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1379 views
02:28 Jun-30-1998
Damon Priestley
Questions on TOFD

I am looking for information on TOFD.
Attached, you will find an image (JPEG)
of an Osciliscope, with an A-Scan presentation, operating in the TOFD mode.

Here is some information on the equipment and setup:

a) NB-2000 AUT system
b) Two (2) .25", 5 mhz, KKB transducers with 60 degree Longitudinal
wedges
c) 25mm thick, 400mm / 400mm plate

A calculated transducer index spacing of 86mm was used.
The block was free of holes, notches, welds, and foreign material.
The transducers were set at least 100 mm away from the sides of the
plate to avoid sound reflections from the side walls.
The objective was to receive a simultaneous response from the lateral wave
and the backwall.

I have two questions which pertain to the attached image:
The lateral wave deteriorates to an unrecognizable state when the
transducers are indexed ~80mm or greater from one another. Applying
more gain just increases the amount of noise level and does not help in
distinguishing the lateral wave from the 'background noise'. How can I
get a clean, high amplitude, lateral wave at increased index spacing?

After indexing my transducers to the proper spacing of 86mm, I receive
a variety of signals. At this distance, the lateral wave has disappeared
and
my first signal is the backwall signal. Several signals of higher amplitude
follow. What are these signals, and how are they applied in the scan?

Your responce is greatly appreciated,

Damon Priestley
nomad@ksamail.com

P.S. I do not have access to the internet, only e-mail (due to
restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
information off of a web-page. Thank-you.

Comments:
We post this message on behalf of Damon Priestley to the Forum.
Please reply anyway to the Forum (we all want to learn from it),
we'll forward all replies to Damon Priestley by email.
The A-Scan presentation (attachment from Damon Priestley) you can access
only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

Rolf Diederichs



 
06:40 Jun-30-1998

Paul Meyer

R & D,
GE Inspection Technologies,
USA,
Joined Nov 1998
47
Re: Questions on TOFD not help in
: distinguishing the lateral wave from the 'background noise'. How can I
: get a clean, high amplitude, lateral wave at increased index spacing?

: After indexing my transducers to the proper spacing of 86mm, I receive
: a variety of signals. At this distance, the lateral wave has disappeared
: and
: my first signal is the backwall signal. Several signals of higher amplitude
: follow. What are these signals, and how are they applied in the scan?

: Your responce is greatly appreciated,

: Damon Priestley
: nomad@ksamail.com

: P.S. I do not have access to the internet, only e-mail (due to
: restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
: information off of a web-page. Thank-you.

: Comments:
: We post this message on behalf of Damon Priestley to the Forum.
: Please reply anyway to the Forum (we all want to learn from it),
: we'll forward all replies to Damon Priestley by email.
: The A-Scan presentation (attachment from Damon Priestley) you can access
: only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

: Rolf Diederichs

Damon,
If I understand your setup, the signal you call a lateral wave is intended to be a subsurface wave running directly from one transducer to the other. You also intend to measure the thickness with the refracted longitudinal wave.
I believe the 70 degree refracted longitudinal wave is generating a very weak subsurface wave which disappears at larger spacings. The signal in your second image might be a mode converted shear wave from the incidence of the L-wave at the lower surface of the block. Therefore, the TOF would be faster than a shear wave but slower than a longidutinal wave. The additiona signals are probably the shear wave and its mode-converted reflections.
Suggestion: Increase the angle of the L-wave close to 90 degrees and use the shear wave to see the back wall. Please calculate the energy into each of these components. Many people use Snell's Law to calculate refracted angles but fail to check the amplitude of the resulting L and S waves. These can change very quickly, especially near a critical angle


 
00:20 Jul-01-1998
Ane Anev
Re: Questions on TOFD not help in
: distinguishing the lateral wave from the 'background noise'. How can I
: get a clean, high amplitude, lateral wave at increased index spacing?

: After indexing my transducers to the proper spacing of 86mm, I receive
: a variety of signals. At this distance, the lateral wave has disappeared
: and
: my first signal is the backwall signal. Several signals of higher amplitude
: follow. What are these signals, and how are they applied in the scan?

