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Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 776 A wire is stretched tight between two pylons, one of which is shown
in Fig. P776. The 1.50kN force is parallel to the xy plane and
makes an angle of 20° with the
yaxis. The supports are
equivalent to a balland
socket joint at B and short
links at A and C. Determine
the support reactions and
the force in each member of this space truss. SOLUTION From a freebody diagram
for joint D: Z? = TAD + Tao + T + F CD D A A _ O _ O _ 0 _ 0
 TAD(cos 45 1 cos 45 E) TBD E + TCD(cos 45 J cos 45 E) A + 1.5(cos 700 1  cos 200 j)
= (0.7071TAD + 0.5130) 1 + (0.7071TCD  1.4095) J (o.7o71TA — TB — 0.7071TCD) E = O D D Solving yields: TAD — 0.7254 RN 9 0.725 kN (C) Ans. TBD 0.8966 RN 2 0.897 kN (C) Ans. TCD 1.9934 RN 2 1.993 kN (T) Ans. ~—N
‘7 ~
— \ .4
!_
DD ) From a freebody diagram
for joint A: ‘ \
1’ 5*
U
\ SE = TAB + TAC + TAD + A at
\
N A 0A 0A
TAB 1 + TAC(—cos 45 1 + cos 45 J) (0.7254J(cos 45° 3 + cos 45° E) + A2 E [TAB — 0.7071TAC+ 0.5129) 1 + (0.7071TAC) J + (0.5129 + A2) E = O ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 776 (Continued)
Solving yields:
A2 = 0.5129 kN
K a 0.513 E kN Ans.
= 0 Ans. 0.5129 kN Q 0.513 kN (T) Ans. From a freebody diagram
for joint C: EF=TAC+TBC+TCD+C A = 0  TBC J + (1.9934)(cos 45° 3 + cos (Cx) 1 + (1.4095  TBC) J + (—1.4095 Solving yields:
= 0
= 1.4095 kN
C a 1.410 E kN Ans. TBC = 1.4095 kN H 1.410 kN (C) Ans. From a freebody diagram
for joint B: 2F=TAB+TBC+TBD+B = (+0.5129) i + (1.4095) J + (0.8966) R + 13x 3 + By 3 + Bz E (+0.5129 + Bx) i + (1.4095 + By) 3 + (0.8966 + Bz) E = U Solving yields:
= 0.5129 kN
1.4095 kN 0.8966 kN B a 0.513 E + 1.410 3 + 0.897 E kN ENGINEERING MECHANICS  STATICS, 2nd.  ' F. RILEY AND L. D. STURGES 780 The space truss shown in
Fig. P780 is supported
by a ballandsocket joint
at A and by short links at
B and C. Determine the
support reactions and the
forces in members AB. BC,
BD. and BE of the truss. SOLUTION From a freebody diagram
'for the complete truss: A (2.75 E) x (B 3 + B 3) + (1 i) x (c 3)
x y Y
+ 2 3) x (500 3  800 E) + (2 3) x (300 i) .758y  1600) 3 + (2.7513x + 800) 3 + (cy + 1100) E = 0 290.9 N 58l.8 N = 291 i  582 3 N A 1100 N 1100 J 1Axi + Ay3 + AZE)  300 i + 500 3  800 E A  290.9 3 — 581.8 3  1100 J (Ax  59019) i + (Ay — 1181.8) 3 + (Az  800) E = 6 590.9 N I Ay = 1181.8 N
800 N K = 591 i + 1182 3 + 800 E N ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. 780 (Continued) From a freebody
for joint C: From a freebody diagram
for joint E: = TDE  300 = 0
300 lb = 300 lb (T) TBE sin ¢E = 0 —TAE  TBE cos ¢E From a freebody diagram
for joint B: 1 .
[———]TBD  290.
