- #1
whatsupdoc
- 1
- 0
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?