Problem 9.3

YI

I

L

Wo

A x

L

A

[)(":L..) j-=-O]

['J(-::L;):1;", 0 1""DX

~ TM

J :5x

'1.IIJOXI

~ ~3'\

A loE~ {JV

[116= L..) .)' ': 0]

(Q.) EiASfi Co C:tJ~lIe-

(6) j @ ?t-= D:

CC) *- @ ~= 0 :

9.1 through 9.4 For the loading shown, determine (a) the equation of the elasticcurve for the cantilever beam AB, (b) the deflection at the free end, (c) the slope atthe free end.

LJ.se ~Neeo.boi t AIa

+.::>L MJ" -::.0: M... w;:. fM -= - ~ 'dr-a

': 0

dZy -= .-.l.. ~x.$cH., 6 L

£.1' ~ - -.L wo')!,'f CrJ?t, - 2"f L + I

,.-- , w.x!"c::I ,,= - - ~ + C ~ + C~ r~o L I 1

r"X-=L 6.: (;) ] .. - ,;L Wo L:! -'- CL' .) h . ,11.." T'" I

c. - J... ,01 L'3I - ~... VVg

EI - - -L hI L'" ..d- ..1 L'1 C -j - I~o ¥Vo +.t 'I VVo + z - 0

C2 - -doWe) Ly.

~r= II~~~.I Co'(~S - -5 1..~?t. + If LS )~

h. .vJ. . \Ik -:: ~~ iI L (-; ~., ~ L ~I'

- - WoL'f - ""0L" -J,I:fA - 30 cI JA - 30EI~

!LI

-=h,..

W L3-2..-'J.~EI e - woL"A - ~tIEX A.. .......

Problem 9.7

L

('1(:;>..) Y::D] [1f:L...Iy =oJ

~J

wiL-

l _un !rk-~

r<J!I

wL

w

~ MA )

V~ X ~

[-x.=L..,) ,]'=oJ

(Q..) EJ)e..s+;c. cVV"\Ie

9.7 For the beam and loading shown, determine (a) the equation of the elastic curvefor portionAB of the beam, (b) the slope atA, (c) the slope at B.

~x U.s~r1j -FV'€eboly ABC

4-) :2M8= 0: - RAL+(wL'(~) -(wLX~) : 0

RA ':: t wL

FDV" porfiC>1I\ ~B (o4!')b4!L)

+, 2M,J"= 0 ~ - RAx.. + (w~"l.1:) -= 0

tv) = t wL~ - ~w'i/-

Er ~ ': ..L w l-x. - J..v.J?~.~~"L'" ~

EI ~ - .lwli - J.wi!. -4 L~ . '8 6 '

I L 3 \ '1ET::f -= ~., W ')( - ;?'i w-x.. + C, ~ -t C-z.

[X= °", j :oJ -: 0= 0-0+0 +Cz. C2, = 0

0::- ii w l 'f - ~\'fw L't + C, L + 0 :: 0 C :: 0I

vJ ( ,. 'f )J =- - L?l-~~L/EI

b - -Y:L. ( 3 Lx.?.- 4';(.3)A' :?~ EI

(0 !; ~~ L).

(6 ) ~ J ,,:::0~

(c.) :tL d ~ = Lch6

..

£IA ~ DG"= 0 ~

:hI

- - wL.3h B - ;<tjEI

e - wL'!g - i4EI ~ 411

Problem9.12 9.12 For the beam and loading shown, (a) express the magnitude and location of themaximum deflection in terms ofwo, L, E, and I. (b) Calculate the value of themaximum deflection, assuming that beam AB is a W 18 x 50 rolled shape and thatWo = 4.5 kips/ft, L = 18 ft, and E = 29 X 106 psi.

'I U"~

~~'""""" L"'"

[1C:O).)'=O] LX: lJ j=O 1 Us; n~ e.1,J..Y'ft be"-",,, o...s 0.. -t t'e~ b~)

fA ~W.Lt ~ sf1:?~ R(\ .