Y ..)2 1112 +12)2 + (2.13 1031.1 lb 2 1030 lb (IT) Ans. ’——_§______—__——_—___E TBD
+ (2.75) (1) +12)2 ——72'—”— (1031.1)
/_____
+1—2.~75)2 (1)2 +12)2 800 1b = 800 1b (C) Ans‘ F. RILEY AND L. D. STURGES ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—87* A pin—connected system of levers and bars is used as a toggle for a press as shown in Fig. P7—87. Determine the force F exerted on the Ican at A when a force
? = 100 lb is applied
to the lever at G. SOLUTION From a free—body diagram
for the lever: + C EMF =100(30) — FDE(8) = 0 FDE = 375 lb From a free—body diagram
for the pin at D: + —» ZFX = —F cos 67° — F cos 78°  375 = 0 BD CD . 0 . 0
FBI) s1n 67 + FCD sin 78 639.5 lb +T2F 0
y F BD F 601.8 lb II CD From a freebody diagram
for the piston at B: T : — , ' o :
+ ZFY Ay 6395s1n 67 0 R3 Ay = 588.7 lb = 588.7 lb T Force on the can: F g 589 lb l L: Ans. £27 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES ?—91* Determine all forces acting on member ABCD
of the frame shown in
Fig! P?910 SOLUTION From a freebody diagram
for the complete frame: lttn + C EMA = F(12}  75(24} = 0 F = 150.0 1b = 150.0 lb T A  75 = 0 X ?5.0 lb = 75.0 lb9 4.
l
5? tﬁm 3
ll 4%; A + 150 = 0
y A = —150.0 1b = 150.0 lb ¢
2 2
{(Ax} + (AV)
2 2'
{(75.01 + {—150.01 = 167.71 1b _ '1 15000 _ _ 0
6A  tan 75.0  53.43 +
_._)
[‘1
I11
‘4
II :1:
ll K a 167.7 's 63.4° Member BE is a two~force member;
therefore, the line of action of
force 3 is known as shown on the
freebody diagram for member ABCD: 433 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES TQl {Continued} + c EMC 2 B cos 45° (6) ~ 75413)  T5i6} = o B = 424.3 lb B = 424 lb 5 45.o° Ans. 0x + (424.3) cos 45° + 75 — 75 = 0 300 1b = 300 lb é + T EFY = 0V — {~424.3) sin 45° + 150 = 0 150.0 lb = 150.0 lb ¢ /{Cx}2 + (Cylz = /{30012 + (150.0)2 = 335.41 lb _ 1 150 ~ _ o
3C — tan —§55 — 26.57 C = 335 lb 1 26.6” 43‘! »ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—103* The fold—down chair of Fig.
P7103 weighs 25 lb and has
its center of gravity at G.
Determine all forces acting on member ABC. SOLUTION From a freebody diagram
for the chair: + C EMA = 25(3)_ E(21) = o
E = 3 571 1b = 3.571 lb +— A  3.571 = 0 X 3.571 lb = 3 5 = 25 lb = 25 lb T l
$ = /<A )2 + (A )2
x Y
/(3.571)2 + (25)2 = 25.25 15 _ 1 25
ex ‘ tan 3.571 K = 25.3 1b a 81.9° = 81.87° Member ED is a two—force member;
therefore, the line of action of
force E is known as shown on the
freebody diagram for member ABC: + C ZMC = B cos 39.81° (9)  3.571(21) 0 B = 10.846 lb E = 10.85 lb 8 @ 39.8° = c + 3.571 — 10.846 cos 39.810 = o c = 4.761 lb X X = Cy + 25 — 10.846 sin 39.810 = 0 Cy = 18.056 1b‘ = ﬂex)2 + (cy)2 = /(4.761)2 + (—18.056)2 18.673 15 _ 1 18 056 _ _. o _ o
 tan "ZT761 _ 75.23 C _ 18.67 lb 8 75.2 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7105 Determine all forces
acting on member DEF
of the frame shown
in Fig. P7105. SOLUTION For the distributed load: §t14o)(5  2 tan 30°) = 269.2 lb §I5  2 tan 30°} = 2.564 ft From a freebody diagram
for the complete frame: + C EMA = Fxtzl  269.2(2.564} = 0 FX = 345.1 lb = 345.1 lb * H. Members BE and CD are two~foroe
members; therefore, their lines of
action are known as shown on the
freebody diagram for member DEF: + —% 2Fx = 345.1 + E cos 75°  D cos 600 = 0
+ T ZFY = FY + E sin 75° + 0 sin 60° = o
+ C EMF 2 E sin 75° (2) + 9 sin 60° (5) = 0 Solving yields: = 319.5 lb D a 320 1b 5 60.0° = ?15.1 lb E u 716 lb r r5.o°
415.0 lb /(Fx 2 + try)2 = /(345.112 + (415.0)2 = 539.7 lb F a 540 lb 2 50.3° ...
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 Spring '08
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