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17_ch 17 Mechanical Design budynas_SM_ch17

# 17_ch 17 Mechanical Design budynas_SM_ch17 - θ = 180° K 1...

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436 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Use Table 17-12 to interpolate for H tab . H tab = 1 . 59 + 2 . 62 1 . 59 2000 1000 (1696 1000) = 2 . 31 hp/belt H a = K 1 K 2 N b H tab = 1(0 . 944)(2)(2 . 31) = 4 . 36 hp Assuming n d = 1 H d = K s H nom n d = 1 . 25(1) H nom For a factor of safety of one, H a = H d 4 . 36 = 1 . 25 H nom H nom = 4 . 36 1 . 25 = 3 . 49 hp Ans . 17-19 Given: H nom = 60 hp, n = 400 rev/min, K s = 1 . 4, d = D = 26 in on 12 ft centers. Design task: specifyV-belt and number of strands (belts). Tentative decision : Use D360 belts. Table 17-11: L p = 360 + 3 . 3 = 363 . 3 in Eq. (17-16 b ): C = 0 . 25 363 . 3 π 2 (26 + 26) + 363 . 3 π 2 (26 + 26) 2 2(26 26) 2 = 140 . 8 in (nearly 144 in) θ d = π , θ D = π , exp[0 . 5123 π ] = 5 . 0, V = π dn 12 = π (26)(400) 12 = 2722 . 7 ft/min Table 17-13: For
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Unformatted text preview: θ = 180°, K 1 = 1 Table 17-14: For D360, K 2 = 1 . 10 Table 17-12: H tab = 16 . 94 hp by interpolation Thus, H a = K 1 K 2 H tab = 1(1 . 1)(16 . 94) = 18 . 63 hp H d = K s H nom = 1 . 4(60) = 84 hp Number of belts, N b N b = K s H nom K 1 K 2 H tab = H d H a = 84 18 . 63 = 4 . 51 Round up to ﬁve belts. It is left to the reader to repeat the above for belts such as C360 and E360. ± F a = 63 025 H a n ( d / 2) = 63 025(18 . 63) 400(26 / 2) = 225 . 8 lbf/belt T a = ( ± F a ) d 2 = 225 . 8(26) 2 = 2935 lbf · in/belt...
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