Answer:
a) 34 mm
b) 39 mm
c) 93.16%
Explanation:
power transmitted P = 28 kW 28000 W
angular speed N = 440 rpm
angular speed in rad/s Ω = 2[tex]\pi[/tex]N/60
Ω = (2 x 3.142 x 440)/60 = 46.08 rad/s
allowable shear stress τ = 80 MPa = 80 x [tex]10^{6}[/tex] Pa
torque T = P/Ω = 28000/46.08 = 607.64 N-m
a) for the minimum diameter of a solid shaft, we use the equation
τ[tex]d^{3}[/tex]= [tex]\frac{16T}{ \pi}[/tex]
80 x [tex]10^{6}[/tex] x [tex]d^{3}[/tex] = [tex]\frac{16*607.64}{3.142}[/tex] = 3094.28
[tex]d^{3}[/tex] = 3094.28/(80 x [tex]10^{6}[/tex]) = 0.0000386785
d = [tex]\sqrt[3]{0.0000386785}[/tex] ≅ 0.034 m = 34 mm
b) For a hollow shaft with outside diameter D = 40 mm = 0.04 m
we use the equation,
T = [tex]\frac{16}{\pi }[/tex] x τ x [tex]\frac{D^{4} - d^{4}}{D^{4} }[/tex]
where d is the internal diameter of the pipe
607.64 = [tex]\frac{16}{3.142}[/tex] x 80 x [tex]10^{6}[/tex] x [tex]\frac{0.04^{4} - d^{4}}{0.04^{4} }[/tex]
3.82 x [tex]10^{-12}[/tex] = [tex]0.04^{4} - d^{4}[/tex]
[tex]d^{4}[/tex] = [tex]\sqrt[4]{2.56*10^{-6} }[/tex]
d = 0.039 m = 39 mm
c) we assume weight is proportional to cross-sectional area
for solid shaft,
area = [tex]\pi r^{2}[/tex]
r = diameter/2 = 34/2 = 17 mm
area = 3.142 x [tex]17^{2}[/tex] = 907.92 mm^2
for hollow shaft, radius is also gotten as before
external area = [tex]\pi r^{2}[/tex] = 3.142 x [tex]20^{2}[/tex] = 1256.64 mm^2
internal diameter = [tex]\pi r^{2}[/tex] = 3.142 x [tex]19.5^{2}[/tex] = 1194.59 mm^2
true area of hollow shaft = external area minus internal area
area = 1256.64 - 1194.59 = 62.05 mm^2
material weight saved is proportional to 907.92 - 62.05 = 845.87 mm^2
percentage weight saved is proportional to 845.87/907.92 x 100%
= 93.15%
