Question

Q.2) A 60-mm diameter solid shaft with a 100-mm diameter collar A is subjected to an axial load of 50 kN and a torque of 20 kN-m as shown below. Determine at the contact between the collar A and the support, (a) the average shear stress and (b) the bearing stress. 2.5 mm 601mm 100 mm 50 kN 2.5 mm 15 mm

Answer 1

Given data Diameter [d] = 60mm solid shaft with a[D] = 100 - mm diamtere collar 'A� subjected to axial load 50 kN torque = 20 kN - M Average shear stress = ? Bearing stress = ? Diameter collar D = 100 mm \r\n Max shear stress tau_max = 16 T/pi D^3[1 - k^4] equation (1) where k = d/D k = 60/100 k = 0.6 load = [p] = 50 kN: T = 20 kN - M A = pi/4 [D^2 - (d + 2c)^2] where c = clearance = 2.5 mm The above values are substitute in the above equation (1) tau_max = 16 times 20 times 10^3/pi[100]^3 [1 - (0.6)^4] times 10^9 \r\n tau_max = 3,20 times 10^3/3.14[1,000,000][1 - 0.1296] times 10^9 = 320 times 10^3/3.141[1,000,000][0.8704] times 10^9 tau_max = 117 Mpa sigma_Avg P/A = 50 times 10^3/pi/4([100]^2 - [65]^2)10^- 6 P/A = 50 times 10^9/0.785[1,0000 - 4,225] P/A = 11 Mpa \r\n (a) Average shear stress between collar and support