Question

DEFINE GIVEN: A simply supported beam 150 11/ 200 lb TTTI - - FIND: Support reactions N V, M equations at cut between 6 (origin starts at 1) Shes and moment diagrams (abel all changes) Absolute maximum shear and moment values SOLVE -- M-- VERTEX

Answer 1

720016116coolb 150 16147 I pou sino=45 (oso = 3/5 Ekteek * 344 to 31 - oft EBD RO Reaction force. RatRB = 150*6 +%* 5046 +500+ 45 RAPOR = 1450 KN 7 sollipt goo*sino EMB=0, Ry*10 - 150*6(3+6)- **50*6 isolat (679+6)-500x465*3 = a piece 500 XCOSO Ra= 87515/ RD=57516 Re shear force section considering x Bending moment Sx = RA - 150X2 – Ys+SOXX Hukte te pet tisoille SA (2-0)= 875 76 ) TRA Sc(2=6ft) = 875-15046 -0.5*50*6 15C = -50 Kad 157 Mcc2=614)=875*6 - ISOXG-0.5*36*7*6* Sx=0,875 - 156XL - 6.5*****=0 Tmc = 2450 lblfte 22=875,2=5.6747 7 =y=Soux * x x My =Roxx-150X4, -0.5450242R Makezo) = 0 154.16 Mmare(x=5-674)=875*6-67 – 150*5:672 -0.5use xa x(567 Tmmad = 2460.78 168ft), at x=5.67# the shear fource sign changing s section CD 172501b7th sy-Rosta- My =R9l6+7) – 150*6*43+x) Sx=R4 – 15046-4750X6 6 selmo) =875-150x6-0.5*50X6 Sa=-50 lb -**xely -Y*80*6*(4/35+2) McC220J=87876-150%6*3 - 1 g 45026*2/46 I Meaght - 1980) MO6t=3H)= 8754971504646-Xatsoko luotes [MD=1995 blft 50(7=34)= - solb

5x=-RB SB(2=0 ) =-575 15 150 (x=376) = -5.75 lb tx P 2=515 16 My = fgxx M2(x=0) = 0 mp(X=346)= 575# 3 Mo = 1125 15747), arial force 500 00so EFX 20, RB = $00*3$ = 360 lb SF0875 AL (+) 5-671- 57515 1995 BMID (4)| M mar=auto auto .D - NORD AL (+) 0 15