WORLD FOR CIVIL

Topic 6.3c: Power Transmission - Example 3

Example 3:
We have shown a compound shaft with 500 horsepower being applied at point B, and with power being taken off the shaft - 100 hp at point A, 240 hp at point C, and 160 hp at point D. Shaft AB has a diameter of 1", shaft BC had a diameter of 3", and shaft CD had a diameter of 2". If the shaft is rotating at 1200 rpm, determine the maximum shear stress in each section of the shaft.

Solution:
Step 1. We first determine the amount of horsepower being transmitted through each section of the compound shaft. In section AB, we can see that 100 hp must being transmitted internally through the shaft from B to A. In section BC, with a little thought, we realize that the amount of power being transmitted through BC must be 400 hp - of which 240 hp is taken off at point C, and the remaining 160 hp. continues through shaft CD, and is taken off at point D.

Step 2. Now that we know the horsepower in each section of the shaft, we use the horsepower equation to determine the amount of torque which is being applied to each shaft.
Power _hp = [2 pi T n / 550 ft-lb/s/hp]
Solving for T = (P_hp 550 ft-lb/s/hp) / 2 p n; then putting in the values, we have:
T_AB = 100hp (550 ft-lb/sec/hp)/[2 (3.1416) (1200/60 rev/s)] = 438 ft-lb. = 5256 in-lb.
T_BC = 400hp (550 ft-lb/sec/hp)/[2 (3.1416) (1200/60 rev/s)] = 1752 ft-lb. = 21024 in-lb.
T_CD = 160hp (550 ft-lb/sec/hp)/[2 (3.1416) (1200/60 rev/s)] = 701 ft-lb. = 8412 in-lb.

Step 3. Finally we use the shear stress formula, = T r / J, to find maximum shear stress in each section
_AB = 5256 in-lb * .5 in / [pi(1")⁴ / 32] = 26,770 lb/in².
_BC = 21024 in-lb * 1.5 in / [pi(3")⁴ / 32] = 3966 lb/in².
_CD = 8412 in-lb * 1 in / [pi(2")⁴ / 32] = 5355 lb/in².