Question

6-B16 Using the Method of Sections, determine the force in HG, BC and GF. 2 kN 1.5 m D 0.8 m A o E F G H 1m 1.5 kN -1 m 2 m 2 m

Answer 1

Fulce body Drågoum Method by Section 1.5 m 2KN 120.8m7790° KAx m ><lm balm- IRE Ray 1.SK section cact 2 for finding force in Sechon Cuti for finding force in Ho, and BC GF o Re- Reachon face at E Raxo Horizontal Reaction face at A Ray = vertical Reaction force at a Take moment at A=0 RE XAE = 2XAB + 115 XAG REX4 = 2x0.8 + 15x2 TRE = 1.15 KN /^ Consider section cut o and take Right hand side boot

_section cut , 1,5 m an obat gê , the DC OS Te 2018mons-t7 KTG I SKN RF = 1.15KN ILI Coo Fim Fim I= 2 = JP=0.82 x=0.6 = BB1 Take moment at C =0 FHG X GG = RE XGE FHG X 1,5 = 1.15X2 Tension / = foste is FHG = 1.53333 KN Take moment of HEO Fec X BH + 115 X HG7 = RE X HË FBc x 0.6 + 105X1 = 1'15X3 IFB C = 3,25 Kn Compression =force in BC

consider section cut @ 1 sechoncus 2 II FCD 1,5m I RECE. and take Right hand side past Take moment at Czo FGFX CG = RE XGE FGFX1.5 - 1.15X2 FGF = 1.53333 KN L- Tension force in GIF 'RE G EF Kimyelm