strngth of materials

8/8/2019 Strngth of Materials

1/39

594dM

JTRENGTH FMATERIALSE= r shear orce Se e ag e319)

F=EI z "(123)dxt

I ' l i l l l ' ( 'l l( )N OF BEAMS 595

Differentiating equation (12.4) w.r.t. .r , we gert=u t!dx d_fdF,!*=n the rate of loadingw=EI dl ,d-fHence the relation between curvature, slope,at a section is given by :Deflection :y

Slope =qdxBending moment :EI2 dxzShearingorce =EI +dx3Therateof loading= I +.CTfUnits. In the above equations,E is taken n N/mmz/ is taken n mmay is taken n mmM is taken n Nman d r is taken n m.

I2.4. DEFLECTION OF A SIMPLY SUPPORTED BEAMCARRYING A POINT LOAD AT THE CENTRE4 simply supportedbeam 48 of length L and carryinga pointload W at the centre s shown n Fig. 12.3.As the load is symmetrically pplied the reactionsRa andRBwill be equal.Also the maximumdeflectionwill be at the cenrre.No w Ra:Rs:YConsidera section at a distance from A. The bendingmomentat this section s givenby ,Mr=Rnxx

_W,. , . (Plussign s as B.M. fo r left=;-X,fz portion at X is clockwise)

rLr- 7-

Fig. 12.3l lut B.M. at any section s also given by equation (12.3) as

Bu t

Bu t. . . (12.s)

deflect ion etc.M=EI *.v.dxzl:quating the two values of B.M., we ge ts14!=Y*dxz 2On integration,we get

nr dl =w yla6," ' d t -T z

. .( i)

. ( i i )wlrcrcC1 s the constant f integration.And its value s obtainedf ornboundary onditions. he boundary ondition s that at x:!,2'/ rlv \rlopc | ? | =0 (A s th e maximum deflection s at the centre, hence- \dxlrlope at the centre will be zero). Substituting hi s boundary conditionrn equation (ii), we ge t

. i i i )

() r

w tl- \zo=; ' l; j . ' 'cl=- y!16

Substitutinghe valueof C1 n equation ii), we ge t^- dv Wxz WLzLI t--dx416The above equation is known the slope equation. We can findtrre slope at an y point on the beam by substituting he values of x.Slope is maximum at ,4 . At A, x= 0 an d hence slope at ,4 will beobtained by substituting :0 in equation (iii).. .EI +\ =Y*o-Y!\ tu)^,o 4 16| (+ j , , thestope t / and s representedy ,alL \ dt / . ,^or EIxi wLzo:- 16

i----t


2/39

STRET.ICTHOF MATERIAIS

, WL,n=-T6EIThe slope at point B will be equal to r,a, since the load is

symmetrically applied.

t rllt.t F.(:noN oF BEAMStttettio f the beam(i.e. I) is givenas equal o 78x106mma. If E forrhr nmterialof the beam=2.Ixld Nlmm2,calculate (i) deftectiontt lltt centreof the beam and (ii) slope at the supports.

Sol. Given :Length L=6 m=6x1000=6000mmPoint oad. IY=50 kN=50.000NM.O.L, 1=78x1ff mm aValueof E=2.1x1ff N/mm2Let /c=Deflection at the centre andie=Slopeat the support.(r ) Using equation 12.7) or the deflectionat the centre,weP.cl

. , _wL3,,- 48EI50000x600G=a8r2l r 1F;t8" 1ou=13.736 m. Ans.(ii) Using equation (12.6) for the slope at the supports, we get

wLztB=tA=- r6EIwL2= r6EI50000x60002

. ; - . ' - wL2'B- 'A- - 16EIEquation (12.6) gives the slope in radians.. . . ( r2.6)

Deflection at any pointDeflection at any point is obtained by integrating the slopeequation (iii). Hence integrating equation (rii), we getErxy=Y!- f f ,*c, . . .( iv)where C2 is another constant of integration. At.4,.r :0 and thedeflection (y) is zero.Hence substituting hese values n equation (iv), we getEIx0:0-0*Cz

or Cz=OSubstituting he value of C2 n equation iu), we get

EIxY=Yt -wLz ' x72 16 "' (v)The above equation s known as the deflectionequation, Wecan find the deflectionat any point on the beamby substituting hevalues of -r . The deflection s maximum at centre point C, whereLLx:=. Let yc fepresentshe deflection t C. Then substitutingr= iand y=y. in equation v) , we get

Erxy"=Y(+\ ' -ry ' r ! )n\r)- 16^\T)wL3 wL3 WL3_3WL3= 96- !z: 962WL3 WL396 48wL3v, - - 4gEI

(Negativesign shows hat deflection s downwards)Downward - wL3deflection,,= qg m ...(12.7)

hobfem 12.1. A beam 6 m long, simply supportedat itsends, s carrying a point load of 50 kN at its centre.The momentof

(180\f 1 radian:? degreeJ=3.935o. Ans.Problem 12.2. A beam4 metre ong, simply supportedat itsends,carriesa point load W at its centre. If the slope at the ends ofthe beam s not to exceed1", find the deflectionat the centre of theheam.

(Numerically)radiansl6x2.l x 1Fx78x 10e=0.06868 adians

1R O=0.06868xj: deereefre

Sol. Given :Length, L-4 m-4000 mmPoint load at centre=WSlope at the ends' '^==tf;]' A Alq,E r:'ffi=o'otz45 adians


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598Let y,=llsflection at the centreUsingequation 12.6), or the slope

. wLzte=T6EI() r O.O1I11,S=Y-!:16EINow usingequation (12.7),we get

wL3V-=- 48EIWLz L=_x_16Er"

=o.ol745xoooJ

STRENGTHOF MATERIALS

at the supports,we get(Numerical ly ' i

equation i) i=23.26mm. Ans.

