F =4 

 TOT 



lb 



, 2. , , 226 in." 

 (4.7 m. )(3) + 



= 56 lb + 452 lb = 508 lb 



bearing resis- frictional 

 tance component component 



The frictional component of the total force has a moment arm of 

 Do/2 = 2 inches, and the bearing resistance moment arm is from the center 

 of the auger blade face, or 1.5 inches. Assuming that the normal force 

 from the bearing resistance component is modulated by a coefficient of 

 friction of 0.5, the required torque is 



T = y F_ r^ + F r 

 f f s s 



= [(0.5) (56 lb) (1.5 in.) + (452 lb) (2 in.)] 



ft 



12 in. 



= 78 ft-lb 



The required removal rate is 



V = (4 in.) (36 in.)(l.69 ft/sec) 



ft 



144 in. 



= 1 .69 ft /sec 



The volume of soil trapped by one pitch length of the auger is 



V = 2 A^ p 



(2) (4.7 in. ^) (3.6 in. /rev) = 0.02 ft"^/rev 



Therefore, the required angular velocity is 



(^ = 1 /£j|_^fd^ ^ Q Q2 ft^/rev 2 it = 530 rad/sec = 5,000 rpm 

 V V rev 



Therefore, the power required is 



P = T 0) = (78 ft-lb) (530 rad/sec) 



P = 75 hp 



hp 



550 ft-lb/sec 



69 



