Question

a) Identify all zero force members

b) Use the METHOD OF SECTIONS to determine the member forces in menbers EF, GP, OP. Indicate if the members are in compression or in tension.

2. (a) Identify all zero force members. (5 pt.) (b) Use the method of sections to determine the member forces in members EF, (15 pt.) GP, and OP. Indicate if the members are in compression or in tension. F3 [kN] Fs [kN] = 144 [kW] Fu = 143 [kN] b=6 [m] c = 15 [m NNNN F3 [kN] Fe (kN] (m), 5 (m), (m), (m), 5 (m), 0 [m), o (m), (m, (m), o (in,

Answer 1

Fz=144 KN b=6m fu=143 KN C= 15m Fs [k] F3 (kN) F KN * NNNNN F3 kN! FA (KN) (m) ! (m) 5 (m), b (m), (m) (m) $ (m), (m), (m), b (m), a. Criteria for zero force member: 1. If only two non-colinear member exist at a trus joint and no external force or support reaction is applied to the joint, the members must be zero force members. 2. If three members join at a point and out of them, two are colinear and also no external load acts at joint, third member is zero force member. According to criteria “2”, members : BT, QE, FP, OG, HN, MI, and JL are zero force members

b. By removing zero force members for calculation: F3 [kN] F4 [kN] Es kan MW B Rk RA 'F3 (kN) FA (KN) ” [m], [m], (m), (m), (m), 5 (m) 0 (m), (m), (m), º (m), Vertical Reaction at support “A” and “K”: EMA Taking Moment about support “A” =0 F * 2b + F * 2b + F2*3b + F2*3b = RK *10b putting the values, 144*2*6 + 144*2*6 + 143*3*6 + 143*3*6 = R *10*6 Solving the equation: Rk = 143.4 KN Now, ΣF, = 0 RA+RK = 2*F; + 2*FA R + 143.4 = 2*144 + 2*143 A R. = 430.6 KN A

cutting section "XX" and considering left side of the section F3 (kN) F4 [kN] AW B Rk RA 'F3 [kN) F4 (kN) (m, (m), 5 (m), ø [m], [m], (m), o (m), (m), (m), (m), * *Note : Assuming force on the members to be tensile if the value comes out to be negative member force will be compressive. F3 kN) F4 [kN] Esk •Feo FPG N (u ? 0 FEF RA 'F3 (kN) F4 (KN) [m) $ (m) (m), 5 (m), 5 (m) }) (m) 0 [m), 6 (m), 0 [m)(m), Taking moment about “O” EM, = 0 RA*6b = F3*4*b + F *4*6 + F2*3*b+F_*3*b + FEF *C 430.6*6*6 = 2*144*4*6 +2*143*3*6 + F 15 F = 229.44 KN O EF EF

EFX = 0 Fpo + Fe = 0 Fpo = - FEF Fpo = - 229.44 KN (-ve sign means compressive force) F3 (kN) | FA (KN) Epo FPG PG = 152 +62 PG=16.16m Cos O= 6 16.16 FEF RA F3 [kN] Fe (kN) ♡ (m), (m), 5 (m), 5 (m), ) [m], $ {m}, {m}, (m), (m), 5 (m) EF, = 0 RA = Fz+ Fz + F + F + Fpc cos e 430.6 = 2* 144 + 2* 143 + Fpc * (6/16.16) F = -386.224 KN (-ve sign means compressive force) PG so, וד GP PG = -386.224 KN or 386.224 KN (Compressive) F = F = -229.44 KN or 229.44 KN (Compressive) F = 229.44 KN (Tensile) = OP PO - EF -