ies conventional civil engineering 2009

8/13/2019 IES Conventional Civil Engineering 2009

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Sl. No. 18401 A-FTF-J-DFA

CIVIL ENGINEERING Paper

Conventional)

Time Allowed: Three Hours j Maxi;1Wn Marks : 200INSTRUCTIONS

Candidates should attempt any FIVE queStions.The number o marks carried by eaCh

subdivision o a questio11 is in dicaJ.ed at the e11do the subdivision.The total number o marks for eO.ch question

will be 40. l Vherever a question attempted, all its

subdivisions are to be attempted. Notations used are standard and will have their

usual meanings.Assume suitable d t ~ i found necessary, and

indicate them clearly. Newton may be convertedto kg using the relation 1 kilonewton(1 kN) = 100 kg{, i found nece.o:s:ary

Answers must be written in ENGLISH.


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1) Draw a sketch showing the typicalcreep strain-time curve under uniaxial_compression for concrete. 4

ii) Draw a neat sketch of macrostructure ofexogenous tree. 3

iii) Draw neat sketches showing vanoustypes of shakes. 3

b) i) Why is seasoning Of timber required?

ii)

List out the methods of seasoning. 4

Write briefly on the compositiOnproperties of refractory bricks. and6

c) Briefly describe the admixtures generally usedin concrete and the properties they impart tothe concrete. 10

d) i ) Give a detailed account o the cylindersplitting test of concrete. 8

ii) What are the limitations of the above testin evaluating the real tensile strength ofconcrete. 2

AFfF-J-DPA 2 Contd.)


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2. (o)

Tlm 20 mm x 20 mm

0Q S m \ 0-5 m { kN

lm JOmrn x JOmm

c 04m \ 02 n1 6kN

lm1 20 mm ' 20 mmD. 5kNA Stepped vertical steel bar ~ C D is fixedat the top end A Each segment of the barAB BC and CD is l m long and has crosssections 20 mm x 20 mm, 10 mm x 10 mm and20 nun x 20 mm respectively. A 5 kN load isapplied direi::lly at D and 6 kN loads are appliedon, the levers attached to the stepped bar atB and as shown in the above figure. Find thevertical displacement of D and the change involume of the bar. E = 2 x to MPa and Poisson s ratio J l = 025.Connections between the levers and bar atB and C. are hinged. 15

A-FfF-J-DFA 3 (Contd.)


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(b)

A cantilever beam with circular cross-section Ofradius 100 rnm is subjected to a uniformlydistributed loud over the enti e span t is giventhat the deflected shape of the beam has amaximum curvature of 101&592 x I


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b)

600 kgf/m-t- 1 kg - ' 7 ' c 2 = 6 c ' ; D ~ 1-2m

B E 4mF_J _m- -- A

m 4 --.r-- 3 ~

A portal frame ABCDEF with inclined legshas hinges at B D and E .as shown mthe above figure. Joint at C is monolithic.Supports at and F are fixed. Calculate allcomponents of l eactions. 20

A-FIF-J-DFA Contd.)


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5.- a) An unequal angle section 200 mm x 150 mm x15 m m is to be used in a truss as a strut of

~ n g t h 45 m. The cross-sectional properties of.the section are as folloWs :Area of cross-section = 5025 rnm2

= 2 x 107 mm4 = 97 x 106 mm4x x YY= -83 x 106 mm4xy

using the table of pennisSibJe compressivestresses given below, detennine the safe load onthe member.

Slenderness 100 110 120 130 140 150 160 170 180Rmio

Pennissible .compres 80 72 64 57 51 45 41 37 33sive stress

MPa20

(b) A mild steel T section has the following crosssectional dimensions :Total depth = 200 mmWidth of flange = 120 nunThickness o'( flange = 20 mrnThickness of web = 20 nunl the yield stress. a = 250 MPa detenninethe plastic moment tapacity of the section.Also calculate the shape factor for the section.

20

A-F 1F-J-DFA 7 (Contd.)


