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Old 30-03-2012, 11:27   #16
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Re: Engineering Calculation

Calculations aside - few designers would specify aluminum in this application. While the stiffness in bending may be comparable to stainless steel, the durability as a bearing, the toughness under duress, the reliability, etc. are all negatives. Aluminum is used where weight savings is important. Saving ounces here is not important. The proper stainless part is easily obtained or fabricated. My engineering opinion.
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Old 30-03-2012, 11:27   #17
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Re: Engineering Calculation

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Originally Posted by noelex 77 View Post
Can someone calculate the size of stainless steel rod that would have the same yield strength as a 16mm diameter aluminium rod
The aluminium is 6061 T6 (yield strength from Wikipedia 35,000 psi)

My guess is 16 mm diameter rod in aluminium would be similar to about 12mm stainless steel.
Ok, for the sake of learning etc i've gone through a step by step thought process below.

Initial assumptions:
You've been using 6061 T6 and it works.
The length of the rod is no more than 2x the diameter
You're planning on replacing it with AISI316L (preferred) or equivalent.
The rod is supported on both ends.

Firstly, don't use ultimate tensile (UTS) in your calculations as you're not wanting to know when it breaks, but you're wanting it to operate in its elastic region and not go plastic. For this always use the 0.2% offset yield (sometime referred to as "yield strength"). Most PEs will usually use a material factor too, somewhere in the region of 15%, this is to take into account any material variations in the stock that you have. This way you keep the actual stresses within a workable repeatable limit without causing any permenant damage to the material.

So, if your rod length is 2 diameters or less then you don't need to think about bending moments.. only shear strength.

Your allowable stresses for each would work out as:
Aluminum 6061-T6: 275MPa/1.15 = 239MPa
Stainless AISI316L: 355Mpa/1.15 = 308MPa

The allowable shear stress, i.e. trying to snap the pin in a clean break across its section, from Von Mises, would be:
Aluminum 6061-T6: 275MPa/(1.15*root(3)) = 138MPa
Stainless AISI316L: 355Mpa/(1.15*root(3))= 178MPa

Remember that 1 MPa is equivalent to 1 Newton per square mm.

You mentioned that you aluminum rod is 16mm diameter. Remember that the rod's supported on both ends, so you have 2 shear (or break) regions through the section of the rod.

The maximum allowable shear load for the Aluminum would be 2*x-sectional-area*allowable-shear-stress:
2*((16)^2)*PI/4)*138MPa = 17664PI = 55493N = 5657Kg

Knowing that the rod can take 5657Kg in shear, we can then look to see what happens if it's made from AISI316L - so we reverse:

55493 N /178MPa = 311.76 sq.mm.
311.76 / 2 areas = 155.9 sq.mm.
root(155.9*4/PI) = 14mm diameter rod.

If you looked at the problem as a simple scaling of the two yields you'd end up with an answer that didn't take into account that area changes as a square and not as a linear.. in that respect you'd get an answer of 12mm which would be incorrect...

Remember that a 16mm rod has 77% more cross-sectional area than a 12mm rod, eventhough it only has 33.3% more diameter!

If it turns out that the rod you're looking at is longer than 2 diameters i'd be happy to show you the appropriate strength calculation if you give me a sketch of the loading scenario, supports and geometry of the rod etc.

If you want to have a stab as this yourself:
You calculate the shear stress.
You calculate the stress due to bending moments (remember the correct moment of inertia of a rod calc and sectional W-value).
You then combine the two using Von-mises which is Tot-stress=root(((bending-stress)^2)+(3*(bending-stress)^2)))

As long as your total stress comes out below your allowable stress (in this case 308MPa) then you're in good shape..

Hope this helps.. good luck!

Phil
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Old 30-03-2012, 11:55   #18
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Neat thread. I have a 5/8 s.s rod on my roller. It is severely worn were it rides against the frame of the s.s. Anchor frame.
Was not considering aluminum but bronze either silicone or manganese 60/30 not the 90 manganese. Because I am considering what the metals will do in salt water. I have seen bad stuff with SS particularly 305. On a roller 316l seems reasonable. I saw tested manganese bronze blow apart in salt water conditions from stress crack corrosion. It did this in very short order. Where naval bronze doesn't text as strong it seems to do much better in salt water. Just remembering that pure numbers need to also account fir a corrosive environment and the load
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Old 30-03-2012, 12:09   #19
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Re: Engineering Calculation

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Neat thread. I have a 5/8 s.s rod on my roller. It is severely worn were it rides against the frame of the s.s. Anchor frame.
Was not considering aluminum but bronze either silicone or manganese 60/30 not the 90 manganese. Because I am considering what the metals will do in salt water. I have seen bad stuff with SS particularly 305. On a roller 316l seems reasonable. I saw tested manganese bronze blow apart in salt water conditions from stress crack corrosion. It did this in very short order. Where naval bronze doesn't text as strong it seems to do much better in salt water. Just remembering that pure numbers need to also account fir a corrosive environment and the load
You usually account for the corrosive environment through proper cathodic protection design. What most people forget is that Stainless Steel isn't a magic bullet - if it's in electrical contact with the wrong materials then it will act as the anode an degrade rapidly in salt-water.
When placing items subsea, the whole cathodic protection system of a design has to be considered, especially where Stainless is being used as it can degrade the protection system faster than initially calculated (compared to carbon steel with a proper coating system and anode) and end up degrading itself.
A lot of the 'naval bronze' that's used is actually JM7-15 or equivalent which is actually an oil-impregnated material which helps it self-lubricate in water but also stop some of the component metal leeching that can take place as part of the electrolytic action.

