ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune ...

: 2 : Civil Engg. _ ESE MAINS


01. (a)

Given Data:

Borrow Pit: Embankment:

1 = 18 kN/m3` 2d 15 kN/m3

w1 = 8% w2= 10%

Volume = V1 Volume = V2

08.01

18

w1 1

1d1

= 16.67 kN/m3

The volume of embankment per metre

length, V2

3m2442

102

Using the relation:

1

2

d

d

2

1

V

V

67.16

15

24

V1

V1 = 21.60 m3

Volume of earth required to be excavated

from borrow pit = 21.60 m3 (Ans)

Remoulded state (Embankment):

e1

Gwd

e1

67.281.915

Void ratio, e = 0.746 (Ans)

To find degree of saturation, S:

S

G.we

S

67.21.0746.0

S = 0.358 or 35.8% (Ans)

(b)

Computation of DWF:

Average water supply = 200 100000

= 20 MLD

= 0.23148 m3/sec

Assume that 80% of the water supplied

appears as sewage,

Average sewage discharge = 0.8 0.23148

= 0.1851 m3/sec

Assume the maximum sewage discharge to

be 3 times the average discharge, we have

Maximum sewage discharge

= 3 0.1851

= 0.556 m3/sec

Strome water discharge computations,

Time of concentration tc

= time of entry + time of flow

= 10 + 20

= 30 minutes

Hourly rainfall for the area R = 3 cm/hr

= 30 mm/hr

Rational formula the discharge

360

AIRQWWF

5360

305.0120

m3/sec

Optional

10 m

4 m

2 m

1:1

: 3 : Test – 6


The combined maximum discharge

Q = QDWF + QWWF

= 0.556 + 5

= 5.556 m3/sec

Now the sewer to be running half at the

maximum velocity 3 m/sec at the time of

maximum flow.

Area required =Velocity

Q

3

556.5

= 1.852 m2

Area of the sewer running half 2d8

A

2d8

852.1

d = 2.1716 m

Hence, use a sewer pipe of 2.1716 m dia

(c)

Given:

BG track, V = 75 kmph

Axle load = 24 tonnes

No. of driving wheels = 4 pairs

Total weight of driving wheels

= 4 24

= 96 tonnes

Hauling capacity 6

1 total weight of

driving wheels

= 966

1 = 16 tonnes

Let, W = total train load in tonnes

Total tractive resistance

C321 RRtRtRt

Where,

1Rt = Resistance independent on speed (or)

rolling resistance

Rt2 = Resistance dependent on speed

Rt3 = Atmospheric resistance

Rc = Resistance due to curve

Rt1 = 0.0016 W

Rt2 = 0.00008 W 75

Rt3 = 0.0000006 W 752

RC = 0.0004W 2

To find (W):

Hauling capacity = total tractive resistance

16 = 0.0016 W + 0.00008 75 W +

0.0000006 W 752 + 0.0004 W 2

W = 1358.81 tonnes

02.

(a) Preliminary analysis data given:Ultimate settlement, Sf = 250 mm

Time, t = 6 yearsSettlement, S = 50 mm

Ultimate settlement, Sf = mv. H. In the present case, the coefficient of volumecompressibility, mv is assumed to beconstant. Sf depends on H and Thickness increased by 20%. Hence theactual thickness is 1.2 HDue to lowering of W.T, the effective stressincreases.



The increase in effective stress = ( – ) h

Where h is the depth of lowering of W.T.As the is not given, let us assume that

≃ sat

Hence = (sat – ) h = wh = 9.81 1 = 9.81 kN/m2

Initial condition:Thickness = H1

Increase in stress: 1 = 24 kPaFinal settlement = mm250S

1f

Changed condition:Thickness, H2 = 1.2 H1

Increase in stress, 2 = 24 + 9.81 = 33.81 kPa

Final settlement =2fS

H..mS vf Sf . H(Assuming mv constant in the present case)

1

2

1

2

f

f

H

H.

