ME350_W10_HW6_Solutions - 29 2 2 7/8 2 1.202 in 4 s A π =...

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ME350 Fall 2007 Homework 1 Solution Problem #4. a) min / 40 50 2000 rev n = = in l 25 . 0 = min / 10 ) 25 . 0 ( 40 in v = = b) screw lbf F / 5000 = in d m 875 . 1 125 . 0 2 = - = 033 . 1 ) 2 / 29 cos( / 1 sec = = α From EQ (8-5) in lbf T R - = - + = 17 . 686 ) 033 . 1 25 . 0 1 . 0 ( ) 875 . 1 ( ) 033 . 1 875 . 1 1 . ( 25 . 2 ) 875 . 1 ( 5000 π in lbf T C - = = 1250 2 10 05 . 0 5000 perscrew in lbf T TOT / 17 . 1936 1250 17 . 686 - = + = in lbf T Motor - = = 8 . 96 8 . 0 50 2 17 . 1936 101369 ) / ( 2 * min) / ( 2000 * ) ( 12 / 8 . 96 * = - = = rev radians rev lbf ft T H ϖ c) Note: Electrical horsepower can be found using the relation: 5252 ) ( RPM ft lbf Torque P hp - = hp P hp 07 . 3 5252 2000 12 / 8 . 96 = =
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Problem #5. Members: S y = 71 kpsi, S sy = 0 . 577(71) = 41 . 0 kpsi. Bolts: SAE grade 5, S y = 92 kpsi, S sy = 0 . 577(92) = 53 . 08 kpsi Shear in bolts
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Unformatted text preview: ( 29 2 2 7 /8 2 1.202 in 4 s A π = = 1.202(53.08) 35.44 kip 1.8 s sy s A S F n = = = Bearing on bolts 2 2(7 /8)(0.75) 1.313 in b A = = 1.313(92) 54.89 kip 2.2 b y b A S F n = = = Bearing on member 1.313(71) 38.84 kip 2.4 b F = = Tension of members 2 (3 7 /8)(0.75) 1.594 in t A =-= 1.594(71) 43.52 kip 2.6 t F = = min(35.44,54.89,38.84,43.52) 35.44 kip F = = The shear in bolts controls the design....
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This note was uploaded on 05/12/2010 for the course MECHENG 350 taught by Professor Shih during the Winter '10 term at University of Michigan.

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ME350_W10_HW6_Solutions - 29 2 2 7/8 2 1.202 in 4 s A π =...

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