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Unformatted text preview: Use the method of sections to determine
the forces in members
BC, CF, and EF of the
truss shown in Fig.
P749. SOLUTION For this simple truss, the
member forces can be determined without solving
for the support reactions. From a freebody diagram of
the part of the truss to the
right of member BF: + C EM C = —TEF{8 sin 30°}  1500(3 cos 30°) = o TEF = 2593 1b a 2600 lb (0) 0 . 0 . O 0
+ C EMF  TBC cos 3O {8 sin 30 } + TBC s1n 30 {8 cos 30 l
— 250013 cos 30°) — 1500(16 cos 30°) = 0 TBC = 5500 lb = 5500 lb (T) + C ZMD = TCF cos 300 (8 sin 30°} + TCF sin 30° {3 cos 30°) + 2.500(3 cos 30°} = 0 CF = 2500 1b = 2500 lb (C) Use the method of sections to determine the forces in members 5‘: 10"
CD and FG of the K’ 10“
u . G al'F
K—truss shown 1n F13. ‘L ‘1» ‘L :1:
Isn lsn um Isn 5n un_ P7—53. (Hint: Use a . . 20mm 30mm 500mb '
sectlon aa.) Fm.P%53 SOLUTION From a free—body diagram
for the complete truss: + C EMA = 3190)  2000(30)
 3000(45)  5000(75} = 0
B = 6333 lb = 6333 lb T
From a freebody diagram of the part of the truss to the
right of section a—a: + C EMU = TFG(20) — 5000(15} + 6333(30) = 0 TFG = 5?50 lb 2 5750 lb {T} Ans.
+ c EMF = TCD(20} — 5000(15) + 6333(30) = o r = 5750 lb 3 5750 lb (C) Ans. CD CHI." Find the maximum load P
n that can be supported by
the truss of Fig. P7—55
without producing a force
of more than 2500 1b in
member CD. Use the method of sections. SOLUTION For this simple truss, the
member forces can be determined without solving
for the support reactions. From a freebody diagram of
the part of the truss to the
right of member CG: + C EMF = TCD{4)  19(10): 0 P = 0.400TCD 2500 lb: P = 0.400I2500) = 1000 lb max ...
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This homework help was uploaded on 04/16/2008 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.
 Fall '06
 Anderson

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