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ProbSet9_Solutions - _‘ 3m Joint FBDs Joint B “Etc...

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Unformatted text preview: ._‘ 3m Joint FBDs: Joint B: “Etc BJ:1§'N PROBLEM 6.1 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. {2M3 = 0:(6.25m)CJ, —(4m)(315N) = 0 Cy = 2401“? IZFy=0:By—315N+Cy=0 By =75Nf HEFI=Or B =0 X FAB =125.0 N c 4 By inspection: F AC = 260 N C 4 PROBLEM 6.7 Using the method of joints, determine the force in each member of the truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: szy=0:Ay—480N=o Ay=480NI ./ KEMA=0:(6m)DX=O D,=o —+E.Fx=0:—Ax=0 A.=0 1 Joint FBDS: JointA: 55mm»; fl = 55—3 : fic— FAB = 200 N c 4 M—w 4379.“! 6 2'5 6‘5 Jam?" ii: A g; f“ PAC = 520N T4 is 154; 200N=h=i F =480NC< 2.5 6 6.5 ”5 FBC =520N T4 PROBLEM 6.7 CONTINUED 3“ ‘ By inspection: FCD = Fag = 520 N T 4 {2010 519ml c fix: art: a" 2 5 “2}? =0. I 6|5(520N)_FDE=0 FDE=200NC4 1.5m 1.5111 1.5111 1.5m 1.5m L5rn PROBLEM 6'12 Determine the force in each member of the fan roof truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: *ZFX =0 : Ax = 0 Bysymmetry: Ay 21y =12kN ? and FAB = Fm; FAC = F61; FBC : FGH F313 = Fm; FDC = FFG; Fee = FEF Fee = F56 Joint FBDs: JomtA: FM @4— : 3% = 1% FAc = 22.5 kN T 4 2:4! q].— 1:526 ”HIV so F5! = 22.5 kN T 4 {:15 Tu W FAB = 2.5JsfikN FAB = 24.6kN c 4 so Fm = 24.6kN C 4 Joint B: 9 _. 2F): : 0: 22.5m — 79—;(FBD + FBC) = 0 1sz = 0:10m —4kN+—J%(FBC —Fs.o) = 0 2 57%“? EN Solxring'. F30 = 19.70 kN C 4 so FFH =19.70 kN C 4 and FBC = 4.92kN C 4 Ioint D: so FGH = 4.92 kN C 4 7%” F' By inspection: FDE = 1930 m C 4 ’” 0‘ so FEF =19.70 kN c 4 and FCD = 4.00 kN C 4 [170 [ZN Feb so Fm = 4.00m C 4 PROBLEM 6.12 CONTINUED Joint C: —. :F, = o: —22.5 kN + 3%(4‘92 m) + £17“ + Fm = 0 $21: = 0: —4 kN - $91.92 kN) + flFC = 0 ’ J97 5 5 (1%” E5? Solving: FCE = 7.50 kN T 4 WM’RI .J" 30 Fm = 7.50m T 4 " .3 ' and FCC = 13.50 10“ T‘ PROBLEM 6.15 Determine the force in each member of the Pratt bridge truss shown. State whether each member is in tension or compression. SOLUTION FBD Truss: fly em em e‘kN fl). FBC = FFGQ Fan = FDF By inspection of joint D: FDE = 0 ‘ FBDs Joints: 4 TEFy=0:9kN-§FAB:0 FAB=1125kNC4 m za=0:FAc—%FAH=0 E4C=6.75kN T4 ——EFx=0:FCE*6.75kN=0 FCE=6.75kNT4 53,, IEFy=0:FBC—6kN=0 FBC=6.00kNT4 C F M " 4 wskN “‘5 125V =0;3(11.25kN)—6kN+—FBE = o FEE = 3.75 kN C ‘ B 3 - 3 q 3 £130 ”21-; = 0: FED - £01.23 kN) — §(9.75 kN) = 0 q __|.+ was-kw (Am 585 F50 = 9‘00kN T 4 From symmetry conditions above FFH = 11.25 kN C 4 FGH = 6.75 kN T 4 FEF =3.75kN C4 FDF = 9.00m T 4 PROBLEM 6.23 m Determine the force in each member of the truss shown. State Whether each member is in tension or compression. SOLUTION FBD Truss: Q EMF = 0: (10 m)Gy — (7.5 m)(80 1m) — (8 mm kN) = o Gy=84kN1 ~2Fx=o:—Fx+30kN=0 Fx=30kN ~— szy=0:g,+84kN~80szo Fy=4kN1 By inspection of joint G: F50 = 0 1 FCG :84kN C‘ fltiici idosm FCE:82.01<NT4 sJfizs FAG = 56.5kN c 4 12F :11: iFAE+i(82.0kN)—80kN= 0 y J3 J5 FAE = 19.01312 _. 2F: = o: 4‘05 — i(19.013 kN) + %(32.0 kN) = 0 J5 J— FADE = 43.99 kN FDE = 44.0 kN T 4 FDF = 67.082kN FDF = 67.1kN T 4 PROBLEM 6.23 CONTINUED 1213. = 0:1(67082m) — PM — 4m : 0 J5 FBF=56.UOkN FBF=56.01CNC{ —— 212x =0:30kN+—%1~FBD———— @sz =0:56kN——§6TFBD-——— Soiving: F30 = 42.956 kN F30 = 43.0 W T 4 6 2 2F = 0: ——-— 42.956N —- F — 67.082N = 0 T y {H61( )+ \g( AD ) ...
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