8_ch 16 Mechanical Design budynas_SM_ch16

# 8_ch 16 Mechanical - budynas_SM_ch16.qxd 17:45 Page 403 FIRST PAGES 403 Chapter 16 16-9(a Counter-clockwise rotation 2 =/4 rad a= r = 13.5/2 = 6.75

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Chapter 16 403 16-9 (a) Counter-clockwise rotation, θ 2 = π/ 4 rad, r = 13 . 5 / 2 = 6 . 75 in a = 4 r sin θ 2 2 θ 2 + sin 2 θ 2 = 4(6 . 75) sin( π/ 4) 2 π/ 4 + sin(2 π/ 4) = 7 . 426 in e = 2(7 . 426) = 14 . 85 in Ans. (b) α = tan 1 (3 / 14 . 85) = 11 . ± M R = 0 = 3 F x 6 . 375 P F x = 2 . 125 P ± F x = 0 =− F x + R x R x = F x = 2 . 125 P F y = F x tan 11 . 4 = 0 . 428 P ± F y =− P F y + R y R y = P + 0 . 428 P = 1 . 428 P Left shoe lever. ± M R = 0 = 7 . 78 S x 15 . 28 F x S x = 15 . 28 7 . 78 (2 . 125 P ) = 4 . 174 P S y = fS x = 0 . 30(4 . 174 P ) = 1 . 252 P ± F y = 0 = R y + S y + F y R y =− F y S y =− 0 . 428 P 1 . 252 P =− 1 . 68 P ± F x = 0 = R x S x + F x R x = S x
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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