10 - Prob 8.3-4 The state of plane stress at a pdint is...

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Unformatted text preview: Prob. 8.3-4. The state of plane stress at a pdint is given by the stresses 0', = —48 MP3, 0', = —24MPa, and 7-,, = —60 MPa, as illustrated on the figure below. P834, P8.5-4, and- PBS-35 ET: gibrigm : Wedje 5 <2Fn=o: cm): + (4% MPaMAsinzAls" h '2((,oMPA)AA Sin‘lS" C0345. UnAA + (NMPQAA (108345" = O \ MALE-a. 48 P \ Ax. . ‘——( M AA x YZFJOVEMAA +(laoMPoOAA sin245‘ TntAAf *(HSMPoO AASl'n‘ls" Cos‘lS' A (look/190A“ “(24mm AA Cos4s' SAW-ls" / WMP’OAA '(looMPaO AAC08245° 'O 4; 5 Int: 12.51194 @4MPaAA5 AAx =‘ AASm 45" AA: = AAcos4s° Uh : 24.0 MPa. Tm: ‘l2.ooMPaL. Prob. 8.3-20. The state of plane stress at point A on the surface of an axially loaded bar shown in Fig. P8340. Deter- mine the comprem've axial load P (in kips) if the cross- sectional dimensions of the bar. are as indieoted in Fig P8.3-20. flc/z/m,;,g /04;/ fl P8340 and Pits-30 flim 5/3 £64, wi/A x'y'fif/z: ftfl/cncc'mccr an/ h/lfl 6 mam/ex [ad ,4», x ’ 7% 1C. 6,-(1’7—2 ”‘Lfi m) )(ax—ZJ”) m 3'25: 257' ”’7 2'91 :( Ml)¥( 451;, "/5é”)[or/-60) +/z,gé;,')/m/—éo‘) i = " 6 00/7 if; p: -054 =(;, @a/7é;,-)/z,2.)/z/;,.) .— 21,007 a,» I 7 Prob.8.4~4.a',=8.8ksi,a',=—1.2ksi,r,,=—4ksi, 1 lab f4) Dc/r/ ”11.46 ’9',» 5" ¢/ J/rtrrtr . 618er Jhé/f/ét/éé f/o (I d—J/fes‘fd/(mtrz/ 1T4krl fi’flm E’5'//I fl: fffly f 7“]? ; [/53IZ')‘/’/124ri)’]2 , (—4.0é1) = 6/?0/[r/ a flflm 7', 3.0% $7 _. 422/0; : /€€érf)+Z/—/.2én‘) =fiflér/ 5am 57;,5/3/ 0/“ = 0;” f ,9 -' 5849' ia4p’0/ér/ * /0, 203/4” 0]: /0. ZOéf.‘ J; ’ $7 45 = ifiérz'—6.4ia/éxx = —2, éoz/ér; a ‘- F/pm fag/Z, y /flL rm Z , 12L 2 ’4 /Z.604;, . 7’ ,Q 0‘ 3.430/ é/h/fiflfio \ ,9’37.’ («W 2%, = LL [9” X 2/3 ,42’0/ \ {5) ”1—K ['mum [Agar Jflcr/ final anger/d, / ”Join/5 From .67 g, liq, 7);.“ -' I? -' 6. 430/ it; fi’flm E/mzza 0;, =JZ = 0;} = 36%;; Fm" t} 5/7 9, 4f” ”643297 Tm,“ -' 6. 434)?’ 01': =42 =j5’é;; Problems 8.5-! through 8.5-4. Use Mohr's circle to solve for the normal stress and the shear stress on the indicated inclined plane NN. Prob. 8.5-4. Use Mohr’s circle to solve Prob. 8.3-4. 6x465 ~48MP4—24MP4. Go: 64V = 2 g -.