SUMMER _ 15 EXANtrNATIONSSubject Code: 17604 Mode! Answer- Design of R.C.C. Structures Page No- ol t aflmportant lnstruction to Examiners:-. .

1) The answers should be examined by key words & not as word to word as given in the model answersscheme.

2) The model answers & answers written by the candidate may'vary but the examiner may try to accessthe understanding level of the candidate.

3) The language errors such as grammatical, spelling Lttorc should not be given more importance.

4) While assessing figures, examiners, may give credit for principle components indicated in the figure.

5) The figures drawn by candidate & model answer may vary. The examiner may give credit for anyequivalent fi gure drawn.

5) Credit may be given step wise for numerical problems. ln some cases, the assumed contact values mavary and there may be some difference in the candidate's answers and model answer.

-6) ln case of some questions credit may be given by judgment on part of examiner of relevant answer_ based on candidates understanding.

7) For programming language papers, credit may be given to any other programme based on equivalentconcept.

lmportant notes to examiner

I ) lN Q.NO-2(a) the span is not mention clearly weather it is effective or clear spahence student may assume any of this two so accordingly two solutions havebeen provided which should be asses and marks should be given accordingly.

2) lN Q.NO-3-A-(b) in this probleni unsupported length off column is not mentionhence the checks for minimum eccentricity are not possible so if the studentsassume it and given the check for minimum eccentricity full credit should begiven. ln this model answer one sample calculation by assuming unsupportedlength have been given for reference. Also students may go for finding axialload directly without giving check for e.''ln in that case marks should be given.

3) lN Q.NO-S(a) in the given problem M6 is less than M;;,n hence the section cannoiL be design as doubly reinforced and it is not possible to calculate A"tz and Asc.So calculation up to Ast{ only given full credit.

4) lN Q.NO-S(b) for two alternate solutions are provided one by assuming givenspan as clear and other by effective.

5) lN Q.NO-6(e) in this problem the given value of factored load is not sufficientand giving negative value of Asc therefore though students calculated Asc fullcredits given to students

Subject Code: L7604

:e

SUMMER - 15 EXANIINATIONSModel Answer- Design of R.C.C. Structures PageNo{U gF

Q.NO SOLUTION MARKSQl. A} Solve any THREE T2

(a) Solution: Limit state may be defined as "the acceptable limit for the safety andserviceabiIity of the structure".Types:

1) Limit state of collapse.2) Limit state of Serviceability.

02

0101

(b) Solution:1) A normal section plane before bending remains plane before bending right

upto collapse.2) The maximum strain in concrete at the outermost compression fibre is taken

as 0.0035 in bending.3) Concrete under tension is ignored. Tension is to be taken by reinforcement

only.4l The distribution of compressive stress in concrete across the section is defined

by idealized stress-strain curve of concrete.5) Perfect bond exists between steel and concrete right upto collapse.6) The design stress in steel reinforcement is obtained from the strain at

reinforcement level using idealized stress-strain curve for the types ofreinforcement used.

7) According to lS code, the maximum strain in steel in tension shall not be lessthan:

0.002 + fyl (1.15E) at collapse.

Any four01 foreach

(c) 1) Building with lateral load resisting system comprising l)A ductile momentresisting space frame or ll)A dual system consisting of ductile momentresisting space frame and ductile flexural {shear) wall, qualify for low seismicinduced forces.

2) A frame of continuous construction, comprising flexural member and columndesigned and detailed to accommodate reversible lateral displacements afterthe formation of plastic hinge.

4M

(d) Advantages:1) This provides a type of construction which is always free from cracks under full

working load. Due to this reason, such type of construction is suitable wheregreat danger of corrosion.

2l Sufficient horizontal compression due to prestress reduces principaltensilestress and shear resistance is developed without heavy reinforcement or largewebs.

3) Less quantity of steel is used as high compressive strength concrete is used.4l This type of construction saves material as well as dead loads and hence

considerable saving in supports and foundation is also achieved.5) Smaller sections are used hence can be used for large span.6) The cost of shuttering and centring in large structure is reduced.when

prestressed precast eleme4ts are assembled.7l Deflections of structures arb reduced.

,8) lt saves time and money. '

Any two01 foreach

dr

g L Subject Code: 17604

al

I

SUMMER - 15 EXAIVIINATIONSModel Answer- Design of R.C.C. Structures Page No- e$l 6V

Q.NO SOLUTION MARKSDisadvantages:

1) lt requires high tensile steel and high strength concrete; hence it will becomemore expensive for smalljob. '

2l Prestressing is tedious job and also requires skill supervision.3) lt does not increase ultimate strength df concrete.4) lt requires special equipments such as jacks, anchorages, high tensile cables

etc.Different types of losses are occurs like losses due to creep in steel, friction, slip ofanchorage etc.

