Factor of safety/principal stress

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SUMMARY

The discussion focuses on calculating the revised factor of safety for a shaft subjected to an additional thrust force of 5000N. The previous factor of safety was determined to be 2, with a shear stress of 250MPa and a diameter of 36mm. The principal stress formula was applied, leading to an estimated principal stress of 252MPa for σ1. The maximum distortion energy theory was recommended for a more accurate approximation of Von Mises stresses.

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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...
 

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