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HW8 - Problem 5 2 5.2 For the beam and loading shown(a draw...

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Unformatted text preview: Problem 5 2 5.2 For the beam and loading shown, (a) draw the shear and bending-moment ' diagrams, (5) determine the equations of the shear and bendingwmomem curves. R’eaa‘i‘a'on s, GEMC3'Q: = 5.5 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (5) determine the equations of the shear and bending-moment curves. CaIJCuj-av‘é’e feasiTl'owS crH'EIr repjaca'naj aisi'flko‘i’Efl! pearl by an equwafem'i‘ Cancemfr‘afesi ,pnaai. QanHovxs are A = D = éWCL—Qah Ff‘om AbB: O¢x<a fi‘j‘" +72% = o: 2iw(L-2a.\-v:o x v V: ;ITW(L-2Q) 1:! fiace sec-hum co‘} ax" >6. Rapjace OII'S+TIILu+EcJ p009? Ly egoiwl. con-s. )ngL “ff-3:0: éw(L-2aU-W(x—a\_vg O V=w{‘_|;__x)‘ 9M3: o: —i2w(z_-2a3x 4‘ Mkaxxéflcw + M = O M = §WEL~24JX -.(r—o.)1]“‘ *TEF}: O: V + éWLL-QQ) : o I‘L‘x ¢%W(L'2a} V =-¥ (L-Qq) 92mg: 0: 'M + %W(L~2a)(L-x) = [3 M: yum—mum « Problem 5.10 5.9 and 5.10 Draw the shear and bending-1110mm: diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. REac'howS. 92M; 0: «A +£smw—(e)(3c>‘s= o A: -6 Mrs ie. MP; 4, 92Mn=0: 6C~(s}(:sfl-(fi)(30)=o C 3 hrs 1 50 Odocé 6?}. 3x M £3 -6 -3><—- V: 0 V 3' -é-—3>< kl}; A c B x = -é§x- {.511 k-pfi C. +0 8, 6P} ‘ ‘2 4 q M 30 “a” —9D Ibis-{3+ G J V \_ c’._x HEP; : O: V — 30 = C) V = 30 Hips DZMK—"o: —M —(q.-x)(sou = o M: 30x -— 270 kin-H From Hr? Amara-4% (OJ JV'lnwrzop Lops "'3 (b‘r IM Ln, : 70.0 k‘-P.Ft 4 _f \ Problem 5.20 I _ 1 1‘5 25 it] 1:] it: LN irN k5: 1135 m - 5 5200 X 27.4 [email protected]=2.25m 2525 A " (1.8751(251- (Lb—YRS) - CIJQSWJD) 40.753003 40.379001 = o M A :- {+7.5 kN AFCD Use PorHo-A AC as 3%: boob/a 1' 40-3 5 (Liv.st - 2 o = . 415 V 7) + M M 17.8125 kid m For s 290 x 27.14J s = 235 M03 m3 : 23mm" m3 r 3 J) E .-= L7 .r'LSIQS'xID __ r. Norwr .s'JHres , C S ———-—-—__.235_MI'D“g _ 7519:10 Pa 6-: 15:3 MP0. 4 l Problem 5.22 5.22 and 5.23 Draw the shear and bending-moment diag . . . rams for the beam loadmg shown and deterrnlne the maXImum normal stress due to bending. QQKCJI'IIL‘V‘ J A. 433‘M3 : o: I —3.5 Fig + 25’ +(2.‘f)(2.4]($o) - r5 = o w310x38.7 RA : g;,77g m Qflkcjl-Lr'ow. wt? 8' +3549: 0; 2,3 —(r.2)(2.4)(4o}— I5 + 3.5 23 = o 129, = M222 W 1 I ‘f-0 ogxsmm g” H_jM_.__ HEP; : a: 2 x“: 95,??3 V 66.778 - “(05: —V :0 421122 v : 6.6.773- ‘fo 1a.. MJ kN-m V' = 0 gift '16: LC—fi‘f‘im 30.719 +3 '2' MT: 0: 25 +(Lrofl§ — gsjmac + M -= o ’b M = 42036.1 + £6,778 94 —:ZS -25 ‘45 M =-(10)O.6W4)1 4- (66.778 }(:.é=wnfl — 25 30.728 kl‘J-M (j 2;?”ch “A 2-“! 5964 3.6 v.— -71222 LU M = man (3.;— x}- :s M: 20.05? kN-m ruz‘ 25mm MW: 30.722 WW = 30.7mm? New For WSFOX 38.7 Pop/Fe) She)? SECiLI'OMj 87' 511193303 “'13 = flaw/0* m3 30.728yro3 51+?! to" “‘16 S 1-... 55. 97 :«(o‘ Pq ll 6:. 56.0 "4 and Problem 526 5.26 Knowing that W= 12 kN, draw the shear and bendingmmomem . . Elia beam AB and determme the maxzmum normal stress due to bending. gran“ 1'" E3 Exfmme+p\J A = E 443%,:0: E.) w3|0x23_‘3 A = E r :2. Shea"? I A +:: a: v = sum ‘3 0+» D? V = - (a w 4 CF +2 E: V r 6 km “a E‘+‘=B. v:—-2L5N q BEL/flaking. rfiomgw . M c '- JOEML = o.‘ A C T“ M: mm = 0 54.”