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b)

-

7. a)

p, per unitlenS;thf -L ~ p. e

1-b-- { .I

A simply supported hi gh strength concretebeam of rectangular cross-section b x D ,shown in the above figure, supports uniformlydistribu ed dead ~ n live loads of intensitiesp and q per unit length respectively. The beamis prestressed by a straight lendan carrying aprestressing _force a t an eccentricity e Showclearly and neatly the stress distributions,through the beam depth, due { 0 eccentricprestressing. dead and live loads at a crossseCtion where maximum stresses occur. IS

i) Briefly explain the different types of

ii)bulldozers according to their uses. 6Explainsketch

the Derrick crane with a neat4

A-FIF-J-DFA 9 Contd.)


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. (b) i) Enlist PERT. and explain Time estimales 6(ii) Explain Re:;;:ources Allocation. 4

(c) A construction work consists of.activities withJ'ERT durations in daYs as given below :

Activity . p Q R s T u w y zPredece uor - P, T. Q - s - s s u, w

ttp

3 4 4 3 8 I 2 4 66 8 5 3 14 4 5 7 159 9 9 3 17 7 14 13 30

Detennine:j) The probability of completing the job in

32 days andi i) The cOmpletion time With SO probabHity.

z ProbabiUty-15 007-13 010-10 016

10A-F l'F-J-DFA 10 C , o n t d ~ )


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(d) Calculate optimum cost ~ n optimum dUrationfor jobs of network given in the table below :

Nonn al rashActivity Duration Cost Duration CostDays Rs Days. Rs

1-2 5 4000 4 50001-3 7 8000 . 3 100002-3 6 6000 2 8400

Indirect cost = Rs. 1000/M per day. Sketch pro:je ?t time-cost diagram. 10

AF I F-1-DfA

.


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4.ill1P.illff'1 ~ ir J q-;r \t:n,i P r . r i n ~ s.. vice018725

A F T F J D F B

C IV IL ENG INEER ING P a p e r I I

Co_nven tiona l)

Tirne Allowed Three Hoursj jMaxim.um. Mark.s 200 JIN S T R U C T IO N S

C a n d id a te s sh o u ld at te1np t a n y F IV E ques t ion s . E a ch ques tion .ca rr ie s 40 m a rk s .T he n u n ~ b e r oT 1n-arksc a r r ie d by ea chsubd iv i s ion o a ques t ion is in d ica te d a t th e e n d o th e s u b d i v i ~ i o n / q u e s t i o R

W herever a ques t io n is a i : t e r n p t e d ~ a l l i t s subd iv i s io n s nl.ust eatte1np ted .

A n sw e r s m.ust be w r i t te n in E N G LISH .Assu;rne su i ta b le da ta i fo u nd necessary a n d

in d ica te t he sa m e c lea r ly . U nless in d ic a te d othen..v ise, n o ta t io n s a n dsynz o ls h a ve th e ir u su a l ~ n e a n i n g s

N ea t. s k e tc h e s to be draUJn, UJherever requ ired .

-FTF-.J-DFB IContd .]


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1 . (a) A ine ta l l ic cube 30 em s ide a n d weigh ing 450 N1 s lowered in to a t a n k conta ining a t w o f lu idl ayer o f w a t c r a n d mercury . Top edge o f t he cubeis a t w a t e r surface . D e t e r m i n e t he posi t ion o fblock a t w t e r ~ m e r c u r y i n t e r face when i t h a sr eached equi l ib r ium.

