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HW22 - Solution

# HW22 - Solution - PROBLEM 9.32 Determine the moment of...

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Unformatted text preview: PROBLEM 9.32 Determine the moment of inertia and the radius of gyration efthe shaded area with respect to the x axis. SOLUTION Filst note that A = A1 — A1 — A} =[{5}Eﬁ} - {4H2} *{4ltIJJ in.2 ={3t:--:5-4)in.1 =13 in.1 I: = {fell _ (Ix-)2 _ {Ix}?!- where {L}. =é£5 in.}{ﬁ in.)3 =91?! in.1 {II}; = %{4 mm inf +1}; in.2}[2 in.)2 2 . =34— ”1.4 3 (:33 =33: man in_}3 +[4 in_1}|[E m]— 1 =9— 'n.‘1 ‘5 {I = [90 44% _ 9%] in} 0r ti = 45.0 1.1.4 «I or k: = LSQ‘Q in. 4 PRGPMETARY MA TERML -L‘ 20m The MeGmw-Hill Curr-patties. inc. All rights reserved. Nu part ufrhls Munmd may In: dammed. repmfuma’ ar dfsn'fﬁma'a' in aqrﬁm I?! hjt' {my «man; IL'ﬂﬁatrf the prﬂclr “Titre" pcrhrt'ﬂr'wn (if the ﬁrm-ﬁsher, m- used {Him-rid the Nmﬂt‘d' dim-Mm» to teachers :rndﬂinmmn'permr'm’d by Mtﬂmlr—ch'ﬁtr their r'mft'lu'idmd' t-tmrsepre;mmticm. {ﬂu-IN areasmdc‘m Ming ”tit .‘ﬁ'mwt'. _1'rm tire mating if “when: mrmt'xst'mt. Ml?| PROBLEM 9.35 Determine the mum-nu of inertia of the shaded area shown with reapect tn the x t andy axes when a = 20 mm. ﬂ‘. SDLUTIDN We have II = {II}, + 21:11]: where [J'Jr }. =é1f4ﬂ mmﬁrlﬂ mm}3 £13.33me mm“ {L h = [gun mm)“ gm mmf[4:“ mm m' +£{2i} mrnf|14xm +20] mm]2 2 321‘ =52149x|03rmn‘ k I: = [213.3% 2{52?_49)]><ID3 mm“ or I: = I.268x1[:‘*mm‘ 4 I}. = {1}.)1 + 2U}- )2 u_,.}1= \$910 mm‘ﬂdﬂ' mmf’ = 2I3.33x1ﬂ] mm“ null My}1 = gm mm}4 =62.33><m3 mm" ill .3. = [213.33 + 2{62.E3}]>< 11:13 mm‘ or I? = 339x In} mm“l ‘1 r; - F PRGPHIETARF MATERIAL ~23 2D“) The McGraw-liill Companies. Inc. All rights menrcd. M1 par: Of this Manual may be difpa'ayed. repatriated or distributed in ﬁlm-ﬁlm: or by (my mm}; n'frhuulf rFrc‘ prior “Huff?” wmm-Jm (If Ike! pawl-Sher. ﬁr used {161le the limited distriburm FE rem-hm: and am-ammmmm by Mﬂmw-Ha'ﬂﬁar their Judi-Mm! 1wrseprrpumn'ou. Emu are a 3mm using Hm Mann-5!. K you an“ Ming I: without permm'lkm. Nil] PROBLEM 9.33 U Determine the momenta of inertia of the shaded area ehewn with " respect to the x and y axes when a = 21’) mm. SOL LITIDN Given: Area = Square — 2(Semjeirele5} T" 3am 4. 1 I i | i 5 Here: x = 1"]. = \$1260)“ =103ﬂx103 mm‘i Semieirele (D: 14. = 31120)“ = 62.83 x 14::3 mm“ '.'I " I“. =r .+Ad-; gum“ =11. +[gueﬂaeaaf J. I}; = 15183le - 45.2?x1ﬂ3 T}, = 1155x103 mum4 i I], = E. + A‘{3ﬂ— 3.433}? =I156x103 +%{20}2[21.512}2 r1. =303.3x103mm4 PROPRIETARY MATERIAL. ﬂ EDIE The MuGruw-I-Iill Campaniﬂ, Inc. All rights Tamed. No par: of ﬂu; Magma! may .52 display-gal . repredrmed (Jr dish'iﬁrm‘ed r'n anyfam m' I53.- any mans, wiring“! the prior marten permirsz'pn of {he publisher, m- used MJ-ond mp: ifmx're-d 'J dismﬁhﬂﬂn H} {ﬂickers endednenmrs pa'rnrr'hl‘e'd by Merw—Hﬂ'fﬁ'r Ifrer'r r'rra'r'vidua! enunepmfmmripn. {fyou areasma'em wing .I'hJ'S mung}, ,. wm are using :‘r wr'rhou: purmjsﬂm. I41] PROBLEM 9.3E {Continued} Semicircle (2): Same as samicircle (I). m: II =1ﬂ3f§lxlﬂ3 —2{52.33x103} =9543x103mm‘ I: =93:me mm" «1 I). = 1030x103 —2{303.3x103} =463.3x1:13 mm“ Iv = 463le m“ «I -III II. pgupmgmﬂr MATERIAL 4:31 20") The McGraw-Hiﬂ Companies. Inc. A11 righls reserved. M: pan of m.- Munua! mu he dispu’u}m repaicﬂrf‘ﬁd {1r 5&5”:me in- awﬁmu or by "H.“ means, without me prior written pea-minim?! ofrhe pubﬂshe-r. or used hgwnd Hm h'mJ'Ie'd dmm'mm'an {a rem-1m: and edll'Cﬂ'ID-I‘I permuted {ry mﬂmwhfﬁﬂfm- their Indiw'a‘wf mun-e preparation. {fyuu are n madam“ using this Haunt yum an: wing I: withers: pennimiarr. 1422 ...
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HW22 - Solution - PROBLEM 9.32 Determine the moment of...

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