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Which tensile strength should I use?


PabLo Dal Buoni

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Steel with heat treatment (for example AISI 4140 Q+T) has different tensile strength with different diameters of the material, and also different tensile strength in a same diameter according to distance to the center of the material, like Jominy test demonstrates (hardenability).

since the bar has different YTS along the diameter, which yield tensile strength should I use for the calculation of a shaft?

Moreover, why this image attached shows differents curves with different specimen diameters ?

 

Thanks.

 

4140 jominy.jpg

95841652_chart14140.jpg.e9b063ddcac34500289d6f0f923e5a8e.jpg

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That's a loaded question...(please excuse the pun!)...it depends!

If you are building a structure subject to repeated loading, or safety critical, you should use the yield...NOT the UTS.

If you are designing fastners, the UTS is more relvant...but then again, if subjected to vibration or "impulse" loading, again the Yield becomes far more important.

It is not a straight forward answer to what seems a simple question.

The differring UTS values are generally due to the heat-sink aspect of the material undergoing heat treatment. 

Basically when an item is hardened (quenching for steels). The item is heated up passed the crystalisation temperaure and then quenched in either oil or water. This takes the heat out of the material so quickly that the crystal-growth is restricted and they remain small. The thicker the item, the more heat remains within the material allowing the crystals to grow to larger (unwanted) sizes....basically for hardness, smaller crystals are better (in steels al least - be careful...other metlaic materials differ quite significantly!)

I would always err on the side of caution and use the lowest value. This is especially the case for a shaft...the hardening may only affect the surface!

For the diagram....for me there is insufficient detail to answer the question....there would be accompanying information to expain further what it is showing...the only answer I can give is that it is two differing samples! (although...as I think about it further, it is possible my previous answer may actually be helpful here!)

 

 

 

 

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1 hour ago, G B Reid MIMechE, SIMarEST said:

That's a loaded question...(please excuse the pun!)...it depends!

If you are building a structure subject to repeated loading, or safety critical, you should use the yield...NOT the UTS.

If you are designing fastners, the UTS is more relvant...but then again, if subjected to vibration or "impulse" loading, again the Yield becomes far more important.

It is not a straight forward answer to what seems a simple question.

The differring UTS values are generally due to the heat-sink aspect of the material undergoing heat treatment. 

Basically when an item is hardened (quenching for steels). The item is heated up passed the crystalisation temperaure and then quenched in either oil or water. This takes the heat out of the material so quickly that the crystal-growth is restricted and they remain small. The thicker the item, the more heat remains within the material allowing the crystals to grow to larger (unwanted) sizes....basically for hardness, smaller crystals are better (in steels al least - be careful...other metlaic materials differ quite significantly!)

I would always err on the side of caution and use the lowest value. This is especially the case for a shaft...the hardening may only affect the surface!

For the diagram....for me there is insufficient detail to answer the question....there would be accompanying information to expain further what it is showing...the only answer I can give is that it is two differing samples! (although...as I think about it further, it is possible my previous answer may actually be helpful here!)

 

 

 

 

Hi GB REID

 

thanks for your response.
I know what is YTS, UTS and when to use each one. I also know why is the difference in hardness (crystalisation, martensite creation, and so on)

Let´s see again the same AISI4140 example.....

We have a chart with different hardness according to the diameter of the material. Is it an average?
And we have different curves of hardness in Jominy tests when we change the sample diameter. I think where water hits, the hardness should be the same.

Furthermore, the final diameter of my piece is smaller than the originall diameter of the material.

Then, which YTS should I use for my calculation?

 

I am really lost.

Thanks !

 

 

 

 

chart1 4140.jpg

chart2 4140.jpg

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Actually for 4140 Alloy, the hardness achieved by heat treatment in that chart was excessive, and it is far too brittle. The values you see in graph are suitable for AISI-4340 but not 4140. It will crack in service. But 4340 has more alloying elements allowing the hardness values to be achieved without cracking.

The reason that the bar at 3 inches has lower tensile strength is that the official figures for tensile strength of an alloy are normally based upon a 1 inch test bar, however, as is well known, tensile strength drops gradually with diameter increase of test bar, this is a well known phenomena not related to quality of manufacturing. The cause of this is more errors that occur at the grain boundaries, so the larger the size of test bar or actual part, the larger the amount of microscopic defects, thus lower tensile yield and ULT.

