Geometry: Due to symmetry only one half of the pole is simulated.
yaverage = 0.12 m
Lz = 1 m,
S = 0.08 * Lz m2
Given: Young's modulus of steel E = 207000 N/mm2 Density of steel ρ = 7800 kg/m3 Poisson's ratio ν = 0.3.
Rotational speed ω = 50*2π rad/sec.
Winding mass m = 600 kg;
Task: Determine the mechanical stress distribution in the salient pole.
Solution: The internal centrifugal force acting on the steel pole can be calculated as: Fr(r) = ρ*ω2*r [N/m3]
Winding adds a vertical force which is Fy = mω2*yaverage [N].
Since we simulate only half of the pole the force value is divided by factor of 2.
We consider that Fy is uniformly distributed along the contact surface S.
Results: Mechanical stress distribution in the pole is shown on the picture