Problem 12,2. (a) A beam 3 m long, simply supported at itsends, s carrying a point load W at the centre. f the slopeat the endsof the beamshouldnot exceed ', find the deflection t the centreolthe beam. [AnnamalaiUniversity,1991]Sol. Given :Length, L=3 m=3x1000=3000 mmPoint load at centreSlopeat the ends,

f f , l , l | , ( ' I ION OF BEAt\, lS II, 1. I)IIII,ECTTONOF A SIMPLY SUPP'ORTED EAM WITH ANI.]CCENTRICPOTNTLOADA simply supported eam AB of tengthL and carryinga pointl,,,r,l W at a distanced from support A and at a distanceb fromaul, l)r)rt is shown n Fig. 12.4.

Fig. 12.4'fhe reactions at A and B can be calculated by taking moments

nlxrtr l A.We find that reaction at A is given by

ne=Vf and rY(a) Now consider a section X at a distance r from .4 in length'l ( 'l'he bending moment at this section is given by,Mr:R4xx

Wxb:--L-*,Bu t B.M. at any section s alsogivenby equation 12.3)as

,t2,,M=EI " t=dxzEquating the two values of B.M., we get

- . &v WxbEI *: ,-xxIntegrating the above equation, we get

, , dy_Wxb .xz n" ' dr= L xz+Lrwlrt'rc Cl is the constant of integtation.Integrating he equation i), we getEI.t=Y! . !+c,.r+c,wlrere C2 is another constant of integration. The values of(', lrc obtained from boundary conditions.

(i ) At A, x=0 and deflection y= 0Substituting these values in equation (rd), we get

(. . wL 3 wL 2 L\|- :\ 48Er 16Er' 3 ir WLz| . .-=0.01745 fron'L loc/

w.bT-

-wi^=it=1o

1v r= .#=0.01745 radianst6 u

(Plus sign due to sagging)

. ( i )

. ( i i )C1 and

Let yc=Deflection at the centreUsing equation 12.6), we get. wL2,N:T6fI

0.0174s=1%Now usingequation12.7),we getWL' WLz Ly-=-=-A-48EI 16EI 3I=0.01245x*J

:0.0r?45x39F3:17.45 mm . Ans.

. . .( i)

ffi*'ot''ot)( ' . ' L=3000mm )


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0l6m0=0+0+Cz

Cz:O

STRET.ICTHOF MATERIAI-S

. (ii i)of i6 is

value of C2 in equation (ii), we get

ff ."+c,.*an d slope *=,r. (Note that value

I rl l.t l(:lloN oF BEAVISlrquating he two valuesof B.M., we get

Pr * l =w'b--dxz'L' '-wv-altntegratinghe aboveequation,we get

-, dy =w,.bt^ *rr:o), *r ,'E= L'T-- 2 ' 'wlrcrr' lt is the constant f integration.Integrating th e equation (vi) again' we ge tEI =H.+-+++c3x+C4wlrcrc C4 s another constantof integration' Th e values ofCa are obtained from boundary conditions'

(i ) At B, x=L an d y=0' Substituting hese values n equation(r,,r) ,we ge t o=ry+-I.";"rc.xL+co.W.b_.LzW.b3+C\.L+C4 ( ' . " L-a=b)66

c*wlt -* 'or ' " -cjxL "'(vi i i )66(tt) At C, x=aand slope*=,r' (The valueof i6'isunknown)''a xThe value of Cr is obtained by substituting hese values n

ctluation vi).Hence, we get from equation vi)

E i =w -b:o' ) @ a) z c=w-l;o'-o*c,cr="'''r-Yf

Substituting the value of C3 in equation (vfii)' we get,*yf _Yy-(r,,,-uff)=Y! -w'a-'u", ,., V!4=ry @z-L\-E,.tr.r*Y!4=ry p, rryw'!r'o'E L. i

SubstitutingheEI'Y=

(tt) At C, x=a. .(v i)

. . .(v i i)C3 an d

. . . ( ix )

unknown).The value of C1 is obtained by substituting these values in

equation (i). Hence, we getEI.ic=Y*t*r,

ct=Erxic-V#Substituting he value of C1 n equation (i) and (iii), we get

, , dy _W.b W.b.az-, dr-;.x2+Elxig-? . . . ( iv )Ery=ff*'*( r.b-U#)- ...(,)oL\zThe equation (iv) gives the slope whereas equation (v) gives

the deflection at any point in section AC. But the value of i6. isunknown.

(b) Now consider a section X at a distance x from A in lengthCB as shown in Fig. 12.5. Here x varies from a to L. The B.M. atthis section is given by,

Fig. 12.5Mr=Ra.x-W(x-a)

w. b= , .x-W(x-a)B.M. at this section s also given by equation (12.3) as

,12.,M=EI ;l


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STRENGTHOF MATERIAI .

L=a-rt, l

DEFLECTION OF BEAMSThe deflection at the point C can also be obtainedsubstituting=c in equation u). Hence,we ge t

613by=ry [bzL2 Jaz]- .L. i

=ry [b2-(a+o1za3azJ-Ei.L.i6.=# fbz az bz zab 3a2l- .L. ic

wh, O- I2a2-2ab)- EI.L. ig=ff*zog-b)-EIL. ic=*{u [a_b]_ErL.k

'Ihe stope ( , " #) ", "n, point in CB is obtainerlby\ arlsubstitt-tinghe value of C3 n equation r,i). Hence,we get fronrequation vi),

t, *=y.,r-!


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@4 STRE}JGTHoF MATERIAISor . __W.a.b ^. .[Neeativeign",Ji;.g"1j,l]','n",,r"*"",", ,;"t*,"] makesan angle n the anti_clockwir" i n"g"tive direction].