(b) W a t e r IS p u m p e d f rom . a wel l t a p p i n g a nunconf ined aqu i fe r a t a r a t e o f 2400 m 3/day. Ano-d rawdown b o u n d a r y exis ts a t a d i s t ance o f5 k m frolTI t h e well centre . A s s u m i n g t he well tobe ful ly pene t ra t i ng , c o m p u t e the s t e a d y s t a t ed r a w d o w n a t the well face. Given In i t ia ls a tu r a t ed th ickness = 50 m , hydrau l icconduct ivi ty 20 m/day, effective well r a d ius =

1

1 m. 1(c) A l a rge s t ream h a s a reoxygena t ion c ons t a n t o f

04 pe r day. A t a veloci ty o f 085 m/s; a n d a t t h epo in t a td i scharged .10 mg /L D 0

which a n organ1c po l lu t a n t l.Si t 1s sa tura ted with oxygen a t

= 0). Below the outfal l , t he u l t im a ted e m a n d for oxygen is found to be 20 m g L a n dt he deoxygena t ion cons tan t is 02 per day. W h a t is t h e dissolved oxygen 483 k m d o w n s t r e a m ? 1

d ) (i) E x p l a i n t.he t e r m 'opt.inlum mois tu reconten t . How is i t affec ted by compac t ing e f fo r t ?

(ii) S t a t e t h e fac tors affec t ing field compac t iono f soil .

A-FTF-J-DFB 2 Contd


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2. a ) f h e veloci ty dis t r ibut ion in a p1pe is given b y

U = U n 1 n x 1 - ~ rwh e re u is t h e velocity a t any r ad ius r, r 0 is t heradius of pipe and un1nx is t h e m a x i m u m velocity,f ind i ) average velocity a n d {ii) k ine t i c ene rgycorrect ion factor.

(b) T h e r ecorded a n n u a l ra infa l l f rom five ra ingaugcs ta t ions 1n a cat. .chmcnt an d t h e cor re spond ingTh. s I g a f I I s es e n po y on r e a s are a s 0 ow-

Th ie s s en polygon a re a s e m ~ ) Rainfa l l ern)-25 125

30 17530 225-10 275 ~5 325

T h e scale of the m a p is 1 : 50,000. Est imat . e t h ~volUme an d th e .m(Jan dep th of t h e r;;i.infall.Es t ima te the average an n u a l discharge a t t h eoutfct , i t t h e runof f coeffic ient o f t h e c ~ h m e n t is

1

03. 1(c) W ater t ab le in a cana l command rcce1ves a

recha rge a t t h e ra t e 25 mm./day. Sub-sur facedi tch d ra i n s a t a spac ing o f 2 km a rc prov ided fort h e sub-surface dra inage . E s t i m a t e t h emax1mum r i se of t h e w a t e r t ab le a t s t e a d ys ta t e . Given : hydrau l i c conduc t iv i ty o f the soil =10 m/day, d ep th o f t h e impe rv ious l aye r be lowin i t ia l w a t e r t ab le posi t ion = 20 m . A s s u m e t h edi tches to be ful ly penet ra t ing . 1

A-FTF-J-DFB 3 Con td . ]


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d ) iJ

( i i )

A deposi t of fine s a n d h a s a poros i ty o f45%. Es t ima te t h e cri t ical hydrau l i cgrad ien t to develop quicksand condit . ion i ft h e specific gravi ty o f g r a in is 27.De:, ign s low s a n d f i l ters for a popula t ion o f40,000 with a n average r a t e o f w a t e rsupp ly o f 150 l i t res p e r capi ta p e r day.

3. (a) A 100 m wide r ec tangu la r channe l w i th bo t toms lope o f 000016 carr ies a discharge of 2292 m 3/sa t a norma l dep th of 20 m and crit ical d ep th o f08 m . T h e dep th immedia te ly u p s t r e a m o f d a mis 100 m. C o m p u te t h e lenb' th o f t h e surfaceprofi le be tween 100 m an d 6 m u s in g Chow'sor s tep method . Take s tep of 20 m a n d a s s u m eM = 30 a n d N = 333 and M anning s n = 0015.T h e leng th of surface profi le by Chow's m e t h o dbe tween lwo dep ths is given by

nX 2 - x l = 80 [ u 2 - u 1 IF u 2 N F (u 1 , N))

U f JN /F(v 2 J F (v 1 J)J]u F u , N ) v F(v, ,J)5 001 9 00274 0017 8 00313 0034 7 0038

