Resolução do Capítulo 10 - Método dos Trabalho Virtuais
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Resolução do Capítulo 10 - Método dos Trabalho Virtuais

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PROBLEM 10.1 
Determine the vertical force P which must be applied at G to maintain the 
equilibrium of the linkage. 
 
SOLUTION 
 
Assuming 
 Ay\u3b4 
it follows 
120 1.5
80C A A
y y y\u3b4 \u3b4 \u3b4= = 
1.5E C Ay y y\u3b4 \u3b4 \u3b4= = 
( )180 3 1.5 4.5
60D A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = = 
( )100 100 1.5 2.5
60 60G A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = = 
Then, by Virtual Work 
( ) ( )0: 300 N 100 N 0A D GU y y P y\u3b4 \u3b4 \u3b4 \u3b4= \u2212 + = 
 ( ) ( )300 100 4.5 2.5 0A A Ay y P y\u3b4 \u3b4 \u3b4\u2212 + = 
 300 450 2.5 0P\u2212 + = 
 60 NP = + 60 N=P W 
 
 
PROBLEM 10.2 
Determine the vertical force P which must be applied at G to maintain the 
equilibrium of the linkage. 
 
SOLUTION 
 Link ABC 
 
 Link DEFG 
 
 
 
 
 
Assume 
clockwise\u3b4\u3b8 
Then for point C 
( )5 in.Cx\u3b4 \u3b4\u3b8= 
and for point D 
( )5 in.D Cx x\u3b4 \u3b4 \u3b4\u3b8= = 
And for link DEFG 
15Dx\u3b4 \u3b4\u3c6= 
 5 15\u3b4\u3b8 \u3b4\u3c6\u2234 = 
or 1
3
\u3b4\u3c6 \u3b4\u3b8= 
Then 44 2 2 in.
3G
\u3b4 \u3b4\u3c6 \u3b4\u3b8\uf8eb \uf8f6= = \uf8ec \uf8f7\uf8ed \uf8f8 
Now cos 45G Gy\u3b4 \u3b4= ° 
 4 2 cos 45
3
\u3b4\u3b8\uf8eb \uf8f6= °\uf8ec \uf8f7\uf8ed \uf8f8 
 4 in.
3
\u3b4\u3b8\uf8eb \uf8f6= \uf8ec \uf8f7\uf8ed \uf8f8 
Then, by Virtual Work. 
( ) ( ) ( ) ( )0: 80 lb in. 40 lb in. in. 0E GU x P y\u3b4 \u3b4\u3b8 \u3b4 \u3b4= \u22c5 \u2212 + = 
 10 480 40 0
3 3
P\u3b4\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 
 or 40 lb=P W 
 
PROBLEM 10.3 
Determine the couple M which must be applied to member DEFG to 
maintain the equilibrium of the linkage. 
 
SOLUTION 
 
 Link ABC 
 
 Link DEFG 
 
 
 
 
Following the kinematic analysis of Problem 10.2, we have 0:U = 
( ) ( ) ( )0: 80 lb in. 40 lb in. 0EU x M\u3b4 \u3b4\u3b8 \u3b4 \u3b4\u3c6= \u22c5 \u2212 + = 
 10 180 40 0
3 3
M\u3b4\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 
 or 160 lb in.= \u22c5M W 
 
 
 
 
 
PROBLEM 10.4 
Determine the couple M which must be applied to member DEFG to 
maintain the equilibrium of the linkage. 
 
SOLUTION 
 
Assuming 
 Ay\u3b4 
it follows 
120 1.5
80C A A
y y y\u3b4 \u3b4 \u3b4= = 
1.5E C Ay y y\u3b4 \u3b4 \u3b4= = 
( )180 3 1.5 4.5
60D A A A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = = 
1.5 1
60 60 40
E A
A
y y y\u3b4 \u3b4\u3b4\u3c6 \u3b4= = = 
Then, by Virtual Work: 
( ) ( )0: 300 N 100 N 0A DU y y M\u3b4 \u3b4 \u3b4 \u3b4\u3c6= \u2212 + = 
 ( ) 1300 100 4.5 0
40A A A
y y M y\u3b4 \u3b4 \u3b4\uf8eb \uf8f6\u2212 + =\uf8ec \uf8f7\uf8ed \uf8f8 
 1300 450 0
40
M\u2212 + = 
 6000 N mmM = + \u22c5 6.00 N m= \u22c5M W 
 
 
 
PROBLEM 10.5 
An unstretched spring of constant 4 lb/in. is attached to pins at points C 
and I as shown. The pin at B is attached to member BDE and can slide 
freely along the slot in the fixed plate. Determine the force in the spring 
and the horizontal displacement of point H when a 20-lb horizontal force 
directed to the right is applied (a) at point G, (b) at points G and H. 
 
