 Constant force applied to a long bar 137 Now let the bar be of finite length L, as shown in Fig. 6.13. At x = L the strain has to be zero. Therefore at any time E = -f, + gk-, = 0 S:-1 = fn or At x = 0 the force, X, is constant and therefore X = -EA ( rn + gl) = EA (frz - A-1 1 (6.42) and v = c ( fn + gl) = c ( fn +fn-J (6.43) From equation (6.42) X f =- n EA + A-I Thus X o EA f =- X 2x f1=,,+f,=- EA A = EA Hence (n + l)X Substituting into equation (6.43) (n + 1)X nX +-I EA v=c ( EA CX EA - (2n + 1) PAE Fig. 6.13 13 8 Impact and one-dimensional wave propagation now time t = n2LIc so the average acceleration is v cx C - (2n + 1) - t EA 2nl _-_ X PAL - (I + 1/2n) As n tends to infinity vx t PAL - So we see that the result is that which would have been given by elementary means. From this we learn the very important lesson that rigid body behaviour may be assumed when the variation of force is small compared with the time taken for the wave to traverse the body and return. After a few reflections the body behaves like a body with vibratory modes super- imposed on the rigid body modes. The wave method is most suitable when dealing with the initial stages which, in the case of impacting solids, may well be when the maximum strains occur. As mentioned earlier a vibration approach will require a large number of principal modes to be included. 6.8 The effect of local deformation on pulse shape In the previous analysis for which impact occurred between plane surfaces it is seen that the leading edge is sharp leading to instantaneous changes in strain and velocity. Although these are not precluded in continuum mechanics, in practice some rounding of the leading edge occurs largely due to the impacting surfaces not being plane. We shall assume that in the immediate vicinity of the impact point the material behaves as an elastic spring with linear or non-linear characteristics. Referring to Fig. 6.14 we see that the impacting surfaces are convex and the separation of the two reference planes is denoted by (so - a), a being the compression. It is assumed that the compressive force deflection law is of the form X= ka". The rate of approach of the two reference planes is a = Y + c,g' - CJ (6.44) and the contact force X = -(EA),g' = +(EA)# (6.45) Fig. 6.14 The efect of local deformation on pulse shape 139 Eliminating g' andf we get xc, xc, m a = V - - - - = V - ka (l/Zl + l/Z2) (E4 (E42 (6.46) Let h = k(l/Z, + l/&) so that equation (6.46) becomes a + ha" = V (6.47) If m = 1 then the interface behaves like a linear spring and the solution is, with a = 0 at t = 0, and (1 - e") ZlZ2 4 + z2 =v (6.48) from which we see that the maximum force is as given by equation (6.41) with n = 0. The Hertz theory of contact for two hemispherical bodies in contact states that where R is the radius and p = (1 - u)/(xE). (u = Poisson's ratio). We may write x = ka3'2 where -1 3x k = [ 4 (PI + P2)\ ($, + i2)] Equation (6.47) now becomes 6 + ha3/' = v or Using the substitution leads eventually to (6.49) (6.50) 21 213 +p+1 2p + 1 I= (!) 3v h [iln(71 - ) - ,3 arctan( T) +'$](6.51) Now 140 Impact and one-dimensional wave propagation x = h3’2 = - kV p 3 h and because as p + a, t + 1, V X,, = kVk = (6.52) (l/Z, + l/ZJ Thus X - = p3 XI, Introducing a non-dimensional time 213 113 - t=th v (6.53) leads to a plot of NX,, versus 7 being made. Figure 6.15 shows the plot. Equation (6.53) can be rearranged as - t=( 2,2 r( 33( :) (6.54) and taking u = 0.3 the constant evaluates to 0.478. Note that from equation (6.52) X,, is proportional to V. 3~(l - u’) Also shown on Fig. 6.15 is a plot of - X/X- = (1 - e-‘) (6.55) and this shows a reasonably close resemblance to the plot of (6.54). Equation (6.55) is of the same form as equation (6.48) which was obtained from the linear spring model. Thus by equating the exponents an equivalent linear spring may be obtained. Therefore (6.56) 33 113 h,t=?