Select a key for a 4-in (10.2-cm) diameter shaft transmitting 1000 hp (745.7 kW) at 1000 r/min. The allowable shear stress in the key is 15,000 lb/in 2 (103,425.0 kPa), and the allowable compressire stress is 30,000 lb/in 2 (206,850.0 kPa). What type of key should be used if the allowable shear stress is 5000 lb/in 2 (84,475.0 kPa) and the allowable compressive stress is 20,000 lb/in 2 (137,900.0 kPa)?
Calculation Procedure:
The torque acting on the shaft is T = 63,000 hp/R, or T = 63,000(1000/1000) = 63,000 lb·in (7118.0 N·m).
The shear force Fs lb acting on a key is Fs = T/r, where T = torque acting on shaft, lb·in; r = radius of shaft, in. Thus, T = 63,000/2 = 31,500 lb (140,118.9 N).
When a key is designed so that its allowable shear stress is approximately one-half its allowable compressive stress, a square key (i.e., a key having its height equal to its width) is generally chosen. For other values of the stress ratio, a flat key is generally used.
Determine the dimensions of the key from Baumeister and Marks-- Standard Handbook for Mechanical Engineers . This handbook shows that a 4-in (10.2-cm) diameter shaft should have a square key 1 in wide X 1 in (2.5 cm X 2.5 cm) high.
The length of a 1-in (2.5-cm) key based on the allowable shear stress is l = 2 Fs/(wkss), where wk = width of key, in. Thus, l = 31,500/[(1)(15,000)] = 2.1 in (5.8 cm), say 2 1/8 in (5.4 cm).
The length of a 1-in (2.5-cm) key based on the allowable compressive stress is l = 2 Fs/(tsc), where t = key thickness, in; sc = allowable compressive stress, lb/in 2. Thus, l = 2(13,500)/[(1)(30,000)] = 2.1 in (5.3 cm). This agrees with the key length based on the allowable shear stress. The key length found in steps 4 and 5 should agree if the key is square in cross section.
When the allowable shear stress does not equal one-half the allowable compressive stress for a shaft key, a flat key is generally used. A flat key has a width greater than its height.
Find the recommended dimensions for a flat key from Baumeister and Marks-- Standard Handbook for Mechanical Engineers . This handbook shows that a 4-in (10.2-cm) diameter shaft will use a 1-in (2.5-cm) wide by 3/4-in (1.9-cm) thick flat key.
The length of the key based on the allowable shear stress is l = Fs/(wkss) = 31,500/[( l)(5000)] = 6.31 in (16.0 cm). Use a 6 5/16-in (160-cm) long key.
Checking the key length based on the allowable compressive stress yields l = 2 Fs/(tsc) = 2(31,500)/[(0.75)(20,000)] = 4. in (10.7 cm). Use the longer length, 6 5/16 in (16.0 cm), because the shorter key would be overloaded in compression.
Related Calculations: Use this procedure for shafts and keys made of any metal (steel, bronze, brass, stainless steel, etc.). The dimensions of shaft keys can also be found in ANSI Standard B17f, Woodruff Keys, Keyslots and Cutters. Woodruff keys are used only for light-torque applications.