Question

A solid steel shaft supports loads PA = 230 N and Pc = 660 N as shown in the figure. Assume a = 475 mm, b = 705 mm, and c = 615 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. If the allowable bending stress is 100 MPa, determine the minimum diameter that can be used for the shaft. D a b PC d= mm

Answer 1

Given support loads P 230N, Pc = 660N distances, a - 475mm - 0.475m b- 705mm 0.705m c-615 mm -0.615 m Allowable Bending stress (6)= loompe = 10 N/mm2 find minm dia of shaft (dmin) = ? A B 0 .0.615 Tours 0.705 m 230N RB 660N Ro for equilibrium condition Efuso, (RB and Ro are support Reaction) RBtRo = 230+660 RB + Rp = 890N Emo =0 230 x 1.795 - RBX132 +660 40-615 =0 RB = 230 X 1.795+660 x 0.615 1.32 6 20.2 N RA 890-6202 269.8

1269.8 BMD Mr = 269.8% at a =0, Mp=0 atx=0.615, Me = 165.9 Nm Imm SER section DC cx SFX = 269.8 at x=0, SFR = 269.8 GCON* at x = 0.615, SF = 269.8 as at there is point doast section CB B uc She = 269.8-660 Stx = -390.2 660 at a=0.615, SF = -390.2 at n = 1.32m, SFB = 390.2 Ma= 269.8x 660 (X-0.615) MB = -10g. 16 Nm 1269.8 atx=132, c section AB AX T RB Geo N SEE = 269.8-660+620.35 230 N at a = 1:32, SFR 230 N SF₂= Ma= at x = 1.795, SFA = 230N 269.88 – 660 (2–0.615) +620.2 ( x-1-32) at a = 1.795, MAE 0.086 Nm no know from eqr of flexural er formula b = M = 2X1 у I X / 64xd → we M dy x2 of x ñ d M = ob I d 32

d3 32 м то) а - 3) 32 г Here m=109.16 Nm - 109. 16x to° Ју» oЬ а - 32 x 109.6 хус" их (ол) т 22.3 mm)