| UT1 Level | ||
| 1 | The point in time when the instrument pulser generates the initial pulse voltage is: | |
| a) interface zero | ||
| b) acoustic zero | ||
| c) electric zero | ||
| d) the peak point | ||
| ANSWER | Electric Zero (left hand side of the Main Bang). | 1) |
| 2 | Determine the wavelength when straight beam contact testing aluminum with a 2.25 MHz transducer: | |
| a) 0.05" | ||
| b) 0.11" | ||
| c) 0.9" | ||
| d) 0.46" | ||
| ANSWER | Wavelength = c/f = 0.25 x 106 in/sec /2.25 MHz = 0.11 inch. | 1) |
| 3 | If a straight beam delay line transducer is used to inspect a 1/2" thick steel plate, the How long should be the Plexiglas delay line at least: | |
| a) 0.5" | ||
| b) 1" | ||
| c) 0.25" | ||
| d) 0.75" | ||
| ANSWER | Based on the velocity ratio of Plexiglas (.11) to steel (.23) = 1:2 Ensures no delay multiples to the left of the backwall echo. | 1) |
| 4 | Why do most angle beam inspections utilize the refracted shear wave than the refracted longitudinal wave? | |
| a) Using the refracted shear wave because of less sound attenuation than longitudinal wave. | ||
| b) Using the refracted shear wave allows the refracted longitudinal wave to be moved to the surface of the part thereby eliminating confusing echoes for both beams. | ||
| ANSWER | "b" | 1) |
| 5 | What are two advantages of using piezocomposite materials in the design of immersion transducers? | |
| a)Lower acoustic impedance for better matching to the water | ||
| b)Higher coupling factor for improved sensitivity | ||
| c)flexibility to form elements to shape | ||
| ANSWER | Three possible correct ANSWERS (there are probably others): | 1) |
| 6 | Which of the following factors has a major effect on the length of the zone adjacent to a transducer in which fluctuations in sound pressure occur: | |
| a) acoustic noise levels | ||
| b) selection of test frequency | ||
| c) sound scattering | ||
| d) all of the above | ||
| ANSWER | This is another way of asking what controls the length of a transducer`s near field?, which is affected by frequency, element diameter, and the sound velocity of the material through which the wave is propagating. Frequency is the only one of those three factors available as a choice here. | 2) |
| 7 | Plate waves cannot be generated by air-coupled transducers because shear waves do not exist in air. | |
| a) true | ||
| b) false | ||
| ANSWER | Air-coupled transducers can generate plate waves, provided the sound pressure is sufficiently high, and provided the incident angle is properly adjusted. | 3) |
| 8 | Which of the following materials have the highest compression wave speed. ? | |
| a) Aluminium Nitride | ||
| b) Silver | ||
| c) Aluminium | ||
| ANSWER | This engineering ceramic has extremely high elastic moduli with a compression wave velocity of about 10800 ms-1, compared to 6400 ms-1 for aluminium and 3700 ms-1 for silver. | 5) |
| 9 | Which part of exciting pulse influences the bandwith most of all? | |
| a) amplitude step pulse is used, bandwith expands into lower frequency region. Aplitude can also affect the bandwith, in case exciting generator slew rate can not ensure high amplitude pulses. | ||
| b) width step pulse is used, bandwith expands into lower frequency region. Aplitude can also affect the bandwith, in case exciting generator slew rate can not ensure high amplitude pulses. | ||
| c) front | ||
| ANSWER | The front raising speed is most important for wide bandwith. Pulse width, if square pulse is used, can cancel a "tail" and improve the bandwith a bit. If step pulse is used, bandwith expands into lower frequency region. Aplitude can also affect the bandwith, in case exciting generator slew rate can not ensure high amplitude pulses. | 7) |
| 10 | How high is the pulse repetition frequency of the ultrasonic single-shot flaw detector during scanning if the scanning speed is 200 mm/s with an index of 0.05 mm ? | |
| a) 4000 Hz | ||
| b) 400 Hz | ||
| c) 8000 Hz | ||
| ANSWER | The pulse repetition frequency f can be calculated by the number of measuring points per second: An index of 0.05 mm means 20 points per mm. F= 1/t= 200mm/s * 20/mm = 4000/s = 4000 Hz | 8) |
| 11 | Why is the sound damping of composites much higher than that of metals? | |
| a) sound scatter | ||
| c) acoustic impedance | ||
| ANSWER | Composites are inhomogeneous and unisotropic materials which produce scatter of ultrasound. Therefore the damping is much higher than that of metals. | 8) |
| 12 | Does the focal length of a 50 MHz-transducer change if the test frequency f is shifted to 10 MHz ? | |
| a) no | ||
| c) yes | ||
| ANSWER | Yes, the focal length will be much smaller because the near field length N of a transducer is proportional to f. | 8) |
