Ultrasonic TestingNotes from the lecture from Mr.RamakrishnanReflectionBasic principle of ultrasonic testing is reflection. Reflection of sound is called echo. Reflection is because of the change in acoustic impedance (Resistance of material to the transfer of sound waves)Acoustic Impedance, Z=Velocity X Density (xV)Percentage of energy reflection, R= (Z1-Z2) / (Z1+Z2)Steel and Air reflection is 100%, thats why we use couplantPrinciple of ultrasonic testing is anomalies in acoustic impendence

Crack suddenly bursts. So there is vacuum in the gap. Acoustic impedance in vacuum is 0,ie, Z=0. So reflection is . For same are of defects the echo amplitude defers. So nobody can say the exact area. Ultrasonic will locate the defect.Q1 A steel sample is immersed in water. The acoustic impedance of steel and water are 46 and 1.48 respectively. What will be the percentage of sound wave received from the back wall?

R1= (46-1.48) / (46+1.48)X100=88%T1=100-88=12%R2=12x0.88=10.5%T2=12-10.5=1.5%R3=10.5X0.88=9.3%T3=10.5-9.3=1.2%So transducer receives 1.2% from the back wall.SoundSonic means sound. Sound is a form of mechanical energy. Source of sound is vibrations. All vibrating bodies execute Simple Harmonic (Repeating) Motion {SHM}. Whenever a body vibrating with SHM will have amplitude, period, cycle and frequency etc

Y=A Sin (t)Where A=Amplitude, =Angular velocity-2n/60 (ex-Fan motion, n=rpm)=Phase angleAt maximum displacement, phase angle will be 90. Maximum displacement is called amplitude.Y=A Sin 90Y=AIn ultrasonic testing, when the amplitude is very high the probe has lots of energy.Velocity, V = Distance/Time (T)Here distance= Velocity, V = /TFrequency, f=1/TIe, Velocity, V=f This is common for any type of wave. No two elements have same velocity. If the medium is changing there is a change in velocity. So there is a reflection and reflection of sound is called echo. So ultrasonic testing is echo principle.The amount of reflection not only depends on the velocity but also on the densityAcoustic Impedance, Z=Velocity X Density (xV)If Z1=Z2, 100% of sound will be transmitted. But in practically Z1 cannot be equal to Z2. Ultrasonic testing depends on the anomalies in acoustic impedance.Frequency below 20Hz is called infrasonicFrequency between 20Hz to 20 KHz is called sonic (Audible range)Frequency above 20 KHz is called ultrasonicResonance FrequencyResonance will be at /2. Below this there will not be any resonance. High frequency is used to detect small defects.For 4MHz probe in steel. The minimum discontinuity can be find out is /2 = 5.9/(4x2)=0.75mm.Why /2 ?Assume 0.5mm wire

10mm length

30mm length

60mm length

If they are pulled and released, the longer wire will make more wide vibrations. So as the length decreases the vibrations reduces. So the defect should have at least /2 lengths to vibrate.If a tuning fork is vibrated on one end, the other end will also start vibrating. This is due to resonance. We can consider one end as job and other end as defect. If one side is damped, other side will also stop vibrating. So if an ultrasonic sound hits a defect, the defect should also vibrate. Ex:- /2 for 4MHz probe is 0.75mm. If the defect size is 0.5mm, it will not vibrate and defect cannot be found out.High frequency ultrasounds obeys the laws of opticsLight can be reflected-Sound can be reflectedLight can be refracted-Sound can be refractedLight can be diffracted-Sound can be diffracted

Pattern of the reflection is an indication of the nature of the defect. 1 and 7 will show the same, no defect will be shown. Same size of lack of fusion will show different echo heights.Sound travels in the form of wavesProbe is compared to mouthJob is compared to mediumMachine is compare to nervous system (brain)Energy transfer in physics is wave, ie energy in motion. Energy only will transfer, the particle will remain in the same place.There are four types of wavea. Longitudinalb. Transverse (shear)c. Surface (Raylegh)d. Lamb wave (Plate Wave or guided wave)Longitudinal waveLongitudinal waves are called principle mode of vibrations. In all probe only longitudinal waves are there, by mode conversion only if converts in to other forms. Partilce moves in the same direction of sound wave. Why the velocity is changes for different material because of difference in density and elastic modulesV (longi) =E/, E is the youngs modulesUltrasonic waves other name is elastic waves. It travels in solid, liquid and gas etc. Velocity increases wavelength also increases and penetration also increases.Most of the codes suggest to use 2-2.5MHz probes, because higher the frequency lesser the penetration.Why higher frequency high loss?