Any two01 foreach

(e) Solution:1) Vertical stirrups2l Bent up bars along with stirrups3) lnclined stirrups

4M

d' \:.

4i!*

SubjectCodet il6OtSUMMtrR - 15 EXAMINATIOi\

Model Answer

SOLIJTION

PageNo,!,u{MARKS

**ir**t V"e" c**f\ #Y*) e-

e' *slc "2**I

t*eX }-.s*fi$i*A

J.'rcr,.rn ? <> C, i"l-l rn >< SclO rn t-?- e;rj-€-a&r'ar<-,fl6:r* *rrt* L"b*'-v-e-c{ B, t*T * F-

I 7 6 k N,,rn " i-€ d cJ nc-?e-4-e- )<1 20 3- s;te-ct-l

*s,-ax**$e "fTe ffie>c-) #-'3-*e- rJ:;ec*? , d&'"nci CR.?r"-e eL

6 W -5f-e""*'| ,

*#i:> Iul-b*'c.rY) **j"

<3p l'\re 1-'r !> t: '3 e O vrnr ro, ol = 5C:O rarr fir

f4Lj =H4t r*- = -z- *' lar"{. ' mL ,. 'F' = 6 c'o N / fn rn 2-

-ffi n.l l-*1 u."

&, 133&X# #' 133 X 2**X ?C *x5*e2*

fa:: f Sg g "Xlcr* r..5 rr1t-4 u" 1 ir'r l3S'6 FqF{,rr)rn }

,'D sinc-e.LJ nrJ e :r-*

l*3

6, % -lr-*-K-

€7Fs

="o"*''*-#'*-**"'*

'*ctc bdz

€ 4*- 6x tY Bxl$6"

4.!

SubjectCode: 1'l 60tJ

a.No

SUMMER - 15 EXAMII\ATIOI{Model Answer

5l 3?Page No: ---/ ft

a

SOLIJTION MAf{I(S

in€* r-flcp t*)c) t* r?r-e gis"

t11

-#u,c_g{r*ya I _* LJ ndsg f.3'ff in {a"ar"C e C scc$g{:L I-lYl --itl

t\1in fc, ?rL e *l

-'o u

>a-->yierl*e.$ i.=; 4 [=q-'r:s <=F. ld nnrr: J ?qrl*l-ra'

in t=ea

i5cs I u-u*hJ'*rrr -*2,1

c*i v-e'rr - A*l. = 4- >< ,fl] * t <-* = 8o4-'24 rnni

b 5 SClc)-bry: Y?r"1 a = 6** ii"*r")t-"-l

-{* w r 7* ** lf"frr**l*\LiJ Tb {".'n* e*a"brA*J t*{,A" ?t L{

c'g.7 €- As*#, 3G Fce b

o,87X6aaXAa4*'z+ : f<1'qe nr

r}r\& >r 2-"*>< SclC)

*T*c> 4"{ ngl}

d-tr fxr*'3c ,-- Q-'&6 A : n'4*€\6CJQ

r* u rrn#->< ": ?*7& fT\ TY)

A* &LL 1- -X*u rfio'"x'

SUMMER * 15 EXAMIi\ATIOI{ o6lavPage I{o:

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NSubjectCode: t-.t 60tl

a.No

Model Answer

ff lc>**ry* 6\ n is' I'r -:

*}* <F g' .-eJ **rJ cs"**C* t> e si5 'n fc"-v1 st<a"vtK-

# ,/+-8 J r # "/t"-gXJ

c*J

MAR.I<S

dI*><

r *?*tc:,

d

d

*}"#

"#r* !V€".-

"fi g>irae Lalm-j -f3*: I I &q-i; in

rfl 3 1-*6YT L";t,-€ l*o-c* =4 kr"lfrrr

q tcucq1; "l'lj{ sl"r :---l h< I Txr'z*r Cclnc

m,F : I '4*. -gk-e*-*'-n f/-g oF*

s t **"fut "" F-c, {"'-*} i n

c h.ec K-sJ

n a-t->rre. r:"ac, b)€ fvr | *- f-s na t: cl eq-'t'\

err figz.r ed c'= ecq-tl^r t -r- ' --*-in e' -s Pcr"-rr i S cl €cr"

