Question

6 in. 8 in 8.22 Assuming that the magnitudes of the forces applied to disks A and C are, respectively, P, = 1080 lb and P, = 810 lb, and using the expres- sions given in Prob. 8.21, determine the values of ty and tx in a section (a) just to the left of B. (b) just to the left of C. 8.23 The solid shaft AB rotates at 720 rpm and transmits 50 kW from the motor M to machine tools connected to gears E and F. Knowing that fall = 50 MPa and assuming that 20 kW is taken off at gear E and 30 kW is taken off at gear F, determine the smallest permissible diameter of shaft AB. 10 in 10 in. Fig. P8.22

Answer 1

solo 8.22 Given; Assumed load P = 1080l6; P2 = 810lb. Shaft diameter = 1.75 in. R = 1/2 1-75 0.875 in. Polar moment of inertia J- Td4 32 T (1.7504 0.92077 in 4. 2 32 (a) Now Torque on portion ABC; TP 6x1080= 8x8 10- 6480 lb. fo.in just to left of point B; V Shear force= 1080 Lb. , Moment M = P, x3 - 1080x3 = 32406. fe ß2900, T = 6480 lb. in R M² + T2 combined loading Then Ty= 0.875 3240² + 64802 0.92077 6884.73 lb/in2 → CK= Jmcospjº+ (3 vR+T) - 0.875 0.92077 X 1080X0-875 +64898] TK = 6756.57. Cblis?

Bending in horizontal plane: - 40546 810lb 40546 ko → 4050lb.in Bending in vertical (Transverse plane) 1080lb 1242 lb, 8106 64846. 1 t 1 324016-ün.

b. Vz just to the left of c: Vz |(162)2 + (405)* = 436.198 46. de tail (162) = 21.80 M = √(1620)² + (40502 4361.98 lb. in. ~4362 ra tan (1620) = 21.800 ß= 90- 21-8° - 21,80 = 46°24' Now Ty = 0.875 6480² + 46322 43622 0.92077 7611. 37 661 is? VR + T x 436.198X0-875 + 6480 6734.45 MLOSBE 4362 Cos 46.240 3008.12 0.875 (3008.12)* +(6434.45) 0.92077 700g. 1079 iblis