Homework7asol.fall11

Homework7asol.fall11 - ME 352 - Machine Design I Name of...

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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 Z-direction. 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 slider-crank mechanism. The vectors for the mechanism are as shown in Figure 1. Figure 1. The vectors for the offset slider-crank mechanism. The vector loop equation (VLE) for the slider-crank 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 first-order 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.

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Homework7asol.fall11 - ME 352 - Machine Design I Name of...

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