*-d eFtr<qt:i v-e - he'-n c*e et+ul*d e-"*t caln

*kul v*e* *f-{-. i := r*v-p i;-lc"r*v*r

I b4q-6- k5 *s l^'r *"^l d Uz L g \ ^'e--r) .*-,pCs*d i

Sol tr,.bJc.rr- 4 - T"€ cn-i@

* L*Sfr * 4* "F"ffi

L i rre l*:e "** # k"r--r t rt- L*

5*c, Nf w"*: =, &l s f.-l I rnYY5

f*1 ,ff :* I 't

T}{ 1'4-:1 6o "7

Subject Code: 1T b 0 L\SUMMER * 15 EXAMINATIOI{

Model Answer

cFlsrPage No:

--/ I\

MARI(Sa.No SOLI.JTION

e?r *J 1 Je.p*tr J" -{* +

&-5.*'*e*/-rff1 ing lL + f'rrc-J'n t*=o-"'fl3

= \ 6o-7

Ja''ril = P-:Ja-*

1 $ry a- 70 *- 12* <+ 1 C + r?ntra

A a..,n,lo*,{ot - a l-€+YT-J]fT)

= 4 5e* y?nru?

: 2 SX O 'lSo= 4'75i

L f \re- tcpa d A* tr-t"+ t Hn'z*

f, lc>clr {i1 n \skr teruf r^L

o**d vu = L,L

=- 4,7 5*t-- ,7 g kr* lrn

ex-"r6Eye*l Loe.*d: Tf K\nl

knJ

'tF*_Ctgl--)--q Cj

\i.

Subject Code: 1"7 6 $qSUMMER* 15 EXAMINATION

Model Answer

0E lgrPage lYo:

-l

I\

SOLUTION

37,olx1 o 8 = o'l3gX"-Xlte,txd2-

& eft rqc*i n stT {

F +'6X?"1huu = o'v 4*k- f;"-4c-fi:- b4^

+ '(X 37, olxrei-4 o' 6x 2- c)

zafi looOX \6&

= 6A+'dC r^ry?

sp>*-CtnT df* l2--nnrda cp [=o-rfl.5

c F- &€b xtoclov

= l l 3' oJ rn1n2-

MARI{S

6F4 'Ge

44 *'f"t'

SUMMER _ 15 EXAMTI{ATIONModel Ansrver

S A r?ra.cl->C S +U; e lrr\ Y1-l

?J e*. 6 rraTn F f",s-,r @ _S_?_

o5 lstPage No: _-_/ N

Subject Code: t? 6 01a.No

t

\j'

MAITI(SSOLUTION

a.riJ e, \ ? vt..rn $

/\--.c & eW srfe-*4 ?*r1".ry}C>c>x tr 3a:

1$clox 113'(Js,

6 g r 'zg '{.n Ylrr 2*

r *.* Or\5 bD €w # 2-6c'd

= c>' l6 r.z. loooX ldO : Zg-*, rntnL

_e *i:J otr 6Y- l-?n nn, 5 'lo*-Y.s.

&tb- b = TA-xC\-r c:*e)A=oa

:95zo5;ry g5 rn.M

s r-Xl 6 1- crt+LT)

?m.* Cl (

SubjectCode: t-l 60Ll

a.No

SUMMER- 15 EXAMINATIONModel Answer

tglgfPage No: --/

N

st

MARKSSOLUTION

C>€4::e; I t r'1

6 nn{ J is*. b+:r= 6 vTIrT) C/C ,j

t%rn'"n & nn ry n bcn'..rl @ f qf .T.'- {5

+5sc> rytl?

qff Co f1 std eY{r.r

SC}* rr\ nn

x.lfdffi eg-* y 3*3c?T\rr)

pe sTg

***4 c* *^-,,1 c*fe- s Ic*J: *ftaTck?aes*s

:-exl ',+

€-Gn--p?J e*=u'- *d"i &, c3 p rry1 *J t\ b$g&_

I l"t

Subject Code: 1f b 0 \SUMMER, _ 15 EXAMII\ATTOI\

Model Answen

nb+Page No:

