CHAPTER 12 PROBLEMS AND EXERCISES

Transcription

1 CHAPTER 1 PROBLEMS AND EXERCISES Problem 1: Consider a PWAS transducer of length l 7 mm, width b 1.65 mm, thickness t 0. mm, and material properties as given in Table 1.. The PWAS is bonded at one end of a 3 1-mm thick 00-mm long aluminum strip with E 70 GPa, 700 kg/m, The PWAS length is oriented along the strip. The PWAS is excited with a 3.5-counts tone burst. (i) Calculate the first frequency at which the S0 wave propagation is dominant. (ii) Assuming that the frequency is adjusted to the value at which the S0 wave propagation is predominant as calculated in part (i), calculate the time taken by the wave packet to travel to the other end of the strip specimen and come back. Sketch the wave pattern. (iii) Repeat part (ii) assuming that a through-the-thickness crack reflector is present at 400 mm from the PWAS. Sketch the wave pattern. (iv) Superpose the effects of (ii) and (iii) assuming that the energy is equally partitioned between the waves reflecting from the end and reflecting from the crack. Sketch the wave pattern. Discuss. Solution (i) To calculate the tuning frequencies, use the tuning theory developed in textbook Chapter 11. For rectangular PWAS, one refers to Section.4.5, specifically Eq. (.1), i.e., S S S A 0 0 A a ( 0 ) ( 0 ) ( 0 ) A a NS i x t aa NA i( 0 xt) x ( xt, ) i sin a e i sin a e S A DS( 0 ) DA( 0 ) (.1) (1) One could code Eq. (1) or could use the software programs posted on the LAMSS website In particular, download and activate the following program: After activating the program, one obtains the chart below. 1 1

2 Since the problem asks us to find the first tuning frequency, the chart is restricted to lower frequencies. In the considered frequency range, only two Lamb wave modes exist, S0 and. In the chart, the S0 mode is presented in solid line ( symmetric ), whereas the mode is presented in dashed line ( anti-symmetric ). Examination of this tuning chart indicates that the mode goes through a minimum (mode rejection) at 180 khz. When one mode is rejected, the other becomes dominant. Hence, the first frequency at which the S0 mode is dominant is S 0 f 1 =180 khz. ============= (ii) Assuming that the frequency is tuned at 180 khz (i.e., when S0 is dominant), we calculate the group velocity for this mode at this frequency. By plotting the group velocity, we obtain the chart below

3 The chart shows that, in this frequency range, the S0 group velocity (solid line) varies very S 0 slowly with frequency. The exact value is calculate separately as c 5387 m/s. For convenience, we express the speed in mm/μs, i.e., c S 0 g f 180 khz g f 180 khz mm/μs. Using group velocity, one calculates the round-trip time of flight (TOF) to the end of the strip as L 00 mm tend S s () cg mm/ s To calculate the wave reflection process, define the incident forward wave and the reflected backward wave as the following continuous waves (CWs: S 0 i( x t) u ( x, t) Ae incident forward wave (3) i S 0 i ( x xr ) t S 0 i ( x x ) it R ui ( x, t) Be Be e reflected backward wave (4) S0 where x R is the reflector position and is the frequency dependent wavenumber defined as S 0 frequency-dependent wavenumber (5) S 0 c S 0 where c is the phase velocity of the S0 mode. At this relatively low frequencies, the S0 mode is very slowly dispersive, hence we will take a linear approximation of its variation over the frequency range of interest. The total wave is the summation of the incident and reflected waves, i.e., S 0 S 0 i x i ( x x ) (, ) (, ) (, ) R it uxt u xtu xt Ae Be e complete wave (6) i r For a generic situation, we take A 1. Assuming perfect reflection at the x R L end of the strip, we take Ba 1. For a given excitation signal st (), e.g., a tone burst, the solution is obtained by performing the convolution between the excitation signal and the CW definition of Eq. (6). The convolution is performed in the frequency domain as a multiplication. The resulting wave pattern is shown below; the transmitted incident wave package and received reflected wave packet are clearly defined. The TOF can be easily estimated. The t 371 μs value calculated with Eq. () matches well the estimation obtained from the chart. end Initial bang End reflection

