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HW10 - Tltlmgh'm fin Problem7.3...

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Unformatted text preview: Tltlmgh'm Fwthggivmmofmdcnerminedmmalmdsheafin Problem7.3 Mannedugflruhliqwfiweofdnshadedfiangfludmm Use: meflmdofmalysisbasedonflncequifibfimoflhatelemenhaswasdmeinun derivatimsofSec.7.2. I 123MB] mo Hag A I20A “520' ”:5 MPa._ 45A c.0525 Asm20° / _ §+______r~255es .59 ”P‘- 20.5 m 2:; ‘12—? SF = O: 6' A 420A caszaocos 20° - LEA Cos 30.33» 20°— 45430. 20.5% 90' '5QA sin 20°5a‘n20' = O -. 5 3' IZOCoszZO" +95c0520°s£w 20' + 4,155,“..2020520" +655""2209 ‘35.: " MPq.‘ .A ‘5 F = O 7 1A MI’DAcOsZD'siu 320' - 45Ac9320"c95.?o°+ 45A sin 20’5u‘n.20° —EDA sin 201:0; 30‘ = 0 f 3 "+336 as 20.5,}, 20°+ 55(ch 220°- Sl.n1209) + 605M 200035 20’ 7:15;}? MPa.‘ Problem 7.6 -- - - - - 11.:3111:1121]:Egg-Smglvemslateofstress,detmnme(a)thepnnc1palplanes,(b) l2 Illa: 5;: 2H5: 6'3. = — t2 ks.’ 2;: r 451:5.- 2'2; gay-.15) 3 _..._...._L :- = _ {a 'i'au 29: 6.“ 63 Li + :2 L375 29,.- - 6:33” GP: 41.02510" 4 6+6” 6—6- 2 = g 1 (b) 6m,” “*1; 4T1) 73:1 _ a _ a 1' (&%2")1+ ’51 = -‘f:.‘: I? Fm = 13.09 ks; III 61,-" =- -2J.o Its: «a Problem 7.10 7 9 through 7 12 For the gmn m ofstress, determuw (a) me onenmuon oflhe planes of maxunum III-plane sheanng mess, (b) the: correspondmg nouns! stress ,12 lea: s; -416. 63=-12 ks. rvnrsks. (at) h» 295 = _ 3%? .- — (iii) - 0.53333 29; 22.07" Q= ”Lo" lama” 4 no} 6'=6‘ = 4—4—2 21m: :7. 00 ks: «It 6‘" —H.oo ks: 4 Problem 7.15 ,duenmncllwmrmal shear”,g [asbemrmmdtl‘lroush(a)25° clockwise {b}10° comm-clockmse. 9mm 305,113,. 6* : -60 MP“ g: 5’0 MP6. T5 =39 MPG. mMPn 153%:3.) : {SHPA g'VZLg-J—u-ZSMPA 6",: £1ng .4. EXQLQ cosza + 11!? SH'IZB 235,- —§l;—G'l 5:n29+?‘5ms25 63'! Z? 6"; 6'; " 62‘s.)? C03 29 ‘_ x! 5!.“ 29 {a} 9:475" 29=~$o° 5*. : (g .. 75 “5650“) + 30 s:‘n(—50°) Ql= ‘5‘.2 MPI: 4 11.3; = + 75 sin (-50“ ‘} + so 003 (-50”) 1:3,: = —32,2MP¢ .. 63- = r: + ”:5 :05 (-so') — 3:: sin 650°) 6}! = 34.2 MPo. -I (b) 8 = w" 29 = 20° 6} f 1'5 -'?.S' to: (20°) + 30 53.420") 6",: ‘95.2MPQ .3 fly. 2 + 75 5M (10') + 30 “5620“ fi-y‘ = 53.8 “Pa " 63. 2: :5 +73 Cosmo”) ~ 30 ss‘ufflo“) 63 = 75.2 HPo. 4 _ -. _ 7.22 Two wooden members of 80 x 120-11111: uniform rectangular era secti are In .. | - I I 58 -‘. .