-/

i\

SOLUTION

J's*v*^Jol :f g*-J*cP**6* \64*|-=rv"Tlt

c-"fo*f ""r*fl* g Y:el''"'?-l

)"-r : *.1 + t : 46oq *ZZo -= 477o vnv>-t

p)-4- :: -l+ A = +sa <>+ l3l = 46 G 4 Yv:"-"t

*T*cx.-ts:-e* Jep{: 4,€6+1"}1

o*d cp**-4 losd = PEr> ::76Vo'l:1 :4'75F4Lf rr-e t *a-cl

f t#c3-*?'- Fi'n I *kr

\rr/ : 4'7.5 trrJ

W*l r l'5x-S'ZS : 14'625-Htrr

ex-*{::r*a*e d tt*-.x

1-{ J

d{ e{"*

SubjectCode: \T 6OI

a.No

SUMMER.15 EXAMINATIONModel Answer

SOLIJTION

: 3-}{ 16+ dv ?#e rn l..rl

} a e rrt \rra

lzl ETPage trt{o:

-/

I\

rrll:

rj? 16()'l\

Ast-f s753'33rnrn> \i

MARI(S

Re<{\^-j =€*+ e:ffq1 *&J -lf€* ;e"P'X}

: F"f *, t d r^ -: ce, lffit*f - 1* ff"-fie+

>g .T 6X\o

ffio"cr2 m' <aPra-avidc4 s14fl

ff-rrect* aff* ftr c..-i n €t-e

&sy = a'5'f*K l-r **.? bAz

r* O'5 X Zee

P luY'-?-*+ kte-?rs

5 = Cf,sb ooCI : T/4xl>Ash 7 + l'7*

: 16L"55 Y I 5A fI1 r?1 S

5 mn* = ZA a"€ 7aa rnr^ +f i<h Zuu''

SUMMER _ 15 EXAMII{ATIOI{Model Answer

\"lerPage No:

-/

NSubject Code: \1 6 0t

a.No

lI

iI

.**-,{

MARKSSOLUTION

d &-- F* ?*5<3

= O' )g xr loOO X 14O = >PErnrriL

T.-T _g k'cq {y *spq*c*-lu1r\a a? 6 Fr-lr'*r

TT/ox 8*#>&5

: $3,2** rlnYr.l g5 rnm clc4qxiTtn u,4^rr "j Qq*"n"-,r. n 5 *n of + 5* rnrn

= *gX 16+w. A_SC

vr)rz- 6nnrn & ba^r(.b 15 rntn cl t

6 nn"'r & bo-r( )-{sf, \ G.l I I nn:n cl

5o TnTn c

+fuOw1'ry-)

SUMMER _ 15 EXAMII{ATIOI{Modetr Answen

tAl lsqPage tr{o:

-l

lo{Subject Code: l?607

a.No

e'-F rrl I r' v-€ I <e cq*cl ? P-*l

5S Taa i Y*\ nnda.,-il n bcuos'

iry1 1 fl c*"$ C{;. ;*-(3 "fY1 YT-\

.II i --1 -;-- -,-- i r, D = |95,71 Y \gornryl 5-?J

es,l qn ar'vef nft-zrceS Co-r-''c-"*e-,f€ E I 4-02 Pc\ne

% Y-r-l 5;f r?rptY ,S

d-t 1 -+--Le- -{bcfrf sid es' ,l+ ha9-}-c' Caar-Y-i e-S

i v^e- t aod o{: + errlr'{Lf n .^JJ d+lo-rr -f* t+s

loa-d , ( rse t{ 2s conc-ffi-J'€*f-e 4-15 o

e-f - le*=i--h -tt-* c/-s o&:td *l tn

nlo che6p-;'t Use- p(r : o'oG2*'PdY:0"060

LV 8' 6'3 .*? I *

t

L*?c* 4 r Q

lim z O'17& *c e adL *#a-r tre +ls

fn c.^-l c '--0

€#T ffe,+ \S

*|FxL,lb - \ao .rt n

SubjectCode: t-+6otlSUMMER* 15 EXAMINATION

Model Answer

lslg+Page No:

-l

h{

MARI(Sa.No

=*{.rVJA x €-y* : O,0C Lsl*,6rsy,

'___| l'14 = -\ g'73-q N " ttl , ----* i7 '. '' i

r;l| | i 1P -----. l.

.---f 1-1-;-= oan.wj-*it;;_-z J

1"1y4 = |g.oSC trN, 'yol1 l

'((

SOLUTION

-z*ffi 4d f a)Aen-

7t

'a.*<e- +1. ic Hhes.-g d Psc^-t:

J*' s ),^+f s{,s+o'Lb = 4,79 nD

q = -)oc-l-J = 4'-<tO'165 = 4'6(5 rn

Tc.L}+<-* Sr^*; I ezr a F- a--,1-,.rV--e* fcq:O

Je* = +'6e 6rn

c'o--$ c-c.-* c. c^--Fe, +*^

o a A 1le^A

Lt' \r€- | r,qdtoozfinlsh,: I bt"\ FnL CAss'{""-t€*

I *a-d : ,T s kt*."i1m

,t^**'--<4 loa-d = ff1 XW : l'SX-9'79wd | 4 '6 %6 tr-t-::l_rr?

