HW5 Solution

HW5 Solution - 42%.“): RA+Re-amlm(HmE-Hkm o: fi+P~a~lbLM...

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Unformatted text preview: 42%.“): RA+Re-amlm(HmE-Hkm o: fi+P~a~lbLM =5 P-A'HEBT H0 “N QEMAzO: 31LNIMC'4MXZM)“ pa£bmijk~(8m) O: -R3Uam\ +60 kN-m -—‘> 129,: sum-m: 61.33 kN Lam P4»: Mg m- qaaw = Lam m * -m'm '3 3 C) I‘— l/mzCOCIo-ALHO m) - 1}IZ(-OQOM'Z‘E)(-HOM“ 2t} m 3. g. ) $12: 5"“"‘ 058'" “.000005 m." =5 szLOGIOrnK-HOrnf' V1"(1(-OCICDIVWEJtX-“Om-41')3 1‘ .DOOOOB mL1 ' 50mm “we, JFDUr-kh order thafl-abm, “t: .0OLM‘52. m - £2.10 52 m is m+ possible, Or“ 1:: .looszm :0 ‘tmin: .OOHU5ZMR LL53 mm “mow: [WU-I KN A: (m mourn); (.09: «at Iomhoofll (using {7"- 1-1.5 mm - --0t t=10mm HOmm - Find @ sectlom OK— OK _ Cl: (+0?” rcd‘Ormfldor SCLfiDh) v: gm; bum.qu .5 61: Ag = (-qom5(-01m)(.05m3= -000‘4‘5 m I = ‘h 1(.oqm)(.um\3~ ‘memxm ~03: .owomo “ __ (mm 1L103N)(.OOOHS m‘) _ N/ ‘1‘" .ooooou m‘i ' 500750 m To aka-min a PS. 0-? 3-, five fircngi—h 0-? fine, we Id mU‘b'fi' be ’5fim-L6 c‘ . =5 minimum shear sfrcng-H-x 0"?- mm}: l‘SQg'Iso Nlm I. ii? . " 5h. it'sfi.’ 1) fiche. Q€6L0+10h5 .. 20%. " (TERI: O: P"*+R‘5—2P ,{3—-—u‘ r.» 125+ R5: 29 2;. Pa grim: o = 95% 3+ pow.)— Eafizom 2:") pg : all! P Bk: 5/H P 2) WOMEN? T Shear Diagrams Shim” Mame/ni— a) qmoq: Mrgvc For NZ‘XHH I I: 8”?) inl'l C": \03'5 in how: Wm“ = 2—5 1103 psi 75 0.3- 26~on3953= gqgif‘“ => P= 212025 \b. b) “t: 5% V=S/HP: moo-qu (3F 31: git-5ih)(o.u5in)(lo.‘osan3= 261.5511 1.45 1:- ‘EfiH in '1? (9-5 in ’5 : (augozg‘g 1! 231.5511 if ) :: Leg: kg; (6% wk [95 1n) c) T = —"' \fr 'BLIO.02.0H* W“ Aw A“: (204919 HAL-35m) = d-tw: 7-2m in’ __ mamflk __ . . C: . ’5‘”— 723‘ in; _ umozfifib km H”) Ks. a: mam 521cm 2M 5‘ sownms Ir/‘zmacf problem 5 160 5.168 Dcmfinetheiu'gest pgrmissiblevalm of? formebeamandloadingshown, ' knoulingthalthcafiowablenormalsh'cssis+80MPaintensionandv-J4OMPain mpressiou. DZMc=o: —o.1sA +0.59—OJ5'P = c.) A" 0-“(61 P +95MA=O: 0.7SC - o.2§P- 0.‘1P=o C. ‘-' [-53333 P ‘ib B. v: can“? ‘5’ V: o.qac7?- P - —- 0.53553 P = —a.$asss‘P+ I- 53333? = 'P += ‘3. (As-mum”) .- o.ngc? ‘P a (0.5”‘(15'3‘533‘P n. 032‘“? 1:» (ans) P s 0.15? momen‘fs: MA =0 MB: :2 +- O.U‘C7P -r 0.13667? M.L 2 0. new P - 01:40? 2 - 0.;5’ P Ma -.--'o.rsP . 0.5? = a Cenh‘m'o‘ and Winn“ a? Mei-kn. I= EAJ‘+ Z'I :- xnoors m: up: 3: :4 m g- : 5795—“ = nan-ta“ m? a 21.37am)" 1..” 89H”: 3:4va §= “712°” e magma“ u? =-IO.C&2 um" I Bendin honed! J’infis «' M ‘= "g—G: ‘3 3 Tension of B. s(—ID.611‘ED"\.I(ZDWO‘-) = 954.56 N-M <1 8 (“9. at B. dams“ unfit-mum") = tum; x103 N-n Tana!“ dc. —(1fi.37£xlo"\(loxto‘ -' - 135x12)" N-In C"?- fil C. - Flasuvto“ )él'io -£0‘\ =‘ - HHS-'3‘ xto’ N-M *3 C Afifiwabfie Pam-l: omsn P = 354.56 P = Iszxto‘ N —- 0.5? = - Lqewno’ P = 177'10’ N The smaJJEf me r: P:- 7.32 W «In 6.1? Fmthebcamandloadingmdcmineflieminimmquimdfidflzb, Problem 6.17 _ . mmmwmmofnmwmm=12upundm=szsm 2.4%! . 7. 48"“ 2k“ Zak-{cpl «3m 7.1M - 3 A +C’Ufi2-‘H + (13cm) -L95K7. 2) = o A = 2 kw 'r 3.6 w-m = 3.5m? N.“ era-E __ 1H1“, _ 3.0m“ - .3 Shin 5- " [2x105 1 300"10‘h " For a. M Janjugq-F Seca+fomJ S 2 "L b ['11 _ S _ (6]{3oono’ _ 3°- 9F * W14 * 3° N" SLIGQV“: Mu;~u3n shear-Ina 54'mss mauM 5+ W l Hue. nzfiwafi auqu :5? b3uakr‘.3 $91» a P: J—mnjufimfi Sacha-n. ‘ (3 ' Ron ‘ SISMOSM ' _ C2\(:5'oxlo")(825xl03) b = 37-3 1w”! The begruirecf Udue at? b {5 “Hue Pgrjer one. 5’ 87.3mm "‘ ...
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This note was uploaded on 01/05/2011 for the course MAE MAE 131A taught by Professor Vladolubarda during the Spring '10 term at UCSD.

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HW5 Solution - 42%.“): RA+Re-amlm(HmE-Hkm o: fi+P~a~lbLM...

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