Two Dimensional FEM Simulation of Ultrasonic ... - COMSOL.com

Presented at the COMSOL Conference 2010 IndiaTwo Dimensional FEMSimulation of Ultrasonic WavePropagation in Isotropic SolidMedia Using COMSOL ®Bikash Ghose 1 * , Krishnan Balasubramaniam 2 #C V Krishnamurthy 3 , A Subhananda Rao 11 High Energy Materials Research Laboratory, Sutarwadi, Pune - 212 Center for Non Destructive Evaluation, IIT Madras, Chennai - 363 Department of Physics, IIT Madras, Chennai -36E-mail: * ghose.bikash@hemrl.drdo.in , # balas@iitm.ac.in

Motivation Application of Ultrasonic wave propagation• Under Water Acoustics• Medical Diagnostic• Structural Health Monitoring (SHM)• Non-Destructive Evaluation (NDE)• Material Characterization Non-Destructive Evaluation• Bulk wave Longitudinal wave (Pulse echo technique, Throughtransmission etc) Transverse Wave• Surface Wave• Guided Wave / Lamb Wave etc. Guided Wave• Many different modes of ultrasonic vibration(Symmetric, Anti symmetric) Numerical Simulation of Wave propagation• Fluid (Only Longitudinal wave)• Solid (Longitudinal, Transverse, Surface, Lambetc) Use of COMSOL for simulation of wavepropagation in solids

Transient Analysis for WavePropagation Transient Analysis always pose difficult problems Length of element (∆x)( Time steps (∆t)( Type of Elements (Triangular, Quad) Integration scheme for integration over time Issues• Instability• Numerical Dispersion• Convergence

Numerical Simulation of UltrasonicWave Propagation using COMSOL ®Thumb Rule ?•Length ofElement (∆x) = λ/12•Time Step = ∆x/C phInitial DisplacementDisplacement at diff positions with timeSolution at 7.0×10 -5 sSolution at 1.7×10 -4 s

Materials & Ultrasonic WaveProperties Young’s s Modulus (E) = 2 × 10 11 Pa Poisson’s s ratio (ν)() = 0.33 Density (ρ)() = 7850 kg/m3 Ultrasonic velocity (C L) (longitudinalwave) = 5850 m/s Frequency of incident ultrasonic wave (f)(= 20 kHz = 20×103 s -1 Wavelength of the longitudinal ultrasonicwave (λ(L) = C L/f = 0.2925 m Source Length : 0.04 m (40 mm)(located at the middle) Source Excitation : On a line and pointson line Simulation Domain : 5.0 m (L) x 2.5 m(H)D isplac em ent (m )6420-2-4-6-8-100 0.5 1 1.58 x Time (s)x 10 -4|Y(f)|54.543.532.521.510.5x 10 -7Single-Sided Amplitude Spectrum of y(t)(Hanning windowed signal)0 2 4 6 8 10 12 14 16 18 20Frequency (Hz)x 10 4

Meshing and Application Modes Triangular Elements Elements AutomaticallyGenerated Approximation : Plane Stain Element Type : LagrangeQuadratic Analysis : Transient Solver : Time Dependent

Effect of Length of Element (∆x)(As per CFL Criteria∆tcritical=∆xCph∆xCmax max0 −6= = = 3.1×10L.0182m5850m/ ssThe time steps chosen initially for simulation for different ∆x maxis2.5×10 -6 s

Onward Propagating Wave at variousDistances for Different λ L /∆x max at t = 4×104-4 sDisplacements (m)1.00E-078.00E-086.00E-084.00E-082.00E-080.00E+000-2.00E-080.0001 0.0002 0.0003 0.0004 0.0005-4.00E-08-6.00E-08At 0.3 mRatio = 2Ratio = 3Ratio = 4Ratio= 5Ratio = 8Ratio = 9Ratio = 12Ratio = 16Displacement (m)6.00E-084.00E-082.00E-08-2.00E-08-4.00E-08At 0.5 mRatio = 2Ratio = 3Ratio = 4Ratio = 5Ratio = 8Ratio = 9Ratio = 12Ratio = 160.00E+000 0.0001 0.0002 0.0003 0.0004 0.0005Displacement (m)-8.00E-085.00E-084.00E-083.00E-082.00E-081.00E-08Time (s)At 0.8 mRatio = 2Ratio = 3Ratio = 4Ratio = 5Ratio = 8Ratio = 9Ratio = 12Ratio = 160.00E+000 0.0001 0.0002 0.0003 0.0004 0.0005-1.00E-08Displacement-6.00E-084.00E-083.00E-082.00E-081.00E-08Time (t)At 1.0 mRatio = 2Ratio = 3Ratio = 4Ratio = 5Ratio = 8Ratio = 9Ratio = 12Ratio = 160.00E+000 0.00005 0.0001 0.00015 0.0002 0.00025 0.0003 0.00035 0.0004 0.00045-1.00E-08-2.00E-08-3.00E-08-4.00E-08-5.00E-08Time (s)• Oscillations remains after passing by of signal• Convergence of the solution for the ratio (λ L/∆x max) ≥ 8• Time steps are taken from solver or exactly what has beengiven does not make any difference to the propagating signal-2.00E-08-3.00E-08-4.00E-08Time (t)