‘H . Problem 7-22 Joined by the snnple glued scarfspliee shown. Knowing than? = 22° and thoalt1 the V maximum allowable stressm in the joint are, respectively, 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel: ' mine the largest centric load P that can be applied. 0 the 5pm de 1)’ N. 1% rces Areas 1: _._..._.. L ..;_- " A= (soECIafl r acxgo‘ m: = as “0-“ w ._.-1, N4 = GwA/es‘nfi = W = 10.25: “0’” _ s-‘n 21" i;" +/ 25,:01 N—Psmfinp P: 3% = %5?27.¥3!01N 1“ $4? = fill A/‘fi‘fl’ :- Coo 2:33213-6xmd) T 15. 37$ >401 N +362 F» = o: 5 _ Pwsfln= o P, a??? : 15:7:303 7 I6.53v(o’ u T!” ”J!“ «the PM Paw-ems. P :— 1&5: w 4 '~ .‘r.-. J_I-u . L" 7.24 7.14 A 19.5-kN force is applied at point D ofthe cast-ironpost shown. Knowing I .' PrObIem than the poet has a diameter of 60 mm, determm' e the pnnclpal stresses and the maximum shearing stress at point K. in; :1/{30011 :- (125? = 325 MM Resofiae Hm [ms kN an: F- M— Ford: 1: iw+o )4) j) 0-614“ Z comfohtn‘i's, are F, 2-33%: 07.5] =43 W ? "3““33” F31 = 1%? (I‘r-S) :—7.sw = —7.5‘n:o" N I I 53¢}:th 'Hu: game «mafia. {3.5km a?! 1 "We. Poimi‘ an HR: -va.I'S wheme H’ infersaci": ‘Hne, Prime canfar‘mu'na efamen‘f‘s E. H MJ’ K. YIN”: Ff: 01 Fix: ‘— tam? N J (32,: —7:5vuo’~ Mx’: ”(7.5- kN )(300Mm— [Gaunt-1)": 4- LS'KIDS Ne!“ Mf {7.5 kumsa m1 r l. mswo“ w-m Hg. Mg: —U$ kN‘)( {sol-m) = -2.7'~cto3 N-m \V/ Prob/Few. 7. 2'1 COW+|-HUEG‘ ——-—-————-—.._._________________ ELEEEFiI-CS) O'F SE¢I+FOV1 . (Cuflcfe) C = 'Ea' =' 30 mm = 30 MID-1 WI A = Fc‘ = VCSo)‘ = 2.3279 xro3 m“ - $35.!7x203 m" : agsm no“ w." J = ¥c1 = £610? = mmzsxro‘m“ = r.27235x/o"m" (SeMl'cl'r‘cIPe\ %(3C")‘ = I'S’HI'ba mm-a ~'-‘ 1"? “KO-chug M .c - [Skid5 (- 2.7x105)(3oxm-3) _z_= IS A + I 2.227%*!o‘3+ manna-9 ———+. ’1’ = —:33,ngro‘ Pa. = was.“ ME; I J ___F2Q _ ;_ (-ZSHO‘IIMO") (:.':2s*~{a‘)(30wlo‘3J I I f” It + _ (£36.f7*z°=‘i(6mro"§+ r.2?235‘~ra-‘ “l “ = 3.5353xro‘ + 2g.szcuo‘ = 30-05::{0‘ Pa = 30.06 MPa. E ‘ ‘ _ 5+6} ( gg—g |"Inc-I “/0 S+P€S§E$ Gina‘JM-‘n _ *1 i C 2 )2. + f3; _ 9 - [33.6% { |33_6 61.49am“ ‘ _— 2 I." (g%———q)z ‘F (39- 05]; neasqs i 73.2% mpg. 61w: 6.45 MPq 4: 6"...“ =-I€+a‘.1 MPG. "' Maximum skeécflng aim-.55 fW=/(5—-*—;—xar+ 2;; 11..., = 73.3 MPQ, «4 7.33 Solve Prob. 7.11, using Mohr‘s circle. 7.9 through 7.12 Forthegivenswe ofsirus, determhm(a)fl1eorientation nfthe planeaofmaflmmninwplancahmingm, (b)!heoorrespondingnormal stress. 6;;- fiksi, (gr-Ska?) Ely-"fhi Saw. '-' §%§* 2- l-Sksi I’m-‘- 1 3.5-0 k3: ‘ (5"r L50 Its: in II ‘ Problem 7.44 thesplloe) determme thelargmtcenu-icloadl’dmcanbeapplied. '3' P E’ ~—] 3: 3L ‘_i '4. 