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  • Chapter 10

  • Preprietary Material. 2009 The McGraw-Hill Cempantes, Inc, Ali rghts reserved. No part ofthis Manual may be displayed, reproduced, or dlstributed in any form or by any rneans, withour the prior written permssion ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student usng this manual is using it without permisslon.

    L..et 9 b~ i~e c:11Y1.,1~ c.h"WJe cJ bav- A8. F -=- kx = k L s1'11S

    10.2 Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the critica! load P cr-

    1

  • Me, .,.. ke -+:) M~ ~ o ; l< e - i=',. L -::- o

    '3etV' Ac O Me. = o; ~"' ~ L e - { L ~ =- o

    10.4 Two rigid bars AC and BC are connected by a pin at C as shown. Knowing that the torsional spring at B is of constant K, determine the critieal load P cr for the system.

    --T ~L 2

    A

    Proprietary Material. 2009 The McGraw-Hill Companles, Inc, Ali rights reserved. No part ofthis Manual may be displayed, rcproduced, or distributcd in any forrn or by any means, without the prior written permission of the publisher, or used beyond the lmired distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it wthout permission.

    Problem 10.4

    3f$ + 3PS s: 11~ '; o L. ~L (_'.l .E - k) ~ = o ~L

    ?c ... ::: -:

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    fc. .. = 3G3 N 1T-i(=1ltf1~) e - '&l .. ~ N L:i. ( l.oo ) ,_ e,,.,

    10.9 Determine the critical load of a wooden meter stick whch has a 7 x 24-mm rectangular cross section. Use E= 12 GPa. Problem 10.9

  • Proprietary Material. 2009 The McGraw-Hll Companles, Ine, Al! rights reserved .. No part of'this Manual may be dlsplayed, reproduced, or distribured in any fotm or by any means, wthout the prior wrinen permisson ofthe pubtisher, or used beyond the lmited distribution to teachers and educators pcrmiued by McGraw-Hill for their individual course preparation .. A student usng this manual is using it withoul permission ..

    -rr?.(7o >;lo q) (IHo;. )< l) (15)7.. e-

    o = 15/t N le..- SteeQ shv+:

    -rrz. l:I 'Pcv-::: Ll.. I;; ~ J..s 't = tk ( (Mi/1.) 'f ~ J1Zt >< /12 M"

    -W"L(~{JO lo: lo") (/1t.fi )< /11.) ( 15)7..

    I = 1k J't =- it"(tHl.Zol)'t -= 1362 >

  • Proprietary Material. l!:l 2009 The M cGraw. Hill Companles, Ine, Ali rights reserved. No part of this Manual may be dlsplayed, reproduced, or dlstributed in any form or by any means, wthout the prior wrtten permisson of the publisher, OT used beyond the lmited dstrlbuton to teachers and educators permltted by McOraw-Hill for their individual course preparation. A student using this manual is using t wthout permission.

    (e) Wei~~+ f'

  • "3 l? .. 17 ~10 N

    1 - (~ l -= 1 - (~) 'f e" - e/ = e"

    Is-!f'(ts)1= 39. 76fx/o3rvtwe"'-= 3Cf.7C.{>

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    Po)b ::' 1f'EI ::- 1q rr E J" Le._ &lf Le. t..

    B .. ,ll. 1 ~ J2 ~'1-

  • Proprletary Meterla!. 2{)il9 The McGraw-Hll Companles, Inc, Ali rights reserved. No part ofths Manual may be displayed, reproduced, or distributed in any fonn or by any means, without the prior wrnen permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it wrhout permssion.

    17/500 1-S

    lT'?.E I...,,..., L'l.

    Ix:;:;. z[ {17 'Sto"+ (/100 )(3.t,)l.]-:: .ft,536J'l.IO, (htr." I_y= 1.[o>6?YJ~"] = .z,y.1/' ,.,,.," -= I..,.,"l

    - y = 31' ,,.,... A t: /{O() Jn..,_ ..)

    l~}!_ll]ll.1_.1}(!21111~1 T" .=

    7(i nni '11 r- 10.16 and 10.17 A compression member of 3.6 m effective Jength is

    made by welding together two 102 X 76 X 6.4 mm steel angles as shown. Us- ing E = 200 GPa, determine thc allowable ccntric load for the member ir a factor of safety of 2.5 is required,

    Problem 1 0.16

    p ::: / 0 lt\\)

    E' -::.. ~oo >< IOtt PA Pe. r = ,. =r~,., ::: r(2. (ZDO \( 10" )( 8 10. ~lf )i 1 o-") . L2- (

  • 303 ~103. N :::;- S9Cf.'1j)o.lO'?> - 1.~

    [ - el -l] I .Y = ; A -= ;z no ........... 2. L ::' 13.it~io~ -il'.'t'f) IJ = o.53~ .,.,ot ........ ,/ j, = tl.f.Cf ........

    Fo,,,. +tt~ +~6~.e .. :f td c.oiv.,.,,,v-i J

    10.18 A single compression member of 8.2-m effectve length is obtained by connecting two C 200 x l 7. l steel channels wth lacing bars as shown. Knowing that the factor of safety is l.85, determine the allowable centric load for the member, Use E= 200 GPa and d = l 00 mm.

    ]~[ 1~-d--1

    Problem 10.18

    p~-::: //5 kfJ

    f:ov' /o:J. >< 11, >< 64 mh:t s+e() t::lvtt4 Re) .. A -: /100 1n,.."I.. .J f)l t: ,17x10" mm~ ) y = 31 , ,...,.. J -:.:. O 5(,.5 '/.I o' hJm ;. . X. -:: I ' , 111 "' :: >

    10.16 and 10.17 A compression mcmber of 3.6 m effective length is made by welding together two 102 X 76 X 6.4 mm steel angles as shown, Us- ing E = 200 GPa, determine the allowable centric load for the member if a factor of safety of 2.5 is required.

    Pe.- = F.S.

    L ::: .s- 6 M o TT"f I ... ;;, [ Glf' :::- l 'l ::;:

    :::1 i(l rnml76 mml , .. T_

    "E: l... ... ,,, .... ,, .. .,.,y .. ! ;

    Problem 10.17

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    T'Sa.tf-= 2.77 ldJ

    ~------ 2.25 111

    ,J_,,.,.., C' ... 2r-- o: h -2~- - , . r-)( ;;: Se. 't ,08 IA~ - o f="sc 7k T ... c. := l. 8 lfq~G. F&:. Z-Fj z o: 3 5" + ~o - ~ J;, "'- o T,.(., T;c. = l. 1883 Fcp

    3.5m

    10.19 Members AB and CD are 30-mm-diarneter steel rods, and members BC and AD are 22-mm-dameter steel rods. When the turnbuckle is tightened, the diagonal member AC is put in tension. Knowing that a factor of safety with respect to buckling of 2. 75 is required, determine the largest aHowable tension in AC. Use E= 200 GPa and consider only buckling in the plane of the structure.

    Problem 1 0.19

  • o

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    'P ~ -= lf. oo IOJ

    4. 00 l(tJ

    AJ Jo .. )-.. tP( ve.}.;~ r.. 'P. ?

  • Proprictary Material. 2009 The McGrawHill Companies, Inc, All rights reserved, No part of this Manual may be displayed, reproduced, or distributed in any form or by any rneans, without the prior written permission ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this manual is using it without permission,

    'P = 5. 3 7 ldJ ......

    13.43 OG '1""+"'J. r2~o11~. (\..e.)=-- ; "" ~ -e \_ ~ .... I,;:: 1~olb:?>-= ~(3(.)(;zo)~::: Z'f>

  • Bvdc(>; "5 '"" 1-z. -p P~ ....
  • Prnpretary Material. 2009 The McGraw-Hill Companles, Ine, All rights reserved. No pan of'this Manual may be displayed, reproduced, or distributed in any form or by any means, wthout the prior wrtten permission of the publisher, or used beyond the limted distribution to teachers and educators perrnittcd by Mcflraw-Hill for their individual course preparation, A student using this manual is using it without permission.

    le::- 0.5 Lec. La.:. = ob~ 12.l. m o.z ~ Le. -= 2 Leo L= -= ob~ 0~15 h. T =

    -:::: IT2(7:tll/O") (~JCS"l-Y-.1t/i) (3_ :L )( {.],y-

    (b)

    I -=- ~lo h s. = U- (.is)(.zs- V,,, 32.~s 2 '1.10~ '""' '* (a.l hv~vc..2e..r+ ..Pe..,JH.s :- AB Le.:: O*?G.s.,,. 063,.,...

    BC T f,All

    .. _J.

    -r Lcu

    _l_

    Problem 1 0.24. 10.24 A 25-mm-square aluminum strut is maintained in the position showu by a pin support ar A and by sets of rollers at B and C that prevent ro- tation of the strut in thc plane of the figure. Knowing that [,AE1 = 0.9 rn, de- termine (a) the largest values of Luc and LcD that may be used if the allowable load Pis to be as large as possible, (b) the magnitude of the correspondng al- lowable load. Consider only buckling in the plane of the figure and use E = 72 GPa.

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    L =

    l - L .... ]: ITT - ~ - i-Yfr.:" lo. i:;

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    (1C.){3.o)(4.4 >'/o3)(r)l. '=" 1fi. (:M

    (a)

    bol~ I=-yr ;r'l.E t;Jl 1a.( :n )2.. 71'-fh'11

    lf8 PL l.

    ;J b 3 L.> I:: ~ lr'l f.d b3 /~ l '1

    ('Et:Pb l:tPL~

    ! F. s ) ':' ( 8 .. ) 1. \ 2. p

    (~S.),

    10.26 Column ABC has a uniform rectangular cross section and is braced in the :xz plane at its midpoint C. (a) Determine the ratio bid for which the factor of safety . is the same with respect to buckling in the xz and yz planes. {b) Using the ratio found in part a, design the cross section of the column so that the factor of safety will be 3.0 when P = 4.4 kN, L = 1 m, and E= 200 GPa.