4 (iii) Assuming a through-the-thickness crack reflector at L c=400 mm, we calculate the time of flight from the crack reflector as Lc 400 mm tcrack S s (7) cg mm/ s The wave pattern of the transmitted and received wave packets is shown below. The t crack can be easily estimated. The value calculated with Eq. (7) matches well the estimations obtained from the chart. Crack Initial bang reflection (iv) Assuming equal partition of energy between the end reflection and crack reflection we take B B A/. The superposed plot of the two reflections is shown below. end crack Discussion: One sees the reflections from the crack and the reflection from the end. The t end and t crack can be easily estimated. The values calculated with Eqs. () and (7) match well the estimations obtained from the chart. Initial bang Crack reflection End reflection

5 PROBLEM 1.1 SOLUTION Units khz 00 Hz MHz 6 Hz krad 3 rad ms 3 s s 6 s pf 1 F nf 9 F m 6 m 6 MPa 6 Pa GPa 9 Pa TPa 1 Pa MN 6 N GN 9 N ms 3 S S 6 S ns 9 S k 3 M 6 G 9 SETTINGS AND CONTROLS fs 1.5 MHz dt 1 fs dt s s fnq fs tmax 5000 s log tmax log dt s s im ceil im 13 NN im NN 819 log( ) NN NQ Nt NN Nf NQ T NN dt T 5461 s 1 df df 183 Hz T Display controls: Tmax 6 6 T floor 00 Tmax 5000 s fmax 00 khz 00 Frequency range: nf 0 Nf f nf df f 750 khz nf Nf nf f nf Nf 471 krad s 5

6 definition of the space domain dx 1.0 m Nx nx 0 Nx 1 x nx dx nx x 00 mm 1 Specimen properties Aluminum strip E 70 GPa 700 kg m d 1 mm L 00 mm Lc 400 mm L L Lc Lc 6

7 Hanning smoothed tone burst Nc 3.5 fc 180 khz c fc Signal amplitude p0 Window size function of central frequency and Ncount 1 1 a Nc a 9.7 s fc Define time range t0 a nt 0 Nt 1 t t0 nt dt t nt Nt s Ht ( ) p0 if 0 t t0 a 0 otherwise Ht ( ) p0 sin t t0 if 0 t t0 a a 0 otherwise h Ht HH FFT ( h ) nt nt HS HH nf nf square window Hanning window and its FFT ct ( ) sin c ( t t0) sc ct SC FFT ( sc ) nt nt ss( t) sin c ( t t0) Ht () s_ ss t S FFT ( s_ ) nt nt carrier signal and its FFT tone burst signal and its FFT s_ nt t nt time, microsec 0.01 S nf f nf 7

8 S0 and axial wave speed in aluminum strip E c0 1 cs0g 5387 m s cs0_0khz 5393 m cs0_300khz 5387 m s s c m cs0 cs0_0khz s cs0g 5387 m nf s cs0_300khz cs0_0khz Nf nf Nf cs0 nf f nf khz 8

9 S0 waves analysis Basic wave propagation analysis nf wave numbers nf cs0 nf medium transfer function for the Nx locations x in space domain G e i nfx nx nf nx excitation spectrum x medium transfer function at all x locations U S G nf nx nf nf nx IFFT to return to the time domain u nx IFFT U nx plotting offset control End reflection analysis End reflection TOF estimate: cs0g mm L t_end t_end 371 s s cs0g End reflection mean the whole signal is returned, hence A=B A 1 B A U S G nf nx nf nf nx u nx IFFT U nx G nf nx A e i nfx nx B e i nf x nx L

10 Crack reflection analysis Crack reflection TOF estimate: t_crack Lc cs0g t_crack 149 s Ac A G nf nx A e i nfx nx Ac e i nf x nx Lc U S G nf nx nf nf nx u nx IFFT U nx G nf nx End reflection and crack reflection of axial wave (half-half energy split): B A Ac A A e i nfx nx B e i nf x nx L Ac e i nf x nx Lc U S G nf nx nf nf nx u nx IFFT U nx