Material strength calculations rarely take into account degradation apart from in the pressure vessel type scenarios where the material's yield reduces with temperature - it's always assumed that the material will be protected properly as part of the design. Stress-cracking is usually avoided by avoiding high-stress regions, or generally by not permitting the Utilization Factor of the material to go over 80% of yield. In addition, low hardness materials are preferred to that the H2S which occurs as part of the foundry process can diffuse from the surface of the material easily rather than collect in micro-annuli in the granular structure of the steel and create phenomenal point-pressures (and therefore stresses) in the material - this leads to pre-stressing and therefore failure with repetitive loading..

There are standards for the dynamic factors which you can apply to load-scenarios based upon wave-height (Hs), so you can end up with 3-5times more load than in a static environment just by adding a couple of meters to the wave-height. Though that comes down to the load scenario and initial assumptions which must be correct before you start an analysis.
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Old 30-03-2012, 16:07   #20
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Re: Engineering Calculation

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Originally Posted by daddle View Post
Calculations aside - few designers would specify aluminum in this application. While the stiffness in bending may be comparable to stainless steel, the durability as a bearing, the toughness under duress, the reliability, etc. are all negatives. Aluminum is used where weight savings is important. Saving ounces here is not important. The proper stainless part is easily obtained or fabricated. My engineering opinion.
Yeah.... it just doesnt sit well in the gut does it?.... but easy to keep an eye on I guess. With both ends closely fixed they are in shear... permanent deformation may occur very close to the shear stress.... as there will be very little bending able to occur. It's a big pin though ... about .5 i nch right?
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Old 30-03-2012, 16:37   #21
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Re: Engineering Calculation

The one thing that I think has been forgotten is durability. Sand/mud will find its way between the shaft and the roller and being aluminum it will be a mess in no time.
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Old 30-03-2012, 16:37   #22
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Re: Engineering Calculation

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Originally Posted by Cheechako View Post
Yeah.... it just doesnt sit well in the gut does it?.... but easy to keep an eye on I guess. With both ends closely fixed they are in shear... permanent deformation may occur very close to the shear stress.... as there will be very little bending able to occur. It's a big pin though ... about .5 i nch right?
Closer to 5/8"
But I agree. He's got plenty of material just make up a few extra and keep an eye on the roller. If it starts to look sloppy then change out the pin. I'd say just keep it well greased to avoid galling and wear. Heck, my bow roller is plastic and it's holding up really well.
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Old 30-03-2012, 16:49   #23
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Quote:
Originally Posted by Cavalier

You usually account for the corrosive environment through proper cathodic protection design. What most people forget is that Stainless Steel isn't a magic bullet - if it's in electrical contact with the wrong materials then it will act as the anode an degrade rapidly in salt-water.
When placing items subsea, the whole cathodic protection system of a design has to be considered, especially where Stainless is being used as it can degrade the protection system faster than initially calculated (compared to carbon steel with a proper coating system and anode) and end up degrading itself.
A lot of the 'naval bronze' that's used is actually JM7-15 or equivalent which is actually an oil-impregnated material which helps it self-lubricate in water but also stop some of the component metal leeching that can take place as part of the electrolytic action.

Material strength calculations rarely take into account degradation apart from in the pressure vessel type scenarios where the material's yield reduces with temperature - it's always assumed that the material will be protected properly as part of the design. Stress-cracking is usually avoided by avoiding high-stress regions, or generally by not permitting the Utilization Factor of the material to go over 80% of yield. In addition, low hardness materials are preferred to that the H2S which occurs as part of the foundry process can diffuse from the surface of the material easily rather than collect in micro-annuli in the granular structure of the steel and create phenomenal point-pressures (and therefore stresses) in the material - this leads to pre-stressing and therefore failure with repetitive loading..

There are standards for the dynamic factors which you can apply to load-scenarios based upon wave-height (Hs), so you can end up with 3-5times more load than in a static environment just by adding a couple of meters to the wave-height. Though that comes down to the load scenario and initial assumptions which must be correct before you start an analysis.
Just guessing this supports my idea that silicone bronze or x low yield manganese is a good choice. You guys are to smart for me. Aluminum would be last choice.
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Old 30-03-2012, 16:53   #24
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Re: Engineering Calculation

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Just guessing this supports my idea that silicone bronze or x low yield manganese is a good choice. You guys are to smart for me. Aluminum would be last choice.
Wouldn't that be like using bronze thruhulls on an alum. boat.
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