ACE Engineering Academy. Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru Chennai Vijayawada Vizag Tirupati Kukatpally Kolkata

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2 : : Civil Engg. _ ESE MAINS 0. (a). At th wall, irrspctiv of whthr th boundary layr is laminar or turbulnt, µ ( u/ y ) y = 0 = dp/dx. This mans that th prssur gradint dp/dx dtrmins th curvatur of th vlocity variation u(y) at wall.. Brnoulli quation applis just outsid th boundary layr, p + ρu / = const. and sinc th prssur variation across a boundary layr is ngligibl, dp/dx = ρdu/dx. A dclrating outr vlocity, du/dx < 0, mans that th prssur incrass in th flow dirction.. Boundary layr sparation occurs for zro wall shar strss, τ w = µ ( u/ y) y=0, and this situation can occur only with dp/dx > 0 sinc it implis a positiv curvatur at th wall with zro gradint. (b) Soil A: (c) Soil passing through.75 mm siv = 9% % gravl = 8% % passing through m siv = % % fins = % % sand = 9 = 78% (> 50%) Soil A is ssntially a sandy soil C C u c D D 60 0 D0 D D W L = 6%, Wp = 8% % fins > % I P = W L W P = 6 8 = 8 > 7% Soil A is classifid as SC (clayy sand) Th approximat mthods of dirct masurmnts ar listd blow:. Pacing. Masurmnt with passomtr. Masurmnt with pdomtr. Masurmnt with odomtr and 5. Masurmnt with spdomtr. Pacing: Th survyor walks along th lin to b masurd and counts numbr of stps. Thn th distanc masurd is qual to no. of stps avrag lngth of a stp. Avrag lngth of a stp can b found by

3 : : Tst 9 walking along a known lngth. A normal man taks a stp of lngth 0.75 m.. Passomtr: A passomtr is a watchlik instrumnt which should b carrid vrtically in th shirt pockt or tid to a lg. Mchanism of th instrumnt gts opratd by th motion of th body and rcords numbr of pacs. Thus, th problm of counting pacs is liminatd.. Pdomtr: It is a instrumnt similar to passomtr, but it rcords th distancs instad of pacs. In this bfor walking zro stting is mad and lngth of pac is st dpnding upon th prson.. Odomtr: It is an instrumnt which is attachd to th whl of a cycl or othr vhicl. It rcords numbr of rvolutions mad by th whl. Knowing th circumfrnc of th whl, th distanc travlld may b found. (d) By dimnsional analysis it is known that th prssur cofficint is function of Rynolds numbr of th fluid in th pip P V f.v.d For dynamic similarity, th rsistanc is du to th viscosity. Th Rynolds numbr for both modl and full scal pip must b sam..v.d 000 V 0 mod l mod l.v.d fullscal V modl =. m/s (Watr vlocity in modl pip) Sinc R is sam, ( P) watr watr.vwatr 000 (.) ( P).V oil oil oil ( P) oil 800 (.8) ( P) oil.05 kpa 5. Spdomtr: Odomtr may b calibratd to giv distanc dirctly, if it is usd for a particular vhicl. This is calld spdomtr. () Dsign of Foundations on Expansiv Soils: A fw rcommndations ar as follows.. A saf baring capacity valu shall not xcd 50 kn/m (0.5kg/cm ). A minimum dpth of foundation of m

4 : : Civil Engg. _ ESE MAINS. Th bottom of th foundation trnch to b filld with sand or moorum or brokn stons. Sand filling on sids of trnch was also rcommndd.. Rinforcd concrt bands to b usd at th foundation, plinth and lintl lvls. Mthods of Foundations in Expansiv Soils: Th various mthods can b put undr thr catgoris:. Dsigning th structur to withstand th ffcts of swlling and shrinkag of strata.. Eliminating th swlling. This can b don by (a) stabilizing th moistur, (b) loading th soil mor than th swlling prssur, (c) trating th soil so that its bhavior is changd, and (d) rplacing th soil by suitabl nonswlling soil.. Isolating th structur by taking down th foundations to a stratum which is not affctd by swlling (us of th undr ramd pil-bam construction). (f) N 55 P 5 Sinc th forbaring and backbaring of RS diffr by ( ) = 80, local attractions at R and S ar th sam. Local attraction at R = W Local attraction at S = W Corrct baring of RS = = 09 Corrct baring of SR = = 9 Corrct baring of RQ = = Corrct baring of QR = 80 = Corrct baring of QP = + 00 = Corrct baring of PQ = 5 Corrct baring of PS = Corrct baring of SP = Obsrvd baring of PQ = 55 Hnc local attraction at P is 55 5 = W Local attraction at S = W Q R