Which tensile strength to use: First of all, for machinery design,  must be using Yield strength, not ULT. Second of all, use tensile strength corrected to size of actual shaft, for example 6 inch shaft has lower strength compared to 3 inch test specimen. Finally, you can use the internal hardness as the indicator of shaft strength, not the outer hardness.

Abdulrahman Alkhowaiter

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The hardness of higher value is brittle in nature. More grade in number means more alloying element are present. In performing tensile test of the above specimen, first start with high grade heat treated alloy material ( AISI 4340) which is 3 inch shaft test specimen. As, the load on it, more stress is applied on it. It will give one set of ultimate and breaking stress. Then go for the (AISI 4340) 6 inch shaft test specimen, it will show low ultimate and breaking stress, because stress decreases as area of the cross section increased. 

After this, go for low grade alloy (AISI 4140) as followed above diameter specimen (3 inch followed by 6 inch) to determine the respective stress (ultimate and breaking) values.

To have high strength, the specimen has to undergone heat treatment process such as aeging or annealing for respective Re-crystallization tempertature. The material which has higher hardness indicated that it has high strength.

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Before I answer this , I have many question for you. What do you mean calculation of design shaft ? . If you want design from early you must study about machine design first. I suggest you to reading books about machine design , author MS.Khurmi. It's has clear explanation and has many equations which has more variable to count it.

So if you want to know exactly connection between hardness and wear design calc, please read this book first

dcabc9bbca04f83c221f4e243b643d45.png

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6 hours ago, Andika Dinata said:

Before I answer this , I have many question for you. What do you mean calculation of design shaft ? . If you want design from early you must study about machine design first. I suggest you to reading books about machine design , author MS.Khurmi. It's has clear explanation and has many equations which has more variable to count it.

So if you want to know exactly connection between hardness and wear design calc, please read this book first

dcabc9bbca04f83c221f4e243b643d45.png

 

Hi Andika,

thank you for your response.

NO, I do not want to know the connection between hardness and wear design calc. I did not say that.

I do not need the equations either.

My question is, what YIELD TENSILE STRENGHT I have to use for calculation, since the bar has different YTS along the diameter. Just that.

Regards.

 

 

 

On 6/28/2021 at 1:07 AM, Abdulrahman-1 said:

Actually for 4140 Alloy, the hardness achieved by heat treatment in that chart was excessive, and it is far too brittle. The values you see in graph are suitable for AISI-4340 but not 4140. It will crack in service. But 4340 has more alloying elements allowing the hardness values to be achieved without cracking.

The reason that the bar at 3 inches has lower tensile strength is that the official figures for tensile strength of an alloy are normally based upon a 1 inch test bar, however, as is well known, tensile strength drops gradually with diameter increase of test bar, this is a well known phenomena not related to quality of manufacturing. The cause of this is more errors that occur at the grain boundaries, so the larger the size of test bar or actual part, the larger the amount of microscopic defects, thus lower tensile yield and ULT.

Which tensile strength to use: First of all, for machinery design,  must be using Yield strength, not ULT. Second of all, use tensile strength corrected to size of actual shaft, for example 6 inch shaft has lower strength compared to 3 inch test specimen. Finally, you can use the internal hardness as the indicator of shaft strength, not the outer hardness.

Abdulrahman Alkhowaiter

Hi Abdulrahman,

thank you for your response.

Yes, I think I have to choose the minimum YTS, corrected to the final diameter size, like you said.

On 6/28/2021 at 3:09 AM, Balaji Nidaganti said:

The hardness of higher value is brittle in nature. More grade in number means more alloying element are present. In performing tensile test of the above specimen, first start with high grade heat treated alloy material ( AISI 4340) which is 3 inch shaft test specimen. As, the load on it, more stress is applied on it. It will give one set of ultimate and breaking stress. Then go for the (AISI 4340) 6 inch shaft test specimen, it will show low ultimate and breaking stress, because stress decreases as area of the cross section increased. 

After this, go for low grade alloy (AISI 4140) as followed above diameter specimen (3 inch followed by 6 inch) to determine the respective stress (ultimate and breaking) values.

