sr"r. >.*# t*ril*J

Example: Analyse the continuous rearilEED snown in figure

method. The support B sinks by 1Smm'

,.*" t =raa'1d KN/ m'z andl = 120x10rma

lb --[.lor}3by slope deflection

aO h^r lrn

Solution:

ln this Problem

FEMs:

/oow

0e=0, 0e+0, 0c*0, 6=15mm

Wab'

l(a

D

c --rAB- f ,

Wa'bFro = +11-

= --'q=-41.67KNM,Bc_ lF. - -wL' = +41'67 KNM

r^^=-zo,1.s] *ott*,rE"t'a aue to Yietd ol suPPort B

= -44.44 KNM

= +88.89 KNM

For sPan AB:6El -

mau = ms6 =--L^O

_ _ 0"300 *1ou x120x10-6 * f?= = -o fUrtn-62

6El ^nt=mtb=+7-o

= a612P r1o''12sv1s* xJ1= +8'64KNM'tr

3

q;

I

't

+

Slope defl ection equation

M *= F *+4(2e A+e B35.-7)

6EI6-- r;,+ffQe,+e,)

=-+q.q++Iile u-6

= -s0.44+:EIe,

u;=r*+ff(ze,+e,)-ff

=+rt.w+!ue#

=+82.89+?EIoB

a a-r *+ffqtfe")*ry= - u.at +! u(zo,+o

"\ + 8.64

=.r.or*!rue"*!erc"

M.;r*+ff@r.e)*ry

=+ +r.fi +luQei+o )+s.il

=+so.t+tue"*!aeuM---30 KNM

L2

--------, (1)

---:----r(z)

--------r(3)

.i-t,

--------'(4)--------, (5)

There are only two unknown rotiatiors Orand 0"' Accordingly the boundary

conditions areMBA +MBc = 0

M", +M", = 0

Now, 1r4ro 4rr" = +s.ao + ff etee+f rre"= o

McB+McD= 2o'3r*f ere"+f rre"= o

Solving these equations we Petere"=-ii.ss Anticlockwise ': o-otlo '

,.

ElO"= -9'71 Anticlockwise :- o '0 o \ |

Substituting these values in slope deflections we get the final moments:

M*= -so.aa +][et.ss) = -oo.8e KNM

M*= +s2.8e +f, (-sr.as) = +or.ee KNM

n,l,"= -33.6 a 1 6 sr.es)* f (-s.71) = -61.ee **,M""= +50.31 +3 1- s.zr)* f (- ar.35) = +3o.oo KNM

Mco=-30 KNM

6t,qq 30

Consider the free body diagram of continuous beam for finding reactions

IOOk.et9q

-2cl1(^'.!o I

r€r-I o

I Rnro*'{.= 3&/slor

Reactions:

Span AB:Rgx $ = 100 x4 + 61.99-60.89

Rs = 66'85

Rn=100-Re

=33.15 KN

Span BC:

Rox5=20x5x 9+61.99-302

Ra = 56.40 KN

Rc=20x5-Rs

=43.60 KN

\

1'.

t

-..-i,ti:€]ra

BMD

I

2iv,4a uo

a2ial

Span AB

Span BC

td *r,

SF D

,!

R"x6 = 1gPr4*67.1i-61Ra=67.@ KNRr= 100:R"= 32.31 KN

ER"x5 = 20xi*5+30_6i.1f

Rc=42.58 KNRs= 20x5-&= 51.42 KN

Maximum Bending Moments:

Span AB: Occurs under point load

Max = 133.33 -61 -[91_1L!!x4 = 68.26 KNM)Span BC: where SF=o, consider SF equation with C as reference

Sx= 42.58 - 20x = 0

*=a?58 =2:te ^20

... M,o=42.58x 2.13-20x2.132 _30=15.26 KNM

lkN f!+L

+{

4*SElkrl

@'

gttg

3a.ls kr ,a4ok!

