A solid circular shaft has a uniform diameter of 5 cm and is 4 m long. At its midpoint 65 hp is delivered to the shaft by means of a belt passing over a pulley. This power is used to drive two machines, one at the left end of the shaft consuming 25 hp and one at the right end consuming the remaining 40 hp. Determine:

a) The maximum shearing stress in the shaft.
b) The relative angle of twist between the two extreme ends of the shaft. The shaft turns at 200 rpm and the material is steel for which G = 80 GPa.

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

J = π(0.05)⁴/32 = 6.136 x 10^(-7) m⁴

ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

Similarly;

θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

θ = θ2 – θ1

θ = 0.0591 - 0.0363

θ = 0.0228 rad.