Question

5. For the truss shown below, what are the reaction supports at the pin support A, and the roller support D? Determine the forces in all of the members and state whether they are in compression or tension or zero-force members. (20 pts.) 3 ft 3 ft 3 ft F E 8 kip 5 ft B 5 kip

Answer 1

3 ft 3 ft 3 ft E 8kip RAX B RAY 15kip PRO { Freo 8 - RAX=0 Rax= 8 kip K & Fyzo Ray & Ro-5=0 Ray tro= 5 -i) § MA=0 Rox (3+3+3) 8x 5 – 5x 3 = 0 - Ro= 8k5+ 5x3 = 6.11 kip Ro= 6.11 kip RAY tro = 5 RAY= 5 - 6.11

Ray= -l'll kip Ray = I'll kip > Reaction at A RA=J RAY +RAY 82 +1112 RA= 8.077 kip Ro= 6ill kip Forces in each meambers using method of joint at Joint A FAF 53 tano = 5 3 Q = 59.036 8 = RAK FAB bay=-1.11 & Fy=o. AF sind & Ray=0 Far & sin 59.036 – 1.ll=0 (tension FAF = 1.2957 kip

& Ex=0 FAF coso + FAB FAB - RAX=0 1.2957 cas 59.036 + FAR - 8 = 0 FAB=7.33 kip I tension) at Joint D FED FER Ro=6'll kip T Boa & Fy=0 FER sino + Rp = 0 FED sin 591036 et 6.ll=0 FEDE - 7.125 kip FED= 7.125 kip (Compression) & Ex=0 FED caso & FLD = 0

- 7-las cos59.036 + Fcp=0 Feo= 3.665 kip (tension) Joint c A FEC FBC L > FCD at EFyzo FEL=0 & freo FED - Focao FIDE FBC = 3.665 kip (tension) at Joint B FBF FBE FAB FBC B 5kip

e Ex=0 Foc & FBE caso FAB=0 3.665 + FBE cas 59.036 7.33=0 I FBE = 7.123 kip (tenston) z Fyzo FBF + FBE wino -550 For +7.123 sin 59.036-5=0 For=-1.107 kip For = FBF = 1.lotkip (compression) at Joint FFE La Fark VFBF & Fxzo FFE - FAF Caso = 0

FIE - 1-295A cos 59.036-0 TFFE 0 666 Kip (tension) member. force (kip) 1.2957 Kip ł T) AB 7.33 KIRP BE 1.107 Kip (T) (C) (T) (T) BE 7.123 KIP 3.665 KIP BC O CE (p 3.665 KIP (T) (T) EF 0.666 kip 7.125 kip (C) ED
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