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?