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Metallurgy/A286 Tensile/Yield strength

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QUESTION: Dear Sir
we have a stud-bolt M8 , LENGTH 54MM
raw material : A286 , NiCrTi 26 15 , EN. 1.4980
solution anealed + Aged (720C , 16 Hour)
the stud is tested by SGS Laboratry for Tensile & Yield Strength.
the test report Tensile strength = 1072 Mpa
however, they said they are unable to do the Yield strength test because of the short length of the stud. they said for Yield test the stud must be longer than 90mm.
Anyway, we need a standard or a reliable document to know the relationship between Tensile and Yield strength so that we can obtain Yield strength.
some people say that Yield is 0.92 Tensile.
but we need a standard or document that we can refer our customer to it.
please kindly advise how we can obtain Yield strength by having Tensile strength ?

Yours sincerely

ANSWER: Dear Ali,
in this regard, I will say that we can't correlate the properties by formulae.Still I am attaching two of my searches which may be acceptable to your customer.
No.1
Calculating Yield and Tensile Strength
In most cases, the strength of a given material used to make a fastener has strength requirements or parameters described as pounds per square inch (psi) or thousands of pounds per square inch (ksi). This is helpful when analyzing what grade of material should be used for a given application, but this doesn’t tell us the actual strength of that diameter of material. In order to calculate the actual strength values of a given diameter, you would use the following formulas:
Note: the formulas below do not depend on the finish of the fastener.
Ultimate Yield Strength:

Take the minimum yield in psi of the ASTM grade (see our Strength Requirements by Grade Chart for this value), multiplied by the stress area of the specific diameter (see our Thread Pitch Chart). This formula will give you the ultimate yield strength of that size and grade of bolt.
Example: What is the ultimate yield strength of a 3/4″ diameter F1554 Grade 36 rod?

This is the minimum requirement for F1554 grade 36. In other words, a 3/4″ diameter F1554 grade 36 anchor rod will be able to withstand 12,024 pounds force (lbf) without yielding.
Ultimate Tensile Strength:

Take the minimum tensile strength in psi of the ASTM grade, multiplied by the stress area of the diameter. This formula will give you the ultimate tensile strength of that size and grade of bolt.
Example: What is the ultimate tensile strength of a 3/4″ diameter F1554 Grade 36 rod?

This is the minimum requirement for F1554 grade 36. In other words, a 3/4″ diameter F1554 grade 36 anchor rod will be able to withstand 19,372 pounds force (lbf) without breaking

No.2
 
LIMITS TO Y/T RATIO IN ACCORDANCE WITH VARIOUS DESIGN CODES
REFER BS:5400 STD
Regards,
RP SHARMA



---------- FOLLOW-UP ----------

QUESTION: Dear Dr.Sharma
Thank you for your last email .
I realized that UTS and YIELD do not behave proportionally for A286.
I gave my stud to heat-treatment ( solution anealed 900C , 1 hour followed by aging 720C 16 hour ) ,and UTS  became 920-930 Mpa and Yield 620-700 Mpa in two different test in laboratory.
unfortunately , the minimum limit for the manufacturer to accept my stud is UTS:1000 mpa and YIELD : 900 Mpa.
so my stud was rejected.
I need to know if my aging process at different temperature can provide higher Yield strength. my major problem is that my stud Yield is 620-700Mpa while the acceptance limit is 900 Mpa
so , I need to increase the Yield at least 200-250 Mpa .
I also found a graph. I attach it.
Please advise .
Thank you
Ali

Answer
Dear Ali saab,
The success of any heat treatment depends upon
   1.The composition of the alloy.
   2.The type and character of the quenching medium.
   3.The size and shape of the specimen.
What is composition of your steel, if it is as per A286(verify from spectra lab analysis and let me know the sulfur and Phosphorus percentage, better total composition), then your heat treatment cycle 899*C/1hr-720/16 hrs is ok.Following are the precaution while performing heat-treatment:
1) Quench oil should be at 40*c
2) Material transfer from furnace at 899*C to quench tank should be instantaneous.If there is any delay, it will affect the properties adversely.
3) Soon after material is removed from quench tank, it should go to ageing immediately.You ageing cycle is ok but soon after ageing, it should be cooled by forced air-fans etc.
Check it again and I am sure if composition is ok then you will get YS above 900 mpa.
All the best and happy new year.
Khudahafiz!
Regards,
RP SHARMA

Metallurgy

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RAJINDER PRASAD SHARMA

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Questions that i can answer:- 1-sand casting of cast iron,S.G. iron,steel and aluminium alloys 2-Gravity die casting of aluminium 3-Low pressure die casting of aluminium 4-High pressure die casting of aluminium 5-Failure analysis of ferrous and non-ferrous castings Questions that i can`t answer:- 1-testing of iron and steel 2-Problem related to welding 3-Problems related to corrosion 4-Amorphous metals,rubbers,nylon,plastics and plastic moulding

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1-Sand casting and ferrous and non-ferrous alloys 2-Gravity die casting,low pressure die casting,high pressure die casting of aluminium alloys.With a specialisation in auto-mobile components castings. 3-Alloy steel making on arc furnace,AOD,VOD,Con-cast root of alloy steel making. I have been working as Foundry man since last 27 years.

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Hindustan motors Ltd.

Education/Credentials
I am a Metallurgical Engineer graduated from National institute of Technology,warangal(India) in 1984 and in 1991 i completed my masters degree in business administration(M.B.A.) from Institute of Management Technology,Ghaziabad(India).

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