Question

11: 2. For each truss below, determine the forces in all of the members marked with a checkmark () using the method of sections. b) KN kN 90 kN CNA NIz 60 KN 100 60 KN KN

Answer 1

Page.No.2 --> 40 KN k 4m x 14m I gokn 40 KN EME=0 - FOF X 5-740*5 = 0 FDF = 40KN FEE y gokN {MD=0 + FLEX 5 + 90X4=0 FE = 72 KN fano = 54 EV=0 8-51.340 FDESino-go=0 >> FOE = 90/sin 51-34 = 115.296 FDE = 115.256 Note: -- Negative sign represent Positive sign represent compresion axial force and tension axial force. in the member. OTCC

o Page 2 boka 02 80KN D um 3m Sho 3m x 3m Hand = 4.__ => 9=69.444° QUAN Cut section -. (Take upper part) EME=0 - Fro coso x4 -Frosinox3 +60 caso X3.-60sino x4 =0 >-FFD (46050 +1.5 sino) +60 (1.5cos® — 4 sino) –o 3m

{mg=0 - Page 4 - FDE X4 -136667X3 (38:33 3 ) x8 +60 sino x 8-0 >> FOE = (-136667x3 +38.333 cos69-444x8 +60 sinc9-444) X Fox = 367785 KN um 60KN 100KN 180KN SOKN 60KN , {MA=0 - Gyx (4x6) -60X6+100x4 +60X12 = 0 Gy = 31-667 kN &H=0 SV=O Ax=60KN Ay + Gy = 100t60 - 160 Ay = 28-333 kN Cut section 0.1 (Take Left part) {MI=0 +FB C X 6-10084 - Ax*6+Ay x8=0 → FBC X6=400+60X6–128-333x8 OON FB C = -44.444 kN] Pero V om Fillo um Чm 0=5631

Pages EMB=0 -FHTX6-Ayx4 =0 FHI --Ay x 4 = -12833384 - = -85.555 FHU = -85.999 KN {V=O tay - FB 1 sino - 100=0 Agers FBI = (ny=100) = 128:353-00) – 34052 sino sing631 + FB 1 = 34.052 KN Cut section (2-0 {'Me=0 - EFX6–31667X4=0 [ GOKN For = 21.111 kN] 31.6676N <h=0 FGF + FLK-60=0 Tak = 60+TGF = 60-211111 = 38.889KN | Elx = 38.889 KN) LV=O FFL & Gy=0 LFFL= -31667 kN)