35_ch 03 Mechanical Design budynas_SM_ch03

35_ch 03 Mechanical Design budynas_SM_ch03 - (a n = 2000...

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48 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design T 1 / 8 = 1 3 (9 . 6)(10 4 )(5 / 8)(1 / 8) 3 = 39 . 06 lbf · in Ans. τ 1 / 16 = 9 . 6(10 4 )1 / 16 = 6000 psi, τ 1 / 8 = 9 . 6(10 4 )1 / 8 = 12 000 psi Ans. (b) θ 1 = 9 . 6(10 4 ) 12(10 6 ) = 87(10 3 ) rad/in = 0 . 458 /in Ans. 3-37 Separate strips: For each 1 / 16 in thick strip, T = Lc 2 τ 3 = (1)(1 / 16) 2 (12 000) 3 = 15 . 625 lbf · in T max = 2(15 . 625) = 31 . 25 lbf · in Ans. For each strip, θ = 3 Tl Lc 3 G = 3(15 . 625)(12) (1)(1 / 16) 3 (12)(10 6 ) = 0 . 192 rad Ans. k t = T = 31 . 25 / 0 . 192 = 162 . 8 lbf · in/rad Ans. Solid strip: From Eq. (3-47), T max = Lc 2 τ 3 = 1(1 / 8) 2 12 000 3 = 62 . 5 lbf · in Ans. θ = θ 1 l = τ l Gc = 12 000(12) 12(10 6 )(1 / 8) = 0 . 0960 rad Ans. k l = 62 . 5 / 0 . 0960 = 651 lbf · in / rad Ans. 3-38 τ all = 60 MPa, H ± 35 kW
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Unformatted text preview: (a) n = 2000 rpm Eq. (4-40) T = 9 . 55 H n = 9 . 55(35)10 3 2000 = 167 . 1 N · m τ max = 16 T π d 3 ⇒ d = ± 16 T πτ max ² 1 / 3 = ³ 16(167 . 1) π (60)10 6 ´ 1 / 3 = 24 . 2(10 − 3 ) m ± 24.2 mm Ans. (b) n = 200 rpm ∴ T = 1671 N · m d = ³ 16(1671) π (60)10 6 ´ 1 / 3 = 52 . 2(10 − 3 ) m ± 52.2 mm Ans. 3-39 τ all = 110 MPa, θ = 30 ◦ , d = 15 mm, l = ? τ = 16 T π d 3 ⇒ T = π 16 τ d 3 θ = Tl J G ± 180 π ²...
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