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HW16

# HW16 - 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 “22.1! = 55‘xIO‘ Pa P: louo‘ W, {3:15 Hi _ P _ IOXlOS _ _ T—m‘m- [0&IONM £12 (2) (a 09.12)) 3 = Lamina" m — = was " - ‘ITSS’HD‘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 1m: 5&5; = ‘ ﬂeece) = 3300 15% /s 3300 _ . 2,3160 xto’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 6560x103 25.5,, /s _ 5503.121. = 5.152: “0’ 144.. 21T(7~03 _. (23(5.2521x103)_ o , . 3 ' “(75.003 — .‘f‘lé8 m 0’ = 2C. = L528 :ﬂ. 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 211/ = 3910‘ Pa. 13 = 0.379103 w = 262 Hz _. P _ 0.37S‘Xlo3 __ 2T 3 _ 27’ _ ‘(2)(2-O§80\ __ -‘l 3 ms ’ TT’Z ' 1T(3§>(IO") - 37-43“) "‘ 3.34nco’3m = 3.345 WM 0!: 2c. 7 6.99 m PROBLEM 3.71 SOLUTION ’2’." = So not Pa. P = 250v :03 w 27F 2'" (30) J = L; (cs-cw) = so «em: 7: = 1:95 = l2;— (:3: J L073: c,“ 1 PROBLEM 3.72 \T - Ca 'F=-LG-§- = 2:75 Hz — S T= 3— = ————-—-2"°"° = 13233 N-m ll 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. ¥=30H1 1.0738 at“ T nus: .— ).0738 7: ' {n.mszxsoxto‘) d1: 2C2. = 58.2 "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= 11.2 X106 SOLUTION 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: 11E(c,_"-c,")= Eater“): 8031.8 in" = I59. - T: J’f =L___—X————)603l°88 959° -: 6,40583‘10‘1b-l‘h (CL) P = ZWFT = 21:02.75 )(GJfoSBv‘IO‘) = Marine,“ 11,4». /5 “DIMKD‘ Ib-ia/s 3 6600 lh‘in/S hp = |£.78 1‘10 hP' ‘ l." 125' ‘H‘ '3 I500 in TL (6.4088 x10 " )g I500) . 7 w" 1" - 0.19 M! \ ¢ {337 (ll.2>¢lo°)(6031-8) 2% p = 8.IS° d 3.137 and 3.138 A 750—N'm torque T is applied to a hollow shaft having the cross PROBLEM 3.137 section shown. Neglecting the effect of stress concentration, determine the shearing stress at points a and b. Thick "P5 = 6 mm- /\ SOLUTION 63W” \ Area. boondeal by carter finer, Q = Z—Ecas)‘ + (com): 738] mi“ : 733) >110" m" t = 0.006 .5 a.+ ball» am 5. 750 W = 8197*"? 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 PROBLEM 3.138 section shown. Neglecting the eﬁ‘ect of stress concentration, detemiine the shearing stress at points a and b. Thickness = 8 mm. a SOLUTION 0 {"22" Darla-'9 0? Corn er b = 90-26, .— 76./4L/ m, Area looonsleci lo] center 9.3.2 (1: a)? L {\$195 “ff—104.1%? = 2\$I0.6mm1' ; Zé’ltléxio‘e m1 t = 0.008 m 3*; 7 7'50 ' _‘ s l8.g€7>€lo‘ P4 220/ (Z)(O-008)(2510.éylo) » , =13.€7 l’lPa. , 4 3.139 and 3.140 A 50-ldp-in torque T is applied to abollowsbaﬁbaI/I‘ngtliccross 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 L.) . {\% 3.5m. 0.2m. CW+8¢IW° T— ‘_._\|» L a = %(n.q)‘+(3.3§ )(u) 5.35 b : 18.‘-Io if t: 0.21") N 50 — . ”th (2X0-2)(12.wo\ = 67’" *5" ‘ , _ . _ l- _ SO _ . A‘l Poln‘l‘ l: t - 0.3 m, 2/- Zia. .(ZYO.3)08.‘IO) - thgg ks. ‘ 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. hednd’u" B7 ceq‘l'erfu'ne T 0.43M) = 12%} H5 = mm W i le—zé, —al J- . N _ _ 5‘0 _ . L ‘ 2m. ’(2\(o.25)(:2.7m) ‘ 7'86 “5' “ A+ Pom L t = 5:15. r 0.37:... 1,: T 50 27:0, : (axosmmmq) = 5. 2". k5; “ 4.3 The wide-ﬂange beam shown is made of a high—strength, low-alloy steel for which PROBLEM 4.3 V a, = 345 MPa and 0,, = 450 MP3. Using a factor of safety of 3.0, determine the largest couple that can be applied to the beam when it is bent about the z axis. Neglect the eﬂ‘ect of ﬁllets. SOLU TION “'T 360mm 7 1,: ﬁthAa‘ 7 lomu 7. .329 m». : 7%(250X183) L—mﬁ—J 18mm i 4(250)(18)(17I)‘ 1_______l : :3;.7vexlo‘ m." 11: 7'5 003(329? = 22.344xlo“ mm 13/. 706 >40“ mm" 29L-7ex/o‘mm“ = 29/.74x/o“ m G : MC where C : gig—Q: [80 mm = O. '80 M 150 HO: ’Pa. C —L _ (150,40 )(2‘7L7éxlo ) = 243x103 N-m C 0.!80 : Z43 kN-m ‘ 4.9 thnough 4.11 Two vertical forces are applied to a beam of the cross section shown. PROBLEM 4.10 Determine the maximum tensile and compressive stresses in portion BC of the beam. 3 in. 3 in. 3 in. SOLUTION 36 = 3 I." Neu'i'vm/ man‘s ﬂies 3:14. aJnoVe ‘H\e base. - ﬁsh? A,J,2 = g (33(€\3+ (19>C2)’ r was zn" 3551525 A”); :45 (W233 4 :mm‘ = 73 w \26 + 78 20"} in” ysd = ‘3M ‘P M M- Pa. = O ____|§f> M= Pa. = (15x40): 600 k.‘,=.;n. P a. —>l 6;, “%‘ﬁ°‘> = _M = -I’+.7/ ksr 20‘4- _ (coax—3) 20L! 7 8.82. “Sf ...
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HW16 - 3.67 Using an allowable stress of 55 MPa design a...

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