Homework9bsol.spring11 - ME 352 - Machine Design I Spring...

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1 ME 352 - Machine Design I Name of Student___________________________ Spring Semester 2011 Lab Section Number________________________ Homework No. 9 (30 points). Due at the beginning of lecture on Monday, April 4th. Solve Problem 14.30, see pages 681 and 682.
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2 Solution. The input link 2 is rotating with a constant angular velocity 2 ω 10 rad/s = counterclockwise. The angular velocities of links 3 and 4 are 3 ω 1.43 rad/s = clockwise, and 4 ω = 11.43 rad/s clockwise, respectively. The angular accelerations of links 3 and 4 and the acceleration of the center of mass of link 3, respectively, are 2 34 84.8 rad/sec counterclockwise αα == and 2 3 G A 310 i 295 j in/sec =+ + In the given position, the spring and the damper are both parallel to the X-axis. The stiffness of the spring is k =12 lbs/in and the free length is O R4 . 5 i n s . = The viscous damper has a damping coefficient C = 0.25 lbs.sec/in. There is also a known external force acting at the coupler point C, that is, C F 125 j lbs. =− The data given in Problem 14.20, page 679, is O 2 A = 6 in, O 2 O 4 = 18 in, AB = 18 in, O 4 B = 6 in, AC = 24 in, AG 3 = 12 in, W 3 = 10 lbs, 2 24 I I 0.063 sec GG in lb , 2 3 I 0.497 sec G in lb , and g = 32.2 ft/ sec 2 = 386.4 in/sec 2 (i) Figure 1 shows the vectors that will be used in the kinematic analysis of the four-bar linkage. Figure 1. The vectors for the four-bar linkage. The first-order kinematic coefficient of link 3 can be written as 3 33 3 2 1.43 0.143 rad/rad 10 ω θθ ′′ === = + (1) The first-order kinematic coefficient of link 4 can be written as 4 4 2 11.43 1.143 rad/rad 10 θ = + (2)
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3 The angular acceleration of link 3 can be written as 2 33 2 3 2 α θω θα ′′ =+ ( 3 a ) Rearranging this equation, the second-order kinematic coefficient for link 3 can be written as 2 2 3 2 2 rad/rad θθ ω == ( 3 b ) Therefore, the second-order kinematic coefficient of link 3 is 2 3 2 84.8 ( 0.143)(0) 0.848 rad/rad ( 10) +− = + + (4) Similarly, the angular acceleration of link 4 can be written as 2 44 2 4 2 ( 5 ) Therefore, the second-order kinematic coefficient of link 4 is 2 4 2 84.8 ( 1.143)(0) 0.848 rad/rad ( 10) θ + + (6) Since the mass centers G 2 and G 4 are located at the fixed pivots O 2 and O 4 , respectively, then the first and second-order kinematic coefficients of these mass centers are 0 2 = G x 0 2 = G x 0 2 = G x 0 2 = G y 0 2 = G y 0 2 = G y 18 4 = G x in 0 4 = G x 0 4 = G x 0 4 = G y 0 4 = G y 0 4 = G y To determine the first and second-order kinematic coefficients for the center of mass of the coupler link. Using the point path approach, the vector loop for the center of mass of link 3 can be written as 323 3 G R RR ( 7 ) where R 2 = 6 in, θ 2 = 60º, R 33 = 12 in, and θ 33 = θ 3 = 321.8º. The X and Y components of the vector equation for the center of mass of link 3 are 32 2 3 3 3 cos cos 6cos60 12cos321.8 12.4303 in = ° + ° = + G xR R (8a) and 2 3 3 3 sin sin 6sin 60 12sin321.8 2.2247 in = ° + ° = G yR R (8b) The first-order kinematic coefficients for the center of mass of link 3 are 2 3 3 3 3 sin sin 6sin 60 12sin321.8 ( 0.143) 6.2573 in =− °− G R (9a) and 2 3 3 3 3 cos cos 12cos321.8 ( 0.143) 1.6515 in = ° + ° = + G R (9b)
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4 The second-order kinematic coefficients for the center of mass of link 3 can be written as 2 32 2 3 3 3 3 3 3 3 3 cos cos sin
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This note was uploaded on 04/23/2011 for the course ME 352 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Homework9bsol.spring11 - ME 352 - Machine Design I Spring...

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