Question

(11th May 2020) The end gear D connected to the shaft is subjected to the loading F, as shown. If the bearings at A and Bexert only y and components of force on the shaft, determine the equilibrium force F, applied on the gear D when the torque T is applied at gear C, and then determine the required diameter of the shaft to the nearest millimeter that will support the loading by using the maximum-shear stress theory of failure. The radius of gear C is 45 mm, and the radius of gear D is r = 120 mm. The torque at gear C is T = 110 Nm. The allowable shear stress of the shaft materials is Tallow = 75 MPa. The radius of the solid shaft can be obtained by using c = (2_VM2 +T2) Assumptions: The force produced from the gear mesh at gear is in the y-direction, and the frictional force at the gear mesh is ignored. The stress concentrations such as at keyways are ignored in determining the radius of the shaft. Marking criteria: 1) Correct formula applied 2) Correct values and units 3) Neat and clear diagrams/sketches 4) Detailed logical solution procedures 5) Easily readable handwriting

Answer 1

Torque, T = 110 N-M Radius of Gear C, Re = 45 mm Radius of Gear D, RD? 120 mm = 0.045 m = 0.12 m. the force produced from gear mesh at (, Fy - I : 110 = 2444.44 N 0.045 - Torque along shaft remains same. Fzx Rd = EU XRE F2 = 24.44.40 x 0.095 K150-K 250 — 100—+ 0.12 2A1 RB2 F2 = 1527.775 N @pt +- 1594.475 N 75 mm EMB=0 F2 x 500 = Raz x350 Raz = 2182.52 N Roz=- 654.75 N BMD : x-2 plane: May = Fz x 150 Mey = 65,475 N-m : = 1527,77x150 N-mm Mey= RBzx100 = 654.75 X 100 Nl-mm May: 229.165 Nm

x-y plane: Fly = 2444 44 N K 250 — 400 - > By At support A, Mazzo At E ME. Fy ab = Fyx 250x100 250+100 2444.44 X250X100 350 = 174.602 N-m MAMay' + Max = 1224.1652 702 = 229.165 Nom (Mmax) MEV Meu?+ M62 - V65.4752 + 144,6022 = 186.474 N-m section - A is critical. shaft diameter: Using MSST, V shaft radius, c = [noi avead morz]% · [axts Vazquesta no x 103) : 12.9 mm Shaft diameter, de 25.8 mm Take [d=26mm