Answer:
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.