5 : 5 : Tst 9

6 (g) Intrnal stability: Two main failur mchanisms nd to b invstigatd ar: () Tnsion failur () Pull-out failur Tnsion failur Pull out failur : 6 : Civil Engg. _ ESE MAINS considring both pull-out capacity of individual layrs and th quilibrium of planar wdg mchanism through th rinforcd zon. For ach layr of rinforcmnt, th bond or anchorag lngth (L a ) byond th point of maximum tnsion must b sufficint to prvnt pull-out failur. Th maximum tnsion in a givn rinforcmnt layr will occur at som distanc from th fac, dpnding on th lvation of th layrs. ) Tnsion failur Tnsion failur will occur if th rinforcmnt strngth is insufficint to carry th horizontal loads. It is chckd for ach layr considring th ffct of vrtical loads of wall backfill and th bnding momnt causd by xtrnal loading du to th rtaind backfill/original ground. Each layr is assumd to carry th local horizontal strsss, acting ovr an ara qual to half th vrtical spacing on ithr sid of th layr pr mtr run of wall. ) Pull-out failur: It occurs whn th lngth of rinforcmnt is lss and sufficint intrfacial frictional capacity is not mobilizd to hold th rinforcmnt in position. It is chckd by (h) ET at GMN = m 0 s Sinc th ET is incrasing aftr GMN., its valu will b lss than m 0 s bfor noon. Now, 0 h 0 m AM occurs h 0 m bfor th noon. Chang in ET in h 0 m = sc.5 =.5 sconds Equation of tim at 0 h 0 m AM = ( m 0 s.5 s ) = m 8.5 s Now GAT = GMT + ET = 0 h 0 m m 8.5 s = 0 h 5 m 5.5 s

7 (i) (j) For a stp slop NDL lis blow CDL. From th fig at th outfall (or) ntranc to channl th dpth of flow is y c Rsrvoir Bd y c q g / 0 q = 0 y c / =.68 m Th dpth at outfall =.68 m Spcific Spd: Th spcific spd of a cntrifugal pump is dfind as th spd of a gomtrically similar pump which would dlivr unit quantity (on cubic mtr of liquid pr scond) against a unit had (on mtr). It is dnotd by N s. Th spcific spd is a charactristic of pumps which can b usd as a basis for comparing th prformanc of diffrnt pumps. N s N y c Q / H Stp CDL y n NDL : 7 : Tst 9 Drivation: Tangntial vlocity of runnr vlocity of fluid from impllr U V DN 60 DN H gh H D () N Q = AV Q D. H () Q H N H. H Q N H Q k N / By dfinition Q = m /sc H = m N = N S K = N s H Q NS. N N S N. NS H Q / / N.Q / H. H

8 : 8 : Civil Engg. _ ESE MAINS Typ of pump Spcific spd Slow spd radial flow 0 to 0 Mdium spd radial 0 to 50 flow High spd radial flow 50 to 80 Mixd flow 80 to 60 (or scrw typ) Axial flow 60 to 500 (or propllr typ) K For chang of Tmpratur from 5C 0C: k Chang (0 5) (5 0 o C = mm/s for = 0.7 is for = 0.85 is? 5 0. (a) 0C and = 0.70 is 0. 0 m/s and =.0 is.80 0 mm/s K CD 0 w K for sam tmpratur At tmpratur, T = 0 = 0.7, K =? =.0, K =.80 0 K K K K = mm/s = 0.7 is 0. 0 mm/s = 0.7 is mm/s = 0.7 is? (b) (i) For T = 0C K K K = 0.6 0, = 0.7 K =?, = 0.85 K = mm/s Givn data Gat AB dimnsions = Lngth = m Width = 0.9 m Spcific gravity of oil = 0.8 Mass dnsity of oil = 80 kg/m = wight of th watr = 9.79 kn/m

