Ch A - A-l Determine the lncnlion 5' of the cenlraid (' for...

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Unformatted text preview: A-l Determine the lncnlion 5' of the cenlraid (' for Ihc hcanl‘: cross-senioqu area. I‘m: beam is symmch will: re spect l0 Ihc y‘ax'u. 2y?! = (2M6) <4>~(o.s>(1)(1)- (2.5mm; = 40 in’ m =6(£l)—“0-30)=1’Aflli‘n2 A-J Determine 3-. which 10031;: the oanrnid. and Lhcn find the moments of inenia L- and 1,,I'urlht‘l'ibearn. 2 314 = 1255:0050} + 275900360) = 5687500 mm’ z A = 50(250) + 300(50) = 27500 mml £33k 5537500 _ = _2 = M 27500 2068 mm 20711111: Ans L = figoxzsof +50(250)(206.82 - 125)1 + I—lzcmcof + 300(50)(275 — 200.82)2 = 2200‘) mm‘ Ans I, = YIE(250)(50’)+-113(50)(3003}= 115(10‘) mm‘ Ans A—J Delerminc the localion (i. f} of Ihc ccnum'd (Alhcn find Um mammlan! Incrlia I.’ and 1,]. i=2fl=Wflm m 1‘1 lawman f=@=W,—_2m Am 51 (6X2) + (23(6) if = 1—;(6)(23)+ (6mm — n1 + l—liaxsan + 2mm — 2); = 64 'm‘ Ans 1;- = 712(4J(2’1+(4J(2)(3— I? + gum?) + (2)c8)(4— 3? = 136 in‘ Ans ‘A-d Determine lhe_cenlmid ? [or the beam‘s crassvscc‘ uonal area. than find If. 1.1 mm 2, 25 ,..... m = 125(150x25) + (75)(]00)(25) = 234 375 mm’ 24 = 150(25) + 100(25) = 5250 mmJl #5.”; 33—433 =37.s mm Ans 1:4 6250 1,- = (300)(2s’)+3oo(25)(375— 12.5)2+2[é{2§)(1003)+25(100)(75—37.S)2] _1_ 12 =16.3(10‘)mm‘ Ans 'A—d Dcmmm thc_cenlro.ld P {Or the bczm‘s amm‘ Iiunal area. flleu find .I',‘. 15 m V 25 mm 50 mm 1'5 llII'lI 5“ mm 2334 = 125(150)(25) + (75)(100)(25) = 234 375 mm“ 24 = 150(25)+1oo(25)= 5250 mm2 §_%_234375 2A 6250 = 37.5 mm Ans 1;. = 1—12900x25’n 300(25)(375— 12.5)2 +2[T12(25)(1m3)+2s(100)05 —37.5)21 =16.3(10‘)mm‘ Ans A-S Damn-Jinn 1,. area shown. for H1: beam havng Ihc floss-mellonal 1, = i(25)(3003)+2[é(100)(253)+1000516752)! = 9-1-8006) mm“ A” [2 A45 Determine i whtch Iogaws ll'lI? centroid C. and then find It: mamcals a! inertia Jr and I; for the shaded area. _——-——=68.0mm Ans 160(40)+2[120(40]} f: a _ (20)(160)(40) + 2[100(120) (40)] 224 L. = fiaaoxtso)’ ~ T'Eazoxsof = 49.500“) mm“ Ans 3,,«=[1—‘5(160)(160)3 + < 160] ( 160)(80—68.0)2]- [Ii—2:30) [ xzo)’ + 30mm 100-430)? = 36.9(10‘)mm‘ Ans A-‘I Dclcnmnc Um moments ol incrua J'. and i, 0! Ihc 2‘ section. The origin of coordinates is at the ocrllroid C. I, = imam)” + zil—‘z-(zoxzzo’) + zomoxlocrzn = 123.89(10‘)mm‘ = 124 (10‘)mm“ Ans I, = 1—12(20)(6003]+2{1—lé(220)(20)3 + 220(20x310fl = 1205.97(10‘)mm‘ = 1.21(1o’)mm* Ans 'H Dclenrnnc the lacmion (E. y) Dl'lhe can Lroid C at UN: cross-sectional area ior :hc angle. Ihm find this pmdua of ,v y- mth wilh respcct lo [11: r and y axes and whh {expect to _ the 1' and y' am a i: _z£_A = (1)0)(4)+(4)z2xs) = 3 in m 2.4 2(4)+<2>(s) j: 