Question

Find GERF and then using the method of Joints find the forces in all members. Make sure you provide all the angles and magnitude of all the forces in a table stating Compression or Tension. You can use cm to draw the structure. 12-5kN 12:5KN 12:5KN 12.5kN kam am 2m WA VB Vo 2-5 m 0 Chatarrhad .. .

Answer 1

C- 2m - 2.50 6 : 22.5 . 0.8330) 27 A .S similarly k 6m - 25) 425 - 1.667 2- Joint D 12, SEN 6m 2.5m - 25 or (6 01 22 . 62°

na cryo FGpxsin ( 22.62) - 12.5=0 | FGD = 32.50 ri cl ans 7 ZFx = 0 FGDYCOS (22.62) - FCDEO 32.507 COS(22.62) - FCD=0 Feo: CokNT ans at Joint G 1 FCG FGD: 32.50Kn G FFG FCG= O KN FGO=FFG: 32.50kN c Ang At Joint C 12.5KN +2m 30kn 02 ☺ 02 72 = 1.667m/ ICE 11.6642 02: täh ( 1,667) 02: 02- 39.811 1+ EfY0 fcrusin ( 39.811-121550 FCF = 19.52kN c 7 Anis

ñ are so 30+ 19.52X COS (39.811 - FBC=0 TFBC = 45 KN T| Ans 12.5KN 12.5KN 12.5KN 12.5KN Sam - le am - /e 2 m / AXA Ау 2.5m of pointe A AdEMA.o Ex x 2.5= 12.582 +12.5% 4+ 12.546 Еу Со к

EFB20 Ax = 60 KN MA EFYzo AY -1205-12.5" 12.5 -12.520 Ay: 50 KN 1125KN D FAB At point A & 6 OKN ↑ Ay=50KN NEXO FAE FAB: 60 KN 1 AnS pot EFY EO - FAE +50 -125 0 FAE = 37.5 kN T Ans

at point B 12.5KN 60114 x > 45 EN 2mg 215ml AFBF Lasotan ( 22 ) FBE 03: 546349 Ey 20 FB EXCOS (51.34) +45-6050 15 FBE: T: 24 KN cosc51-34)

LFBE = 24 kN c) Ans Af Erya FBF † 24sin ( 51.34) -12.5 =0 [FBF = 6.24 KN TI Ans aut point E 37.5 KN 24 KN AF FEF 51.340 04 corto k 04 a 4an ( 0.93) 043 22.610 PEFxzo FEFY COSI 22.611 - 24X Cas ( 51.34 1 4 60:0 FEE: 45 COS(22.61) FEFE 48175 KNC ans