9 : 9 : Tst 9 8 m =.678 ( ) = m Oil * h h m m A X F m G B 50 o A = Th inclind distanc X dos AB lngth = 0.56 m sin 50 Whr h = vrtical distanc form fr surfac to cntroid of gat surfac AB h sin sin = +.5 =.5 m Ara of contact surfac = 0.9 = 0.9 m Hydrostatic forc (F) = oil.h. A = s oil watr h A = ( ) (.5) (0.9) = 9.6 kn Cntr of prssur (h * ) from fr surfac I.sin h A.h 0.9 (sin 50) = =.678 m Th vrtical distanc X form and A = h * ( + sin 50) (ii) y =.50 m R A P.5 6 Vlocity of uniform flow at sction () V n V m / /.R. S /.5 / ( ) 0.05 z c V = 0.5 m/s Discharg, Q = A V m 6 = (.5 ) 0.5 =. m /s As channl Bottom raiss E = E + z.50 m

10 : 0 : Civil Engg. _ ESE MAINS At critical condition, E y E y y c y c c c E = E c + ( z) c V g =.5 m 9.8 / q g ; / q Q B ; q =.7 m /s y c = 0.58 m E c 0.58 = E = E + ( z) c (c) ( z) c = = 0.75 m RL of swrlin at A = = m 0 At X, 0 m from A RL = At Y, 0 m from A = m RL = m 0 00 Hnc dpth of trnch at X = =.75m Dpth of trnch at Y = =.9m Backsight Intrsight Forsight Ris Fall RL RL of Distanc Rmark s w r A X

11 : : Tst Y Point B BM (a) (i) BS FS =.0.07 =.9 Ris Fall = =.9 Last RL First RL = () in th arc =.9 K v, bcaus rot v 0 r K K () v man dr r r r r r r r ln r r Bcaus of continuity: v man = v o v K v A man K r (r r ). r ln r = m/s, K v B = 6. m/s r From th Brnoulli s quation: (ii) p A p B (v B v =.5 kpa A ).50 Sol: Th maximum valu of Rynolds numbr Pa at which laminar flow can xist is 000. That is R vd = 000 Whr th avrag vlocity is givn by v 8 p L R Substituting, th Rynolds numbr is thrfor R 8 p d L d

12 (b) p d L Or rarranging in trms of pip diamtr R d p L = 0.08 m Th maximum insid diamtr is found to b 8 cm. : : Civil Engg. _ ESE MAINS sat kn / m.59 o (upto middl of clay layr) = (09.8) + ( ) = 58.6 kn/m Initially i 9 8kN / m Aftr rplacmnt () f = 60 kpa C f o i = = 96.6 kpa o f = = 8.6 kpa m =9.5 kn/m m Sand sat =9.5 kn/m C R C C 6 m clay w.c = 60% G =.65 W L = 75% o =58.6 c =96.6 f =8.6 Saturatd unit wight of clay sat G.S r = w.g = =.59 w S f C C = (60 0) C = 0.6 R.C 5 Sttlmnt CR o C H log c o Cc o H log 0 f c

13 : : Tst 9 S f H o CR.log log.59 c o C C log 0 f c log 58.6 Sf 0 0 S f = 0.08 m S f = 0.8 mm Diffrnc in lvation btwn B and A = (.70.85) (.50.5) = = 6.80 m, B bing highr Gradint from A to B i.., in (c) A N B. (a) Applying Brnoulli nrgy quation btwn and T Lt th station of obsrvation b T. S = =.70 m S = =.5 m Distanc TA = cos 5 = ( ) = 7.8 m Distanc TB = 00.5 cos 0 = 00.5 ( ) = 5.9 m Angl ATB = Baring of TB Baring of TA Distanc AB = = = TA TB ( 7.8) = m Vrtical componnt for A V = 7.8 tan 5 (5.9) = =.7 m p v g z z h L p v g p p v g p p v g From th manomtr p ( 0 )(y 0.080) [(0.87)(9.79)] (0.080) ( 0 p p = kn/m v g ( 0 v =.5 m/s )y p v ) [()(9.807)] ( )(0.050) Q = A v = m /s = 6.8 L/s