=h v t or k,(l/Z, + l/ZJ = (k(l/Z, + l/Z2))z3 V1’3 (6.57) where k, and 1, refer to the linear spring model. Figure 6.16 shows a plot of the rise time to three different fractions of the maximum ver- sus the product of impact velocity and nose radius. It is seen that as the nose radius tends to infinity the rise time tends to zero as was predicted for a plane-ended impact. Also as the impact velocity (or the maximum force) increases then the rise time decreases. Fig. 6.15 Prediction ofpulse shape during impact of two bars 141 Fig. 6.16 Rise time based on Hertz theory of contact 6.9 Prediction of pulse shape during impact of two bars We shall consider the impact of two bars having equal properties. One bar is of length L whilst the other is sufficiently long so that no reflection occurs in that bar during the time of contact. If we assume a plane-ended impact then the contact will cease after the wave has returned from the far end of the short bar, that is the duration of impact is 2Llc. Because the rise time, in practice, is finite several reflections will occur before the con- tact force reduces to zero and remains zero in the long bar. The leading edge profile has been predicted in the previous section using the Hertz theory of contact where it was also shown that this could be approximately represented by an exponential expression. To simplify the computation we shall adopt the exponential form. Figure 6.17 shows the x, t diagram (which is similar to Fig. 6.10). At x = -L, E = 0 and thus fn - gb-, = 0 or fn = gb-1 &'-I =O) (6.58) At x = 0 the difference in the velocity of the reference faces is a, (V + CA + Cg;) - cF~ = a,, or VJC + fn + g: - Fi = U,/C (6.59) Also, by continuity of force, -(EL4g', - EAfn) = -(-EAF;) = ka, 142 Impact and one-dimensional wave propagation Fig. 6.17 or k X - g: =gan (6.60) and k EA F:, = - a, (6.61) Adding equations (6.59), (6.60) and (6.61) gives V ci, 2k - + 2f, = - + - an c c EA (6.62) From equations (6.58) and (6.60) k fn = g' = f - - a,,+l n-' EA n- I As fo = 0 k f;= EA a' and k k f; = f; - E a, = - - (a0 + a,) fn=-EC a, EA so (6.63) k n-' 0 Substituting equation (6.63) into equation (6.62) gives .PDrediction ofpulse shape during impact of two bars 143 V 2k 1 dun 2k - - - Ea, = - - + - a,, c dt EA c EA We define the non-dimensional quantities 0 - 2kc a =a- EAV and - 2kc t=t - EA Thereby equation (6.64) can be written as n-l _- (6.65) (6.66) (6.67) 0 As C has to be continuous tin (0) = tin-1 @I (6.68) The parameter p is the value of 7 when ct = 2L, that is the time at which the wave in the short bar returns to the impact point. From equation (6.66) 4Lk P=E - Multiplying equation (6.67) by e' gives d n- I e;(l + E;,)= e; (3 + ti,,) = (e' a,,) 0 and integrating produces - - n-l - a,, = e-' [ J e' (1 - 5 di, ) d'i + constant] Carrying out the integration - - - a. = e-' [ J e' (1 - 0) dI + constant] - = e-' [e' - 11 = (1 - e-') (the constant = 1 as Go = 0 when i = 0) - - - - a, = e-' [ J e' [ 1 - (1 - e-')] d? + constant] (6.69) (6.70) (6.7 1) - = e-' [I + tio@)] (6.72) the constant being determined by the fact that GI (0) = Eo@). Continuing the process - - a, = e-' [ J e' (1 - e-' [e' - 1 + 'i + Eo@)]} dI + constant] - -2 t = e-' [ e ' - e-' + I - - - I - ti@) +constant] 2 [...]... z + 2 (4 (f rz2 + B2 ] (6 .81 ) where and e-’LL’L c [ - - [ -4p+ 2p L e (2 - 4p) + 312 (6 .82 ) where B, = e-6p + e-4’ (1 - 8p) + e-’’’ (1 - 8p + 8p’) + 1 At the impact point, x = 0, the velocity and the strain are given by vo = cfn + Cg; = C(fn - A-1) (6 .83 ) and = - f n + gl = -K + A-1) At the fixed end the velocity is zero but the strain is EO EL = -fn +gh+l = -2fi (6 .84 ) (6 .85 ) Figure 6.2 1 gives plots... frequency and o is the circular frequency Figure 6.24 is a similar plot but this time versus x An increase of 21.r in the argument corresponds to a change in x of one wavelength h Thus kh = 2n (6 .88 ) k = 2dh (6 .89 ) or k is known as the wavenumber Fig 6.24 150 