E(ampte: Three span conlinuous beam ABCD i, fu,ffi and continuous over B, cand D. The beam subjec{ed b bads as stloryn. AnalyEe t "

Oo, by slope deflectionmethod and draw bending nnment and shear brce t bgran.

Solution:

Since end A is fr(ed 0^ =Oe *0, 0" +0, eD =OFEMs:

-t-\

4s6o' @Cx,a,

- 30 KNU

+ 30 KNM

/

-13.33 KNM

+13.33 KNM

r - W d)x4 -

E -. w 6ox4,BA---6-=il_=

Fac=+f =+12.s!q\lM

-MFcB = +; = +12.5 KNM

E - wf 10x42'---lt=-- 1Z-=,n=*Y=*roio" =

Slope defl ection equations:

tI

ts-

' nlt*= r^, *f (zeoro" )

= _so*f {o*e")

= _30+ 0.SE|(L

t,t*=r.^*f€er*e^)

= so+f(ze"+o)

=+30+El0a

m*=r**f{ze"*e")

= rz.s*?€er-*e")

= i2.S+EtOa+O.SEt%

u""=ro+f@e"ro")

= 12.5++eoc{o!)4'= 12.O+EDc+O.5EIS

n_=r-*f{ze"*e,) .

= _re$*T@".o,)1

=_13.33+Et0c+0.5EtS

n*=r**f{a,*e")

= isss++eeD{€c)4'1 l3.3it+O.SEt0"+EtQ,

ln lhe $otle Equadons lhere are thee unknoirrns,

bondary condilions are:

i ttfBA+UBc= 0ii lfu+ll|"o=Oiii lroc= 0

Norlrao+ lt ac= 030 +E10"+12.5 + ElOBr{).sEDc = O2EleB.{.5El0c{42.5 = O

t

(f.'hinged)

--------, (r)

--------,(2)

--------,€)

--------, (4)

---------- , (5)

----------' (6)

El0B,EPc & EDD, accordingly the

------r(7)

ircB+t\r'c= 0

, +12.5 + EOc+0.SEteB_13.33 +Etec]o.SEloo= O

0.5E10"+2Et0"+0.SEt0o{).83 = O

Mo"=0

13.33 +0.5El0ci€FD= O

By solvins (7), (8) & (9), we get

------r(8)

------, (e)

ed' l""oq

1ElO"= -24.94ElO" = al l .'1 5

El0o= -19'99

By substituting the values of OB, e" and 0o in respective equations we get

Mrs= -30 +0.s (- 2a.04) = -42.92 11Nt

It o= +30+(-Za.Oa)= +5.96 KNMMsc= +12.s+(-zl.o4)+0.5(+1 1.rs)=-s.s6 KNMMo= +12.S+11.1S + 0.S(_ Ze.Oa) = +11.63 KNMI!to= -13.33+t 1.ts+O.s(-1S.90)= _ii.63 KNMIt oc= +13.39 +0.s(11.is)+(-18.90)= o KNM

Rs-actions: Consider the free body diagram of beam.

Beam AB:

Beam BC:

Beam CD:

ry2o*ry"8,D2

Re=60x2+5.96-42.02

= 20.985 KI{

.'. RA = 60 -RB = 30.015 KN

n" = 11'61'!9:!'99 = ,r3.s2 KN

.'. RB = -Rc = -13.92 KN .'. RB is downveard

{ou!-\ ilre"r& -- C..

no =10I j'?:JJ€l=

17.oe.KN !,

.'. R" ='l$:1tf -flo =22.91KN

t. ' .1,

6

I

t

l!,42

SOtrNrf|lo kxr!

J /{^-

!6toto .,,1', D ',

Mr*ry:- 6o -Q z'oa--33.30 hx,

SFP

E<ample: AnalyEe fte @ntinuous beam shorm using slope deflection method.dr bending moment and shear force diagram.

