HW 19 Prob extra 6_28

# HW 19 Prob extra 6_28 - w a“ y y“ I l[‘,y‘r 34:1“...

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Unformatted text preview: w a“, y, , y“ , I, ,,, l [‘ ,y‘r 34:1“ PROBLEM 6.28 4 “173164 mgr-4 m'Ij‘T-él m Determine the force in each member of the truss shown. State l whether each member is in tension or compression. 5 . \m SOLUTION Free body: Truss ZFx = 0: H, = 0 +§2MH=Oz 48(16)-G(4)=0 G=192ka +1211, =0: 192—48+Hy =0 Hy =~144kN Hy =144kNl' Zero-Force Members: Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC, BE, DE, DG, F G Thus, FBC =FBE :FDE :FDG = FG = Also note: FAB =FBD =FDF =FFH (1) FAC=FCE=FEG (2) Free body: JointA: ﬁ=ﬂ£ =ﬂ 8 J73 3 +8 m E *z ~ .~. :a3 A: r A 31 {1" 48%! 3L3 3 \.._FA¢ - .— 31 g as FAB =128 kN " FAB =128.0kN T 4 FAC = 136.704 kN FAC =l36.7 kN C 4 From Eq. (1): FED = FDF = FFH =128.0 kN T 4 From Eq. (2): FCE = FEG =136.7kN C 4 PROBLEM 6.28 (Continued) F 144 kN Free body: JointH Jigs" = 9 FGH =192.7 kN C 4 F I44- kN Fm 4:»; 8 /H {44%} Q m 4M'SL/Jq F Ecm/ 8 -G“ FFH _144kN Also 8 — 9 FFH=128.OkN T (Checks) ...
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