Question

4. Here is a similar problem with pulleys and shafts, complicated slightly due to the fact that there are forces in two planes, not just one, as previously. Dimensions in the figure are in mm. Some additional dimensions are: dahaft 18 mm 200 mm300 mm 300 400 Ts Te 150 d Assume bearings 0 and C are frictionless. Assume tensile forces are T = 200 N and Ti 40 N. Further assume steady-state operation. (a) On pulley A, suppose the slack-side tension is 1/9th of the tight-side tension. Determine the values of T and T (b) If the rotation rate is 250 rpm, determine the power delivered in SI units (e) Solve the statics problem in the xy-plane and draw shear and bending mo- ment diagrams in the xy-plane, labeling extreme values of shear and bending moment (d) Solve the statics problem in the xz-plane and draw shear and bending mo- ment diagrams in the xz-plane, labeling extreme values of shear and bending moment (e) Moment is a vector. Your moment diagrams give orthogonal components of the moment vector. Find the largest magnitude of the total moment vector 3 by combining components appropriately. At what value of z does this peak moment occur? (f) Draw a stress element from the candidate trouble spot, showing bending and torsional stress (K) Draw a Mohr's circle (h) Find the safety factor assuming the shaft is 1018 CD steel by the Maximum Shear Stress Theory and by the Distortion Energy Theory

Answer 1

a)Palley A T4 T3 Pulley B But T2 = 200 N TI - 40 N diamere dg- 300 mm. Mt s) =) Torauo on MG) T2 -T) 11 200-4200 24 Nm 2 4000 Nmnm But Mt (B) M a)24000 Nmm 3-7) DA Mt (A) PA) =200 mm. T3= T4 T) 20 24000 T4 30 N ax30 2 70 N 13 = qx T4 2 7o N T4 30 N Powel N=250 3Pm 2 x 250x 24 b0 2 N TMt 24 Nm 628. 31 w P 6a8.31 W

FBO )n 45 Cos45 tz+t RO too- 150- 300 x-Y FBD on Plamo Sin45= 212-13 N Ro Ro 300 40 0 50 C A o Ro t Rc =+212- 13 Moment about a' Rc C300 +400 150) 212.13 x 300 Rc 7+ 87 Ro -137 2 59 SFD 74 87 C

1Bond ing Moment C end Pont) MB(e) -1 M CA) - 13. 25a x 300 = H) 177 .96Nmm= 41.177 Nm. s 2305 Nmm 11-23 Nm Momont dva gam Bondung A 123 Nm 4) Nrn Plana FAody diaqum 212-18 C A C 300 -> +400 - 150- 'RC Ko +212.13 +R¢ = 240 240 - 212.13 = 27.86 N Ko t RC Nament about Co) RC 35) 00)- a40700) 212.13 Rc 122.2 N

7 21 N C 22 7 N Bordung Moment M M ) 28.433 Nm Ns (A) 28473 Nmm ta155N 122 3IX10 - 184155 Nmm = = 18.415 Nm 18. 415 Nm C 28.413 Nm.