    Problem 10.26

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    (1) Le = (r)(;>.D) = :"" ': lo.~/ kN ...... (4) Le-:- (D.7 )(2.o) ~ Ll/""') PoJ.t ::- S3Gj2 N P...u:::: s. 3C} ktv -- (G"\ [e:-(f.o)(7.6) =- :

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    q -(~ 'B:"' ::- 1T~[2.(J-o"td o H 3C:(J61 ll w ) = 3, bz.ft, 1::-AI fo"' "'~ cvlvlM'll (::z,4}'- irrr ? ::- ~w =- llf 7 h-N. ::: ~ f F. s. Pe..- ts- f.cZ.h fo CJ f: -: :: :: 'P 'P p 7

    7..Er 112E k 'Pc..-.(1)

    TI - 'Pc(~Y

    ~ .... (d\ :' z, L 2. = 1 ( ~=)) ~ ( ~) 1- ::. R~

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    .) M .... ~. - 'P (y~~ el ": ( 3o )(010/,f; _. OOD8:J.91) = 6'!/9{ ftN.m G"..,~ ': E. + t!"!!~ 3oo~ul 6 '/ F{, X/OJ "'""' ': i5Mf1>. ::: + A s..J /,l6o Ko-6 -{, /]5X lb

    W200 X 52 e [ ~ec(~~) - 1] (o..) s- "" -::.- (IS) [ s ce (~-,/oda 14 ) t] = (IS) [ s e e (o87o

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    - (., "?. 5l)(l0 ~ !. '?. s l >< 1 o ~W\ .,,.

    (a.)

    X

    . . " . .. . .

    "2.. . . : -(. z, w ~OO>

  • Propritl11ry Materi11I. 2009 The McGrawHill Companes, Ine, Ali rights reserved. No pan ofthis Manual may be displayed, reproduced, or dstributed in any forro or by any means, wthout the prior wrnen permisson ofthe publisher, or used beyond the lirnted distribution to teachers and educators permitted by McGraw-i-lll for thelr individual course preparation. A student using this manual is using ir without perrnisslon,

    (t) M .... "..t-$ f'(e+ j.J =- ()5" 10") ( Lf >

  • Proprletary Material. 2009 The McGraw-Hil\ Cornpanles, Inc, Ali righrs reserved, No part of.this Manual may be displayed, reproduced, or . distributcd in any form or by any mcans, withoui the prior wriuen permisson of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGrawHill for ther individual course preparation. A student using ths manual is using it without pennission.

    (b) MM..,., z: 'P (e+ 'i~-' ::::;(qo)( l~4z.-+ B) = 'lq .. ~ NWi s .... ...,., ""' _E -t M~c ':" qo>(/O~ + e qq,. 8)( i6)(ro"'>) z: 1;,2.0'3 1v1-Pir A I 1011'1 ~2- 44 s = I -; 2 / .. Afl1 .....

    e =

    (o) J..._-:: e [se.:. (~/f )- 1 J ~ e[se.c oS'7b1 - 1] z: o~S"b2( e.

    P'

    --iz. Z[wox1oqll ~'?44$ tto ) := - (o n.5")2 J_

    B j 1'-~1

    l' t 750 mm e

    36-mm ,,.,.diameter

    10.32 An axial load P is applicd 10 the 36-mm-diamcter steel rod AB as shown. When P "" 90 kN, it is observcd that the horizontal dcflection of he midpoinr e is 0.8 mm. Using E = 200 GPa, determine (a) the eccentricity e of the load, (b) thc maximurn stress in tbc rod.

    Problem 10.32

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    :: = if?.8>1to''Pc;, (1'!.3. Z't)(o.016) ~ 7 . 3 s t 3 l

  • Propretary Material. 2009 The McGraw-Hill Companies, Ine, Ali rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission ofthe publisher, or used beyond the limited dstribution to teachers and educarors pcrmitted by McGraw-Hll for their individual course preparaton. A student using this manual is using it without permission.

    e- Y~ _ s D.7U-&i0 - o. 78~""4 o

    c.b) M-"-:.-'?(e+y.,,...,.'! ""(350.,,.o")(r;,.:,7x{o_.,-f_'),,[o-1).,,.. 3.978> tJ.11v1 s.. -=- A'f + Ms"'"'""'- =- 350>'to3,.. + 3:ng )( /~'!. G1L7Yto' R.

    ~ .-:;' l'f~O'i'{O-"' l8Sx/O-

    JI"' (200 'll /01)(13.8 )l. l') (b~'t)1.

    W :{SO ">t.5~ A== .... - G. t 7't~o """' ~ 7'-t~O to -.., I -=o l'S.'S Y to'~ ..... 'f ':: 1'8_ 8 )/ /0-G M -r d s..t 3 l . -c. _; l '85 )( /O ""'"" "" 185 )( io M X Le. -:::- ZL ~ (z )(s.z! = ' Lf n-.,

    e

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    M.....,..c. z: I _E.+ A

    ('b) t--'lM..,.. -=- 'P (e.+ y ... _) z: ('!7z"? ) ( e +lf) ::: J.q ' lc.N ~M. A -=- 1T (c0 z - e/) -:- .71 (ib

  • Prepretary Material, 2069 The McGraw-HiU Companes, lnc. Ali rights reserved. No part ofthis Manual may be displayed, reproduced, or distribured in any form or by any means, withoui the prior written permission of the publisher, or used beyoud the limited distribution to teachers and educators perntted by McGraw-Hill for ther individual course preparauon. A student usng this manual is using il without permsson.

    + (:i5ofi&oo)( bz-.5) _:s.

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    y ... 'tr:t e e

    L~ ... :ti = liH3J-;. (;,#\, if~ E'I _ 17t(zo-o"Kto'l)l1R,&1no-'-)

    Le~ - l')~ .;:::. 1 o 3 o 8 /LN,

    X

    10.37 An axial load P is applied at a point located on the x axis at a distante e = 12 mm from the geometrlc axis of the W250 X 58 rolled-steel column BC. Using E = 200 GPa, determine (a) the load P for which the hor- izontal deflection of the top of the column is 15 mm, (b) thc corresponding mximum stress in the column,

    (b)

    1f

    Problem 10.37

    -::- ' 'fq1.s1 (ct1 p = (O~if'1'1S7).0o3i-) :- .J.(,39 kN

    'P (e + :t-) =- ( lf l>> ~){ ts- 1 '2-) ~ /J. .. .l'J. bN lh. ? M- ..... -;: ~ ::-

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    (t.) M ..... ..,.. ~ 'P (e -t" J ..... ) ":" (poSS~) (t>102 +O /2) z: q,17.q w s"" ...... J: ~:::- p()Ot.3 1bJ1lJ ':!' + + ~ z: 523 /v1f(). A ~ _, -b 7$~D Y,./O f1Cf Y.IO

    (et)

    12.+Jo ==--- :: ' 625 20

    X

    .; :::- :to'tr.a/' 1n1rt" S.:1:- f11/xrtJ?. "'"'~ jr.. = ti,),.,,..,

    1(2"u~10 'i )(.zo,.+x1~) = qz41f-25 ., ( '1h ) ,_

    w200.,. ::1 A=- 75to 1:: '= JaohPa

    e :: t()Mlr

    10.~8 The line of action of an axial load P is parallel to the geo- metric axis of the columu AB and intersects the x axis at x == io mm. Using E = 200 GPa, determine (a) the load I' fer which thc horizontal deflection of the midpoint C of the column is 12 mm, (h) the corresponding mximum stress in the column.

    z

    y 20 mm"--. .P

    Problem 1 0.38

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    Co~ e: /k)-: 'j~+ e_ o_LJCf'1S7

    -=- T37. 'J.. -x fo?. N -::. 737.7... kW

    :: 1T ic:) ::: Cf'14 'I N ""'

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    L_, Me_ A "T - ~1.2.7Cf>'[0"6 'P (e-+ :-... j :o (_:z::o. 7'1 >< /~ )(11. + ts )(10-!t) ;<

    1:,. M = A Ss

    (t'.t,)

    4le -s- ?: L -::: q .o "".; rrt.(:

  • Propretary M11terial, 2009 The McG raw-Hlll Companles, lnc. Ali rghts reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is usng it without permission,

    s = o LoL.(Ar)

    ($. ?oS"&' >

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    14.D'i >

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    PY '=" .5J.~ ~ xo~fl)(t%tro~) -:- L-qsos,S p ..u :: .J:i... -= 4qsor6~-

    F. S. ~. s

    i., L 3m L 366 ts.is = :: - : -= V- 4-14 (o.) e= 9 /nfl1 g.s = (.9 )(7o) 02l-q2 - t~ (47.4 )2..

    F11d 1 o. :'.l'-1 .B- ~ /{,O Mfq A fr = (7.!:Lt>B y; 10-6) (160 llrt1') =- -f01t118 N ? .u -= p.; ifotf f:Z.8 p ..1l ~ .. / lf lf. j k-!l .,., F. s 2.8

    (b) e =4mm ec (4-)(7o) {J, /).;-b '( ..... -;::. (414) 2. -:-

    Fi:d IO. 2.4 fr z: J 'f b "1f'4 A

    1 e--, ;; I = ~(c. 't_ e, 'f) = $6e6bl51r11Yi '4 V' ':" ll = 47/r t!"i

    /tfo e ~ ~ .J = ~ : 10,.,,.., c1 -= c. - t -=- 7o - b ':! 64 mm A~ rr( c.7.. - e 'Z.)-:- 2t25B ml>i,_ '

    e-- .. 1 P

    10.43 A pipe having the cross section shown is u sed as a 3-m column. For the grade of steel used ur = 252 MPa and E = 200 GPa. Knowing that a factor of safety of 2.8 with respect to pennanent deformation is required, de- termine the allowahle load P when the eccentricity e is (a) 8 mm, (b) 4 mm

    (Hint: Since the factor of safety rnust be applied to the load P, not to the stress, use Fig. 10,24 to determine /'y).

