WORLD FOR CIVIL

Topic 6.3b: Power Transmission - Example 2

Example 2:
A car has a hollow drive shaft with a two inch outer diameter and a 1/8 inch wall thickness as shown in Diagram 2. The maximum power transmitted down the shaft is 185 hp at a shaft speed of 4400 rpm. What is the maximum shear stress in the shaft?

Solution:
Step 1. We first use the horsepower equation to determine the amount of torque which is being applied to the shaft. Power _hp = [2 pi T n / 550 ft-lb/s/hp]
Solving for T = (P_hp 550 ft-lb/s/hp) / 2 pi n; then putting in the values, we have:
T = 185 hp (550 ft-lb/s/hp)/[ 2 (3.1416) 4400/60 rev/sec] = 221 ft-lb = 2650 in-lb

Step 2. Now that we have the torque being applied to the shaft, we use the transverse shear stress equation for circular shafts to determine the maximum shear stress in the shaft.
= T r / J ; where
T = 2650 in-lb., r = 1 in.; J = (pi/32)(d_o⁴ - d_i⁴) = (3.1416/32)(2"⁴ - 1.75"⁴) = .65 in⁴.
= T r / J = 2650 in-lb.* 1 in / .65 in⁴ = 4080 lb/in².