F 1s*>( T.}.l{

SOLUTION

= o,13

l3173y?od0 ,l3g Y 2 6X 1o on)

35r"cL q nq e? €1€.g

A*fu;r- 5

5 (}'5X2b # lrb4'€Xlg'T3xlo

LA.tx- = 3*3'zc"-mr"-_

*5 Ct'r}g u'7 le rn'nr) + W*.q TwxloL

**<'ar

Subject Code: \"601a.l.{o

SUMMER- 15 EXAMINATIONModel Answer

rn"#

lelsTPage 1\o:

-/

I\

MARKS

X LLS rr)Y*n s L</cx p*4e s+

$oLE> A*t-"

a4 *t-t

SUMMER * 15 E,XAMIX\ATIOhI

Subject Code: t'-t 6 btl Model Answer

SOLLJTIO}J

1*"z*,qo -tc dft -s{cx-Jc> *; gn *.shcs

a.No

C}'SJ( 2S

ffiss,x l6fiT-

sfa c-{'i:1 ** t- w y-}ny'?n + .d { a- k=a-er-s

Tax s\r ooc)3ffi+'89

= I +l' 6 +|arlq tul f

c/s -? sla*z -sl.rosstn3 tsZf de-t+l'l-s

s hn rn 6 ($ t'4o m rn c/c- t o rnrn f @r s;so nnt' Ve

%.LE'r*Y)rr)

t+l 37Page No: .-/ l{

MARI(S

z-i"* t-

jii

1

_*L_.-__ "*- - -j

u'

SUMMER, * 15 EXAMIIYATIOI\Model Answer

SOLUTION

le *s* c"Fa G.* L*&:etff. t 1o*":j

ls lsryPage tr\o:

-/

l{Subjcct Code: t" 60 \

a.No

es;\

MARI(S

s f '# 11n__* s,*- ^L:I .*r*FNIIlY)

nish -- (P"5 rN In.?=s ry:-a- \iT\*qJ 3

2-gO rr-l'ff\ X 2*:7 a rn )-CI.

ls syc.4 , gts<+=lr -ft-e cls aF c hajJa'sh acs:i ng g-fc" eJ -lScel c*s*b4#n I

e., WAz i*'rsexfrE bdl

ffi=*,e.t rn*-4r*:"yr u & -5r I *uJ5, +tfi T*k- 14 tr p5,

*1*.#*-4*e T-1 , P =. t't -trc.rtr fg1 41S

ffiU a \a6+',2o-t- V:|e.::: | 3# rT1 r-rl

tr {k *-W: ar'-t-

J -e : so ffi2-; s; fo^r1 y".l f!-#\*r# l' C: ffi3 r"am"J

ifl

Subject Code: \T k O 1a.No

SUMMER- 15 EXAMINATIONModel Answer

SOLt-lTIOl'i

te)s?Page lrlo:

--l l{

MARI<S

# pen***} tpsyd -s" 3s* p : ?*-m"Fr c3q tbC"}L*€3 *-*& * c{ ****fi' *.-*fi C' a'J

#?

p.- L *'3'2*S -le*t-":l.n

, t-i#lpm : l X 1>r C!-** ; rli* .S 1crs lYYr

-fabe4 1at.,.r-{ W

4\YS X\'{ * -('12*S htslnf

F{*e tr eJ}d >,'' e**r* y,|ruffiW I's-$%

'tr"TA *:- ?.ffi*,* tsf-m

AP ?*1\rerd *-t | "*f*r

s '1?g'Y t)*eu v'l srclg

Subject Cocle: t? 60 1

SUMMER - 15 EXAMII\ATIOI$Sfodel Answen

SOLUTION

Page*o,Sl

T':t*"€.e, " *k # #'?** n*'*"1 vr 53'

e-s-)i-tie h €"V-e""-i.} e-# rYa l"Tt i s slnal lq:f

: "9,"{ I P 5 €"r"-t-* ? d} c} 'rY*\l'-'"f

+.6X 3"g2-X lodec o*-gX?*ft,j

;LeX f CIesXluffi

s l*z 'Se {T1 Fn ry

irn i- &tI2*-

r lg& rlntufllZ**G

h* A.v 4 Asr*rnrn F>vfAs* Avt'=l{<n6

-6 c:{ T9 *& # trnr?rl eF k}dq'?rs

0** ..# {rlq-xg&-X toc:>m

Asu 15*€

SUMMER _ 15 EXAMII{ATIO}$Model Answer

Ll lzrPage No:

