Question

(20 points) Problem 2: A sheave with a 7.90 inch pitch diameter delivers 8 hp to the shaft at location D. The sheave at location A has a 12.40 inch pitch diameter and delivers that power to its mating sheave. The shaft rotates at 1200 rpm. a. Determine the magnitude of the torque at all points of the shaft and draw the torque distribution diagram. Determine the magnitude and direction of the forces acting on the shaft at the power transmission elements. Calculate the reactions at the bearings B and C in both the horizontal and vertical shaft directions. Determine the magnitude and location of the maximum bending moment in both the horizontal and vertical shaft directions. X-X XIN

Answer 1

Solutions to 10% Pitch diameter of prill ort location. I . do = 7.90 Tuch Pitch diameter of sheave at location A da = 12:40 Inch. speed of Shaft, N= 1200 rpm. Power shaft. transmitted p= to the 8 hP

от 96 P= 815 Со 15 - ft РИчоо . 1- Sec sec Then Torque. Ţ is given by: - Б3025 P(he ) (Ib-In) speed (RPM) Т- b7o 2 SXR | 2 оо - Ч2-0, 17 ц, - jи

This tofane T is received by the shaft at location D and tone off from the thast at location. A. (a) Tosque magnitude ont D = 420011 lb i Torque magnitude at c=420117 db-in Toique magnitude at B = 420:17 lb-In losque magnitude at A= 420 17 lb-in 420117 4:20.1.7 :HE 42007. lbin Torane Distribution diagram of the shaft (6) As it can be seen that sheave out location D transmite torque in horizontal direction. Also the theave ont location A transmite torane in veertical disectioni. E be the maximum force acting the shaft at location Dino norizontal alirention.

Then Foy do - T So Fo = Tx2 - 420.17 22 1 F = 106.37 lb 7 Ans C (In horizontal direction & Similarsly, let FA be the s marimum force acting on the shaft at lolution A but in veertical direction Then FA= TX2 = 420117x2 12:40 IPA=67077 lb | Ary (In veertical direction] (c) Reactions at B and c in horizontal a direction - Let RBH and RH be the reactions occurring in horizontal direction tot 15" on , 106,37 KRL IRCH force diagram in horizontal direction

The forces in conilibrium will be ... Ron & Ron = 106.37 lb . 6) Then taking moment about point 106,374 33 = Rox 25+ Roux 10 1 10 RBnt 25 Ron = 3510.2) 05 RBH 4 2.5 Rem = 351.821 sowing foo RBM and Ron from equation 0 and equation ) Subticuting Equation Ô from Equation @ giver 1.5 Ron=244.651 | Ren= 163110 alb And RBM = 106.37 - Ren Rein = 106.37-163.10 R$*= -56.73 lb . Anda Negative Sign shows that Ron is in veortically downwand direction al hesite to what we thought initially in the force diagram in honzanted t direction

so Corrected force diagroom in the holizanted direction will be por los mita 15 156.73 metros 163.10 lb Reactions at B and c in veertical direction let Rev and Rov be the reactions at location B and C in the veentical direction. 167.79 lb D is - AR CA D RBU 'Rev from force equilibrium Rout Rev=67.77 lb Now Taking moments about D e 67.774 33 = Roux 23 + Reux.8 tolo - 23 RBv + 8 Rev=2236.4) 2.815 RBV + Rex = 279.55 6

solving froos RRv and Rev from equation (in) and equation (IV) Subtracting equation (n) from canation (W) 1.875 RBv = 211.78 RBV = 112.95 Ib And Putting RBV in eonation (1) gives 112.95 + Rev = 67.79 Rev= -45118 lb . And So again, we need to draw the correction force diagram in veertical direction 67.77 us 18 lb to sa hlon a is 112.55 . lb (a) maximum Bending moment magnitude in horizontal and vertical direction First we will drown the bending moment a diagram of horizontal direction. Again draw the corrected force diorgsen y hasizontal direction.

106.37eb 56.713 lb hlou A B C 163.00 lb Bending moment calculatious B. M. cut D= o lb-in B. M. aut c=1-106.3788) =-850196 lb in B.M. at B=(-106.37423)+(163.10x15) = -2446.5) + 2446.5) o lb-in - B.M. at QA = 106.37x33)+(163.10825) +(-56.73810) =(-351012) + 4077.51+(-5673) = o lb-In

850.gb lb in So Maximum Pending moment ocurris at location c and site mangritinda is Mmax =-850.36 lb-in in horizonted direction And Now for similar analysis will be done veertical direction. . 67.77 45.18 Ain_10 1 5!? Somos - 112.95 Bending moment calculations B. Me art D=0 B.m. aut c= o Rom. ait B= (-45.18 x 15) =-6777 lb-In Rom out A = (-45,18 x 25) + (112195710) = (-112915) + (1295) Olb-In

to lo so 15" to 8 4 - 67717 ebrin So, again the maximum bending menent occurs at the location c and its magnitude is Mamax=-6777 lb-in in veertical direction