    Probiem 10.43

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    (b) ec. -= (4 X7o) Y'.... (4-71f-)7. J3L = //,.f Mfa A

    P'( -::." (7525lf'f..lO-b)(135'Kl0b) ::- J,Jtoq8'?- /.J P. ::- _B__ ":"

    -...lf F:S.

    = 024q-:z (a..) e = $mm

    1 e-~ l'' I= ~(c..'t-c,"') =.!;t,e1>1>tsm,.,-1 \f' -:::- -ff ::: /t71t mt.i

    /t.o C. ""~J: -A.- -= 10ml") c1 -= c. ~ t = ?o - b ~ 64 mm

    Problem 1 0.44 10.44 Solve Prob, l0.43, assuming that the length of the colurnn is increased to 4.2 m, 10.43 A pipe havng the cross section shown is used as a 4.2 m column.

    For the grade of steel used ay= 252 MPa and E= 200 OPa. Knowing that a factor of safety of 2.8 with respect to permanent deformation is required, de- termine the allowable load P when the eccentricity e is (a) 8 mm, (b) 4 mm (Hint: Since the factor of safety must be applied to the load P, not to the stress, use Fig. 10.24 to determine Py).

    e - .... 1 l'

  • 1 87 i< 1 o-.., /V "" 78 7 k N (61 IT2 (~oc.n
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    S65 ktv

    ~ '~"' ::- 135~ F.S. ;(. l.j ~ 1.. t. 7 kp.J

    11~ (~oo ~10"' )(7. 03. ~ 1 u-e (3. 2. )'2-

    -= .s (,5 lttJ (b)

    P...tt -:l.lf 7 kN

    ctt) F;~ .. '~ '1..tf Cb) Py z: (S.n.)-Lo')-::- 59~.,.fD"'!. ""' S'l~ ~rv

    L~ 3.~ r = ':' e_J,,g .3S. 1 )( to: 3 ~ "' ~\2.}(7'-i) z: o. 7?..077 'f'4 (3S,. 1 )1

    fx_ =- 103.S MP~ ~

    ,- t. 'i -e 't 4-j ::' 7.0i'l

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    L =- L337...., .....

    O. :l.873211' "1,(ioo )(/0'1)(1.'A..C. x 10') =- lo0 )C {O"!>

    o. 2$7 ~2.

    o.::dl731 71-L EI 'P

    Jl__(A..-.1- - 1) .=- (;(3.s)'[U1..1oy10-f.{'80 -,tlO~ - '1 e e p (6)(51) roo i o L 5015'1

    X

    1. ::Z' Jo"'. 'Wt 'f .'9 :- .. 1:'/d5 N.

    w ISO )l 18 b.p .::- ''- "'1"1 Iy ~ 1. ~' "'',,.

    1().47 A 100-kN axial load Pis applied to the Wl50 x 18 rolled-steel column BC that is free at its top C and fixed at its base B. Knowing that the eccentricity of the load is e "" 6mm, determine the largest permissible length L if the allowable stress in the column is 80 MPa. Use E= 200 GPa. y

    Problem 10.47

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    l ""' ~.17 ....,

    o. '3ill4S' .7i'tEr ::: p Le. ~ 4. '34- W\ = ;tl

    o. 312.. 45rr7.(1.00 >< to'i )( 7 .{.'f jt lo"") z: lSD >-IOr....,...,_-. "" 1.f>'l~/0-< ..._'f 'l'J =- lfO.g,......, ()~:::- sox 10' 'P~ P::. :

  • J. rt 13 l l/02.(Jo)(84) _ '] ::- l;oo o o-e

    l 'l - e. '

    k. -::: [ t, (ot>SS-q )] :l z: 0 12 52 'P -- 7T 7. s r ?t;v- -e O 12 5'2 Le"' 0 12~2 li't. EI

    ?

    2 lv\'

    le~ ::- 05''jTr%(';i.o-e K(Oq)(4-31x_10D) =

    r-0-0 0-0-0

    ....

    (6S)'2. . [ { 102-60) { 04-) _ '] ( b ){ IZB ) be.o oo e: J[ fP :- 1. 1 3 2 ~1~

    Is =- 43 1 x '" '""" 't G :::- es: ""'-M.. 6JJ ~ o_ = S4MPtl e- - f [ 1 + ~ se,-../ 7r /P.) 1 u ,,,_ - Y' 1.. - \ :t 1 f[,.. J

    10.49 Axial Joads of magnitude P = 600 kN are applied parallel to the geometric axis of the W250 X 80 rolled-steel column AB and intersect the x axis at a distance e from the geometric axis. Knowng that CT a1i = 84 MPa and E = .200 GPa, determine the largest permissible length L when (a) e = 6 mm, (b) e = 12 mm.

    X

    !/ "

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    ~ =-p~- C'i' o.0'140& 3 - Le 1.

    o.oc+'4oa3. n~ El p

    O. OLJLJ083

    ( - ts {b) e=- 10...,wi. sec ~'/t) tos ( ~lt J ~ O. qt(, 11

    _f_ -= r ~(.3Z't8) ]z. = Peo- \. tt

    le-= 8.'31 ...,.

    0.47101

    (a_) e:- S- mm. s ec, ( ~J{ J-: (;~;~~.:) l (1o2~~~;~s-xJo') - 11 = z. 113q co~ (~![) ~ O. 47305 ~ /{ .,. l.0780it f :- . I ~ ( i. 7 8_0 ~ ) r = o. y 7 l ,

    ec ( TT fE ) 'r} sec il/~ -= IJ_ ( A._ - t ) ec l5

    Ar;> .... _ - 1 -=- ?

    sec. ( lf fE) 1. Y?.:

    ~ , ~ IJ -= 143. I l( to(, M""' ':::" 43. i ,. 10-~ IN"" S...v ~ 6""" ....... ':'.:' 7.S M Pcl.. ~ 7~ -io: ?~ e: ..... z: r I , + ~~ see-(t1[ )J

    X z e = J .,. 1 Zl .5 ,,.,""

    w 2.SC> )l. 80 6.p. z: :Z5S ,,..,,.,

    A: 10:00

    y "

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    Sh~pe A (10.. ...?) ~ br(...,"") ~(o-'+'l'lq) f'(hl,._) ?o--(k)'. ~ 6..."11(MPa..) V'l.. ~bO X 'ti. 7 S31D llb c:r.01 'l 1.2 131~ 0.1.'13tl SG.S 2.DO :t 2(,. '- 33'l0 133 3.30 31. ~ So~.G O, t.f o'1q 11 7. l.\ 1 ~::~.?' -:J ";( 8' o ID2 !. 4~ 2:

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    ec r- r. '' 'f_.. v-1,.MPc..) q34 en.o

    L/oCfq ~& Z'17'1 l / 0'1. s 3"U.{. 172. G -~----. --""" .......... ~--

    ShC!re A (/6

  • d::- 5~ 1'1-\W\ ~ 0.05b WI A ~ ~ e:iP ":" Z, seso 10-3 rn '- I -= ~of., :::- !f82. 7S>t /o_., ~"' V',_ -=- ~ ~ 1 -~G -x [o_, ec. .,,,iQ. 010){.o__._ozfl..l = 1 'l z 86 P... v-... 1 qc, "-.fo.,,.. g _ -.11"z2oO'KlO")+J>'l.1$"v.ft:f"8 s: ... ~ ~ ::::;;~03 [ 1 .. L~~86 sec. (f J o.~g3o"3 )] == 77i5"' fo' e:

    l.f 't 4-!6 s~ ${; .Go (O'-f "i '7 7..

    - .J.S~'\'q (?!?.: .. > 1) -;($ ~ u t~b T1~S ~

    'S7. S '15.3 37 .. 0 2d . O

    Tl~ !2 tikf e.'!>+ see.h'u...., wi.\-;~ .S' ... ~

  • ded I=f.~(o.t>,-f)" = 81.3.SS">'\., P. :::- ir"L(200>
  • F. s.= 3.oo

    e.e -= ti(:. ){7

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    e

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    e . ': 6., r \ - .l (Lf!Jr)'l.J ,. z,5"0 I , . - J.. ( 0814"}> Y-] ~ q Mo< f.S. L 2 e,. 1.~11_ . 2.

    P ...u -e --" A : C 8-0 8i..)l t;t&t}) = 4-tz_. z. kN

    6" ::: 2~Mfet Ce. =1 :z~:E -:jr;n~(~c;;:,coo)::::. 12G~7 l=.S.:: + i("87'f3)-'f (o.f7'93.)"-= 1.111.;

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    ~! V'_y = 53. :Z > 147. o~ (G\.) w ?. 00 )( '35". 9

    .SteeP:

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    F. 5. :: f + -f ( o q ;q-) - i (o ! 3 4 )'3 -: 1 # 9 J S S..M ~ f~. [1 - *(~Y]: ,;~e;, [ l - -4: (oq34Y-] ~ 7?,r hMfq, PJJ ~ fi'JJA : (7~,_t,)(5?6o)=431,.3ft-tJ.

    (b) C:. a.t1 ~ it(L/r-t :- 71f-,z Mf4.