-/

NSubjcct Code:

a.No SOLUTIO}J [MnrtKs iit

?^f."6$-- o? "{ rs +di

;rlg€nnrn'z''\ F gvn''..r & P:.2-a'g er-sed @? -l bmi-

5 L z2g*-'2-"1 rh'-r)

4 r-ancs>< s 6 A cy.v,' 4 6e Fn w cs)FarcFr€v^c.--

: "6 >( lC S Gltr

: 4 {b k-n ,v--1 r

--ur1,ce- p6zvf Lo- plnt"t{ @ }>o^- 4c

fA s €e- -4 l3"hr1' bL"41 S

a.lS s;fiec-{-i \ry ?/ P *A e*."J l_s

*l c

g fnrt1

l;ili

I

.J, - -._---L--_* -- ---:

I

S{JMMER - 15 EXAMINATIONSSubject Code: I W"4 ffig gl,Aplwer Design of RCC Strnec{ure$ Page No TALT

Q 03. Attempt SlfKf,$q/K *f f*Hfio

Conditions of formation of flaxrs

1. The slah shall bre cast tr

effective ly bonded toget

2. If the main reinftrrcen

reinforcefflent shall be p

percent of the rnain reinl

Effective flang€ wiCth f*n 1, ftlFor T beanr:

For L beam.

Wher", bf = ef f ert-ive wtrJLtr

Lo = d"tstunce hetweebw = widttt of webDr = thi"cltn€ss of f Ia

To find rnoment of nesistanse

t.. ,,[^ t' [-b[.f *'j6

L",ht'= , ". +- h

'tl./-t- l.*

Data Given:Df = I20mm, b1

fcn = 25 lrlP{r,

Xu*o* = 4.46Assum e, xu

x 1100mrn,

fv * SsoMx 45{} x Zt}7wtffif

"' xu = 105. 45 TTTryt* ffi

... Mu,473.54

Mur" =

= 0.36 x 25x tS6 Nwnm

Mu, :

'*" fu(u, tr 0'.}7 X j

**"J&r, Y *3fr

SOLUTIOI\ R{ARKS

wing:(04x04:16) t64pgpd beam-:

ntegrally with the web, or the weh ancJ the slab shall be

ther in any other manner; anct

:rent of the slab is parallel ta t}"r* beam, 'transverse

rrovided such that reinforc€ment shall not be less than 60

iforcement at mid span of the slah,

nd T beam: ,.

+ 6Df < Actual" f lange widLtt (tr)

w -{- 3Dr

cf f lange?n points of zero moments

'nge

2

1

1

qf T-beaJn:

b* = 275ffiffi,{{)sm

" 0,87 x fy x Astt\r''r 'L7t 0.36 x frrc x bf

0.87x500x2400Y r'uu 0.36 x zs x 11oo

u dso

. .A.ssumptton is correct.= 0.3 6frrcb7x" (d - 0 .42xu)x 1100 x 105.45 x (450 0.42 x 1.05"45)

Y& = 423.54kNm

= 0.8I fyA tt(d 0.42xu)500 x 24A0 x (450 0.42 x l"{}5.45)3.56 x 106 Nmm = 423.56ftffne

d = 450ffirr1, Asf = 240Ammz

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' SUMMER- 15 EXAMINATIONISSubject Code: 17604 Model Answer Design of RCC Structures Page No J

ztl*r

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c) To find development length in tension and compression.

Data Given:

A = L6mm, fy : 4LS MPa,

rbd = L.2 MPaf orplatnbarstntenston

Development lengf,h (Le ) in tension:

0.87x frtALd=ffi...(raa has been increased by 60% due to HYSD bars.)

. r _0.87x415x15.. ucr - 4 x L.2x r^o = zs2.L9ntm

Development length (Lo ) in Compression:

0.87x frxA,o:@...(rna Has been increased by 60% due to frySD bars) and

...(raa Has been further increased by 25% due to bars in compression.)

0.87 x 4tS xL6La=ffi=601.75mm

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d) Development Lensth: Development length is the length required to develop the stress

in the bar from zero to maximum by transfer of stress from concrete to steel.

