ch 12 - thhm 111-? Calculate the mm of inertia I: and I!...

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Unformatted text preview: thhm 111-? Calculate the mm of inertia I: and I! wifla rcspctl [u the x and y axes for Em: L-sl-upcd aura Wu in the. figure for Prob. [2.3-1 J.“ ............................................... __..._ .................................................................................... Il'l uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu nu Snlullnn 12.4w? Moments of inertia firming-nit: am 1' 1' 1: =1. + I? I ® ® =?35)[t}.5}3 +%{u.5](6)1 = 3m in.‘ 1- fi'm IElia. Iy=f3+fi _ 1 3 1 : (ED @ — iflljflfi} + 3(5.5}{fl.5} = Iii-.9 in.‘ 1— fl 1; ‘ _ I 4 mgl 4m Thickms I = I}.5 in. PrflMElfl 115-4 The mutant of inania with Impact to axis l-l of the scalcnc Iii-Laugh: sham: in Ill: figun: i5 90 K l0"-I mm‘. Calmlau: its mom! of inertia Ll with [esp-act In axis 12. Eulltiun 12.54 Moment of inertia —l— b=4flmm II=QflXlfl3mm4 fl=bh3f12 l2! 3 I h b g. ‘ Ic=bh3flfi=3flxlfl3mm‘ 1 I _ r_ 2_ |-—40 f-~I,+M r1r+ bhfld-flflxlflz‘ 2 HM Min—“'1 2 3 1 i I + 3:40) am 1251! =41}: x mlmfi «— Prnhlum 122-2 Dczcrn'lim nu: diam j- m I11: acme-id C of a. trapcmid having hascsa and b and altitude in {see (235: &Appendix D}. .................................................................................................................................................. ...... ..."... . . ......_.___.__............... ... .... "......" ........_.......... ".....u uh”... Salutinn 12.2-2 Emu-aid of: trap-amid a h I d 91: Irfl=1ylb+{a-bwhldy I F: J" J" Jr‘- _L =—{3a +b} 3 I I5 WidIh ofelernent = h + (a -- bum F .__ g = “2" +1]. ...._ fl=lb+{a-b}flhldy “‘ 3'31”) Ma + b} 2 {I A=Idfl=J [b+{a—b}yfh]d}r = El thlcm 115-5 Fur the beancross sanction dram-Elm! in Pruh. [2.36. calculaie th: centroids] manual: of inertia I If and I, will] Impact H:- was mm the cent-Did c such um the 1: axis is pmnél m flux: 1 axis and the y: axis coincides with the y axis. Sulufinn 12.5—5 Moment of inertia HIE. ~¢I n .r FFEIITI Prob. I235: E —' 13.94 in. W 24 3:162 d = 251]] in. A = 411' in.2 I d3? = 11.5 in. I. = 5m} m“ I; = I), -~ 443 in.‘ IL: = fl +2441}? - dill" = 51m +{4?.?m.44}1 ‘= 5269 ill.“ Fn = I: 2 443 "1-4 PLA'I'E I; = whammy} + {anamw + 333 — i}: = 0.2313 + fiifim + 11375 — 13.9412 2 0.21m + 6:11.44? = T85 in.“ I; = mzqamuaf = 323 m.‘ ENTIRE muss SECHDN 3,, = I; + n: : 52:59 + T35 = man in.‘ :k =1; + I; = 443 +32 =41: Ln.‘ #— #— Problem 114-! A semicircular area of radius 150 [mu has a rummagular cumin! of dimnaima 50 mm X HI] mm {339. figure}. Calmlan: the mnmcnts nl“ I'mm'a I, and 1'Jr with [esp-3m to 1h: 1: and 3' axes. Also, calculate the canespumjing radii flfgymlim rx and rr. 0 |— 50 in " mm mm 150 mm :51] mm Sulllifln 12.14 Mummtsnf inertia of composite am J, All dimmsiflns in nfiflimflers r=lfiflmm b=1mnun b=5flmm J- 3 m H: I! = [Illmmk ‘F = ? — Y = [94.6 H 1&5 mm‘ fyzfx ‘— 1 TIT 4— fl — an an x mm mm A ="——HI =3fl.34 X [Harm-.11I tfifl rum IS-D mm 2 n: VLM = 30.] mm ry=rx 1— 4.