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# HW11 - Problem 8.1 8.1 A W10 x 39 rolled-steel beam...

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Unformatted text preview: Problem 8.1 8.1 A W10 x 39 rolled-steel beam supports a load P _ _ as sh 45 laps, a = 10 .m., and a." = 18 ksi, determine (a) that:- IVI-m ‘ 90 {ﬁrs ilk/Illa... : (%I(f0] = [115.0 Liv-in Foe W IO 3’3“? ﬂied 5+9} SEG‘fl‘OIﬂ.) a: 932:”. bp = 1735“"- tr = 0-530“ -415- . M (km-[‘0 t”: 0'3’5‘“ I: '“ 2"" in” 5x = 42‘ | m? - _I _ _ : Y ' g6! - LL36 an. ya : C _ tF -_- 4-143;.“- = ‘ML-su ’45?) “50 “F“ m 6'“ 5.. ' “ﬁst:— 5-»; 10.9: L1. 43 ’1‘. ‘i’é A4. = ht? : 4.13205- M 3+“ ~' *(myn = Hanan. K QB: AFR; G" = ( )(I0.G'?)'-' c1255" ks!" : H.866: m3 ,2; ._. IVLHGE : (gamma) J ' Lt, Ezquosrs] R =1/(‘5ihT‘ + ti; : Hid-+0 ks: 09) Sun:- 1 g; + R :- lq. '7 k5; 6 : [q [7 Its. May. ’ I (g) ssm 6H,, > 6;.(= ts lei)J w 10 5: 3e rs mi ”mum, = {3.5—8 *5:- Problem 3_15 8.15 Neglecting the effect of ﬁllets and of stress . . . . concentrations smallest permlsmbfe dlarneters of the sohd rods BC ' andCD. Usetu':6° 180 mm 160 mm Behslina Home-«+5 magi Torgues Jus-l' +9 Hue Age-Ff EC: M = (.b‘ooﬂmm = 30 N-m T = (\$0010.18) = 6:0 u-m JMZ+ T1 = 120.4% N-m Smaﬁjﬁsf‘ eV‘Ml.\$'5£b’e C‘I‘nme‘i’eV‘ O! c.“ : (21(120.~ne] —r, 1T(CD>‘IO‘) — 1.277GS'KKD M c = amassm = 10-95mm d3; 2r.7mm "‘ Sm'aﬂes‘l Eerma‘ssfble chancel-w 0'53. 2 as .32.) - cs: (7T 60:10‘) T ”.6613 “0‘ mg C = 0.61670 m = 16.? Im- clw-"BBﬂmm “' Problem 8.19 . .. 2% C- CD1 T: {4)(500) = 20:10 ' DE: T = o " h 3 C ‘D E .500 5'00 .500 .500 35a; in-JL. Forces in how'zau‘fa.’ In at: 533.33 EGG-‘7 335.35 I l CWHCAJF seal-Fons are ef-Haw side ”4‘ disk C. T= 2000 .‘nJL. M2 = 350:; 9.1}, " M} = 9667 in J5 3 Ho”- _ ic _ 2;”: 2 466? +35m 1- boo“- : (177083th .2 LS7? in. Problem 8.27 __ 240 FEM FF _ QOSQc/nin : 4 ”2' = P - #ono‘ _ Z1T'F ‘ (21')(‘l) ‘ Benéinﬂ mane-ff a?" 8: M8 = TA’E—W + “I" m 311nm}:- _ (2)J9#‘2.l‘+3?7.8ﬁ" 2' 'n 1;” ' 1r (cowo‘) = Q3103 no ’- MS I_ a c'IRLC ’5 C' = ”3.47? no" in A: 2: -— 37.0 :10" m a! = 37.0 mm 8.40 Two forces are applied to the pipe AB as shown. Kno inner and outer diameters equal to 35 and 42 mm, normal and shearing stresses at (a) point a, (6) point I). 1:._f_blem. B40 Wing that the pipe has . . resPectively, determine the Cu: in ' 21mm C-‘-‘A" a [7.5mm ) a "f A 1' "1617c:a "‘33) -- Has-.33 m:- J" = EEGJ— q") -: IS'SJGC no“ mﬁ‘ [2135; I = -,';J'- 77.033: 10‘ rm" .‘I For sea-unﬁafmﬂa w\'+L| samﬂcfr-cufv cﬁaﬁ — - x Q=%(C:_C;’) Q: 2.50:: x103 m3 -'._ A+ ‘H'Ic Janina-n Con+a£wﬂn3 Pal-VA a ‘3' My 13 -.'__ — lsoo N V==-—moow Vx=o :5";- t =—(LLS'xlo")(|\$t>o\ = ~57“; N-m [1'- ' =-(75x.'o")(lzoo) : —eo N-w " ~—- (4702:6310sz - IOB N-m -.: 3— MILE : -(F_Sbca _ [-9OKZIXJD‘3) _-.--- A I 1423.33.45: 7.1033“ to-.. 6‘ 20.4 Mpg .— .= , TA- L/zQ _ (108\(2rx:o'3) _ ,7: I + It ‘ 153.:66xm"? +0 249.3% NF” "" 5 _ .E. Mic. .— "5°° (-GZSKRIMD'S) _ 6' - A + I H23.35¥f0"+ 7?.083xucy'l ' ‘ -2l'5MPGL "‘ N Tc IVIIQ (:03 Kama") (I200) 2.50:: :00") L = ‘— + = -1 + -ﬁ -3 .3“ It lss-lccnlo ("measure )0on ) 1321??? MPO; “'3 Problem 8.45 8.45 Three forces are applied to the bar shown. shaming mm at (a) point a, (6) point I), (c) point c_ ctjcujzd'e ‘FchS and coup?“ rd 52:47 __ POI-“+5 g, b) 3—1) 9, 1.1-10 5-n. COIL. ﬁlPur/I‘\ M1 Fortes C0019)“ P: 5'0 kilns m = 6 1“I“ V: 7 2 {ﬁrs -' M; = (10.5— 21(9) —— SI L594“ Mu =(I0.5)(2) = 2| kip-in Sad-fan meW‘I-Fes. A? (LSMH‘.83 = 8.6.?le I, = Tg-(q “(L 81‘ = 2 3322 :n' I, = 73-09(41)" = rag-sag m; c -= --..E. M 76 M Z _ YQ 5+» axes 6‘ A + 4—1; 4. J—L _ L? (a) m: 76 = OF) 2 = 0.? in.) Q:— {f_3j(1_q)(r_1) .1. 5‘ [34' In: — _ ﬂ (21mm) _ ”2'”; 6 96‘! + O " 2.3313 6 - 2.3: k... V .l " V (5M: :84) .. I' = ' _ O. X s 2/ (I6.\$38'8)(!.3] 'r: - LOIIZqu‘ (b) Po€n+ l: 1 2o: 1.2m.) Z=0.‘?u'n. , Q:- (1.310.236.2211 w 3.935 in‘ = so (sham gmcom ._ y 2’51; 6— 2.54 16.5888 + 2.3322 62" 6' " [.5 __ {555.8831 __ . ﬂ '. 3‘ Z' " (iéﬁsssXLg) ~ 0.73! ks‘ ._ a 0.73”“ (c) Poin‘f‘ C 3 70" 2.965.) 2=0_q{n. QT- O J 5—... ..._-5I°_ + (SHCZﬂZ + £20031 5; g 1.4“" “3.5% 16.5833 2.53.21 ...
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