Value of Maximum DeflectionSince a' is more than .b , hencemaximumdeflectionwill ben fengthAC ' The deflection", "" f poi"t-in tengtnAC isgivenbyequation v) asELy=Y-br + ( u.r.-Y.b.o' \ - 6L ' ( " ' ' ' t - - 2L I '=#".Luffio-b)-yltl.

| ,,=rowi@-b)rom.o.,r.r;I=#".[ry@-b)-w]4:. ,=y ff +2a(a-b)x-3az.x]=Y [xt+ 2az- za - 3a2f=+* ff -a2x-2ab4=Y! fu3-x(a2+2ab1l,=##[r-x@2+2abyr ...(D)

The deflection ill be maximumO#=OBut *=#h pxz-(az+zab)lFor maximum eflecrion, #=O

#* rt,r-(a2+zabsl=s':=:fr'*3

,=fo'*uzot1t '

DEFLECTION OF BEAIT'IS 605Substituting his valueof r in equation D), we get maxrmumdeflection.

w. b | ( a2+2ab 3n ( a2+2ab 1nrm=utul l , , / -1, 3 )w.b | 1az+zab\tn 1az+2ab)3n=6ErLL-l;6----E-- J=ffi.{o'n2ab)3nt# #]=ffi{or*2ab)tnH= --!'b-(az+2s6fn9{ 3 EI.LNegativesignmeans he deflection s in downwarddirection.

Downward,,^",=##@2 +2ab)3n

t. =ffita-a@2+zab1l=ffi[ot-ot-Zazb]

@r+zaqf

. . .(12.10)Deflection under the point loadLet y.=psflection under the point load

The deflectionat any point in lengthAC is givenby equation(D), as w.b- .. yr:ffiVt-x(a2+2ab)lThe deflection under the point load will be obtained bysubstituting =a in the aboveequation.

w.bff i cannotezero)

W.b Wazbz:TEILX\-/a-D1=- 3EILNegative sign means he deflection is downward.

Downward,e## ...(12.11)Note. The above method for finding the sl ope and deflection isvery laborious. There is a simple method of finding the slope anddeflection at any point in a beam. This method is known as Macaulay'smethod which will be discussed ater on.Problem 12.3. Determine : (i) slope at the left support,(ii) dcflection under the load and (iii) maximum deflection of a simply

or


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606 STRENCTH F MATERIAT^\supported beam of length S m, which is carrying a point load of S L.tat u distance ?f 3 m lrom the teft end. Taie E=2xrd Nlmiz antiI : l x 10 "mm".

So!. Given :Length, L= 5 m=5000 mmPoint load, W=5 kN=500C,NDistance between point load and left end,a=3 m=30C|0 mb=L-a=5-3=2 m=2000mn rE=2x1F N/mm21=1x108mmaie=Slopeat the left support,y6=Deflectionunder he load, and/*=Maximum deflection.(l) Usingequation 12.9),we ger. W.a.b,e=-dEIZ\a+zD)

= __!900:!000x:?OOO6x2 x lF x 1Oosooo (30002 x 2000)(radians)= -0.00035 radians. Ans.

. -Negative ign means ha t the anglemadeby tangentat,4 isanti-clockwise.(ii) The deflectionunder he load s givenby equation t\.tt),ASWaz.bzr -=- 3EIL

=-5mo1?T',1zooo'0.6mm.3x2x1SxlSx50O0-"(iii) The maximumdeflection s givenby equationw.blnat=_=7_(az+2a6yn9,l3EI.L5000x2000=ffi(30002 +2x30002000)32

(9000000+2000000)32

Vatue ofM.O.r . ,Let

I fr,l l l ( l l( )N OF BEAtvtS ff l

ll.(r. |)F]}'I,ECTIONOF A SIMPLY SUPPORTEDBEAM WITII AIINIFORMLY DISTRIBUTEDLOADA sirnply supportedbeam AB of length L and carrying arrrrrlormlyisirlUuteaoadof w per unit engthover he entire-lengthr,,..lrownn tsig.72.a.The reaciions t A and B will be equal'Alsorlrc nlaximurn- eflectionwill be at the centre' Each verticalwx Lrr' ; tcl1Ofl=J

Ro=P^-II!L

Mr--Raxx-wxxx:w. L w. f= z'x- zBut B.M. at any section s alsogiven by equation (12'3)' asM=EI#

Equating the two valuesof B.M., we get-- *v w. L w'xzr- i *=jx- z

Integrating he aboveequation,we get",*=+-; *',where C1 s a constantof integration.Integrating he aboveequationagain,we getErry=wu+- i .!+c,r+c,

where c2 is anotherconstantof integration.Thus two constantsofintegration(i.e. Cl and C2)are obtained rom boundaryconditions'The boundary conditions are :(d )at x=0, /=0 and (ii) at x=L, Y= 0Substituting irst boundary ond[ion i'e' x=0,./:0 in equation(ii), we get

Ans.(12.10), s

. ( i )

. (rr)

LENGTH

x2 Fig.12.6Consider a sectionX at a distance from A'

rnomentat this sections givenby ,

=0.6173mm. Ans.


8/39

6m

in equation (i i) , we ge to:w.L L, _y !+c,.tv- 4 3 6' 4 -' -(Cz is alreadY zero)

w,La w.L4,n t= n- U -t-(- l .L^ wL 3 . wL 3 wL 3or c1=--5-+ 24 =- aSubstitutinghe valueof C1 n equationsi) and(ii)' we get

-, dy w. L .. 2 n *r-r . \ t . . .( i i i )EI fr= 4 .r ' - 6r 24

or Ery=+i-# f-#' "'(iv)The equation iii) is knownasslopeequation'We can find theI dv \stope { i.e. the value of # | " any point on the beamby substitut-

ing the differentvaluesof x in this equation'The equation rv) s knownas deflection quation'We can indthe deflection (i.e. the value of y) at any point on the beam bysubstituting he differentvaluesof x in this equation'

