Homework12csol.fall11

Homework12csol.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. 12 (30 points). Due at the beginning of lecture on Friday, December 9th. Problem 1. Solve Problem 17.7 on page 820, using the analytical procedure. Then check your answers using the graphical procedure. Problem 2. Consider the shaft shown in the figure below with a distributed mass system rotating counterclockwise with a constant angular velocity 50 rad s ω = /. The forces acting from the system (denoted as 2) on the foundation (denoted as 1) at bearings A and B are 21 A (F ) 60 N = and B 21 80 N, = respectively. Determine the magnitudes and the locations of the correction masses to be removed in the correction planes (1) and (2) at a radius C1 C2 C rrr 1 5 c m . = == Figure. A distributed mass system on a rotating shaft.
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2 Solution to Problem 1. The rotating system shown in Figure P17.7, on page 820, is repeated here in the figure below. Figure 1. A rotating shaft with three mass particles. (i) The Analytical Method. The inertial effects of the three rotating mass particles are, in ounce-inches 1 11 2 (4 oz)(5 in) 20 oz-in == = F mR ω (1a) 2 22 2 (3 oz)(4 in) 12 oz-in = F (1b) and 3 33 2 (4 oz)(5 in) 20 oz-in = F (1c) Therefore, the inertial effects of the three rotating mass particles can be written as 1 2 20 oz-in 90 20 oz-in F j ° = JG G (2a) 2 2 12 oz-in 210 10.39 6.00 oz-in F i j ° = G G (2b) and 3 2 20 oz-in 300 10.00 17.32 oz-in F i j ° = G G (2c) The sum of the inertial effects of the three rotating mass particles can be written as 2 2 0.39 3.32 oz-in 3.34 oz-in 263.30 m R m R m R i j =+ + = = ° G G KK K K (3)
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3 Therefore, the inertial effects of the balancing masses can be written as ( ) 11 2 2 33 0.39 3.32 oz-in LL RR mR mR mR mR mR i j += −+ ++ G G KK K K K (4) The sum of the moments about the left correction plane L is () 0 = L M (5a) which can be written as 17 8 27 35 0 km R kmR R kmR −× × × = K K K K (5b) Substituting Equations (3) into Equation (5b) gives 17 20 8 ( 10.39 6 ) 27 (10 17.32 ) 35 0 kj
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Homework12csol.fall11 - ME 352 - Machine Design I Fall...

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