Question

Answer 1

BKN Rox 3m 9/18 10kW B बार E Roy sta 6 A RAY z Fx = 0 10 KN - Rox=0 Rox = 10KN Fy = 0; Roy+Ray = 8KN 0 X EMA= 0; 10x 6 + 8 x 6 3x3 Roy = -0.667 kN (1) RAY BKN + 0.667 KN 8. 667 kN At joint A FAB FAE & Fre= 0; FAECO S45º = 0 FAE = O(T) 45° Fy = 0; FAO+8.667 KW + FAE Sin45º =0 8.667KN FAD +8.667 KN +O = 0 FAD = -8.667 kN (C)

At joint D. Roy : 0.667 KN {Fy = 0; - Roy - Foe sin 45º = 0 -0.667 - FoE Sin45º = 0 Foc Ron=10kN FDE = 0.667 Sin 450 KFDE FoE = -0.94378 KN Frage = 0; => - Foc-Ron-FoE Cos45° 0 =) - FDC-10-F-0.94328) (0545°-0 F> Foc =-9.33 KN (C) At Joint C 8kN {Fx=0; → Fco - Foc = -9.33 KN 450 45° -9.33 + FCE COS 450 FCB COS450 FEB AFCE Fce - FCB = 13.194617 EFy = 0; - 8- Ece Sinusº - Fcesin 45º = 0 FCE -Fee = 873 from both equation FCE = 0.943KNI (T) Fcb = 12.357 KN (e)

At joint B -13.257 KN E Fx = 0; JOKN -17.257 C0545° + FBE = 0 10KN 450 FBE FBE = -1.333 KN (C) - 8.667 KN Results:- Member force (kn) 8.667 (C) AB C- Compression T - Tension BC 12.757 (C) CD 9.333 (C) DE 0.943 (0) BE 1.333 (C) AE o 1C) CE 0.943 (T) please no vote

Computer Analysis for check given below

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8.000 KN 90° 00 0.667kN 4 T2 9.353kNC DEFAULT 10.000KN %0.943kNT DEFAULT 12.257kNa DEFAULT 0.943KNC DEFAULT 10.000 KN 4 180° 3 T4 1.333NC DEFAULT DEFAULT T6 8.667KNC 0.000kNT TO DEFAULT 8.667kN -2