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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 65(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 __.+ waskw (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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 Spring '08
 VALLE

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