3 A 5'96 G'x=-EJSLY=O 1:9,:0 ) .Ppatka‘ Fania Fur Motif-‘1- ¢|IVV=IE. X: (’P/AJ o) J r: (0,0) C: (P/zAJO) _--_._P_ F2~c><_2A CDOV'Jl'no-Jges of POD-1' Y! G=§PECI~GHS 2‘8) ‘Z: %5Fn2/§ D911»: A: (soxrzo) . €1.4x103m‘r 9.910" vi“ LC 6- Lioo kPa. - #Ooxio’ Pa, P _ M...— : (2)(°r.cxlo“3)(‘loouo’) ‘ [mm W ’1 .-- 279+vzojw - 2m Ma .7 «a s 1 _ _ s _ 3M:- _ (239$.sz ggecowzo) 31-? 1t: - 600 kPa — mono (’41 P - at,” 2’3 . 5"“ W" (6.58::{05N =- 16.58 w i TLIE sun-Jpg; vague o‘F Tags-ferns- ,i' '. P —— {9-59 m 4 Problem 7. 83 7.83 The stare of plane sum shown occurs in a machine component made of a Steel with or = 36 ksi. Using the moimum—disionionwcnergy criterion, determine yield occurs when (a) 1,, = 15 ksi, (b) by = 18 ksi, (c) 1;, = 21 ksi. If yield does not occur, determine the corresponding factor of safety. (int—Ilka?) 6J=~3ks£J 81:0 6'” = Ji(5}+§3) 2' -7.55 ks; (at) ’Z" : I'S' ksf Q = W "‘ 16'. Q6 ks.‘ 6}} 6;" 1- R= SJG‘ksI 6's Gan—R - — 23-16 Ge=¥éf+ 5? — 62.61, -.- m = 13-”! hrs? Since 23.14 ksr' <' 36 ICSIJ wide! class no+ occur. Es,= g1=_a_§_ E3: m7? is Q (E21 2;! = I3 ks: R = #20451» 033’- = [8.55:4L {4'57 Gargan-l-R ‘9 H-059 Its; 61):.- 6:“: — R, :- ...2;_05q 1“,; 6;: 7r 6‘: + 6],“ -6;61 =- (J:.os«+)‘+(-2s.osq)‘-(mos-u)(-zc.osq) -‘ 33.0 ks; Since 33.0 k5; < 36 {(55 R=—/20.25+C2I)1 = 219477 leaf 6““: ELM-F2 - 8.977 ks; Sb : 6;“ — R = — 28.977 ks: 6;: J6: + 61‘ - m6; = J 0337?}H(—23.977)‘—03.977[—223775 = 37,615 La..- Since. 157.95 5251' > 3G hi) yhhfliq ammo; .. «d . I n 33 are Solve Prob. 7.3? using the maximum~distortion~energy criterion. ;[. m 7. 7.87 The 36-m-diameter shaft is made of a grade of steel with a 250~MIPa tensile yield stress. Using the maximmn-shearing-stress criterion, determine the magnitude ofthe torque T for which yield occurs when P = 200 kN. P7: QOOkNT- 200x103“ c=skal=13mmt 19,6153“ A = 1m“: Fear/go's)" -.- |.o.'733 xzo'3m‘ _ ‘, E : _ 200x103 _ a. 6:? .. A Lows-axle" _ [93‘433 x10 P‘ 1'I 1515.498 M‘Pa 67m 0 5m=L1<¢z+gH= £5} =98.qu H8: 5“ )‘+ a; = m Sufi *3 6; = 6m - r2 . ”Ir-'i'ian aw» mam". '2. r 6: — 645; —— 6,, mm? + (aw—:2)1 - (gamma—m = (if 4- 3R“ = 6'? 'i'1'f'-_. 2%)1' + {3\[(‘?8.2%)2 + 2.1;] = (2:30)" = = 3a. 2'12 MPq J": E6." = gCI‘Bwo-EV -" 169.34Gx1‘D-q w." "1% T 2 J1}! t e . ero'q)(‘8?.2*rZ>-Jo‘) C 1%Xlo'3 ? El's N-m ngngU-m 4:1 —---——.—.._______ Problem 7.113 ...
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