Calculating the Factor of Safety

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Discussion Overview

The discussion revolves around calculating the Factor of Safety (FOS) for a bolt subjected to tensile and shear forces. Participants explore the appropriate values for maximum allowable stress and how to apply them in the context of the given stresses.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the direct stress and shear stress for a bolt and proposes using the ultimate tensile stress to determine the FOS.
  • Another participant questions whether the FOS in shear is smaller or greater than in tension, prompting further exploration of the implications.
  • A participant suggests calculating the FOS in shear using the ultimate shear stress and the calculated shear stress, arriving at a different FOS value for shear.
  • There is a discussion about which FOS is more critical based on the calculated values and the potential for failure under increased load.
  • A participant revises their calculations and expresses uncertainty about the high FOS values obtained, indicating a lack of confidence in their understanding.
  • Another participant confirms that the calculated FOS values appear correct based on the provided information and emphasizes that the smallest FOS should be considered as the determining factor.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating the FOS but express differing views on the implications of the results and the significance of the different FOS values for shear and tension. The discussion remains unresolved regarding which FOS is more critical in practical applications.

Contextual Notes

Some calculations depend on assumptions about the material properties and the conditions under which the bolt is used. There are unresolved questions about the implications of the calculated FOS values and their practical significance.

Who May Find This Useful

Individuals interested in mechanical engineering, structural analysis, or materials science may find this discussion relevant, particularly those focused on safety factors in design and analysis of mechanical components.

Brinkley23
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Homework Statement
Ultimate Tensile stress = 500NM/m2 and Ultimate Shear stress = 300MN/m2, determine the FOS
Relevant Equations
Direct stress worked out to be 34.47mPa
Shear stress worked out to be 199mPa
Hi,

I need to work out the FOS for a 16mm diameter bolt with a force of 8KN exerted on it.

I have already worked out:
direct stress = 34.47mPa
shear stress = 199mPa

Information given:
Ultimate Tensile stress = 500NM/m2
Ultimate Shear stress = 300MN/m2

The FOS calculation I have been given to work this out is - Maximum allowable stress / stress

Would I use the Ultimate tensile stress (500) as the maximum allowable stress value? and the direct stress value (34.47) giving me a FOS of:

FOS: 500/34.47 = 14.51

Many Thanks!
 
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Is the factor of safety in shear smaller of greater than in tension (the one you calculated)? What does this tells you?
 
Thanks for replying,

Would I be calculating that by taking the Ultimate shear stress (300N) and the shear stress (199mPa)?

Assuming yes, FOS 300/199 = 1.51

FOS tensile - 14.51
FOS shear - 1.51

It asks me to determine the factor of safety in operation?
 
So you have two FOS, one in shear, one in tension. Based on those values, do you think your bolt has more chances of breaking in shear or tension? If you increase the load on the bolt, which FOS will be going below 1 first? Answering those questions will tell you which one is your crucial FOS.
 
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That makes sense...but I think I made an error with original workings out.

My full revised workings are below:

Tensile stress (direct force/area)

6928 / 2.01 x 10∧-4 = 34467661.7 Pa (34.47 MPa)

Shear stress (shear force/area)

4000 / 2.01 x 10∧-4 = 19900498 Pa (19.90 MPa)

TASK - "Determine the Factor of Safety in Operation, assuming that" -
- Ultimate tensile stress is 500MN/m2
- Ultimate shear stress is 300MN/m2

My answers based on the values above and using the calculation (max allowable stress/stress):

FOS (tensile) = 500/34.47 = 14.5
FOS (shear) = 300/19.9 = 15.08

...they just seem too high to me? but I KNOW NOTHING! haha

Thanks again if you do cast your eyes over this!
 
Can anyone please say if he got this write?
 
According to the values given, those are the FOS for each case. The smallest one would be the determining FOS.
 

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