Homework4asol.fall11

Homework4asol.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. 4 (30 points). Due at the beginning of lecture on Wednesday, September 21st. Consider Problem 3.30, see Figure P3.30, page 162. For this homework assignment, only solve for the angular velocities of links 3, 4, 5, and 6 using the method of kinematic coefficients. Recall the procedure is: (i) Draw your vectors on a sketch of the mechanism and write the vector loop equations. (ii) Differentiate the X and Y components of the vector loop equations with respect to the input position variable. (iii) Use Cramers rule to solve for the first-order kinematic coefficients of the mechanism. (iv) Determine the angular velocities of links 3, 4, 5, and 6 using Table 3.2, page 137.
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2 Solution to Homework 4. Position Analysis. For the input position o 2 180 θ= , see Figure P3.30. For the triangle DBC, the law of cosines can be written as 22 2 43 3 3 RD B R2 ( D B ) R c o s = +− θ (1a) Rearranging this equation gives 222 34 3 3 DB R R cos 2(DB)R (1b) Then substituting the known values into this equation gives 3 342 cos 0.8750 2x3x4 = (1c) Therefore o 3 28.96 (1d) From the law of sines 3 R BC 4 sin sin = θ β ( 2 a ) where BDC. β =∠ Substituting the known values into Equation (2a), and rearranging, gives 1 4 sin 28.96 sin 104.44 2 ⎛⎞ == ⎜⎟ ⎝⎠ D D (2b) Therefore, 4 180 104.44 75.56 θ = °− °= ° (2c) You could check these answers using the program that you wrote for Part (v) to Homework Set 2. You could also check the answers using the four-bar linkage program that you wrote for Project 2. The orientation of link 6 and the location of the point of contact between link 5 (the wheel) and link 6 (the rack) can be obtained as follows. For the triangle BEH, the law of cosines can be written as 2 EH BH BE 2(BH)(BE)cos = φ (3a) where 3 HBE 90 ( θ 36.87 ) 90 (28.96 36.87 ) 24.17 oo o o o o φ = + = + = Substituting the known values into Equation (3a) gives o 2 EH 1.5 2.5 2(1.5)(2.5)cos24.17 inches =+− that is EH 8.5 6.84 1.657 inches =− = Therefore the distance from point E to point H is EH 1.287 inches = (3b)
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3 Also for the triangle BEH, the law of cosines can be written as 222 BE BH EH 2(BH)(EH)cos = +− ψ where BHE. ψ =∠ Rearranging this equation gives BH EH BE cos 2(BH)(EH) ψ= Then substituting the known values into this equation gives 2.25 1.657 6.25 cos 0.6067 rads 2 x1.5 x1.287 + =− Therefore, the angle BHE 127.35 o = Check: The law of sines for the triangle BEH can be written as BE EH sinBHE sinHBE = (3c) Substituting the known values into this equation and rearranging the equation gives o 2.5 sin24.17 0.7953 radians 1.287 == Therefore, the angle o BHE 127.31 = (3d) The two answers for the angle BHE are in good agreement (the difference is due round-off error). Using the Pythagorean theorem, it can be shown that the orientation of link 6 is oo o 6 127.31 22.77 104.54 θ= = (3e) Similarly, it can be shown that the location of the point of contact between links 5 and 6 is 9 R 1.187 inches = ( 3 f ) where the vector R 9 is shown on Figure 1 on the following page.
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Homework4asol.fall11 - ME 352 - Machine Design I Fall...

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