6 00485 0062 0087

1

A-FTF-J -D FB 4 (CoAtd.J


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-------- ---------------------------------------------------------(b) Follo \.Ving observa t ionS w e r e m a d e for conduc t ing

a w a t e r b u d g e t o f a reservoir over a period of onem o n t h ;Average su:rface a rea = 10 Jun 2 M e a n surfaceinf low ra te = 10 cuinec, Mean sur face out11owr a te = 5 cumec, Rainfal l = 10 em, J ~ a l l i n t h e: reservoir level = 15 m P a n evapora t ion = 20 em.Assum:ing the pan- fac tor asaverage seepage dischargeduri11g the month .

07f rom

es t imate thelhe rese rvoi r

(c) Des ign a r ec tangu la r g r i t chamber for a Oow o f40 MLD. Spedf ic gravi ty = 265 a n d s ize to b eremoved is 02 m m .F i n d the(a) se t t l ing veloci ty o f 02 m m pa r ti c le s(b) cri t ical hor izonta l veloci ty of flow, an dc ) s ize o f t h e gr i t chamber .

Assume k inema t ic viHcosity o f t h e l iquid =10 x 1 0-2 cm 2/s.

{d) Diffe rent ia te be tween t h e fol lowing t e rms ; (i) Time m e a n speed an d Space m e a n speed(ii) W et dock an d D r y dock(iii) S t o p wa y an d Clea rway o f r u n w ay l en g th(iv) Hydroph i l i c aggrega tes a n d Hydrophobic

aggrega tes(v) B as e t u n n e l an d Sadd le t u n n e l

1

1

1

A-FTF-J-DFB 5 [Con td . ]


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4. (a) A pipe ne twork in the form o f a t r i ang le A B C h a sin f lows of 5 m 3/s a n d 4 m 3/s a t A a n drespect ively. The . outf low a t C is 9 m 3/s. GivenKAB = 10, KHC 50 a n d KAC :: 20, computed ischarges in each pipel ine [h,.. = K Q 2]. 1

(b) A 120"6 m long s tu rm s e w e r collects wastewB.terf rom a c a t c h m e n t a r e a o f 50 hec ta re , w h e r e 35a rea 1s covered by roof -I = 09), 20% a r e a lscovered by p a v e m e n t s 1 = 08) and. 45% a r ea i scovered by open l and I = 013). D e t e r m i n e t h eaverage I, a n d d i a m e t e r o f s to rm scwer l inea s su m i n gi ) the t ime of e n t ry = 3 m i n

(H) veloci ty o f fuH f low = 150 rn ls(iii) n = 0013 a n d slope = 0001

I = runof f rat io.(c) A r e t a in ing wal l 10 m high is proposed to hold

dry s a nd o f voi< -ratio o f 06. The va lue o f ang le o fi n t e rna l frict ion p = 30 a n d speci f ic-gravi ty o fsoi l gra in is 27. T h e back face o f wal l is vert icala n d smooth . Top sur face o f backf i l l is horizontal .

C ~ l k u l a t e the mab'Tli tude o f the to ta l ac t ive e a r t hth : rust aga ins t the wall a s su m i n g the wall is f reeto move. Also s h o w the d i s t r ibu t ion of e a r t hpre s s u re a nd poin t of apphcat ion o f the r e su l t an t .

1

A s s u m e un i t weight o f w a t e r = 10 kN/ma. 1(d ) (i)

A-FTF-J -DFB

W h a t a r e the a s su m e d condi t ionS o fr u n w a y leng th for s t a n d a r d e n v i r o n m e n twhich dcide the basic r u n w a y l eng th ?