14 : : Civil Engg. _ ESE MAINS

15 : 5 : Tst 9 (b) Wight of soil + wax = 700 gm (i) Instrumntal Errors: Th following typs Mass of soil = 685 gm of rrors aris in compass survy du to Mass of wax = = 5 gm dfcts or faulty adjustmnts of instrumnts: wt of wax Volum of wax= Dnsity of wax. Pivot of ndl not bing in th cntr of graduation circl. = 5 = 6.85 cc (or) 6.85 ml 0.89 Volum of soil = =.6 cc mass Dnsity of soil, = Volum d = =.056 g/cc g / cc w 0.5 G w G d w. S r = w.g 0.57 S r = 5.7 d = Blunt pivot point. Th ndl not bing straight. Ndl bing sluggish 5. Sight vans nto bing vrtical 6. Lin of sight not passing through th cntr of th graduatd ring. 7. Graduations not bing 00 prcnt corrct Not: Corrction is distributd in proportion to th lngth of sids and it has opposit sign of rror. Corrct valu = obsrvd valu + corrction coordinat of nd point of a lin = Coordinat of first point of lin + Latitud (or dpartur) S r = 77.7% (ii) Prsonal Errors: Th following (c) (i) Errors in Compass Survy Th rrors in compass survy may b groupd into th following thr: (i) Instrumntal rrors (ii) Prsonal rrors, and prsonal rrors ar also likly:. Inaccurat cntring. Inaccurat lvlling of compass box. Inaccurat bisction of ranging rods. Errors in rading, and 5. Errors in rcording. (iii) Errors du to natural causs.

16 : 6 : Civil Engg. _ ESE MAINS (c) (iii) Errors Du to Natural Causs: Du to natural causs, th following rrors crp in compass survying:. Local attractions du to magntic substanc nar th sit of obsrvations.. Variation in dclination. Irrgular variations du to magntic storms. Magntic changs du to cloud and (ii) storms. Total Northing and Southing = cos cos cos sin sin80.6 sin788 = = 8.9 m Closing rror = Baring of closing rror = 0.00 m 8.90 tan.58 = Which is clos to baring of BC, Hnc it can b surmisd that a mistak in lngth of 0 m has bn mad in th lin BC. Th mistak is du to tallis bing similarly looking for 0m and 0m, a diffrnc of 0m. (d) Laminar flow D For laminar flow, th rat of flow is givn by Hagn-Poisuill quation 8.Q. 8Q P d r Ovr th lngth of th taprd pip, th diffrntial prssur P incrass with dcras pip radius. Hnc from gomtry r D D d x Lt D = 0 mm, d = 5 mm L = 500 mm r = ( x) P 8Q.dx [0.005( x)] 8Q P (0.005) Lt u = x; 8Q P (0.005) (0.005) 8Q (0.005) x L dx ( x) 0 du = dx du u u ( 0.5) = 8.5 N/m dx l r u ( x) ( 0.5) d

: 7 : Tst 9 5. (a) Corrctd valus of S.P.T for ovrburdn prssur @.5m, @ m, o No 0.77 NO 8.5 7 905 log o 905 No 0.77 0 log =. 7 o No 8 5 kpa 905 0.77 5 log 5 = 7.60 8 @.5 m, 8 0.

