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

This preview shows pages 1–12. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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%,- ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern