Calculating Minimum Shaft Diameter: Factor of Safety vs Maximum Shear Stress?

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The discussion centers on calculating the minimum shaft diameter for a maximum shear stress of 80 MPa with a factor of safety (FoS) of 1.5. Participants clarify that the correct approach is to divide the maximum shear stress by the FoS, resulting in a working stress of approximately 53.33 MPa, rather than multiplying the maximum shear stress by the FoS. This method ensures that the designed shaft can safely withstand the applied loads without exceeding material limits. Additionally, the conversation highlights the importance of accurately applying safety factors in engineering design to prevent structural failures.

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  • Understanding of shear stress and its significance in mechanical design.
  • Familiarity with the concept of factor of safety (FoS) in engineering.
  • Basic knowledge of material strength and failure criteria.
  • Experience with shaft design principles and calculations.
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  • Study the principles of mechanical design, focusing on shear stress calculations.
  • Learn about the application of factor of safety in various engineering contexts.
  • Explore material properties and how they influence design decisions.
  • Investigate the effects of combined loading on shafts, including torsion and shear.
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jmiersy
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im working on a question that requires me to calculate the minimum shaft diameter for a maximum shear stress. the value is 80mpa. The factor of safety is 1.5, now I am aware that i can't simply increase the shaft diameter by 1.5 to do this correctly. Should I calculate a suitable diameter for 1.5x the 80mpa that is required?

Thanks, I hope its ok to post this here, i don't need guidance on how to work the diameter at all just where to apply the FOS.
 
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Should I calculate a suitable diameter for 1.5x the 80mpa that is required?

Not quite.

The factor of safety is applied to the stress, but we reduce the working or maximum stress by this factor, ie use a stress of 80/1.5 (80 divided by 1.5).

This is important.
We do not design for a higher stress (1.5 x 80) than we know the material will support - we design for a lower stress and the factor of safety is the difference.

Remember also that for a shaft, the max shear may arise as a result of combined torsion and direct shear.
 
Studiot said:
Not quite.

The factor of safety is applied to the stress, but we reduce the working or maximum stress by this factor, ie use a stress of 80/1.5 (80 divided by 1.5).

This is important.
We do not design for a higher stress (1.5 x 80) than we know the material will support - we design for a lower stress and the factor of safety is the difference.

This is a procedure thing, the company I used to work at used to do exactly this, we designed for incrased load rather than applying a fos to the working load. I assume he OP meant design for 1.5 x load but said stress, otherwise it doesn't really make sense.

Working pressure x 1.5 = testing pressure
Test pressure x 1.5 = design pressure.

So for a 10,000psi differential we used to design to hit yield at 22,500 psi.

Short of nomenclature, this achieves exactly the same thing as designing to 10,000 and applying a factor of safety of 2.25. You may get some slight differences in results between the two methods but they will be close (assuming you are designing to yield strength and not ultimate tensile). As long as it's incredibly clear what method has been used (so OP, record exactly what you are doing) it's fine.
 
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Chris,

Modern structural design does not work that way.
The complete story is quite complicated so I did not think it appropriate to delve into partial safety factors and limit state design.

To cut a long story short, we apply different safety factors to different aspects of the design, based upon statistical analyses of the risk of failure. All of these factors are combined in a global or overall factor for the design. And yes they are generally multiplicative, some less than one for material strengths, and qualities, some more than one for loads, impacts and so on.
 
Studiot said:
Chris,

Modern structural design does not work that way.
The complete story is quite complicated so I did not think it appropriate to delve into partial safety factors and limit state design.

To cut a long story short, we apply different safety factors to different aspects of the design, based upon statistical analyses of the risk of failure. All of these factors are combined in a global or overall factor for the design. And yes they are generally multiplicative, some less than one for material strengths, and qualities, some more than one for loads, impacts and so on.

So what? The methods achieve the same goal (it's acutally the same calculation, just approached slightly differently). You can't say 'it's not done that way' because in some places it is. I'm not saying your method is wrong, just that the statement 'it's not done that way' isn't true. It may be that 'it shouldn't be done that way', but in reality is.

Designing a shaft so that max stress < yield. Is not a complicated problem, infact this is textbook difficulty.
FoS = Strength / Load
For a simple problem such as this, whether you incraese the 'effective load', or decrease the 'effective strength' is irrelevent.

PLEASE NOTE: OP the above equation is what you are using. So do you see why saying FoS*strength (1.5*80 as you did in the first post) is wrong?
 
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So what?

Is this argument or rational discussion?

Do you know anything about partial safety factors?
They are embodied in most national codes of engineering practice and in many cases codified in national laws and regulations.
Even for those that are not so codified, any designer that flouts them lays his or herself wide open to civil action if things go wrong. It is always a useful defence to prove that the codes have been followed, even when failure occurs, as has happened on a significant number of occasions.
 
Studiot said:
Is this argument or rational discussion?

Do you know anything about partial safety factors?

How is this relevant to the OP's question? We aren't asked to consider reliability levels, or potential failure modes or anything even remotely complicated. This is like launching into a discussion on fracture mechanics or fatigue when you were asked the very simple question 'will this yield'?

It's a simple problem, and appears to be one designed to test basic theory on factor of safety.

To be honest I don't really know as much as I should on partial safety factors as it seems to be more a civil/structural thing. Also because I have never needed to apply them day to day in he jobs I've done I've not had the inclination to learn more. Basic safety factor calcualtions have sufficed for all standards I've worked to.

I take it you are a civil/structural engineer?


This hasn't really gone as smoothly as I would have liked as it's kind of hijacked the thread, it wasn't my intention to antagonise.
 
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How is this relevant to the OP's question?

jmiersy may just be starting to consider safety but will meet the conventional approach if his(her) studies continue in the engineering direction.


It's a simple problem, and appears to be one designed to test basic theory on factor of safety
I thought my initial response recognized that.
 
ok well i suppose i need to be more specific to the factors of this question. It is a first year engineering question that i have been studying, i will post the details of the question for you guys to comment on. thanks for everyones time.
 
  • #11
The main thing is, do you now understand what Chris and I have both said?


PLEASE NOTE: OP the above equation is what you are using. So do you see why saying FoS*strength (1.5*80 as you did in the first post) is wrong?
 
  • #12
"calculate a suitable diameter to the nearest millimetre for the secondary drive shaft for a FOS of 1.5, if the shear stress in the shaft is not to exceed 80MPa when the shaft is under maximum load. (Hint – be careful about the point in your calculations where you apply the FOS. It isn’t just a matter of increasing the final diameter of the shaft by 50%.)"

to be honest i don't actually understand why 1.5x80mpa is incorrect. would that not result in designing a shaft that will withstand 1.5 times the maximum shear stress?


regards.
 
  • #13
or am i thinking about this in the wrong way?
 
  • #14
or am i thinking about this in the wrong way?

Yes you are, so here goes.

You tell me that the maximum shear stress is 80 MPa. That is the shear strength of the material is 80 MPa.

Then you suggest you design for 80 x 1.5 = 120MPa.

How would any shaft of your material support this?

Surely if you actually applied 120 MPa to this material you would break it?

Where is the logic in this approach.

Sorry to be so blunt.
 
  • #15
ok i think i now understand. Your bluntness is appreciated haha. I worked the question with both values and now see how designing to 80/1.5 delivers a greater diameter. I've attached the working if you could look over it and see if i have done the correct thing? thanks for your time once again. the actual question is in a link a few posts above.

Regards
Jonathan
 

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  • #16
Yes I will go along with this.

I would recommend explicitly showing the FOS calculation however.
If you do so but cock up the arithmetic you will only loose the arithmetic marks, but if you don't show the step and cock up the arithmetic you will losse the rest of the marks for the question.

For the future:
As soon as gears A and B on the secondary shaft are engaged in transmission they will apply torques to the shaft and the torque will vary along the shaft. So you would have to consider each section individually to find the max.
 
  • #17
Thanks! And yes I will show the working more clearly. And I also understand how the torque will vary along the shaft. Thanks again for your time.
 

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