| 13 | For a pulse-echo inspection instrument which clearly shows the difference in reflections from discontinuities that are 1/4 inch apart in depth one can say that this instrument has a good | |
| a)sensitivity | ||
| b)resolution | ||
| c)sensitivity and resolution | ||
| ANSWER | the answer is resolution | 9) |
| 14 | To test a specimen and if you need to know the depth and size of any discontinuities found we would select | |
| a) through-transmission | ||
| b) pulse-echo | ||
| c) both setups | ||
| ANSWER | pulse-echo would be the best setup to use | 9) |
| UT2 Level | ||
| 1 | The term reference level refers to: | |
| a) the number of decibels of calibrated gain required to place the signal from a known reflector at the reference line on the screen | ||
| b) the number of decibels of calibrated gain required to place the signal from a flaw at the reference line on the screen | ||
| c) a percentage of full screen height to which an operator compares signals | ||
| d) none of the above | ||
| ANSWER | "a" | 1) |
| 2 | Calculate the transit time for a straight beam contact test in a 4" thick piece of copper ? | |
| a) 34.4 usec | ||
| b) 4.4 sec | ||
| c) 44.4 usec | ||
| d) 0.0044 sec | ||
| ANSWER | TT = 2T/C = (2 x 4 in)/0.18 = 44.44 usec | 1) |
| 3 | Calculate the near field length of a 2.25 MHz - 0.5" straight beam contact probe when testing in nickel. | |
| a) 0.88" | ||
| b) 0.64" | ||
| c) 0.64 mm | ||
| d) 1.27" | ||
| ANSWER | NF = D2F/4C = (0.5in)2 x 2.25/4 x 0.22 = 0.25 x 2.25/0.88 = 0.5625/0.88 = 0.639 in | 1) |
| 4 | Lamb wave inspection techniques are most commonly used for testing: | |
| a) thin sheets or plates | ||
| b) surfaces of forged or cast materials | 2) | |
| c) metallic materials only | ||
| d) braze joints | ||
| ANSWER | By definition, Lamb waves exist only in thin sheets or plates where thickness is less than one wavelength. | 3) |
| 5 | Guided plate waves can be used for the non-destructive testing of bulk material. | |
| a) true | ||
| b) false | ||
| ANSWER | Plate waves exist only in plates having a thickness of less than a wavelength. | 3) |
| 6 | Determination of the true weld centerline ( root ) is performed utilizing primarily: | |
| a) A straight beam ( 0° ) probe | ||
| b) 45° shear from both sides of the weld | ||
| c) 60° shear from both sides of the weld | ||
| d) correlation of CE1 and CE2 | ||
| ANSWER | "a" | 4) |
| 7 | In verifing the axial location of the crack, which signal will usually provide a more accurate result? | |
| a) CE1 b) CE2 c) 60 degree shear waves d) 70 degree shear waves | ||
| ANSWER | "b" | 4) |
| 8 |  Given the frequency response of the 10 MHz transducer shown in the figure, which of the following three time domain excitation pulses would give the highest transducer output with the best bandwidth performance ? It is
assumed that the peak amplitudes of the signals are the same. | |
| a) A 700 ns gated burst at 10 MHz. | ||
| b) A 10 ns wide pulse. | ||
| c) A 50 ns wide pulse. | ||
| ANSWER | The first signal gives a high output but is narrow band. The second signal will give high bandwidth but not much output since energy is wasted exciting the transducer beyond its passband. The third signal gives the best performance because the pulse width has been selected to put most of the pulse energy in the transducer passband. | |
| 9 |  The Figure shows two transducers immersed in a water tank in pitch catch mode. These excite oblique incidence ultrasound compression waves which are used to scan the interface between two slabs of material. Which of the following probe types would reduce the sensitivity of the system to angulation errors and/or curvature of the surface ? | |
| a) Apodised probe. | ||
| b) Focused probe. | ||
| c) Planewave probe. | ||
| ANSWER | The use of a focussed probe (with the point of focus set at the interface) will reduce plane wave interference which is sensitive to changes in angle. | 5) |
| 10 | The TOFD technique augments but cannot replace pulse-echo techniques determinations of nonfusion defects | |
| a) the lateral wave prevents inspection near the OD | ||
| b) hi-lo conditions cause the root region to be masked in TOFD | ||
| c) TOFD techniques cannot provide accurate vertical extent | ||
| d) both a) and b) | ||
| ANSWER | "d" | 6) |
| 11 | Which is not an advantage of mechanised UT on pipeline girth welds? | |
| a) no radiation hazards | ||
| b) improved defect characterization compared to radiography | ||
| c) speed of inspection allowing potential for process control | ||
| d) ability to determine vertical position and height of defects | ||
