Question

Given the magnitudes of the applied forces on the truss shown below are 0 = 7 kN, P = 3 kN, Q = 1 kN, R = 6 kN, S = 4 kN, T = 7 kN and U = 5 kN. O KN PKN TO KN JR KN IS KN IT KN U KN 0.4 m 0.8 m ** 0.8m →+0.8m → 08 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m How many zero force members are there in this truss? 01 2 or more Determine the forces in members EF, CF and EG. Answers (tolerance 0.5 kN, observe tension positive sign convention): FEF = Number Units FCF = Number Units FeG = Number Units

Answer 1

7 KN 3 kN 6 kN 4 KN 7 kN 5 KN B NAT 0.4 m 0.8 m +0.8m 0.8m → 0.8 m 0.8 m 0.8 m 0.8 m 0.8 m

FBDvof the system .

from equilibrum equation, EM₂ = 0 - Pax 4.8 – 7x4.8 – .3x4.- 1x 3.2 - 6x2.4 - 4x1.6 - 7x018 = 0 1. Ron 15:667 kW Now : Consider FBD of cut section :- I len 0.4m corim t 0.80 15.667 kn from geometry tanca= 04 0 = 26.565°

from equilibruin equation, EMELO Pax 1.6 – 7x1.6 - 3x0.8 - E x0,4 = 0 CA . 15667 x 2.6. – 17X16 = 3x0.8 - fefxolk zo Fur 28667 kW Hensies) - Now, Sty=one 15667 - 7 - 3-1 - Fof Sm 0 = 0 =0 15:667 - 7-3-1. Fer Sin 26.565 . for 10:436 kw Flemasan) - Sfx=0 Fea + FER Coso + FCF = 0 TEG A 10.436 40 26.565 +28.667 =0 FEG = - 38 KN 1 li-e compression)

final Answers zero force member - 2 or more Link A. CR. IK are zero force members forces in members - 7 FEF fer = 10-436 . kN 28.667 kN - 38.00 kn FEG =

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