Question

Question 2 The shaft shown below is rotating at 650 rpm, and it receives 7.5 hp through a flexible coupling. The power is delivered to an adjacent shaft through a single helical gear B having a normal pressure angle of 20 degrees and a helix angle of 15 degrees. The pitch diameter for the gear is 4.141 inches (De) and the Tangential (Wu), radial (W) and axial/thrust (W) are shown below on the drawing. The shaft is to be made out of SAE 1020 and will have the following strength characteristics: Su = 61 ksi, Sy = 51 ksi, Sn = 22 ksi (normal stresses), Sn = 16 ksi (Shear Stresses) The fatigue factor is kr = 1.6 and Bearing A is designed to take the thrust load.

Answer 1

Solution:- The given problem has been solved using standard formulas for calculating the shear forces and bending moments. Usual notations have been followed.

Solutions Given - P = 7,5 HP na = 650 R.P.M. Db= 4.141. Inches W+B= 351 lb Wro 132 lb Wył gulb I - su=61ksi Sy=51ksi INsg = 132 lb ' Bo- 3"_ - AC st 2 RoA Roc Reaction and forces in vertical plane Root Roc= 132 lb - Taking moments about Rocas = 13243 Roc= 386 - 75,2 A. lb REA= 132-7812 = 52.8 lb Shears fosce calculations: - sofort A= 52.8 eb - S. f. ort just lyr of B = 52.8 lb sif. aut just right of B= 52.8 - 132 3-79.2 lb S.f. aut just left fc = -73.246

Sifat com 79.247.2- Bending moment calculotilaus Bom, at Azo Romi aut B=5218x3= 158,4 lb-in Romicat c= 5218x5-13272 = 0 lb-im 1132 lb ok-2- K 3"- Roa= 52.8 Rap=79.2 Reaction forcel ding som 152.8 с L 79 Shear forus dingsom NIS8.4 lb-in Bending moment diago Reactions forcy, shears for us and Bending moment dingsans in vertical o plane

351 lb A1 R ſc REA 1. Reaction foses in horizontal plane Rzat Rec = 350 lb Taking moments about A Rtc X5= 3543 Rtc = 210.6 eb RtA= 351-210,6 = 14o.4 lb Sif Calwelations Sof. cet A = 140,u eb Sif. out just let of B= nou eb Sof- out just sight of B = 140.4+ (-351) = -210.6 lb sof, at just left of c= -210.6 lb Sif. aut just right of c= -210.67 210.06 Bending moment calculations - Rime cat Azo B.M. at B = 140.4x3 = 421.2 lb-In D. m. cat c = (140.4x5+ (-35), )

- 351 lb RtA=140.4 Rt c= 210.6 Reaction rosu diangsom a yo.y eb I shear force diagram 21016 eb 421-2 lb-im Pending moment B č aliang sama Reaction forces, swear forces and Bending moment ding som in horizontal Bending plane (b) Resultant loads ont bearsings A and beasing B Resultant load at A= √ R2 R2 = 1(140,4)}7 (52872 - 150 eb Au

Ruultant load cut B = WER't work = V (351) 27 (132)2 = 375 lb Au (6) Maximum Stress location It can be seen that maximum bendin moment and maximum shar ocul at point B. so maximum stress location will - be at Bo Also the diameter of Shaft will be determined at point B.

@) ab'ameter of Shaft Col d= [ 32NIKE Mmar 12 momento nax = N= duign Safety Jautor= 2 (given) Kp = fatigue feutor = 1.6 ( given) Mman = /[Map]²+ (Muz)² = Maxisnum bending MMB = Maximum bending moment in hosizontal plane = 421.2 lb-In Mur = Maximum bending moment in vertical 1 . plane = 158.4 lb-im - Mmax = √ (421.2) 37 (158.432 B = uso lb-im Si= Estimated endurance strength Taking 85% og Su Sn'a 0.85 (6) = 51.85 ksi = Tiero РК T= Torque = 63ooox P. a 650 = 726.3 lb -In Sy = 5llesi = 51x103 Psi (given) in hre

T d=0,723 ... In am