    P u '!" 6--" A = l 71.1-~ 7 z_) ( 9-6' o) ~ 4 5 7,, 9 /L N

    ~Ti :t ( 2-oo X 1 o ) ZSt>

    E= "J.o~6fq lct-= ,111

    .(a,) 6"., '= Z~MAt

    W '20-d X lf-6~ 1 A ~ 58 -o l>1 w 7.- le /r ::- t17~4

    10.60. A W200 X 46. l rolled-steel shape Is used to form a column of 6-m effcctive length. Using allowable stress design, determine the allowable centrc load if the yield strength of the grade of steel used is (a)

  • t. (,7 M/lCf. 4Cf-. 4- k-A(

    6c Cp ~ ( r.s- )(O/? {,Cf-) : -:- 6~ A :o ( , 1o1)('i{;,b u'::" f '-h ~: = O :t. 70)2.. oc. - u ::- 1 + 6

  • A ::: (lo )(7~) + (~)(loo)(llf) = 35:1- 1r>m" . ,.. :2.~3'B~3 r .... J.A ... :" = ::. :zs.791nrtJ 35J.o

    L ';:: 15m L - JSbo - 58.185" r - ;:;:::; - t;;o(;.I re;. Ji..1-..-"'"'"" ~Jf,,y w1+t-i L/tr < ''

    f;o.l.f::: 13'1 - (o86S)( 58/8-S') : 884-"I nr

    P Ji -:- ()JI A '::' (8B41i1o")(?s.zo in')

    Ii-= t(7J.)(1o)s 55' ;:- :::- '3'72,.f" - ~n.:.olo3 -::: \07.'3 MPct -..;,..1.i (l /r) ,_ - (S8. gg) '- 'P ..11 :- 6 J,I A '::" ( 1 o 7. 3 )( I o' ) t soo .,. 1 o" ) e- 3:< I o' N

    (o..) b06 l - T~ L /r -e 6~ 6JJ-: l~'l - o. g,s (L Ir- 1 -:: 13'1 -(o. 86g)(5g_g~ > ::- 87. t::t M?c:t. P...u::: fi.JtA ~ (87.ct)t10')('300,.o-c;) = Z'.4 k /O'J IV

    P...u:: l~. lf kN _.

    A -= (.3o1(ro 1 : 300

    10.63 Bar AB is free at its end A and fixed at its base B. Determine the allowable centric load P if the aluminum alloy is (a) 6061-T6, (b) 2014-T6.

    Problem 10.63

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    p ~ = icco klV 4111

    JO 4_ '-t I MPc..

    11'3.Llo > qS.S'P>:l S~ee-J: ~

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    p a..l( ::: fu 0.87.?~. -z: 5'1. ,7 MP.q F. s. ~ l {. '1 D4A = (5'1. f(,7 K/0')(15. 3>

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    rir;; -= if2 (,_00 -io: --= 6e. = (L2lry1. (tocr .::ro)~ l {. 3.43 MP, OM( -:: ~s '::! 1.16} O.f:.~S2s-o/f.3.'f~J (2.b'P) : 're, q 1 (.. MP" fJ4 ~ 6.,.A ~ (78.'1J">

  • 'P....tt :::- to t, I k~ 6...u -= cf Se ~ (O .. Lf':l9DS)(8.1:.) -:' 3.S
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    31Zfl~ (Ldr)'1

    ? SS f..

    A -=

    h0 :- z.Uo)+5o h..:. = 5lJ 1-'L /vt..

    bo..,. 0)(~)~] )( l:t o .. ~ ~ I:{.

    = llf6 Y..ri~ M,,.,...*- ~ [ h..b.,~ -h..; b;] ::.- i"i[l7o)(bz)3- ('1>)('3-)>] =- ,:7f '!-to '"'""q.-: I ... ,., ff 7

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    Acsubte.J.. G pq Se1.'\m Ju~ be,. ;

    oc. = 1 #?. /VI P~_ A"::! :

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    (b 1 Foir P :dvSktJ use n z: 5 Al

    :w~s 7 .. ~ 3. G'

    \r cP ( I~~~~) J

    Lid ~e- ,; Cr "" (W'' (Mf>o..) (l-1Pi:

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    d ~ IZ3. 1 -~

    e = o 443 + ~-~(o oos) = 0.4'145" F .015'3 ~ o...a = (0_4q4s-)Cc:t.~) = 4_osqq MPQ.

    Cp 6J,t rJ Lid ~~ 6'al~ Cr /JJ!.p ( ,..~" .... eJ (MP4\ mwi (M PC\..) (c':-/'

    1 + Sct'I~ _ f (t+ ~e~-;)1.. _ ~!_&;. :le. 1 2 c. e

    cet V'w" ,- e A "'-f 1"' t J..e foil o__...,,; ~j

    ~ (O. 8~~ )(!:.-700) ( L lzf)'l.

    Le. I cJ ~ l.f 000 I J p ...u = G:ZOOO N

    2111 r

    p

    10.73 The glued laminated column shown is free at its top A and fixed at its base B. Using wood that has an adjusted allowable stress for compression parallel to the grain r1c = 9.2 MPa and an adjusted modulus of elasticity E= 5. 7 GPa, determine the smallest cross section that can support a centric load of 62 kN.

    Problem 1 O. 73

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    71- q A~s~""'rf ,"' ,:. no-f ve,,., .. f.-J. SSJ>c./o!. $Sl> 6b.

    r s:

    h 2. - 2. 706/(, b - .07';.. = o ~. 706/6 -t /(1-. Jo(, l& )~ -('/- ){:(.78)(-0. 070...)

    (2)(~78)

    ;ns

    b.,.J~ :::- 13'1 - O. 8'8 (4./r-):: 1'39-(0.8b8~/.S-S"f8S) z: ~q - /. ~~S8 M Fo..

    P.,..u ::- 6.pA -=- ( 13'1 - L ~~3~ )(10')(.lh~) 79.. Y (Os z: 'J..7'2 )(.fo' 4,2 - '2. 70(; IG )i( ID' b

    0.45 .111

    l.SSS95 b

    Assu>ve. L; < Gb..

    HP"l.

    10. 7 4 A 72-kN centric load must be supported by an aiumnum colwnn as shown. Using the alumnum alioy6061-T6, determine the minimum dimensin b that can be used,

    Problem 1O.7 4

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    b - 44. q .... .., ...... < 5S

    3. ;(,1-l 6 - 1.80 )(/o"3 = o 3. z ~4 + .J (3. '-'1'f r + (l-1 )(:tp >(~80 )< '_.,)

    (~ J( ~ 1 '.l)

    Pe-tt= 6r.J.f A z:

    As~vMe 6'...t1 t: .112 - f.$85(Ll1') ~ ~11-(l.SB'r")(:?.07BS lb) = (21~ - 3biq't) MPec. ::(~1:- 3t''l](/o'} ?e.

    [ :

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    t:::: q mw. vse Sine.e.

    t 1/'- A 1 V' L/V" 6o11 . P..u ... ~ \o(,"""" 't M?~ kf.) \IYiM ~M1 "'"" mm G 3q l.S-83 l.tlOl.f ;?.C, _ 78 75 . .SG GS.1' /03. \ q 3, 'J."J.C[O t. .:rol Z8,'02 78. 08 G 1.01 l~/t. 7 IZ 33 2 rr 4 1 2. zs q ll. qo 80. 65 61 _'20 IC.8~2 IS 30 3S3'1 Z.5~ SS

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    d ~ .3.3. S l'lllWI

    s.5v""'e. Lre ? Ce. (t.t)(1.tt2)(llb)< to J - r .... 9.'-&~'1$"\c/o~

    d -::' 1 ""'"1 (b) f -= I 66 >C I 03 t-J

    L__.jl/::: (4)(1.~~)'P Le 12. ~ 0 11 :_E d ~ aa. sq3,. 10-3 Pi

    ("+)(l. i:ti )(\ 08 )f '~ 2( \. l.f )\. z: -r?.(200 }t 10"')

    r -= 8. G. '16 >< /O - :!. )to\

    Ce. ..

    10.77 A centric load P must be supported by the steel barAB. Using allowable stress design, determine the smallest dimension d of the cross section that can be used when(a)P~ tOS'kN, (b)P= 166kN. Use uy=250 MPaandE= 200 GPa.

    (!J)(1_q2)'Ple'l. _ 't -

    So. 1 ~s )1 1 o-?.V>'\

    A

    s: 1 l.11H u j (o.)

    Problem 10.77

  • .....

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    -:::: /003 kN > ctoo klV ( a.ce. epf c..1.Je )

    Use VJ 'Z. O >< b 7,

    p .{ = bev- A l.~7 =:: (195.'2.0)c/0")('&5'80')(/0-') f. G 7

    W '2So ')r G7

    = 8.SL/ kN < croo k N ( Y\o+ ~c.ceptJ,/~)

    (1'12.11 )

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    Ose W 3 6 o :X I z, z.

    ::- -rr E _ n-t (~o-e o) 1.12.(Llr)a. - {L"12.}{/o'1'j)a.

    =- 6JI A :: cg g .. 3 J (l.P' Jo b

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    A.U J,j~+~,,.. seJ\o.S +a..,P tlie ....,.,vi; n.ivW'I t'Y\OW'lev;f ~ .'vte.rfr~ c.ll'ifevio ... ~. Use. W ~00'1< '16. 1, ..,..

    '> S:' J 0 M ::: 1 O. 7 '.( )( l 0 IMwi .I 1.~1 PLf!. ..... -=-(l.G.7)(5~S">lf0"!.)('f.G)z- _ ...... ;\'\,. D.877 rr-aE o.377 '.1f4(7..(!iO"'LO"')

    t"v-i:t. "'"',\--e""' L, Lf. 7 J J ~ -= 4-. 11 :

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    'I')( = 1 B. f "'IV\ -= 18. r )1 I o - ~ ,...., Le IV' -:- :< 10-3 e- 13:{. S17 -e 133. ~2 El: x: TI1.:._ -= noi. :

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    :;;- 3~5 ktJ 334 kN MPtt. [ ;so/~'.l.'i,t'?.] ( -, tkr1 Tr'1 6"e. ::::: (4/ r- ) l. e:

    Le V'

    A "' (Z.) (17&0) -=- SS~o ...,....., 't -e 8S60,,c{~...,}, I>

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    PM!( =

    Tv-'f t :::- q/11 Ir) b = 13.).. /._.,,_ A -;:- J:01b vi.,1..,_z.., ,(. I-= h. 1050)'# - (132-)'+] ~ /6,R-gR X10"' 1k1vt 't 0 =-/f = !.;7. 6~ hlft? le ': 101) -= 13~-23 > C: V S7b9

    == /Lfo~3

    T-t'j t -o:: i 2 1vi h" I = -k_ [(/-->V)'' -(n .. b)., J -=- .:L t. I 8 X r i, /v.J.-4-. V'"" lf = 6~b /Y'TYI- ~ = 1ffe!O V' !;!." b

    1f'2..E" S...u = '-'~?{l./r)2. = 5.J~ .i.3 MPcr P-..u = r:;W\W\ \:,.: b -'l.t: () ... o r = ~ (h .. 'I - b..'.

