Answer:
External diameter = 158.15 mm mm
Internal diameter = 59.31 mm
Explanation:
We are given;
Diameter ratio; d_i = ⅜d_o
Where d_i is internal diameter and d_o is external diameter
Power;P = 375 KW = 375000 W
Rotational speed;N = 100 rpm
Max torque is 20% greater than mean torque; T_max = 1.2T_avg
Shear stress;τ = 60 N/mm²
Length; L = 4m = 4000 mm
Angle of twist; θ = 2° = 2π/180 radians
Modulus of rigidity;G = 0.85 X 10^(5) N/mm²
Formula for the power transmitted by the shaft is;
P = 2πNT_avg/60
Plugging in the relevant values, we have ;
375000 = 2π × 100T_avg/60
T_avg = (375000 × 60)/(2π × 100) = 35809.862 N.m = 35809862 N.mm
Since T_max = 1.20T_avg
Thus, T_max = 1.20(35809862) = 42971834.4 N.mm
Checking for strength, we'll use;
τ = Tr/J_p
Or since r = d/2
It can be written as;
τ = T(d_o)/2J_p - - - (1)
Where T is T_max
But Polar moment of inertia of hollow shaft is;
J_p = [π(d_o)⁴ - π(d_i)⁴]/32
Now, we are told that d_i = ⅜d_o
Thus;
J_p = [π(d_o)⁴ - π(⅜d_o)⁴]/32
J_p = (π/32) × d_o⁴(1 - 3⁴/8⁴)
J_p = 0.0926 d_o⁴
Plugging this for J_p in eq 1,we have;
τ = T(d_o)/2(0.0926d_o⁴)
Making d_o the subject gives;
d_o³ = T/(2 × 0.0926τ)
Plugging in the relevant values to give;
d_o³ = 42971834.4/(2 × 0.0926 × 60)
d_o³ = 3867155.7235421166
d_o = ∛3867155.7235421166
d_o = 156.96 mm
Thus, d_i = ⅜ × 156.96 = 58.86 mm
Checking for stiffness, we'll use;
T/J_p = Gθ/L
Again T is T_max
Plugging in the relevant values, we have;
42971834.4/0.0926 d_o⁴ = (0.85 × 10^(5) × 2π/180)/4000
464058686.825054/d_o⁴ = 0.7417649321
d_o⁴ = 464058686.825054/0.7417649321
d_o⁴ = 625614216.5028806
d_o = ∜625614216.5028806
d_o = 158.15 mm
d_i = ⅜ × 158.15 = 59.31 mm
So we will pick the highest values.
Thus;
d_o = 158.15 mm
d_i = 59.31 mm
