About Through thickness tensile testing (Z Direction)

Date 01-08-2008 20:31

I would say, shootin from the hip, that your concerns may be warranted, though I would hope some of our API guys would comment on this one.
The min Z direction tensile requirement is 30% (15 ksi) and you got 82% (41 ksi) . Thats a substantial difference.
However, the material is also required to have a 25 ft/lb min Charpy result at -40F. And one of the things that help your Charpies can help your Z direction tensiles as well. Clean steel.
Maybe that high of a result is not that uncommon. I don't know.

By QUOCVUONG (*)

Date 01-09-2008 02:43

Thank for your replies,
Z quality determine by reduction of area value. As API 2H specification requirement minimum is 30% .Also, as per EN10164 divided Z quality in three classes Z15, Z25, Z35 only. Therefore, could I believe in this result (Z82%)?
Could you please tell me some good explanation on this.
I think that result was wrong. What do you think about this?

Thank a lot,
B.Q.V

By js55

Date 01-09-2008 14:34

This is as good one for the new metallurgical forum. But its here so we'll deal with it.
In answer to your question though, why not?
Metallurgically its not impossible. But understand, I have no experience with this material. I am doing the best I can since there doesn't seem to be any API guys as yet stepping up to the plate.
But, the reason the Z direction is problematic is grain orientation and non metallic laminations (sulphides for example) that have been extended due to the rolling process.
If the material is clean then there will be a smaller volume percent of laminations and the Z direction mechanicals will improve.
If the material has undergone a heat treatment above the upper transformation temp (which I actually doubt) then grain orientation will be essentially null and void since the microstrucutre will recrystallize. And again the Z direction mechanicals will be improved.
If you have concerns verify it with additional testing. Contact the manufacturer of the material. Communicate with your customer who has probably dealt with this material before.

By QUOCVUONG (*)

Date 01-10-2008 01:54

Thank for your explanations,
I already understand what affect in Z qualify of steel. But I want to survey who has ever met too high value of Z% like me and with that result can believe or not, that is logical or not?
As your recommendation must be test more specimens for verification I think it is very simple that why I would like to listen your explanations without carry out testing.
Thanks,
B.Q.V

By DaveBoyer (*****)

Date 01-10-2008 03:51

I am far from an expert, but it is My understanding that extremely low Z axis strength is a result of insuficient reduction from billet to finished thickness when rolling heavy plate. There is heavy plate available with improved Z axis properties. I am not at all familliar with the material You are working with, but I would expect from the seat of My pants that the Z axis tensil strength of a plate only 40 mm thick would be well over 30% of the x axis strength. Might I suggest that I am misunderstanding Your question. The 30% specification You mention, is that the percentage of X axis tensile strength, or is that the minimum reduction in area? Is the 82% result You got from testing the percentage of X axis tensile strength, or the reduction in area?

By QUOCVUONG (*)

Date 01-10-2008 09:08

Thank for your concernment,
Z quality of steel determine by reduction of area value. It is achieve from through thickness tensile test (Specimen was get from through thickness direction of the plate).
Z=(Ao-A/Ao)x100%
Where: Ao is original area of specimen
A is area of specimen after test measuring at broke location

By DaveBoyer (*****)

Date 01-11-2008 04:20

Once again, I stress that I am no expert and I am not familiar with the material, but 82% reduction in area seems extremely high to Me.

By QUOCVUONG (*)

Date 01-11-2008 07:11

oh that why I wonder & I am not familiar with this matter also. so I would like to listen who can be positive about this result if it was wrong.

By Noel Tan (**)

Date 01-11-2008 09:21

through thickness properties include through thickness tensile strenght and area of reduction.
Some standards limit material through thickness tensile strength shall have min. 80% of its minimum specific tensile strength.
area of reduction is about 30%. The 82% in the report is refer to tensile strength or area of reduction?

82% area of reduction means the area after fracture is only 18% of the original area before fracture.
I do experienced tensile test of API 2W materials which is thermomechanically rolled and normally used for offshore structure, very good toughness which can have more than 200J @ -40 degree C. i was wondering when we did the transverse weld tensile test and it shown fracture at parent metal with relatively large elongation and reduction of area (i did not really measured, only by visual). i can't even notice when the test specimen broken (specimens size is about 25mm x 25mm), it broke quitely... however, it was not about Z direction.

82% reduction of area in Z direction... i think when i have this result from the test, i don't really know i should be happy or worry.

Best Regards,
Noel Tan

By kipman

Date 01-11-2008 09:02

Dave,
This would not be an 82% reduction in area. It would be 82% of the original area. For this formula (which I am somewhat familiar with) the lower the value the better the ductility - i.e. 72% would indicate a material that is more ductile in the through thickness direction than one that results in an 82% value.
The material spec for the project that I am currently on requires that the Z direction value be 80% or less. The material that Mr. QUOCVUONG describes would not be acceptable for the project that I am on. It may be acceptable for his project - clearly he needs to look at the material spec and see what the acceptable values are for this test.
Regards,
Mankenberg

By QUOCVUONG (*)

Date 01-11-2008 17:21

Mr kipman you are right,
That material is API 2H Gr50 , it is unacceptable for through thickness property.
I really need an API 2H Gr50Z but our vendor have offered an API 2H Gr50 with additional through thickness tensile test (I didn't accept) Then they provived us report with result of 82% in reduction of area. I have suspected that it has something wrong and I have asked them about that result. of course they shall be protected their results.
I want to reject this one but I didn't witness during test.That why I hope in this forum if I can verify the result. if it was wrong. I'll have a lesson to learn.
Thanks!

By DaveBoyer (*****)

Date 01-12-2008 04:17 Edited 01-12-2008 04:22

Mankenburg, So You are working with the area at the failure as a percentage of the original area. Doesn't the formula QUOCVUONG mentioned in His post give the reduction in area in the usual sense as oposed to the way You are working? If I understand all this then the 82% figure is actually 18% reduction in area.

By QUOCVUONG (*)

Date 01-15-2008 08:52

They also have explained us with correct equation [(A-Ao)/Ao]x100% (=82%)?
Thanks

By DaveBoyer (*****)

Date 01-16-2008 05:00

Have You actually seen the test specimine? Is the area where the specimine broke 18% of the original size [82% reduction in area] or is the area where it broke 82% of the original size [18% reduction in area] This is an obvious difference. Some fully anealed steel alloys have a reduction in area as high as 70%, I don't recall ever seeing one higher, but I am not saying it isn't possible. I agree with You in doubting the 82% reduction in area, it seems out of the ball park, so to speak.

By QUOCVUONG (*)

Date 01-16-2008 11:16 Edited 01-17-2008 09:32

But here is not a steel alloys, it is API 2H Gr50.
I think that the laboratory used wrong formula as (A/Ao)x100% = 82% that means % reduction in area is 100%-82%=18% only.We have the same result with correct formula as [(Ao-A)/Ao]x100% =18%
Where A0 is original area of speciment measured and caculated before test
A is area of speciment measured at broke location and caculated after test
But they have protected their result by correct formula as [(Ao-A)/Ao]x100%=82%
Can the above material be ductile like that?
Do you agree with me?

By DaveBoyer (*****)

Date 01-17-2008 06:14

I agree that they are inacuratly representing the test and that that material will not be nearly as ductile as they would have You believe. The suplier either doesn't understand the test, or is hoping You don't. They specify 30% reduction in area as a minimum with this material, the actual results will not be 273% greater than the minimum, it will only be a little higher. I suggest You get a sample tested by an independent party if You can not test it Yourself.