All the molecules in front will be pushed by the sound wave, ie compression. For every wave length compression is there. So in 4MHz compression is more. So more heat will be produces. Heat will be produced by the probe and job is the sink.So by the time the sound energy reaches the defect, sound energy will be very less for higher frequency compare to lower frequency.So for more frequency more compression, more compression more heat and more heat more loss of sound energy. This is called absorption. Sound energy is easily converted in to heat; heat is easily absorbed by any metal. Whether we use 2MHz or 4MHz, the back wall echo should be set to 80%Attenuation is proportional to the square of frequencyWhen scatterer is smaller than the wavelength then practically there are no consequences. As size of scatterer increases to size of wavelength, scattering increases. Objects that have sizes smaller than the beam will scatter the waves.Signal to noise ratio

4MHz requires more db to bring the defect echo to 60%. But more noise will be created. Signal to noise ratio is 2:1 for the above example. But the code may specify the signal to noise ratio as 3:1. Then we cannot use the probe according to the code.Shear wave1. Particle vibration is perpendicular to the sound wave propagation2. Velocity of transverse wave, Vt=G/, where G is the bulk modulus/Modulus of rigidity or shear modulasShear means break in to two pieces. Air and water cannot be sheared, ie, fluids do not have shear, hence, G=0. So shear wave cannot travel in fluidsVt =1/2XV (longi) =V/f, So for a constant frequency, ex-4MHz shear wave will have less wavelength. So it can detect smaller defects than detected by longitudinal wave. So shear wave is more superior. So all weld inspection is done by shear wave. Shear wave in steel, =3.23/4=0.8mmLongi wave in steel, =5.9/4=1.48mmPrinciple of reflection and refraction

Sound is a form of mechanical energy. Mechanical energy will push the molecules in all directions. Longitudinal wave will automatically mode converted in to transverse wave.Snells LawSin a / V (longi) = Sin b / V (longi) = Sin c / VtIf a = 20, b will also be 20 so c = 11Walking signal

When we move the probe the signal will also moves. This walking signal is because of mode conversion. If the defect is oriented at an angle, mode conversion will happen.RefractionBending of wave from one medium to another. For probe Ist medium is plastic

Sin a / V (longi) = Sin b / V (longi) = Sin c / VtFor an incident angle a, there will be refracted longitudinal angle b and refracted shear wave angle c with different velocity. It is not possible to do any analysis while two waves are existing at same time.

Both indications from 50mm longitudinal and 100mm shear wave will be shown at same place on CRT. One echo will be on top of another and the defects cannot be separated and identified.Whenever sound travels from lower to higher sound velocity, the refracted angle will be away from normal. Whenever sound travels from higher to lower velocity, the refracted angle will be towards the normal. First Critical angleThe incident angle at which the longitudinal wave touches the surface of test object.

b = 90Sin a / 2.76 = Sin 90 / 5.92a = 27First critical angle in steel is 27 Shear wave angle produced is, c = 33If the refracted shear wave angle is reduces to below 33 longitudinal wave will come inside the object.For making 45 probeSin a = Sin 45 x 2.76/3.23a = 37So if angle of incidence is 37 on steel, the refracted angle will be 45Q2 A 45 degree probe made on steel is kept in water. A thick weld plate is immersed in water and 45 degree probe is immersed in wateri. Wrong selectionii. Due to mode conversion lots of noise will comeiii. In the steel sample 45 degree refracted angle will be producedHere answer is iii.

Surface Wave

Second critical angle is the angle of incidence at which the refracted transverse wave touches the surface.Surface wave is a component of shear wave. Particle oscillations are elliptical in motion. It can penetrate only one wavelength. It can be damped by a drop of oil or any obstructions. It is more sensitive than penetrant and magnetic particle testing. Velocity of surface wave is 09xVsPlate waveIt will travel in thin plates. It travels through thin plates up to 3xWavelength thickness. Plate wave is generated at first critical angle (check). Plate wave will not have any specific velocity.

