Homework8bsol.fall11

Homework8bsol.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. 8 (30 points). Due at the beginning of lecture on Friday, October 28th. Solve Problem 13.6, page 603. The recommended procedure for performing a static force analysis of this linkage using the Graphical Method is as follows: (i) Draw a scaled figure of the linkage and measure the orientations (i.e., the angles) of links 3, 4, and 5. Then use trigonometry, such as, the law of sines and the law of cosines, to check the accuracy of your measurements. (ii) Draw free body diagrams of each of the moving links. (iii) Identify all two-force members, three-force members, and/or four-force members. (iv) Draw all the force polygons. Clearly state the scales that you use for your polygons. (v) Determine all the unknown internal reaction forces. Specify the magnitudes and the directions of these forces. (vi) Determine the magnitude and the direction of the external force P acting on link 6 at point E. (vii) Specify whether the two-force members are in tension or in compression. The assumptions are: (i) gravity acts into the paper (i.e., the negative Z-direction); and (ii) the effects of friction in the mechanism can be neglected.
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2 Solution to Homework Set 8. A scaled drawing of the six-bar linkage is shown in Figure 1. The figure is drawn one-quarter full size. Figure 1. The Six-Bar Linkage. (The figure is drawn one-quarter full size). The known position variables and link lengths of the six-bar linkage, see page 603, are o 1 0, θ= o 2 90 , 24 1O O R R 60 mm, == 2 2A O R R 100 mm, = = 3A B RR 1 5 0 m m , = = 4 BO R1 2 5 m m , =
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3 4 4C O R R 200 mm, == and 5C D R R 400 mm. The unknown position variables 3 θ , 4 θ and 5 θ can be obtained from trigonometry. Using the Pythagorean theorem gives 4 22 2 2 2 12 AO R R R 60 100 =+= + ( 1 a ) Therefore, the distance from pin A to the ground pin 4 O i s 4 AO R1 1 6 . 6 2 m m = (1b) Consider the triangle 24 OOA Δ then 2 1 R 100 mm tan O O A 1.67 R6 0 m m ∠= = = ( 2 a ) Therefore, the angle O O A 59.04 (2b) and 42 O AO 90 59.04 30.96 ° ° = ° (2c) Using the law of cosines gives 4 44 2 33 A O 4 BO AO RR R2 R R c o s B A O =+ (3a) or 4 2 3 AO BO 4 3A O R cos BAO 2R R +− (3b) Substituting the numerical data into this equation gives 2 4 (150) (116.62) (125) cos BAO 0.585 2(150)(116.62) = (4a) Therefore, the angle 4 BAO 54.18 (4b) Therefore, the orientation of link 3 is 34 4 2 90 BAO O AO 4.86 5 θ=− °+∠ +∠ =− °≈−° (5) Using the law of sines gives 4 BO 3 R R s i nA OB s i nB A O = ∠∠ ( 6 a ) Substituting the numerical data into this equation gives 4 4 BO Rs i n BAO 150 sin54.18 sin AO B 0.973 R 125 ° = = (6b)
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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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Homework8bsol.fall11 - ME 352 - Machine Design I Fall...

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