# HW17 - PROBLEM 6.4 Using the method of joints determine the...

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Unformatted text preview: PROBLEM 6.4 Using the method of joints, determine the force in each member of the V ’ M truss shown. State whether each member is in tension or Compression. SOLUTION FBI) Truss= Q 2MB = o: (1.5 In)(,"‘. + (2 m)(1.8 1m) — 3.61n(2.4kN) = 0 C} : 3.36 kN ‘ ‘ 25;. = 0: 5.. +336kN —Z.4kN : 0 B", = 0.96 kN , Joint FBDs: 2 b , _ — JointD: ~>3Fy = 0» ER”? - 2-4 kN — 0 Fm _ 3.48 kN T 4 \‘FLUD Z: ‘I/e/J 21 2" 4’ 2F\':O:FCI)_EF»AD :0 ~27» . F‘ D 2.1 - ~m> Fm : 58.48 kN) F”, : 2.32 kN C 4 By inspection: Ff = 3‘36 kN C ‘ PM. : 2.52 kN c 4 JoinlB: Afng .L/ F5 3“l 4 t X; ’ 2 1 \$21". =02TF13—0.9kN:O Fm=i200kNT4 2:32. 4// ' 3 ‘ ' h [,\ h ',_\ in :1.\ Ex. AV ' ‘ * 4:43; (-0 ha (:a H. ' 1+4 Tun-J-wt m—Je-x IIIJ’4I m—— PROBLEM 6.9 132 m l _,__L lv\\ ’ V1: " an Determine the force in each member of the Gambrel roof truss shown. ,9 . . . . , ‘ i i State whether each member ls 1n tenswn or compressmn. i. 3;“ | SOLUTION FBD Truss: —» 25‘. =0:Hl‘, =0 By symmetry: A = H‘, =12 kN ‘1 b )3 4m 4m-‘L-4m —‘r. /’l \ ism By inspection ofjoints Cand G, FCE = Fur and FBC = 04 Jim—J”! M FEG = FGH and FFG = 0‘ \$213,:0:12kN—3kN—§F/,3=0 F113=15.00kNC< 5 aEFr=0=FAc-§(15kN)=0 FAC=12.00kN T4 4 10 4 >ZF:0:—15kN *7}? _5F :0 " 5( ) 10.44 B” 5 BE 3 3 3 2F,—0: 15kN 6kN F + F 0 T .\ 5( ) 1044 ED 5 BE Solving yields *> 21"} = 0: by symmetry FED =11.93kN c 4 FEE : 0.714 kN c 4 Fm. :11.93kN c4 FDE = 0.856 kN T 4 FEF = 0.714kN c 4 FFH = 15.00 kN c 4 FGH =12.00 kN T 4 FCE = 12.00 kN T 4 JointD: 6"“ 121i,=0:—FDE—6k1\1+2i(11.93kN)=0 D ' 10.44 ///Y‘\‘\ F / \. _ ”.93 I’cN N Br NEE By symmetry: FEF = FEE so FFH = FAB FGI] = FAC From above FCE = FAC FEG = FGII Fm = 12.00 kN T 4 PROBLEM 6.23 SOLUTION FBD Truss: [Cm/11; /I_§Z 2:\1 575.5";an I ‘ H.013 Iw (EMF : 0: (10 mm), — (7.5 mm kN) — (8 m)(30 kN) = 0 Gy=84ka >2F}:0:—F\.+30kN=0 Fx:30kN <— fZﬂ,:0:ﬁ,+84kN—80kN=0 F),:4kNl By inspection ofjoint G: 84 kN _ FCE FAQ. _ 7~~=10.5kN FV-=82.OkNT4 8 J5 J23 Ct FM 2 56.5 kN C 4 1211,: 0:11?” +i(82.0 kN) — 80 kN = 0 ' J3 J5 FM 2 19.01312 FAB = 19.01 kN T4 _. 2R, : 0; —FDE —i(19.013 kN)+ i(82.014N) = 0 ' 13 m FDE = 43.99kN FDE : 44.0 kN T 4 EF—o-iF 750m~0 x 'ﬁ Dr 5 FDF = 67.082kN FEG=O4 Fm =84kN C4 FDF = 67.1kN T4 PROBLEM 6.23 CONTINUED , 2 126‘, : 0: f(67.082 kN) — FBF — 4 kN = 0 FBF : 56.00kN FBF : 56.0 kN C 4 2F~060kN+iF —iF 7 .\‘ ' ‘ m [1D m AB TZF *O-SﬁkN—LF ~iF 70 x y ' M ED 5 AB Solving: FED = 42.956 kN FED = 43.0 kN T 4 FAB = 61.929 kN FAB = 61.9 kN C 4 1 ER : 0: i(42.956 N) + imp 7 67.082 N) = 0 ’ J5 \/§ FAD =30‘157kN FAD :30.2N T4 ...
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