4_ch 16 Mechanical Design budynas_SM_ch16

4_ch 16 Mechanical Design budynas_SM_ch16 -...

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Chapter 16 399 16-4 (a) Given: θ 1 = 10°, θ 2 = 75°, θ a = 75°, p a = 10 6 Pa, f = 0 . 24, b = 0 . 075 m (shoe width), a = 0 . 150 m, r = 0 . 200 m, d = 0 . 050 m, c = 0 . 165 m . Some of the terms needed are evaluated as: A = ± r ² θ 2 θ 1 sin θ d θ a ² θ 2 θ 1 sin θ cos θ d θ ³ = r ´ cos θ µ θ 2 θ 1 a ± 1 2 sin 2 θ ³ θ 2 θ 1 = 200 ´ cos θ µ 75° 10° 150 ± 1 2 sin 2 θ ³ 75° 10° = 77 . 5mm B = ² θ 2 θ 1 sin 2 θ d θ = ± θ 2 1 4 sin 2 θ ³ 75 π/ 180 rad 10 π/ 180 rad = 0 . 528 C = ² θ 2 θ 1 sin θ cos θ d θ = 0 . 4514 Now converting to pascals and meters, we have from Eq. (16-2), M f = fp a br sin θ a A = 0 . 24[(10) 6 ](0 . 075)(0 . 200) sin 75° (0 . 0775) = 289 N · m From Eq. (16-3), M N = p a bra sin θ a B = [(10) 6 ](0 . 075)(0 . 200)(0 . 150) sin 75° (0 . 528) = 1230 N · m Finally, using Eq. (16-4), we have F = M N M f c = 1230 289 165 = 5 . 70 kN Ans. (b) Use Eq. (16-6) for the primary shoe. T = fp a br 2 (cos θ 1
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