Solution manual for vector mechanics for engineers statics and dynamics 12th edition ferdinand beer by wilfred.nelson623

Solution Manual for Vector Mechanics for Engineers: Statics and Dynamics, 12th Edition, Ferdinand Beer, E. Russell Johnston Jr., David Mazurek, Phillip Cornwell Brian Self

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Solution Manual for Vector Mechanics for Engineers: Statics and Dynamics, 12th Edition, Ferdinand Visit TestBankBell.com to get complete for all chapters

PROBLEM 9.1

Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe y axis. SOLUTION At   0,2 :0 xybbka  2 b k a  then  2 2; b yxadAydx a      22 22 2 0 4322 2 0 5423 2 0 2 111 523 y a y a a dIxdAxydx b Ixxadx a b xaxaxdx a b xaxax a                13 30 y Iab  ◀

PROBLEM 9.2 Determinebydirectintegrationthemomentofinertiaoftheshadedarea

withrespecttothe y axis. SOLUTION 1/3ykx  For: xa  1/3bka  /1/3kba  Thus:1/3 1/3 b yx a  22 y dIxdAxydx  21/37/3 1/31/3 7/310/3 1/31/3 0 3 10 y a yy bb dIxxdxxdx aa bb IdIxdxa aa            33 10 y Iab  ◀

Determinebydirectintegrationthemomentofinertiaoftheshadedareawith respecttothe y axis. SOLUTION 22 2 22 For: thus ykxxabka bb kyx aa   Now22 22 2 [()] y dIxdAxbydx b xbxdx a           Then24 02 a yy b IdIbxxdx a           3311 35 abab     23 or 15 y Iab  ◀

PROBLEM 9.3

PROBLEM 9.4

Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe y axis. SOLUTION 33 3 3 ,for, ykxxabka b yx a   then dAydx  2223 3 5 30 6 3 0 6 y a y a b dIxdAxydxxxdx a b Ixdx a bx a                  13 6 y Iab  ◀

PROBLEM 9.5

Determinebydirectintegrationthemomentofinertiaoftheshadedarea

-HillEducation.Allrightsreserved.Noreproductionordistributionwithouttheprior writtenconsentofMcGraw-HillEducation. 1422

withrespecttothe x axis. SOLUTION At   0,2 :0 xybbka  2 b k a  then  2 2 b yxa a  or1±,takethenegativerootsothatwhen0. y xaybx b            22 252 12 0 372 12 0 1 121 37 x b x b dIydAyxdy Iayydy b ayy b                             13 21 x Iab  ◀

PROBLEM

9.6 Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe x axis. SOLUTION 1/3 1/3 b yx a  33 31/3 1/3 332 0 111 333 11 332 x a xx bb dIydxxdxxdx aa bba IdIxdx aa            13 6 x Iab  ◀

PROBLEM

9.7 Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe x axis. SOLUTION SeefigurefromsolutiontoProb.9.3: Now, 22 2 22 For: thus ykxxabka bb kyx aa   Now33 3 36 6 11 33 1 33 x dIbdxydx b bdxxdx a   Then 3 36 06 37 3 6 0 1 33 1 37 a xx a b IdIbxdx a bx bx a                          37 3 6 3 1 37 11 321 ba ba a ab               23 or 7 x Iab  ◀

PROBLEM 9.8

Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe x axis. SOLUTION 33 3 3 ,for, ykxxabka b yx a   then 3 39 9 3 9 09 310 9 0 310 9 1 33 1 3 1 310 1 310 x a x a b dIydxxdx a b Ixdx a bx x ba a                                13 30 x Iab  ◀

PROBLEM 9.9

Determinebydirectintegrationthemomentofinertiaofthe shadedareawithrespecttothe x axis. SOLUTION At3,: xayb  3(3)bkaa  or3 8 b k a  Then3 83() b yxa a  Now3 3 3 3 3 9 9 1 3 1 () 38 1536() x dIydx b xadx a b xadx a       Then 3333 910 99 3 10 9 1 ()()153610 1536 [(3)0] 15,360 aa xx aa bb IdIxadxxa aa b aa a       13 or 15 x Iab  ◀

