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HW5_solution - Prob 7.4-4 The uniformly loaded beam in fig...

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Unformatted text preview: Prob. 7.4-4. The uniformly loaded beam in fig. P7.4—4 is completely fixed at ends A and B. (8') Use the second-order integration method to determine the reactions RA and MA and to determine an expression for the deflection curve v(x). (b) Sketch the shear’c'liagram, V(x), and the moment dia— gram, M(x). ‘, 00 WW NW) P» I 2 B . . ; A K : Egr'libt'dm F’fir—r-fi i EBD: beam Beebc‘n} 02x 2 L 2 E N mm P flit/0m: Mm= Migmflg} 3 a” M“ W 2 E "X z . MAA L+1fiil M00 EIV"=MA)L*E%"+ b +0! E 2 2 MAKE BAX} fix}! RA V“) BTW: 2 + e + 24 +Crx+cz W: Wok \r’m = V-(L) aflm = 0 EC: Cl>O Prob. 6.10-8. A we beam nas me mlativc cross-sectional hown in Fig. H.108, and it is subjected to a force V. (a) Dalerminc an expression for TwO’). b, and (b) sketch the dimensions 5 vgnical shear me (vertical) shear stress in the we function My) obtained in Part (a), (a) 1/6” {frail/L; #061, ZOKfl/t (Kwflég: _, fl' —: I‘A! 7—1, 7, 4/, 7““ ‘ 729 m; ., (mm/m f f//»;)( jfl’f/CLZJ = V 7.3/7)- / W (1%W 2“ +3 my) 77%;; , a.” _ 91 (721),“; ‘ zrf‘ 71/ man may» : 9 fl! H9; 1" : fff/J/lW Z/ffH/J *1 = H WWW z J J: 2/1; £44,?) = {£354 Marla/4m] 4 ’5?” ma ¢ : 7/ M5 ffltz/ f/Zfl: [L i t ffj) L {26/2/2215 I) f /ff}/f}/A5/) Z a ‘lx — .11 fL 'LL «1, [76“::8¢4,,' [gum/1r; Prob. 6.10-16. The beam shown in Fig. P6.10-16a is a W12 X 35 wide-flange beam. Determine thc web shear sucsscs Tc and ‘n: at locations C and D in the cross section just to the left of the support at B. fZ.l41'/-é/I.Z{m" ”l 3! < 3|" [’4 m/imfiao: . ‘ F r P “ 6}. {cm »{/gz,,',,)(,> [1“) DUWOWW) 1‘02 41/9; )0 ff) =0 C(cenlroid) ,4} = 547;! 6r FM 0/ 6’49 : MEI? =0; ’4] flab/‘9’ ’ T‘s-fl =0 74 -’ '5 5W 6273“ It’d/6'54 4/ {/((t arapgx fikf.’ , 8 FM» M/r (2/, 4%" M A flflr/ may Mia 44‘ {Am/n an ’4] 5’,“ 3H ’1 74 14/5} ia/flu/ n'f/é/f 75: fl [1‘ . 7:174ng 0 III!“ 572’ I 07 0” : (6'56'5'Wf3'U/5m.Jazz/a.) : 20. 433 1}); f i 7“ ’ W WINZarw/m g ‘ am, a (MI/MHMM ‘W T 47‘ 1 Q0 * ”Ii/”JKWI/m)7fl ‘ 13796”fo , 57mg, “/8" at f/fL/«IJKZL Diff/la” ‘ ‘—l_ :_—"—W=Mrmz, . 7; I "w @3124} /am) ’ _y{ from”. Prob. 6.10-18. For the W12 X 35 wide-flange beam loaded I w as shown in Fig. P6.10-18, 151 w = 2 Rips/fl, (a) Determine , , the maximum flexural stress. am“, and (1)) determine the maximum transverse shear stress, 7“,". A A ,,,—L.. B ,. F—6f1—+—4fr—-{ P6.10-18 m mflgfim. 4-111 9% grai CA: r593 WEE—Iii? 1? MAC- “ W407i) w fiizmm: Ajup'Fti‘ momma) =0 5 A1: W(§ft)= L50 Epfi => Mm: no MFR = 2 L12. 