S

S

1

2

1

1f

H

H2.1

24

81.33

250

S2

63.422S2f mm

Ultimate settlement under the actual siteconditions = 422.63 mm (Ans)

To find settlement at 2 years aftercompletion of building:Consolidation settlement time is reckonedfrom the middle of construction period

years5.12

3

Initial condition:t1 = 6 yearsS1 = 50 mm

100S

SU

f

11

%20100250

50

Changed condition:t2 = 1.5 + 2 = 3.5 yearsTo find settlement, S2

degree of consolidation = U22

1V 100

U

4T

1

0314.0100

20

4

2

We know that2

vv d

t.CT

2v d

tT

2

2

1

1

2

v

v

d

d

t

t

T

T

1

2

2V

H2.1

H

6

5.3

0314.0

T2

0127.0T2v

2

2v 100

U

4T

2

2

2

100

U

40127.0

U2 = 12.72%

100S

SU

2f

22

mm76.53S

10063.422

S72.12

2

2

Consolidation settlement after 2 years ofcompletion of building for the changedconditions = 53.76 mm (Ans)

: 5 : Test – 6




(b) (i)

At a certain temperature, the rate of

deoxygenation is assumed to be directly

proportional to the amount of organic matter

present in sewage at that time.; i.e.

tt KL

dt

dL

Where Lt = Oxygen equivalent of

carbonaceous oxidisable organic matter

present in sewage after t days from the start

of oxidation, in mg/l.

t = time in days

K = rate constant signifying the rate of

oxidation of organic matter and it

depends upon the nature of organic

matter and temperature. Its unit is per

day.

Integrating we get

dt.KL

dL

t

t

Loge Lt = –K. t + C

Where C is a constant of integration, and can

be evaluated from the boundary conditions

at the start i.e.

When t = 0 ,Lt = L (say)

Substituting in equation we have

Loge L = K(0) +C

C = loge L

Substituting this value of C in equation we

get

loge Lt = – K.t + logeL

loge Lt – loge L = – K.t

KtL

Llog t

e

Where KD is the De-oxygenation constant or

more strictly, the BOD rate constant

t.KL

Llog D

t10

tKt DeL

L

Now, L is the organic matter present at the

start of BOD reaction, (expressed as oxygen

equivalent) and Lt is the organic matter left

after t days; which means that during t days,

the quantity of organic matter oxidized

= L – Lt

If Yt represents the total amount of organic

matter oxidized in t days (i.e the BOD of t

days), then we have

Yt = L – Lt

We have

L

L1LY t

t

L

L1

L

Y tt

L

Y1

L

L tt

Substituting this value ofL

L t in equation we

get

: 7 : Test – 6


t.kt DeL

Y1

tKt De1L

L

tKt

De1LY

(ii)

Depletion of oxygen is (DOfinal – DOinitial)

= 4 ppm

5-day BOD at C20o oc20

5y

DFDODO initialfinal

DF = Dilution factor

= 2% of total sewage sample

502

100

5-day BOD = 4 50

= 200 mg/lit

We know the 5-day BOD at 20oC

kto

205 e1y

200 = lo (1 – e–0.23 5)

Where lo = ultimate BOD

Ultimate BOD lo = 292.67 mg/lit

03. (a) (i)

Ans:

Sleeper density:

The number of sleepers per rail length is

called sleeper density.

The spacing of sleepers is generally indicated

by a formula (n+x) where n is the length of

the rail in meter and x is a number 4, 5, 6 or 7

depending upon importance.

The sleeper density varies according to the

following factors.

1. Speed and axle load

2. Type of rail section

3. Type of sleepers

4. Type and depth of ballast

5. Bearing area of sleeper on ballast

6. Nature of formation

(ii)

Sol: Given BG, track

e = 12.5 cm

Vmax = 95 kmph

Do = 4o

As per Indian Railway

The length of transition curved based on

following

(1) Based on arbitrary gradient 1 in 720

L = 7.20e

L = 7.20 12.5

L = 90 m

(2) Based on rate of change of cant deficiency

L = 0.073 DVmax

D = cant deficiency in cm

D = 7.5 cm [Assumed for BG track]

L = 0.073 7.5 95

L = 52 m



(3) Based on rate of change of super elevation

L = 0.073eVmax

L = 0.073 12.5 95

L = 86.68 m

Take maximum of three

L = 90 m

To calculate offsets at every 15 m interval

oD

1720R

Where, Do = Degree of curve

Do = 4o

m4304

1720R

Cubic parabola equation for transition curve

RL6

xy

3

Taking offsets at 15m interval

m0145.0904306

15y

3

15

m116.0904306

30y

3

30

m392.0904306

45y

3

45

m93.0904306

60y

3

60

m817.1904306

75y

3

75

m14.3904306

90y

3

90

Shift of the circular curve

R24

LS

2

43024

90S

2

S = 0.785 m

03.(b) There are two types of piping failures

1. Backward erosion piping failure2. Heave piping failure

1. Backward erosion piping failure:

Figure: Backward erosion piping

When seepage takes place from U/S to D/Sthrough the foundation of hydraulicstructures, if the exit hydraulic gradient ismore than the critical hydraulic gradient, thesoil is removed by percolating water. Whensoil at the exit is removed, the flow net getsmodified. There will be more concentrationof flow lines in the remaining soil mass,resulting in an increase of the exit gradient.This causes further removal of the soil. Thisprocess of backward erosion continuestowards the upstream side and a sort of pipeis formed as shown in the above figure. Assoon as the channel approaches the reservoir,a large amount of water rushes through thechannel so formed and the hydraulicstructure may fail. This is called backwarderosion piping failure.The backward erosion piping may also occurin the body of earthen dams, when the

D/SU/S

Piping

: 9 : Test – 6


phreatic line cuts the down stream face of thedam and seepage pressure is high.

