Ut testing add03 q&a mistakes

Addendum-03Questions & Answers

Collection of My Pitfalls

Uncertain Questions

21. Which type of calibration block is used to determine the resolution of angle beam transducers per requirements of AWS and AASHTO

a. An IIW block b. A DSC block c. A rompus block d. An RC block

24. Resonance or standing waves are a result of:

a. mode conversion b. interference from reflected waves c. beam divergence (spread) d. attenuation of the sound waves

Make mistakes now, not during exam!

RC- Resolution Calibration Block

30. On an A-scan display the dead zone refers to:

a. the distance contained within the near field (incorrect)b. the area outside the beam spread c. the distance covered by the front surface pulse width and recovery

timed. the area between the near field and the far field

40. The second critical angle is the angle of the incident beam at which:

a. the angle of the refracted compression wave is 900 b. the angle of the reflected compression wave is 90°c. total reflection occurs d. surface waves are produced

--------------------------------------------------------------------------------

17. Surface waves are used to detect discontinuities in the test materials:

a. At half the depth. b. Above the lower surface. c. On the surface where the probe is in contact.d. None of the above.

26. Which of the following probes is most commonly used for testing welded metals for laminations before angle beam inspection.

a. Surface wave probe. b. Twin crystal 0° probe. c. Single crystal probe. d. An angle probe.

29. Artificial flaws can be produced by using:

Side drilled holes Flat bottom holes EDM notches (http://www.phtool.com/pages/edm.asp)All of the above

31. As the acoustic impedance ratio between two materials approaches 1 the amount of sound reflected at an interface:

a. increases. b. decreases. c. is not affected. d. varies depending upon the velocity of the materials.

34. Significant errors in ultrasonic thickness measurements can occur if;

a. Test frequency is varying at a constant rate. b. The velocity of propagation deviates substantially from an assumed

constant value for a given material. c. Water is employed as a couplant between the transducer and the part

being measured. d. None of the above should cause errors.

45. When examining thin materials for planar discontinuities oriented parallel to the part surface, what testing method is most often used:

a. Angle beam b. Through-transmission c. Straight beam - single crystal d. Straight beam - dual crystal

7. The ultrasonic test method in which finger damping in most effective in locating a discontinuity is:

a. shear waveb. longitudinal wavec. surface waved. compressional wave

15. Which type of test block is used to check horizontal linearity and the dB accuracy per requirements of AWS and AASHTO?

a. Distance/Sensitivity blockb. A DSC blockc. A rompus blockd. A shear wave calibration block

Mistake Made --------------------------------------------------------------------------------

Question: Which probe will be used for critical examination in a forged component with a curved surface.: Your answer: 1 megahertz, 10mm dia. Correct answer: 10 megahertz, 25mm dia.

Question: A general term applied to all cracks, inclusions, blow holes etc, which cause a reflection of sonic energy is: Your answer: a refractor Correct answer: a discontinuity

Question: On an A-scan display the dead zone refers to: Your answer: the distance contained within the near field Correct answer: the distance covered by the front surface pulse width and recovery time

Mistake Made --------------------------------------------------------------------------------

Question: Dead zone size depends on: Your answer: construction of the probe. Correct answer: All of the above.

Question: The second critical angle is the angle of the incident beam at which: Your answer: total reflection occurs Correct answer: surface waves are produced ---------------------------------------------------------------------------------

Mistake Made --------------------------------------------------------------------------------

Question: When a longitudinal wave encounters an interface between two material with different accoustic impedances, what occurs when the Your answer: Reflection and refraction Correct answer: Reflection

Question: In an ultrasonic instrument, the number of pulses produced by an instrument in a given period of time in known as the:Your answer: pulse length of the instrument Correct answer: pulse repetition rate

Question: Which probe will be used for critical examination in a forged component with a curved surface.:Your answer: 10 megahertz, 10mm dia.Correct answer: 10 megahertz, 25mm dia.

