Question

Using the method of joints determine the force in each member of the truss shown. Indicate whether the member is in tension or compression. F1=3550lb 

Write the final answer here and upload the file with step by step calculations. 

Answer 1

1.

2.

Preaction forces 8 Ft- .B SH=0 >F = 355016 RAH +3550 -0 RAH = -3550 16 -@ Since negative change the assumed Rau 2oft duection. RAN Ev=o RAY + RC -0. RAV=-Rc B 2M@A=0 (Rex 20) = (355045) do. Re: 17750 Rc-887.5 16 :-) Substitute into San Trava -887,15161 Since negative change the assumed direction.

3.

B > 35solb AD From Triangle ABD tano = opp BD 5 ads 8 tan o= 0.625 Os toni(0.625) 0=32 5fl- 3550lb 70 А D. 8ff- 1241 887.5 lb 88151 From Triangle BCD tanda opp 12 adj tano 0.4167 Os tan co.u167) Os 22.62 BD OC Joint Air FAB FAB. Sin32° ZV-O 3550 lb 320 3550 * FAC. FAB Cos 32° FABE bin 32 - 8875 FAC 887-516 FAB = 887.5 Sin32º FAB = 1674.78 1b Elo FAS + FAB COS 32 - 3550 FAC + 1674.78XCOS 32°= 3550 FAC = 3550 - 14420.29. TFAC = 2129.7115

Joint a FEB Sin 22.62² FB Vco FCB Cos 22.629 22.12 FEBX Sin22.62 +887.5.0 FEA FCA 1887.5 887:5 8815 FEB - 8815 Sin 22620 TECB: ~2307.4816 BH=0 FA + FEB Cos 22.62 to 2129.71+ FBC8522.62*- FEB- 2129.71 Cos 22.62 FEB - 2307.48161 Member Force Nature AB 1674.7816 Tensile BO 2307.41816 Compressive. AC 2129.711b Tensile.

Answer 2

softi 12ft F = 3ssolo 03 08 A 30 Body Diagrams) with attest I knows B 20ft Metod Method of Joints i} Draw FBD (Free Joint B has I known FABsinger FAB FCB By force equilibrium -, {foto & Elyo we need to find angles O, O2, O3 first § - %, = sint (**) = 3668° 38-68° 5/2) Oz= sint (32) Fi=35solf You 031---F sino, a sin Oz 24.62° 180° . Efx=0 »7 03 = 180°-90°-0, 180°-90° - 38.68° = 51.32° 04 180° – 90°- Oz = 90° 24.62 65.38° This clirection is taken positive Fi & FcB sin Ou FAB Sin Oz = 0 Fus FAB sin Oz .fi FAB x 0.78 0.909 3550 こ sino 4 » Fee 0.416 {fy=0 pt -FcB cos O4 FAB Sos O3 = 0 FCB cos Os FAB Cos Oz - FAB X 0.624 on equating ® & © we get 0.78 FAB - 3550 0.624 FABB 0.909 0.416 0.324 FA1 -1476-8 = 0.567 FAB FAB = 1476.8 +1657.46 lb tre indicates Tension pour direction 0.891 indicates & Fib 1657.46x0.624 =-2486.19 lb arrow assumed should be reversed compression 0.416 - ve

is the answer FBD =) of B B Fr=3550 lb FAB 20 F CB Fcb was negative it will point towards alane Fee Now pt. c has 1 known so let's choose pt. c FBD for point C Rc is reaction force pt./joint c { Fe = 0 Fuecos FAC 2260-17 lb tre indicates Tension & our Sign Larrow is correct, Here Fee the value is taken as its sigh is already considered while drawing F80. FAC t - AC & Rc cos Oz FCB cos Oz i 2486-19 x cos 24.62 B 2486-19 lb Compression 1657.46 lb tension A A Tension 2260.17lb

The answer is in the last diagram