: Your responce is greatly appreciated,

: Damon Priestley
: nomad@ksamail.com

: P.S. I do not have access to the internet, only e-mail (due to
: restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
: information off of a web-page. Thank-you.

: Comments:
: We post this message on behalf of Damon Priestley to the Forum.
: Please reply anyway to the Forum (we all want to learn from it),
: we'll forward all replies to Damon Priestley by email.
: The A-Scan presentation (attachment from Damon Priestley) you can access
: only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

: Rolf Diederichs

Damon,

I used 10 mm, 5 MHz, Krautkraemer TOFD transducers
with 60 degree Longitudinal wedges on a hollow
shaft with 325 mm wall thickness.The index separation
was about 800 (eight hundred) mm and the lateral wave
was clearly discernible on the B-scan. I found that
the amplitude was very dependant on the pulse repetition
frequency, possibly because of some internal averaging
in the equipment. Try to increase PRF and may be some
averaging will help to reduce the noise.

The signal after the longitudinal wave back wall echo
is mode-converted back wall echo. It coincides quite
well with the theoretical calculations.

Good luck,
Ane Anev
Technical University - Sofia


 
01:46 Jul-01-1998

Udo Schlengermann

Consultant, -
Standards Consulting,
Germany,
Joined Nov 1998
172
Re: Questions on TOFD oes not help in
: distinguishing the lateral wave from the 'background noise'. How can I
: get a clean, high amplitude, lateral wave at increased index spacing?

: After indexing my transducers to the proper spacing of 86mm, I receive
: a variety of signals. At this distance, the lateral wave has disappeared
: and
: my first signal is the backwall signal. Several signals of higher amplitude
: follow. What are these signals, and how are they applied in the scan?

: Your responce is greatly appreciated,

: Damon Priestley
: nomad@ksamail.com

: P.S. I do not have access to the internet, only e-mail (due to
: restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
: information off of a web-page. Thank-you.

: Comments:
: We post this message on behalf of Damon Priestley to the Forum.
: Please reply anyway to the Forum (we all want to learn from it),
: we'll forward all replies to Damon Priestley by email.
: The A-Scan presentation (attachment from Damon Priestley) you can access
: only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

: Rolf Diederichs

++++++++++++++++++++++++++++++++++++++++++
Damon,

TOFD means Time-of-Flight-Diffraction method, so you have to test your set-up using obstacles which generate diffracted signals, e.g. a notch, a small area internal defect, etc.. The backwall gives only a reflected signal (like a mirror).

To be sure that all diffracted signals from possible internal defects will be caught by the receiving probe, wide beams must be used, i.e. transmitting and receiving probe must be sensitive in a large range of beam angles.

The intersection point of the beams of the two equal transducers at fixed distance (one transmitting, one receiving) has to be in the centre of the plate, not on the backwall. (Your calculation of 86 mm probe distance for two 60° beams is valid for an intersection point at the backwall of a 25 mm plate).
An intersection point at a depth of half plate thickness (12 mm) means a probe distance of only 42 mm.

With this probe setup - on a flawless area of the 25-mm-plate - there must be constant signals of the longitudinal lateral wave travelling direct from the transmitter probe to the receiving one (42 mm sound path), and a reflected longitudinal wave from the backwall (65 mm sound path) . This gives the time gate for expected diffracted (longitudinal) signals from possible internal defects.

If you get no signals of the lateral wave and the backwall using this setup with 42 mm distance of the probes, the beam spread is too small. You have to use smaller transducer diameters or lower frequency.

I regret that you cannot receive a sketch of the setup.

Kind regards
Udo Schlengermann



 
05:55 Jul-01-1998
Ivan Silva
Re: Questions on TOFD reases the amount of noise level and does not help in
: : distinguishing the lateral wave from the 'background noise'. How can I
: : get a clean, high amplitude, lateral wave at increased index spacing?

: : After indexing my transducers to the proper spacing of 86mm, I receive
: : a variety of signals. At this distance, the lateral wave has disappeared
: : and
: : my first signal is the backwall signal. Several signals of higher amplitude
: : follow. What are these signals, and how are they applied in the scan?

