Question

A countershaft carrying two V-belt pulleys is shown in the figure. Pulley A receives power from a motor through a belt with the belt tensions shown. The power is transmitted through the shaft and delivered to the belt on pulley B. Assume the belt tension on the loose side at B is 15 percent of the tension on the tight side. Yield of the Shaft ?? = 560 ???

a) Determine the tensions in the belt on pulley B, assuming the shaft is running at a constant speed.
b) Find the magnitudes of the bearing reaction forces, assuming the bearings act as simple supports.
c) Draw shear-force and bending-moment diagrams for the shaft. If needed, make one set for the horizontal plane and another set for the vertical plane.
d) At the point of maximum bending moment, determine the bending stress and the torsional shear stress.
e) At the point of maximum bending moment, determine the principal stresses and the maximum shear stress. f) Determine the Factor of Safety using the Maximum Shear Stress Theory (MSS).

400 ce 45N 300 N 150 250 dia.- 300 dia. 20 dia. All dimensions are in mm.

Answer 1

@ T2 = 0.15 ET=0 = (300-45)125 +CT2-TI)150 = 0 · 31875 +0.1TI-T12150 TT = 236.11 N T2 = 23.61N 3456845=244N B7' Roy 300, 400 : 150. kly EMO =0 at - Rcy x 850 + 244x300 =0 Rcy= 86.100 V EFyzott, 244 - Roy - Reyzo Roy = 157.902 259.72 Ak Roz 300, 400 i 150 Rcz EMO =0 +; 244x300 - 259.72x700 + Rcz(850) =0 RC2 = 127.770 1 . Eryto it, Roz + Rcz +244-259.12 =0 RO2=-112N = | Roz= 112N

o zy plane shear froce diagram SF out o, Vo =- Roy =-157.90 SF at A, VA = -157.9+244 = 86.10N S Fat C, Ve= 86.10 N Bending moment diagram BM at O, MODO Bli at C, Mc=0 BM at A, MA = - Roc x3 = -157.9X013 = -47-37N.m 2441 hood diogram 8 • 157AN VOND, 881 Shear fire of -x. diagram -157,9 (N.m) / A PE Bending Moment Diagram -4737

12 plane Shear froce diagram SF at O, VO = -112 N . SF at A, VA =-112+246 = 132N. SF at B, Ug = -112+244–259.12 = -127.72N SF at C, VC = -127-72N Bending moment Diagram BM at o, Mozo BM at A, MA=-112x0,3 = -33-6 N.m BM at B, MB = 127.77X0.15= 19.158 Nim Bm at C, Mczo Rc2= 127-77 Rozza112 ok I - 300 244 259.72 t + 400 + 150 Load diagoan V(N) , Ex shear fore diana -112 127.77 MCN m) 19.158 c. a Bending moment Diagram -336

a A, where both the planas critical location 28 at have maximum BM M = JE 47.37) C-27.3692 = 57.94Nom T= (300-45) 0.129=31.88Nim. reme - 320 = 32C57.947 [ Tide T200203 = 73.778100 pa o=73.77 MPa tate = 10 = 16*3188 = 20.29x106 Pa e = 20.29 MPa. Les Oi = £+ J©)372 = 57,94 +557.94) ? 20.242 61 = 64.93 MPa] . 62=81) 72² = 57.945 - 57.943²+20.257 [52 = -6.39 MPa] Zmax = = = = 587.94)7 20,297 Remax = 35:36MPa] -6.39 M pa 2

& Sy = 560m pa Tonax = Sy = 560MPa = 280 mpa By maxinsum shear stress there Fos= ((max) yold Emas - 28ompa 1. 35-36mma For = 7.9