Homework7asol.spring11 - ME 352 - Machine Design I Spring...

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ME 352 - Machine Design I Name of Student ______________________________ Spring Semester 2011 Lab Section Number ___________________________ Homework No. 7 (30 points). Due at the beginning of lecture on Friday, March 11th. Consider Problem 14.20 on page 679. Solve the dynamic force 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. You must include clear free body diagrams of each moving link with the three scalar equations. Clearly identify all of the unknown variables in this problem. Note the following assumptions: (i) the cranks 2 and 4 are balanced, that is, the centers of mass of links 2 and 4 are coincident with the ground pivots 2 O and 4 , O respectively; (ii) gravity is acting into the paper (that is, in the negative Z-direction); (iii) the effects of friction in the mechanism can be neglected; and (iv) the only torque acting on the mechanism is the driving (or input) torque on crank 2 at the crankshaft bearing (or ground pivot) 2 . O 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.
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2 Solution to Homework 7. The free-body diagram of link 2 is shown in Figure 1. Figure 1. The Free-Body Diagram of Link 2. The sum of the external forces in the X-direction acting on link 2 can be written as 22 XG X EXT Fm A = (1a) Since the center of mass of link 2 (that is, 2 G ) is coincident with the ground pivot 2 O then Equation (1a) can be written as 12 32 0 XX FF + = (1b) The sum of the external forces in the Y-direction acting on link 2 can be written as yG Y EXT A = (2a) which can be written as 12 32 0 YY + = (2b) The sum of the external moments acting on link 2 about the fixed pivot O 2 (which is also the center of mass of link 2) can be written as 2 GG EXT MI α = (3a) which can be written as 2 23 2 2 1 2 2 += XY YX G RF T I (3b) or as 2 2 2 32 2 2 32 12 2 cos sin G T I θ θα (3c) Therefore, there are three equations and five unknown variables for the free body diagram of link 2. The unknown variables are the four internal reaction forces F 12X , F 12Y , F 32X , F 32Y and the crank torque 21 2 . = TT
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3 The free-body diagram of link 3 is shown in Figure 2. Figure 2. The Free-Body Diagram of Link 3. The sum of the external forces in the X-direction acting on link 3 can be written as 33 XG X EXT Fm A = (4a) which can be written as 23 43 3 3 XX G X FF m A + = (4b) The sum of the external forces in the Y-direction acting on link 3 can be written as YG Y EXT A = ( 5 a ) which can be written as 23 43 3 3 YY C G Y FFF m A + −= (5b) The sum of the external moments acting about the center of mass of link 3 (that is, G 3 ) can be written as 3 = GG EXT MI α (6a) Taking moments acting about the center of mass G 3 will couple the equations (preventing the use of the method of inspection). Therefore, the sum of the external moments will be taken about point A on link
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This note was uploaded on 04/23/2011 for the course ME 352 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Homework7asol.spring11 - ME 352 - Machine Design I Spring...

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