Question

The bearings at A and B exert only vertical reactions on the shaft. Take d1=0.1 m, d2=0.4 m, d3=0.2 m, d4=0.2 m, F1 = 572 N, F2 = 320 N, and F3 = 193 N. Draw the shear and moment diagrams for the shaft if it is subjected to the vertical loadings and determine the maximum bending moment in N-m. Be mindful of the sign convention. Saite d2 - d3 1 24. F3

Answer 1

FBD 15721 F3.193N GA IF:20 VE 193 193 342.715 VO - SFD. VE:0 - - - - -22826 34.32 BMD -38.596 - 56.14

First Calculate the Reaction Rn and le Taking moment of all forces about A, we get REX0-7 = 0[193X0-9]+(572X0:1) – (320x0.5) RB = 173.7757-2 -160 07 TRB: 101.285N Then we have RA= Total Load -RB = 1937572-320 - 101-285 TRA = 343.725 NI Sheas fosce Diagram (SFD) Shear force at A = RA IVA= 343.7150

Sheas force at BC Vo VCE VA - 572 = 343.715-572 i Ve = -228.285 N shear force at D VDVc +320 - -228.285+320 [Vo = 91.715N] Shear force at B VB = VD+RB - 91.715 +101-285 TVB = 1930] Shear force at E VE = VB - F3 = 193 -193 I VE =O

Bending manuni Diogeam BIM at A is zoo. B.M at hoc Mc = RAXOI = 343.715X0-1 MC: 34•32 N-mil Berding moment cet D MD- (RAX0.5)-(F4x0.4) = ( 343.715x0.5) -(572X0-4) MD. 17186 - 228.8 i Mps -56.94 Nom Bending moment at B MB = CRAX0.7) - (FIXO:6) +(F2X0.2) = (343.72x0.7) – (572X0-6)+(320X0-2) MB= -38.596 Nom

Bending moment at E Me= ( Raxo.9)-(FjX0.8) + (F2X0.4) +(RBX0:2) = (343-72x0.9) #(572X0.8) + (320X0-4) +(101: 285X0.2) I ME = 0 Maximum Bending moment. Mmax =-56.94 N.m]