6 Col"l td.f


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i i ) DcterTil ine t he ac tua l runway l eng th a f te rapp ly ing necessa ry cor rec tions for elevat iona n d t e m pe ra tu re a s per ICAO a n d gra d ie n tcorrect ion a s per FAA spec i f ica t ion fur thed a t a given below. 5 5

Bas ic runway l eng thElevat ion o f Ai rpor ts i teMonth ly m e a n o faverage dai lyt emperu tu rc fort he ho t t e s t m o n t hof the yearMonth ly m e a n o fmax imum da i lyt e m pe ra tu re for thes a m e m o n t hF.ffect.ive brradient

1800 m e t r e s

6 0 0 m e t re s

06o/o

. a) D r a w i nd i ca to r d iag ram for t h e fol lowing cases o fa rec iproca t ing puffip :i) W h e n no a i r vesse l i s ins tal led .ii) \ V h ~ ~ > n a i r vessel is ins ta l led on suc t ion s ide

close to the p u m p .ii:i) W h e n a i r vessel is ins ta l led on de l ive ry

. s ide close to the pump.iv) Whei1- a i r vesse l s are ins ta l led on boths ides Of tlnii p u m p .

10

7 [Cont .d . ]


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(b) How m a n y days would be r equ i red by a clays t r a t u m 5 m th ick dra in ing a t both ends wi thcoffic ient of consol idat ion = 5 0 x 1 0 -4 cm/sec toa t t a in 50% of i ts u l t ima te se t t l ement ?G i v e n : T 50 = 0197. 1

(c) De tc nn ine the l eng th of t rans i t ion curve a n d()ffscts a t every 15 m e t re s for B.G. curvedt r ack hav ing 4 cu rva tu re and c a n t of 12 em.T h e m a x i m u m permiss ib le speed on curve is85 k m p h

(d) W a t e r emerges f rom a sp i l lway wi th a vell lci ty of15 mlscc a n d a depth o f 0-5 m . Calcu la te t h enecessary subcr i t ica l depth a t the toe of thespi l lway for the Occurrence of a hydrau l ic jump .

1

Calcula te the assoc ia ted energy loss. 16 . (a) W h y i s ver t ica l shaf t necessary in case of long

t unne l ?A ci-rcular t unne l o f 3 m e t r e s d ia m e te r a n d3 ki lometres long is to be dr iven by full faceexcavat ion. T h e proposed tunnel a l ignmen t i spass ing t h rough rock. T h e ra te of excava t ion perblas t is 5 metres . Calcu la te t h e n u m b e r ofhaul ing t rucks requi red for muck ing operat ionf rom t h e daLa given below.

A-FTF-J-DFB

Capaci ty of haul ing t ruckAverage haul ing a n dr e tu r n speed of haul ingt ruc kAverage hau l ing dis tance

8

1 0 t o n n e s

25 k m p h3 k m

Contd


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Dens i ty o f m u c kswel lover -breaking

Opera t ing eff iciency ofh au l in g t ruckLoad ing t ime , d u m p i n gt ime, accelerat ion an ddece le ra t ion t ime etc.

= 1 6 0 0 kg m3= 15c 5

- 75

= 4 6 m i n u t e s(b) A Pe l ton whee l i s produc ing 300 kW work ing

u n d e r a h e a d o f 1800 m with discharge of02 m 3/s. Compute hydrau l i c eff iciency an dveloci ty o f whir l ttt. in le t and ou t le t and t he m e a nbucket. speed . Take coefficient. o f veloci ty = 0985,angle of def lec t ion o f j e t = 165" a n d re la t iveveloci ty a t e}(it = re la t ive veloci ty a t in le t .

(c) Th e o r d in a t e s o f h o u r un i t hydrograph o f aca t ch men t are a s fol lows.

T i m e hour s ) Discha rge 3 /sec)0 06 10

12 4018 5524 453 0 3 03 6 742 0

1

9 [Contd . l

'


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(d) Th e fol lowing a r e t he observed va lues o f a n anglean d t he i r w-eightage :

.Angle Weigh tage

30 24 ' 20 230 24 ' 1 8 230 24' 19 3

F m d( i) Probable e r ro r o f s ing le obse rva t ion o f un i t

weight .(i i ) Probable er ro r r weighted ar i thmet ic

m e a n .(i i i) Probable e r r o r r s ingle o b s e r v a t i o n r-eight 3.

A FTF J OFB 11

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