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Unformatted text preview: ME 352  Machine Design I Name of Student ____________________________ Fall Semester 2011 Lab Section Number ________________________ Homework No. 7 (30 points). Due at the beginning of lecture on Friday, October 21st. Solve Problem 14.15, page 679. For this homework, solve the dynamic force analysis problem by the method of inspection; that is, solve one equation for one unknown variable, or in the worst case scenario solve two equations for two unknown variables. Note the given assumptions: (i) the crank is balanced, that is, the center of mass of link 2 is coincident with the ground pivot O 2 ; (ii) there is no friction in the mechanism; (iii) the external force F B is acting on the slider (link 4) at point B; and (iv) gravity is acting into the paper, that is, in the negative Zdirection. If you would like to check your answers for this homework assignment then you could write a computer program in Matlab which will use matrix inversion to determine the unknown variables. 2 Solution to Homework Set 7. (i) Kinematic analysis of the offset slidercrank mechanism. The vectors for the mechanism are as shown in Figure 1. Figure 1. The vectors for the offset slidercrank mechanism. The vector loop equation (VLE) for the slidercrank mechanism, see Figure 1, can be written as 2 3 4 1 ? ? I R R R R − √ √ √ √ √ + − = (1) The X and Y components of Equation (1) are 2 2 3 3 4 4 1 1 cos cos cos cos R R R R θ θ θ θ + − − = (2a) and 2 2 3 3 4 4 1 1 sin sin sin sin R R R R θ θ θ θ + − − = (2b) Substituting the numerical data into Equation (2b) gives 3 4 0.1sin 120 0.38 sin sin 0 0.06 sin ( 90 ) R θ ° + − ° − − ° = (3a) which gives 3 sin 0.3858 radians θ = − (3b) or 3 3 22.69 or 337.31 θ θ = − ° = + ° (3c) Substituting Equation (3b) and the numerical data into Equation (2a) gives 4 0.1cos120 0.38 cos ( 22.69 ) cos 0 0.06 cos ( 90 ) R ° + − ° − ° − − ° = (4a) Therefore, the displacement of the slider is 4 0.30 m R x = = (4b) 3 Differentiating Equation (2a) and (2b) with respect to the input position variable 2 θ gives 2 2 3 3 3 4 4 sin sin cos R R R θ θ θ θ ′ ′ − − − = (5a) and 2 2 3 3 3 4 4 cos cos sin R R R θ θ θ θ ′ ′ + − = (5b) Writing Equations (5a) and (5b) in matrix form gives 3 3 4 3 2 2 3 3 4 4 2 2 sin θ cos θ θ sin θ cos θ sin θ R c o s θ R R R R ′ − − ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ′ − − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ (6) The determinant of the coefficient matrix in Equation (6) can be written as 3 3 4 3 3 4 3 3 4 3 3 4 3 3 4 sin θ cos θ sin θ sin θ cos θ cos θ cos ( θ θ ) cos θ sin θ R DET R R R R − − = = + = − − (7) Using Cramer’s rule, the firstorder kinematic coefficient for link 3 can be written as 2 2 4 2 2 4 2 2 4 2 2 4 2 2 4 3 3 3 4 sin cos cos sin sin sin cos cos cos ( ) cos ( ) R R R R R DET DET R θ θ θ θ θ θ θ θ θ θ θ θ θ − − − − − − − ′ = = = − (8a) which can be written as...
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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.
 Fall '08
 Staff
 Machine Design

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