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Factor of safety/principal stress

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data
    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


    2. Relevant equations

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


    3. The attempt at a solution

    σx i assume is 5000N which has to be converted into MPa so ive divided the force by the area .... 5000/ ∏x (18/1000)^2 = 4.9MPa ( i dont know if ive 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 doesnt 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?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 28, 2012 #2
    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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