| ANSWER | "b" | 6) |
| 12 | What for time variant gain(TVG) is used in ultrasonic NDE? | |
| a) to compensate the acoustic noise | ||
| b) to compensate the shading effect of defects in the ultrasonic beam path | ||
| c) to match the reflected signal and acquisition channel dynamic range;d-for ultrasound signal attenuation compensation. | ||
| ANSWER | The scattering, attenuation and absorbtion of acoustic signal derates the intensity of ultrasonic signal as it propagates deeper. The TVG is used to compensate these factors. In such way dynamic range of received signal and acquisition channel are mached so acquisition channel dynamic range can be narrower than if there was no TVG. As well, shading effects can be compensated as well, if digital TVG is used, so (b) is partly correcct. Of course, TVG is equivalent to multiplication of ultrasonic signal with TVG function and not always means the compensation of losses in material, especially when there is significant attenuation change versus signal frequency. So, answer (d) is not exacly correct. | 7) |
| 13 | Assume you have the task of testing a rolled plate with plate waves for surface defects. How many surfaces should you have to access? | |
| a) one surface | ||
| b) two surfaces | ||
| ANSWER | One surface since plate waves cover the whole thickness of the plate | 9) |
| UT3 Level | ||
| 1 | When working with flat bottomed holes for calibration or evaluation purposes, what is the relationship between a #3 FBH and a #6 FBH ? | |
| 5% | ||
| 25% | ||
| ANSWER | Using the Law of the Ratio of Squares you multiply the diameter of each hole by itself and take the ratio. 32/62=9/39=1/4=25% | 1) |
| 2 | What water path is needed when using a straight beam transducer with a 4 inch (101 mm) focus to place Y0 at a bondline 17 mm deep in copper? | |
| 4,12in | ||
| 2,06in | ||
| ANSWER | Convert 17 mm in copper to 0.67 inches in copper. This has a water equivalent of: Cl (copper) = 0.18 x 105 in/sec Cl (water) = 0.584 x 105 in/sec | 1) |
| 3 | What is the decibel difference between 2 signals on the display with one at 85% full screen height and the other at 17% full screen height? | |
| 7db | ||
| 14db | ||
| ANSWER |  | 1) |
| 4 | When testing a part with a rough surface, as compared to a similar part with a smooth surface, one should generally use: | |
| a) a higher frequency transducer and a more viscous couplant | ||
| b) a lower frequency transducer and a more viscous couplant | ||
| c) a higher frequency transducer and a less viscous couplant | ||
| d) a lower frequency transducer and a less viscous couplant | ||
| ANSWER | When testing on a rough surface, a lower frequency transducer will tend to smooth out some of the surface ringing caused by relatively thick layers of couplant at some points under the transducer. This happens because the period of the transducer`s waveform will be greater than the round trip transit time in the couplant layer. A more viscous couplant will tend to fill the gaps rather than running off, and high viscosity couplants generally offer a better impedance match and hence more efficient sound coupling than lower viscosity couplants. | 2) |
| 5 | Asymmetrical plate waves are | |
| a) faster then shear waves, but slower than compressional waves | ||
| b) slower than shear waves | ||
| c) as fast as Rayleigh waves. | ||
| ANSWER | At low values of the product frequency times plate thickness (f*d), asymmetrical plate waves are slower than the Rayleigh waves. At large values of f*d, symmetrical and asymmetrical plate waves both asyptotically approach the surface wave v | 3) |
| 6 | Satellite pulses are unique to | |
| a) IGSCC | ||
| b) notches | ||
| c) fatique flaws | ||
| d) Satellites can be expected from any planar flaw | ||
| ANSWER | 'd' | 4) |
| 7 | When attempting to size a crack with 1/2 Vee tip diffraction, the tip signal should: | |
| a) move with the signal | ||
| b) be 180° out of phase with the corner signal ( usually ) | ||
| c) should appear ahead of the corner signal | ||
| d) both a) and b) | ||
| e) a),b) and c) | ||
| ANSWER | "e" | 5) |
| 8 | The concept of vertical zones in pipeline weld inspection allows: | |
| a) application of ECA criteria | ||
| b) precise sizing of defects | ||
| c) characterzation of defects | ||
| d) all of the above | ||
| ANSWER | "a" | 6) |
| 9 | What sampling frequency of 8 bit analog-to-digit convertion has to be used with fo central frequency ultrasonic wide bandwidth transducer? (More Answer) | |
| a) fo/2 | ||
| b) fo*10 | ||
| c) as high as possible | ||
| d) fo*4. | ||
| ANSWER | "b and c" | 7) |
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