Question

QUESTION 1 It is required to design a square key for fixing a gear on a shaft of 27 mm diameter. The shaft is transmitting torque of 195,844 N.mm to the gear. The width of the key is equal 7 mm. The key is made of steel 50C4 (sy - 577 N/mm2) and the factor of safety is 3. For key material, the yield strength in compression can be assumed to be equal to the yield strength in tension. Determine the minimum required length of the key for shear (mm)?

Answer 1

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OF SQUARE KEYS IS BASED THE DESIGN ON TWO CRITERIA viz. 0 SHEAR STRESS CRITERIA 2 M dol - (0) Met r - shear stress torque diameter of chaft l length of key b> width of key COMPRESSIVE STRESS CRITERIA = 4 Mt dbe tiig compressive stress b = 7mm Given d = 27 mm, M+=1955844 Nimm, Sy = 5577 Nimm 2 FOS - 3. 3 SHEAR . ACCORDING TO MAXIMUM STRESS CRITERIA - SHEAR 0.5 Sy = 0.5x577 STRENGTHY = 288.5 Nimm2 Ssy Scy Y = Ssy 288.5 fos 3 T = 96.167 N/mm2 Syc Strength in compression = strength in tension Syc = Sy 577 N/mm2 0 =192.33 N/mm2 6 c = Syc 3

From equili X = 2 Mt db l il = 2 X 1959844 96.167 x 27 x 7 1 l = 21.55 mni from eq" lilil l=- 4 M+ se db l=21-55mm DPA LENGTH OF FROM BOTH CRITERIA, THE KEY SHOULD BE 22 mm (Aus)