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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