17 : 7 : Tst 9 5. (a) Corrctd valus of S.P.T for m, o No 0.77 NO log o 905 No log =. 7 o No 8 5 kpa log 5 = 7.60 m, = 5+5 = 69 kpa o 905 No log =.9 69 Corrction valus for W.T (dilatancy) No No =. (No corrction ndd, no W.T) No No Final SPT No is Avrag= (tak lowr) Pck-Hansn quation S = Sttlmnt D w = Dpth of watr tabl blow G.L q nt = 0. N S C w C w D w 0.5 B D f q nt = = 7.79 kn/m

18 : 8 : Civil Engg. _ ESE MAINS (b) m 0m : Th quantity of spag/mtr lngth of dam q = K.S Lt AB is upstram fac. Its horizontal projction L = 0 = 0 m BC = 0.L = 0. 0 = 6 m Bottom width of dam = () + 6+ () = = 00 m With rspct to focus F as origin, th coordinats of C, th starting of bas parabola x = = 56; y > 0 m Equation of parabola, x L y C 0 m x S S 0.L B 56 6 m m 00 m 0 56 S = = 5.05 m Quantity of spag pr mtr lngth of dam 90 q K.S Lngth of dam = 00 m 5 Total quantity of spag = F : S 60 m m = m /s Dirctrix /s / m (c) W know that th ara of a triangl whos sids ar a, b and c is givn by th quation: A s(s a)(s b)(s c) (i) Whr x = a + b + c Ara of ABC: a = 65, b = 0 c = 50, s = 55 s = 7.5, 7.5(7.5 65)(7.5 0)(7.5 50) A A = 9.9 sq.m Ara of ACD: a = 55, c = 6 d = 0, s = 57 s = (7.5 55)(7.5 6)(7.5 0) A = sq.m Ara btwn irrgular boundary and sid CD (.50.5).0 5 = 0 ( ) = = 8.5 sq.m Ara btwn irrgular boundary and sid DA

19 : 9 : Tst 9 A (.6.0).5 A = 0 ( ) = =.6 sq.m Ara of plot of land ABCDA = Ara of ABC + Ara of CDA + A A = = sq.m (d) Givn: U = 5 m/s, V u = 8 m/s, V f =.5 m/s, Q = 0.8 m /s i. Eulr s had: gh = E = (U V u ) /g as V u = 0 E = 5 8 / = 50 J/kg = gh H = 50 / 9.8 = 5.87 m iii. Intl blad angl ( ) V Inlt V f V u U / V = V V 8.7 m / s f u V f = V f = V f = V =.5 m/s tan = V f / (U V u ) =.5 / (5 8) = 9.65 o (iv) Dgr of raction (R) H R 5.87 R = 0.6 V V H V r V =V f / g Outlt U V u = 0 V f / 9.8 ii. Powr dvlopd (P): P = QE = Q (gh ) = ( ) / 000 P = 60 kw

20 : 0 : Civil Engg. _ ESE MAINS

21 6. (a) H H-h a From similar triangl x a y a f = X a Ya H h a x X h a b b A X a = Y a = y Y b b (H h f (H h f f H h a a ) x ) y Lngth btwn A & B = a f X a x a a a b ; X b H h f H h ; Yb f ( X b b a X b ) (Ya Yb ) x a = 9.50 mm, X a =? h a = 60 m x x b = 6.70 mm, X b =? h b = 780 m y a =.60 mm, H = 00 m, y b = 0.80 mm, f = 0 mm X a = = 6 m X b = m x b Y a =.60.7 m b X b y B b H-h b h b b : : Tst 9 (b) Y b = m L a X b ) (Ya Yb ) = (X ( ) (.7 7.8) AB = 5.5 m Scal = h av f H h av = 70 m 0.0 = Initially, A o d L = cm, L = 9 cm Ara at failur, tst A f =.08 cm = cm Ao Af for unconfind A 9.65 o L L 9 Unconfind comprssiv strngth, q u failur load A f = tan f + C tan f In unconfind tst: = 0, = q u q u = C tan f. kg / cm. = C tan 5 ()