""ox

IAp:1~"'WI) l "

-0 ~Me = 0:

RA = iwol

- RioL + (k 'No l )( ~) ~ 0

Us;~~ A:r c:...S Gl~4! boJy.> V21'1j : {):

- * Wo Lx+ ( ~~X~)( ~) + M ~ 0

M -= ~ 'No l)( - i Wo X. 3... b L

EI ~ - .J..WLx - .1.~ x3~ - 6 0 (. L

ET h. .L L 2. J..Wo" C-..Ix ~ ,~ Wo )( - ~&IT: X -t I

fLy::' -kwoLX3 -I~O ~xS" + C,X ~C2

[x:o)y=o]

[x-:=L.)::J=O]

EP",-st-ic. C uN"! &

0 :- 0 - 0 ... 0 + CZ Ca. = 0

0 '= 3" WOL" - ~ Wo l" + e,l 4 0 C, =-

- Wo f ..LL 3 - ...L~ - ~ L3 1j - EI l 31:. X I~o L 3,"0 X ~

~ - ~ f...L 2. - -'- ~ - ..L L31Df'C' E rl~ Lx :1'1L .3'0

10 ~;Y'lJ jQC.#../t1'0"'1 of ~~~MU""" Je.-tJeG-tiC>\I\" set ~: 0

\.5)(: - 30 e~ .. 7 L'I = 0 x ~:- ~L I.- -I qoo l" - 9:l~ ~. ~ 3.0

>< 2. =- (, - ~)p. ~ o. ~'ln L' x...,:' O.SIQ3 L

~ - ~ f ..L l/C $"'"3 L) $ L (D.s""3 '}..$) -.2 L~(o S/Q3 L ) 1.~M'- EI l?f:. \.'. ,;to L 3(,.0' J

= - O.OO6.s-~ wol" OV' OOOG.S2. WoL" I~r . Ez f

- 7 woL.33GO

~

~

DQ..t Q.: Wo -= 4.5: k:p$/-Pt 4500 '= 375 it j;., .)= ,~ L::: 18ft ~ ~Jb ;~-

I = 800 i.., If ~... Wlg,.so

j~ = . (0.006$2 )(3 ?s)(;t,'- )"(2.Q.,,-/D' )(soo) ::-

o. ~~Cf ;V\. ~ ~

Problem9.15 9.15 Knowing that beam AE is a W 360 x 10I rolled shape and that Mo= 310 kNm,L = 2.4 m, a = 0.5 m and E = 200 GPa, detennine (a) the equation of the elasticcurve for portion BD, (b) the deflection at point C.

:Mo Mo

Use c.Q~i; VIu; -l "I

aVlJ SjWlYhef~1

F~oW\ Sft,)lh'c.s j

bDVV\J~"1 Gc,"IJd i O"" c<.,-t BbovV\4e.."'1 c.:;>"".,..f1t ,'O"" ~ C.

RA = 1<& = 0(~o \/:01

J,J I

[\(=~~Ir)' J()I.~(A...)~=~ ]H.h.

[)(::L~ :0]

o~%-~Q.

M~O

EIZ::: 0

£L~ ,., c.£1:J = C, 'X.J + C1.

0. ~~~ L-Q..

M'" Mo

£1 ~ ~ Mo

Er~ ~ Mo~+C~

Elj = ~~)(~+ C.!~ + ~

LX= °..., j:: D ]

C~ ::: ~;) *:: 0 ]

[x=aJ~: ~][x ~ Cl..) .:J "" :J]

0 = () + CL Cz. = 0

0 = fVlD L + C3 C - - L t1 L3- ;;{ C>

c., = Mo0. + C3 : -t0o(~L - do)

-l-i'IpC.,*-t:~o.)~ + 0 :::- ~~ - ~Mol~ + Cot

Cy- '::" i Mo 0..,"2.

(a) E.P4.Sfic. c;.uYVc. (Q. ~?(,~ L-o..). Elj :" -f t-'I~'"X..1.,!,"tHo Lx +., Moo.:~

,,':: Mo.

('t,1.- L~ -+ a.2. )\oJ 1.e:I

-4

(6) De!'.Ped; 0tA ~t x.= ~.

:Jc.:::2~ [(;)L-(L)(~)+a.~1 = -lir (L2.-lI~)

'DcJ",: Mo:- 310 x IO?#.M J L:: '-JI,/ "" ~ Ct.J' o. S M ..) E ~ ~oo ~/d'l 'P~

I .::: go~ ')I'Oc; WlfIf#\'t ::: 3o:()( )O-c. M" E I,:: bO. Lf-Y 10" tV. M '2.

3/0)(10'!.[(

&.

( )f. "2.J -:!.yc.= - (g)(bO..~)lrotOf R.If) - &f ,.0.5) =- 3. oS)( 10

Jc. = 3..DS ",,",IV>~....

