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Unformatted text preview: ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 635* A beam is loaded and
supported as shown in
Fig. P6—35. The beam has a uniform cross 500 lb 250 lb section and weighs 425
lb. Determine the reactions at supports
A and E. SOLUTION A freebody diagram of the beam is
shown at the right. The reaction
at support A is represented by
force components IX and Ky. The reaction at E is a force E normal
to the support surface. The weight
w of the beam acts through the
center of gravity G of the beam and
is directed toward the center of
the earth. + C EMA = E(9) — 500(1) + 250(4)  425(4.5)  400(7) = 0
E = 468.1 lb 8 468 lb lb = 468 lb T +—>2F :0
X X + C ZME = Ay(9) + 500(8)  250(5) + 425(4.5) + 400(2) = 0 A = 606.9 lb 3 607 lb 607 3 lb = 607 lb T ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 637 A structural member is
loaded and supported as
shown in Fig. P637. The member has a uniform
cross section and weighs
208 lb. Determine the
reactions at supports A
and B. SOLUTION A freebody diagram of the beam is
shown at the right. The reaction
at support A is represented by
force components xx and Ky. The reaction at B is a vertical force 5. The weight w of the beam acts
through the center of gravity G of the beam and is directed toward
the center of the earth. Lid1 + dez + Lads 6(3) + 4(6) + 3(7.5) G = 4.962 ft Ax + 200  300 = o 100 lb = Ay(2)  208(2.962)  200(2)  100(7)  300(2) = 0 = 1158.0 lb = 1158 lb {(100)2 + (1153)2 = 1162.3 1b a 1162 lb _ tan—1 1162
100 X = —35.03° a 35.1° A = 100 T  1158 J lb = 1162 lb E 85.10 + C EN = 3(2)  208(4.962)  200(2)  100(9)  300(2) = 0 A
B = 1466.0 lb = 1466 lb 3 = 1466 3 lb = 1466 lb 1 ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 643* Determine the reactions at supports A and B of the curved bar shown in Fig.
P6—43 SOLUTION From a freebody diagram
for the curved bar: + c 2MB = A sin 45°(6 + 6 sin 30°) + A cos 45°(6 cos 30°)
50(14 + 6 sin 30°) = o
A = 84.68 1b a 84.7 lb K = 84.7 lb 1 45.o° Bx + A cos 45°
Bx + 84.68 cos 45° = 0 59.88 lb 8 —59.9 1b = By + A sin 45° — 50
= By + 84.68 sin 45°  50 = o 9.878 1b 8 9.88 lb B: + B: = /(59.88)2 + (9.88)2 = 60.69 1b a 60.7 lb ‘1 _9l88
an . 0 O
_59.88 — a 170I6 — 9.88 3 N = 60.7 N r 9.37° ENGINEERING MECHANICS  STATICS, 2nd. Ed. 660* The coal wagon shown in Fig.
P660 is used to haul coal
from a mine. If the mass of
the coal and wagon is 2000
kg, determine the force P
required to move the wagon
at a constant velocity and
the forces exerted on the
front and rear wheels by the inclined surface. SOLUTION W = 2000(9.807) = 19,614 N From a freebody diagram
of the coal wagon: + /” ZFX P — w sin 30°
P  19,614 sin 30° = 0
9807 N a 9.81 kN 9.31 kN a 30° Ans.‘ W. F. w cos 30° (1) + w sin 30° (1) s FR (3) RILEY AND L. 19,614 cos 30° (1) + 19,614 sin 300 (1)  FR(3) = 0 8931 N a 8.93 kN 3.93 kN a 60° 0
FF + FR  W cos 30 F + 3931  19,614 cos 30° = 0 F
8055 N 3 8.06 kN = 8.06 kN 5 60° D. STURGES an“..___._... . ..._~._...i‘._.c..—. ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 662* A frame is loaded and l—ﬁJ'W—T03mi I mmNm. supported as shown in 4uonm, Fig. P6—62. Determine
the reactions at supports A and E. SOLUTION
The distributed load can be
replaced on the free—body diagram by a resultant force
R at a distance dx from joint C. R 400(0.6) + %(400)(0.6)
240 + 120 = 360 N = A‘Id‘1 + Azd2 400(0.6)(0.3) + %(400)(0.6)(0.4) a = 0.3333 m From a freebody diagram
for the complete frame: + c EMA = E cos 30° (0.6) + E sin 30° (0.6 + 0.6 tan 15°)
 360(0.3333 + 0.6 tan 15°) = 0 E 197.63 N a 197.6 N E = 197.6 N L 150.o° Ans. + —+ EFX A  197.63 cos 30°: 0 A = 171.15 N a 171.2 N X X Ay + 197.63 sin 30°  360 = 0 Av = 261.19 N z 261 N /(Ax)2 + (A )2 = {(171.15)2 + (261.19)2 = 312.3 N z 312 N 261.1 0
171.1 K s 312 N A 56.8 Ans. = 56.76° a 56.3° ENGINEERING MECHANICS  STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 686 Beam CD of Fig. P686 is
supported at the left end
C by a smooth pin and
bracket and at the right
end D by a continuous
cable that passes around
a frictionless pulley.
The lines of action of
the force in the cable
pass through point D.
Determine the components
of the reaction at
support C and the force
in the cable when a 5kN
load W is being supported
by the beam. SOLUTION’ Since the cable is continuous: I TAD=TB =l
TAD: 8/0 v. [ ’i “1+3E 2 8
/( )2 + (8)2 + (3)2
9117T 3 + 0.3419T E
2 8 3L 2
= 0.2279T i  0. r
= r
= T n
T A
TBD = Tee ’1‘— “1432 ]
ID
/(2)2 + (8)2 + (3)2 = U.2279T i  0.9117T 3 + 0.3419T 2 From a freebody diagram
for the beam: For moment equilibrium: 2“c = Cc + [(rD/C) x (TAD + TBD) + (rs/c x a) A = J + MCZ E + [(3 j) x (1.3234T j + 0.6838T 2)]
+ [(4 j) x (—5 E)!
= (5.4704T  20) 1 + Cy) J + (Mcz) E = 0
Solving yields:
J+MCZIE=U T = 3.656 kN E 3.66 kN m r .‘l ) “aw—.1 »:_wungym—Hé "maum ENGINEERING MECHANICS ~ STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 6—86 (Continued) TAD = 3.656(0.2279 i  0.9117 3 + 0.3419 2) = 0.8332 3  3.3332 3 + 1.2500 E kN TED = 3.656(0.2279 t  0.9117 3 + 0.3419 2) = 0.8332 3 — 3.3332 3 + 1.2500 E kN For force equilibrium: SE = RC + TAD + TED + W = RC)‘ 1 + RCy 3 + RCz E + 0.8332 1  3.3332 3 + 1.2500 R  0.8332 3  3.3332 3 + 1.2500 R — 2.5 E 1 3 + (Rey — 6.6664) 3 + (RC2  2.500) E = U 1 + Rev J + RCz E
= 6.666 3 + 2.500 2 RN 3 6.67 3 + 2.50 2 kN RC = /(6.666)2 + (2.500)2 = 7.119 RN 2 7.12 kN ‘1’??? ...
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 Spring '08
 McVay

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