Consider the prop (drive) shaft of a RWD car spinning at 5000 RPM. If the car ha
ID: 2993486 • Letter: C
Question
Consider the prop (drive) shaft of a RWD car spinning at 5000 RPM. If the car has 200 Hp and if we neglect stress concentrations, find the required shaft thickness to achieve a factor of safety of 3 for static loading if the shaft is made out a) aluminum and b) carbon fiber. In both cases, the shaft has an outer diameter of 2 inches. Be sure to indicate which specific material you chose (including material properties with citation). Note that the hollow shaft's outer diameter equals its inner diameter plus twice the thickness.
Explanation / Answer
Power = Torque*Ang.velocity
N = 5000 RPM
w = 2*pi*N/60 = 523.34 rad/s
Given P = 200 Hp = 200*746 Watts = 149.2 KW
149.2*10^3 = T*523.34
T = 285.09 N/m
Dout = 2inch = 0.051 m = 51 mm
Dout = Din + 2*t
Shear Stress Is max at Outside cylinder Surface
Shear Stress = T*(Dout/2)/J = 285.09*0.025/J
Shear Stress = 7.13/J------(1)
J = polar moment Inertia = pi/32*(Dout^4-Din^4)
a) Aluminium 6061 T6
Syt = 241 MPa
Sys = 0.55*Syt
For FOS = 3
Shear Stress= Sys/3 =44.18 MPa------(2)
from 1 and 2
7.13/J = 44.18*10^6
J = 1.62*10^-7
Pi/32*(0.051^4-Din^4) = 1.62*10^-7
0.051^4 -Din^4 = 1.645*10^-6
Din^4 = 5.12*10^-6
Din = 47.57 mm
t = (Dout-Din)/2 = 1.72 mm
b)Carbon fiber
Hexcel IM9
Syt = 6000 MPa
Sys = 0.55*6000 = 3300
Shear stress = 1000 MPa
J = 7.13*10^-9
Pi/32(0.051^4-Din^4 ) = 7.13*10^-9
Din = 50.8 mm
t = 0.1 mm
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.