Sound waves to test for faults in girders

[sgstudent]

hi i have a physics questions about sound Sound waves are used to detect faults in girders. a cro is used to detect these faults, a pulse producer and a detector is placed on opposite sides of the detector.

The safe detector has a upwards spike at 0microseconds and at 10 microseconds.

The unsafe detector has a upwards spike at 0 microseconds and 10 mircroseconds and also one additional but smaller upwards spike at 4 microseconds.
the image i drew of the CRO is here http://postimage.org/image/4ure9j3sr/

why is the the unsafe detector considered unsafe?

my understanding: there is a crack in the girder so when the sound waves is sent in, some of the waves will have to go around it so it causes it to be slower than the sound waves which do pass the crack. because it is slower, so it is shown in the 4 microseconds. meaning that if another pulse is sent out one of the recorded one will be at 10microseconds from the previous one and another which is 4 microseconds ahead of just detected pulse

 

[Studiot]

Hello sgsstudent, welcome to Physics Forums.

there is a crack in the girder so when the sound waves is sent in, some of the waves will have to go around it so it causes it to be slower than the sound waves which do pass the crack. because it is slower, so it is shown in the 4 microseconds. meaning that if another pulse is sent out one of the recorded one will be at 10microseconds from the previous one and another which is 4 microseconds ahead of just detected pulse

Is English your first language? No offence but this doesn't make your meaning clear.

The basis of acoustic testing for cracks is that there is a reflection of a an acoustic wave, pulse or signal traveling through the girder at any boundary or discontinuity.

This includes the outer surfaces of the girder.

So I would interpret your diagrams as follows.

On both diagrams, the peak at zero is the reflection given by the pulse entering the girder, the peak at 10 is the reflection of the pulse from the far wall and the small peak at 4 is from a crack between the two walls of the metal. It is smaller (less strong) because a crack is (hopefully ) smaller than the full outer face of the girder.

Most probes include both the pulse transmitter and detector in a single unit so that you only have to work from one side (often you can only get at one side) so I have shown this in the attached sketch.

If you know the speed of sound in the material you can calculate the thickness of the section or the depth to the crack by multiplying by the time read off the CRO trace. Remember that this time is for the double journey and must be halved.

 

[olivermsun]

I interpret the OP as stating that the pulse generator and detector are on opposite sites, so the transit time would not be halved. Anyway we would expect more than one spike in the middle if it were a two-way TDR.

 

[sgstudent]

hi sorry about my english in the other post. i was using my phone to type all of that so i didnt really pay much attention to the usage of english there.

i don't think u understand what i was trying to mean over there (my fault) i drew it up the same way as my school's physics worksheet over here.

http://postimage.org/image/5cgkpiwv7/