Question

Problem 2: (Problem 3.79 in the book) 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. (a) Determine the tensions in the belt on pulley B, assuming the shaft is running at a constant speed. marines act as simple (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. 500 IN 75 B-india. 14-india. 10-india Problem 3-799

Answer 1

(a) % of the tension on Belt tension on loose side at B is 15 vight side -tie. Td = 0.15 Tz For the bully-shaft system shown, apply equilibrium of torque- ET=O: ( Tz - T.) (5) - (500-75) (+) = 0 (0.85 Tz) (5) = 1700 Tz = 400 lbf » Ti = 60lbf ity n by (b)? Now applying equilibrium of forces to find the Bearing reactions in- About boint 'o'. E Moro- 460 (10+18) - 575(10) = Rc (10 +18+12) > Re = 198.25 lbt, An EF=0; [18:25 - 460 +67s + Ro = 0 -> Ro = -293-25 lbfin notif

with these calculated bearing reactions, we can draw the loading diagram, shear force and bending moment diagram - Loading diagram in x-z blandi 293-251b7 575164 460 lbf 178.25 lbt lo in 18 in I sin ON TA 13 293.25 178.25 -281-75 2932.5 Ibf.in 2139 (d): The bending moment diagram shown above states that the maximum moment occurs at 8:10 in. wie. Mmax = 2932-5 lbf.in

& The torque at the maximum bending moment : B/w A and B - T= (som -- 75) * 4 = 1700 lbf-in with these values of M&T o we can find the bending & Torsional shear stress ad? Do = 32M d3 = 15294 Psi = 32* 2932-5 Tx (1.252 out o 5 = 15-3 ksil, tu = 15-3ksil = 16T - = 44 33 PS! (6 x 1700 (4-25)3 097, asr [T = 4-43 ksi! T=4 -43 ksi (e)! For the stress element at max•bending moment position, the principal stresses is calculated as a + 754.43 ksi Hºt = I 15-3ksi w 0,102 = + (?)2+(2)2 = 6 = 15.2 + (15:3.12+(+143)2 9 = 16-5 Kbsilon, [= - 1-19 kpsi Anne -4 Max- Shear should.. {max = V(15:3)4 (4.43)2 = (max=8-84ksi