= ’6‘; / - 2 z W2 1 1% (Ea—m") "”193 ‘\/'( 8 2 )+('LOMP9L ‘ \.I g 2exn ‘ 2<-45°3 = in: _ 'lbOMPA _ 8" {am l(%MP¢—5(OMP4>‘}EI-LBO a 29m». = “30' — Q,— lzexnl r 190° -=t<a.L°ro°-~=io° = 1. lo Oh ‘ 01+ RCO‘B 29np| = ”SLR/“74.4%“. IQKMP‘JQOS H.510. = 24 MP4. ”Eur - Rsinaenp. :10leng MPabin H.5lo" = — \2 Meg —— On ‘3 24.0 MP4. Tm: "lZooMPav Prob. 85-30. Use Mohr's Circle to solve Prob. 83-20. chmnm izr'a/ )Zd/cc P W ‘27-'46: di-‘O mu/ 7;] =0 {and /44/)/ Y—a P] am] Pas-20 and 93.5.30 ' Y4 P2 or: Mair? (/rz/c, [am/t Va/ {0, 0}. [oat/é Tld/ (-4.5; 2,5) flaw Mo 1/8 4/ 60" /o V410,?) xic. [9,“, = 60°; 0-{kf/J [)nu/ aka/r k/lfl (III/47 (0; 0) 1/ YEP! fi/K'ZZJ rap/M ’ 1?: C)". fax; xfloq/J pm flraufi Yam) (-4.3;z.6)X’ TYk Ii} 71‘, 251,1, = 5. 002214;,- fl'néo" fl"! 60‘ E J; = 0;, f 19(0150" . ~45/I/ f/iflozz/L') (av/60' |= —- quti‘éil. 0;.- 0;" ,9.- — 2, 97571;;— $002245 = —g. OOH/r; P:|a; |,q = {6.oouénfl4/A’) - 29600517/7’“ [0‘ Z¢10 Aliaf Problems 8.6-1 through 8.6-12. For the given m 3mm, (4:) Sketch Mollr’s sin-la for thee-dimensional muses, like the circla in Fig.- 8.23, and (b) determine the absolute maximum steam 7... . ' Additional pmbIm-zn the topic of absolute nutri— ulmn shear m may be forum! in Chapter 9. (a ) ”706/1 (Ike/ff. ac“: 4;ng = //Z 49') + (42%») _ 0:41" tor/'ncrflz/ :frrr/cr.’ 0; = 0; = Z4 ks: U; = 0; + Q= Oéqi I 204:; = 20%: 0,3 0: “e: om — zoérz = ’Zoérz‘ /6 } flém/u/e max/mam r460 fit/cm P" _ 030‘ _ {z4é’iJ'l—20éfl.) ’33: ' 2 z a f 0; _ 0441; ,4 /— 2011/] 0;, ‘ Z, ’ )2, = [2 [if “3.6-9 7‘63! = 2211'; hoblems 8.3-1 through 8.3-4. For each oftheee pmblem you are to sketch a [tee-M dim like the one In Fig. 2 of Example 8.1, and use gumbrlmn gallons to solve for the uni-mu stress and shear em on the Indicated Inclined pique NN. ' Prob. 8.3-3. The mmotplanesmesau'poimisgiven by thestrmca=3000p¢m= -5000pxi,an'dr.,=-3000 pfiuflumtedqnthefigui’ebclow. fin/”[f/km; I7 1({000/vh) My iO'OOOpJI )dfl) 606 ‘Lf (30007;!)019‘ [?000,0n'.)dl93( n / TIM/3‘ A14. " 4/9 car-1‘0> A14, ‘ 49 {I}: W° d; "5260 pm" 7;} = ’4‘4é0 ,0]; "3-3, P8.5-3, and PBS-35 ’72P; :0: 0,314 f {imam/MA} (MM. {— [3000 mi) Aflx frhfi' - (3000 pri) A").‘ (of f0.” 2“ (Jam/01") 4/9) rlh¢0 ‘ = 0: =FZ/3000/n') nhflvw M ° {-6 0&0pfi ) (OI ‘VO 0 —(<wop// )9?) 2/00 f. = — 9257.5 gif \25 =0: 7:! 0’9 f6000/fl'100’9‘ {I}! ¢0 0 f6000fl1f)4/7k (of-Mo 2"(5'000pfiwé’, [0/ 4% ° v/a’ooo/pfi )419, m ¢a° 7.} = — /M00/m') Maw/Ma” .