This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. Use the method of joints to
determine the force in each
member of the truss shown in Fig. P77. 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 C: + T 2Fy = TBC = 0 T = 0 BC Ans. From a freebody diagram
for joint B: 0
AB cos 45 + TBD , 0
AB Sln 45  TBD + —+ 2F = —T
X T + T 2F =
v Solving yields: TAB = TBD = 141.42 lb 8 141.4 lb (C)
989.95 lb 3 990 lb (C) From a free—body diagram
for joint D: ' + —9 XFX = T  T CD BD CD T = CD 700 lb = 700 lb (T) From the free—body diagram
for joint C: + T = _T c0 AC
700 lb = 700 lb (T) T + 700 = 0 +—)2Fx= Ac TAC = Ans. Ans. Ans. Ans. w. F‘ cos 45° + 600 = 0 sin 45°  300 = o RILEY AND L. D. cos 45° = —T  (—939.95) cos 45° = o STURGES ENGINEERING MECHANlCS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Use the method of joints to
determine the force in each
member of the truss shown in Fig. P79. 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: + “\ 2F  T sin 30° — 1000 sin 60° = 0
y 30 TBD 1732.05 lb a 1732 lb (C) ' Ans. 0 °
+ /” 2Fx TCD  TBD cos 30  1000 cos 60 TCD — (—1732.05) cos 30° — 1000 cos 60° = 0 TCD 1000 lb = 1000 lb (T) Ans. From a free—body diagram
for joint C: + /” EFX = TAC + TCD = TAC + 1000 = 0 TAc {1000 1b = 1000 lb (T) + “\ EFy TBC = o TBC = 0 From a freebody diagram
for joint B: 0 0
+ + 2Fx  TAB  TBe cos 60 + TBD cos 60 0
TAB  (0) + (1732.05) cos 60 866.03 lb 3 866 lb (C) Ans. 5)7 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Determine the forces in
members BC, CD, and DF of the truss shown in Fig. P716. State whether
each member is in tension
or compression. SOLUTION From a freebody diagram
for the complete truss: + C EMA = Ey(8)  10(4)  8(6) = o Ey = 11.00 kN = 11.00 kN T From a free—body diagram
for joint E: + T 2F T sin 30° + 11.00 = o
y DE. TDE = 22.00 kN = 22.0 kN (C) BKN ll! for joint D: T g l
c 6 ’
= T cos 30°  8 cos 300 = 0 D ’
DF I
8.00 kN = 8.00 kN (C) Ans. To; ,’D
\ ,go/ 706‘” From a freebody diagram "  ‘ ’X . 0 . O
 TDE — TCD + 8 sin 30  TDF sin 30 (—22.00)  TCD + 8 sin 30° — (8.00) sin 30° = 0 14.00 kN = 14.00 kN (C) Ans. From a freebody diagram
for joint C: 0 0
+ —+ ZFX — TBC cos 30 + TCD cos 30 TBC cos 30° + (:14.00) cos 30° = 0 14.00 kN = 14.00 kN (C) Ans. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES w A A r r mass1:1"; m 7—43* The length of each truss
member in Fig. P743 is
5 ft. Use the method of :l ' sections to determine the itf‘""" forces in members CD, DE, ' lOOOlb 8001b 6001b 8001b I
a) and EF' Fm.PL43 SOLUTION From a free—body diagram
for the complete truss: o goo 1.00 300
")1 n, )1. )L
A = 0 A = 0 X X C D
= Ay(25) + 1000(20) ' + 800(15) + 600(10) + 800(5) = 0 = 1680.0 lb = 1680 lb T From a freebody diagram
of the part of the truss
to the left of member DF: “£34? “an A'a,».;,.,,sm',.r..:,.«.J;n «em “a ,5 , t t1 . z + c 2MB = 1000(5)  1680(10)  TCD(5 sin 60°) = 0 TCD = 2725 lb a 2730 lb (C) .14.,ng Lyn.1) y + C 2MB = 800(2.5) + 1000(7 5) — 1680(12.5) + TEF(5 sin 60°) = 0 TEF = 2656 lb 2 2660 lb (T) T + EFy = TDE sin 60° — 300 — 1000 + 1680 = 0 "DE = 138.56 lb 3 138.6 (T) 545 759* Use the method of
sections to determine
the forces in members
CD, GK and LK of the
Howe roof truss shown
in Fig. P7—59. SOLUTION From a free—body diagram
for the complete truss: + —e ZFX A = 0 A = 0 X X + C EMG = Ay(36) + 1000(30) + 600(18) + 800(12) = 0 AV = 1400.0 lb = 1400 lb T From a freebody diagram of
the part of the truss to the
left of member DK: 29.05° 48.01° r
‘
'n
i
n
V 1400(18) + 1000(12) . O O _
TCD s1n 29.05 (6)  TCD cos 29.05 (6.667)  0 1510.0 lb = 1510 lb (C) 1000(6)  TCK cos 48.01° (6.667)  TcK sin 48.01° (12) = 0 448.45 lb E 448 1b (C) Ans. TLK(6.667) + 1000(6)  1400(12) = 0
1619.9 lb 2 1620 lb (T) m ENGINEERING MECHANICS  STATICS, 2nd. Ed. 768 Use the method of
sections to determine
the forces in members
CD, DF, EF, and FG of
the transmission line truss shown in Fig. SOLUTION From a freebody diagram
for the part of the truss
above member CG. + C EMD = 5 cos 30° (6)  TFG(3) = o TFG = 8.660 kN = 8.66 kN (T) Ans. O
+ C ZMG = TCD(3) + 5 cos 30 (9)
— 5 sin 30° (4) = o TCD = 9.657 RN 3 9.66 kN (C) From a freebody diagram
for joint F: 1 4 e = tan' 5 = 53.13° sin 51.13  8.660 = 0 + T XFY : TEF TEF = 10.825 kN E 10.83 kN (T) Ans. T  T cos 9 BF EF
0
 TDF  10.825 cos 53.13 — 0 6.495 kN z 6.50 kN (C) Ans. WI F. RILEY AND L. DI STURGES ...
View
Full
Document
This note was uploaded on 07/16/2009 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.
 Spring '08
 McVay

Click to edit the document details