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Unformatted text preview: PROBLEM 6.4 Using the method of joints, determine the force in each member of the V ’ M truss shown. State whether each member is in tension or Compression. SOLUTION FBI) Truss= Q 2MB = o: (1.5 In)(,"‘. + (2 m)(1.8 1m) — 3.61n(2.4kN) = 0
C} : 3.36 kN ‘ ‘ 25;. = 0: 5.. +336kN —Z.4kN : 0 B", = 0.96 kN ,
Joint FBDs: 2
b , _ —
JointD: ~>3Fy = 0» ER”?  24 kN — 0 Fm _ 3.48 kN T 4
\‘FLUD Z: ‘I/e/J 21
2" 4’ 2F\':O:FCI)_EF»AD :0
~27» .
F‘ D 2.1 
~m> Fm : 58.48 kN) F”, : 2.32 kN C 4 By inspection: Ff = 3‘36 kN C ‘ PM. : 2.52 kN c 4 JoinlB:
Afng
.L/
F5 3“l 4
t X; ’ 2 1 $21". =02TF13—0.9kN:O Fm=i200kNT4
2:32. 4// ' 3 ‘ ' h [,\ h ',_\ in :1.\ Ex. AV ' ‘ *
4:43; (0 ha (:a H. '
1+4 TunJwt m—Jex IIIJ’4I m—— PROBLEM 6.9 132 m l
_,__L lv\\
’ V1: " an Determine the force in each member of the Gambrel roof truss shown. ,9 . . . .
, ‘ i i State whether each member ls 1n tenswn or compressmn. i. 3;“
 SOLUTION
FBD Truss: —» 25‘. =0:Hl‘, =0 By symmetry: A = H‘, =12 kN ‘1 b )3
4m 4m‘L4m —‘r.
/’l \ ism By inspection ofjoints Cand G, FCE = Fur and FBC = 04
Jim—J”! M FEG = FGH and FFG = 0‘
$213,:0:12kN—3kN—§F/,3=0 F113=15.00kNC<
5
aEFr=0=FAc§(15kN)=0 FAC=12.00kN T4
4 10 4
>ZF:0:—15kN *7}? _5F :0
" 5( ) 10.44 B” 5 BE
3 3 3
2F,—0: 15kN 6kN F + F 0
T .\ 5( ) 1044 ED 5 BE Solving yields *> 21"} = 0: by symmetry FED =11.93kN c 4 FEE : 0.714 kN c 4 Fm. :11.93kN c4 FDE = 0.856 kN T 4
FEF = 0.714kN c 4
FFH = 15.00 kN c 4
FGH =12.00 kN T 4 FCE = 12.00 kN T 4 JointD:
6"“ 121i,=0:—FDE—6k1\1+2i(11.93kN)=0
D ' 10.44
///Y‘\‘\ F
/ \. _
”.93 I’cN N Br
NEE By symmetry: FEF = FEE so
FFH = FAB
FGI] = FAC
From above FCE = FAC
FEG = FGII Fm = 12.00 kN T 4 PROBLEM 6.23 SOLUTION FBD Truss: [Cm/11; /I_§Z 2:\1 575.5";an
I
‘ H.013 Iw (EMF : 0: (10 mm), — (7.5 mm kN) — (8 m)(30 kN) = 0 Gy=84ka
>2F}:0:—F\.+30kN=0 Fx:30kN <—
fZﬂ,:0:ﬁ,+84kN—80kN=0 F),:4kNl By inspection ofjoint G: 84 kN _ FCE FAQ. _ 7~~=10.5kN FV=82.OkNT4
8 J5 J23 Ct
FM 2 56.5 kN C 4
1211,: 0:11?” +i(82.0 kN) — 80 kN = 0
' J3 J5
FM 2 19.01312 FAB = 19.01 kN T4
_. 2R, : 0; —FDE —i(19.013 kN)+ i(82.014N) = 0
' 13 m
FDE = 43.99kN FDE : 44.0 kN T 4
EF—oiF 750m~0
x 'ﬁ Dr 5 FDF = 67.082kN FEG=O4 Fm =84kN C4 FDF = 67.1kN T4 PROBLEM 6.23 CONTINUED , 2
126‘, : 0: f(67.082 kN) — FBF — 4 kN = 0 FBF : 56.00kN FBF : 56.0 kN C 4
2F~060kN+iF —iF 7
.\‘ ' ‘ m [1D m AB
TZF *OSﬁkN—LF ~iF 70
x y ' M ED 5 AB
Solving: FED = 42.956 kN FED = 43.0 kN T 4
FAB = 61.929 kN FAB = 61.9 kN C 4
1 ER : 0: i(42.956 N) + imp 7 67.082 N) = 0
’ J5 \/§ FAD =30‘157kN FAD :30.2N T4 ...
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 Spring '08
 Jenkins
 Statics

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