PROBLEM 9.10

areawithrespecttothe x axis. SOLUTION 1/2 121/2 ; bab yyx xa  Now 33/233 33 213/23 111 333 x bxbx dIydxydxdx aa             Then 33 3/23 3/230033 aa xx bb IdIxdxxdx aa   35/234 3/23 0 3 5/2433 21 1512 a bxbx aa ab                      13 or 20 x Iab  ◀

Determinebydirectintegrationthemomentofinertiaoftheshaded

PROBLEM 9.11

Determinebydirectintegrationthemomentofinertiaoftheshadedarea withrespecttothe x axis. SOLUTION At,: xayb  / or aabbkek e  Then//1 bxaxayebe e  Now3/13 33(/1) 11 () 33 1 3 xa x xa dIydxbedx bedx   Then 3 33(/1)3(/1) 00 1 333 aa xaxa xx ba IdIbedxe       133(1) 9 abe or0.10563 x Iab  ◀

9.12

Determinebydirectintegrationthemomentofinertiaofthe shadedareawithrespecttothe y axis. SOLUTION At3,: xayb  3(3)bkaa  or3 8 b k a  Then3 83() b yxa a  Now2223 3 23223 3 ()() 8 (33) 8 y b dIxdAxydxxxadx a b xxxaxaadx a      Then 3 543232 3 3 652433 3 652433 3 652433 (33) 8 1331 86543 1331 (3)(3)(3)(3)86543 1331 ()()()() 6543 a yy a a a b IdIxxaxaaxdx a b xaxaxax a b aaaaaaa a aaaaaaa                              or 3.433 y Iab  ◀

PROBLEM 9.13

Determinebydirectintegrationthemomentofinertiaofthe shadedareawithrespecttothe y axis.

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SOLUTION 1/2 12 ; ymxykx  12 1/21/2 For: or/ or/ xayyb bmamba bkakba    Thus,1/2 121/2 ; bb yxyx aa  1/2 11/2ybx a              2221/2 211/2 21/2 01/2 5/23 1/200 7/24 1/2 0 3 7/24 21 74 y a yy aa y a y y bbx dIxdAxyydxxxdx aa bbx IdIxxdx aa bb Ixdxxdx aa bxbx I aa Iab                                              Then3 1 28 y Iab  ◀

PROBLEM 9.14

Determinebydirectintegrationthemomentofinertiaoftheshadedarea

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HillEducation. 1431

withrespecttothe y axis. SOLUTION At,: xayb  / aa bke  or b k e  Then//1 bxaxayebe e  Now22 2/1 () () y xa dIxdAxydx xbedx   Then2/1 0 a xa yy IdIbxedx  Nowuseintegrationbypartswith 2/1 2/1 xa xa uxdvedx duxdxvae   Then2/12/1/1 0 00 3/1 0 ()2 2 aa a xaxaxa a xa xedxxaeaexdx aaxedx        Usingintegrationbypartswith /1 /1 xa xa uxdvedx dudxvae   Then   3/1/1 0 0 322/1 0 2()|() 2()| a xaaxa y xaa Ibaaxaeaedx baaaae                   32221 2()baaaaae    or0.2643 y Iab  ◀

PROBLEM 9.15

Determinethemomentofinertiaandtheradiusofgyrationofthe shadedareashownwithrespecttothe x axis.

1432

SOLUTION 2/3 ykx  For2/3 2/3 b xabkak a  Thus, 2/3 2/3 3 3 2 2 1 3 1 3 x b yx a dIydx b xdx a    3 2 20 33 2 3 33 a xx b IdIxdx a ba a    13 9 x Iab  ◀ 5/3 2/3 002/32/3 3 5/35 aabba Aydxxdxab aa     12 9 22 5 327 5 x x Iab kb A ab           5 27 x kb  ◀

PROBLEM 9.16

Determinethemomentofinertiaandtheradiusofgyrationoftheshaded areashownwithrespecttothe x axis.

1433

SOLUTION 22 22 1 xy ab  or 2 2 1 y xa b             2 dAxdy  Then 2 222 002 221 bb xx y IdIydAyxdyaydy b              Setsin,cosybdybd  /222 22 02 sin 2sin1cos x b Iabbd b                  /2 322 0 2sincos x Iabd       /2/2 323 00 111 2sin1cos4 422 x Iabdabd      /2 33 0 111 sin4 4442 x Iabab                13 8 x Iab   ◀ 2 002 221 bby AdAxdyady b                /2 2 0 21sincos Aabd     /2 2 0 2cos Aabd    

PROBLEM 9.16 (continued)

  /2 0 1 21cos2 2 Aabd     /2 0 1 sin2 2 Aab       1 2 Aab   3 2 2 1 8 14 2 x x Iabb k Aab    2 x b k  ◀

PROBLEM 9.17

Determinethemomentofinertiaandtheradiusofgyrationofthe shadedareashownwithrespecttothe y axis.

SOLUTION FromthesolutiontoProb.9.15,2/3 2/3 b yx a  and 3 5 Aab  2222/3 2/3 11/3 28/3 02/32/3 () 11 3 y a yy b dIxdAxydxxxdx a bba IdIxxdx aa    33 11 y Iab  ◀   33 11 22 5 311 5 y y Iab ka A ab           5 11 y ka  ◀

PROBLEM 9.18

Determinethemomentofinertiaandtheradiusofgyrationoftheshaded areashownwithrespecttothe y axis.

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SOLUTION 22 22 1 xy ab  or 2 2 1 ybx a             dAydx  Then 2 22 2 1 a yy a x IdIxdAbxdx a              Setsin,cosxadxad  /2 222 /2 2sin1sincos y Iaaad     /2 322 /2 sincos y Iabd        /2/2 323 /2/2 111 sin21cos4 442 y Iabdabd       /2 33 /2 111 sin4 84822 y Iabab                 13 8 y Iab   ◀ FromthesolutiontoProblem9.16;1 2 Aab     13 8 22 1 2 1 4 y y Iab ka Aab   1 2 y ka  ◀

PROBLEM 9.19

Determinethemomentofinertiaandtheradiusofgyrationoftheshaded areashownwithrespecttothe x axis.

1437

SOLUTION Wehave 1/2 1 x dAydxbdx a                Then 1/2 0 3/2 0 1 21 33 a a x AdAbdx a bxxab a                      Now 3 1/2 3 31/23/2 11 1 33 133 3 x dIydxbxdx a bxxx dx aaa                                   Then 31/23/2 0 3 3/225/2 3/2 0 133 3 232 325 a xx a bxxx IdIdx aaa b xxxx aaa                or 13 30 x Iab  ◀ and 13 230 1 3 x x Iab k Aab  or 10 x b k  ◀

PROBLEM 9.20

Determinethemomentofinertiaandtheradiusofgyrationoftheshaded areashownwithrespecttothe y axis.

1438

SOLUTION Wehave 1/2 1 x dAydxbdx a                Then 1/2 0 3/2 0 1 21 33 a a x AdAbdx a bxxab a                      Now 1/2 222()1 y x dIxdAxydxxbdx a                         Then25/237/2 00 112 37 a a yy IdIbxxdxbxx aa                or 13 21 y Iab  ◀ and 13 221 1 3 y y Iab k Aab  or 7 y ka  ◀

PROBLEM 9.21

Determinethepolarmomentofinertiaandthepolarradiusofgyration oftherectangleshownwithrespecttothemidpointofoneofits(a) longersides,(b)shortersides. SOLUTION (a)Wehave3 1 3 x dIadx  Then   3 3 4 1 3 2 3 2 3 a x a Iadx a a a     Also   22 y dIxdAxadx  Then2 24 3 a y a Iaxdx a    Now44 22 33 oxy JIIaa  or 44 3 o Ja  ◀ and 4 2 2 4 3 2 o o Ja k Aa  22 3 a  or 2 3 o ka  ◀

PROBLEM 9.21 (continued)

(b)Wehave 3 2113 3212 x dIadxadx                       Then   2 3 0 3 4 1 12 2 12 1 6 a x Iadx a a a     Also   22 y dIxdAxadx  Then 2 2 0 3 4 2 3 8 3 a y Iaxdx a a a     Now44 18 63 oxy JIIaa  or 174 6 o Ja  ◀ and 4 2 2 17 6 2 o o Ja k Aa  172 12 a  or 17 12 o ka  ◀

PROBLEM 9.22

Determinethepolarmomentofinertiaandthepolarradiusof gyrationofthetrapezoidshownwithrespecttopoint P. SOLUTION 1 22 av xaav a          00 2 0 2 2 11 22 1 4 3 4 3 2 aa a dAxdv AdAavdv avv a Aa                       Then 22 0 3 4 0 4 4 4 11 22 1 38 11 38 5 24 5 12 a x a x IvdAvavdv v av a a Ia                          

PROBLEM 9.22 (Continued)

  3 0 4 0 4 4 4 4 1111 2332 111 2 342 1 62 5 32 5 16 a y a y Ixdvavdv av a a a Ia                                              Now44 55 1216 oxy JIIaa  354 48 o Ja  ◀ and 354 22 48 34 2 35 72 o o Ja ka Aa  35 72 o ka  ◀

PROBLEM 9.23

Determinethepolarmomentofinertiaandthepolarradiusof gyrationoftheshadedareashownwithrespecttoPoint P. SOLUTION 1:y At22,: xaya  2 1122 2 1 ()or akak a  2:y At0,: xya  ac  At22,: xaya   2 1 22 aaka  or2 1 4 k a  Then22 12 22 11 24 1 (4) 4 yxyax aa ax a   Now222 21 22 11 ()(4) 42 1 (4) 4 dAyydxaxxdx aa axdx a      Then 22 22232 00 1118 2(4)4 4233 aa AdAaxdxaxxa aa       Now 33 33222 21 6422466 33 642246 3 11111 (4) 33342 111 (644812) 3648 1 (6448127) 192 x dIyydxaxxdx aa aaxaxxxdx aa aaxaxxdx a                        

PROBLEM 9.23 (Continued)

Then 2 642246 3 0 1 2(6448127) 192 a xx IdIaaxaxxdx a    2 643257 3 0 643257 3 44 112 6416 965 112 64(2)16(2)(2)(2)965 11232 12812832128 96515 a axaxaxx a aaaaaaa a aa                  Also2222 1 (4) 4 y dIxdAxaxdx a     Then 22 222235 00 23544 1141 2(4)4235 141321132 (2)(2) 23523515 aa yy IdIxaxdxaxx aa aaaaa a                   Now44 3232 1515 Pxy JIIaa  or 644 15 P Ja  ◀ and 644 15 22 82 3 8 5 P P Ja ka Aa  or1.265 P ka  ◀

PROBLEM 9.24

Determinethepolarmomentofinertiaandthepolarradiusofgyrationof theshadedareashownwithrespecttoPoint P.

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SOLUTION Theequationofthecircleis 222 xyr  Sothat22 xry  Now22dAxdyrydy  Then22 /2 2 r r AdArydy   Letsin;cosyrdyrd  Then /2 22 /6 /2/2 222 /6/6 3 22 26 2 2(sin)cos sin2 2cos2 24 sin3 22 22438 2.5274 Arrrd rdr rr r                                                 Now   2222 x dIydAyrydy  Then222 /2 2 r xx r IdIyrydy   Letsin;cosyrdyrd  Then /2 222 /6 /2 22 /6 2(sin)(sin)cos 2sin(cos)cos x Irrrrd rrrd           Now221sin22sincossincossin2 4  

PROBLEM 9.24 (Continued)

Then /24/2 42 /6/6 42 263 4 1sin4 2sin2 4228 sin 2228 3 2316 x Irdr r r                                                     Also  3 32211 33 y dIxdyrydy  Then223/2 /2 1 2() 3 r yy r IdIrydy   Letsin;cosyrdyrd  Then /2 223/2 /6 2 [(sin)]cos 3 y Irrrd      /2 33 /6 2 (cos)cos 3 y Irrd      Now42222 1 coscos(1sin)cossin2 4   Then /2 422 /6 /2 4 /6 21 cossin2 34 2sin21sin4 324428 y Ird r                          2 433 2266 4 4 21sinsin 1 324224428 21313 3416124248322 293 3464 r r r                                                                  

PROBLEM 9.24 (Continued)

Now 4 4 3293 23163464 Pxy r JIIr         44 3 1.15545 316 rr             or1.1554 P Jr  ◀ and 4 2 2 1.15545 2.5274 P P Jr k Ar  or0.676 P kr  ◀

PROBLEM 9.25

(a)Determinebydirectintegrationthepolarmomentofinertiaoftheannular areashownwithrespecttoPoint O.(b)UsingtheresultofPart a,determine themomentofinertiaofthegivenareawithrespecttothe x axis.

(b)FromEq.(9.4):(Note bysymmetry.)

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SOLUTION (a)2dAudu   22 3 (2) 2 O dJudAuudu udu     2 1 2 1 3 4 2 1 2 4 R OO R R R JdJudu u      44 221 O JRR     ◀

xyII  44 21 2 1 24 Oxyx xO JIII IJRR      44 421 x IRR     ◀

(a)FromProblem9.23:

PROBLEM 9.26

(a)Showthatthepolarradiusofgyration kO oftheannularareashownis approximatelyequaltothemeanradius12()2 m RRR  forsmallvalues ofthethickness21 . tRR  (b)Determinethepercentageerrorintroduced byusing Rm inplaceof kO forthefollowingvaluesof t/Rm:1,1

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2,and110 SOLUTION

44 221 O JRR       44 2221 22 21 O O JRR k ARR       222 21 1 2 O kRR   Thickness:21 tRR  Meanradius:12 1 () 2 m RRR  Thus,12 11 and 22mm RRtRRt  22 222 1111 2224Ommm kRtRtRt           For t smallcomparedto: mR 22 OmkR  OmkR  ◀ (b)(exactvalue)(approximationvalue) Percentageerror100 (exactvalue)      2 221 14 4 222 11 44 11 P.E.(100)100 1 m m t mmR mt R RtR Rt      

PROBLEM 9.26 (Continued)

For1: m t R  1 4 1 4 11 P.E.(100)10.56% 1     ◀ For1: 2 m t R      112 42 112 42 11 P.E.(100)2.99% 1     ◀ For1: 10 m t R      112 410 112 410 11 P.E.(100)0.1250% 1     ◀

PROBLEM 9.27

Determinethepolarmomentofinertiaandthepolarradiusofgyrationof theshadedareashownwithrespecttoPoint O.

1451

SOLUTION At,2: Ra  2() aak   or ka   Then1 a Raa             Now()() dAdrrd rdrd     Then (1/)(1/) 2 0000 1 2 aa AdArdrdrd              23 22 0 0 3 232 11 11 223 17 1(1) 66 Aada aa                             Now22() O dJrdArrdrd   Then (1/)3 00 a OO JdJrdrd     (1/)4 44 00 0 55 445 0 11 1 44 11 4511(1) 20 a rdad aa                                                  314 or 20 O Ja   ◀ and 314 22 20 72 6 93 70 O O Ja ka Aa    or1.153 O ka  ◀

PROBLEM 9.28

Determinethepolarmomentofinertiaandthepolarradiusofgyrationof theisoscelestriangleshownwithrespecttoPoint O.

1452

writtenconsentofMcGraw-HillEducation.

SOLUTION Byobservation: 2 b h yx  or 2 b xy h  Now 2 b dAxdyydy h          and23 2 x b dIydAydy h  Then3 0 2 2 h xx b IdIydy h  4 3 0 1 44 h bybh h  Fromabove: 2 h yx b  Now2 () h dAhydxhxdx b          (2) h bxdx b  and22(2) y h dIxdAxbxdx b 

PROBLEM 9.28 (Continued)

Then /2 2 0 2(2) b yy h IdIxbxdx b   /2 34 0 34 3 11 2 32 11 2 322248 hb bxx b hbbb bh b               Now33 11 448 Oxy JIIbhbh  or 22 (12) 48 O bh Jhb  ◀ and 22 222 48 1 2 (12)1(12) 24 bh O O Jhb khb Abh   or 1222 24 O hb k   ◀

PROBLEM 9.29*

UsingthepolarmomentofinertiaoftheisoscelestriangleofProblem 9.28,showthatthecentroidalpolarmomentofinertiaofacircularareaof radius r is4/2. r (Hint: Asacircularareaisdividedintoanincreasing numberofequalcircularsectors,whatistheapproximateshapeofeach circularsector?)

PROBLEM 9.28 Determinethepolarmomentofinertiaandthepolar radiusofgyrationoftheisoscelestriangleshownwithrespecttoPoint O.

SOLUTION

Firstthecircularareaisdividedintoanincreasingnumberofidenticalcircularsectors.Thesectorscanbe approximatedbyisoscelestriangles.Foralargenumberofsectorstheapproximatedimensionsofoneof theisoscelestrianglesareasshown. Foranisoscelestriangle(seeProblem9.28):

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22 (12) 48 O bh Jhb  Thenwithandbrhr   22 sector 1 ()()()[12()] 48 O Jrrrr    142(12) 48 r     Now   sectorsector42 00 1 limlim[12()] 48 OO dJJ r d                       14 4 r  Then   2 442 circlesector0 0 11 () 44 OO JdJrdr     or 4 circle() 2 O Jr   ◀

PROBLEM 9.30*

Provethatthecentroidalpolarmomentofinertiaofagivenarea A cannotbesmallerthan2/2. A  (Hint: Comparethemomentofinertiaofthegivenareawiththemomentofinertiaofacirclethathas thesameareaandthesamecentroid.)

SOLUTION

FromthesolutiontosampleProblem9.2,thecentroidalpolarmomentofinertiaofacircularareais

Twomethodsofsolutionwillbepresented.However,bothmethodsdependupontheobservationthatasa givenelementofarea dA ismovedclosertosomePoint C.Thevalueof C J willbedecreased 2 (;

JrdA

as r decreases,somust). C J

Solution 1

Imaginetakingthearea A anddrawingitintoathinstripofnegligiblewidthandofsufficientlengthso thatitsareaisequalto A.Tominimizethevalueof(), CAJ theareawouldhavetobedistributedasclosely aspossibleabout C.Thisisaccomplishedbywindingthestripintoatightlywoundrollwith C asits center;anyvoidsintherollwouldplacethecorrespondingareafartherfrom C thanisnecessary,thus increasingthevalueof(). CAJ (Theprocessisanalogoustorewindingalengthoftapebackintoaroll.) Sincetheshapeoftherolliscircular,withthecentroidofitsareaat C,itfollowsthat

wheretheequalityapplieswhentheoriginalareaiscircular.