Factors affecting development lensth:1. Grade of steel,2. Grade of concrete,3. Diameter of bar,4. Design bond stress,

2

2

e) Design of column:Data Given:Factored load Pu: 3000kNI,Concrete M20 and Steel Fe415, unsupported length :3m, Assume pYo: lYo.

p: pWl00: 1/100: 0.01,. pu = to.+f,u + (o.ez6 - o.+f,y)plen

... 3000 x 103 = [0.4 x20 * (0.67 x 475 - 0.4 x 20)0.}L]As

" An = 280360.73mm2

Side of square column : rl7FF36L = 529.49 * 530mmProvide column of 550 mm x 550 mmArea provided An: 550 x 550 = 302500mm2Check for minimum eccentricity,

,,3000 550 t€mina

500 *

S0 =24.33mm

0.05h:0.05x550 :27.Smm i

Here, e^in < 0.05h, therefore the column can be designed as axially loaded,. pu = lo.+1uen + (o.oz1, -.0.4fck)A.,"1

'

... 3000 x 1000 : [04 x 20 x 302500 + (0.67 x 415 - 0.4x 20)Asc]:, Aon = 2!47.7Smmz

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Subject Code: 17604SUMMER _ 15 EXAh{INATIONSModel Answer Design of RCC Structures Page No I z4J*y

Provide 8-20mm bars area provided :25l3mmDiameter and Pitch of lateral ties:Diameter of lateral ties )20 14 - 5mm " and,

Pitch < 550

< 300mm I

Provide 6mm lateral ties at 300mm '/.

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Qoa+q,1) AttemptANY THREE of following: ( 03 x 04 : 12) 12

a) Methods of prestressing:There are two major methods of prestressing

1. Pre-tensioning2. Post-tensioning - (a) Intemal prestressing , b) External prestressing)

Pre-tensionins:In this system, the tendons (steel wires, cables, strands used to impart prestress toconcrete) are first tensioned between rigid anchor blocks then concrete is subsequentlyplaced and compacted to the required shape and size. After the concrete hardens, thetendons are released from the prestressing bed transferring the stress to concrete.

ORPost-tensioning:This method of prestressing is classified into intemal and external prestressing.

In intemal prestressing, the untensioned wires are placed in the duct of beam duringits casting. After the casing and hardening of concrete the wires are stretched againstthe member itself. Once desired stretching is over they are anchored at the two ends.

After which the space between wires and sheathing is filled with grout.In external prestressing system, wires are wound around the existing structure and

prestressing operation is done to protect the existing damaged structure. In this case

extemally wound tensioned wires, cables are encased by dense concrete secured to themain concrete.

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b) ifv of colLoadoad carrvrng caDaclw oI column:

(Note: In given problem unsunported leneth is notmentioned, and hence checks for minimum eccentricityare not possible without the value of unsupported length.So if students have assumed it and given the check forminimum eccentricity full credit should be given.)Data Given: Column size 300 mm x'450 mm, Reinforcement 4-l6mm0& 4-l2mm0,Materials used: M20 and Fe415.

As = 3Q0 x 450 = 135000ntm2 ,

fi I^ TE

Ar, : n " ;x !62 * 4 x nx L22 : L256.64mm2

. Pu:o.Afa,An + (O.eZ6 - 0.af"x) X A,"Pu= 0.4x20 x 135000 + (0.67 x 415 :0.4 x 20) xt256.64

' "' Pu= L4l9.35kNoR ''r

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Subject Code: 17604SUMMER - 15 EXAMINATIONSMqdel Answer Design of RCC Structures page No

zsl3h/

Here one sam le calculation is given by assumins unsu orted length of columnas 3m.

As= 300x450= l^3iOOOmm2

Ar, = o "Ix L62 * 4 xf,x n2 = 1256. 64ntmzLh,

€min = 500 +

30 or 20mmwhi"chever i.s greater

3000 450€min = .500 * EO

:ZLmm or 2\mmwhi.chever i"s greater

Also,e^in - Zlmm 1o.oun = o;u?t ; '='ffi^*Pu : o.Afrpl,n + (o.oz 1n - o.afrrc) \ A$

pu= 0.4 x 20 x 135000 + (0.67 x 415 _0.4x20) xL256.64

"' Pr, = L4X9.35kNPqrtial safetv factors:Since the safety ofprincipal design factors (viz. load and material strengths) which are not dependent oneach other, two different safety factors, one for load and other for material strength areused. Because each of two factors contributes partially to safety, they are teried aspartial safety factors.