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    ~ -~ ~ Tr:J L acr~6'1J1t '-S'. A~ (~)( IU.) = ~7~0 .......... ~ 27~o '>A/O .,..... I"'-= (;(.)(Lo?)(: 10~ ) -e- ~- l-1,. 10" ...,.. .... 04 .; =- {:l) [o.~3)(10 +-(1'!60 )(IG.q)' J -:- t. 70~85' ~ 10~ ...._" -:- L.; .. \'"' :: r;,'!l =-J t. z;~:~ >< ' = JS.oj.,f- ... ,. zs. ou "to-!>,..,

    ~ ,,. ~~~' "''-~ ... . er 1tt ~ Ce r.r = o. 76 330 F.s .. -: g + ~(a.7l'33o) -;(o~7

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    (J~'17i.:, .,,. ( 47o'f )( 0.65.8) ('-Go) Oqo7f> A (o.esa )~~G"'(

    b"" go/nM b..:, = l~o-J.(~) ;:; /{;L..nfvi L /- h. .,,.. l/lL /vi h? flo -:- 70 /h.1'1. T A -=- uf u Ju 3..,, > - U b 'f J( '' :.:::: it 1 o 4- 1n trv 2 I=- ,~ [ob>l1~)LOl4-)l11'-!}] = ti.1()12>;L fl'\,Ml.f f''="n ~ f;/~

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    l"'o Po +- ~ Fi_ = tj) Po l.~ P0 + l. l ,:;!:>- ~ (o.9o)('2.1-7SI 1e10(.)

    7T 'Z. E" Tf l.(;< 00'1< (

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    11S i.5P 7"2 J,,,... \;z. ::- o7t7 ~c. =- 71 ..:l-Oooeo -:: o- ""' e:: -~ "~ g, z: (4S11-4--)(o.Gss)'7~7 (i.5?J) ~ fJJ-b.J- kJJ > 7'-fJld 1-N

    O se t ~ S- fv"I M

    Tv-y t = J}1nm A = .lf 5' '1-tr m ti; 2--:)

    t.; X J. I In In 71-9~:>. bB< bC. b..: :- /jt- /v'l M

    S&t'\c~ 'P., +h.c.b--.e~~ is

    is ~prol)'t-(139)''] = I 1 (-,(ft, J').. /nM'f

    Y' ~ ~~ 5g.,g'f- M/v Le - 4-0'D ~ 7(;4-P -:: V' - ?;"B 8 4-

    =

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    Use W 310 x 7'1

    w1-f ~ :i; .,,.. rs. 3 >. /Q~O'>

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    P=?._73 ktJ _. ( 1/i;iO io:: )(38. ~~ >< lO') = \

    e ': 4: ( 1 Sl ) : 7 {, m""' -:: 7 (, )t. {o - ' K-\ ... f{i.JO)lfO-J)(76'1C{Q-~) ::

    (Zc;.oz "IL/0-~ )(~:

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    P= ? f> ( 'iO l< lo-~) (7, )

  • Proprletary Materia!. 2009 The McGraw-Hll Companles, Inc. Ali rights reserved. No part of this Manual may be displaycd, reproduced, or distributed in any form or by any means, without the prior written permsson of the publisher, or used beyond the limited distribution 10 teachers and educators penuitted by McGraw-Hill for their individual course preparation, A student using this manual is using t without permission,

    (\,)

    ..L+~ A Ill (a,) e :;: /.1 /tl m

    1+ Gc,:Jo,,. -/(.L~-~/~-Y-- ~,;:/6;. x: o- s'f~J ~G ic e

    6JJ. ~ 5c. C f -= ( f')( O CJ'J ) '::" 4 J'tJM f'q

    :~ "' b J :- (/Cfo) u.)-~) ::: J. 3.7 s ,,.. IV)2- ~ K'c;;tcl~ ce 7 -::- (UJ )'L I ~ '"6,,Z.lbl Jj -=- T7YI

    X

    E~ 8"3 ~Pq k'c.E " 0.300

    ~"""' l'v""b el': 6c. "; /j M Pq e= o. 8

    10.91 A sawn Jumber column of 125 X 190-mm cross section has an effective length of 2.5 m. The grade of wood used has an adjusted allowable stress for compression parallel to the grain a-e= 8 MPa anda modulus of elas- ticity E = 8.3 GPa. Using the allowable-stress method, determine the largest eccentric load P that can be applied when (a) e = 12 mm, (b) e = 24 mm.

    Problem 10.91

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    ' (alf} l ; J-) B = . + ., __

    U.J.{.t'r, )(tf.-1q3) (71 ltlrS- f.f(} b) ( 1) 1

    + BP =

    - o, E:qq / 0-cr/Oc. e Se Cr ':' C8' ico- 5cq 1) ==- q.. 7q 3 M f t:t

    J / lf4-3' X I b r.. ,,_/f- A~ bd -=-(/qo)t).).5) = ')315 ""' . .,.;i. Ix= {-OZ-5)(1'16)3 ::- 6: = Kc.e t:-: ~ \

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    ..E... + )~\ ~ (~\ .E+~ . + Pe., ?(J!:+~T~) A = k? ~ :::- A .s; Ss kP : p ~ 6'AA' k' K = + tfo v t.o:": 1. O&>'!;Sx/oJ;. - 2. ISC(O ~10-' 11. 3x {D~G. + o ':"' '1-1 ? -e: 3G. '-\ 1 }t '( t o & 35": 3 )( 10 P = 3S. 3 kN ~ ..--. - N l".P303S.,.. to?

    =- 'l ( ~ )'I t ott (IG.i! .. 71l):i.

    -::- ( C). & 7 7 )( t;.q. !S) "::' /.(;.7

    s, -::. e; ::- e.:t ;:-

    's \3o x Fs: A'::' !8'70 ...,. /-:::. IPlfo >110-'" """2: c., ~ 'J- =- lt;. 3 w...., t; ff;..3 ){ 1 o' ""'

    7'1.a -1' {0$. WIW.! ::- 7'1-8 >' /Or.. "'7 z, o 'fo ,..""' ,., 'to K{O-s 'r"'1

    SJ.:30 X 15

    .L. + L .. v-A"Si' '"" \" :

    .J0.93 A steel compression member of2. 75-m effective length supports an eccentric load as shown. Usng the allowable-stress methd and assuming e = 40 mm, determine the maximum allowable load P. Use or zx: 250 MPa and E= 200 GPa .

    Le ";:: J.~ 75 Y' ... :., 16.31

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    3G>-'41 s )( {o"- 1~no:g > + ~ + Pe)( _ 'P ( ..L + . + Sk.) -;' kP A A si( v- A s; ~ K? ~ 6,.f p ~ s:. K l< j - + (;O-.C.(O_g + o = I, ~

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    6CE '::' !(t;;iE L z, "" ~Ecll.. w'1ev-e Ke ~ o.scc (Lid }i. Oci' J~M L -:: "-" (o. \So>f-3oX.111 oo) -:= S. '#8 ,., ~~ 3.S"1'8

    W}- y(f~X) +'ji_ ~ e;tl- t x z: J ( 1 - e.y) =-( o. sBs:n.) 1 - { 0(~(o. ~~~~ -: o. qgaso

    l ~ :J l - o. 3SS.:t2.. 60~ = Se (o.43~5"\J) ~(g_3)to. 43350) = 3 .. S'r8 HPet.

    MPa..

    C. '= _h e- O. llO '"'' ~ gs-x 10~ -i (8Sxlo!.Xo.o~s Xo.1~0) ::- 3_,411b ,.,0e. ?~ 4~.:.? )l l

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    , " k'co:: E' L :z z: k'ce E &2. ..,_,;'~e./'(" K:...: ~ o. 300 (L/cl)z. De..: L = djt< (o:o, )(o. o~s-)Co.1 :l) = 3. 911 ~G .,,.r.:f pq_ :.(07. 3; )< 10-t.. ""?a.

    \ OS'i 103 4-3. l y: 10-~ +

    ::- 3.~49G

    C = ~ :: 0. IQO t't'! e= 2s wi wi = o. oz5 WI

    biJ. 'i3. :() lo-s 'l E-= 11100 MP4 = Yl'I ....L d b~ ~ 1~ (o.rsc )(o. 2t.:>)'.!. I~

    W1 'I :::' Zo7. 3tS. Y /0-4

    J '=' ISO;.,,....., -: O. \80 -o A -= I)I .,..

    10.95 A rectangular column is made of a grade of sawn wood that has an adjusted allowable stress for compresson parallel to the grain Fe= 8.3 MPa and a modulus of elasticity E= 11.1 GPa. Using the allowable-stress method, determine the largest allowable effective length L that Cal) used.

    z P"" 105 kN

    10.96 Solve Prob. l0.95, assuming that P = 105 kN. Problern 10.96

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    L-= r.01r; M

    o::: t s. ~3-S ,(to-" ?41 A~suw..e_ ~ ~ L = f 3.SI 'I< !0"' V' l OJP. e- ., L -::- 8 1. ".!.J J(J_o)({Or.)(25')(-{0-?._2J lf'l>-,.( to:?> (30h.8 ~1"r )(1 'fS 11 o1/o.,_,..,,,W\ ... I:: 1fc'I :- so"-. '8"''.,. """"' ~ '

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    >SS

    43.~36 MP~ L 7- -;:. SS. A ssu""e..

    5: - ! l 1 - IP,;~,;-J ~ MI,: v ~.

    _e_ + p~-~ = l A

  • Proprlctary Material, 2-009 The McGraw-HiU Companles, Inc, Al! rights reserved. No part of this Manual may be displayed, reproduced, or distrlbuted in any form or by any means, without the prior wruen permission ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGrnw-Hill for their individual course preparation, A student tising this manual is using it wlthout perrnission.

    r, ~3 6 -'' )( /(.) n-i

    .E + Pe,. -+ p~ - 6~; P_e.,, _ ()~ - 'P Pe, A S.: s>' - A - s~"' .::iy i1-[ S,.;.1 ? P~., s.1 [ . .J.... ~l e)I ::' - - . r--"J 7 A ~1' p A s, J

    e b 781. ';($'Y{D- ? ~ 170 )l. /03 N

    -/'.:[ 110,03 -to' o] (a..) !q_=- o e)( =- 31.';lS >'[O'~ -= J l. ;.;,9 )(/;, ,,..- e)(= 1 J.~ q :-.-. r.- ....

    le =- ? L -:: (';nfo.ss)"" 1.10 ""'

    A-=US'>

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    e = 7. 78 l'l'llYI .... I - 0.41')..0I 75, SS(. e =

    J7C> x[og } - (375c..:> >-le. )(l!O,D3 l< fo"

    110. o~ t--tPC\

    .I, ::- I~ (S>

  • ....

    Proprietary Material. C 2009 The McCniw-Hill Companles, Ine, All righis reserved, No part of this Manual may be dsplayed, reproduced, or distrlbuted in any form or by any means, without the prior written permsskm ofthe publisher, or used beyond the lmited dstributon to teachers and educators permitted by McGrawHill for their individual course preparation. A student using this manual is usng it Without permission.

    ((t) 'P ..u :- :no ~N e 'l:' 4'!2.Xfo .. b I bS122Xtob - J.i.DOO ]::::()001~39' tn ( oo> 'f..z.J.oooo) 3/,ao~10_,

    (b ) P ...tt :- llf-o k.tI

    < \'33 .t:Z. B

    ~="htb ...... b/.)~ ..lf'IZ.1

  • Proprletary Material. O 2009 The MtGraw-Hill Companles. Ine, Ali rights reserved, No part ofthis Manual may be displaycd, reproduced, or dlstributed in any form or by any means, without the prior wrtten permission ofthc publsher, or used beyond tite lmited dstrlbutton 10 teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using t wilhout permlssion,

    e= 4t mM (b) p ._u ~ 7 /l. N

    e -:- 4)(101Joo) 3~ooxio-6

    {a) 'P...u -=- flb l.lJ

    Se..,... -: O.S'771i"E .:; f l 1/f-6 Ml'a. 6M..l-= 6' ... = 4?5q Mf->4 . (Le../r)'i ''7 P.u 'P...u ec P~ec 'P..,,p . . P. ) -\ ':" 6'..u "' 6'-.u- - e. -=- L ( 6..,1 - _jJ A .l. r A c.'P.Jt A

    '---100 mm- ... 1

    r- lOOrnm J ~~

    t

    ::> ;; . z z

    E -=- 200 bflf ~...,, .. i+it:.V\ ~: lf.11./[;. ::- IH.2Z r

    10.102 Solve Prob. 10.101, assuming that the effective length of the column is increased to 5.4 m and that (a) P = 110 kN, (b) P = 70 kN.

    10.101 A column of 4.2-m effective Jcngth consists of a section of steel tubing having the cross section shown. Using the allowable-stress method, deter- mine the maximum allowable eccentricity e if (a) P = 220 kN, (b) P = 140 kN. Use O-y = 250 MPa and E= 200 GPa. -

    5.4m J p

    10 llllll l

    Problem 10.102

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    U.sE_' b-=O.No~ ':::' '"~"" .... - p-= ~, ktV

    q,(w.) d(.-...) S'!e/6;. Cp P...u (k) .OCJo o.oCJo o.S.t\ft.\3 0.4'f3G 7 18.87 c>.! l'io o. 1'-1 o IL':i'-1480 o. 7G081 so.~ c:tt:to o .. t'1'o ~ .. 2'f Zl O.'Bt.78 79.-, o.~io 0.1~0 s. 2"t':l 7 0.&87~ too. 7

    ?JJ + feaec -:: h I)( ~"::' 80Jvv- vJu~s. of b. See ~. b,JvW".

    E= s..e sP~ ~ ~ o,goo

    s..;. .... "' Pv ...... be.,-! ()e:..""' Z.I MPo.. t:.. :::- ."8

    le.= ~ .. ~....,

    4lkN

    10.103 A sawn lumber column of rectangular cross section has a 2.2-m effective length and supports a 4 l kN load as shown. The sizes available for use have b equal to 90 mm, 140 mm, 190 mm and 240 mm. The grade ofwood has an adjusted allowable stress for compression parallel to the grain Uc = 8.1 MPa and E= 8.3 GPa. Use the allowable-stress method to determine the lightest section that can be used.

    Probfem 10.103

  • Use b-=- O. Mo ""- ==- /i.fOmlVI ......

    p-e 41 ktJ

    t

    b( ..... ) U(iot1) \Sc.;/6';. Cp 'P-11 (klJ) O.pcfo 0.0't() 0.51'Ff3 .t.f't~f.7 1..7 # 1. O. {'f D O. l'to l .14'l8o 0.7~08\ 1{ # '+ ~ O~ICf o o .. ieio ~ e, =o.8 Le."' :

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    J..d X: .L c;f S.9v; "~ X ::- ,._, ~ zns ~ - ' IJ)

    =

    d-=- -1.... :;: 70_0 X fO-?. ...., X '/'=- J .., 11.so~ tci3 n-. y

    (3~)(8S)

  • ~:

    Proprietary Miiterial. 2009 The McGraw-Hill Companles, Inc. Ali rights reserved, No part ofthis Manual may be dlsplayed, reproduced, or distributed in any form or by any means, wthout the prior wrirten permisson ofthe publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparaton. A student using this manual is using it without permssion.

    7'1.S>S~ L - J. S- r - ;{O, IK/O-'> d ,... ~

    G't 'PL ~ x " + (16)(~a.. -:::-- (L /r)" .r- 7T (! 'f - Jr. J

  • S,;j., ntl .h,,.. X_, X.::- t> o;i,~b'J ....-f d; t-= 4Lf./ lnM... lc/V'l(..,. (.J. )(CJD)/lfl.f.J 70~7 >SS

    le+ X-=} ..

    6_ _ 37zooo _ l1ioctJr:;J _:S71000,::>fL_ . ..,g ?.Ji. J,c.-{L~/r~)1- l~l~i. a~)(1cc)' -OO.;, .2-

    13,ooa o + (l -::l..)(1>c ooo )(5)( ~el) (57, J)( o. o3Si.7 eF) (S1:, c;P) (lbS-)

    A$~VW>.e '(< :: Y' .... .- .... .J i.e. J < !;b fr. rn lelr .... .- ... ::- f. Le/J

    ? -e- /?ro ~ ./ J V'( """ fli...

    A ~ s-0 e1 L -= -k_ cJ iC:"" "' l 6S )trf v..o>i,. I>< -= 1-i: (~) J?.

    10.108 Solvc Prob. I0.107, assuming that e= S mm. 10.107 A cornpression rnember of rectangular cross section has an

    effective length of 0.9 m and is made of the aluminurn alloy 2014-T6 for which the allowable stress in bcnding is 165 MPa. Using the interaction method, determine the smallcst dimension d of the cross section tbat can be used when e= 10 mm.

    +

    IJ

    P = 130kN

    Problem 10.108

    s .. J111'nj .fe,,,.. X_, x =-0:0.)107) tJ-, ~ ~ 47,4-0 ~ s e-;; L/ric : ({i H 'Jco )/ ( tf1 't6) z: s.s 7 >SS J -=- 'ns 1-

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    -~ ;:.- 40)J.._k/O') ( 3. '3'?~'3>t 10' )3 J

    10.109 A compresson member made of stcel has a 720-mm effective length and must support the 198-kN load P as shown, For the material used O"y= 250 MPa and E= 200 GPa. Using thc interaction mcthod with an allowablc bending stress equal to 150 MPa. determine the smallest dimension d of the cross sectin that can be used.

    Problem 10.109

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    d2. - ' 7Sl,1#/o~" = o J = f f b . .756 '11/' +j(C.AJSt;. ~101)1' ~(l)(:?.30'1 ... /6!>) 1 d;:: 87. ~> !fo )..J;~,~ ::- !So MP~

    o.877 TI1 E ( LeJv.. )t

    o.t.17 If-i.(':. ;l q ) l.

    As~v....-.e d '> t#O ""'"\ = l./O">'{s ...,,. T4e"" I,..;.., -:- lit, v"-=/f ~/~.~;~:~~~~ =ll.S'11)(1o"Wl L;.= l'io.~q > 1:.3.Z1..

    J...._ 1 L l v-av o\1\5 1 vi A : 'io >'-Jol J 1,. = ;i ( qo,, t"' js el :: 1,y ': ll ( qo /O-!;) J ~ '=

    10.11 O Sol ve Prob. 10.109, assumi ng that the effective length is 1.62 m and that the magnitude P of the eccentric load is 128 kN.

    10.109 A compression member rnade of steel has a 720-mm effective length and must support the 198-kN load P as shown. For the material used "r = 250 MPa and E = 200 GPa. Using the interaction method with an al- lowable bending stress equal to 150 MPa, determine the smallest dimensin d of the cross section that can be used .

    Problem 10.11 O

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    g5_0 MPc.. < c17. 3 MP~ Use. t = I~ ""'W\ ..-ti

    .e_ + Pee. A J:

    : X VV\Pvfe. 'f'e~ vil/'2.) ~~""-= ( IOL#. 2. )(~007) ;:: ".(o SS"" W\W\ 1. t loo.g

    T""' t =11...,wi. Le. - :I""" ) (1'1'11\-t ") 3 37 7'2.r;; 0.53Cf -n.i'I b 34 13'lS o. ctf. 1 zc.is '1 '3 1 "l.oc.7 l. zs s 15'.3 I '~ '2 $? ~ 5& tJ f. 52. 8 "2 '-#. 4 / IS' 25" '?,o,3 l. 7'-i :2 3_5q -- ---

    80-mmouter diameter 2.2m

    r. - t o I : ~ ( t,," - r.:' )

    t o :- a ct., -:: " ..... ..,. ) 't':. :- A" n(v\,i.- rf'~'-)>

    e" .....