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4 cir() 2 C Jr   Theareaofthecircleis 2 cir Ar   Sothat 2 cir[()] 2 C A JA  



()Q.E.D. 2 CA A J   ◀

PROBLEM 9.30* (Continued)

Solution 2 Consideranarea A withitscentroidatPoint C andacircularareaofarea A withitscenter(andcentroid) atPoint C.Withoutlossofgenerality,assumethat 1234 AAAA  Itthenfollowsthat cir1234 ()()[()()()()] CACCCCC JJJAJAJAJA  Nowobservethat 12 ()()0CC JAJA 34 ()()0CC JAJA Sinceasagivenareaismovedfartherawayfrom C, itspolarmomentofinertiawithrespectto C must increase. cir()() CACJJ  2 or()Q.E.D. 2 CA A J   ◀

SOLUTION

PROBLEM 9.31

Determinethemomentofinertiaandtheradiusofgyrationofthe shadedareawithrespecttothe x axis.

Assignarea1tobetheflangesandarea2tobetheweb. Firstnotethat 12 2 2 2 [2(10)(40)(10)(90)]mm 1700mm AAA    Now12 2()() xxx III  where 322 1 34 1 ()(10mm)(40mm)(400mm)(25mm) 12 303.3310mm x I   334 2 1 ()(90mm)(10mm)7.5010mm 12 x I   Then334 (2(303.3310)7.5010)mm x I  34614.1610mm 34or61410mm x I  ◀ and 34 22 2 614.1610mm 361.27mm 1700mm x x I k A    or19.01mm x k  ◀

PROBLEM 9.32

Determinethemomentofinertiaandtheradiusofgyrationof theshadedareawithrespecttothe x axis. SOLUTION Assignarea1tobetheflangesandarea2tobetheweb. Firstnotethat 12 2 2 2 [2(3)(0.5)(5)(0.5)]in 5.50in AAA    Now12 2()() xxx III  where324 1 1 ()(3in.)(0.5in.)(3in.)(0.5in.)(2.75in.)11.3750in 12 x I  3 2 4 1 ()(0.5in.)(5in.) 12 5.2083in x I   Then   44 2(11.3750)5.2803in28.030in x I  or28.04 in x I  ◀ and 4 22 4 28.030in 5.0964in 5.50in x x I k A  or2.26in. x k  ◀

PROBLEM 9.33

Determinethemomentofinertiaandtheradiusofgyrationofthe shadedareawithrespecttothe y axis. SOLUTION Assignarea1tobetheflangesandarea2tobetheweb. Firstnotethat 12 2 2 2 [2(10)(40)(10)(90)]mm 1700mm AAA    Now12 2()() yyy III  where 3234 1 334 2 1 ()(40mm)(10mm)(40mm)(10mm)(40mm)643.3310mm 12 1 ()(10mm)(90mm)607.5010mm 12 y y I I      Then33434 (2(643.3310)607.5010)mm1,894.1610mm y I  or 641.89410mm y I  ◀ and 34 22 2 1,894.1610mm1,114.21mm 1700mm y y I k A    or33.4mm y k  ◀

PROBLEM 9.34

Determinethemomentofinertiaandtheradiusofgyration oftheshadedareawithrespecttothe y axis. SOLUTION Assignarea1tobetheflangesandarea2tobetheweb. Firstnotethat 12 2 2 2 [2(3)(0.5)(5)(0.5)]in 5.50in AAA    Now12 2()() yyy III  where 324 1 1 ()(0.5in.)(3in.)(0.5in.)(3in.)(1.25in.)3.4688in 12 y I  34 2 1 ()(5in.)(0.5in.)0.052083in 12 y I  Then   44 2(3.4688)0.052083in6.9897in y I  or6.994 in y I  ◀ and 4 22 2 6.9897in1.27085in 5.50in y y I k A  or1.127in. y k  ◀