STRET.ICTHOF MATERIAIS0=0-0+0*CzCz:0

Substitutinghe secondoundaryondition'e ' at x:L' y=O

Slope at the supportsLet ie=Slope at support'4' This is equal to

and ia=Slope at suPPortBAt A,x=oand#=rn.Substituting hesevalues n equation iii)' we getEL;a=!xa-i"o-ff

wL, WL2=---24 24 ('. ' w.L=W=Total load)...(12,12)

f r'1t,1j'('noNOF BEAtvts 6J)(Ncgative sign means ha t tangentat A makes an angle withl l rr r hc anti-c lockwise irect ion)l ly symmetry io=-WL'. 24EI

l l , t r trrtttrnDeflectionl lrc maximum deflection is at th e centre of th e beam i.e. atl" ' | f rt ( , where =1. Ury6=deflect ion at C which is also maximum

'f r fr (r tr . . Substi tut ingl=16 and.= rL i " th e equation (iv), we ge rEr.y. : ! ! . ( L t ' y I L \ ' _*Lt ( t \tz. (zJ_n \r)_u l ; l

:n.Lo wL a wLa ;* . ; , \ ' I96 3U 48 ---384/c_5wLaSW.L3- :s4'Er=-3u i( .. w.L=W=Total toad)Nr.liariveign ndicateshat deflections downwards.I)ownwarddeflection.

.,_5 wL 3Y7-- 384 Er

. (12.13)

. . . (12.14)

(dv\\d ' )^^

f'r olrlcm 12.4. A.beam of uniform rectangular section 200mmri,te rtttrl In mm deep,k:iygly supported at its ends. It carria a"'ttt"tttllv ri.rtributedroad of g'*r,t/rn'*, or", tlr" "ntir"'rpo"';i'; ^.t th ' rrltt*f E fo r th e beam material s Ixt01 Ut*,*i,;;;: "(tt ttrt slope at the supports and (ii) mori^Vi-.i"yt"rtion..\ol. Given :Wrrlth, b=200 mmlrr 'p l lr , d=300 mmf\ l () l,I | ,l I' .p . lr , L= 5 m=5@0 mm| 'f ;1 11;s61, =w.L*=9000x5=45@0 N\' :r lrrr .o[ E=1X10a Nimm2|, l4r,,r t=lloP: ar thesupportyg = Maximum deflection.

,_b& 200x30G'-T=--Ir-=4'5x108 mmap=9 kN/m=9000 /m

wL2 ' l l , r,I u rr r . t l, r rr r / rh,rrt


9/39

The maximum bendingmoment or a simply SUpportedrt ""carryinlg uniformly distributed oad is givenby'w.L2 W. LM=-{=-gNow using he bendingequation'

Mf- : -Iy fxl 8x/ ( . . , - ' lor M=7= 16t4 \ iu=YtEquatinghe wo values f B'M'' we ge tw.L 7618d

,,,_16x81 1281 ... r , ,or n=-Lrd =TrdNow usingequation 12't4)' we get

5 wL3Yc= 3g4x-EI

i 'F* i ts t a l t t I r l lt l AN,ISI t t I l)r'flcction of a Simply supported Beam with an

-.. .+rfrl y, i l t krurl. A simply supported beam ,4.B of length L an d, . , i i r rs q l ' . , r r t loatl W at a distance'c'from left support and at af: f 4rf iF /, lr ' lr right support is shown in Fig. 12.7. The reactions=! | ar,, l l l ,rr1 given by ,

.. W. b W. a/ i . r= L andRr=---f ', I ' r ' r rr lrrrg oment at any section between A and C at ai i .r?,. . . r lr , ' r rr z{ is g iven by ,M '=Rnxx=w.b *

r .Ef r.rr o;111fl 'a ' .Th e B.M. at any sectionbetween C an d B at a' la;t irrr r l rom A is g iven by ,M,=Ra.x-Wx'(x-a)wh:_::.x_W(x _a\Lllr t 'above equation of B.M. holds good fo r al l values of x16lr+rr 1 r :O an d X=b.

l l r t ' tt.M. for all sectionsof th e beam ca n be expressed n a=i,,g lr r lr ;r t ion writ ten asM,:ff t ': , w (x-a)

Stopat the dotted ine for any point in sectionAC. But for4r r l,r)mt n sectionCB, add he expression eyond he dotted ineal=, '

lh c ts.M. at any sections alsogivenby equation 12.3)as

STRENCTHOF MA]T1l{l \l

( ' . ' W=w.L)

613

goodsL. t2.7M. hol

_L_Fig;r lrovssqurt ion of B.

l -

l,I l r r ' IIRo -W.ol" --rfor the values of .rn=**H"+r( )c=10mmand =; l '

5 728xLz--Y- 38p--" xEs 178xLz 5 .. 128x5ooo2cr d- - t*'-to* = 384 to;t;ld

=347.2mm:M'72 cm' Ans'I2,7, MACAULAY'S METHODThe procedureof finding slope and deflection for a sirrrl'I1support;J 6eam wittr ui "tt"nitit point load as mentioned

n Ar12.5, is a verY tau-iou'' There is a convenient methotl l"rdetermining he deftections f the beamsubjected o point loatt''

This nethodwasdevised y Mr . M'H' Macaulay nd s ktt"'r"asMacaulay'seth;' rni' t"lnoa

mainly::T1:::-t","t1*::ll" '*unn", in which th e bending moment at an y section ls XpIt"""t

"rt l" ,rt" manner in whichihe integrations are carried out'

M=Er#l lcnce equating (i ) an d (ii), we ge tu,#u=ry.,-w1x-a7

. . .( i)

. . .( i i )

. . ( i i i )


10/39

t{ Ft H ni ; s: s$ls"stS+r:5i t : =nSiIi is lS\ t l ' i qr *-Fx** a's" i fE:lS+!siE;!=l*t l5F$ -Ets$tEtS i: i : : \-=i i f -

