The image of this electric motor with its shaft is missing. I have attached it.
Answer:
Minimum Diameter = 40.8 mm
Explanation:
From the question,
Torque(T) = 800 N·m
φ = 1.45° = 1.45π/180 = 0.0253 rads
τ = 60 MPa = 60 x 10^(6) Pa
G = 77.2 GPa = 77.2 x 10^(9) Pa
From the image attached, length of shaft (L) = 0.4 + 0.6 + 0.3 = 1.3m
Now, the polar moment of inertia is calculated from,
J = πc⁴/2 where c is the radius of shaft
To solve this, we will find the diameter based on the angle of twist and also based on the shear stress and we will choose the smaller one.
Based on angle of twist;
Formula for angle twist is given as;
φ = TL/GJ
Now J = πc⁴/2
Thus, φ = 2TL/G(πc⁴)
c⁴ = 2TL/φGπ
c⁴ = [2 x 800 x 1.3]/(0.0253) x 77.2 x 10^(9) x π)
c⁴ = 0.00000033898
c = ∜0.00000033898
c = 0.024m
Now based on shear stress;
Formula for shear stress is given as;
τ = Tc/J
Putting πc⁴/2 for J, we have;
τ = 2T/πc³
c = ∛(2T/πτ)
c = ∛(2 x 800)/(π x 60 x 10^(6))
c = ∛0.00000848826
c = 0.0204m
So, comparing the two values of radius gotten, the one based on the shear stress is bigger and it's c = 0.0204m
Diameter = 2 x radius = 2 x 0.0204m = 0.0408m which is 40.8 mm