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. 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’??? ...
View
Full Document
 Spring '08
 McVay
 mechanics, Force, lb, L. D. Sturges, W. F. Riley

Click to edit the document details