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Sol HW4 - 3.7 The solid spindleAB has a diameter d = 1.5 in...

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Unformatted text preview: 3.7 The solid spindleAB has a diameter d, = 1.5 in. and is made ofa steel with an ' allowable shaming stress of12 ksi, while sleeve CD is made ofabrass with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A. Problem 3.7 SJI‘J sFMaUe AB ‘- c = Ji&¢ = 57:04:?) = 0.75 in. J 2 Ed = §(o.7s)‘ = 0.14am! M 21min: 75:9“ J'z’ (0.4470001) H - . ‘ Tm = c J: Wis‘fl— — 7.953? «mm-m Sfleau‘e. CD 3 Ca: $501 = 13(39‘) = l—5 Eh. C. = C‘;L —t = r-5_o.25 -= !-25 m. J: 'iLCJ- cl”) 2 l{(l.5“-L2s“}= 4.:172 m" ~ It: ($11300) __ ' . . l%‘: “—01-“— :"--—-——’-:-§——— — lq,'213 klp-IV‘ Alwwa/Q‘pe Vat/“J2 6‘? +0Vfiue, T is +Le smdflefi. T: 7.?5' LhYx-{n 4 Problem 3_ 1 3 3.13 Under normal operating conditions, the electu'c motor exerts a torque of 2.8 kN - m on shaft AB. Knowing that each shafi is solid, determine the maximm sheeting stress in (a) shaft AB, (1:) shaft BC, (0) shaft CD. T1; = L4 ICN - m T12: = 0.9 L'N - m " 48 mm To = 0.5 kN - m 46 mm (0.] SLA'H‘ AB: TAB“: 2,8 kH‘Mn -: 23x10? N-MJ C=Ji91= 23mm= 0.028m ,1 To 21‘ “ (2.}(2.8><IO$) : L = -—- - ._ M J" Fr.“ 17 (0.01213 2110:4069; 215311 Mme-i (m 5qu 8C: Tear LL; Wm = LLJXIog N-th agar: 2%“ 0.024% 7: _ 31 q mason?) k" we“ " w(o-oz=n3 (cl 3315+ CD: 725:0-5kumn : 0.5xlozN.m) C='§a’r Z'f'mm: 0.02‘f-m ,‘5, _ QT __ (23(0.5’xl03)_ 176* ' “(0.02433- ‘ G4-‘+7>«lO‘Pa fat: Q‘lfi'HPot 4" 93.053 xnf Pa 11:. = 2%. 0 Met ‘4 'LLD " . r 3.72 The design of a machine element calls for a 40 mm—outerl-diameter strait to : PrOblem 3‘ 72 transmit 45 kW. (0) Ifthe speed of rotation is 720 rpm, determme the maxrmum ' shearing stress in shafi: a, (b)1fthe shafi of rotation can tae Increased 50% to 1980 rpm, determine the largest inner diameter of shafi b for Whlch the rnaxrmum sheanng stresswillbethe sameineaeh shaft. m F= 7:29 = :2 Hz P= 95%}: Hex/o" w 13 ass/03 r. = w———~—- 6.8 ‘ C‘: at”! = 20 mm": CLOZOM-e 1’: I? r 37%- : —--——(27rx(gf’jg:?§ = LEILF/H’V/Oc Pa. 2;”: 47.5MP0..4 (H ‘P’ [23:0 =18 Hz Trfié¥w T 397.3% um T T T}; r #1255sz C'L}: C’j' 27:32 cf: 0-0201—W ‘3 $3.333}be c, r razmzo’g m = {5.20 m Problem 3.35 3.35 The torques shownfi'e exerted on shafts are solid and made of brass (G = 39 (am and B, (b) A and C. pulleys A, B, and C. Knowing that both GPa), determine the angle of twist between 120033111 (a) [Mafia 0'? “(LWI‘S'IL between A emof B The = Ltoo M-MJ L“; L2 m 'c = 9+9! : 0.0L?“ 6 = 37:40" fa er%rc"- 79.52%” M“ _ TL - (HooML'z) , __ _ _.___,__________ (pm 5: (sfixroficflsleo‘?) - 0.154772 m! f) 90M; ‘— 3-37 a f) “ (b5 Amjflg a‘P dean's" Le‘llweem A a”! C T85: goo N-M 3L“: 1.8m) c=iot €0.03: m, G = 3956/59: Jag gt} =E(o.0203"r 25:.327xto" m” ‘ : (SOOKLM __ 59'3" m _ 0--*‘*““2”t ’3 so”: = (pmfflu cpacfi o.15477290.Hmmzapovgtaomt? qflmc 0350"? 4 ...
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