2t+ rf rrt

-.\\''lY

1o

Solution: ln this problem e^= q ...end A is

FEMs:

fixed

. wl2 10 xB2.Dtr=- A=--12 = - 53.33 KNM

F*=*#= +s3.33 KNM

."=-#=_Y= - 22.50 KNM

3?@h,,

- (s'rt -\2d+-i-)"u

9p ' tr'o4-lf..,o9"\f

l- '-Z

tt*f', \1'{ 1t:)/

Y">-

.ts

',

)t' ' 'r' E +$= +53'$ KNrtA, I EA-

FEc=-ry=-q9= - 22.s0 KNM

r*=* $= + 22.50 KNME

, Slope def,eclion equations:

t M* = r^"*f (ze^+er)

TU

--:-----, (3)

--------, (4)

Now, lL+lutEcl24 = ss.as+iero,-22.s*f, rn"*f, ere"*za

=+x.as+f;erq*f,ere"=o ------r(s)

7

l -,1-A_, \ -

i

Fli'-1

E

= -s3.$+E;31(o+e")

m*= r-*f{ze"*e") .

= -o-u*fge"re.)

= -z..s*!aU+?,

ucs=r""*f{ze".o")

= +25+T@c+o")

= +2zs+;E%+3 EC

'ln the &o!re eqllalin ttlee ate UD uford,n qard q. accodngly lhe boundary

conditi:ns arer

i -tuL^-r&-24=Oi ttlcr=O

and rrtca= z2.s +f, en"+f, ere"= o

"' |en"=-tt'zs-f ere"

SuH'hrting in eqn. (5)

-_-------_>(6)

sl.as+f; ere"-rr.es -lem"= o

+ne.sa+f;em,=o

... =,tr= -

*'i3 "

u = -17.432 rotation antictockrvise

.'. frcm equation (6)

ere"= |[ r r.zs-iCrr.sr,1

= -8.159 rdalion anffockyise

substiMing EIQ= -17.432 and Eloc= -€.lsgin the stope d€fledi,r eguatbn we getFinal iloments:

,*= -ua.* *l(-tz.asz)= €6.40 KNM

u*= +ss.ss+!(-17.€2)= +27.18 KNM

rr,"= -zz.u *f, {- t.esz1*f,1- a.,rce)= -s1.18 KNM

rrc"= +22.s +;Fa.rss1*f, {-rz.lsz) = o.oo

@!!9!s: Consider free bodv diaoram of beams as shown

RA 2A,te

Span AB:

Rs=27.18 - 66.40 +10x 8 x4

=35.13 KN

.'. R^= 10x8-RB= 44.87 KN

F

[:

- {,&ri:.! . .

2'f .18

BG:

n"= {Jt:39Ix = 20.s3 KN

Rc= 30-R8e6.47 KN

ax BM

Span AB: Max BM ocanrc ruhere SF={, corlBirer SF eqrilirt wih A as o.igin

S, =r14.87-1(X=Ox = 4.rlE7m

' :.M *= 44.tt x4.qEt -.tor4'0f -6{ = 36.67tGMSpan BC: Max BM occurs under pdnt ld

Bc i4,-= +s - 51j8

= te.41KN M

I

er",r,pnr A*l5e fte sfrpb hrE CEr - f$re. H A b frGd etd aG B & C GhhtgEd. DrnltE b€nfig tlEtEtQr'tt

Sotutlon:ln this problem .e^ = 0,08 16 0, 0c + 0, eD * q

FEMS}_ Wab2 1 20 , 2 ,4

= _t oo.oz xt ritr-r8=--T--=-- 6'

- wafu . 12ox22 x4 - *u3.33 KNMI-8 =+lt- =+--r-

- wl2 2ox 42F*=-h=-'E =-26.67KNM

- wl2 20x42e-=+ft=+=ff- =+26.67KNM

WL 2Ox4rco=+f =+f =+10KNM

r*=-$=-roxrunr

'Sbpe defrections are

u^" = r^" * ?Ea! 1ze^ * e")

