Question

1. A motor delivers 120 hp at 250 rpm to a radial pump through a flanged coupling mounted on a 0.375-inch diameter steel shaft. The pump is mounted on a 0.3125-inch diameter steel shaft and the two shafts are connected by flanged couplings. The flanges are connected using 2.000-inch long, 0.25-inch diameter, SAE Grade 4 steel bolts, equally spaced along a 3.5-inch diameter bolt circle. The coupling flanges are 4 inches in diameter and 0.75 inch thick. The bolts are installed with 0.5-inch diameter, 0.031-inch thick steel washers under the heads and nuts. Determine: (a) The minimum number of bolts in the coupling required to transmit 120 hp to the pump. (b) The bolt and member stiffnesses in the coupling (a=30°). (C) The design factor of safety in shear) of 0.75 in the bolts. 0.25 IN DIA EQUALLY SPACED 10% 0.75 in Ans: (a) Nbalt= (b) k = kn= (c) na =

Answer 1

5252 Aus- Given, Transmit lower, P= 120 hp - ppeed = N=250 rpm. . H.P = Torque XN 5252 120 = Torque x 250 → Torque = 2520.96 Torque, T = 2521 lb-ft = 30251.52 lb-inch " This Torque is transmit one shaft to another by coupling too (which is connected by bolts equally spaced along 3.5 inch diameter circle). D = 3.5 inch . Assume f= Shear force on each 1. bolto n = Number of bolt. : Torque, 1 = (nf) x2) 30251152 = nF x 3.5 = nF = 17286 → PF = 17286) 0 Bolt Materials — SAE Grade 4 steel bolt Bolt diameter, d= 0.25 inch Yield strengths to y = 100000 psi. Cross section area of bolt - a - ģd? *0,25?? a = 0.0491 Basteln in? According to manimum shear stress Theory They = {oissyja ce a distance between 1 Centre) fos

for Minimum bolt. Taking factor of Safety = 31 from eqn-® - & - cose syge os 10.5 /e 0000 17286 - n x 0.0491 stress are where, A= Tensile ki = ALADE K = In = 7.04 8/ Ans. .: Minimum Number of bolt = 8 DAMS - Bolt Stiffness, 4 - 1 of the fastner = 0.049111 Æld + Aa4 . Ad = Major dranch of fashnin = 0.0491 in 0.0497x0.0491 x 29x10 4 = length of Threaded portion (0.0491x0.6) +(0.0497X1-4) 06 inch 1 kg = 4.665x103 An. la= length of Unthread portion of bolt - 1.4 inch E = Young Modulus . = 29007 ksi = 29 x 106 psi Member stiffness ned tand . 12ttanat D-d 1 (0+d zt tane +D+ CD-d) where, a=30 t = Thickness of washer = 0.02 D = Diameter of Washer =0.5 inch a = Deameter of bolt = 0.25 inch 1x29x10x 0.25 x fan 30 - S[(2x0.631 x tauja) +0.5-0.253 (0.5+0.7597 112 0.031 x tan30 + 0.5 +0.25) (0.5-0.15) 5.268107 Ki 8.712 X162 = 6.033 x 108 Tkm = 6.033 x 108 Design factor of Safety in shear of bolt, - Max. Shear stress - 0.5 x Sy nd = Working Shear Stress 17286/420.0491) 0.5 x 100000 md= 132807 K na = 1136 As 77286/L8X0.0490) =):136