Question

We think we have the right answer but we are unsure and having second guesses.

Ray=-55.55N

Rby=55.55N

Raz=-2777.78N

Rbz=2777.78N

A 50:1 worm gear reducer is shown in figure P3 below. It is bolted to the frame at points A and B. The input torque shown on the vertical (worm gear) shaft is in the same direction as the direction of rotation of the shaft. The direction of the output shaft rotation is shown in Figure P3, and the driven load exerts a resisting torque Mo in the direction opposite to the rotation. Assume the gearbox has 100% efficiency (power in- power out). Determine the reaction forces at bolts A and B that hold the reducer housing in equilibrium. You may assume that Ro-Rx = 0. Parameters Mi = 15 Nm L = 270mm L1 = 160mm H = 80mm Figure P3: Worm Gear Reducer Show your answers in the space below. Use the following page to show your support work. SOLUTION RAY = RAZ = Raz =-

Answer 1

The FBb of the gear box is for 100 % efficiency m_i = m_ 0 = 15 NM R_Bx = R_Ax = 0 Applying C.O.E we get sigma MY = 0 (> +ve) R_Bz times 0.27 + m_i = 0 R_Bz times 0.27 = - 15 therefore = -55.55 N (uparrow) sigma M_2 = 0 R_Az notequalto R_Bz = 0 therefore R_Az = -R_Bz = 55.55 N R_Az = 55.55N (downarrow) sigma M_z = 0 (> +ve) - R_By times 0.27 + 15 = 0 - R_By = 55.55 N (uparrow)\r\n