Question: Which type of screen presentation displays a profile or cross-sectional view of the test specimen? Your answer: A-scan Correct answer: B-scan

Question: When a longitudinal wave encounters an interface between two material with different accoustic impedances, what occurs when the Your answer: Refraction Correct answer: Reflection

Questions & Answers

Table 1.2

Chapter 1: Physical Principles

Q1-10 The acoustic energy reflected at a plexiglass-quartz interface is equal to? Answer: R= (Z1-Z2)2 / (Z1+Z2)2 = (3.2-15.2)2 / (3.2+15.2)2= 42.53%

Q1-11 The acoustic energy transmitted through a plexiglass-water interface is equal to?Answer: R= (Z1-Z2)2 / (Z1+Z2)2 = (3.2-1.5)2 / (3.2+1.5)2 = 13%, T= 1-R = 87%

Q1-12 The first critical angle at a water-plexiglass interface will be?Answer: ϴ = Sin-1 (1483/2730) = 32.9°

Q1-13 The second critical angle at water-plexiglass interface will be?Answer: ϴ = Sin-1 (1483/1430) = Error!

Q1-14 The incident angle need in immersion testing to develop a 70 shear wave in plexiglass is equal to?Answer: ϴ = Sin-1 (1483/1430 x sin70) = 77°

Q1-20 Two plate yield different back-wall reflections in pulse-echo testing(18dB) with their only apparent difference being in the second material void content. The plate are both 3” thick. What is the effective change in acoustic attenuation between the first and second plate?Answer: Sound path – 2 x thickness = 6” Attenuation = 18dB/6” = 3dB/in.

Comment:The answer could be confused if the pulse-echo testing, 2-ways path length was not considered, arriving with the incorrect answer of 6dB/in

For evaluating material properties always remember to divide the result with the actual sweep distance if necessary! It was not a one-way–trip!

Q1-15 At a water-Aluminum interface, at an incident angle of 20°, the reflected and transmitted wave are?

Answer: 60% transmitted and 40% reflected.

Q1-22 The beam spread half angle I the far field of a I” diameter transducer sending 5MHz longitudinal wave into Plexiglas block is?Answer: ϴ = Sin-1 (K λ/D) Assumed K=1.2 for null beam edge,

ϴ = Sin-1 (K λ/D) =Sin-1(1.2V/DF)= Sin-1[1.2x2730x103/ (25.4x5x106)] =1.478°

Q1-23 The near field of a round 1/2 “ diameter contact L-wave transducer being used on a steel test part operating at 3MHz is?Answer: Z= D2/4λ = 12.72 3x106 x / (4x5900x103) = 20.5mm

Chapter 2: Equipment

Q2-5 A 5MHz 0.5” diameter flat search unit in water has a near field length of approximately?Answer: Z= D2/4λ = (12.72 x 5x106) / (4x 1480X103) = 136mm = 5.36”

Q2-7 A 10MHz,0.5” diameter transducer placed on steel and acrylic in succession, the beam spread in these 2 material is?ϴ = sin-1(K λ/D). ϴFe = sin-1(1.2x5920x103/10x106x12.7) = 3.2°,ϴAcrylic = sin-1(1.2x2730x103/10x106x12.7) = 1.48°

Q2-12 An angle beam produce a 45° shear wave in steel, what is the incident angle? (Vs for steel=0.125in/ms, VL for plastid=0.105in/ms)Answer: Snell’s Law; ϴincident = Sin-1[(0.105/0.125) xSin45] = 36.43°

Q2-13 Aluminum rod 6” diameter being examined in immersion technique, what is the required offset to generate a 45° refracted shear wave?Answer: First find the incident angle using Snell’s Lawϴincident = Sin-1[(1.5/3.1) xSin45] = 20°Offset = rSin20 = 3Sin20 = 1.026”

Q2-14 What is the offset required, if 45 refracted longitudinal wave to be generated?Answer: First find the incident angle using Snell’s Lawϴ incident = Sin-1[(1.5/6.3) xSin45] = 9.69°Offset = r.Sin9.69 ° = 3.Sin9.69 ° = 0.505”

Q2-16 In a longitudinal wave immersion test of Titanium plate, an echoes pulse from an internal defect is observed 6.56μs following front echo. How deep is the defect below the front surface?Answer: Sound path travel= 6100000 x 6.56 x 10-6 = 40mmThe actual depth = sound path / 2 = 20mm

Q2-17 A change in echo amplitude from 20% of FSH to 40% of FSH is a change of how many dB?Answer: ΔdB= 20log(20/40) = 6dB drop or -6dB.

