Factor of safety/principal stress

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In summary, the client has informed of an additional thrust force of 5000N on a manufactured shaft with a diameter of 36mm. The previous factor of safety was found to be 2, with a shear stress of 250MPa. Using the principal stress formula, the estimated peak stress was found to be 252MPa. However, it is recommended to use the maximum distortion energy theory for a better approximation of the Von Mises stresses present. The axial stress on the shaft can be calculated by dividing the force by the area, and the material of the shaft also needs to be considered.
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Homework Statement


After the shaft is manufactured you are informed by the client that the shaft is subjected to an additional thrust force of 5000N. Provide a revised factor of safety for the client.

Found the previous factor of safety to equal 2
Shear stress τ = 250MPa
Diameter = 36mm


Homework Equations



Principal stress formula: σ1,σ2= 0.5(σy+σx) +/- 0.5 x ( sqrt{(σy - σx)^2 + 4τ^2} )


The Attempt at a Solution



σx i assume is 5000N which has to be converted into MPa so I've divided the force by the area ... 5000/ ∏x (18/1000)^2 = 4.9MPa ( i don't know if I've done that conversion correctly for starters)

τxy I'm assuming moves in a clockwise direction so its positive = 250MPa

gonna assume σy = 0 as it doesn't give us any value for that.

Already its starting to look a bit strange, i'd of assumed σx would be larger than τxy but i plugged it into the formula anyway to see what i'd get 252Mpa for σ1

I got to this point and got stuck, the values are most likely wrong but i think the steps are along the right lines, but how do you get a revised factor of safety from this?
 
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Found the previous factor of safety to equal 2
Shear stress tau_xy = 250MPa

I assume some calcs were done to get these values...

Principal stress formula: σ1,σ2= 0.5(σy+σx) +/- 0.5 x ( sqrt{(σy - σx)^2 + 4τ^2} )
This value is an estimate for peak stresses and is a more conservative approach.
From my experience it is better to use the maximum distortion energy theory:
σ_1,σ_2 = ((σ_x-σ_y)⁄2)±sqrt(((σ_x-σ_y)⁄2)^2+τ_xy^2 )
this gives you a better approximation of the Von Mises stresses present.

You don't have to convert the force you simply calculate the stress associated with this force on a specific area. You're calculation is however correct.

σ_x=F/A. This however is just the addisional stress. Is this the only axial stress working in on the shaft?

What material is the shaft made of? I am in quite a hurry, having a meeting in 5 minutes. Will help you further tomorrow, but you have a fairly good idea...
 

What is the factor of safety?

The factor of safety (FoS) is a measure of how much stronger a structure or material is compared to the maximum load it is expected to withstand. It is calculated by dividing the ultimate strength of the material by the maximum expected load. A higher factor of safety indicates a safer and more reliable structure.

Why is the factor of safety important?

The factor of safety is important because it ensures the safety and reliability of a structure. It provides a margin of safety against unexpected loads, variations in material properties, and other factors that could potentially cause failure. A higher factor of safety also allows for a longer lifespan of the structure.

What is the relationship between factor of safety and principal stress?

The principal stress is the maximum stress that a material experiences in a given loading situation. The factor of safety is directly related to the principal stress, as a higher factor of safety means a lower principal stress. This is because a higher factor of safety indicates a stronger material that can withstand higher stresses without failing.

How is the factor of safety determined?

The factor of safety is determined through various methods, including analytical calculations, computer simulations, and physical testing. It is typically calculated by dividing the ultimate strength of the material by the maximum expected load, but different industries may have specific guidelines and standards for determining the factor of safety.

What is a safe factor of safety?

The appropriate factor of safety depends on the application and industry. In general, a factor of safety of 2 or higher is considered safe, but it may vary depending on the consequences of failure and the level of risk that can be tolerated. Some industries, such as aerospace and medical, may require a factor of safety of 5 or more for critical components.

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