# Safety Factor of Shaft :Machine Design

1. Sep 5, 2009

### speedy21cvb

1. The problem statement, all variables and given/known data
An 8” long, 2.5” diameter, solid bar of Class 40 cast iron is subjected to both a torque of 75,000 lbf-in and an axial compressive load of 165,000 lbf. If these two loads can increase at a constant ratio, calculate the safety factor.

2. Relevant equations
$$\sigma$$x=F/A
$$\tau$$xy=Tr/J
nf= Allowed/Actual
For class 40 cast iron:
Sut=42.5 kpsi
Suc=140 kpsi

3. The attempt at a solution
$$\sigma$$x=165000/4.91=33.6 kpsi
$$\tau$$xy=75000*1.25/($$\pi$$*2.54/32=24.5 kpsi

Take the higher stress of 33.6 kpsi.
nf=140kpsi/33.6kpsi= 4.1667

I did it this way, however afterwards I thought about the principal stresses. Would I need to solve for the principal stresses present in the shaft under this loading? We used mohrs circle in class, does this apply? This is my first post, I'm finally going to stop being proud and ask for help :) Please let me know if my formatting is incorrect.

-Chris Van Buren, Mechanical Engineer Student, Georgia Tech

2. Sep 5, 2009

### nvn

speedy21cvb wrote: "Would I need to solve for the principal stresses present in the shaft under this loading? Does this apply?"

speedy21cvb: That sounds like a good plan.