Symmetrical oscillations

Symmetrical is called because of mirror image.Asymmetrical oscillations

Guided wave principle is that both the surface will guide.

So 160 meters will be covered in one set up.Beam geometryBeam geometry depends on probe size, frequency and material velocity.

First wave converges and then diverges. The portion it converges called near zone (frehnel zone). In frehnel zone the intensity is maximum-minimum and in far zone the intensity exponentially decays. So at the end of near zone all the sound intensity is concentrated making a lens called frehnel lens. After that all the intensity will be distributed infinitely. Near Zone = Df/4VSo if the diameter is increased by two times, the near field will increase by four times. If velocity increased, near filed decreases. In near field, we cannot size the defect because of maxima-minima.

For example, if a crack where the sound intensity is very low, will not show any indication and if there is a porosity where the sound intensity is very high will show high indication. So it wrongly estimate the defects.Each crystal has molecules, each one will vibrate it will send ultrasonic sound waves in to the material. These waves will meet each other and cancels each other forming a point called nodel point due to the phase lag. The intensity at nodal point is 0. Hence intensities with maxima-minima will be produced

Note:- When ultrasonic waves are coming from material having lower acoustic impedance and incident on a material of higher acoustic impedance (Z2>Z1) the reflected wave exhibits a phase reversal.I2 / I1= e d, =attenuation coefficientIn the far field, geometry is perfect and loss is accountable and the loss have mathematics. So we can size the defect and to correct the loss in far field we make DAC (Distance Amplitude Correction Curve)

Beam SpreadSin a = 1.22 X (Wavelength / Diameter) = 1.22 X (V/fD), 1.22 is for null intensity, 1.08 to be used for 10% intensity and 0.7 to be used for 50% intensity. So 100% intensity will be at the center

So if diameter increases near field will increase by square but beam spread will decrease. So if we increase the diameter we reduce the frequency and if we increase the frequency we reduce the diameter.

For the same defect the echo height defers as the probe moves. This means the intensity of sound beam increase as moving towards the center. So for sizing the defect we need to maximize the defect echo. This means the defect is at the center of sound path.Q3 On a 100mm thick unknown material a 4Mhz, 10mm dia probe was placed. The elapsed time was 32.1 microseconds in pulse echo system. What is the material?Velocity = Distance / Time= 200 / (32.1 x 1000000)= 6230 m/sIe, AluminiumQ4 What will be the percentage of energy received by the transducer after back wall reflection from and Aluminum block immersed in water? Z for Al = 17, Z for water = 1.48

R1= (17-1.48) / (17+1.48)X100=70.5%T1=100-70.5=29.5%R2=29.5x0.705=20.8%T2=29.5-20.8=8.7%R3=20.8X0.705=14.7%T3=20.8-14.7=6.1%So transducer receives 6.1% from the back wall.Q5 Determine the frequency of the transducer from the figure?Probe Dia = 10mm, Velocity = 5900m/s

Answer;Near Field = 16.8916.89 = 10x10xf/4x5.9x10^6F = 4MHzQ6 What will be the beam diameter at 100mm from the transducer?

Answer:Sin a = 1.22 x 5.9x10^6/4x10^6x10a = 10.4Tan 10.4 = x/100x = 100 x tan 10.4x = 18.352x = 36.7mmSo beam diameter at 100mm is 36.7mmQ7 What is the focused beam dia?Answer:Tan 10.4 = a/16.89a= 3mm2a = 6mmFocused beam dia is 6mmQ8 The refracted angle was found to be 55 degree in steel for the probe immersed in water. What is the angle of incidence? V(lw) = 1.48x10^6mm/s, V(ls) = 5.9x10^6mm/s, V(ts)=3.23x10^6mm/sAnswer:Sin a/1.48x10^6 = Sin 55/3.23x10^6a=22 degreeQ9 Following DAC drawn on steel for steel for 2 and 4 mhz probe. Identify the probes from the figure?

Answer:Probe A is 2MHz and B is 4MHzQ10 Determine the dia for 2mhz and 4mhz?Answer;For 2 Mhz30 = Dx2x10^6/4x5.9x10^6D = 18.8mmFor 4mhz30 = Dx4x10^6/4x5.9x10^6D = 13.3mmQ11 20mm dia steel shaft is immersed in water. It is desired to produce 40 degree refracted shear wave in steel. What shall be the offset?