Ni

ddl :ql .o l J| - ' l \ l Z! Sl- te.l -l >, -+ | suN ll )Pa:!!+ S H.ENq!;\ - i. ;Sl-E. . . Z.==:

a i .3 i ; e ! E i F .a .g e .E : p:; : E*o E i i r r : ' l i^$ E\- " l : -E ; b qr .8 *1 6 gls ; l i i l -g=; E ; i i l - : j f r - i

oC)'o5oF9g:o

F i3. ; g 6 F s.=.E G': g; : ; * 5;o S lr i t' e.: t x i I : , l [ . r iei : :sF Fr E:: i s :, i l^ .1 s' l t a; - o. : .= +gl . i ; i . q.e F pE! i- i Rg.F ie E s : - 'Y "J"; ig[- f *r 'E{ sr *e, = I . .E:g oi^ ' cg=.E ,- .T o ! . f r l i .o\ ) 'a- d ; "E'g Sl '; ,FEr r I t ig I S; ldT:g ! { E -r : iE i . : . r- i -gi ; 'oY' s- a.L 'E e id i . t E i ;1r $ { i$l* : i . *S,u-rs;i l ; t -+ i ga *" 3; sl* l* l*T *l'+ + r-ni;?, F=d :! i : l s

-- : tEpl io>l->l-s iFg E - l" .T :Eg*,1- I Ii .E Po eEr Z blESeCc S= :,z i E r oo-) \| =t=_ i i =- =.c_

; !: -3 t: ; : i i=


11/39

:\ !: '1)-s.a a

.n vvc )6E

._4eF-OY^Ftw-' O ,1 .)

- , U }c. ) 'i ia ai. \ ts I. , \J ! . ,2 ; -1C* ,T C *+ r + 1 g 0? l . 3o- ;_E y. !z} e 7'n, : . : \ j E: : + EEs- - ' i i6, = >E_.c !-9.5 !2-qyl l oo l l l l 9 o Y 'c '3 - q Y'; q) tE i ! : l l t r f .E.e j4=EE kJ 'r : P' x : : ' ! -b ,E!; ; dEou 6iae - E.:^ -T; .^ i== '- : - '3 ' -=r . 3=.

:=: - : - :

rqI--riI laJ., I- ;=O

a? E -: lt r ;s- : }F. f .3? H C\


12/39

^l ?lI l c . lxtXrnIRX

g. t,1|TIl l c . l.! k: :sbdr i 3" i i X HF;?-=.E h iz e i : .az;; T f ;EEAi: q F-


13/39

ie l e =pgc?^i I :; f F 'E lq+T:E , leq : .oB ?,p i ; ! * r++l*: l ; l ;i t l iSz: i l is : ; F:qe5giF$' ;F ara$E El;= s[ :: Ei i i : : * idr s : r -s-- i-re i?: .3,8; rdrsnnies;nf i ; : E 3E *AEi l r ; :Ei tFE;I f i f i f i l i r i f : B sE !6f =, i E aEs -s: ;sp: i i ;b s is;P- =83a; ! s* -g *-r ' f :Ef+J's5: ="Ti$: E i5 : ii i !L = : ;+rr+ ^. ilE*B+f i#;s $ipar 4; :ErrsqqFi g6 g. _T f i i ! " ' "" [T . l ' i ii i lE: i f r :55d$E=Si q 9E5g i1 o' : i '=" ; ===Ee= 3 q' ;a.3, lTsa rEii i lA:2 Frm:*E;J- " ' r ! . " i iEr sr lEEE;0;EEa;;E ^i l l i : . , r iTs iE;.e oE:: :E=;; ; : ;Eo$e";?ja5'*- ' i9glz FFE* rI I I* ITT+FseE5.! [ i i l= ,;8EE f;Si i1 lg=i !8g,E$Set i . :s' F c c.F -:EJ E F,. -F: s a:n r; E !s guq

S : S H E a 1,c,F ir ' g $ g s g:g ?i* i ; ! i :_ : i i f ; i Ar e] b i f s . .g- h:- a r E= z


14/39

!9: : . :0 )o:l hovv h: : $n^fJo=

SEEac ll 'b.=rtt!(gng'^J;EiUEEEoog +g t- eoR *e aQ ?: 39. ' .Ecd .f ,^.\ C,{ . :> Y::X qE JI ll v7 a 'Z = i-' ; ' ) > ' i r E3 3i : 6a, r +5 =F E':F v b,6 ro i I q. ,d0 l l R'o . : '=: \J2.. \ :9 oE :d[ ; E '= !^q o EE!q?! E: s . ;-

- ' .S r l FE i ,r hoE E); Ea t j ; - ;Er:6E+;E Ji ,^1r gEe b?xLl iE G33l EF.9^ i ;u \o.- i X. I l 'oH.: ? ^^.: + oo 9l- n ?l a. ): i l : :Jc o c- t l - | +l EYE >'.Eo i l ' ;c r r r r -.18;Eo; i . t Pv-.

r r ; c i -9 '?=k " E'= ; ?FE E A.V ' |h i A^! -=-: : : - >2 . : . : 6 --- 'J?a: : - .=

ulh^.3 E*{d!P t"-5< d= s

?3tl


15/39

b8-. 9,- : c

--t e c!- } :V EP"o6 ooE{:.o\ i \ q. =x ' l :E,qE s *J.8I i - ' - 3 *"E ^color S! i -\

0

;

>

F-C.lItl

Xrra.lI

'5 o,9t lo .f 'ExQ)eo'5un6O earOFt* 7

:F, , . i , . , a\ ' ' . l -6 - Isv.S c9 9 =R!o d-5E IE r. tc. r r - i \ l.O.Yf\Ol1'H; I + r ic) E ^ o; 'E ; TE F FE$--,--r1A*O,YO9-\.1 J+ lg+ V r-of :3;- i -R

^ . 9t = v | | |i ' !ur t t t t t l._ rQ)a.=f*: - : - : i.:= i- - ' - : .>====- :-

