Answer:
The minimum diameter is 1.344 in
Explanation:
The angular speed of the driveshaft is equal to:
[tex]w=\frac{2\pi N}{60}[/tex]
Where
N = rotational speed of the driveshaft = 2900 rpm
[tex]w=\frac{2\pi *2900}{60} =303.69rad/s[/tex]
The torque in the driveshaft is equal to:
[tex]\tau=\frac{P}{w}[/tex]
Where
P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s
[tex]\tau=\frac{73700}{303.69} =242.68lb*ft[/tex]
The minimum diameter is equal to:
[tex]d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}[/tex]
Where
T = shear stress = 6100 psi
τ = 242.68 lb*ft = 2912.16 lb*in
[tex]d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in[/tex]