Onward Propagating Wave for Differentλ L /∆x max at t = 4×104-4 sDisplacement (m)2.50E-082.00E-081.50E-081.00E-085.00E-090.00E+00-5.00E-09-1.00E-08Ratio = 2Ratio = 3ratio = 4ratio = 5Ratio = 8Ratio = 9Ratio = 120 0.5 1 1.5 2 2.5 3-1.50E-08-2.00E-08-2.50E-08Distance from Source center (m)Line profile (Displacement Vs Distance from source) at0.0004s for different wavelength to element length ratio

Effect of Time Steps (∆t)( As per CFL criteria the time steps should be less than ∆x/Cx/C Lfor example, for the maximum element length of λ L /10 thetime steps should be ≤ (λ L /10×C L ) that means if Signal so far obtained is not as expected Major change in the shape of signal Time steps further decreased and solution was checked forconvergenceλL∆xmax=10∆x⇒ ∆t≤C⇒ ∆tmaxL1≤10×fλL=10×CL

∆t= 10×10Case: I λ L /∆xmax= = 5−6s∆t= 5×10−6s∆t= 2.0×10−6s∆t= 1.0×10−6s∆t=0.5×10−6s∆t= 0.2×10−6s

∆t= 6.26×10Case: II λ L /∆xmax= = 8−6s∆t= 5×10−6s∆t= 2.0×10−6s∆t= 1.0×10−6s∆t=0.5×10−6s∆t= 0.2×10−6s

Case: III λ L /∆xmax= = 12∆t= 5×10−6s∆t= 2.5×10−6s∆t= 1.0×10−6s∆t=0.5×10−6s

Effect of Excitation on Line and Points onLine for λ L /∆x max = 8 and ∆t t = 2.5x10 -6 s(Line source)At 1.0 m(Points on Line source)At 1.0 mAt 0.5 mAt 0.5 m

Effect of Excitation on Line and Points onLine for λ L /∆x max = 8 and ∆t t = 2.5x10 -6 sLine profile at t = 4x10 -4 s(Line source)(Points on Line source)

Time and Frequency Domain Signal forForward Propagating Waveλ L∆xmax= 12λ L∆xmax= 8λ L∆xmax= 58.00E-088.00E-088.00E-084.00E-084.00E-084.00E-08D isplacem ent (m )0.00E+000.00E+00 2.00E-04 4.00E-04 6.00E-04-4.00E-08Displacement (m)0.00E+000.00E+00 2.00E-04 4.00E-04 6.00E-04-4.00E-08Displacement (m)0.00E+000.00E+00 2.00E-04 4.00E-04 6.00E-04-4.00E-08-8.00E-08-8.00E-08-8.00E-08-1.20E-07Time (s)-1.20E-07Time (s)-1.20E-07Time (s)Single-Sided Amplitude Spectrumx 10-910Single-Sided Amplitude Spectrumx 10-914x 10 -9Single-Sided Amplitude Spectrum98121271010|Y (f)|654|Y(f)|86|Y (f)|86321424201 2 3 4 5 6 7 8 9Frequency (Hz)x 10 41 2 3 4 5 6 7 8 9Frequency (Hz)x 10 41 2 3 4 5 6 7 8 9Frequency (Hz)x 10 4All signals are for ∆t t =0.5x10 - 6 s

Time and Frequency Domain Signal forBack Wall Reflected Waveλ L∆xmax= 12λ L∆xmax= 8λ L∆xmax= 54.00E-084.00E-084.00E-082.00E-082.00E-082.00E-08Displacement (m)0.00E+000.00E+00 4.00E-04 8.00E-04 1.20E-03-2.00E-08Displacement (m)0.00E+000.00E+00 4.00E-04 8.00E-04 1.20E-03-2.00E-08D isplacem ent (m )0.00E+000.00E+00 4.00E-04 8.00E-04 1.20E-03-2.00E-08-4.00E-08-4.00E-08-4.00E-08-6.00E-08Time (s)-6.00E-08Time (s)-6.00E-08Time (s)Single-Sided Amplitude Spectrumx 10-97Single-Sided Amplitude Spectrumx 10-98x 10 -98Single-Sided Amplitude Spectrum677566|Y(f)|43|Y(f)|543|Y(f)|5432221110 1 2 3 4 5 6 7 8 9Frequency (Hz)x 10 41 2 3 4 5 6 7 8 9Frequency (Hz)x 10 41 2 3 4 5 6 7 8 9Frequency (Hz)x 10 4All signals are for ∆t t =0.5x10 - 6 s

Simulation Results for λ L /∆x max = 8∆t t = 0.5 x 10 -6 s at different timeAt t = 1x10 - 4 s

Simulation Results for λ L /∆x max = 8∆t t = 0.5 x 10 -6 s at different timeAt t = 4x10 - 4 s

Simulation Results for λ L /∆x max = 8∆t t = 0.5 x 10 -6 s at different timeAt t = 5x10 - 4 s

Simulation Results for λ L /∆x max = 8∆t t = 0.5 x 10 -6 s at different timeAt t = 8x10 - 4 s

Conclusions Wave Propagation can well be modeled usingCOMSOL with excitation on line segment and / oron points on line (Difference only in themaximum displacements) λ L /∆x max ≥ 8 irrespective of any time steps lessthan calculated by CFL criteria ( For triangularfree meshing) Time step ∆t should be about T/100 even if λ L /∆x≈ 16 No substantial difference in frequency content offorward as well as back wall reflected ultrasonicsignal observed

E-mail:ghose.bikash@hemrl.drdo.in , balas@iitm.ac.in