SOLUTION 
First note: 
3 3G D G Dx x x x\u3b4 \u3b4= \u21d2 = 
4 4H D H Dx x x x\u3b4 \u3b4= \u21d2 = 
5 5I D I Dx x x x\u3b4 \u3b4= \u21d2 = 
(a) Virtual Work 0: 0G G SP IU F x F x\u3b4 \u3b4 \u3b4= \u2212 = 
 or ( )( ) ( )20 lb 3 5 0D SP Dx F x\u3b4 \u3b4\u2212 = 
 thus, 12.00 lb SPF T= W 
 Now SP IF k x= \u2206 
 or ( )12.00 lb 4 lb/in. Ix= \u2206 
 Thus, 3 in.Ix\u2206 = 
 and 
1 1
4 5D H I
x x x\u3b4 \u3b4 \u3b4= = 
4 
5
\u2234 \u2206 = \u2206H Ix x 
 ( )4 3 in.
5
= or 2.40 in.Hx\u2206 = W 
(b) Virtual Work: 0: 0G G H H SP IU F x F x F x\u3b4 \u3b4 \u3b4 \u3b4= + \u2212 = 
 or ( )( ) ( )( ) ( )20 lb 3 20 lb 4 5 0D D SP Dx x F x\u3b4 \u3b4 \u3b4+ \u2212 = 
 thus, 28.0 lb SPF T= W 
 Now SP IF k x= \u2206 
 or ( )28.0 lb 4 lb/in. Ix= \u2206 
 Thus, 7 in.Ix\u2206 = 
 From part (a) 
4
5H I
x x\u2206 = \u2206 
 ( )4 7 in.
5
= or 5.60 in.Hx\u2206 = W 
 
 
 
PROBLEM 10.6 
An unstretched spring of constant 4 lb/in. is attached to pins at points C 
and I as shown. The pin at B is attached to member BDE and can slide 
freely along the slot in the fixed plate. Determine the force in the spring 
and the horizontal displacement of point H when a 20-lb horizontal force 
directed to the right is applied (a) at point E, (b) at points D and E. 
 
SOLUTION 
First note: 
3 3G D G Dx x x x\u3b4 \u3b4= \u21d2 = 
4 4H D H Dx x x x\u3b4 \u3b4= \u21d2 = 
5 5I D I Dx x x x\u3b4 \u3b4= \u21d2 = 
(a) Virtual Work: 0: 0E E SP IU F x F x\u3b4 \u3b4 \u3b4= \u2212 = 
 or ( )( ) ( )20 lb 2 5 0D SP Dx F x\u3b4 \u3b4\u2212 = 
 thus, 8.00 lb SPF T= W 
 Now SP IF k x= \u2206 
 or ( )8.00 lb 4 lb/in. Ix= \u2206 
 Thus, 2 in.Ix\u2206 = 
 And 
1 1
4 5D H I
x x x\u3b4 \u3b4 \u3b4= = 
4 
5
\u2234 \u2206 = \u2206H Ix x 
 ( )4 2 in.
5
= 
 or 1.600 in.Hx\u2206 = W 
(b) Virtual Work: 0: 0D D E E SP IU F x F x F x\u3b4 \u3b4 \u3b4 \u3b4= + \u2212 = 
 or ( ) ( )( ) ( )20 lb 20 lb 2 5 0D D SP Dx x F x\u3b4 \u3b4 \u3b4+ \u2212 = 
 thus, 12.00 lb SPF T= W 
 
PROBLEM 10.6 CONTINUED 
 Now SP IF k x= \u2206 
 or ( )12.00 lb 4 lb/in. Ix= \u2206 
 Thus, 3 in.Ix\u2206 = 
 From part (a) 
4
5H I
x x\u2206 = \u2206 
 ( )4 3 in.
5
= or 2.40 in.Hx\u2206 = W 
 
 
 
 
 
PROBLEM 10.7 
Knowing that the maximum friction force exerted by the bottle on the 
cork is 300 N, determine (a) the force P which must be applied to the 
corkscrew to open the bottle, (b) the maximum force exerted by the base 
of the corkscrew on the top of the bottle. 
 
SOLUTION 
 
 
From sketch 
4A Cy y= 
Thus, 4A Cy y\u3b4 \u3b4= 
(a) Virtual Work: 
0: 0A CU P y F y\u3b4 \u3b4 \u3b4= \u2212 = 
1
4
P F= 
( )1300 N: 300 N 75 N
4
F P= = = 
 75.0 N=P W 
(b) Free body: Corkscrew 
 0: 0yF R P F\u3a3 = + \u2212 = 
 75 N 300 N 0R + \u2212 = 
 225 N=R W 
 
 
 
 
 
PROBLEM 10.8 
The two-bar linkage shown is supported by a pin and bracket at B and a 
collar at D that slides freely on a vertical rod. Determine the force P 
required to maintain the equilibrium of the linkage. 
 