    10.I A steel tube of 80-mm outer diameter is to carry a 93-kN load P with an eccentricity of20 mm. The tubes available for use are made with wall thicknesses in incrernents of 3 mm from 6 mm to 15 mm. Using the allowable-stress rnethod, determine the lightest tube that can be used. Assurne E= 200 GPa and av = 250 MPa.

    Problem 10.111

  • Use t =- IS""'"". :: 0.C}58 < 1

    Le t:1 T"'J t-=15W\'V1. ~::: -3.1C 5"e.::'2?..6.((SMPa.. 1&5"K./o:. + C1~5 ... 1o"")(t5'cfv!x4o)A.to~3)

    (3oc:.3"1o"')(q"l/to1e{o6 (l.1c.>L.lx[o~')(IS'o1t toG)

    e::: ;(Y3.0S" MP~

    p + 'P'1-c. A Q,,JcJ,.,L I 0.J,(1 !:,.,.JI;~

    S...al~;G :' q7_ 33 MPG\. 1'-5'~ fo":l (1c,s-occ})(l5'l"lo-'!>)(4o>'h46< 1v-c:.) (1.s-0 "'10") o.cs t + 0, Lf ?J2. = 1. OC( 3 > 1 (nof JI ON,.l~...t )

    Pee. -:: 1 r,,5 '(ro-:. + (!b.S:i-10~ )(IS"xc>3)(4o ... ,L>-~) I 6:.,1, b~,,J.,.~ ('oo7Y 10-1.)(100. ~"" '') (1. 1.$5'-:lo -e)( l 5D"' toe;)

    = o. '8'20 + o.Sii./ ;::= L ?,~'t >: 1 iot o.l.loweJ.) -"--p---+ A f::~,c._J,;c

    Lf. 7r ~ ~ 4-.11 1

    Lt. - :i"'ID~

    t V', A .I V' (!:!:~ ......_(!:.M) (..-. .. /) (ID',...,..,) ( ....... )

    3 37 7'2G 0.53t_ r/)

    1 p e , r: , 20 mrn ....... +-

    !;l

    10.112 Solve Prob. 10.111, using the interaction method with P = 165 kN, e= 15 mm, and an allowable stress in bending of 150 MPa. 10.111 A steel tube of 80-mm outer diameter is to carry a 93-kN load P with an eccentricity of20 mm. The tubes available for use are made with wall thicknesses in increments of 3 mm from 6 mm to 15 mm. Using the allowable-stress method, determine the lightest tube that can be used. Asswne E= 200 GPa and qy = 250 MPa.

    Problem 10.112

  • Use W ;,t,o X 2/{,

    W .>be~ rZ..2- L ~ ~ ll'f.3.

    .E. + Pe.. A S><

    A::: ~roo fnrn .... ,, s,1( =:LdOOO lnM~_, fJ z: b> ir.Ir} lLfr_ -:: ') o (J'l F..S. =- t. 9 lf-3 O..J,f = 71 lvl fct

    Ce..

    A::- :i.76ot>1n,,,i-:, Sx ::- g6'OO(\ fv\tH',, ~ z: I o/ n; f>\ J_/~ '::- -tb

    Assu~:ltd +J,J o...._ tS ~fN'Kt'pvJel( p~~ov+/o.,..J fo *J H.e '('e1uif"lfef c:t~~ Is

    A "="(,qoo) ~o ::::: r:..ln.; ..."'-:.. 7b.3

    r-F cvPI) IV) n ,-; Jo ... j C\.vi.J .RoJ;nj is Ce.V>+~; c...) F~ s. = 1.Cj.:? o < Pe"' -e- ir'-E.:f) I .._ l.t:i~ 'PL i __ (Lq~')(4Yoooo){1z.ovl (, 4 1 - - " ..- 2 = 24- 2- Y.ro ;...-. 1.e:t:.t 1 .. qz Lz. J ..i T1 E T~(2o-oooa)

    ...e_ + Pee z: P + Pe < s:.d A r, A S)f ~ T"'j W3i'.-Jic/ A::-/zt}o0..'-'"'2-..> ..:i;.=Sob)U;n,.,lf..> ('J.,,.~7..b"'"' Sx:::f(,?oDtJ-Oht-.3

    L 77-cC T JIS

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    Use W 3~ Xlo t

    wioo )(11 25""+)(iflb W'?6o )( w l. So' b Ho6

    W3boXlol s +1~ );~~1e"S-f :s h4pe "'' + ~ (!> )l,,J..-t.ro"'J...,...if

    +. r.,-, t +e--r ....... _q-;z PL'l. " Tt1..EA"J ...

    o.f \V\.ecivJ,...f t. 'i'A PL ,_ =- ir< E Ij

    ( 1) .J..- Pee < A SJ,c.. I.l( S._~. (/) becoWI e S !L ~n) (lf9oco")t1i.a0l ::- 112( 2

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    Ose. W 200 )t. 59 -1111

    ;u1G "'10~ (1..'1f; ~Lo~KtZS'>~I '>

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    6~ z: (111,qe;;2x10Q)(So.~)>'-(D-(, IS5> 0...tC (rwf Jf ~,p< 1 Q '!. V>o\VI- ~ f'1 -= Lf9. Z .......

    o..1.1=-(l"f.'1'1~~>

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    L 8') fo s;.,. '1' IJ Pe.~-: 6'36 N _.

    + .:)'2Ms -= o: -PS + T = o p=- r K.f 312.$" g:> 'S '= e- s 1 i"' g> O'$ 7 s: s.1'" tf

    L -= o5m GJ" t'77x10"1) (io?ti.-ix ,;'; K -= -L F -"--'-'.......___._ = JJ 13> ~ $ N /I> /rae/.

    0'5

    375 mm T= tL g> -::: G.j G-,,,. 77 6Pq ;r~ fc't~ f(~)'f \(b}' ='2-o~S-1,.....,/+.

    A

    10.117 The steel rod BC is attached to the rigid bar AB and to the fixed support at C. Knowing that G = 77 GPa, determine the critical load P0, of the system when d = 12 mm.

    Problem 10.117

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    B-o Le-t P'

    le..+ (;) b4! +k~ a_11j fe 1

    -- --- ~J1/1/ '< G .... ----------------

    p

    10.1 JB The rigid bar AD is attached to two springs of constant k and is in equilibrium in the position shown, Knowing that the equal and opposite loads p and P' remain vertical, determine the magnitude P" of the critical load for the system, Each spring can act in either tension or compression.

    Problem 1 0.118

  • F:s.

    I..,;., ::- o.:Z2'1 >

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    (L~). ........ ::' O &p h'I 1f7 (12.'1.10'1) {;z~t;z. ~ ro-t-)

    --T t,\f

    - .. J.

    t l.11c .J_

    o - r...., -

    Problem 10.121 10.121 A 25-mm-squarc aluminum strut is maintained in the position shown hy a pin support at A and by sets of rollers al B and C that prevent rotation of the snut in the plano of the figure. Knowing that L,111 = 0.9 m, Lnc = 1.2 m, and Lw = 0,3 m, determine thc allowable load P using a factor of safety with respect to buckling uf 3.2. Consider only buckling in the plano or the figure and use E "' 72 GPa.

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    F.. 5. -= 'i..'i.. 7 3. /'3 F. s. '='

    Lec. t. :: ~ ::- ~_,cr-

    Mew-.be"' 8 C.:

    1.2 m L~ ;Gr~---- i.e lll __

    22-mm diameter

    Joinl J3 . l=v-o"" .f'o..-c.e .+ir.,:;. """:rR e J , _fu__ :: ~ s.~ :' 5 ;'125"' s;n~o .Si" 135"~

    ~8-::- 3. 101q ktV C::i) .... , ) ~n. (c.o~p) i=-ec. ~ '2..S1.5:< k~

    10.112 Knowing that J> = 5.2 kN, determine the factor of safety for the structure shown. Use E = 200 GPa and consider only buckli ng in the plan e of the structure. Problem 1 0.122

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    A fJ4 " ~(3)'l.. 7 06. '1 h..-~ 1. =- ":' (b) (;)...,"le. ';:: .E + Me. '3 s Cr-0 o (-- J !:)"' )( (o"l)(! 5) l/tf-.1'2.I ~Ar ": +----- -: A r 70b '1 3Cf7bJ =- /14-t z: f.1 p,,

    L -: 1 ;u., 12

    rEI -:o iT:[U'f1oc)(3'17b!'f...10) = (1 z.)'Z.

    10.123 An axial load Pis applied to the 30-mm-diamcter stecl rod AB as shown. For P = 35 kN ami e = 1.5 mm, determine (a) the dcflection al the midpoint C of the rod, (b) the maxmum stress in the rod. Use E = 200 GPa.

    3()-mm diameter 1' '[

    B

    Problem 10.123

    (CL} 'f..,,~ ~ e [ SeL (i_/f. ) - 1] ==- /5" [z .. zG11 J ':: _3,4 """~

    1 1 2.D 5- (: "" "i_c::' "' T -c I W) . I -= ~e"'= {f(;s-)'f i::: 3c.nb1 ~wi+ Le. -::

  • S"~ -: :l.J'- - lS-8~(Llr-)"" :>-1:>.. - U!:JlS)(Lfo,.).t) :::: /t.f.,; /vlPq Pi:..tt ~ 6...uA-:- (llf~T?)(32.76) :::: 4~.f,g kA/

    : 55

    ~-100 mm--! b.,. :: / 0-0 "1-M b; "' b,, - :?t .. fz #< "1 A ~ (100)1 - (9-:i..)2 z: 3i7fo (/vi.~ I -= ! [U()o) .. - (Jl-:i.) .. ) -e- 4r- ,.r b.S"2- tn.w.')_ 'f ~ l[ =- J,7.3 ,,., IY1 le-:: t~S:- frl

    D l t = 9mm 100mm J 10.125 A compression member has the cross section shown and an ef-

    fective length of 1.5 m. Knowing that the aluminum alloy used s 2014-T6, determine the allowable centric load.