Q2-20 What is the lens radius of curvature is needed in order to have a 20mm diameter 5MHz transducer focus in water at a distance of 40mm drom the lens face?Answer:R=F(n-1/n), n= V Lens/V water , n= 2.67/1.49= 1.792. R=40(0.792/1.792) = 17.7mm

Q2-18 In Fig.29 what is the rate of attenuation in dB/in of 5MHz transducer in Far Field, the horizontal scale is 0.5” per division and the vertical scale is linear.Answer:ΔI = 20log(1.25/2) D=<2” , Attenuation = 2.04dB/in orΔI = 20log(1.25/2) D=1.85” , Attenuation = 2.21dB/in orΔI = 20log(1.075/2.2) D=3” , Attenuation = 2.07dB/in

Q2-19 What is the rate of attenuation for 2.25MHz transducer?Answer:Δ I = 20log(0.9/2.2) D=2.5” , Attenuation = 3.11dB/in

Q2-21 Two signals were compared to each other. The second was found tobe 14dB less than the first. This change could be represented by a change of?Answer:ΔI = 20Log(I/Io), -14dB= 20Log(I/Io), (I/Io)= 0.2

2 answers could be confused:70% FSH to 14% FSH, a drop of 80%20% FSH to 100% FSH, an increase of 80%

Q2-11 A change in 16dB on the attenuator correspond to an amplitude ration of:Answer:ΔI = 20Log(I/Io), 16dB= 20Log(I/Io), (I/Io)= 6.3

Charter 3: Common Practices

Q3-6 In Fig. 3.7 the respond from 3.23mm FBH at a depth of 25mm is above that detected from 1mm FBH by?Answer:ΔdB= 20Log(2.1/0.6) = 10.88

Q3-7 The half angle beam spread of the reflected wave front from #8 FBH in an aluminum “A” block being immersion tested using 25MHz transducer is?Answer:Focal size = 8/64 x 25.4 =3.175mm diameter.The beam spread is in aluminum block, the wave velocity VL=6300 m/sThe half angle beam spread ϴ= Sin-1(Kλ/D) ϴ = Sin-1[(1.2x6300x103)/(3.175x 25 x 106)] = 5.47°

Comment: Be careful with the unit used, my mistake is:ϴ = Sin-1[(1.2x6300x103)/(3.175x 10-3 x 25 x 106)]

Always Check the units correctly!!!! Only Donkey made such mistake!

Monkey made mistake too!

Smart Engineer do not made mistake with UNIT USED, so do you!

Smart Himba Girl do not made mistake with UNIT USED too, so do you!

http://cn.bing.com/images/search?q=himba+women&go=%E6%8F%90%E4%BA%A4%E6%9F%A5%E8%AF%A2%E5%86%85%E5%AE%B9&qs=bs&form=QBIR





Smart Papua New Guianese do not made mistake with UNIT USED too, so do you!

http://www.tennenaturephotography.com/gallery/papua/Native_Dancer_Face?full=1

Q3-8Answer: The next SDH used will be 5/4T, first SDH after backwall echo.The node is 5/(4x2) = 5/8 node

Q3-11 When using a focued, straight beam search unit for lamination scanning in an immersion test of steel plate, a change in water path of 0.2”will result in the focal point moving in the steel a distance of?Answer: The change in water path=0.2” correspond to 0.2 x 1483/5900 = 0.05”

Q3-12 A search unit with a foal length in water of 4” is used. A steel plate 8”thick velocity 0.230”/ms is place at a water depth of 2” from the search unit, At what depth is the focal point in steel?Answer: Focal depth in steel = 2 x Vwater/ Vsteel = 2x1480/5900 = 0.5”

Q3-13 During examination, an indication of 25% FSH is detected and maximized. Foe better analysis the gain is increase by 12bB and the indication increase to 88% FSH. What value should be reached and what is the apparent problem?Answer:12dB= 20Log(I/25), I/25= 3.98, I=100%