: : Your responce is greatly appreciated,

: : Damon Priestley
: : nomad@ksamail.com

: : P.S. I do not have access to the internet, only e-mail (due to
: : restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
: : information off of a web-page. Thank-you.

: : Comments:
: : We post this message on behalf of Damon Priestley to the Forum.
: : Please reply anyway to the Forum (we all want to learnfrom it),
: : we'll forward all replies to Damon Priestley by email.
: : The A-Scan presentation (attachment from Damon Priestley) you can access
: : only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

: : Rolf Diederichs

: ++++++++++++++++++++++++++++++++++++++++++
: Damon,

: TOFD means Time-of-Flight-Diffraction method, so you have to test your set-up using obstacles which generate diffracted signals, e.g. a notch, a small area internal defect, etc.. The backwall gives only a reflected signal (like a mirror).

: To be sure that all diffracted signals from possible internal defects will be caught by the receiving probe, wide beams must be used, i.e. transmitting and receiving probe must be sensitive in a large range of beam angles.

: The intersection point of the beams of the two equal transducers at fixed distance (one transmitting, one receiving) has to be in the centre of the plate, not on the backwall. (Your calculation of 86 mm probe distance for two 60° beams is valid for an intersection point at the backwall of a 25 mm plate).
: An intersection point at a depth of half plate thickness (12 mm) means a probe distance of only 42 mm.

: With this probe setup - on a flawless area of the 25-mm-plate - there must be constant signals of the longitudinal lateral wave travelling direct from the transmitter probe to the receiving one (42 mm sound path), and a reflected longitudinal wave from the backwall (65 mm sound path) . This gives the time gate for expected diffracted (longitudinal) signals from possible internal defects.

: If you get no signals of the lateral wave and the backwall using this setup with 42 mm distance of the probes, the beam spread is too small. You have to use smaller transducer diameters or lower frequency.

: I regret that you cannot receive a sketch of the setup.

: Kind regards
: Udo Schlengermann

==============================================================


Udo,

I'm starting work with TOFD, and read about some limitations of this technique. One of them are dead zones near the surfaces of the plate, so I have some questions.
What are the pratical values of this dead zone for plates of 25, for transducers of 5MHz and 2MHz?
Another question that I have is what the maximum thickness plate that I can work with only one pair of transducers (5MHz)? And what is the minimum crack heiht that I can measure?

Thanks

Ivan Costa da Silva
Federal University of Rio de Janeiro - Brazil.





 
00:43 Jul-02-1998

Udo Schlengermann

Consultant, -
Standards Consulting,
Germany,
Joined Nov 1998
172
Re: Questions on TOFD ther. Applying
: : : more gain just increases the amount of noise level and does not help in
: : : distinguishing the lateral wave from the 'background noise'. How can I
: : : get a clean, high amplitude, lateral wave at increased index spacing?

: : : After indexing my transducers to the proper spacing of 86mm, I receive
: : : a variety of signals. At this distance, the lateral wave has disappeared
: : : and
: : : my first signal is the backwall signal. Several signals of higher amplitude
: : : follow. What are these signals, and how are they applied in the scan?

: : : Your responce is greatly appreciated,

: : : Damon Priestley
: : : nomad@ksamail.com

: : : P.S. I do not have access to the internet, only e-mail (due to
: : : restrictions in the Kingdom of Saudi Arabia) so I will not be able to pull
: : : information off of a web-page. Thank-you.

: : : Comments:
: : : We post this message on behalf of Damon Priestley to the Forum.
: : : Please reply anyway to the Forum (we all want to learn from it),
: : : we'll forward all replies to Damon Priestley by email.
: : : The A-Scan presentation (attachment from Damon Priestley) you can access
: : : only via the Forum Webpage http://www.ndt.net/wshop/forum/forum.htm

: : : Rolf Diederichs

: : ++++++++++++++++++++++++++++++++++++++++++
: : Damon,

: : TOFD means Time-of-Flight-Diffraction method, so you have to test your set-up using obstacles which generate diffracted signals, e.g. a notch, a small area internal defect, etc.. The backwall gives only a reflected signal (like a mirror).