— /9’000/n‘) 69/sz ° "' {(7500 ,on' ) fl}! ¢0°wJ¢a° + [face [11/ ) f/a' Z¢0° fir = - fliéa: z m Prob. 8.3-2]. The state of plane stress at point A on the surface of a rectangular bar subjected to pure bending' in the xy plane 15 shown in F1g.P8.3—21. Determine the bending moment M (in kip-in.) if the location of A and the (2085-. sectional dimensions of the bar are as indicated in Fig, P8.3-21. (Hint: Consider Eq. 8.7.) "33-21 and P8.S-31 itch/1.7 x’gc‘f’ll = OX + 55 bVL‘L for" £146 beam shOWn In Fla P8 3 l‘i 0‘5: 0 so = ‘ Oglgn— l keI = 4 {55] Now, :For a. beam ewbjccizcd £0 Pw‘o bowl mj) (Y: —% -GxI M = ‘3 _-(—’-lksm (I 5in54in)?’ l2. lIn = I’D-2 [99- in} M = 52,0 kip-In- Prob. 8.4-5. 0,, = —4 ksi, a, = 12 ksi, and 1,, = 6 ksx. 1‘ ,2 k5; COW 9”: ark—k51— ‘8’ ll ‘— Pc (—L-M “’2 I... fir—m 1mg l ¥r0m .2 I4, 03651; -4k’+|2k5' [I :Fr:A\/3_ 5| 1 = . 395528633. ‘8..l5, _ av1=+R 4ksi*lol<su=l_lsgl 02 (Save) R: qkSl IO kSl= ILISSL 8.12, E 355.33% sin 26?: = R ks: Ox' 5 —L-l -_l2 .1 005299.: 2K g 2(loksii from ears. Prob. 85-3. 'Use Mohr’s circle to solve Prob. 8.3-3. flc/flmxhé d," fine/7;}, fir: fl/M’r/ (IPvé'. d; ; 0; Haw/1;) r Amway/:1 . 02 ’ Z ' Z _ L—fi‘ __x = — woo m; - 3000 WI’ E‘ /£2—aj)£ ’7‘? 1 3000 ,v// (’ {000}! if, 3000M") 7‘09”) ”92ml ”ban" %) = ”'57" fl= Ara”— 80’— £57" = 63: a?" : ‘3Zé0/vfl‘ = - 44%0/01; ' 0:= '/000/m‘ — prrfl = -' fifiiflrr' PI 4/? Gt 7,} z _ gjlfifi/f = - ¢¢éa Z/vr.‘ Prob. 8.5-31. Use Mohr's Circle to solve Prob. 83-21. O"+C5’ -5ksi-| 5' . 02"‘1—2—‘L1= 7&4 "2ks1 . o Fromgcp... 37, O} +(TOj/:Cxl*_ 29“ 28:”) lng VJ: = “Sksi'lksi=’_4_ksL ( 3X I ms» 01,0 ('5) Lay) X’ 1? a 1153 m M h 3 Circle, (check O’x _ gx“ GerCOSIoo°=‘2RSI—2kSICOsLo° =“3Lsi \/ x' — i Y5 I_ M=§TI= Mk50y'f1'3H'" /'2 —52kgp.in. M = 52.0 kiP'In. — -—- H-IrA-—'-—- “mm 1 “at". _-__.-w-n.-...-.- . .. ,.. .1 Problem 8.6-1 through 8.642.- For rheglven m states, (a) Sketch Mohr’s tin-let for three-dimensional met, (Ike the clrtles In Fly. 8.23, and (5) determine the absolute mlmunu shear M 7‘, . ' m Addltlonal prvhlems on the topic of absolute max!- nuun shear m may hefimnd In Chapter 9. (4) Mair} (I'm/(J UV J;_g£ f/ééh)f/3£rr‘) 7‘1 /< if) flf/h (tha/ f/l’ffft'f 5 07* 02 +I€= MA; / Séféiifi a; :0; '1? :/Zléf(‘ 'féféfén‘r Q30; =~Zli€ (é) fléfo/w/é ”MK/Imam {Acer .r/rerr. 0', = 07* 03 (/7. 6M7 bib/14w) C Z z —J 4; - 721,43: 073 y __ //76f67 If) ‘ lZlii lg: /{a;;_¢,_ [l/éérr'l- (54_____x;1] ”,4“. =f65'6745.‘ " /7 65-6 7,1,1. 6,3457%? : z 5M¢ 4:,- - 1;) U f = 7,5254%,- ...
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