11 Problem : Consider a PWAS transducer of length l 7 mm, width b 1.65 mm, thickness t 0. mm, and material properties as given in Table 1.. The PWAS is bonded at one end of a 1-mm thick 00-mm long aluminum strip with E = 70 GPa and =.7 g/cc. The PWAS length is oriented along the strip. The PWAS is excited with a 3.5-counts tone burst. (i) Calculate the first frequency at which the wave propagation is dominant. (ii) Assuming that the frequency is adjusted to the value at which the wave propagation is dominant as calculated in part (i), calculate the time taken by the wave packet to travel to the other end of the strip specimen and come back. Sketch the wave pattern. (iii) Repeat part (ii) assuming that a through-the-thickness crack reflector is present at 400 mm from the PWAS. Sketch the wave pattern. (iii) Repeat part (ii) assuming that a through-the-thickness crack reflector is present at 400 mm from the PWAS. Sketch the wave pattern. (iv) Superpose the effects of (ii) and (iii) assuming that the energy is equally partitioned between the waves reflecting from the end and reflecting from the crack. Sketch the wave pattern. Discuss. Solution (i) To calculate the tuning frequencies, use the tuning theory developed in textbook Chapter 11. For rectangular PWAS, one refers to Section.4.5, specifically Eq. (.1), i.e., S S S A 0 0 A a ( 0 ) ( 0 ) ( 0 ) A a NS i x t aa NA i( 0 xt) x ( xt, ) i sin a e i sin a e S A DS( 0 ) DA( 0 ) (.1) (1) One could code Eq. (1) or could use the software programs posted on the LAMSS website In particular, download and activate the following program: After activating the program, one obtains the chart below. 5 11

12 This chart indicates that the first maximum of the is obtained around ~50 khz. At this frequency, the S0 response is relatively weak. Hence, this is the first frequency at which the mode is dominant, i.e., f1 =50 khz 50 khz. (ii) Assuming that the frequency is tuned at 50 khz (i.e., when is dominant), we calculate the group velocity for this mode at this frequency. By plotting the group velocity, we obtain the chart below 6 1

13 The chart shows that, in this frequency range, the group velocity (dash line) varies very rapidly with frequency. The exact value is calculate as c 1355 m/s 1355 mm/μs. g f 50 khz Using group velocity, one calculates the round-trip time of flight (TOF) to the end of the strip as t end L 00 mm 1476 s () c mm/ s g To calculate the wave reflection process, define the incident forward wave and the reflected backward wave as the following continuous waves (CWs: i( x t) u ( x, t) Ae incident forward wave (3) i i ( x xr ) t i ( x x ) it R u ( x, t) Be Be e reflected backward wave (4) i where x R is the reflector position and is the frequency dependent wavenumber defined as frequency-dependent wavenumber (5) c where c is the phase velocity of the mode. At this relatively low frequencies, the a0 mode is very dispersive and varies like ; hence we will take a approximation of its variation over the frequency range of interest. The total wave is the summation of the incident and reflected waves, i.e., i x i ( x x ) (, ) (, ) (, ) R it uxt u xtu xt Ae Be e complete wave (6) i For a generic situation, we take A 1 r. Assuming perfect reflection at the x R L end of the strip, we take B a 1. For a given excitation signal st (), e.g., a tone burst, the solution is obtained by performing the convolution between the excitation signal and the CW definition of Eq. (6). The convolution is performed in the frequency domain as a multiplication. The resulting wave pattern is shown below. Initial bang 0 End reflection The transmitted incident wave package and received reflected wave packet are clearly defined. The TOF can be easily estimated. The tend 1476 μs value calculated in Eq. () matches with CG of the envelope of the plot of the squared dispersed wave signal shown on the graph. 7 13

14 (iii) Assuming a through-the-thickness crack reflector at L c =400 mm, we calculate the time of flight from the crack reflector as Lc 400 mm tcrack 590 s (7) cg mm/ s The wave pattern of the transmitted and received wave packets is shown below. The t crack can be easily estimated. The value calculated with Eq. (7) matches well with CG of the envelope of the plot of the squared dispersed wave signal shown on the graph. Initial bang Crack reflection (iv) Assuming equal partition of energy between the end reflection and crack reflection we take B B A/. The superposed plot of the two reflections is shown below. end crack Initial bang 0 Crack reflection End reflection Discussion: One sees the reflections from the crack and the reflection from the end. The t end and t crack can be easily estimated from the CG of the dispersed wave packets shown on the graph. The values calculated with Eqs. () and (7) match well the estimations obtained from the chart

15 PROBLEM 1. SOLUTION Units khz 00 Hz MHz 6 Hz krad 3 rad ms 3 s s 6 s pf 1 F nf 9 F m 6 m 6 MPa 6 Pa GPa 9 Pa TPa 1 Pa MN 6 N GN 9 N ms 3 S S 6 S ns 9 S k 3 M 6 G 9 SETTINGS AND CONTROLS fs 1.5 MHz 1 dt dt s fs fnq fs tmax 5000 s log tmax log dt s s im ceil im 13 NN im NN 819 log( ) NN NQ Nt NN Nf NQ T NN dt T 5461 s 1 df df 183 Hz T Display controls: Tmax 6 6 T floor 00 Tmax 5000 s 00 fmax 00 khz Frequency range: nf 0 Nf f nf df f 750 khz nf Nf nf f nf Nf 471 krad s 15

