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homework3solution ok - ME 301 THEORY OF MACHINES I FALL...

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Unformatted text preview: ME 301 THEORY OF MACHINES I FALL 2011 SECTION 01 HOMEWORK 3 SOLUTION where n = l since all the joints are revolute, then f i = 1 3.l = 2.j +5 (1) Since the right hand side of the equation is odd, the leftside must be also odd. Thus; 3.l = odd Since 3 is odd, l must be odd. l = odd eq.(2) And for any mechanisms number of links equals to the sum of the binary, ternary, quarternary and so fort, where l 2 is the number of binary, l 3 is the number of ternary and etc. l = l 2 + l 3 + l 4 + l 5 + ......l n eq.(3) The number of kinematic elements in the mechanism will be equal to: 2.l 2 + 3.l 3 + 4.l 4 + 5.l 5 + ......n.l n =# of kinematic elements eq.(4) Since two kinematic elements, when paired, will form a joint; the number of kinematic elements is equal to twice the number of joints in the mechanism: 2.j = 2.l 2 + 3.l 3 + 4.l 4 + 5.l 5 + ......n.l n eq.(5) Substituting Eqs. (3) and (5) into Eq.(1) 3.l 2 + 3.l 3 + 3.l 4 + 3.l 5 + ......3.l n = ( 2.l 2 + 3.l 3 + 4.l 4 + 5.l 5 + ......n.l+ ....
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This note was uploaded on 01/08/2012 for the course AAEC 2005 at Virginia Tech.

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homework3solution ok - ME 301 THEORY OF MACHINES I FALL...

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