- posted on March 1, 2006, 11:09 pm

Hello everyone,

I would like to ask if anyone could please help me with the following situation.

I am using a standard hardened steel "pull out" dowel pin from www.mcmaster.com as a bearing shaft. I need to know the maximum load
the dowel / shaft can support without taking a permanent set and/or
becoming permanently deformed or bent. I need the shaft to always
spring back to its original position after the load is removed.

I emailed Mcmaster, but they were not able to give the maximum Yield strength of the dowels.

Does hardening increase the maximum Yield stress? If so, is there a way to calculate or estimate how hardening affects the yield stress ? If I know the yield stress, then I can compare the maximum yield to the bending stress given by my beam design program, and I think this will tell me if the dowel can support the load without taking a set.

Here is what Mcmaster said about the dowels and material...

"Hardened Steel- Made from hardened steel such as C1541, or 4037 and 4140 alloy steel. Core Rockwell hardness is C47-C58 (surface hardness is RC 60). Shear strength is the amount of force that the side of a pin can withstand before breaking. Single shear strength is the amount of force applied against a fastener in one place causing the fastener to break into two pieces. Single shear strength is 130,000 psi. An internally threaded tapped hole in one end of these pins lets you pull them out with a removal screw or a threaded puller such as 92330A (see page 3083 ) and reuse them. All meet ASME B18.8.2. Length tolerance is ±.010"."

I would appreciate any advice or suggestions on how I can get a close estimate on this, and what would be a reasonable safety factor to apply. Nobody could get hurt if the device fails, but I just need it to be reliable. I have to consider several factors when choosing a shaft size, and everything fits in a tight space.

There are tradeoffs and space constraints when going to a bigger shaft, so I need to know how to estimate this in order to make the best compromise. It's desirable to use the smallest shaft diameter possible, that will support the load with a reasonable safety factor & not take a set.

Thanks for your help. John

I would like to ask if anyone could please help me with the following situation.

I am using a standard hardened steel "pull out" dowel pin from www.mcmaster.com as a bearing shaft. I need to know the maximum load

I emailed Mcmaster, but they were not able to give the maximum Yield strength of the dowels.

Does hardening increase the maximum Yield stress? If so, is there a way to calculate or estimate how hardening affects the yield stress ? If I know the yield stress, then I can compare the maximum yield to the bending stress given by my beam design program, and I think this will tell me if the dowel can support the load without taking a set.

Here is what Mcmaster said about the dowels and material...

"Hardened Steel- Made from hardened steel such as C1541, or 4037 and 4140 alloy steel. Core Rockwell hardness is C47-C58 (surface hardness is RC 60). Shear strength is the amount of force that the side of a pin can withstand before breaking. Single shear strength is the amount of force applied against a fastener in one place causing the fastener to break into two pieces. Single shear strength is 130,000 psi. An internally threaded tapped hole in one end of these pins lets you pull them out with a removal screw or a threaded puller such as 92330A (see page 3083 ) and reuse them. All meet ASME B18.8.2. Length tolerance is ±.010"."

I would appreciate any advice or suggestions on how I can get a close estimate on this, and what would be a reasonable safety factor to apply. Nobody could get hurt if the device fails, but I just need it to be reliable. I have to consider several factors when choosing a shaft size, and everything fits in a tight space.

There are tradeoffs and space constraints when going to a bigger shaft, so I need to know how to estimate this in order to make the best compromise. It's desirable to use the smallest shaft diameter possible, that will support the load with a reasonable safety factor & not take a set.

Thanks for your help. John

- posted on March 1, 2006, 11:34 pm

Make sure that the loads never get very close to yielding the shaft, if you only knew the yield strength of this catalog article.

You might be learning a good lesson here.

You picked a part from a catalog.

The catalog dealer tells you he doesn't know what material it will be made of.......

He tells you that the surface hardness is evidently fairly well known.

However, he tells you that the interior hardness can vary quite a lot.

Since there is an empirical correlation between ultimate strength and the hardness, you know that the untimate strength can vary a lot.