: : To be sure that all diffracted signals from possible internal defects will be caught by the receiving probe, wide beams must be used, i.e. transmitting and receiving probe must be sensitive in a large range of beam angles.

: : The intersection point of the beams of the two equal transducers at fixed distance (one transmitting, one receiving) has to be in the centre of the plate, not on the backwall. (Your calculation of 86 mm probe distance for two 60° beams is valid for an intersection point at the backwall of a 25 mm plate).
: : An intersection point at a depth of half plate thickness (12 mm) means a probe distance of only 42 mm.

: : With this probe setup - on a flawless area of the 25-mm-plate - there must be constant signals of the longitudinal lateral wave travelling direct from the transmitter probe to the receiving one (42 mm sound path), and a reflected longitudinal wave from the backwall (65 mm sound path) . This gives the time gate for expected diffracted (longitudinal) signals from possible internal defects.

: : If you get no signals of the lateral wave and the backwall using this setup with 42 mm distance of the probes, the beam spread is too small. You have to use smaller transducer diameters or lower frequency.

: : I regret that you cannot receive a sketch of the setup.

: : Kind regards
: : Udo Schlengermann

: ==============================================================

:
: Udo,

: I'm starting work with TOFD, and read about some limitations of this technique. One of them are dead zones near the surfaces of the plate, so I have some questions.
: What are the pratical values of this dead zone for plates of 25, for transducers of 5MHz and 2MHz?
: Another question that I have is what the maximum thickness plate that I can work with only one pair of transducers (5MHz)? And what is the minimum crack heiht that I can measure?

: Thanks

: Ivan Costa da Silva
: Federal University of Rio de Janeiro - Brazil.

: ++++++++++++++++++++++++++++++++++++++++++

Reply to Ivan Costa da Silva (ivanco@metalmat.ufrj.br)
by Udo Schlengermann (uschlengermann@krautkramer.de)
to ‘Questions on TOFD’

Ivan,

You asked three questions, here are my answers:

Diffracted ultrasonic signals are much weaker than reflected signals (more than 26 dB). Therefore you need high gain when doing TOFD-tests, e.g. reflected pulses during that tests are pulses with high amplitude and greater length than diffracted signals.

1. The display of TOFD results is a B-scan image, i.e. it is a diagram showing the time-of-flight over the position of the pair of probes. The amplitude of the unrectified ultrasonic pulses is coded in grey from black to white.
A lateral wave travelling on the coupling surface and the backwall echo on the opposite surface are marking the top and the bottom of the plate, generating dead zones close to both surfaces.
The depth of these dead zones is determined by the pulse length, e.g. it is independent of plate thickness. As you know, pulse length depends on frequency and on damping.
If for example the pulse at high gain shows 4 oscillations (wavelengths) and the beam angle is 60° this means for a longitudinal wave in steel (5920 m/s):
with 2 MHz: pulse length = 12 mm zone depth (with 60°) = 6 mm = dead zone,
with 5 MHz: pulse length = 5 mm zone depth (with 60°) = 2.5 mm = dead zone.

2. Which sector of the plate is covered by one pair of probes depend on the beam widths (divergence) of the two beams. The width depend on transducer size and on frequency:
Smaller transducer means wider beam,
Lower frequency means wider beam, and opposite.
If the beams do not cover the whole plate thickness, you have to use two different pairs, e.g. a first one to cover the upper half of the plate and a second one to cover the lower half. This is the same as with tandem testing of thick plates.
The best is to verify the coverage by using a set of sawcuts with different depth, to look for the amplitude of the diffracted signals from the bottom of the cuts.

3. If you talk about a crack open to the surface, it has to be deeper than the dead zones to be detected and sized by TOFD. Then it has to be deeper than 2 mm for a test with a frequency higher than 5 MHz.
If the crack is internal you have to resolve two diffracted signals from the upper and lower boundaries of the crack. That means the crack has to be larger than the pulse length to be sized by TOFD. If it is not, the crack may be detected, but it is not possible to size it by time-of-flight difference.

Kind regards
Udo Schlengermann

:




 


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