Question

6. Determine the reactions at supports A and B. (15 points) 700 lb/ft 500 lb/ft oft-6ft 8 ft-

Answer 1

Arctoral Member CO is hinged at both the ends and there is no force or moment acting along the spen of CD so it will developonly axial force (fo) and behave just as truss member. free body diagram of whole structure 700 tbl ft. 500 l/h Foo ko xe 8.14 Ay - tem 0= 16 = 3/4 Sino=2 / Loso=415 Step I Analysing member BD. - 700 lift III Bu max Foto 9ft VB O found above for equilibrium's 6 Net moment about B, EM 2=0 7 > (fco Sin )DB + Foco les ox0-4x700X DOxZDB - resultant of trianguler loading 5f0X3x9+0-4x700x9* (Exg) -o act atzo framb [FcD = 3500 l CW E F 20 7 fo Caso - B₂=0 73500x - Beco Bx = 28001

> Fp Simb-4x700x9 +Byeo 3500x3 - 4x200x9 + By = 0 By = 1050 hp Step II_Analysing member AC. soolb.lft , fco= 3500 lle (stop I) MASAULIN Ax M 12ft & Ay Equilibrium conditions i EFx=0 Au folos0=0 Ax-3500*4 = 0 - An= 2800l 7 cu Efy-09 A3-2x500 XAC-fco Simeo → Ay - 4x 500 x 12– 3500*3520 Ay=5100 lb T 1 cup Net moment at A=0 EMAFOD - MA +ļx500xAC XĄL + Es Sün@ x AC =0 O resultent of triangular loading is at Al from A -MA+X500X12X12 +3700x3,6x1220 → MA= 43200 Watt 1 Answers's Reactions at A's Ax=2800t - Ay=5100 b 1 Ma-43200 lb-ft +) Reaction at Br Bx=2800 bt By= 1050 lei