-fg ; '3

a

, o ., , .F E Q .9'24+HFv.=-

| -r 7a I =l ico\s-s l


17/39

H g s Eg s ES3 E O i' : $ -.,ts 60! g E : e i g$g; s F e6 = 5 F;Fu# F t E$ E gE;Eb! , *9sgi F' ; EE;P*Ft . :EE:E e Eb F E 3 *CFsiq ' i ? gi E B F ;s ict i l j I ?R =q E Ft E E;E;sg;I . ."4 !FR"T E X iE ir fAE grEg: ! ; is=T;s g FFgi l? Es+tti ; : ; ; t gEFl-5s;i i : { l l r ;FF,iE'i Tg ,fal-!f1-r *i:"gFlis5#: : i ; rE s " sF$ ;! * i T I $gp: : ; t = = .: i : i sEsE;: : r3=: ; s; ! i : : :F; i e :- ! S=: = - i; $ E -: j ; st$

dl,-IYI Ntl /\rt lor 9: .Y l I& T 5t t!-. I c.l l AtD ). ';tp + 3-5+.i .No+ 9' '5+rt(E . . . . . . i_ \,rF++ r\ 9XV(D\Jrr+ bo + r iOrul , ' - i vI Jt l x i lc i +S i i lNs : r l .g gl^ Rl-

f = - l r E" r rr= ls l t s=:\ : { \ ::-ar=: '- : :;n-:;llit#

EE

C);o '5onr .9e \

P} oeE^o

2 !t :F ? d Si iFE l*:F=6 : F E!* 's t -6HX0vd


18/39

-E ta2g6=0 ( ' . ' Ct=200'Lor -25x2+200-4

t ' t tu lor ,=Y '; =2x''/2 mNow substituting :2x ''/i i" equation iii) upto dotted

line'we get the maximum deflection.

.' .EI yn - -! " P" { 2 +zoo(zx\/t)= -188.56+565.68=377-72 Nm3=377.12x1012mm3

,^^=ffiu=9.428mm upwards'ns 'I2.8. MOMENT AREA METHODFig. 72.17showsabeam AB carrYingsometype of loading, andhence subjected o bend-ing moment as shownin Fig. 12.17 (a). Letthe beam bent intoAQf$ as shown n Fig.12.17 b) .Due to the load ac-ting on the beam.Le t Abe a point of zero sloPeand zero deflection.Consider an ele'ment PQ of small lengthdx at a distance from B.The correspondingPointson the deflectedbeamarePtQr as shown in Fig. Fig. 12.1712.17 b) .

Let R=Radius of curvatureof deflectedpart P:Q:dl-Angle subtended y the arc P1Q1at he centreOM=Bending moment betweenP and QP1C=Tangentat Point P1

QrD=Tangentat PointP1 .ThetangentsatPlandQlarecutt ingthevertical l inethroughB at points C and D. The anglebetween he normals at P1 and Q1

fftll l,r ' tl()N OF BEAtr,lS 6l l*tll lr t crpral o th e angle between he tangentsat p1 an d e1 . Henceffr6. 1111'11.' lctween he lines Cp 1 an d DO;will be equal to'ag.f or rh e deflected part ptet of the beam, we have

PtQt=R'd0l lrrr PtQEdxdx=R.d0

de=4Rllrrt fo r a loadedbeam,we haveME^EI7:R or n=E\rrlr51i1u1inghe values f R in equation i) , we ge td-r M dxnH:- ( EI \ EIt-l\M )

ll t lrc slope at A is no t zero then, we have . . .(12.15)" linil changeof slope betweenB and A is equal to the area ofF N ltusrom betweenB and A dividcd by the flexural rigidity El,il r ir,-;r=Area of B.M. Eltween 4 and BEI "'(12'1'6)

. \rrrcc he slopeat point .4 is assumed ero,hence otal slopert lf t*brained by integratinghe aboveequation etweenhe imitsll d rrr l If L *r .d* | rLe: I y#=+l u.a,.o EI = EIJ. M.

l lut M.dx represents he area of B.M. diagram of length dr .fLf ferrrr' | ,.at represents he area of B.M. diagram between A"t tr t l l l fe=! gi tArea of B.M. diagrambetweenA and B]

llrrr O=slope t B=iaSlopeat B,

n.ea of n.U. Oiag.a, UetweenA and BR= - EI

STRENCTH OF MATERIALS

(q )

8. M.DIAGRAM

.\

hd 0\o

. (r)

. (rr)

K

AREA- M,dr

1Dt-(-

-#:*.


19/39

612 STRENCTHOF MATERIAI \Now the deflection, due to bending of the portion P1Q1 r:

given bydY=x'do

Substituting th e value of d0 from equation (ii), we ge t. M.d-xdy=x . EI . . .(rt t l

Since deflection at , 4 is assumed o be zero, hence th e totaldeflection at B is obtained by integrating the above equation betweerrth e limits zero an d L.

f L xM'dx- t f t ,u.a,'=Jo EI EI JoRut xxM.dx representshe moment of area of the B'Mdiagram of length dr aboutpoint B.

nLHence I xM.dx representshe moment of area of B.MJo

diagrambetween B and A about B. This is equal to the total areaof B.M. diagram etweenB andA multipliedby the distance f thtC.G. of the B.M. diagramarea rom B.1 -Ax. y=_:rA Xr=:'EIEI . (12.11where A=Area of B.M. diagrambetweenA and Bf :Distance of C.G. of the areaA from B'

I2.9. MOHR'S THEOREMSThe results ivenby equation 12.15)or slopeand (12'17) o'deflection are known as Mohr's theorems.They are stateas :I. The changeof slope betweenany two points is equal tothe net areaof the B.M. diagrambetween hesepoints dividedby EIil. The total deflectionbetweenany two Points s equal t,'the moment of the area of B.M. diagrambetween he two pointsabout the last point (i.e. B) divided by ELThe Mohr's theoremss conveniently sed or following cases1. Problemson cantilevers zero slopeat fixed end).Z. Simplysupported eams arryingsymmetricaloading zer,,slopeat the centre).3. Beams ixed at both ends(zeroslope at each end)'

tiFl,rll('noNoF BEAI,IS 633. 'l 'heB.M. diagramsa parabotaor uniformry istributedoads.I lrr l,ll'wing properties f areaand centroidsor paraboraare given{n

l.ct BC=dAB:b

In Fig. 12.78, BC is a a surroundingDparabola nd .4BCD isNI.\anea Al .\\\\ \\\ d

A.LD IFig.12.18tr 'r i tngle.

l .et At=Area of ABCxr=Distanceof C.G. of A1 from ADAz=Areaof ACDIz:Distance of C.G. of A2 from ADGr=C.G. of areaA1Gz=C.G. of areaA2'fhen Ar=Area of parabolaABC2=t bdAz=AreaACD=Area ABCD-Area AB C=bxd- ? a=luaJJ\rt=i b_1iL=i b.