= -,*.., *f 1e") = -106.67 + 3

E6" ---- > (1 )

,trm,

i M^" =-106.oz+ft-s.ael= -112.s6 KNM

M* = 53.$+f (-8.83) = 41.56 1111y

Mec = -26.67 + !f+.sl * f t-r3.36) = -4e.e4 11NUl

Mcg = +26.67 +]t-rS.sot*|t-8.83) = o

M- = 10+(-8.8t)+f,i+'u.alrt) = 8.38 KNM

M* = -10 + (14.414) + 1(-€.83) = o

J,

Reactlons: Consider fiee body dllgran of edt mell$els

,zole r$.10 zrlY,",o,

[,h =FBA+t(2o'+08)

= +s3.$ +# eos) = +se.aa + J

en"

q* =r*+f €e"+eJ \,4= -za.at *ff ,f;Qe" +e") = -26.67 *

f, ere, * | ere"

uo =r""*f {ze"*er)

= +m.az +f ,f;@o" + o") = a!g.67 *! ere" * | ere,

n- =r-*f {ze" *eo)

=+to+f €e" +eo)= +ro+Eros +lEtoo

16 =r**f@e"*e,)

= -ro+?€e" +e")= -10+Etoo +JEte"

ln the *cnre equations rve have three unknovyn rotations 0B , ec , Oo accordingly we havethree boundary conditions.

t\4BA +l4Bc + ttAD = 0i4" =Olvto" =0

Sotuing equations 7, B, & 9 we get

ElO" = -6.63ElO" = -13.36El0o = 114.414

Substituting these \ralues in slope equations

olS

---->(2)

---->(3)

---->(4)

---->(5)

---->(6)

----> (7)

---- > (8)

---->(9)

Since C and D are hinged

Now

i,L +ir"" +M* =s3.33+lEBB -26.62+;EmB +9Etoc +16+EDB +lEloD

= 36.66 +?EB8 +| ere" *|ere, = o

tvlcs = 26.67 +; Ete" +| en" = o

u* = -ro*lere" +EloD = o

)

Span AB:

Span BG:

Column BD:

n, - 4156 -ll2!6+l2oxz =2t.r7 KI,r

.'. X, =l?0-f,, =91.83 XiV

q = '19'9{+?ox'lx2 = sz48s KN

;.& =20x4-& =27.sisrc{

'L = a-ziq''u

=7.e2*il;.lL =12.781111 [.'x^ +rL =2s]

.\

. l.ri'".l-r1-t

lo

.'rlrt.

F19.

E (ample: Analyse the beam shown

antidoch ise moment cf 12 KNM.

in figure. End support c is subjected to an

l2kNrYr

*mSoluton: ln this problem 0^= Q .'. end is fixed

FEMs:

- wl2 2or4t = _26.62 KNMr*=- n=- 1i

-2F^^= +-= + 2o.67 KNM12

Slope def,ec-tion equations:

c

--------,0

--------, (2)

--------, (3)

--------,(4)0" and 0" , accordingly the boundary

A

dr

u^,=ro*f{zeo*e")

= o*f(oro")=El0g

u*=r**f€e"*en)

= o*ff1ze,ri= 2El0s

u*=r**ffQor*e")

= -ze.at +E!'d(ze,+e")

= -26.67+:EbB+iEroc

u""= r""*f (ze"*e")

= +26.67+;Ercc+; EleB

ln the above equation there are two unknorvns

conditions are

_ * r..u, * ?E.1.5! 1ze"re")

M*+M""= 0

Itl6"+12 = 0

. No, ,, trrle{+t4Bc= 2EOe_26.67+f ete"+f erc"

Jere"*fen"_zo.oz=o ---__-__>(s)and [rc8+12 = 26.02 +ten"+

f eoa+rz

From (S) and (6)=s8.67+;BoB+;EEc=o --_---__>(6)