Q3-23 A air filled #3 FBH 0.5” into the bottom of 4.5” aluminum block, will return to the 0.75” diameter sending immersion transducer ans echo signal equal to ? Of the initial pulse. Assume no attenuation to beam divergence or other causes.Answer:The size of reflector = 3/64” = 0.046875”. For a small reflector used inverse square law;Echo1/Echo2 = Area 2 / Area 1100/x= 0.0468752 / 0.752 , x = 0.39%

Q3-15 In contact testing, the back surface signal from a 2” plate was set at full screen height. Passing over a coarse grained area, the back surface signal dropped to 10% FSH. What is the change in attenuation in this area?Answer:ΔI=20Log(10/100), the drop in dB= 20dB.The sweep distance = 4”The attenuation is 20/4 = 5dB/in.

Comment: Remember that the attenuation is cause by the sound path traversing thru the sweep distance.

Q4-12 Answer:First calculate the principle offset d; ϴ = Sin-1(1483/3250 xSin45)=18.8 °d=R.Sin18.8= 0.323 (Assume R=1).Wobbling ±10%; d’=0.355 ~ 0.290d’=0.355, ϴ = Sin-1(0.355)=20.8 °giving inspection Φ = Sin-1(3250/1483xSin20.8)=51, 13.3% above 45 °d’=0.290, ϴ = Sin-1(0.290)=16.9 °giving inspection Φ = Sin-1(3250/1483xSin16.9)=39.6, 12% below 45 °

Q4-13Answer: PRR = number of pulse per second N/s, Length generated by pulse per second = PRR x DFor effective inspection Vp ≤ PRR x D

Q4-14Answer: Effective inspection Length generated by the PRR x Width = 600in/sFor a defect to be detected 3 time consecutively, the travel speed Vp= 600/3= 200in/s

Q4-15Answer: Offset = T.tan70 x Number of ½ skip.Offset = 3x 1.5 tan70Comment: 1 skip= 2 legs

Q4-16Answer: ?

Q4-16

Q4-17Answer: Total length of axial= 8x12x0.0254mL=2.438m, Sweep distance for a complete return loop =2 x L= 4.876mFor PRR = 2000Distance travel by each pulse Lp= 5920/2000 m Lp=2.96m

Since Lp is less than the 4.876, the next pulse was found to be generated before the previous echo has returned to the receiver, thus reduce the PRR is required.Set PRR=1000, yield Lp=5.92m > L=4.876mWill resolve the problem.

Q4-17 Illustrations

Length of axial 8’ or 2.438m

Complete loop=4.876m

2nd pulse generating

The previous pulse return position when 2nd (next) pulse start to send

Incoming & returning wave meet

₵

0.522m

0.958m0.958m

Q4-18Answer:

8. When testing a 30 mm diameter, 500 mm long shaft from the flat end of the shaft using longitudinal waves from a 20 mm diameter 2 MHz probe, numerous signals are seen on the screen after 500 mm. These are:

a) ghost imagesb) side wall echoesc) internal thread indicationsd) none of the above

Break!

mms://a588.l3944020587.c39440.g.lm.akamaistream.net/D/588/39440/v0001/reflector:20587?BBC-UID=e5203c9d59fef1a79c12d8c601e839f58db16f7d5d6448f55674c540f1856834&SSO2-UID=

Q5-20Answer: None of above

Q5-22Answer: Class C

Q5-22 Table B-1

5. At a solid to free boundary, an obliquely incident longitudinal wave from the solid can result in, at most:a) a reflected longitudinal wave onlyb) a reflected longitudinal and reflected shear wavec) a refracted longitudinal long waved) a reflected longitudinal and reflected shear and refracted longitudinal wave

6. Geometric-optic treatment (?) of ultrasonic waves fails to account for:a) reflectionb) refractionc) diffractiond) normal incidence

34.The most useful range of incident longitudinal wave angles for ultrasonic testing is:(a) Normal incidence to the first critical angle(b) First critical angle to the second critical angle (?)(c) Second critical angle to the third critical angle(d) Above the third critical angle

