Previous Topic Next Topic Up One Level Table of Contents

Solid Shafts In Bending And Torsion

A 30-ft (9.1-m) long solid shaft weighing 150 lb/ft (223.2 kg/m) is fitted with a pulley and a gear as shown in Fig. 2. The gear delivers 100 hp (74.6 kW) to the shaft while driving the shaft at 500 r/min. Determine the required diameter of the shaft if the allowable stress is 10,000 lb/in 2 (68,947.6 kPa).

Calculation Procedure:

1. Compute the pulley and gear concentrated loads

Using the method of the previous calculation procedure, we get T = 63,000 hp/R = 63,000(100)/500 = 12,600 lb·in (1423.6 N·m). Assuming that the maximum tension of the tight side of the belt is twice the tension of the slack side, we see the maximum belt load is Rp = 3 T/r = 3(12,600)/24 = 1575 lb (7005.9 N). Hence, the total pulley concentrated load = belt load + pulley weight = 1575 + 750 = 2325 lb (10,342.1 N).

The gear concentrated load is found from Fg - T/r, where the torque is the same as computed for the pulley, or Fg - 12,600/9 = 1400 lb (6227.5 N). Hence, the total gear concentrated load is 1400 + 75 = 1475 lb (6561.1 N).

Draw a sketch of the shaft showing the two concentrated loads in position (Fig. 2).

2. Compute the end reactions of the shaft

Take moments about RR to determine LR, using the method of the previous calculation procedures. Thus, LR(30) - 2825(25) - 1475(8) - 150(30)(15) = 0; LR = 4580 lb (20,372.9 N). Taking moments about LR to determine RR yields RR(30) - 1475(22) - 2825(5) - 150(30)(15) = 0; RR = 3720 lb (16,547.4 N). Check by taking the sum of the upward forces: 4580 + 3720 = 8300 lb (36,920.2 N) = sum of the downward forces, or 2825 + 1475 + 4500 = 8300 lb (36,920.2 N).

3. Compute the vertical shear acting on the shaft

Using the method of the previous calculation procedures, we find VLR = 4580 lb (20,372.9 N); V5L = 4580 - 5(150) = 3830 lb (17,036.7 N); V5R = 3830 - 2825 - 1505 lb (6694.6 N); V22L = 1505 - 17(150) = -1045 lb (-4648.4 N); V22R = -1045 - 1475 = -2520 lb (-11,209.5 N); V30L = -2520 - 8(150) = -3720 lb (-16,547.4 N); V30R = -3720 + 3720 = 0.

4. Find the maximum bending moment on the shaft

Draw the shear diagram shown in Fig. 2. Determine the point of zero shear by scaling it from the shear diagram or setting up an equation thus: positive shear - x(150 lb/ft) = 0, where the positive shear is the last recorded plus value, V5R in this shaft, and x = distance from V5R where the shear is zero. Substituting values gives 1505 - 150 x = 0; x = 10.03 ft (3.1 m). Then Mm = 4580(15.08) - 2825(10.03) - (150)(5 + 10.03)[(5 + 10.03)/2] = 28,575 lb (127,108.3 N).

5. Determine the required shaft diameter

Use the method of maximum shear theory to size the shaft. Determine the equivalent torque Te from Te = ( M2m + T2)0.5, where Mm is the maximum bending moment, lb·ft, acting on the shaft and T is the maximum torque acting on the shaft. For this shaft, Te = [28,575 2 + (12,600/12) 2]0.5 = 28,600 lb·ft (38,776.4 N·m), where the torque in pound-inches is divided by 12 to convert it to pound-feet. To convert Te to Te' lb·in, multiply by 12.

Once the equivalent torque is known, the shaft diameter d in is computed from d = 1.72( Te',/s)1/3, where s = allowable stress in the shaft. For this shaft, d = 1.72(28,500)(12)/(10,000) 1/3 = 5.59 in (14.2 cm). Use a 6.0-in (15.2-cm) diameter shaft.

Related Calculations: Use this procedure for any solid shaft of uniform cross section made of metal--steel, aluminum, bronze, brass, etc. The equation used in step 4 to determine the location of zero shear is based on a strength-of-materials principle: When zero shear occurs between two concentrated loads, find its location by dividing the last positive shear by the uniform load. If desired, the maximum principal stress theory can be used to combine the bending and torsional stresses in a shaft. The results obtained approximate those of the maximum shear theory.

Previous Topic Next Topic Up One Level Table of Contents