Problem 9.23

tV = wO(xIL)2 Wo

L = 10 ft

[x =0 ) 1.:0 ][x= 0 J~ " 0 1

y ~~I ~ j> -~

rLL-' -;RB~: )( ! : 1.

EI £i..tx.. ~

EI~h- -=

EI::J

[x: LJ Y=0 ]

9.23 For the beam shown detennine the reaction at the roller support when Wo= 1.4

kips/ft.

K'ec..ch oV\S C1..V'e s-f 4"+i ce..JJj iVIe:!d e.1I"""; V\Glt~.

'6C>t)~ok.lI"j C:o",d.-4i"""5 CtV'e SI,GlIoV"\ G\.t left.

Us;,.." -tV'.e~ bod.)' Jg> +;> ~ MJ" = b:L

- M + S 1* ~2 (~- x) J ~ of 1(&(L -)c ) :: D)(

M ~ r~ S: !~('~-)()J.5 - Rg(L-x)

= ~: (~!~- t}(~3)1: - Rs(L-)()

: ~~ (t Lif - t is x + ~)( ") - I?C3 (l - )( ')

~ (t L1/ - ~ ex of ~ x") - Rs (L -)( )

~~ (~ L/IX - t l$)( ~ + &~ x$) - Ra ( Lx - t x2) -to CI

~ ~ CtL")(2--/~'L3X~+3~O)(C) - Rs (~L)(1.- tx3) -+ C.X -+ Ca,

[)(:o.)~:o]

[)(:o~~:= 0]

h=L.)'j=oJ

D~t~:

0 -=

0 ~

0 f 0 + C1 C,::' 0

0... 0 ~ 0 of C.. c~". 0

l .L - .J...+ .J.. ) w LIf - (.1- .L) 0 L! - 0i 13 3',," :a (8 1"\8 -

~ Wo l" - 1.. f?, L3 = 0 R ~ 1J. MI.LIto 3 8 B ~o Q

w~;;- 1.&4 k:fS I~t,)

1<8 ::- fo (1.4-)(10)=

l= Ie> ft"

3.03 kips 3.o~ k;rs t ~

Problem 9.66

IV u>L"

i . ,- ~Y~ ~~--L~J

9.65 and 9.66 For the cantilever beam and loading shown, determine the slope anddeflection at the free end.

Loc..J'V\3 I: Do.vnwt::tv-d J's+v";b.)+~J ioo.Jw ~ppl;e l to portio,"" AB.

CD.5e ~ of Arpa-",'"~"..D t:A.pp.tec1+=>po"..t,o,,", A~.

e '- w(L/:J.)?- _..L.wL'3.Ii . - " E;I - If-csE.r-

~(L/:l)" - --L wL"g E.r - ,~~ EX

'1'-Je. -

C Po"...h'oll\ Be.. t'e.""'~I""~ ..si~c"'jht.

~wl"2./~" r...' - r\ I ~ L. wL3

~) "c - "'s '18 E.I,- I J:. ,,: _-,-wL'f LwL"- _L wLYl~ Ye. - Yf3 +{~}~ 122>EI 't~ EI - . Jog't EI

Loca.Jil1,:tJI: CO"'I'\+e"'c.PoG-kwl~e covpJ4!. ~}L ¥pi..ee1 t:<..tCo.

c.~& 3 Q-t A ppe ,,1;)( 1)

(we/').,,) l -EI -

(wL""h.t ) L'I. -~E.r -

1\ II -°c. -

" "-..JCo-

1 wL~--:?'t E.rJ VIIl"

UEI

8::1 s,)fe"'l-;,o!>;~iu"'J

- 1'\ I 6. II - - -L 'AJL'!. .J.. w L~ - ..L w P ..:::::::::1.9c. - 1::', + 1:7, - LIS E:r + 2" EI - .,~ EI

I ., 7 wL. , \AI L" ,wL" t:1c. ': 1, ... ::Ie. =- - .3£-t EX + IJ~£r :" m EX

-4a ,I

......