2. Heave piping:

Figure: Heave piping

When the seepage force due to upwardflow of water near the bottom of the sheetpile shown above, is greater than thesubmearged weight of the soil above thatlevel, the entire soil mass in that zone heavesup and is blown out by the seepage water.This type of failure is known as heave pipingfailure.Measures that are generally adopted toprevent piping failure are as follows.1. Increasing the path of percolation by

increasing the width of the structure orby increasing the depth of sheet pile etc.

2. Reducing seepage by providingimpervious core in the body of earthendam.

3. Providing drainage filter to change thedirection of flow away from D/S face. Italso prevents the movement of soilparticles along with water.

4. Providing loaded filter to increasedownward force at the exit point.

04. (a)

KD = De-oxygenation constant = 0.2

Do = Initial oxygen deficient of the mix at the

mixing point in mg/lit

= Saturation D.O at mix – D.O of mix

D.O of mix =000,52

7000,40112000

= 5.165 mg/lit

BOD at mixing point

52000

2400005012000y t

= 13.076 mg/lit

Do = 9 – 5.165 = 3.3846 mg/lit

We know 5 day BOD yt = )e1(L kto

13.076 = )e1(L 52.0o

Ultimate BOD lit/mg68.20Lo

KR = Re-oxygenation coefficient

= 0.8 day–1

D.O at 50 km down stream

Time taken by the stream to flow 50 km time

=velocity

cetandis

1.0

000,50

= 50, 000 sec

= 5.787 days

D.O deficit in mg/lit after 5.787 days

tko

tktk

DR

Dt

RRD eDeeKK

LKD

787.58.0787.52.0 ee2.08.0

68.202.0

787.58.0e3846.3

Seepage



= 2.1 + 0.03

= 2.13 mg/lit

D.O at 50 km down

= (DO initial – DO deficient)

= 5.165 – 2.13

= 3.035 mg/lit

D.O at 100 km down stream:

Time taken by the stream to travel 100 km

down stream sec0000,1001.0

0000,10t

= 11.574 days

D.O deficit in mg/lit after 11.574 days

tko

tktk

DR

oD RRD eDeeKK

LKDt

574.118.0574.112.0 ee2.08.0

68.202.0

+ 3.3846 e–0.8 11.574

= 0.6803 mg/lit

D.O at 100 km down stream

= (D.O initial – D.O deficient)

= 5.165 – 0.6803

= 4.4846 mg/lit

The time (tc) after which critical D.O deficit

(Dc) occurs is given by

f

L

D1f1log

1fK

1t o

eD

c

(or)

oD

DRo

D

R

DRc LK

KKD1

K

Kn

KK

1t

KR = 0.8

KD = 0.2

4K

Kf

D

R

Lo = 20.68 mg/lit

Do = 3.3846 mg/lit

4

68.203846.3

141log142.0

1t ec

= 1.185 days

Distance = Velocity Travel time

= 0.1 1.185 3600 24

= 10.238 km

Critical deficient

cD tkoc e

fL

D

185.12.0e4

678.20

Dc = 4.0786

Hence critical D.O deficit equal to 4.0786

mg/lit at 10.238 m down stream of A after

1.185 days.

: 11 : Test – 6




04.(b) Given data:

G = 2.70

= 8.1 10–3 poise =2

7

m

S.kN101.8

Height of fall, He = 30 cm = 0.3 mSmallest particle size, d = 0.001 mm

= 1 10–6 m[The time required for all particles to settle isthe time required for the finest particle, 0.001mm] Settling velocity of particle,

Vs =

2

ws d

18

2

ws

d

18

1GV

7

26

101.8

1081.9

18

17.2

= 1.144 10–6 m/s

Time of fall for all particles,s

e

V

Ht

610144.1

3.0

= 262237 sec= 4370 min= 72 hrs 50 min (Ans)

05. (a)Given data of falling head permeability test:

