How to work out the tensile strength in pounds

[Rustam]

Hi to all.
I have a raw material certificate of drilling pipe - there is a YS=863 N/mm2. I would like to know what is the maximum weight in pounds this pipe can withstand.

 

[russ_watters]

Welcome to PF! This is a complicated question that requires exact details in order to be solvable.

 

[Rustam]

Welcome to PF! This is a complicated question that requires exact details in order to be solvable.

Thank you, Russ.
What data is required for that?I didnt expect the thing was so complicated...

 

[russ_watters]

Sorry, I may be missing something due to a possible mismatch between the title and post/confusing terminology("weight it can withstand" and "pipe" had me envisioning a structure). If it is just the max tensile strength, you can probably look it up in a table based on the material name. Or ask the supplier.

 

[JBA]

Part of this is a units conversion issue. 863 N = 194 lbf and 1 mm ^2 = 0.00155 in ^2, i.e, the result is 125,161 lb/in^2. What is then required is the x-sectional area of the drill pipe.

At the same time, this does mean that the drill pipe's rated tensile load is equal to that result, so you must request the load rating from the drill pipe supplier to insure that the drill pipe thread joint rating is equal to or greater than that of the above calculated load bearing value plus the fact that there is surely a safety factor that is applied to the actual maximum yield stress and the thread failure load values.

 

[Rustam]

Part of this is a units conversion issue. 863 N = 194 lbf and 1 mm ^2 = 0.00155 in ^2, i.e, the result is 125,161 lb/in^2. What is then required is the x-sectional area of the drill pipe.

At the same time, this does mean that the drill pipe's rated tensile load is equal to that result, so you must request the load rating from the drill pipe supplier to insure that the drill pipe thread joint rating is equal to or greater than that of the above calculated load bearing value plus the fact that there is surely a safety factor that is applied to the actual maximum yield stress and the thread failure load values.

Thank you very much, JBA.
I got almost the same result (slight conversion issue). The OD of pipe is 5.5" and ID is 4". The pipe itself is integral (no welds etc.). So, the rating of pipe body is equal to that of tool joint. I calculated 1784100 lbs (converted the 863 N/mm2 to 125200 psi (lbs/in2) and then multiplied the cross-sectional area of pipe (5.5"-4"=14.25 in2) with 125200).
I just would like to check if my calcs are right.

 

[JBA]

You need to recalculate your x-sectional area for the pipe using the given 5.5 and 4 in dia values. My result for the area is 11.1919 in^2 for a total load limit of 1,401,229 lbs using your 125,200 psi value.

 

[Rustam]

You need to recalculate your x-sectional area for the pipe using the given 5.5 and 4 in dia values. My result for the area is 11.1919 in^2 for a total load limit of 1,401,229 lbs using your 125,200 psi value.

Yes, JBA, I didnt apply a conversion...my bad.
I agree with you.
Thanks again for an interesting discussion and help.