SOLUTION 
 
Assume Ay\u3b4 
Have 160 1or
320 2C A C A
y y y y\u3b4 \u3b4 \u3b4 \u3b4= = 
Since bar CD moves in translation 
E F Cy y y\u3b4 \u3b4 \u3b4= = 
or 1
2E F A
y y y\u3b4 \u3b4 \u3b4= = 
Virtual Work: 
 
( ) ( )0: 400 N 600 N 0A E FU P y y y\u3b4 \u3b4 \u3b4 \u3b4= \u2212 + + = 
 ( ) ( )1 1400 N 600 N 0
2 2A A A
P y y y\u3b4 \u3b4 \u3b4\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 + + =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 
or 500 NP = 
 500 N=P W 
 
 
 
PROBLEM 10.9 
The mechanism shown is acted upon by the force P; derive an expression 
for the magnitude of the force Q required for equilibrium. 
 
SOLUTION 
 
 
Virtual Work: 
Have 2 sinAx l \u3b8= 
2 cosAx l\u3b4 \u3b8 \u3b4\u3b8= 
and 3 cosFy l \u3b8= 
3 sinFy l\u3b4 \u3b8 \u3b4\u3b8= \u2212 
Virtual Work: 0: 0A FU Q x P y\u3b4 \u3b4 \u3b4= + = 
 ( ) ( )2 cos 3 sin 0Q l P l\u3b8 \u3b4\u3b8 \u3b8 \u3b4\u3b8+ \u2212 = 
 3 tan
2
Q P \u3b8= W 
 
 
 
 
PROBLEM 10.10 
Knowing that the line of action of the force Q passes through point C, 
derive an expression for the magnitude of Q required to maintain 
equilibrium 
 
SOLUTION 
 
 
Have 2 cos ; 2 sinA Ay l y l\u3b8 \u3b4 \u3b8 \u3b4\u3b8= = \u2212 
( )2 sin ; cos
2 2
CD l CD l\u3b8 \u3b8\u3b4 \u3b4\u3b8= = 
Virtual Work: 
 ( )0: 0AU P y Q CD\u3b4 \u3b4 \u3b4= \u2212 \u2212 = 
 ( )2 sin cos 0
2
P l Q l \u3b8\u3b8 \u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6\u2212 \u2212 \u2212 =\uf8ec \uf8f7\uf8ed \uf8f8 
 ( )
sin2
cos /2
Q P \u3b8\u3b8= W 
 
 
 
 
PROBLEM 10.11 
Solve Problem 10.10 assuming that the force P applied at point A acts 
horizontally to the left. 
 
SOLUTION 
 
 
Have 2 sin ; 2 cosA Ax l x l\u3b8 \u3b4 \u3b8\u3b4\u3b8= = 
( )2 sin ; cos
2 2
CD l CD l\u3b8 \u3b8\u3b4 \u3b4\u3b8= = 
Virtual Work: ( )0: 0AU P x Q CD\u3b4 \u3b4 \u3b4= \u2212 = 
 ( )2 cos cos 0
2
P l Q l \u3b8\u3b8\u3b4\u3b8 \u3b4\u3b8\uf8eb \uf8f6\u2212 =\uf8ec \uf8f7\uf8ed \uf8f8 
 ( )
cos2
cos /2
Q P \u3b8\u3b8= W 
 
 
 
 
 
PROBLEM 10.12 
The slender rod AB is attached to a collar A and rests on a small wheel at 
C. Neglecting the radius of the wheel and the effect of friction, derive an 
expression for the magnitude of the force Q required to maintain the 
equilibrium of the rod. 
 
SOLUTION 
 
 
For :AA C\u2032\u2206 
tanA C a \u3b8\u2032 = 
( ) tanAy A C a \u3b8\u2032= \u2212 = \u2212 
2cosA
ay\u3b4 \u3b4\u3b8\u3b8= \u2212 
For :CC B\u2032\u2206 
sinBC l A C\u3b8\u2032 \u2032= \u2212 
 sin tanl a\u3b8 \u3b8= \u2212 
sin tanBy BC l a\u3b8 \u3b8\u2032= = \u2212 
2cos cosB
ay l\u3b4 \u3b8\u3b4\u3b8 \u3b4\u3b8\u3b8= \u2212 
Virtual Work: 
 0: 0A BU Q y P y\u3b4 \u3b4 \u3b4= \u2212 = 
 2 2cos 0cos cos
a aQ P l\u3b4\u3b8 \u3b8 \u3b4\u3b8\u3b8 \u3b8
\uf8eb \uf8f6 \uf8eb \uf8f6\u2212 \u2212 \u2212 \u2212 =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8 
 2 2coscos cos
a aQ P l \u3b8\u3b8 \u3b8
\uf8eb \uf8f6 \uf8eb \uf8f6= \u2212\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8