Question

Problem 3. (65 points). For the truss shown, (a) Determine the forces in members: AB, AH, BH and BC using Method of Joints. (b) Determine the forces in members: CD, CF, and GF using Method of Sections. н. F 4.5 m 3 m E B DI 6 KN 2 KN 8 KN -12 m, 4 @ 3 m

Answer 1

the G I F - 3m. I Compute the values of Ax, Ay, Ey Solo Step - fi B D. of Truss Draw T 4.5m Az Akk 3m ke 3m 6KN where Ax= Ey= Step- -3m pot-3meste 2 KN ØKN Ay Ey Reaction At A in x-direction Ay= Reaction At A in y. direction Reaction At E in y-direction E Fx = 0 (Net Force in a direction aj | Ax =0 (Nect force in y-direction E fy=0 Ay-6-8-2+ Ey=0 0 Ay + Ey = 16 kN

So tom O=3/5 c) Taka moment EMA-o Moment about point (A) (36) (6) Ey x12 - 2x9-846- 6x3 = 0 Ey=TKN . from Ea. Ay +7=16 Ay=9kn Step- using method of Joints to Find force in member AB, AH ,BH ,BL :> a) In A ABH - 3 m fano = 1 3m. 0-450 B) using method of joint At (AJ So F. B.D. At А FAH Sin45 FAH (A310=450 FAB FAH Cos 450 Ay

9) Efy=0 FAH Sin 45° + Ay = 0 FAH va - 9 FAH = -952 /kN E Fx = 0 FAH 208450 + FAB = 0 FAB = - FAH 12 FAB = - (952) va FAB = 9 KN joint (B) 9) using method of joints at So f. 8. D At joint (8) FBH FBA= 9 » Fвс V 6 KN

Efy FB H-6=0 FB H = 6kN E Fx=0 FBc - FBA=0 FBc = 9 KN So According to method of Joints Force (kN) Nature of Forca Member Tension o AB Compression 912 АН Tension 9 B C Tension の BH

Step-IV Using Method of Section to find the force in CD, CF and GF So I cut the section So the Member GF, CD, CF will be cut.

G I F 4.5m. 3m. Ara AX k3m → BK 3m +3mi (-3m & 2 KN 8KN 6KN Ay Ey

F.B.D of Right hand portion :- FFG FFC E Foc D Ey=TAN (580) So first we need to find So we Take

H 1.5m T F oteam 00 4.5m 3m. Ax- A <3m → Bazm- E हर-3ml 1px 3me 2 KN ØKN Ay 6 KN Ey

So we Jake In A GEO tons = .5 3 6.5 - က D - ၃ 8-57 In A FCO ten o = 3 3 | 6 = - Us s°°

FFG Sin 26.57° FFG 8=26.5707 a FEC POS 45° FFG C08 26.5*45° 450 FFC&in 45° F. FC E Foca TKN 2 KN

O E F = 0 FFC (0545° + FFG C0826.57 + FDC =o Efyzo FEG Sin 26.51º - Fec Sin 45º - 2 +7=0 +5=0 FEG Sin 26.570 - FFC 02 FEC 12 FFG sin 26.51° +5 2

(3 Taking moment about F EME=0 7x3 - FDC x 3 = 0 21 - FDC X 3 = 0 Foc=7KN From Eq. O se 0 FFC FFG Cos 26-57 +7=0 + 2 2 FFC FFG Sin 26.57-5=0 2 0 - FEG COS 26.57+7 + Fra Sin26.57 +5=0 Compression FFG = -8.999 KN After putting FFG value in Eq. 0 we get FFC =-1.413 kw Compression

Natuse of force Member Force (KN) CD 1 Tension 8.944 GE Compression Compression CF 1.413