L = 20 cmA = 24 cm2, a = 2 cm2

h1 = 40 cmh2 = 40 –15 = 25 cm

Stratified soilZ1 = 8 cm , K1 = 2 10–4 cm/s;Z2 = 8 cm , K2 = 5 10–4 cm/s;Z3 = 4 cm , K3 = 7 10–4 cm/s;Average permeability for the flow occurringperpendicular to bedding planes is Kv

3

3

2

2

1

1

321v

K

Z

K

Z

K

ZZZZ

K

444 107

4

105

8

102

8488

= 3.24 10-4 cm/sec

Using the equation:2

1e h

hlog

t.A

L.ak

25

40log

t24

2021024.3 e

4

t = 2417 secTime, t = 40.3 min (Ans)

(b) (i)

Considerations for determining the

thickness of concrete lining:

Dead weight of lining.

Weight and presence of surrounded

ground along with water and

superimposed loading.

Weight of internal structure, if any.

Loads developed due to temporary

construction conditions such as

compressed air pressure and reaction of

shield jacks, etc.

Weight and impact of traffic or internal

pressure.

Given:

Bore diameter = 7.6 m

Generally, for every 30 cm dia of Borewell

2.8 cm lining is provided.

: 13 : Test – 6


Therefore, thickness of lining

cm93.7030

7608.2

(ii)

Sol:

1. Breakwaters:

The protective barrier constructed to enclose

harbours and to keep the harbour waters

undisturbed by the effect of heavy and strong

seas are called breakwater.

Classifications of breakwater:

Heap or mound breakwater.

Mound with superstructure or composite

breakwater.

Upright wall breakwater.

Special breakwater.

2. Beaufort scale:

Beaufort scale is an empirical value providing

wind speed according to conditions of

surrounding.

In Beaufort scale, the wind strength is given a

numerical value ranging from 0 to 12 and the

estimate from this scale have been found to be

quite fair and accurate.

3. Wet dock and dry dock:

Dry dock:

It helps in handling cargo and plays as

platform for passengers.

Wet dock:

It is used for maintenance and repair of ship

vessels.

4. Navigation aids:

Navigational aid is anything which helps the

traveler in navigation. Widely used

Navigational aids are light house, buoys, etc.

Necessity of navigational aids:

To avoid dangerous zones like hidden

rocky outroops and sand bars.

To follow proper harbour approaches, and

To locate ports, especially during night

and bad weather conditions affording poor

visibility due to fog or clouds.

5. Dredging:

Dredging is defined as excavation of bed

below water.

Dredging work can be classified as:

Primary dredging

Maintenance dredging

Improvement dredging.

06.

(a) Given data

(kPa) :100 200f (kPa) : 90 103

Using Coulomb’s equation:f = Cu + tan u

90 = Cu + 100 tan u ----eqn (i)103 = Cu + 200 tan u ----eqn (ii)



Solving the above two equations,–13 = –100 tan u = 7.41o

Cu = 77 kPa (Ans)If the same soil is subjected to an unconfinedcompression test, the unconfinedcompressive strength (qu) is determined asfollows:

qu = 2 Cu tan

245 u

2

41.745tan772

= 175.33 kPaThe apparent cohesion of the unconfined

compression test,2

qC u

u

2

33.175

= 87.66 kPa (Ans)

(b)

Sol: Sewage Flow, Q = 5 MLD

70% of solids are removed at settling tank

Concentration of suspended solids

= 200 mg/lit

inST1 CQMsludgefreshin

solidsdryofamounttotal

day/kg700200100

705

watersolids1slu S

water%

S

solids%

)S(

100

1

98

5.2

2

)S(

100

1sludge

(Ssludge)1 = 1.0121

(sludge)1 = (Ssludge)1 w

= 1.0121 1000

= 1012.1 kg/m3

Volume of fresh sludge

1sludge

1

1f )(

M

)P100(

100V

P1 = 98% (M.C)

1.1012

700

98100

100Vf

= 34.58 m3

lit/mg70MLD5Msludgedigestedin

solidsdryofamoutTotal2

= 350 kg/day

P2 = 94

1

94

5.2

6

)S(

100

2sludge

(Ssludge)2 = 1.0373

(sludge)2 = (Ssludge)2 w

= 1.0373 1000

= 1037.3 kg/m3

Volume of digested sludge

2sludge

2

2d )S(

M

P100

100V

2sludge )(

)350(

94100

100

3.1037

350

6

100

= 5.6235 m3

Assuming sludge digested is linear and

digestion period t = 30 to 60 days (usually)

