Calculating the Factor of Safety

[Brinkley23]

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!

 

[jack action]

Is the factor of safety in shear smaller of greater than in tension (the one you calculated)? What does this tells you?

 

[Brinkley23]

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?

 

[jack action]

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.

 

[Brinkley23]

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!

 

[oyster21]

Can anyone please say if he got this write?

 

[jack action]

According to the values given, those are the FOS for each case. The smallest one would be the determining FOS.