A 2-in (5.1-cm) diameter steel shaft is 6 ft (1.8 m) long between bearing centers and turns at 500 r/min. The shaft carries a 600-lb (2668.9-N) concentrated gear load 3 ft (0.9 m) from the left-hand center. Determine the deflection of the shaft if the modulus of elasticity E of the steel is 30 X 10 6 lb/in 2 (206.8 X 10 9 Pa). What would the shaft deflection be if the load were 2 ft (0.6 m) for the left-hand bearing? The shaft weighs 10 lb/ft (14.9 kg/m).
Calculation Procedure:
When a beam carries both a concentrated and a uniformly distributed load, compute the deflection for each load separately and find the sum. This sum is the total deflection caused by the two loads.
For a beam carrying a concentrated load, the deflection D in = Wl3/48EI, where W = concentrated load, lb; l = length of beam, in; E = modulus of elasticity, lb/in 2; I = moment of inertia of shaft cross section, in 4. For a circular shaft, I = p d4/64 = p(2) 4/64 = 0.7854 in 4 (32.7 cm 4). Then D = 600(72) 3/[48(30)(106)(0.7854)] = 0.198 in (5.03 mm). The deflection per foot of shaft length is D f = 0.198/6 = 0.033 in/ft (2.75 mm/m) for the concentrated load.
For a shaft of uniform weight, D = 5 wl3/384EI, where w = total distributed load = weight of shaft, lb. Thus, D = 5(60)(72) 3/[384(30 X 10 6)(0.7854)] = 0.0129 in (0.328 mm). The deflection per foot of shaft length is D f = 0.0129/6 = 0.00214 in/ft (0.178 mm/m).
The total deflection of the shaft is the sum of the deflections caused by the concentrated and uniform loads, or D t = 0.198 + 0.0129 = 0.2109 in (5.36 mm). The total deflection per foot of length is 0.033 + 0.00214 = 0.03514 in/ft (2.93 mm/m).
Usual design practice limits the transverse deflection of a shaft of any diameter to 0.01 in/ft (0.83 mm/m) of shaft length. The deflection of this shaft is 3 1/2 times this limit. Therefore, the shaft diameter must be increased if this limit is not to be exceeded.
Using a 3-in (7.6-cm) diameter shaft weighing 25 lb/ft (37.2 kg/m) and computing the deflection in the same way, we find the total transverse deflection is 0.0453 in (1.15 mm), and the total deflection per foot of shaft length is 0.00755 in/ft (0.629 mm/m). This is within the desired limits. By reducing the assumed shaft diameter in 1/8-in (0.32-cm) increments and computing the deflection per foot of length, a deflection closer to the limit can be obtained.
For a noncentral load, D = ( Wc'/3EIl)[(cl/3 + cc'/3)3]0.5, where c = distance of concentrated load from left-hand bearing, in; c' = distance of concentrated load from right-hand bearing, in.
Thus c + c' = 1, and for this shaft c = 24 in (61.0 cm) and c' = 48 in (121.9 cm). Then D = [600 X 48/(3 X 30 X 10 6 X 0.7854 X 72)][(24 X 72/ x + 24 X 48/3) 3]0.5 = 0.169 in (4.29 mm).
The deflection caused by the weight of the shaft is the same as computed in step 2, or 0.0129 in (0.328 mm). Hence, the total shaft deflection is 0.169 + 0.0129 = 0.1819 in (4.62 mm). The deflection per foot of shaft length is 0.1819/6 = 0.0303 in (2.53 mm/m). Again, this exceeds 0.01 in/ft (0.833 mm/m).
Using a 3-in (7.6-cm) diameter shaft as in step 3 shows that the deflection can be reduced to within the desired limits.
Related Calculations: Use this procedure for any metal shaft--aluminum, brass, bronze, etc.--that is uniformly loaded or carries a concentrated load.