II.IO. SLOPE AND DEFLECTION OF A SIMPLYPORTED BEAM CARRYING A POINT LOAD ATCENTRE BY MOHR'S TTIEOREM- Fig. 12.19 a) showsa simply supportedbeam AB of lengthI rn


20/39

6Y STREI.TCTHOF MATERIAIJC" The B.M. diagram s shown n Fig. 12.19 b) . This is a caseofsymmetricaloading, ence lope s zeroat thecentre .e. at point C.

Bu t the deflection s maximumat the centre.A

(o )wT

Fig. 12.19Now using Mohr's theorem for slope, we get^rea of B.M. diagram between A and CSlope at A-" ' EIBut area of B.M. diagram between A and C=Area of t riangle ACD

=7,L,WL _WL'22 4 t6Slope at A (or ia ) :W,*

. 'LINow using Mohr's theorem for deflection, we get fromequation 12.17)asA7l: EI

where A=Area of B.M. diagrambetweenA and CwL2=--1,6

x-=Distanceof C.G of area A from A2L L323WLz L_v_' 16 ^3 wLtv =-' EI 48EI'

ltlrl.t ,( ll( )N )FBEAMS 635II I I. SI,OPE AND DEFLECTION OF A SIMPLY SI.JP.I{)R'[ED BEAM CARRYING A UNIFORMLY DI$I'RIIIUTED LOAD BY MOHR'S TTIEOREMI ty. 12.20 (a) shows a simply supported beam ,48 of lengthI e,,l r.rrrying a uniformly distributed road of w/unit rength overlhr rrr trr l 'span. Th e B.M. diagram s shown n Fig. 12.20 D). Thisl: rr , rr\r' r[ symmetrical loading, hence srope is zero at th e centrei t 'r l rr) i l l l C.

(r ) Now usingMohr's heorem or slope,we getSfopcat A-lful area of B.M. diagram between A an d C=Area of parabola IC D)=jxACxCD

2 L wL z w.L3=3^t^ g = 24Slopeat a=!,Lj-24EI

(r ) Now usingMohr's theorem or deflection.:{u.tton (12.17) su-Ai'EI*lrpre /=Area of B.M. diagrambetweenA andCw.L324

mr f r=Distanceof C.G. of areaA from A

we get from

B.M. DIAGRAM

E.M.OIAGRAMFig. 12.20

I -=-.*- t '*Ee


21/39

6iffilfEtrr('tl()N OF BEAMS

(rii) ' fhe section,. or which B.M. equation islr,rrtrt rc taken in th e Iurt p".t of ill;rn.0. I,ormaximum eflection,heslopefr i, ""ro.,',nrf,,,,1*-T;tftff

at pointB if slope t 4 szeroby moment-areai r=At"" of B.M. di"go* .'I'hedeflectionby moment area method s given byAi

v --EIArea of B.M. diagram between A an d Bl)isrance of C.G. -o f ,r. , from B.

HIGHLIGHTS1. The relation betweencurvature,a section s given bY :Deflection=y

slope=9'd r

slope,deflectionetc. at

s.y.=ers.F.=EIIdx"

n=gt trl'd-f2. Slopeat the supports f a simplysupported eamcarryinga point load at the centre s given by :wLzIA=IB=_ 16EIwhere W= Point load at the centreL= Length of beamE= Young'smodulusI=M.O.l.3. The deflection at the centre of a simply supported beamcarrying a point load at the centre is given bywL3Yc = EE I

4. The slope and deflection of a simply supportedbeam,carryinga uniformly distributed oad of rylunit engthover the entirespan,are given bY, wLzta=tg=EEIand ,r*ry5. Macaulay'smethod s used n finding slopesand deflec-

tions at any point of a beam' In this method :(i) Bracketsare to be integratedas a whole.(ri) Constants f integrations re written after the first term'

STRENCTH OF MATERIATS 637to be written,.^ 5 L 5L=rxAL=grt= te

w.L3 5L24"16 5 w.Lal=---n =3U EI

t .

*hrrr ,4rrt l r

*hrt e

EXERCBE 12

. I r)crive^" ,tn,Theoreticareuestions

i'f,f r r r ,, ";i i.;;';tJl;tJltJ#; tneslope nddeflectionof a beaml. f ' rove that th e relation thatt4=Et d'l

,t t tlcnding .no."f',; ,I.ilfl''modurus.,r,F,,'.,",i,ill ',t,|j'j'i'J.1",:jl.J,if..l J,i: upportsfasimprya f r.ve that he d_efl.ectioni ,i. .nrre. of a simply upported|}Bt | | | | l | |v| | | , lapointlo 'a. ' i . t - r ," . l " i l ^ . : , [ ! ,u.nuy, . :WL'ih rrr ll l ' ' r r r r load ' - 48EII l.crrgthof beam.

3g1*,,.,,"1,,1,,.i.1ffi1t::'19tr9,'hesrope..11d_.l:"jionora simpryrnf r ,"'i,",..ifi:f,".,fr'Lj;13.1"#.?".;,""..,; .X.Jr,".,io*.,*: :n, I l i:" ii'^?',f .*f,T,ffitl[;1,, *,y uportedeal: f l frts rrtrre lt 'ngth ar e grven by c toaclof w per unit lengthStr,fx.r uROo.tr=_ff, "nO

| )t . l tcct ionat cent.e=_S . I/L.**glr ll l , r r rr l1r.