To have high strength, the specimen has to undergone heat treatment process such as aeging or annealing for respective Re-crystallization tempertature. The material which has higher hardness indicated that it has high strength.

Hi Balaji,

thank you for your response.

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Hi Pablo,

 

Can see your question really set and specific,  try this case:

I read that your shaft/member cuts across the changing values of YTS.  I advise that you use the lowest value of the YTS.  Your chart indicate 570MPa.  Reasons:

1.  Lowest MPa keeps you in the safe range, and you further need a safety factor applicable to the environment of operation of your member.

2.  Lowest MPa ensures that no diameter of your member will run into plastic deformation.  Varied plastic deformation in a member generate secondary stress within the member and this will reduce on number of load cycles that the member can withstand due to fatigue loading.

Allow me to keep it that simple.  No equations at this stage as requested.

Kind regards.

 

Michael

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9 hours ago, Mucugia said:

Hi Pablo,

 

Can see your question really set and specific,  try this case:

I read that your shaft/member cuts across the changing values of YTS.  I advise that you use the lowest value of the YTS.  Your chart indicate 570MPa.  Reasons:

1.  Lowest MPa keeps you in the safe range, and you further need a safety factor applicable to the environment of operation of your member.

2.  Lowest MPa ensures that no diameter of your member will run into plastic deformation.  Varied plastic deformation in a member generate secondary stress within the member and this will reduce on number of load cycles that the member can withstand due to fatigue loading.

Allow me to keep it that simple.  No equations at this stage as requested.

Kind regards.

 

Michael

Hi Michael,

yes !! that is the best answer I think. Thanks a lot.

Do you think these values in the chart are averages?

 

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Hi,

Yes they are experimental averages and their publication means that they are considered to be good working values.  Allow me to add that source for the values is also important.  Is the source authentic.  If so, you are good to go.

Regards.

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Pablo, you should not think one-dimensionally. I touched upon several errors in your expectations, but you only think of one of them. This is not correct, a mechanical engineer or Polytechnic graduate must analyze all issues, i suggest to read my answer again. 

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Hi AliceIn....apologies for the delay, just read your message!

ALWAYS err on the side of caution and use the lowest value unless absolutely sure!

 

Warmest wishes

 

Bruce

 

 

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49 minutes ago, G B Reid MIMechE, SIMarEST said:

Hi AliceIn....apologies for the delay, just read your message!

ALWAYS err on the side of caution and use the lowest value unless absolutely sure!

 

Warmest wishes

 

Bruce

 

 

Hi Bruce ! thank you anyway.

Do you think these values are an average?

ok now, if my original round material is 100mm diameter and the diameter of the final piece in the risky zone is 70 mm, I will use YTS for 100 mm of the chart.

I think it is ok for me with some of your responses, now I wonder if this is an average of Jominy results.

Best regards Bruce, and thanks again. 

 

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Hi Pablo

Alas, I have no first hand experience with the Jominy, so I will not comment on this. (I only comment where I have knowledge or first hand experience...)

That said, on the other, yes, use the figures for the 100mm, if incorrect, the error will be on the "safe" side.   

Bruce

 

 

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Hi Pablo,

Have to deal with case by case of each shaft diameter,

General rule is use the Yield strength as it defines the onset of plastic deformation.  Two, work with the outer most diameter as for any shaft plasticity start at the outer most layer of the shaft.  Therefore if the shear stress of outer most layer of the shaft is below the yield shear strength, then your shaft is safe. (Assumption; we are dealing with torque loads).

Regards.

Michael

Given the tensile strength is important to get the appropriate shear strength for your analysis.  Different materials have differing relations between tensile and shear stress.

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15 hours ago, Mucugia said:

Hi Pablo,

Have to deal with case by case of each shaft diameter,

General rule is use the Yield strength as it defines the onset of plastic deformation.  Two, work with the outer most diameter as for any shaft plasticity start at the outer most layer of the shaft.  Therefore if the shear stress of outer most layer of the shaft is below the yield shear strength, then your shaft is safe. (Assumption; we are dealing with torque loads).

Regards.

Michael

Given the tensile strength is important to get the appropriate shear strength for your analysis.  Different materials have differing relations between tensile and shear stress.

Thanks Mucugia

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