\...OC..J \ n ~ I: Co'" c.edV'<>-t~d POA.e! CL4'8

Cc...se..I ,,1 AppeV\cl i)( D a..ppilee-i to fe.>.rt.'oVlAt?

e I - - 'PL2 - (3 )(C>.75 )~ - O. 8 tf '37 S-a - ZEI - - J. £I - - EX

vi -=- Pl3 - _(.3)(0.75)"3 -= - O_4~187f"J s 3£ r' 3 EI EI

PortioVi Be V'e\N\Ct,'V\S,st ('6..,,~J..t

t'\ I ::' 6",' :- - 0.8</37S""Dc '" E I

I =Y ' - (0 S )e '~ - 0.8'-1375~ 8 . 8 EI

hoCtcl~1'I1:D::= Co..,c.e",f",&..JeJ jioiAd a--tC. Cc..se I of App'lMciix 1).

e "':- PL~= _(3)(/.'-5)'1.. - - ~.3437SA .<E.J :zEI - EXu- - PL3 - _(3)(I.~sf'- - ,.qS31~S-

'jt-. - 3£I - 3 EI - E~

8 . I . A L--.' L"-" 3. Ig 75"j ~vfe.,..po":!I TIO,,", j °A ::: C'A + c;7A ':' - f"T

v '= v I + ,," ':' - :? TCJ(.&75:J A ...)A .J A EI

Problem 9.73

:3kN

B

~o.75m-Lo.5mj

9- CD~IA

-0.75

L~~o.s--I

~A {3

D4-t4.:

9.73 For the cantilever beam and loading shown, detennine the slope and deflectionat end C. Use E = 200 GPa.

J"NUI1,h: FoV'c.es. "', kN J JeVlj"Hs IVI ~.I

SlOO x 11.5

l3c.

E ~ J.00 )C to <It 'Pe:t:> .I ::: Z.53)( Iv" ~\M '"' ::: 2.53)(' /0'" 'M '#

EI :::(.~oo)C,o<t)(2.53)C/6G) -= 506)/103 N'M" ': SOG klV.~\.

sl~pe 1:).1 c.

DetJec:+.'oV\ate..

e :: - 3.187S"= - G.3D ')(Lo-3~ ~ 6.3D )I.'O~,...) 'q'Co 506

',/ :' - :<.7'16875" ': -S.S3 )lIdS M :::0 S.S3 ho\l'I' J,J<' 50b

...

c

Problem 9.78

20 kN/m .

Bc

30kN

1.6m0.8m

@c. 8

1~<i;(0ktV/...,@

~~ C. 1'/

A

~I

9.771}od 9.78 For the beam and loading shown, detennine (a) the slope at end A, (b)the deflection at point C. Use E = 200 GPa.

Ur>..h : FD"'c.es ;jI'I k tJ. l e Vlj+ ~s. .'" .,..,ei-ell"S-

I=- ,~.4)(/D'mM'I-= 13.'" )(/o.t.\v-.Y

IT =-(20D>lIO'f)(13.1..f></O.6) '= :<.68>1/0' N.lNt:- 2GgO IdJ. YI'\~

W150 x 24FoV' W /5D)i 2..4

LoJ; n<t I : Ce...se S of Appe.,..):)i D.

p~ 30 kN.) L=2.4 w\.) Q...="1.6 M.) b= 0.8......

(9 )-=- £..6 (L2.- h2.) -=- (30 )(D:~ )(Z. <jL-D.S 2 )A I 6 €I L (b)(~-'>80)(2. <f")

= - 3.194 I )/ /0-3

Pc ' b? (:30 ) ( L (, ) 2.( 0. ~n2.

(yc\ -=- '3EI L :: - (3)('Z6gc»(~.'t)

-= - ;(.S47ZG ></o.!. "'"

Loo..J in~ ][ ~ C4.Se G of 'A ppetl\; i X "D.

,tAl:: ~o ktJ/ J L = ~-4 WI ;) ~ :: I.b "" J f'°l-",t G.

te ) = - wL? - - (~r»o..4 )"20 - I -sl' A2. .~l/ EI - (~}{J(Z£'go) - - if. ~1"&5 ></0

v = - ~ (-x,'f - ;(Lx.~+C>'t..).Jz ~'f ~r

(Yc.)t.::- - (~4~~,gO) [Ct. 6 ) ..,- (:{ }(;(. '1)( u:)~ of-(::<.'1)3(1.b)J = - 2. <g0 l'it:j )llo-~ WI

((:t) Slope c...1e",J A.

e,. -:-- 7.48 AG ><ID~

(6) Dd'J>ec>+ioVla-+ po;..t C.

91\ ::-(eA). + (G",)2.

-~ ,e ..". 7.4g ')( 10 ri<.,4c:::::J;A.....

Jc. = ('1C.)1 of-('fJz.

