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Homework5asol.fall11

# Homework5asol.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. 5 (30 points). Due at the beginning of lecture on Wednesday, September 28th. Consider the inverted slider-crank linkage shown in Problem 3.15, see Figure P3.15, page 160. For the given position of the linkage, the angular velocity and angular acceleration of the input link 2 are specified as 2 10 rad s ω = / clockwise and 2 2 5 rad s α = / clockwise, respectively. Determine the following: (i) The first and second-order kinematic coefficients of the mechanism. (ii) The velocity and the acceleration of point B fixed in link 3. Give the magnitude and the direction of each vector. (iii) The unit tangent vector and the unit normal vector to the path of point B. Show the directions of these two unit vectors on a sketch of the mechanism. (iv) The radius of curvature of the path of point B. Indicate your answer on a sketch of the mechanism. (v) The X and Y coordinates of the center of curvature of the path of point B. Indicate the position of the center of curvature of the path of point B on a sketch of the mechanism.

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2 Solution to Homework Set 5. The inverted slider-crank mechanism is shown in Figure 1. Figure 1. The Inverted Slider-Crank Linkage. Position Analysis. The known position variables are o 1 0 , θ = 1 R 125 mm, = o 2 150 , θ = and 2 R 75 mm. = The other position variables can be obtained from trigonometry. Consider the triangle 2 4 O O A, + using the law of cosines gives ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 4 34 1 2 1 2 4 2 AO R R R 2 R R cos O O A = = + (1) Therefore ( ) ( ) ( ) ( )( ) 2 2 2 34 R 125 75 2 125 75 cos150 = + ° (2a) which gives 34 R 193.62 mm = (2b) Also for the triangle 2 4 O O A, + using the law of sines gives 34 4 2 4 2 4 R R sin O AO sin AO O = (3a) Rearranging this equation gives 4 2 4 2 4 34 R 125 sin O AO sin AO O sin150 R 193.62 = = ° (3b) that is 2 4 O AO 18.83 = ° (3c) Therefore 3 34 (30 18.83 ) 11.17 θ = θ = − ° − ° = − ° (3d) Velocity Analysis. The vectors for the inverted slider-crank mechanism are shown in Figure 2. The vector loop equation (VLE) can be written as 1 I ?? R R R 0 2 34 √√ + = (4)
3 The X and Y components of Equation (4) are 2 2 34 3 1 1 cos cos cos 0 R R R + = θ θ θ (5a) and 2 2 34 3 1 1 sin sin sin 0 R R R + = θ θ θ (5b) Figure 2. The Vectors for the Inverted Slider-Crank Mechanism. Differentiating Equations (5a) and (5b) with respect to the input position θ 2 gives 2 2 34 3 3 34 3 sin sin cos 0 R R R + = θ θ θ θ (6a) and 2 2 34 3 3 34 3 cos cos sin 0 R R R + + = θ θ θ θ (6b) Then writing Equations (6) in matrix form gives 34 3 3 3 2 2 34 3 3 34 2 2 sin cos sin cos sin cos R R R R R ⎤ ⎡ = ⎥ ⎢ ⎦ ⎣ θ θ θ θ θ θ θ (7) The determinant of the coefficient matrix in Equation (7) is 34 3 3 34 34 3 3 sin cos 193.62 mm cos sin = = − = − R DET R R θ θ θ θ (8)

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