Homework9asol.fall11

Homework9asol.fall11 - ME 352 Machine Design I Fall...

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ME 352 - Machine Design I Name of Student ____________________________ Fall Semester 2011 Lab Section Number _________________________ Homework No. 9 (30 points). Due at the beginning of lecture on Monday, November 7th. Solve Problem 14.34, see page 684, using the Power Equation and the Equation of Motion.
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2 Solution to Homework Set 9. For the convenience of the reader, the slider-crank mechanism shown in Figure 14.34, page 684, is repeated here as Figure 1. Figure 1. The mechanism modified from Problem 14.5, page 678. The vectors for a kinematic analysis of the mechanism are as shown in Figure 2. Figure 2. The vectors for the mechanism. The unknown position variable of link 3 can be obtained from trigonometry. For example, for the triangle 2 ABO , Δ the law of sines can be written as 3 2 22 R R sin ABO sin = θ (1a) Rearranging gives o 2 3 Rs i n 3sin45 sin ABO 0.1768 rad R1 2 θ ∠= = = (1b) Therefore 2 ABO 10.182 (1c) Therefore, the coupler angle (that is, the angle of the vector 3 R and the vector 33 R ) is 3 10.182 θ =− ° (1d)
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3 (i) The first-order kinematic coefficient of link 3 can be written as 3 3 2 17.96 rad/s 0.1796 100 rad/s ω θ == = + (2) The first-order kinematic coefficient of link 4 can be written as 4 4 2 250.23 2.5023 in 100 = + V R (3) The negative sign indicates that the vector 4 R is decreasing in length for a positive change in the position of the input link 2. The angular acceleration of link 3 can be written as 2 33 2 3 2 α θω θα ′′ =+ ( 4 a ) Rearranging this equation, the second-order kinematic coefficient of link 3 can be written as 2 3 2 2 ′′= ( 4 b ) which can be written as 2 3 22 2 ( 1736.20) ( 0.1796)(10) rad/s 0.1738 ( 100) rad /s +− + + (4c) The linear acceleration of link 4 can be written as 2 44 2 4 2 AR R ( 5 a ) Rearranging this equation, the second-order kinematic coefficient of link 4 is 2 4 2 21361.02 ( 2.5023)(10) 2.1336 in ( 100) R −− = + (5b) Since the mass center of link 2, G 2 , is located at the fixed pivot O 2 , then the first and second-order kinematic coefficients are 0 2 = G x 0 2 = G x 0 2 = G x (6a) and 0 2 = G y 0 2 = G y 0 2 = G y (6b) To determine the first and second-order kinematic coefficients for the center of mass of link 3. The vector loop for point G 3 , see Figure 2, can be written as 32 3 3 ?? G C I R RR = + (7)
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4 where the magnitude of the vector 33 R is given as 4.5 inches and the angle o 33 3 10.182 . θ= This implies that the first-order kinematic coefficient 33 3 θ and the second-order kinematic coefficient 33 3 ′′ θ . The X and Y components of Equation (7) are 3 22 3 33 cos cos = 6.5504 in G XR R θ =+ (8a) and 3 3 sin sin 1.3258 in G YR R = + (8b) The first-order kinematic coefficients of the center of mass of link 3 are 3 32 2 3 3 3 3 sin sin 2.2642 in GG xX R R θθ == = (9a) and 3 2 3 3 3 3 cos cos 1.3258 in yYR R + = + (9b) The second-order kinematic coefficients of the center of mass of link 3 are 3 2 2 3 3 3 3 3 3 3 3 cos cos
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This note was uploaded on 02/13/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue.

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Homework9asol.fall11 - ME 352 Machine Design I Fall...

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