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Unformatted text preview: Use the method of joints to
determine the force in each
member of the truss shown in
Fig. P7—3. State whether
each member is in tension or compression. SOLUTION For this simple truss. the
member forces can be determined without solving
for the support reactions. From a freebody diagram
for joint D: + T 2F? 2 Tan — 3000 = 0 Tan = 3000 lb = 3000 lb (Tl From a freebody diagram
for joint B: I . o_ . ‘o_
+ —a 2Fx — TBC Bin 30 TAB Sln 00 — 0 0 O
+ T 2Fy  —TBC cos 30  TAB cos 60 — TBB — 0 0 0
{1.732 T83) cos 30  TAB cos 60  3000  0 1500.0 lb = 1500 lb (C)
T = 1.?32 TAB: 1.732(1500.01 = 2598 lb 3 2600 lb {C} From a freebody diagram
for joint C: . O
+ —9 SF3  —TCD  TBC 51D 30 . 0 _
—TCD  {2598} sin 30 — 0 TCD = 1299 lb = 1299 lb (T) Ans. From the freebody diagram
for joint D: + —+ 2Fx = TAD + TCD = “TAD + 1299 = 0 TM) 1299 lb = 1299 lb (T) Ans. M Use the method of joints to
determine the force in each
member of the truss shown in
Fig. Pf12. State whether
each member is in tension
or compression. SOLUTION From a free—body diagram
for the complete truss: + C ZMD = 8(2113 cos 45°) + 10:3 sin 15°) — Ayiﬂ cos 450} = 0 8.66? kN = 8.66? kN T .',_'
II 5'
+ T 2F = A — 8 + D
'x' Y Y
= 8.66? — 8 + D = 0
Y
D = —u.66? kN = 0.667 kN i + 4 2F = 10 + Dy = 0
D = 10 kN = 10 kN «— From a free—body diagram
for joint A: + —4 EFX = THE cos 45° = 0
TAE = 0 Ans.
+ T 2F? = T53 + TAE sin 45° + 3.667
: TAB + (0) + 8.667 = O 8.66T kN a 8.6? kN {C} H
II Ans. IIiEIWContinued) From a free—body diagram
for joint B: . 0
+ T EFy — TBC 51n 45 — TAB . 0 __
TBC Sln 45  I8.66?} — 0 —12.257 RN 3 12.26 kN {C} 0
TBE + TBC cos 45 + 10 THE + (—12.2571 cos 45° + 10 = 0 l.3330 kN = 1.333 kN {C} Ans. From a free—body diagram
for joint C: + T 29y —T sin 45° — T sin 45°  3 BC CE —(—12.257} sin 45° — TCE sin 45° — 3 = 0.9433 kN 8 0.943 kN (T) Ans. . o _ 0
Ten + TCE cos 45 TBC cos 45 TCD + (0.9433) cos 45o  (12.26?) cos 450 = 0 9.334 RN 3 9.33 kN C} Ans.
‘ a» \
From a freebody diagram
for joint E: 0
+ /” EFK, TDE — TBE cos 45  TAE 0
TDE  (—1.3330} cos 45  O — 0 0.9426 kN 3 0.943 kN {C} Ans. Determine the forces
in members DE, EI,
and HI of the truss
shown in Fig. P7‘24. SOLUTION From a free—body diagram
for the complete truss: + C EMA = 6(16}  3(4)
— 3(a) — 5(12} = o G = 8.50 kN = 8.50 kN T From a free—body diagram
of joint G: ¢ = tan1 20.550 6 tan" 56.310 0 0
+ —+ 2Fx TFG cos 56.31 ~ TGH cos 20.56  0 1 O . D _
= TFG Sin 56.31 + TGH Sin 20.56 + 8.50 — 0 l3.622 kN H 13.62 kN (C) 8.070 kN = 8.07 kN IT) From a free—body diagram
of joint F: + /; EFx’ = TFH = 0 + E\ SE T + 13.622 = 0
3:" EF TEF 13.622 kN a 13.62 kN (cl 724 (Continued) From a free—body diagram
of joint H: + /” ZFXJ TEH + E\ SE ,
y TR] 0
TEH cos 20.56 — 0 0 THI  8.070 = 0 8.070 kN = 8.07 kN (T) From a free—body diagram
of joint E: + T 2F
y + % 2F
3! E1 DE *6 “'3 ''.
[11 ‘ TDE UIIH'— gum
m
hq 10.557 kN 3 10.56 kN {T} 16.002 kN E 16.00 kN (C) Ans. ' + 13.622 sin 56.310 = 0  13.622 cos 56.310 = 0 Ans. Ans. )LOOZKN E .3
TE! kawi + C3 13.1.2210: ...
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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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