l. Partial safety factor forLoad (y)2. Partial safety factor for Material Strength (7_)

Values recommended in 15456-2000 for concrete and steel.Material Partial safety factorsConcrete

Situations in which a doubly reinforced beam is provided: (ANY FOUR1. When the applied moment exceeds the moment resisting capacity of a singly

reinforced section.2. When the section of the beam

room, appearance etc.3. when the section of the beam is subjected to reversal of bending moment.4. In continuous T beam where the portion of beam over middle s=upport has to be

designed as doubly reinforced.5. When the beams are subjected to eccentric loading, shocks, or impact loading.6. I-f the bending moment exists over a relatively short length of the beam o-nly,

then the cost of additional steel is small as compared 1o the cost of singiyreinforced beam of constant section.

7. compression steel is sometimes provided to reduce the deflection.

is restricted due to the requirements of headI for eachany four

AttemptANY ONE:Ultimate moment capacify of beam:Data Given:b-280mffi,d=Jl0mm,d'fy:415MP&, fr. -353 MPa

:50mffi, Art-2455mm2, Ar.: 402mrr],

t'

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'353x

frrA*

- 393 ... A = 2O6Zmm2

Subject Code: 3#6$4

Mr, ffi {}.t} ',l f'y,,ts*u {d * a.42xu*o*) * f ,r4rr(d

Mu = 0,87 x 4i.5 x {}{,?l x (510 * 0.42 x 244.8) + 3s3 x

SIIMMER - 15 EXAMIi\ATIONSY'',$9.*-eJ.A+,pJypr Design of RCC Strqx*€tlr#$ PageNoLt3fti

* a._

,y*' \e,4- ,i

4{i2{5r0 s0)

"' Mu = 368' 42kNmOR

Mu = A.87 frAstr(d * 0.42x"*o*) * (f ,, - f ,r)Ar"(d * d,)Mu = 0.87 fnAstr{d - A.4zxu*or) * (fr, - 0.446fck)As((d _ d,)

,. Mu = 0.87 x 41"5 x2062 x (510 -0.42x244.8) + (3S3 * 0.446 x 30x 402(510 * 50) '

Mu = 365.21kNmoR'

Mu - 0.36frpbxu*or(d * A.4Lxu*o*) * f ,rAr"(d - d')

'. Mu = 0.36 x 30 x 2fl0 x 244.8 x (510 -0.42x244.8) + 3s3 x a0z(510 - 50)

"' Mu = 366.7kNmOR

Mu - 0.36frybxu*o,(d * 0-42xu*o*) * (f ,, - f ,r)Arr(d - d.'),. Mu = 0.36 x 30 x 280 x 244.8 x (510 -0.42x2a4.8) + (3S3 * 0.446 x 30

x 402(510 *"50)

"' Mu = 364.23kNmNote: Here students may calculate the value of Mu by either consirlering the valueof fcc or neglecting fhe value of fcc, so full credit should be given accordinglData Given:Beam size: 200nrn1 x 3{}t}nrn"} eiibctiveb:200 mm, d='3{}0 mtn, f'*i,,,,,,,,20 MPa' fr,-4 15 MPa.'Mu:74 k].,Im, d'-30mm, ff.rr_=,3si MPa

x a.unrax : 0,48 x

Mrrr*o* -"' M*r*** ffi 0'l-38 X 20 X

300 = L44mmRr*o*bd2200x3002=49.6$ftffneMur*o*

Balanced mofflent" Muz tr Mu -* Mut* 74 * 49,68

"" M"rz ffi ?^e. 33 kNm

Arr.st2

o'o A.;r,'l

Total area of tension stee I",

,A'e ll.sl.

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Total area of cornpressi*l'r st*cl"

/1t-h/ as'[1 *0.B7fv@ - 0.42xu*o*)

49.68 x 106

C"{}7 x 4LS x (300 0"42 x 1,4qi'*"dsrr = 574.4\mmz

Muz

0.87 fr(d * d')24.32 x 106

0,87 x 4LS x (300 30)

'"' Astz = 249'4\tmmz

Ast = Asn*AstzArt = 57 4.48 + 249,48"" Ast = 823.96mrn2

ciA.Lt'j / t.sc

0"87 fyAstz

0.87f,,

x 4LS x 249.48

353= 255.68mm2

Subject Code:

SUMMER _ 15 EXAMII{ATIOFiModel Answer

SOLTJTION

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Tc: o' S{ N lrnflF

Subject Code: \ q-t olSUMMER_ 15 EXAMINATION

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