Jere"*f eo"-zo.oz = o

?=n *t=o_-+-

EB.= +<Ox$ = a1a.72Frcm (6)

ll

ere"=-f[CI.e2* lt"nl)=-33.14 -ve signhdiubs loHi,n anticlocl(wise

Substitl,lhg Ele and ED.is slope &Eeclbnequafirs

M6= EtQ"= 1t{./! t11g[6ilar = 2Et0a= 2(14. till = Zg.i2 KNlnue= -m.ez+!04.74+:G$.14)= 2e.4a Knurr{c'= +26.67 +; (_€:r.rq+}6l.zy

= _tz roru

Reaction: Consider free body dkrgrarns dbearn

g

qSpan AB:

Span BC:

Qe Rg

11.72+29.44=11.04 KN

R^

&=

=*=-1l.oaKN

8.41+12+2Ox4x24

R"= 20x4-R = 29.&4KN

50.36 KN

BMDla! kll ,rt .

ffifl:l* the portal frame subjected to roads as shown. Arso draw bendins

r there

o, oD=

Hence

Q0"*l

unsymmefical.

lem 0^= 0,0"*

rg is uns

problem

t loading

ln this or

;al but

right. I

symmetrical

/ towards rio

reis(

suray

The tame

Assume sr

FEMs:

The

cirI

,Jare is a

t"=0

F - w2 10x42I ^B--7,=----:- = - 13.3i1 KNM

r -- wl' .10x42, aa- -lf = *-lr- = +13.33 KNM

r -- wl 9ox1o. r"--E-=-- 8_= _ 112.5 KNM

E _ wl 90x10'cB-;-= = +112.5 KNMSlope deflection

"qrition", "

rrr* = r* +f(ze^ +e, +!g)

= _13.33.T(r*"_,99j

= _13.33+0.SEq _O.37SE!6

u*=r**f(ze,.r^.f)

rs.sa+f(ze,*_"A)

= 13.33 + Et0"{.37S Et6

M""=rr"+f (zer+e")

= _112.s*#@e,+e")

= _ 1 12.i + 1.2EE+O.6E|O.

M"r=r"r+f (ze"+e.)

--------'(1)

--------, (2)

--------'(s)

tE.

1L-

:;

= +112.s+ff{ze"*e")

='l 12.5+ 1.2EEc{O.6 El0B

rrr",=r**f(ze.*e,-f)

= o.f[ze".o-f)= El0c-0.375E16

u*=r**f[ze,.e"-f)

= o.f[o.zr"-?)= 0.5El0c-0.375E16

Thereare3unknovns E10". El0" and El6, accodingly the boundary conditors arc

M.^$t'*=0ttl6r+fut"o= 6

H^+Flo+40 = 0

Here Hox 4 = M*+ i[o- 10xar. J,, _ M^r+ tV!*-80,,a- 1

and Hox4 Jr/l-+[{g5

,, _ M"o+[4,rrD--

,*.r"1* - ito+r4,.a6-6"4M^"+ttfu+M"o+[furA0 = 0

--------, (a)

--------, (5)

--------, (6)

Now Msa + Mec = 0

MemberAB:

u _ 61.10-1.88-1Ox4x24

=-5.195KN -,/e sign indicabs diec{ion

Member BC:

P" = &'52-61J0+9oxs

=47.3{KN.'. R" = 90-& = aB.34KN

MemberCD

4 = &{'54+54'z

=3.f.olt(N"lCheck

of HA is from right to left

EH=0Hr+Ho+10x4=0-5.20 - 3C.Bl+ 40 = O

Hence okay

BMD

[9

F

___-_-__>(7)

-------->(8)_ 13.33 + 0.5Et0r_0.375Et6 + 13.33 + Et0._0.37SE106 + Et0._0.375Et6+ 0.SEl0"_O.37SEl6 + 80 = 0LSEl0r+1.SEIO"_1.SEt6 + B0 = 0