22/39

l r l , f I I ( i l ()N()1. IJF,.AMS 639638 STRENGTH FMATERIAI57. What is a Macaulay's ethod ? Where s it used ? Fino r.expression or deflectionat any sectionof a simplysupportedbeamwith ar'eccentricpoint load, usingMacaulay'smethod.8. What is moment-areamethod ? Where s t conveniently sedFind the slopeand deflection f a simplysupported eamcarryinga (i ) pcinloadat the cntrean d (u ) uniformlydistributedoad over he entire enst;usingmoment-area ethod.

(B) Numericel ProblemsI. A wooden beam 4 m long, simplysupported t it s ends, r-carrying point oa dof 7.25kN at ts centre.The cross-sectionf the bea'r:is 140mm widean d24 0mm deep. f E for th e beam=6' l0r N/mmt, in, lth e deflection t the centre. [A"s. 10 mm,2, A beam5 m long,simply upported t its ends, arries pointIoad W a t it s centre. f the slopeat the endsof the beam s not to excee{1", ind the deflection t the centreof the beam. [Ars. 29.08mml3. Determine (i ) slopeat the left support, ii ) defiection nderthe load and (rii) maximumdeflectionof a simplysupportedbeamof lengt!-.10 m, which s carrfng a point oad of l0 kN at a distance m from th e

left end.Take E=2x10tN/mm'and =1xl08 mm'.[A"s. 0.00028ad., 0.96mm an d 0.985mm ]

4. A bema of uniform rectangular ection100 mm wide and 240mm deep s simplysupportedat its ends, t carriesa uniformly distributedloadof 9.125 N/m run over he entirespanof 4 m. Find he deflection tthe centre if E=1 .1x' l 0' N/mmz. IAos. 6.01mm]

5. A beam of length 4.8 m and of uniform rectangular ection ssimply su pported at its ends. It carries a uniformly distributed oad of9.375 kN/m run over the entire length. Calculat-ehe width and depth ofthe beam f permissible endingstresss 7 N/mm' and maximumdeflectionis not to exceed0.95 crn.Take E for beammaterial=1.05x10' /mm2.

[Aos. b=240mm and d=336.8mml6. Solveproblem3, using Macaulay'smethod.7. A beam of length l0 m is simply sup ported at it s ends andcarries wo point loadsof 100 kN and 60 kN at a distanceof 2 m and 5 mrespectivelyrom the left support.Calculate he deflections nder each oad.Find also he maximumdeflection.Take I=18x10t mmoan d E=2x105N/mm2.

[Ans. (t -4.35 mm (lr) -6.76 mm QD y^'=-A.78 mm]t. A beam of length 20 m is simply supportedat its ends and

carries wo point loadsof 4 kN and 10 kN at a distanceof 8 m and 12 mfrom left end respectively. Calculate : (i) Deflection under each load(ir) maximum deflection.

l . r l r / I .106 N/mm2 an d /=1* '109 mm4.[Ans. (0 10.3 mm and 10.6 downwards, (rj) 11 mm]

e A rrt 'amof length 6 m is simply supportedat it s ends. It carries4 rffrf ' r r rr lY lrslributed oa d of l0 kN/m as shown in Fig. 12.21.Determineth r ,l ' tt , r r,rr .f th e beam at i ts mid-point an d also the posit ion and thei l ta.r r t i lr i l r , l t . l l t . t . lron.

Fig. 12.21|.rLr. /. / .4.-5 x-l08N/mm:,

[,Lns, -2.515 mm , x=2.9 m, lmat=-2.5g2 mm ]ftt. A beam AB C of length 12 metre has on e support at the lefttr, l , l r , l r r t l r t . rsupport at a distanceof 8 m from the left-end. The beam,4rrrr ,r l r r rrr l load of 12 kN at the right en d as shown in Fig. 12.22.

Fig. 12.22.lrrrrl th e slopesover each support and at the right end. Find alsotf * rL h.r.trorrt the right end.

t , rkc I: . 2 x105N/mm2and l=5 xl0E mmo.fAns. ia:6.tX)364. r;=-0.00128, ic:-0.00224, yc:-7.68 mml

I L An overhangingbeam AB C is toaded as shown in Fig. 12.23.I rFtr rr(. lhc rlcf lection of th e beam at point C.

Fig. 12.23t,rLr. /. 2 xl0s N/mm2and /=5x108 mma. [Ans. yc=_4.16 mm ]A l--:rm of span8 m and of uniform flexural rigidity E/=40 MN_m2,li- rlurl'lv rrrP;xrrtcrl at its entls. lt carries a uniformly distributed load of 15lN4tr rurr ovct l lt t 'cntirc span. t is also subjected o a clockwisemoment oflFlt N'r :r r r rhst;rrrcef _Jm. r.m the left srpport. Calculate he slope of th eE*rr ur r'r ' lxrrrrr f apprrcati.n rf th c momeni. lAns. 0.(x)61 ad.l

tr **= lu F


23/39

13Deflection of Cantilevers13.T. INTRODUCTIONCantilever s a beam whoseone end s fixed and other end isfree. In this chapterwe shall discuss he methodsof finding slopeand deflection or the cantileverswhen hey are subjected o varioustypes of loading. The importan t methodsare (i) double ntegrationmethod ii) Macaulay'smethodand (iii) Moment-area-method. hesemethodshave alsobeenused or finding deflections nd slopeof thesimply supportedbeams.13.2. DEFLECTION OF A CANTILEVER WITH A POINT LOADAT TTTE REE END BY DOUBLE INTEGRATION METHODA cantileverAB of length L fixedat the point ,4 and free atthe point B and carrying a point load at the free end B is shown nFig. 13.1.

Fig. 13.1Consider a section X, at a distance x from the fixed end ,4 .The B.M. at this section s given by ,

M,=-W(L-x) (Minus signdu e to hogging)Bu t B.M. at any sections alsogivenby equation 12.3)as

,t2,,M=EI *, /

strngth of materials

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