Question

a) For the truss shown below identify all zero force members. (Circle) (10 Points) b) Determine the forces in members BE DE, and AB. (15 Points) y (4) 9 (4)-12(8), 9K 12 KN 4 m join 3m • 35.25 KM 0 3m 12 +35.25 + <DE TBy 35.25 АЧ BEST STU

Answer 1

Solution 9KN 4 m 12 KN 4m -> 3KN 3m FBD 12KN am f am 361869A Зm lo V 36.00 53.13070 36.869 RUB Ex=0 RAH+ = 0 + VE) RAH = - 3 KN k-ve) Efy=o ( +ve) RAVTRB - 9-12=0. RAVERB= 21 KN - 1 EMaso (+)) (clockwise is taken positive)

9x4+1988 - Rgx4=0 A Rp=36+96 Rp = 132/4 = 33 KN (T+ve) Rav= 21-33 = – 12 KN (-vel Taking point & 112KN ЗKN 36.869'y ve > 3KN PREB LE EX=0 (→ tve) 3- FED-FEB COS 36.889=0 Pep +0:00 FEB=3 til EX=0 (4)tve -12 - FER SIN 36.069=0 --12-0.60 FEB=0 FEB=- 20 kN (A-ve) Ang then FED-gg ( tve) (Ans) 19 KN Taking point B ABD 53.130 FBE FBA AB 33KN

Ex=0 (→ tve) FBE Sin 53.130 - FBA=0 - 20 sin$3:130 - FBA=0 FBA=- 16 KN & Ang Ey=0 (alive 33+ FBD+ FBE COS53:130=0 33+FBD - 2000953:130=0 33 + F5 D-12=0 FBD = - 21KN ( to-ve) Taking point a 9кл 36.89 CORV FB Efy=0 altue) -CP-9-for-frosim 36.869= 0 -9+91- FAD Sin36-04 = 0 FAD = +20 KN Efn=o tevel ADE-FOC- Fon Cos 36.869=0

19 - Foc - 2080:00=0. - FDC=3 Foc= -3 KN Taking point A FACA FOA 32.661 BA ZFy=0(1)ve Ravit Fac+FOA Sinb 36.869=0 -12+ fac+dosim 36.86920 - AFAC+x=0 FAC=0 FBE = -20 KN FDE = 19 KN FAB = -16 KN And