11.1 All the action is in the horizontal plane: 900 N rearward at A and 
900 N forward at B.  Bearing reactions are 600N forward at the back 
bearing and 600N rearward at the front bearing.  The shear diagram 
starts at -600N, jumps up to 300N, and returns to -600N.  The moment 
diagram starts at 0, ramps down to -120Nm, ramps up to 120Nm, then 
ramps down to 0.  The torque diagram is at 75 Nm between the pulleys. 
So the critical location is at either pulley, where M = 120Nm and T = 
75Nm.  There appear to be no stress concentrations.  This is a huge 
relief.  The text front cover gives Sy for 1080 as 380MPA.  Taking that 
plus the desired Ns = 5 into Eq. 11.17, gives d = 26.7mm.  We would 
probably specify a 30mm shaft.

Rotor Frequency

11.10 We'll ignore the weight of the shaft.  The deflection of the end 
of the shaft to a load at the end is FL^3/(3EI).  Remembering that for 
one mass, omega = Sqrt( g / deflection), radian frequency = Sqrt( 3gEI 
/ (WL^3) ).  The weight of the gear is its mass m x g, so this becomes 
Sqrt( 3EI / (mL^3) ).  I is pi d^4/64.  Filling in the variables 
(including m = 100) gives 318.3 Rad/sec, or 50.7 Hz.  This means that 
any unbalance in the gear would be magnified at a speed of 50.7x60 = 
3040 RPM.
You could also solve this by getting the stiffness of the end of the 
shaft as F/deflection = 3EI/L^3.  Then compute omega = Sqrt(k/m), which 
is the same result.  Maybe simpler, given the mass vs weight thing.