2% = time) +(3)(2](6) :2 in Am "£1 (6)0.) +(2)(6} 4,. = 25A = (—2)(IJ(6)(2)+ (2)(—1)(6>(2>= -48 in‘ Ans A-B Dc Lamb: the produci of inertia of [he mm auctional men with respect $0 the x and y are: that haw [heir migjn healed ll flu: ocnlroid C. 1:, = W = (1.5)(3)(4>(1>+£0)(0)(5)(1>+(—1.5Jt—3x4x n = 35 in? m A-IU menu: (Ill: mntroid E. y) of Ihc cyal\ncl_sccli0n and men determine the momcnls orincrlia If and I,-. f_ g = (52.5)(2x125){15)+ (7.5)(270x15) = _ m 2([25)(15)+270(15) 33.942 mm = 33.9 mm Ans Ducmsymmeu-y y'=150mm Ans L. = [—12-[125)(3003)—«i-1§(110](2703) = 101(10‘) mm“ Ans 3-: 2[$(15)(1253)+15([25)(62.5 . 33.942)13+T‘§(270)(1s3) +2'm(15)(33.9424.5)z = 10.8(10‘) mm‘ Ans a—ll Imam the centroid (i. 1‘») ol Ihc Channel section and Ihcn determine the produn of marlin Ly wim mapch In 1hr. faxes. _ m 615)]! :25 15 .5 2-: x=w=w=31942mm=333mm Ans m 2(125}(15)+270(15) Due bosymmetry i: 150 mm Ans 1;. -, - = 23614 = (625 — 33.942)(150~ 7.5)(151025) + (7.5 - 33.942)(0)(270}(15) +(62.5—33.942)(7.5~150)(15}(125)= 0 Ans 'AP‘IZ Locate the position (i. F) for the ccnlrmd C n! Ihc cross sectimzal area and than dalerrnmc the product of in- ertia with respect to thex’ and y‘ axes. L—mm—J - ZfA 25(50)(50) + 100603900) ‘ 85 x = -~— : ‘————— — mm Am 24 5050) + (SOMZOO) Zia! [50)(100)(50) + 25(150X50) f=—~—= = 35 mm. Ans IA 10060) + 15060) 1,3. = 25.4 : (~35 + 25x75 — asxsoxsm + ( mo — ssx—as +25)(200)(50) = 4.5m 10‘) mm‘ Ans A—IJ Delerrninn the prndnci of inertia of the area with re‘ spec: lo lb: 1 and y axes. 1,; = 1554 = (0-5)(4)(3)(1) + (6){0-5)[10)(1) + U 1-5)(1-5)(3}U) = 90.75 in‘ A115 5-14 Delcnninz \hc mammals ol mania L- and r,- or me shaded arcn. I. = 712(40x2003) + 712(200x403) = 27-733UO‘J m" I, = gem)an figment?) +(40)(200)(1202) = 142.933(10‘) mm“ 1,, =0 (Symmelxy aboutxaxis) I _ 1, = Jig—IL +3—2ic0529-Insin 26 27.733 + 142.933 + 21733-141933 2 =[ c0390“+01(1o")=35.3(105J mm‘ Ans r, = —1—’—;ic0529+f,,sin 26 27.733 +142.933 A 27.733 —142.933 2 cos 90°+01(10‘)= 35.31:?) mm‘ Ans =I A—IS Determine 1h: momonu cl lnenia L- and I, and 1h: product of Inertia If,‘ to: lhe semicilcular arcs. I, = I, = émxao') = 5.0894(10‘mm“ L, = {symmmry aboulyaxis) I, 4, 90529— 1,, sin 29 5.0894 + 5.0894 + 5.0894— 5.0894 2 cos tam—0100‘) : s.09(1o°)mm‘ Ans =1 My Ix—r, ;"=”2__ 2 c0329 + 4,, sin 29 5.0894 + 5.0394 _ 5.0394 — 5.0894 2 2 cosao°+01(lo‘)=s.09(10‘)mm‘ An: =[ 1—! I”- : ’—2”-sin 29+lflcos 29 5.0 —. qwsmmuomomo Ans 'A—Ifi Dmn'nine [he momenls ol Inertia hand I); and the product of inertia If, [at Ihc reelangular area. The x‘ and y’ ues pm lhrough lb: Ctfllroid C. 1; 1—;(4ox1603)= 13.65300“) mm‘ .1. 