22 In tri axial tst: = 0. kg/cm = + d = = kg/cm = 0. tan f + C tan f () Substract quation () from quation (), w gt : : Civil Engg. _ ESE MAINS n = 57.5 kn/m S = tan(9.7) S = 6.9 kn/m (c) m 0.8 = 0. tan f 0.8 = 0. tan 5 = 9.7 Substitut valu in quation () d 9.7. = C tan 5 C = 0. kg/cm G w.7.7 = sat.5 G w sat = 0. kn/m d = = 6.67 kn/m Shar strngth at m dpth S C n.tan d =.7 gm/cc.5 (d) Sol: Gradint, g =.5 % Gradint, g = 0.5% S = 00 m, h =.5 m Assuming hight of obstruction, h = 0.m L= 00 S (g h h = 75 m g R.L of point of commncmnt ).5 75 = 75 = 7. m 00 R.L of point of tangncy = R ud.5 % From th graph C D = 0.8 R.L = 75 m 0.5% (00) (.5 0.5) = 7.8 m 75

23 : : Tst 9 CD u d Projctd ara = C R u A D R Projctd Ara m N 07. (a) H = 60 m; = 5 o ; V r = 0.85 V r ; V w = 0; d = 00 mm; D =. m; C v = 0.97; K u =?; N =?; F x =?; P =? h =? W know that th absolut vlocity of jt is givn by V C v gh m / s u V w = 0 V r u V r V = V w u V = V f Dflction angl Lt th buckt spd b u Rlativ vlocity at inlt = V r = V u = 09.7 u Rlativ vlocity at outlt = V r = ( 0.5)V r But V r cos = u = 0.85 (09.7 u) 0.85 (09.7 u) cos 5 Hnc u = 9.8 m/s But DN u and hnc 60 60u N rpm D. u 9.8 Jt ratio = m 0. 5 V 09.7 Wight of watr supplid = Q = = 8.6 kn/s Forc xrtd = F x = a V (V w V w ) But V w = V and V w = 0 and hnc F x Work don / scond = F x u Kintic nrgy / scond = av = kn/s kn = = kn/s V f = 0

24 : : Civil Engg. _ ESE MAINS Hydraulic Efficincy = Workdon / s h Kintic nrgy / s Downward forc on th gravl layr FOS Upward forc on th Gravl layr du to artsian prssur (b) Silty clay 8 m = 90.7% m Gravl m G =.7 w = 0% k = 0 5 mm/s sat sat (8 ) 0 w w (G ) 9.8(.7 0.8) 0.8 sat = 9.0 kn/m FOS (ii) Aftr Construction: (9.06) 00 FOS (a) S = w G = 0..7 = 0.8 k = = 0 8 m/s Critical hydraulic gradint (c) At point Radius r = 00 mm Hight, z = 00 mm Vlocity, v = 5 m/s G.7 i Prssur, p = 0 kn/m At point h k.a z 8 8 Q 0 Radius, r = 00 mm Hight, z = 00 mm = m /s/m = 0.7 m /yar/m Dnsity of air, w = 0.0 kn/m (b) Factor of Safty: (i) At th nd of xcavation: Prssur P : For fr vortx flow vr = constant v r = v r

25 (d) v = (v r )/r V = 5 0./0. = 7.5 m/s Using th quation: p w v g p w v g z z p w 9.8 p w p w p = = 0.0 kn/m T : 5 : Tst Obsrvations from station A Vrtical componnt TT a = AT tan 05 = R.L of th towr, T =.77 m R.L of B.M + Staff rading + TT a = = 56.0 m Obsrvations from Station B: Vrtical componnt TT b = BT tan 9 = = 5.5 m R.L of th towr, B A 50m T T B T a A.70 m From ABT, w gt BT AT AB sin 60 sin 50 sin(80 (50 60)) T = RL of BM + Staff rading + TT b = = 56.0 m R.L of th top of towr, T = 56.0 m 50 sin 70 50sin 60 BT sin 70 50sin 50 AT sin 70

26 : 6 : Civil Engg. _ ESE MAINS

15. Stress-Strain behavior of soils

15. Stress-Strain behavior of soils 15. Strss-Strain bhavior of soils Sand bhavior Usually shard undr draind conditions (rlativly high prmability mans xcss por prssurs ar not gnratd). Paramtrs govrning sand bhaviour is: Rlativ dnsity Effctiv