Impact and one-dimensional wave propagation Hence kx = at - kx z = ckt - (6.90) and 0 e = - = k (6.91) cp This quantity is called thephase velocity... ‘I2 ] 149 (6 .86 ) A plot of this aproximation is included on Fig 6.22 6.11 Dispersive waves Let us first discuss the sinusoidal travelling wave The argument for a sinusoidal function is required to be non-dimensional so we shall adopt for a wave along the positive axis 2 = k(ct - x) = (ckt - kx) where k is a parameter with dimensions lllength A typical wave would be u = UCOS - kx) (ckt (6 .87 ) Figure 6.23... V/c and V f = - e-’l;’L 0 (6. 78) C This finction is valid until the wave returns from the far end, that is, when z = 2L or t = 2L/c at x = 0 For t > 2LJc = e-P’L - e-@‘ f, [ ( -E Lc [ - -v z + B, ] 2p J e’lLIL e-’li/L - pve-p/L Lc LC Now the velocity must be continuous at x = 0 so that ) d~ + B , ] Impact o a rigid mass on an elastic bar f 147 (6.79) c C and hence (6 .80 ) The same procedure can be employed... be observed Since the amplitude appears in both numerator and denominator small variations will have little effect This is borne out by Figs 6. 28 and 6.29 where the curves for a square pulse and triangular pulse are seen to be very close together Figure 6. 28 shows the three wave velocities versus k for a concave downwards dispersion curve Here the group velocity drops quickest and eventually becomes... case the contact force can be deduced either by measuring the strain ( E ) by means of strain gauges 7 = tK( I , + In, ) I p = tiww o arrival offirst reflectionjvmfiee endof the f impacting baz Fig 6. 18 Pulse shapes for varying lengths of impacting bar (a) Fig 6.19(a) Measured pulse shapes showing variation with bar length Impact of a rigid mass on an elastic bar 145 &) Fig 6 1 ( ) Measured pulse shapes... that a first approximation in this type of problem is to add one-third of the mass of the rod to that of the end mass For this case we equate the initial kinetic energy with the final strain energy 1 48 Impact and one-dimensional wave propagation Fig 6.21 Fig 6.22 Maximum strain at x = L for various period numbers, n 2 12 1 X’L (M + pAL/3) V = - - = - 2 2 k 2EA 1 - 1 1 = - EALE’ = - C’PALE’ 2 2 Hence... impacting bar it is possible to deduce the compression across the contact region and thereby obtain the dynamic characteristics of that region or of a specimen of other material cemented there This is particularly useful for tests at high strain rate where the bars can be sufficiently long for the reflected waves to arrive after the period of interest 6.10 Impact of a rigid mass on an elastic bar In... assumed to be of such short duration that a rigid body approximation is practicable figure 6.20 shows the relevant details; for this exercise the far end of the bar is taken to be fixed At the far end the particle velocity is zero and thus when x = L v = cyn + cg;+, = 0 or g: = L-I (6.76) At x = 0 the contact force is a‘u X = -(-EAYn + EAg;) = - M T = -M(c2fl c’g:) + at Let EA - PAL _ Mc2 ML Therefore... wavenumber) The functions depend on the shape of the pulse and are given by standard Fourier transform techniques We seek the peak of the pulse therefore at the peak a -U C q k , sin(o,t - k,x) = 0 (6. 98) d X Fig 6.27 Dispersive waves 153 In the neighbourhood of the peak and for small dispersion the argument of the function will be small so that sin z may be replaced by z This gives t C U,k,a, = x Z . + 312 f3 = - e-’LL’L - - 2p [ -4p+ c [Le (6 .81 ) (6 .82 ) where B, = e-6p + e-4’ (1 - 8p) + e-’’’ (1 - 8p + 8p’) + 1 At the impact point, x = 0, the velocity. Cg; = C(fn - A-1) (6 .83 ) and EO = -fn + gl = -K + A-1) (6 .84 ) At the fixed end the velocity is zero but the strain is EL = -fn +gh+l = -2fi (6 .85 ) Figure 6.2 1 gives. but this time versus x. An increase of 21.r in the argument cor- kh = 2n (6 .88 ) or k = 2dh (6 .89 ) k is known as the wavenumber. Fig. 6.24 150 Impact and one-dimensional