: 15 : Test – 6


Capacity of digester t2

VV df

302

6235.558.34

= 603.05 m3

(or) (Sludge digestion non linear capacity air

= (Vf –2/3 (Vf – Vd)t)

Depth = 10 m

Surface area of digester =depth

Volume

10

05.603 = 60.305 m2

305.60d4

Area 2

Dia of digester d2 = 76.78 m2

d = 8.76 m

07. (a) (i)

1. Taxiway:

The strip of pavement connecting the runway

to the apron or to the hanger is known as

taxiway. The standards for the design of

taxiway are not as rigid as those of the main

runway as the speed of the aircraft is much

slower along the taxiway. The taxiway

connects the runway ends to the aprons but

sometimes for smaller aircrafts taxiway exits

are provided in the middle of the runway too.

2. Aprons: The airport apron is the area of

an airport where aircraft are parked, loaded

or unloaded, refuelled or boarded.

The details of the apron depend upon the

characteristics of the highest type of aircraft

expected to use it. In the case of holding

aprons the area of the apron should be large

enough so that it can bypass another aircraft

standing ahead with adequate clearance.

(ii)

1. Correction for altitude:

L300

1000

100

7LL1

1800300

1000

100

71800

= 1800 + 420 = 2220

L1 = 2220 metre.

2. Correction for temperature:

Equivalent standard temperature reduced for

elevation

C4.810001000

6.615

Temperature rise = 17 – 8.4 = 8.6

Length corrected for temperature

L2 = 2220 + 2220 100

1 8.6

= 2376 m

Runway

Fillets

Tax

iway



3. Correction for gradient:

Increase at 2% for 1% effective gradient

5.0100

20LLL 223

10

123762376

= 2376 + 237.6

= 2613.6

= 2614 metre

For minimum clearance, length of runway

= 2614 + 120

= 2734 metres

Check:

Total correction for elevation, temperature

and gradient

= L2 – L

= 2376 – 1800

= 576 m

% increase 1001800

576 = 32%

(b) (i)

Sol: Assume the average daily sewage flow 150

lit/head (limit 135 to 200 lpcd)

Quantity of water supplied = Percapita rate

Population

= 200 150

= 30,000 l/day

Assuming that 75% (limit 75% to 80%

lpcd) of water supplied becomes sewage,

we have the quantity of sewage produced.

Quantity of sewage produced

= 0.75 × 150 × 200 l/day

= 22,500 l/d

Assume detention period is 30 hr (limit 28

hr to 48 hours)

Quantity of sewage produced during

detention time of 30 hrs

lit281253024

22500

= 28.125 m3

Now, assuming sludge deposition in the

tank @ 30 l/capita/yr (limit 30 to 40

lit/capita/yr) and cleaning period being 6

months, we have

Sludge volume

cum3it300012

620030

Total required capacity of the tank

= capacity of sewage + capacity of

sludge

Total capacity required = 28.125 + 3

= 31.125 m3.

Assuming 1.5 m depth,

Surface area 75.205.1

125.31 ≃ 2m21

Assuming the ratio of the length to width as

3 :1

L B = 21

3B2 = 21

B 7

B = 2.645 m

: 17 : Test – 6


L = 7.9 ≃ 8 m

Say, provide B = 2.645

Hence, use 8 m × 2.645 m × (1.5 + 0.3)m

sized septic tank.

(ii)

Sol: The quantity of sewage to be treated per day

= 0.8 250 10, 000

= 2 MLD

= 0.023 m3/sec

The BOD content per day = 2 200

= 400 kg

at latitude 20oN, the organic loading in the

pond as say 250 kg/hect/day

The surface area required

rateloadingOrganic

daypercontentBOD

250

400

= 1.8 hect

= 18000 m2

Assuming the length of the tank (L), as twice

of its width (B).

Area = LB = 18000

2 B2 =18000

B2 = 9000

B = 94.868 m ≃ 95 m

Length L = 189.73 m ≃ 190 m

Using a tank with effective depth as 1 m, we

have the provided capacity

= 190 95 1 m

= 18050 m3.

Detention time =flowSewage

Capacity

DT =2000

18050

= 9.025 days

≃ 9 days

Hence, use an oxidation pond with length

= 190 m,

Width 95 m and overall depth (1 + 1) = 2 m,

and detention period 9 days.

We know

Detention period in days

=

e

i

D y

ylog

K

1

=e

i

D y

yn

K

1

yi = Influent BOD

ye = Effluent BOD

ee y

200log

2.0

19

ye = 33.059 mg/lit

Effluent BOD = 33.059 mg/lit

: 19 : Test – 6


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