12(160)(403) = 0.853(10‘mm‘ 1,: I” = 0 (symmetry) L. I — _ 1,-= :1’ +‘—21|'-c0528-!vsm 26 13.653 +0353 + 13.653 — 0.853 2 =i 00s 60°—0;(10‘)= 10.5(10‘) mm‘ Ana; 1, 1 I —1 . 1-=—+-?-——‘—J-00329+L $11129 ’ 2 2 ’ 13.653 + 0.853 _ 13.653 —0.353 =[ 2 cos 60°+01(10‘) = 405(10‘) mm‘ Ans L 4y 2 I . - . = [flan 60°+0] = 5.sa(1o‘)mm‘ Ans 1,, = sin 29+1,,cos 28 A-I'.‘ Dclcn-ninc Ihc principal momcnls uf inctlia ol‘ the ems—sectional urea ebmn the principal am that have their orig-in locach at tho centroid C. Use “1: cqualmm devel- oped Ln Sec. 11.4. For the calcuialion, assume all earners lo - be square. ' :1 = 2%(«03733 “amazing—0.13732] + éwsvsxmzsf = 55.55 in‘ I, = 2r1—12(o.375)(43) + (cannon—0.13732] + fiazsxoms)” = 13.39 in‘ 1,, = 2,554 = (—2+ 0.1373(4—0.1375)(4)(o.375)+(0)(0)(725)(0.375)+ (1.3 1 25x—3. 1325)(4)(o.37s) = 40.73 in‘ _ Isl—52 2 {"33— 2 i ( 2 }+I,, = 55-55 +1339 1 (Mp 4—20.73): 2 2 I’m-x = 64'1 in‘ Ans 1m. = 5.33 in“ Ans A-Il Salve Prob. A-n using Morn": circle. I, = 21%(4x033533 + (4x0315)(4— 0.13733] + écomsxmsf = 55.55 in‘ , = 2%«137595+(o.375)(4)(2 4.13732] + lizafinos-isf = 13.39 31‘ 1,, = m .= {—2 + 0.13799: — 0.18?5)(4)(0.375) + (0)(0)(725}(0.375) + (1.3125)(—3.1325)(4)(o.375) = 40.73 m‘ 1,”, _ 5555 + 13.39 2 2 = 34.72 A(55.55.—20.73) C(34.32.0) R = V (55.55 — 34.72)2 + (20333}2 = 29.337 rm = 34.72+29.337 = 54.13? Ana 1..“ = 34.72— 29.337 = 5.33 in" Ans A49 Duenninc lb: prlndpal moments ol inertia I'm [he angles uosisecliuna] arts with respccl. m a an or principal axes that hart \huir origin localed al the cenltoid C Use the equalions dcvelopcd in Sec AA. For the cakulanon. assume - all corners In be square. l I: = = — 1 4, 12mm?» 30(20)(32.22— 10)” + fiaomoo’)+2ouoox50—322232 ,; :1, = 3.1422(10‘) mm‘ 1,, = 29324 = (—32.22 + 10x50 - 32.22x100x20) + (50 — nun—32.22 + 10)(30){20) = 1.7773(10") mm.“ __Ix+[y L“! "=;.+.=‘ 2 *V‘TL’WZ _ 3.1422+3.1422 11422-31422 I 2 :l: (———2——)3+(1.7?78)21(10‘) zm =4.92(10‘) mm‘ Am rm = 136(10‘) mm“ Ans ‘41-?” Saw: Prob. A-IQ using Mom‘s male, A (3.3442,) +7716) I‘ = 1” = Elitmzos) + “20"(32-22 * 1°? + 113(20X1003) + 20(100){50— 3222? {I = [r = 3Jag-100‘s) mm4 1:) = 253721 = {—32.22 + 10)(50 —32.22)(100)(20) + (60—— 32‘22)(_32.22 + loxsoxzm = 1.7773(106) mm“ A6.l422.-1.7778)(106) C(3.1422,0) R = 1.7778 In." =(3.1422+1.7773)(10°}=4_92(msmm4 Am 1"“ =(3-1422-1-77781(10‘)= 1.36t10‘1mm‘ Ans ...
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This note was uploaded on 01/28/2010 for the course STRUC 201 taught by Professor Ali during the Spring '10 term at University of Guelph.

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Ch A - A-l Determine the lncnlion 5' of the cenlraid (' for...

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