Design the shaft shown in Figure below It is to be machined

Design the shaft shown in Figure below. It is to be machined from AISI 1010 CD steel (yield strength of 43.5 ksi and ultimate tensile strength of 50 ksi). The shaft is part of the drive for a machine system transmitting 1440 in-lb torque (TE). Gear D receives torque (TE) from gear E. Gear D delivers the power to gear A. The shaft rotates at 600 rpm. Pitch radius of gears are shown in the figure. Pressure angle = ? = 20°. Determine the shaft size at the critical point(ignore radial gear loads).

Solution

The torque of the shaft is related to the size as,

T = tau * J / R

The ratio of gear radius is proportional to the angular motion of the shaft,

N2 / N1 = R2 / R1

N2 = N1 * R2 / R1 = 600 * (3/2) = 900 rpm

The power transmitted from the shaft 1 with input torque of Te,

Pe = Te * 2 * pi * N

Pe = 1440 * 2 * 3.14 * 900/60 = 13.564 *10 ^3 lb-ft / s

The pressure angle (phi) 20 deg reduces the power transmission to gear D by a factor of cos (phi),

Pt,D = Pt,E * cos phi

Pt,D = 13.564 * 10^3 * cos 20 = 12.747 * 10^3 lb-ft/s

Now the torque on the shaft holding gear D is given by,

Td = Pt,D / 2*pi*N

Td = 12.747 * 10^3 / (2*pi*600) = 202.97 lb - ft

The torque from the gear D produces a shear stress on the shaft,

tau = T * R / J

The allowable shear stress is obtained from the max shear stress theory,

tau, allowable = Sy / 2 where Sy = 43.5 ksi or 6.264 * 10^6 kip / ft^2

J = pi * R^4 / 2

Rearranging for \'R\' in the shear stress - torque relation,

R^3 = 2*T / pi * tau

R = 0.0401 ft