You can figure out how much the strength variation could be by google searching for "Correlation hardness ultimate strength rockwell c" meaning that you are looking for the correlation between ultimate tensile strength and the hardness as measured on the Rockwell "C" scale. You will be able to see that there is an enormous strength difference between Rockwell C47 hardness and C58 hardness.

You may therefore assume that the tensile strength will be equivalent to that of Rockwell C47 - at a minimum.

There is also an emperical correlation between hardness and yield strength, so you can repeat all of the above to give you an estimate of the yield strength of whatever alloy you end up with.

Remember, the manufacturer told you that you might get any of a number of alloys.

The lesson I would take away is to not buy a "part" and then use it for a different purpose than intended.... unless I had some way to become assured of the minimum mechanical properties of the "part" so one could do an analysis.

Without having the properties of the part, analysis might not be very meaningful.

A number of people make bearing shafts for various purposes, and maybe that is the way to go. At least the catalog from a bearing shaft outfit might have the appropriate bearing shaft data.

-------------------------------------------------------------------- I can imagine a way to use a metallic zipper as a sort of continuous switch.

I wouldn't expect to get much help from the zipper industry with my questions on the electrical characteristics of zippers..... because the zipper industry makes fasteners, not electrical switches.

---------------------------------------------------------------------

Good luck.

I would approach the problem using the bearing shaft catalogs first and foremost.

--

1) Eat Till SATISFIED, Not STUFFED... Atkins repeated 9 times in the book

2) Exercise: It's Non-Negotiable..... Chapter 22 title, Atkins book

1) Eat Till SATISFIED, Not STUFFED... Atkins repeated 9 times in the book

2) Exercise: It's Non-Negotiable..... Chapter 22 title, Atkins book

Click to see the full signature.

- posted on March 2, 2006, 7:27 am

jbuch wrote:

AFAIK there is a correlation between hardness and ultimate tensile strenght for steels with a certain plastic deformability. For hardened steels, it is difficult to measure tensile strength, because fracture is flaw-controlled here. Just curious: Did you ever carry out a tensile test using a C58-hardness-specimen?

Unfortunately, I don't have the standards in my actual office, but I rememeber that the correlation between hardness an UTS is valid only up to a certain maximum hardness.

Michael Dahms

AFAIK there is a correlation between hardness and ultimate tensile strenght for steels with a certain plastic deformability. For hardened steels, it is difficult to measure tensile strength, because fracture is flaw-controlled here. Just curious: Did you ever carry out a tensile test using a C58-hardness-specimen?

Unfortunately, I don't have the standards in my actual office, but I rememeber that the correlation between hardness an UTS is valid only up to a certain maximum hardness.

Michael Dahms

- posted on March 2, 2006, 2:51 pm

Michael Dahms wrote:

The old chart I am looking at for the correlation tops out at Rockwell C of 59 with an UTS of 329 Ksi or 232 kg/mm2.

There is at least one such copy you can google to pretty quick.

I have probably conducted just as many tensile tests using a C58 hardness specimen as the fellow raising the question.

The old chart I am looking at for the correlation tops out at Rockwell C of 59 with an UTS of 329 Ksi or 232 kg/mm2.

There is at least one such copy you can google to pretty quick.

I have probably conducted just as many tensile tests using a C58 hardness specimen as the fellow raising the question.

--

1) Eat Till SATISFIED, Not STUFFED... Atkins repeated 9 times in the book

2) Exercise: It's Non-Negotiable..... Chapter 22 title, Atkins book

1) Eat Till SATISFIED, Not STUFFED... Atkins repeated 9 times in the book

2) Exercise: It's Non-Negotiable..... Chapter 22 title, Atkins book

Click to see the full signature.

- posted on March 2, 2006, 5:05 pm

Hi everyone,

Thanks for your feedback.

Here is some feedback I got from another forum, if it's possible to find the maximum yield stress by knowing the shear strength, this may allow me to get an estimate without worrying so much about the material (at least I hope). The shear strength is almost always given for dowels in catalogs ...