! MC: 22mm .5. ‘1' 2m V A+ D’ + EMD = 0: 8?“ MD A C D '3 MD— (23:22; (2m 7m V M = J} imam 4: By SVMME+P7J M 7 ZM‘J-m M‘E 3 HI {(Mth aCL'ul/‘fi @‘I’ E’- Far W3!0><23.8J 5x— — zgoxroaws : 220M04w" ' 3 M 2 : :flfiyimu g :1;th t c N“ 6‘ 5km” “"7"- S: 235nm)” £1,2qu P4 Problem 5.38 5.38 Using the method ofSec. 5.3, solve Prob. 5.5a. 5.1 through 5.6 For the beam and loading shown, (a) draw the shear and be Rem-Hons. +Tf§=0i A-P«P«- B = nding_ moment diagrams, (6) determine the equations of the shear and bending-mm“em CUWES. A: =1: Over amok Pomen ABJ BC md'CD w:o_ $5 =0 v=¢onx+mt Shem» Arajmm, A15: Bf V‘—' P «as: BJmC: v .— 13.? = o «a. CED. V '-' O“? = —F’ «a Belong mohair 09mgng A +7» 3- 3—: - v: ‘P M= 3mm : Poo-PC, M=O J 9¢=0. Thus c:=o M =Px 41 A+ 3‘ M: 'Pa. 8+0 c j—: = v = 0 M =PCL 4 Problem 5.42 5.42 Using the method ofSec. 13,501“: Prob. 5.9‘ 3 0 ML“ 60 it“ 5.9 and 5.10 Draw the shear and bending-moment diagrams for the beam and loading shown, and detcnnine Lhe maximum absolute value (a) ofthe shear, (b) 01‘th bending moment. Read-clams A and! 8, +5 M3 = of — 5A + (magma + (601(2) = o A = 72 kw 1‘ +31”A ~— 0: SB —{31{sol~<l)(so)cm = c: K b E) :' 5‘8 (“OT A . [ 55 Check: +fEFj = 72 -{:H(3°3- go + a = o I.“ M Show Jiggr-am ._ ka VA T 72 k 72 A41: C. 04x42m w=30kwfi~a \,J'c..\/;‘x = -—S:30g!x,: —Go in.) ‘1 Va 2' TKHGD = 11km A c 3% C+OD. V2V1212I<N b+o B. V =VL-G.o=—’+8ku AVEG—S 0F 5;“:th aha. Mun-L. _‘—————~———-———__.4__ A 4» C. A... = fl’mwz 1C2) =8‘HW-m C in D. A“, r (121(1) = 12 w... D +a 8. Am : (—48)(2) : —% kM-M Bending (Mame/ff}: MA = 0 ML: 0+3? :34“: MD: 3$+IQ = 96k“ M3: 96 - 3‘5 ‘3 O as akpcv'f‘eop. co.) Ivlwr 72.0 m «a \MIMA“: 616.0 kN'M “a Problem 5.67 5-673“ 5.68 Forthe beam and loadin For 3;? uépl’b n".ij 8: E r 2.2 kip: .__ ._ Slhew" all-QEV‘M—a_ Ah. ‘8'. V = ~L},gk: s a-& ft -a a _ .0 2 2 3 2 2 E++oc_ U=_Lf3+2_3__2krs Co++b D: V = *2 + 2 :- o , V -" O 4 2 r 2 L'Ps V’ kq“ V= 11 +2.2 = 9.8 KP, N345 a‘F slhamrv Jig; raw“ ——————__j__ I (QM-LL23 = —? a k.P.J>t EEK—2‘1: “1 Lap-Ft (33(0) = 0 (21(2) = ‘1‘ lap-aft C2X‘f—3\ = ?Q k?! 1D?— :"?.6 kllfl*'Ft 13.; apart : x623 wan _= mzsxxo’aawh _ __ - 3 Elana 6:” — 1.75.0 - $3.15? in For" a. V‘emawan/q» secJHcau) l: :43 5143) II"! (2:: '2. 1. S=%‘ = E: = m(bg(q'5) = 6.04:7 L, E UUE'FM "Hue +M'u a“ Mission; Re- 5 fi fl ,9 2 {5.94:7 I: - 93.25? wagon/L «a 50;] lb Problem 5.79 —-| 5.79 A steel pipe 0f4-in. diameter is to support the loading shown. Knowing that the stock of pipes available has thicknesses varying from l in. l0 1 in. in §—in. increments, and that [he allowable normal stress for the steel used is 24 ksi, determine the minimum wall thickness I that can be used. Sheen»: A+b B. V: —5'00 [L r — 0.514;? B +0 C. V: -Soo—$oo = — Moo 1!»: -J.o tar Arms: A 4138- GHL-CLS) ’ ‘2.0- l1{r.-¥‘l 3 +0 c- (HM—Lo) = — :4. 0 MPH Banall‘hfl momen‘lST MA : 0 ME 1- O -Q. 0' = - 2,.5 REYNH‘ Mn: 4.0-9“) = —G.o tow“? MaklmUM ll’ll" 9.0 -' 72 lfi'p- in 6011 = 2‘4 ks; _ lMl fi 72 _ 3 Sum-n — ' 'fi' ‘ 3 'V' ‘“ c:ch Czrfial= 2.om_ CH“ Clg Zq- C,“II . 3 2CL % 1 = 3 “’1 : 3_3QOG in“! C,’ '= l.700‘} in, "Cr c| — 2.0 _ L7oof-r = 0.2?96 in. JS' In. incremewl‘s ‘For‘ alcsfjnJ t=§' ...
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