Bysotving(7),(B)and(9)weget ------->(9)

ElO"= 72.65El0"= -59.64El6=66.34

Final moments:

Mm= _1 3.33 + 0.5(72.65)_ 0.37s (oo.sa) = _1.66 1111yMsA= +72.65_0.375(66.34)

= 61.10 KNMMsc= _112.5+.t 2(22.6s)= $.6(_ss.oa)= _or.1o KNMMca= 1 1 2.5 + 1.2(_ 59. 64) + 0.6(72.65) = s4.52 KNMMco= _59.64_0.A75(66.g4)

= _84.s2 KNMMoc= 0.5(_59.64)_0.375(66.34)

= _54.70 KNM

5+'to

Reactions: Consider the free body diagrams of various members

13.33EtOB-0.375Et6 _ 112.S + 1.2EteB+O.6EtOc= 02.2Et0"+ 0.6Et0c_0.375Et6 _ 99.17 = O6rd Mcsr.l\Ioc = O (+) + (S)1 12.i + 1.2E10"+0.6E10r+E10"_0.37SEt6

= O

. 112.5+2.2Et0"+ 0.6Et0"_0.37SEt6 = 0also M^, +M* +Ms, +MDc +g0 = 0

${q!

t"

Example: Analyse the portal frame shown and then draw bending moment diagram.

Solution:

It is an unsymmebical problem hence there b a s\,yay be tOvvards right0A= 0,QB;E 0, ec* O eo= g

FEMs:

'*= -# = -'o;:" = - 4r'67 KNM

'*=#='oi:' = +41'67 KNM

Slope defl eclion equations:

u*=r**f[ze^.e"-f)

= o1?(0.-e"-?)

= f ere"-f,ero

nr*=r**f[ze:*^-f)

= 0.f;(re"*-$)

="1ere"-f era

u*=r**f(ze,*e.-f)

= - +t.ez * 2E

!1'51(zer+e")

--------, (1)

--------r(2)

l.f

= -ar.oz+f ere"+fem"

u""=r**f(ze"+e,-f)

= o, .a, * ,.lr,u,(ze"+e,_o)

= 41.47 +1.2Elec{0.6 El(L

,*=t*-.?(ze"*e,-f)

= o.f(ze".o-f)= El0c-0.375E16

r,a*=r**f(re,*e"-f)

= o*3El[o*e^-99)4( " 4)= 0.5El0c-0.375E16 ----------r(6)

ln the above equations there are three unknown 0r,0" and 6and accordingly theBoundary conditions are:

MBA+MB.= 0

M6" +M.o = 0

HA+Ho= 0

i ^ M^"#* . M_#o" ,t..9_-r_=i, 34.'. 4(M^B {tlo ) + 3(Iul- rill* ) = 0

-_-_____,(s)

--------, (4)

----------, (5)

I

L

t.

E

I

,]1

I

I

l

l

l

tF

FI

M*+M"6=0

f ere"-f, ero * f,

ere"*! ere"-ar.oz

2.$Etea+!Elec-3El6-41.67 = 0 -- ------ > (7)

Msr+M66= 0

41.67 +1.2E10"+0.6Et0"+E106{)'375E16 = 0

41.67 +2.810"+O.6EDB-0.375E16 = 0 ------ > (8)

MAB+irB{ +

ltlcp+froc =O34

o ffi

er."-f era*f ere"-f ero].