16 Definition of the space domain dx 1.0 m Nx nx 0 Nx 1 x nx dx nx x 00 mm 1 Specimen properties Aluminum strip E 70 GPa 700 kg m d 1 mm L 00 mm Lc 400 mm L L Lc Lc 16

17 Hanning smoothed tone burst Nc 3.5 fc 50 khz c fc Signal amplitude p0 Window size function of central frequency and Ncount 1 1 a Nc a 35s fc Define time range t0 a nt 0 Nt 1 t t0 nt dt t nt Nt s Ht ( ) p0 if 0 t t0 a 0 otherwise square window Ht ( ) p0 sin t t0 if 0 t t0 a a 0 otherwise h Ht HH FFT ( h ) HS HH nt nt nf nf ct ( ) sin c ( t t0) sc ct SC FFT ( sc ) nt nt ss( t) sin c ( t t0) Ht () s_ ss t S FFT ( s_ ) nt nt Hanning window and its FFT carrier signal and its FFT tone burst signal and its FFT s_ nt t nt time, microsec 0.04 S nf f nf 17

18 and flexural wave speeds in aluminum strip Group velocity at 50 khz Flexural wave approximation cg_50khz cg_50khz 1355 m s m s cf( f ) E d f Phase velocity interpolation using a square root rule c_0khz 954 m s fc 50 khz cf( fc) 699 m s cgf( f ) cf( f) cgf( fc) 1399 m s c c_0khz nf f nf 0 khz 00 c nf cff nf f nf khz 18

19 waves analysis Basic wave propagation analysis wave numbers nf nf c nf medium transfer function for the Nx locations x in space domain G e i nfx nx nf nx excitation spectrum x medium transfer function at all x locations U S G nf nx nf nf nx IFFT to return to the time domain u nx IFFT U nx plotting offset control

20 End reflection analysis End reflection TOF estimate: cg_50khz mm s L t_end t_end 1476 s cg_50khz End reflection mean the whole signal is returned, hence A=B A 1 B A U S G nf nx nf nf nx u nx IFFT U nx G nf nx A e i nfx nx B e i nf x nx L

21 Crack reflection analysis Crack reflection TOF estimate: G nf nx Ac A A e i nfx nx Ac e i nf x nx Lc Lc t_crack cg_50khz t_crack 590 s cg_50khz mm s U S G nf nx nf nf nx u nx IFFT U nx

22 G nf nx End reflection and crack reflection of axial wave (half-half energy split): B A Ac A A e i nfx nx B e i nf x nx L Ac e i nf x nx Lc U S G nf nx nf nf nx u nx IFFT U nx

23 Problem 3: (i) Explain the principles of the time reversal methods. (ii) What difficulties are encountered when the time reversal method is applied to Lamb waves and how they might be alleviated. Solution (i) The principles of the time reversal method are covered in some detail in the textbook Section and will not be repeated here. (ii) As explained in the textbook Section , the difficulties that are encountered when the time reversal method is applied to Lamb waves are related to the multi-modal character of the Lamb waves. Hence, when time reversal is applied, the reconstructed signal presents other wave packet besides the original wave packet. For example, if S0 and modes are simultaneously present in the Lamb wave, then the reconstructed signal will have three wave packets instead of the original one. These difficulties might be alleviated through Lamb wave tuning which may result in a single wave packet after time reversal reconstruction Problem 4: Describe how the wave propagation method and PWAS transducers can be used to detect the location of an impact on a plate. (i) Highlight the methodology. (ii) List the main difficulties to be overcome and explain how they might be alleviated. Solution (i) The principles of detecting an impact on a plate using the wave propagation method and PWAS transducers are covered in some detail in the textbook Section and will not be repeated here. (ii) The main difficulties to be overcome are related to the dispersive nature of the Lamb waves generated by the impact events. Being dispersive, the Lamb waves spread out and make difficult the correct time-of-flight (TOF) identification. These difficulties may be alleviated through the use of a different TOF criterion, e.g., the energy-peak arrival time. 9 3