Question

Problem No. 1: For the truss shown a) Determine the support reactions at A and G b) Using the methods of joints, determine the force in all members and indicate if the member is in tension (T) or compression (C). Assume all connections are pins 100 KN 4 m 4 m B 4 m H .. m

Answer 1

Ans: Given Given truss is Symmetry. about 1OOIN Y-axis a, Apply equilibrium equations to whole um truss E 4m EHEO HA=0. F B I um 5 A HA Н. VA ING =0 um Ev = 0 VA+ q= 100 IMA=0 10064) - VG(8) = 0. um NG= = 50KN VA = 5010 It is the forces in a symmetrical truss. So finding members ino half portion is enough 5011N consider Joint 1, Y FAB 71.5° [u=o. FAH 50+ FAB Sin 71.5 = 0. Jomt A FAB = -52.721(NCC) tano 413) 4 [H=0 FAH & FAB COS11.5 =0 O=71.5°. FAH = 16.72 IN G). FBC consider Joint By IH-0. verso 71-5° FBI FBct FBICOS 71-5° FIBH COS 52-23° 752-72 52.257 Enzo FBH 52.72 FBH ter sin 52-23+ FBI sim 71.5 =0. CS Scanned with CamScanner

consider All B.ID consider Joint H, tomb = 4(2) BI BI = 4622 : tan 71-5 FHF 33.73 BI = 2.67 33.73 y FAH tameloto) Consider All IBH GH IH CH=0 tans = BI + FHB Sim 33:43 = Fast Fof S14 33-13 LIBH = tantum 2-67 FHB-FHF = 0 LIBH 256-27 LABHZ 52-23° Il=0 FHB.COS 33-73 + FAF LOS33.73 =0. ZEHB=LIHF FAB+ FAF -0. =180-90-56.27 =0. = 33.73 : FHB = FAFF Considering this bon sowing Joint B FBc TFBI FB COS 1.5 O FBc & FBI COS 15 = 52.72 Sm56.23 + FBI sio 71.5 zo FBI Substitute in O FBH smstor =0 FBC = 52.721CN G). considel Joint I. Fic FIE = 0. CS Scanned with CamScanner

consider Tointco , FBC In=0. +1.50 & ECE Fec+ FLECOS 11.5 = FLD FD Feepee FeCous FLE COS 71.5 -FCDC 52.72 - Ev=0 Fre Sim 115º = 0 Fle=0 Sub im =4 FD = -52-72 CT). Forces im all members. FABE FGpz 52.721CN CO) FOGH= 16-721CN (1) FAH =o. FHB FBI -FHF =FFICO FIC Fle=0 FBC EFFE = 52.12KN (T) FIE Fcp = FED = 52.721N (7) CS Scanned with CamScanner
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