3 [Ercc-o.37sEl6+0-5E10"-o.375E6] = 0

Ieru-|en.f ere"-f ero+a.semc-2'25E16 = o

8EqBr4.5ElOc-7.53E16=0 ------>(9)

By solving f0, (8) ard (9) vre Set

El0"= +25-46

ElOs=-23.17El6;+12-8

Final moments:

' 125A6-?*12'8=8'44 KNMMo=i: 3

^ "*i.AG -?112.8 = 2s./rc KNMM*=A,

3

um= f , zs.ls + l*1-ze-'v1-

t1 '67 = -25'40 KNM

iics= 41.67 + r.2 (- 23.17)+ 0.60 (20.46) = 28'50 KNM

irco= -23-70 -0.375 (12'80) = -28'50 KNM

lLe 05x -13.T0 =0.3T1x1?.00 i 10'0f, KNM

t ls-

Reactiohs: Consider the free body diagram

llemberAB:

\"t AHr= 25.40+8.44

=.t ,t.28 KN

Member BG:

MemberCD:

E2A.5-2O.3O+20x5xY

22

= 51.64 KN lo.6(Rc=

... RB = 20x 5 -51.&l = 48.36 KN

28.5+16.65Ho=

LL,4= 11.28 KN

C *>-ou* tAD'o

Sato/^'u'kn ryt

t

Example: A portar frame having crifferent corumn heights are subjected for forces asshown in figure. Analyse the frame and draw bending mornent diagram.

I

2t!

It is an unsymmetical proUem0^=O0r+QO"*QO'=9, hence ttrere is a sway be towards right.

FEMS:

- W 30x4=-15 KNM

- Wl 30x4rs^=+8-=+l-=+15 KNM

r-=-#=-ry=-30 KNM

r* =*f =*@Ja =+3oKNM

F6o = Fo" =e

Slope defl ection equations:

r'a*= r**f(ze^.r,-f)

=_rs.ff[0.r"_?)= -15+ E|0B-0.7SE16

,*= r**f(rt.r^-?)

=rc+ff(ze".o-?)=+15+2Et0g-0.75Et6

--------'(1)

----'----r(2)

)l

M"c= FBc+?(2e"+0")

= -3sn..?Er31ze,+e.)

= -30+2El0a+El0c

ruo= ro.*f 1ze"*e")

=.oiElz1ze"rer)= 30+2El0c+El0s

r.r-= F-*?(r**-?)= o.f(ze".o-f)

= f,ere"-fero

u*= F**?(**-?)= o-?(r*"-?)

' = fere"-!en

There are lhree unknorns, El, OB,ElOc & El6 ' accordingly the Boundary conditions

are

MBA+MB. = 0

M"" +M"o = 0

H^ +Ho+30 = 0

,.", ***Foo *McotMoc 139 =g

3(M^" + uo.)+ +(M", +M*)+ 180 = o

Me,n+M""= 15 a 2EleB-0'75E16 - 30 + 2El0'+E10"

= 4ElOs+El0c-O'75E16 - 15 = 0

Mce#co= +30 + 2E10"+E10"+ f,

Ere"- J

ero

= ete"+*Elo"-f,ero* oo = o

.:

--------, (3)

--------r(4)

--------, (5)

-- -----r(6)

------r(7)

------r(8)

;.

3(M*+Mr ) + 4(M"o+Mo") + 1 80 = 3 (- t s + Elee-o.75E16 I 1 5 + 2Eq-O.75E6)

* il lerc^ -?era * ? ere^- 3 ero) + rao\3 " 3 3 -3 )= 9El0e+8El0c-9.833E16+180 = O ------, (9)

By solving (7), (8) & (9) we get

El0"= a$'577

ElO"= -7 '714El8=+20.795

Substituting these values in the slope deflection equations se get

Maa= -15 +9.577 -0.75 (20.795)= -21.01 KNM

Me,t= +t5+2 (9.5n)-o.75(20.795) = 19.55 l$lulirec= -30 +2 (9.577)-7.714 ='18.55 KNM

Mce= 30 +2 (-2.714)+9.577 = 24'15 KNM

u"o= f 1-z.zre >l!o.tss'l= -za.1s KNM

mu= ?s-:t .t ul - f (zo'zss) = -1 e.oo KNM

Reactions: Consider free body diagrams of the members