"israelkk (Aerospace) 1 Mar 06 19:09 If the shear strength is 130,000 psi then the tensile strength is 130000/0.577"5000 psi which is approximately in agreement with the C47 core hardness. At this strength the yield stress is close to the ultimate tensile strength and is is at least 90% of the ultimate tensile strength. This gives ~200000 psi. Safety factor to use depends for what purpose it is used. For aerospace use the factor will be 1.5 to brake."

israelkk (Aerospace) 2 Mar 06 7:35 The theoretical ratio between the ultimate shear strength to the ultimate tensile strength is 0.577. Therefore, if the pin/shaft spec gives the minimum shear strength then you can use this formula to calculate the ultimate tensile strength. However, the relation between the yield stress and the ultimate stress depends on the heat treatment. I estimated the 90% due to the 47RC. Anyway, since the pins are case hardened the outer diameter is much stronger than the core therefore you are on the safe side. On the other hand the pins may be too brittle.

Dowel pins are designed for shear/double shear use and not for bending. For bending I would prefer a custom made and properly heat treated shaft where the properties can be guarantied and verified using tensile test specimens.

CoryPad (Materials) 2 Mar 06 8:08 israelkk provided a good first step, but his information is slightly incorrect.

The 0.577 factor is the relationship between shear yield stress and tensile yield stress according to the von Mises/distortion energy/octahedral shear stress criterion. This criterion has been found to be suitable for nearly all crystalline metals.

The relationship between ultimate shear stress (the value provided by McMaster-Carr) and ultimate tensile stress varies from one material to another, but it is usually near 0.6.

israelkk's 0.9 factor for the ratio between tensile yield stress and ultimate tensile stress is accurate for hardened steels similar to yours. His advice regarding brittleness should be heeded, and so should his advice regarding using a real shaft made for this purpose rather than using a shear pin from a catalog.

Regards,

Cory

Thanks for your feedback.

Here is some feedback I got from another forum, if it's possible to find the maximum yield stress by knowing the shear strength, this may allow me to get an estimate without worrying so much about the material (at least I hope). The shear strength is almost always given for dowels in catalogs ...

"israelkk (Aerospace) 1 Mar 06 19:09 If the shear strength is 130,000 psi then the tensile strength is 130000/0.577"5000 psi which is approximately in agreement with the C47 core hardness. At this strength the yield stress is close to the ultimate tensile strength and is is at least 90% of the ultimate tensile strength. This gives ~200000 psi. Safety factor to use depends for what purpose it is used. For aerospace use the factor will be 1.5 to brake."

israelkk (Aerospace) 2 Mar 06 7:35 The theoretical ratio between the ultimate shear strength to the ultimate tensile strength is 0.577. Therefore, if the pin/shaft spec gives the minimum shear strength then you can use this formula to calculate the ultimate tensile strength. However, the relation between the yield stress and the ultimate stress depends on the heat treatment. I estimated the 90% due to the 47RC. Anyway, since the pins are case hardened the outer diameter is much stronger than the core therefore you are on the safe side. On the other hand the pins may be too brittle.

Dowel pins are designed for shear/double shear use and not for bending. For bending I would prefer a custom made and properly heat treated shaft where the properties can be guarantied and verified using tensile test specimens.

CoryPad (Materials) 2 Mar 06 8:08 israelkk provided a good first step, but his information is slightly incorrect.

The 0.577 factor is the relationship between shear yield stress and tensile yield stress according to the von Mises/distortion energy/octahedral shear stress criterion. This criterion has been found to be suitable for nearly all crystalline metals.

The relationship between ultimate shear stress (the value provided by McMaster-Carr) and ultimate tensile stress varies from one material to another, but it is usually near 0.6.

israelkk's 0.9 factor for the ratio between tensile yield stress and ultimate tensile stress is accurate for hardened steels similar to yours. His advice regarding brittleness should be heeded, and so should his advice regarding using a real shaft made for this purpose rather than using a shear pin from a catalog.

Regards,

Cory

- posted on March 3, 2006, 7:15 am

jbuch wrote:

Thanks! Might be interesting to try a test.

Michael Dahms

Thanks! Might be interesting to try a test.

Michael Dahms

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