Question

The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for a solid steel shaft if the allowable shear stress in the shaft is not to exceed 6100 psi.

Answer 1

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

$w=\frac{2\pi N}{60}$

Where

N = rotational speed of the driveshaft = 2900 rpm

$w=\frac{2\pi *2900}{60} =303.69rad/s$

The torque in the driveshaft is equal to:

$\tau=\frac{P}{w}$

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

$\tau=\frac{73700}{303.69} =242.68lb*ft$

The minimum diameter is equal to:

$d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}$

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

$d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in$