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shaft's Ideal diameter

21 REPLIES 21
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Message 1 of 22
sibarut
4296 Views, 21 Replies

shaft's Ideal diameter

Hi! Can anyone explain/show what's behind the computations of "ideal diameter"? Thanks!

Problems exist because there are solutions to find out.
21 REPLIES 21
Message 2 of 22
graemev
in reply to: sibarut

I've never heard the term before, but my guess would be: the smallest shaft possible without exceeding the material yield stress.

Message 3 of 22
jingyi.liu
in reply to: sibarut

Here is the explaination from online help.

Ideal Diameter

The graph shows an ideal shape of the shaft is based on the bending moment and calculates a diameter of a circular section which results in a 50MPa stress in every given coordinate of the shaft. This graph is independent on the selected shaft material. It shows the overall relations of the shape based on some constant stress.

 

http://wikihelp.autodesk.com/enu?adskContextId=DA_SHAFTCOMPGRAPHS_DIALOG&language=ENU&release=2014&p...

 

 

 



Jingyi Liu

Inventor Product Manager
Message 4 of 22
sibarut
in reply to: sibarut

Thanks for the replies.

I understand that "ideal diameter" is the design proposal. I need a computation to see how it arrived into such values. Also, what does this mean " 50 Mpa stress in every coordinate"?

I trying to find out what is behind the calculation. This is justify if the results are useful.
Problems exist because there are solutions to find out.
Message 5 of 22
timothy.warnock
in reply to: sibarut

Hi

 

Im also looking for information on this, im not really sure what it means by 50MPa in each section of the shaft, what happens if the bending moment is less than 50MPa. also does the 50 MPa include a safety factor or should I include it on my force?

 

Thanks

Message 6 of 22

The ASME code states maximum shear stress along a shaft should be 56 MPa without keyway and 42 MPa with keyway. The average is ~50 MPa. Maybe this is why Inventor looks at 50 MPa. I find it frustrating that it doesn't use the material properties to size the shaft. As a designer that's what you are looking to do. It would be very useful.

Message 7 of 22
Frederick_Law
in reply to: sibarut

The calculation has nothing to do with shaft material.

It calculate stress from the load.

Message 8 of 22
cadman777
in reply to: Frederick_Law

According to the Machinery Handbook, which I've frequently used:

1. Torsion is based on the material's strength ('S').

2. Deflection of a beam is based on the modulus of elasticity ('E') of the material.

Both calculations appear to be material dependent.

... Chris
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Message 9 of 22
Frederick_Law
in reply to: sibarut

But this only calculate stress under load, without any bending.  Assuming the body is rigid without deformation of course.

That's what I see with this "Ideal Diameter" calculation.

Message 10 of 22
swalton
in reply to: richard.spurr

Shear stress, like bending stress of a shaft is a property of the shaft geometry and is independent of the material properties.  Similarly, stress concentration factors are geometric properties, not material properties.

 

It's not until one is interested in the deflection/yield/fracture of a shaft or beam that the material properties are important.

 

@richard.spurr 

Which ASME code covers shafts?  I'd like to better understand available design resources.

Steve Walton
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Message 11 of 22
cadman777
in reply to: sibarut

@swalton @Frederick_Law

 

I never used the Inventor tools for determining a shaft diameter, so I don't know what an 'ideal diameter' is.

But when I've done shaft calculations in the past, I've done them based on the material and geometry, including shoulders, fillets and round, chamfers, etc. for stress and strain, keyways, pin holes, etc.

 

Who sizes a shaft without doing a deflection calculation?

 

So are you guys saying that a shaft made of steel being geometrically identical to a shaft made of aluminum will be no different in shear, torque, deflection, rotational cycles for life span, etc.?

... Chris
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Message 12 of 22
gcoombridge
in reply to: sibarut

The stress it experiences won't matter whether its made of titanium or cheese. It will depend on the load and geometry only - how the shaft responds to the stress absolutely will. 

Use iLogic Copy? Please consider voting for this long overdue idea (not mine):https://forums.autodesk.com/t5/inventor-ideas/string-replace-for-ilogic-design-copy/idi-p/3821399
Message 13 of 22
cadman777
in reply to: gcoombridge

There ya go!

