HW15 - 3.67 Using an allowable stress of 55 MPa, design a...

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Unformatted text preview: 3.67 Using an allowable stress of 55 MPa, design a solid steel shaft to transmit 10 PROBLEM 3'67 kW at a frequency of 15 Hz. SOLUTION Tw=5§*10‘ Pa P: IOXIOzW) {3:15 Hz _ P _ onlo3 _ . T—m‘m- it (2) (a 09.10) 3 = Luann)" m —- = 17;: " - ‘ITSS'XID‘J 10mm;3 m d = 2:. = 21.4 mm PROBLEM 3.68 3.68 Using an allowable stress of 5 ksi, design a solid steel shafi to transmit % hp at a speed of 1725 rpm. SOLUTION fleece) = 3300 154“ /s 3300 _ . 2, 3160 xlo's in; 3.69 Design a solid steel shaft to transmit 100 hp at a speed of 1200 rpm, if the maximum shearing stress is not to exceed 7500 psi. SOLUTION 6560 x10’ 1%.», /s _ 5593123.. -.— 5.2.52! x103 IL-iu 1711203 oi=2c = L528 in. 3.70 Design a solid steel shaft to transmit 0.375 kW at a frequency of 29 Hz, if the PROBLEM 3'70 shearing stress in the shaft is not to exceed 35 MPa SOLUTION = 2‘? Hz 2.0530 N-m -2.E_42_1<_2-_(2_s’§_-2_ -‘l 3 3.3qg WM 0’: 2c. 7 6.69 mm PROBLEM 3.71 PROBLEM 3.72 SOLUTION 3.71 A hollow shafi is to transmit 250 kW at a fi'equency of 30 Hz. Knowing that the shearing stress must not exceed 50 MPa, design a shafi for which the ratio of the inner diameter to the outer diameter is 0.75. SOLUTION ’l’a‘JSOde'Pa. P=250vlosw ¥=30 H1 — s T = 3%; = fizfibfgfig = 13233 N-m J‘ = 5% (C:_C'H) = g (p -(§)”)c: = 1.0738 c2“ 1:135 =__1—_Q!.___ (:3: T : 326:3 J Long c,“ 1 mm 7: -fl.0738)(50¥(o‘) 2C3. = "7M 3. 72 One of two hollow drive shafts of an ocean liner is 125 it long, and its outer and inner diameters are 16 in. and 8 in., respectively. The shaft is made of a steel for which 1;“: 8500 psiand G= ll.2 X106 of the shafi is 165 rpm, determine the one shaft to its propeller, (b) the corresponding angle of twist of the shafi. psi. Knowing that the maximum speed of rotation (a) the maximum power that can be transmitted by J: IiCcJ-q"? E0339“): 6032.8 in" It: IS"!- ~- T: gg— .: : 6-4058xl061b."” -F 60 (a) P z szT = 21102.75 )(e.4082xzo‘) = “0.7%chc Ié‘l‘n /5 “DJMKD‘ Ib-ia/s 3 ‘Téo‘B Jug/s AP = '5'78 “o “P' ‘ l." ‘25 ‘H‘ '3 I500 in _ TL a (6.4oszx/o")§lsc>o) \ ¢’ {337 " (ll.2>¢lo°)(6031-8) _ O'Hzg NJ - 8. 15° d 3.137 and 3.138 A 750—N'm torque T is applied to a hollow shafi having the cross section shown. Neglecting the effect of stress concentration, determine the shearing stress at points a and b. Thick "P5 = 6 mm- PROBLEM 3.137 QR SOLUTION \ Area. bounded by eerie.» finer, Q = Z—Ec’ss)‘ + (60mg): 738] mi“ : 733) >110" m" t = arl‘ a. anal 0. 006 up 750 W = 3~97><I0 Pa = 3.47 MPa.‘ ’l 3.137 and 3.138 A 750-N°m torque T is applied to a hollow shafi having the cross section shown. Neglecting the efl‘ect of stress concentration, detemiine the sheath g stress at points a and b. Thickness = 8 mm. PROBLEM 3.138 902nm l SOLUTION Darla-'9 0? Corn er t = e Tlan 30° 7 t e — 2 ‘l‘an 30" 3 : ———p\+an 30, ‘-‘ 6328 m». b = 90-26, .— 76./4L/ m, ANN. looonsleci center 9.3.2 (1: a)? L {$195 fi-imwfl‘ = 2$I0.6mm1' ; Zé’ltléxio‘e m1 t = 0.008 m l 750 7 ‘ 75 = ‘ ' = 13.3w"; P 22 0, (Z)(o-003)(25/o.é v10“) » ‘ 7 Q =13.€7 l’lPa. ' 4 3.139 and 3.140 A 50-ldp-in torque T is applied to ahOHOWSbafibaI/IbgUICL‘IOSf section shown. Neglecting the effect of stress concentration, determine the shearing stress at points a and b. PROBLEM 3. I39 a 0.2 in. SOLUTION I Am. L....:.J 1., I Cenle‘l‘fls‘ne. \ 9‘ 0.2 in. L a = %(I.q)‘+(3.3§)(s.s) T— 5.35 b t 18.‘-IO ‘m1 L 1:0.2131 N_ l _ 50 _ - L ' ZtCL ‘ (2X0.2)(I8.‘f0‘\ - 6"” ks' ‘ A‘l [$013+ ‘9 t = 0.3 in, 2/ T 50 - ksi ‘ : 27:0. :(ZYOJSXISJIO) 3.139 and 3.140 A 50-kip-in. torque T is applied to a hollow shaft having the cross section shown. Neglecting the efi‘ect of stress concentration, determine the shearing stress at points a and b. PROBLEM 3.140 3. Km. SOLUTION Area. houndeei' By ceq‘l'erfu'ne T (1,4314%) = lat-3 1+; = mm W J" k—zé, —al Al. Poi-Cl a 1:: {Film ’ 0.25. 1’», "’- ‘ --—————-——§° — 78 k‘ L ‘ 2m. ’(2\(o.25)02.7m) ‘ ' 6 5' “ A+ Poinl L t = -§- i... r 0.375;. . L (QXOmnmfn q ) = 5. k5; “ ...
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HW15 - 3.67 Using an allowable stress of 55 MPa, design a...

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