— FI'IIIIIIII'I‘I 12.3-2 Elna qua-In of a square uf side a is rean {see figure}. What an: the coordinates x and} of Ihc mimid Coffin: remaining area? ‘Ill FHflIS. 12.3‘2 an! 12.5-2 9L4 IIIIIIIIIIIIIIIIIIIIIII l—Hllllll I IJ'.u:......_----------... u. . . u . .....n......_.----—--.“... ...m.._............---- --.................._..—---.-.. mu. . .. _...... .. Silurian 12.3-2 Cenlrnid at a composite art-a }' a: _ 3a A 2-— =— I 4 JP: 4 a2 _ a “3 ==z 32 4 _ 3a a!) a a] 54:3 = =— — +— _ =— Q' 2}" 4(4 4(2) 115 Q; 5:: Fruhlam 121-3 Dammit-re the trauma: of incrtia II of a parabolic spandrei of base I: and height h with respect u:- its base [sac Case IE. Appenjix D}. Salutlnn 12.44 Moment a! Inertia of n parabuflc apandrel Width of element =b—x=h-5\E =1!“ -' 'Vyfh} dd =b{l — Vj'arhldf 1: bit]- I: =[f1fl = LEI-b“ — Viv-“IN! =E “'— FI'DNHII 12.5%”, The wide-flange beam metiuu shown in the figure has a total height of 25!] mm and a column: thickness (If 15 mm. Detemnine the flange widfll b if it is mquimd that the mental [hummus {If inertia J" and I: be in lire ratio 3 m L respectively. All dimensiunfi in I'llillimfitflts. l5 mm I I = _ 3_ _ _ 5 _L L [1 (M250) u 4:1: Irv-“22m T = 0.4141 x: 1m: + 132.1 x mfimmr r, = 2 (15}(b}’ + 11—: {2201(1SJ’ [5 mm bl- = 1.55! + ISLEEI} {mm‘} Equate I, to 3!Jr and rearrange: 15 b3 - (MM? X [[1511 - Ill: X 1&5 = I} I—b——~1 r = IS mm b = flange width Salve numerically: b=25l1mm 1|— Prnhlam 12.5-2 Datennine 1h: mnment of inertia I: with respam m an axis through the amuuid C and pamlltl In £111: 1 axis fn-I flu: gummh'ic figurt duscribcd in PEI-h. 113-2. Snlitlnn 115-2 Mm: of inertia me Pmb. [2.3-1 A=3alf4 E y=5um 1 I a 1 a: a’ 35" J+-(-)(-)=- ' 3(2)“ 32 2 us til 5 I,=:,r+.q? 4- 2 I,=I:‘=I‘—Aj3=31_fi(.5_fl) HS 4 12 _Ila" _ 192 Sniutiun 12.3-5 Cantmid or beam mm mm (9 W 24 K lfil A1 =41? in.1 d = limin. y] = arr: =12: in. Puma; 3.0 x {135 in. A2 = {3.01:035} = 5.0 in? i2 = 25.00 + 035;: = 25.3?5 in. A = Edi=Al+A2=51Tflin9 F flg= ERA; =Lm +§1r11=?43.5in_1 §=$=1334m ‘— .I A Pruhllm 12.3-3 Calculate the. distance 3.? to flu: ccnlmid C of up: channcl section shtmrn in lhe figure ifu = E- in.,b I in, and r = 2 in. FEMS. 12.34. 12.34. and 12.5-3 ................ ..,...m.m................_............ I...—..._...,_.....__._._|.|.I__l....u._|.|....|.|_.u._|_.|.l.|.Iqa—h.l._.u.u.._u....lul...l|_|.lululu1n.lul|.|.u|.u|.u|.InLululLululuuu—._l.u.|.u..u.l..d|.....|+".".."".."1"11.._ Solution 12.3-3 Centruid of a channel median Al =bc= 11.11.? §1=b+cf2 =1in. A2=flb=éiJL1 h=g=fl5im A = 2551.4. +A1=lflin.3 Q.= 25.4. =2§m+§afi1= “Jilin-3 g A E: = 1.10m. «— ...
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This homework help was uploaded on 04/17/2008 for the course AEM 3031 taught by Professor Fosdick during the Spring '08 term at Minnesota.

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ch 12 - thhm 111-? Calculate the mm of inertia I: and I!...

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