Nice clarification on the subject.

This conversation reminds me of the difference between ASD vs. LFRD. Seemed to me that the LFRD method was a bureaucratic endeavor to convert common sense engineering into a government operation, esp. considering how the method served the REGULATORS and CODE ENFORCERS.

... Chris
Win 7 Pro 64 bit + IV 2010 Suite
ASUS X79 Deluxe
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Message 14 of 22
Frederick_Law
in reply to: sibarut

Do a simple beam calculation.

Both ends supported, 500lbf at middle.

What is the load on the supports?

Does it matter what shape or material the beam is?

 

Once we know the load, we pick shape and material that can do the job.

 

Usually I'll model the shape.  FEA stress.  Pick material, A36, 4140 etc

If I can't find material, change the shape.

Machined parts are easy, I can make anything.

Building frame require to pick shape I can buy.

Message 15 of 22
richard.spurr
in reply to: sibarut

Inventor's ideal shaft diameter is calculated by finding the shaft diameter that makes the reduced stress = 50 MPa. The reduced stress calculation is here: Shaft Calculation Formulas | Inventor 2019 | Autodesk Knowledge Network. The 50 MPa is an allowable stress and is usually defined in the relevant design code (if designing to a design code). The allowable stress is effectively a property of the material because it will differ depending upon the properties of the material.

 

For example, in the ASME (ANSI) design code, the allowable shear stress is 56 MPa without keyway or 42 MPa with keyway. However, if shafts are purchased under definite physical specifications the allowable shear stress is derived from the materials properties: 30 percent of the elastic limit in tension or 18 percent of the tensile strength - which ever is less. 

 

I think Inventor's ideal shaft diameter can also be approximated using the below formula (sorry its metric):

 

D = 2 * ( ( 2 / (PI()* Tallow * 10^6) ) * ( SQRT ( M * 1.5 )^2 + 0.75 * ( T * 1.5)^2 ) ) ) ^ (1/3)

 

D = shaft diameter in m, Tallow is allowable stress (set to 50) in MPa, M is bending moment at point x along the shaft in Nm, T is torque at point x along the shaft in Nm

 

I'm in the process of designing a shaft. The standard bore into the gearbox is 100mm diameter. The reduced stress at this point is high at 138 MPa. If I increase the shaft to 160mm diameter the cost of the gearbox doubles in cost. I would find the whole process easier if Inventor could utilise a database of materials and corresponding shaft cost and simply ask Inventor to find a shaft material or a range of shaft materials that could take the 138 MPa stress. Or at the very least just have Inventor size the 'ideal diameter shaft' based on the selected materials properties and design code.

 

Message 16 of 22
Frederick_Law
in reply to: sibarut

I would find the whole process easier if Inventor could utilise a database of materials and corresponding shaft cost and simply ask Inventor to find a shaft material or a range of shaft materials that could take the 138 MPa stress.

So when the shaft fail, who got sue?  AutoDesk?

I'm sorry but those decision should be made by the designer or engineer.

 

138 MPa material?

A36 Yield is 250MPa

AL6061 is 276MPa

🙄

Message 17 of 22
richard.spurr
in reply to: sibarut

!!

Message 18 of 22
Frederick_Law
in reply to: sibarut

 

 

And you're designing shaft without a clue on material?
Good thing they're not usually used for shaft.

LOL

Message 19 of 22

I think it is a bit lame to infer that if a software company provided an engineer with the tools to search for materials or carry out a cost comparison / optimum search (very easily) it would some how make them liable for the design - and I'm really surprised to see an Autodesk test engineer liking your post. Of course the design is with the engineer, and the engineer's employer would have the relevant insurance.

 

Most engineers responsible for design with any form of large liability would be well experienced, probably chartered / suitably qualified and would know to be very careful when interpreting the results of a software output - which is only as good as the information it is given.

 

Sizing the shaft based on the stress is only part of the design process anyway

Message 20 of 22
Frederick_Law
in reply to: sibarut

Software is a tool which need to be used properly.

Too many user believe sim will magically solve all their problems.
If the user have no idea what the sim results are, they need training.

Both on the software and the problem they want to solve.

 

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