'a. :rowi Table, D. i) I ‘= 235 ‘il'fi M d 'E IQEOiVL Mmflfig _ figkigviniZSOiHH/g _ F , + 7' Um: I ’ 285 in ’ M +\ 5mm 3 4.2 LE! b3 W33; JF'rom Tabie D. l, {of}: (0.5L; in. if = 0.52 in. 4 '\/ fw = 050 in. ‘YflmXikam shew e—(zreas occursdat NA, where Z 1 NA 2 A A. Om.“ = bgtfidiz 7) + hi??? tFi(J‘“/ét‘€i= w, iniOiEZinXEflfiiM + 3(o.a‘.n)(535m3 G (25/ k x 2&5 - a V M ’mM 5 fps 25.35‘S'm 1 Imus: = rLItw z "' 235w (0.51m - 23mm --\\ Tm = 2.51 ksi .47, - Prob. 7.3-1. For the unflorm simply supported beam shown 9: in F3. 1’7. 3-1 (3) determine the equation of the deflection i curve. u(x); (b) determine the slope, 0‘. at the left end; and (c) determine the maximum deflection, 6... - max Iv(x)[. SoIve by integrating the seeond- order differential equation (the moment-curvature equation). NW ‘1 “’0 W: A WW x %$: beam AB '1’ _ L _____;§ {9(2M33" O i Av"= 1%k A {:jPM—x A: 51 ngFjfiwjmmx onéL WsML-mwx)‘ Aqx= Ml? EIv”= it: Fl law-12% 4 k ,2“1 Vw E’I. Ixr= 3&1?wa WNW/‘10}- flL) ‘0 ML o=Cz 0‘—L’11:L2* CdL 9’ C4 4" +3}, Ivoofiiéfiéi: "(1795 + (—31)! EWZH] 9‘: I thL. )1 A ,L ’(‘x‘ l [3 \r (03 M.L_ * a- QA = EB: cimmmflmfinn A . z _ - +3, Hm ‘Eflélffiziiiiz’r |]‘= UL ‘V'iéi’g‘é‘i‘iaiz J'éi X3=J=5' «L 5.: T63)" 423% Prob. 1.3-3. A W8 x 48 wide-flange beam made of A-36 structural steel (E = 29 X 10’ ksi) ‘3 simply supported, as shown in Fig. P73-3. The beam is subjected to a couple MA that produces a maximum deflection of 0.10 in. (a) What is the value of MA? (b) What is the maximunt flexuml stress in the beam under this loading? Solve by integrating the second-order differential equation for deflection, Eq. 7.8; and see Table DJ for the aos-secfioml properfiee of the ”3’3 mm‘ ' "(—00 ”Wm ’ ’ M ‘3 *Xm—H ‘ 2L JAM“ W K A B . , . lb—v—Sltt—H Egg: beam AB ”M fll—e‘SH—H mil/05:0: Ar‘I—A (Lash) (EMA rrom Table, DJ, '1 =18Ll’1n4 A F:::F x d: 350 in, A” 87 EBDIbeam Sectionmexeak Mx Lj‘.’x__)g tXEML =OZMNRX1: MA _ f w 3 mt MA "t2 MAX A [ Mm x EIxr’= lN/lllxz- PM" 3" (L. 4’ 7‘ AX Atj .Vm EIv’=—£'X“' [0L 4QX+C2 ' ‘Qlaiz \I'Col 1 V'(Ll = 0 MA 2 "MAL o=C2, o=’-3’+C\L $0: 3 1'2, ' Maximum defiechlon OCQWS a}, T'OO: 0 WW lf-é‘El-sll‘flz+ W6) - 21" o * ”xx 0.422 L95 L mm 'E‘EE‘l (1%; + 5%)2- 2%)] = - on In, * 1 MA=QOZH56 kleln. + \ [Mp 75.2 weft I j blmqum‘l stfie‘ss lg (85 . 1 'M I . ~) . MA'd MW , GMM=\ ME Z“ = 2,1 = 2 (I 4 In =ZO-5‘f’7|§) .4 7. 6mm” 20.6ksll .A\ Prob. 7.3-7. For” the (twoimcrval) simply supported bum in Fig.