Questions regarding maximum yield stress for hardened steel dowels 4

[John2004]

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

http://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

 

[israelkk]

If the shear strength is 130,000 psi then the tensile strength is 130000/0.577=225000 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.

 

[John2004]

Hi Israelkk,

Thanks for your reply.

Can I use the method you gave for finding the Maximum yield stress, regardless of what type of steel is used ? That is, as long as the shear strength is listed, can I always get a fairly close estimate on the Maximum yield stress using the method you gave ?

With a 3/8" OD cantilevered mounted dowel /shaft, my beam design software calculated a bending stress of 67.8 KSI
and a deflection at the end of the dowel of .00649". The dowel is only 1.48" long, so I was a little concerned with the amount of deflection and bending stress. I was worried the dowel might take a permanent set.

They make dowels out of different steels, but I am guessing they are close in yield strength (at least I hope). If someone sells the same part number in several different materials, and I never know which material I'm getting, this could cause a problem if I get a material that will not handle the load.

I guess the question is, will standard hardened dowels handle this amount of stress without taking a permanent set, and can I count on the different dowel materials having reasonably close maximum yield stresses ? I did not see anything in ASME B18.8.2 on this matter.

If I can count on all of them having a maximum yield stress of at least 80,000 or 90,000 PSI, I think I will be OK.

I wish I could find a chart that shows how hardening affects yield stress for different steels, but I have not found one yet.

On the other hand, regardless of the material used, the catalogs almost always list the shear strength of dowels, and If I can just use the shear strength to get the maximum yield stress, then I won't need to worry so much about which type of steel is being used.


Thanks again,
John

 

[israelkk]

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]

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

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[zcp]

John,
Can you describe your joint configuration? Are you using the dowel as a pin in a joint (shear) or are you using it as a post with a cantilever load?

ZCP

http://www.phoenix-engineer.com

 

[John2004]

Hi everyone,

Thanks for your replies,

In regards to ZCP's question, I am using the shaft basically as a post with a cantilever load. The shaft /dowel is pressed into a steel support. Then I press two drawn cup needle roller bearings into a housing, then install the housing on the shaft. The housing rotationally oscillates about the longitudinal axis of the stationary shaft.

Since the dowel is a "pull out" dowel and already comes with a tapped end, I just use a screw with a little loc-tite to take up the axial play & keep the housing from "walking off" the shaft axially or lenghtwise. I put delrin washers between the housing and steel support, and between the screw head and the housing, to reduce any friction and/or rubbing noise.

I am also thinking about trying a "Long-Lok" screw with a plastic strip running lengthwise, to see if it will prevent the screw from backing-out due to housing oscillation, without using loc-tite.

That's the basic configuration, and the dowel looks so ideal because it already comes at the right hardness, diameter, length, and with a tapped end. Plus, they are very cost effective & held to ASME B18.8.2 standards. Using this screw method to hold the housing on the shaft is about the only way I had to go, due to space constraints.

Since I am using needle rollers, I have to use a hardened shaft, and I think most bearing shafting is hardened similar to dowels. I think the hardness is very close, if not exacty at what Timken recommended.

What are the problems with the pin being too hard, would it just break instead of taking a set ? I have seen some 1/4" diameter hardened dowels bend alot before breaking. There are no lives at risk or anything here, if the pin breaks it can't hurt anyone. I am just worried about it becoming distorted & taking a set.

Thanks again guys, I really appreciate your feedback.

John

 

[zcp]

John,

I looked over your post quickly and my first thought is you should have a consultant look over your design and provide you the assurance and/or recommendations that you need.

Some additional random thoughts are shoulder screws, what is the load, what is the beam design program and how did you set it up, is this product held to any applicable standards or codes, etc.

Let me know directly if I can help.

ZCP

http://www.phoenix-engineer.com

 

[Philrock]

Since you are loading the pin so heavily, you should check the compressive stresses in the housing that supports the pin.

 

[tlee123]

Don't you have to consider reduction in strength due to the tapped hole?

Is it tapped thru?

By the way I like your idea of using cheap, off the shelf components for this application, considering there is no danger if the pin fails.

 

[John2004]

Hi everyone,

Thanks again for your replies,

In regards to tlee123's questions, the dowel is not tapped through, it's just a standard "pull dowel" with one tapped end. The tap is only about 3/8" deep.

John

 

[Tmoose]

I'd think the needle bearings would probably die at loads way below the dowel would bend

 

[John2004]

Hi everyone,

In response to Tmoose's comment, actually, these full complement drawn cup needle rollers (Timken part # B-66) can handle static loads of 1,700 pounds and dynamic loads of 1,300 pounds, so the dowel / shaft would fail long before the bearings.

After all the responses I have gotten from this and a couple other forums, I tend to think the dowels will handle the stress. I took a look at the stresses again, and it looks like I can get the maximum bending stress on the dowel down to around 47.9 KSI, with a Max. deflection of .00458" at 1.48" from the cantilever support. The Maximum slope will be about 0.228 degrees at 0.995" from support. The reaction force at the cantilever support is 318 pounds and the moment at the support is 248 lb-in.

I got a reply from another forum I listed below, that seems to concur with what CoryPad and Israellk said.

Thanks again for the feedback guys, I appreciate it.

Sincerely,
John


From another forum...

Author: timkom
Posted on: 03/06/06 13:04 PST

"Yes, the hardening process will raise the yield strength when the ultimate strength is raised through heat treatment. The yield strength of alloy steels is usually around 80% of the tensile strength. The 130,000 psi single shear can approximate the ultimate strength by dividing by 0.6. The result is 216,700 psi ultimate. 80% of 216.7 is 173,000 psi. This yield strength value seems to be a reasonably high performance level with the alloys listed by McMaster. A word of caution, your pins are case hardened and the high carbon surface layer does not contribute to the shear strength since the elongation is too low. Also, in your application, the pin may not bend under load but the supporting hole may deform."