38. The angle of a refracted shear wave generated as a sound wave passes at an angle through an acoustic interface is dependant on:

a) The acoustic impedances of the materials of each side of the interface

b) The frequency of the incident sound wavec) The wavelength of the incident sound waved) The hardness of the materials on each side of the interface

22. The three most common modes of sound vibration are:

(a) Longitudinal, compressional, and transverse waves(b) Longitudinal, transverse and rayleigh waves(c) Transverse, longitudinal and shear waves(d) Transverse, shear waves and rayleigh waves

13. An oscilloscope display in which the screen base line is adjusted to represent the one way distance in a test piece is called a:

(a) A scan display(b) B scan display(c) C scan display(d) D scan display

12. Which of the following test frequencies would generally provide the best penetration in a 12 inch thick specimen of coarse-grained steel?

(a) 1.0 MHz(b) 2.25 MHz(c) 5.0 MHz(d) 10 MHz (Incorrect – silly mistake)

48. A more highly damped transducer crystal results in:

(a) Better resolution(b) Better sensitivity (mistake)(c) Lower sensitivity(d) Poorer resolution

6. The portion of a test piece which is represented by the CRT screen area from zero to the rightmost edge of the initial pulse is called:

(a) The dead zone (mistake)(b) The near field(c) The near zone(d) The far zone

17. Transducer focal lengths are normally specified as:

(a) Distance in steel(b) Distance in aluminium(c) Distance in air(d) Distance in water (mistake)

21. An advantage of using a ceramic transducer in search units is that:

(a) It is one of the most efficient generators of ultrasonic energy(b) It is one of the most efficient receivers of ultrasonic energy(c) It has a very low mechanical impedance(d) It can withstand temperatures as high as 700oC

47. When a vertical indication has reached the maximum signal height which can be displayed or viewed on the CRT of an ultrasonic instrument, the indication is said to have reached its:

(a) Distance-amplitude height (mistake)(b) Absorption level(c) Vertical level(d) Limit of resolution

53. An ultrasonic instrument control which is used to adjust the sharpness of the CRT screen display is called:

(a) Astigmatism or focus(b) Pulse repetition rate(c) Pulse energy(d) Gain

63. The purpose of the couplant is to:

(a) Match impedances between the transducer and test piece(b) Absorb stray reflectors(c) Clean the test piece so a more efficient test may be continued(d) Lock the ultrasonic scanner into place prior to testing

Note: by exclude the air between the 2 interfaces.

72. When conducting an immersion test, the water path distance must be controlled so that:

a) Spurious signals are not created by surface waves on the test pieceb) The (water path distance)/(diameter) ratio does not result in asymmetric

standing wavesc) The test piece discontinuity indications appear between the first front

and first back surface echoesd) The second front surface echo does not appear on the CRT screen

between the first front and first back surface echoes (?)

Immersion Testing Method

Standards Answer: C

Standards Answer: B

Standards Answer: A

Standards Answer: A (or C?)

Standards Answer: A

Standards Answer: C

Standards Answer: B

Standards Answer: C

Standards Answer: C

Standards Answer: A?

Arrows shown standard correct answers:Level I Q&A

Arrows shown standard correct answers:Level I Q&A

Study Blueeeeeeee…28th July 2014 17:34

Arrows shown standard correct answers:


Arrows shown standard correct answers:Level II Q&A

http://www.mtv123.com/mp3/45297/326534.shtml



R↑∝ F↑




3-Screen Height Linearity

The ultrasonic testing instrument shall provide linear vertical presentation within ±5% (According to ASME Sec.V, Article 5 T-532) of the full screen height for 20% to 80% of the calibrated screen height.

The procedure for evaluating screen height linearity is provided in appendix 1 of article 5, ASME code Sec.V and shall be performed at the beginning of each period of extended use (or every 3 months, which ever is less). http://www.inspection-for-industry.com/ultrasonic-testing.html

Take a break


Calculation: Incident angle= 7°Refracted longitudinal wave = 29.11°Refracted shear wave = 15.49°



Q2. During ultrasonic inspection of a weld, having a thickness of 28 mm anglebeam search units are to be used. The recommended angle of search unitIs:

a. 70ºb. 60ºc. 45ºd. any one

Documents

Ut testing add03 q&a mistakes