- P7.3-7, (a) detcrmine the equatidns of the deflecfion curve, v,(x) for 0 S x s % and v1(x) for £1 s x s L: and (b) J r. demrmjne the vertical displacement at 111: center of the Lrwm‘wwm w_ 1 12“,! beam, .5. -’|v(L/2)|. Solve by inugxafing the second-order diffcrcndal equation for deflection. P7.3—7 and P7.3-34 magpie}; an cArxes . W0 Rgrfie;g%.g X W”? c.‘ . L 1__ s G “gamma Moazw'T‘LX - Visé ETI=V%¥'V¥)L<4 2/ ”X EIv.’= W12 - $1; 0. L a»: $3 fiaxugwcz hm Balm” Zéx ' L mama) Mz FW'LX ~ WAR — %\ ,/_—w x W1.2 3% y a“ 1m: :6.sz 75¢ L k M Big: 243‘t \szrD. L X—§—’\\Q(x\ Efbfz =_ 1sz + w.qu 13.x * D2 W:W(OW'VL(L\‘O / Confi'nmiq mnd'iz'onsiz \n.("/2\ ‘ \r (V23 ', WW2): \rJL/zW __W-L— _~’~\| 5 o = 0:1: Cl ' 2920 015la+DlL+Dz .9 C2- 2 WW. A g _ 5m.“ D.L *1st 6%0 + 2 ' 5% 2 "D2 D. ’ I 40 w W-_L‘ g welfi U4 + C. - 52 1' D2 D2 H30 “U 1‘ \r‘ m: {gaff Lfigfiflfi QOL XL L . ” Z ‘3 \IZM’ Hfloéfi' 20m? * looqr— 45(m + 5] g 3,3—zmnina.) b) __L_“ L — _L _L —w L" V1 (L§W=2\§VéOEI[‘4‘8 (é?) 4' SAGE) ‘44 (2)] 2 240.51 _ I-“ _ ELI—l m1 Wamc Solwb'onvxovdd have, been reached 55 * NJ}; “ 24051 5‘3 Mm] §a = {WU—)1 ‘2, E 1%,, 4 w , my z“ «w a Prob. 7.3-27. The simply supported beam in Fig. P7‘3-27 is sub- T‘nis load acts downward, so there is a minus sign in the expres— A ‘EL sion for p(x).] (3) Use the fourth-order method to determine an <. expression for the deflection curve of this beam, Ind (b) determine the slope, 9.4- of the beam at end A. 0;) . lj we - . 0A Whom; epho =‘wasin(‘¥) A L RIM” * weir: (1T1?) (EIM' 4%;th (1%) + Ct M = 51V” = wa(%)zsin (165 + Co: * C; EIV’ = ~w. ('%}SCOS(1E—)+%: 4 szz + C5 ELEV = —w. (#Ysiw ¥§+ gt)? * 92239 * Csx + C4 WEEDS; \l'(o\= \r(L\‘ o 0:04 C(:O 5 (1sz o:%* 2 *‘C5L*Cq => Cz=o O’Cz (15:0 0: Cal—4+ CL (14:0 \rm = ~_LE\/Vf("er‘)q5in<Fn—fl bimmlogeifls \r’oo = 3%(7'?) 335(1):) 9 at; lxr’totl w meg ' Led d' 'b edl d f' ‘ty ' N New“) t = _ L”. : )ec toa ism ut 0a 0 m enst p(x) wosm L [ ote V ‘E ', M(o\= MM: 0 Lil Prob. 7.4-1. For the uniformly loaded propped cantilever beam in Pig. I’M-1, use the second~ordcr integration method (3)10 solve for the reactions 3‘14 and B; and (b) to determine an expression for the deflection curve 00:). P7.4-1, PIA-12, P7.6-31, and P11 00W 13 Wm , A . . . le—f—L————7l E4313 beam 658nm!) 05X ’2 L 2 3| “WM 3“ Dflzwfifio: MOVE); [30?sz <. EIV-H = ROE“ DZ 5 A ._*_>< PM“ X E1v’=%l"- “ELw‘ 2(2‘ EH E575 wax“ K Vixx E1w= 1p --2Lr+Cx+Cz haundacj medians; \r(o\= rm: 0 . Mo = o _ D135 C1 = 4% # czmé L RA: 8“ +T2Fq=oz R3 = W.L’ RA = mm» = 'M = RAL‘W°LZ = jmgflx a o. e. z Rs from mama ration and bomdar Condilion‘s (5F problem aeofion (a?) 3an .340 fiwnL 2 — wa 3 1 ...
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