Yr.. -= s: 35 """"'" ~:/e. =- -.5.. ~Lj'12.5)( 163 WI~

Problem9.81

M~,~~,," 2L ", L3" "3

CD

t- ~ Rar~~

~b-4

Mo("~ci) c.

I

ii ~~ )( ~

M.( ff?~ 3.t!J

LL?c

9.81 and 9.82 For the uniform beam shown, determine the reaction at each of thethree supports.

SeCtIM i~ 5+.:..+..",11:1 i~c\delflM\'rle.+e. -+0 {:i",~t Je:1I/'~e.

COVIS I'de.", -Rs to b~ t"'~ \"~clv",d4.",T '(\~o..c.+"o""" CtV\d

~ep.fQ.C:ctl,e lo~"YI' by Jo~;...,s .r ...11d..1I.

Loo..ol;V\~ I:: Co..se.s o-t Appe",J,')I D.

(" ') =_R5a1.b:z. ~_I<e.(~LlBY(L/3)~:: i.-. f?eL3\J81J 3£I L :3EI L ~Y?' E..r:

L"~'\I\d J:r: C4.se 7 ofAr>f€V\Ji)( D.

( '\ - YIo ( ? ;Z) - 1'-10[(.1:.)3 - ~(L.

)\

':18).11- - bE.!L Y. - l 1-. - - ~ELL 3 L '3 j

M L2--~-L-- g1 EI

s\Jr~pDS;.h()1II a..,ol co"'st~I'lI\t.

j8:' 01'6)1 + (YS)ll = 0

- .JL ~ ~ ~L ~21./3 EX 1- 8 , er = 0 1<8= 3 fo1o~ \ .....L

S+~jCS:

t)LMc.= 0;

- J<~l t- Mo + 3 ~o~~ = 0 RA.., 2. Mo1 -L+1'2F) = °:

~~ - 37:+ ec.=o 1') = H.. t ....1'\c.L

Problem 9.83 9.83 and 9.84 For the beam shown, detennine the reaction at B.

w

A c --'It,

~w

~1'1. +Mt b ~B~t;=--d ~J ~/ 7. c: Ve-A

I BII---U2-J-U2 1

PoM-io"" Ac~ S~fe.rf'°-:.r+ioV\ of' C,,-se~ S~,/ I crl Appe J;)( D."" ;: !:1c-(Lh.)' - Vc.(UZ)3 ~ 'M~L"2.- Vc..L~Je.'2:EI .3 ~r 'iJE£r 'lIfEX

e = Mc.(Lh.)- Vr(L/~t:- ~ ~~L3.c: r;I 2 EI ~ EX gE'I.

po",.,h'o,,\ CB:- Supe.rFr..;,,;.~io"" of' c ses .3J I.) e..~ :z cf: ApreJ~~ 'P.

V ::- J"1~(L/..:?)-..+ Vc.(L/~): - w (L/~ )tJ..Jc. ;{E"I .3E.I: 8 E"r

:::- He. L'L +- Vc. L'3..

8EI 'J...lfEX

e ::- - Me.(L/~) - \Ic.(L./:J..)~+- w'(j-!:<.S!.c. EI ;(E~ £ EX

McL \J. L~ wL:!.= =--4--~;;;;r aE'I 'f'8 EX

wL'tI~Er.

MolJ; OJ e.)O!prf?35~OV\5 ~u'" ':Ie.;)

'M~ Iv. L3 MX \L L:!o wL'I~-~ ~ ~ + tlfc:.£:L- ,zgEiX

MJ~it\j ~pt'essi(.>",sfo,- fie.)

Mc.l V~ - Hc.L' V~ + wL~-=ter - ~ - -1..JfL - ~ tt8~I

Usiroj r°.,..ftoV\ CB c:c..s~ -t,..ee. hcJy>~ . wL

Me. ~ M + t 2~1=j ':" 0 : Rg ~ Vc - ~ = b~ ~ e WL. . \3 t'- J R - ;:;.F- ~ IN L ~ '-3 '" .""L

L B- ~ . 3Z ~

\/. t-- '" 12~ "'" '2 M.. "0: M.. - M" -v..~.. 1Uf.;f = 0

Ma~ ~We+~wL):t)- wf

MB -: -..1L.wL ~ Ma :- t~~ wLt ~1'1~

V. :- ~ wlc. 3:t

M ~.J... WL'a.

c. itS

~

~