kin_sol_hw2

kin_sol_hw2 - 2.2 B B 75 O2 O4 The output angle in a four...

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2.2 O 2 B B O 4 75 ° β α The output angle in a four bar mechanism is at its extremes when links 2 and 3 are collinear. From this O 2 B = l 3 + l 2 = 229 ( 1 ) and O 2 B = l 3 l 2 = 229 ( 2 ) Solving the linear equations in (1) and (2) we obtain . l 2 = 63.5 mm , l 3 = 165.5 mm To draw the mechanism to scale we would require the length of link 1 i.e. O 2 O 4 . From Δ O 4 B B (isosceles triangle, hence bisector is perpendicular) B B = 2 O 4 B sin 75 2 = 138.8 mm . Using the Sine rule for Δ O 4 B B B B sin 75 ° () = O 4 B sin β sin = O 4 B B B sin 75 ° ( = 52.5 ° ) Consider Δ O 2 B B , using Cosine rule we have
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O 2 B () 2 = O 2 B 2 + B B 2 2 O 2 B ( ) B B ( ) sin α sin = O 2 B 2 + B B 2 O 2 B 2 2 O 2 B B B = 143.53 ° Now using Cosine rule on we obtain Δ O 2 BO 4 O 2 O 4 2 = O 2 B 2 + O 4 B 2 O 2 B O 4 B ( ) cos β ( ) O 2 O 4 = 154.34 mm The transmission angle is extremum when links 1 and 2 are collinear. The maximum transmission angle can be obtained by applying the Cosine rule to Δ A max B max O 4 , r 1 r 2 2 = r 3 2 + r 4 2 2 r 3 r 4 cos γ cos = r 3 2 + r 4 2 r 1 r 2 2 2 r 3 r 4 max = 31.63 ° Similarly the minimum transmission angle can be obtained by applying the Cosine to Δ A min B min O 4 r 1 + r 2 2 = r 3 2 + r 4 2 2 r 3 r 4 cos cos = r 3 2 + r 4 2 r 1 + r 2 2 2 r 3 r 4 max = 100.79 ° A max B max O 4 r 2 r 3 r 4 r 1 max B min min r A min 4 O 2
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2.19) Given O 2 F = R Δ O 4 EF ≈ Δ O 4 GH EF GH = O 4 E O 4 G R sin θ x = D + R cos C + D x = C + D ( ) R sin D + R cos 2.30) Given: O 2 A = r 2 = 4 in AB = r 3 = 8 in O 4 B = r 4 = 4 in 2 = 120 ° From geometry in r O O 02 . 12 5 . 8 2 2 1 4 2 = × = = and ° = 135 1 .
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This note was uploaded on 09/20/2010 for the course ME 4133 taught by Professor Ram during the Fall '06 term at LSU.

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kin_sol_hw2 - 2.2 B B 75 O2 O4 The output angle in a four...

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