    Problem 1 0.125

    (b) Le.-:::' ~ .5 M. ~::: "1;:~10-.!i s: \4Cf.77 113.tt Ti?. rr':l..(:looic/D'i) =- O;t, ::::(L1./ir)'.-= (l'1'1.7'7)').. BS.o MP~

    (5..._u =~-~Y'-= l.~7 [(o.~77)(~.o)] ::- LfG.~I M?q

    111.0 MPc

    11?:>. 4- (O..')

    +1 ---X A= ras X 10'.J mm2 1'1 t, = 20.0 X JO mm" : : I,1 = [42.0 X IO mm" 11 '

    r ... ,,_.~ ... :' .. ::' .. "--~-~--,~ .. =._. -- _1

    10.124 A column is made from half of a W360 x 216 rolled-steel shape, with the geometric properties as shown, Using allowable stress design, determine the allowable centric load ifthe effective length of the column is (a) 4.0 m, (b) 6.5 m. Use uy"' 345 MPa and E= 200 GPa

    Problem 10.124

  • J0.127 A 32-kN vertical load P is applied at the midpoint of one edge of the square cross section of the aluminum compression member AB that is free at its top A and fixed at its base B. Knowing that the alloy used is 6061-T6 and using the allowable-stress method, determine the smallest allowable dimension d.

    Problem 1 0.127

    T"y Wt50X!fo. A"::' /02()6/rtMi.. IJ ::- 43.y;,g/Vl,.,

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    ~+ A

    e .,.,t ,.,.))ot.1~.+ ) :' bll--1"1.fb }nr-.~

    le /'('f o- 11 I (' - O'?.?tJIO. -c.

    e:! i'e(;< .. { I s 4-]f )( S: ;y b

  • Below the dashed line we have: critical stress> yield strength

    Ptta b.R.11.M ov trvr Radius Critical Allowable of rod stress load

    m MPa kN .006 71. l 2.87 .008 126.3 9.07 .010 197. 4 22.15 - - - - - .012 284.2 39.58 .014 386.9 53.88 .016 505.3 70.37 .018 639.6 89.06 .020 789.6 109. 96 .022 955.4 133.05 .024 1137. o 158.34

    o F . q: A NI) r:f:.r cr

    C~l"TICltL STRESS 112.E.

    ~r ; (Le/r ),_ Le V lf~l/,ctL SMAL.Lfi!

    o: A p(Alt = Fs

    f.f/T. R RAl>IUS AAD) E FF EC t YE L f NGTH Le ANO r ncro OF S!fFTY FS CO MlllTE ~A l>lll' op

  • LE-r C1cr EQVl\L Sf1ALLEIZ S~tSS COMPUIE ALLOWA"Bl .. f' (t=NTRIC LOAD

    po.11 = Vc.r A f5 PRD6RAM OIJTPUT

    b Critcal Critical Allowable stress stress load x axis y axis

    tn Ir) M~ MPPI. .fN ::io ;.3 (,S- ~~. 4-JJ 8/#J ;z.3 (:J3~ ~71 600 1). Jt/-S 2- fe .k>;; (:,S t.g-, o 'f-9 I ~ qg-c 2-CJ .!"> 3 bS" r'1-77b Z3-" _082. 31. ..f;>.. 3 b.S~ 1J-7tJ 2.St-7{) 35 s ;).. 3 b~ ~1('o1o ).1,gs 38 !.J 3 bs t ox- b"Jb 3o~Cf'{;,

    COMPUTE CRITICl\L. $1'R.ESS~.S 712 E

    (OMPV1E' ~AIJJOS OF 4'f~ATIDtJ A==- 38 b T -= .L b 38 3 X {Z 1 :::.flb3 ~ 6

    ENTf R E; LENC-ll l AND FACfO~ Of SAFert l=S FO(Z b ==7..f11fYJ TO 3f3mm WlH 31nm /N(~EMIFPTS

    SOLUTION

    1 O.C2 An aluminum bar is fixed at end A and supportcd at end B so that it is free to rotate about a horizontal axis through the pin. Rotation about a vertical axis at end _B is preventcd by the brackets, Knowing that E = 70 GPa, use Buler's formula with a factor of safety of 2.5 to determine the allow- able centric load P for vales of b from 20 mm to 38 mm, using 3-mm incrernents.

    ~ ~ >1 . l c= o.s L

    1~ 1-< ~I Le..::: 0.7 L /?Ullt'LJN(;, 11 if t PUHJC

    PROBLEM 10.C2

  • 4.00 4.25 4.50 4.75 5.00 5.25 5.50 s. 75 6.00 6.25- 6. 50 6.75 7.00 7.25 7.50 7.75 8.00

    kg m

    mass

    JOIN 13: F - lS''W - ge._ - t-

    4--- '

    f. ~+w1\ -r : JSW AB ~ 1)- h t7: ; .i VI r

    Ytr LDS i.s-

    h

    T~ 3 -- T~ - h T; ~

    JOfNT O: - _,,.

    7854.88 8345.80 8836.74 9327.66 9818.59

    10309.52 10800.45 11291.38 11782.31 12273.24 12764 .17 13255.10 13255.10 13255.10 13255 .10 13255.10 13255.10

    Weight Weight critical cr tical stress stress AB BC kN kN

    455.11 269.7 455.11 286.6 455 .11 303.4 455 .11 320.3 455 .11 337.1 455 .11 354.0 455.11 370.8 455.11 387.7 455.11 404.5 455.11 421. 4 455.11 438.3 455.11 455.1 455 .11 472.0 455 .11 488.8 455 .11 505.7 455 .11 522.5 455 .11 539.4

    h

    PRobRAM avrPuT

    (OMPVTE MAss Wo.11 9.81

    COMPUTE MOMNT OF 11..JR..nll 1 == Jr. ( " + - o.as-4) 1

    Fo~ ., :::- 4- JO g us1f'J(, G.1s 11Jc~eHEtJTS COMPUTE ALLa WAflLf; LOAD~ f OR.. ti? E M85lS

    u2E! ti~tI (F ) - (F ) = A13 a - 3.>(4)'2.; 'fJc (r ~.5'('1)-i l)r;TERMtNE. ALLoWABLE W

    (W ~u ) -:: Z ( FA FJ )(r j (w411 )'2. = ft {F at)~r Wo.ll t t)UAL$ ~MALL~R. VALUE

    SOLUTION

    1 O.C3 The pin-ended members AB and BC consist of sections of aluminum pipe of 120-mm outer diameter and 1 O-mm wall thickness, Knowing that a factor of safety of 3.5 is required, determine the mass m of the largest block that can be supported by the cable arrangement shown for values of h from 4 m to 8 m, using 0.25-m increments, Use E= 70 GPa and consider only buckling in the plane of the structure. 'f

    h

    j

    r -- rA

    f-

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    ' ! ' l too t=x o s: ; q,~71.- t ; ; i f bO Zot-z_ 3.2 /:,9D f 1 ' l 2-0 o ! 2,re,sr !J.. f. t" 60 i J.

    1 :J_lf-D ~ . 3 3'2.'1 I 5c'!J..; ' ! 1 ~ 4.473 b s. Ji.] ' !

    '

    maximum stress Mf'q

    maximum deflection mm

    Load

    PROGRAM OVTPUT

    r;NTiiR LEN(.Tt-\ l) ECCEN'(R.K\1''( e ENTER r~oPEllTtS A,1~, r'l"f (OMPUt CRlitCAL. LOAD

    11' E I "'- p = ~ cr L '2. FOR p = ICHJ kN TO 3o-ottJ /(\/ /NCREM,N1~ of 2.oll{

    (oMPL'l'E f.10R1?.atJTAL t>EFt..ECT1'1N AT c. ~ e = e ( 5EC (-1{_ /f) - J. 0 )

    SOLUTION

    1 O.C4 An axial load P is applied at a point located on the x axis at a distance e= 12 mm from the geometric axis of the W200 X 59 rolled-stcel column AB. Using E = 200 GPa, write a computer prograrn and use it to calculate for values of P from 100 to 300 kN, using 20-kN increments, (a) the horizontal deflection at the midpoint e, (b) the maximum stress in the column.

    PROBLEM to.C4

  • CONTINUED

    .p = O:\, A LU 1 ""

    !F L/r;~ z C o.S77nz E l. (,7 (L/r'(J~

    /r L/r'I L. e 1T-:; E.

    " (L/ 'ri. )'Z [

  • Allowable centroid load: P = 638.0 kN ---------~--------------------~---------

    6.50 m 13800.0 rnm**2 43.4 mm 345.0 MPa 200 GPa

    Effective Length A ry

    Yield strength E =

    Problem 10.124 b

    1532 k.N Allowable centroid load: P =

    4.00 m 13.8 rnm**2 43.4 mm 345.0 MPa 200 GPa

    Effective Length A ry

    Yield strength E

    Problem 10.124 a Allowable centroid load~ P = 858.637 kN

    6.00 m 11000.0 mm**2 53.2 mm 250.0 MPa 200 GPa

    Effective Length A ry

    Yield strength = E =

    Problem 10.60 b Allowable centroid load: P = 217.727 kN -----~----------------------------------

    6.00 m 4580.0 mm**2 40.8 mm 250.0 MPa 200 GPa

    Effective Length = A ry

    Yield strength E

    Problem 10.60 a Allowable centroid load: P = 431.SS'"I ~N

    Effective Length A ry

    Yield strength E

    """ 2. .t;G 60 "'"" 515 mm JI/-& fvlft 2oOOD Mf'q

    Problem 10.60 b Allowable centroid load: P = 43, z~b 1.

  • CONTINUED

    COFF :=- A ~11

    METl-lo"f>

    ~11. (Otf

    FOR. INTCttltGllJN

    f"'1T&l2- L) E)

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