Ski pole three point bending-calculation of load at yield

[pd2905]

Homework Statement

A ski pole is bent(plastic deformation) in three point bending by applying a load 2F perpendicularly to the ski pole. The forces at the end points are of magnitude F. If the pole yields due to plastic deformation at the mid point longitudinal span of the ski pole which is 90 cm long and which is a hollow tube with inner radius Ri and outer radius Ro give the formula to calculate the minimum load for which the pole will fail, given that the pole is made of an isotropic 5086 Al alloy with E=71GPa modulus of elasticity and that the yield stress of the alloy is Y=230 MPa. We assume a static load.

Homework Equations

Stress at tip=Yield stress=Moment of cross-sectional area*Distance from the centroid/Second moment

I calculate moment of cross-section by M=F*length/2

I would like to know if any other more suitable yield criterion can be used
such as Von mises or Tresca. I think Al alloys are considered ductile materials.

So I get M=Yield stress/Distance from the centroid*Second moment
2F=4M/length=4.

The Attempt at a Solution

Yield stress=230Mpa
Distance from centroid=9mm
Ri=16.5mm
Ro=18mm
length=900mm
2F==4*230/(900)/9*(3.14/4)*(18^4-16.5^4)=2751N is the load to necessary bend the pole.
It seems kind of ridiculous although I have never bent a ski pole.

 

[nvn]

Nice work, pd2905. Your answer is correct. Al 5086 is a ductile material. For ductile materials, von Mises theory is preferred over Tresca, and is more accurate. For ductile materials, you generally use von Mises stress, but in your particular case, von Mises would reduce to give you the same answer you already obtained.