ace engineering academy · ace engineering academy hyderabad|delhi|bhopal|pune|bhubaneswar|...

: 2 : Civil Engg. _ ESE MAINS


01. (a)

Sol:

112

x DxL

DDD

= 3.0x3

3.06.0

3.0x1.0

Convective acceleration as = Vxdx

dVx

2xx

3.0x1.04

Q

A

QV

23.0x1.0

405.0

23.0x1.0

06366.0

1.02

3.0x1.0

06366.0

x

V3

x

33.0x1.0

01273.0

Total acceleration at x = 2 is zero

Local acceleration + convective acceleration = 0

0x

VV

t

V xx

x

0

3.0x1.0

01273.0

A

Q

t

Q

A

13

xx

33.02.0

012732.005.0

t

Q

secsec//m100928.5t

Q 33

= 5.0928 lit

(b) (i)

Ans:

A geogrid is geosynthetic material used to

reinforce soils and similar materials. Geogrids

are commonly used to reinforce retaining

walls, as well as subbases or subsoils below

roads or structures. Soil pull apart under

tension. Compared to soil, geogrids are strong

in tension. This fact allows them to transfer

forces to a larger area of soil than would

otherwise be the case.

Geogrids are commonly made of polymer

materials, such as polyester, polyvinyl

alcohol, polyethylene or polypropylene. They

may be woven or knitted from yarns, heat-

welded from strips of material, or produced

by punching a regular pattern of holes in

sheets of material, then stretched into a grid.

These girds are formed by material ribs that

are intersected by their manufacture in two

directions: one in the machine direction (md),

which is conducted in the direction of the

manufacturing process. The other direction

600 mm

x

3 m

xx

300 mm

: 3 : Test – 14


will be perpendicular to the machine direction

ribs, which are called as the cross-machine

direction (CMD).

(ii)

Ans:

Liquefaction describes a phenomenon

whereby a saturated a partially.

Saturated soil substantially loses strength and

stiffness in response to an applied stress,

usually earthquake shaking or other sudden

change in stress condition, causing it to

behave like a liquid. Liquefaction

phenomenon generally occurs in the case of

saturated fine sands or silty sands.

Liquefaction Mitigation Techniques:

1. Soil Improvement Methods:

Dewatering

Relief well (to reduce porewater pressure)

Excavation of poor soils and replacement

with compacted fill grout injection

In situ densification (e.g vibroflotation,

Terraprobe, impact densification, dynamic

compaction, compaction piles, etc.)

Placement of additional fill (to increase

overburden pressures and soil strength)

2. Structural Fortification:

Strengthen structural connections

Add grade beams and tie beams

Extend pile support into deeper stable

soils.

(c)

Sol:

Population = 50,000

Discharge, Q = 50, 000 200

= 10 MLD = 0.1157 m3/sec

80% of water converted into sewage

Sewage factor = 0.8

Dry weather flow QDWF = 0.8 0.1157

= 0.0926 m3/sec

Assume maximum dry weather flow factor 3

Maximum dry weather flow

(QDWF)max = 3 QDWF

= 3 0.926 = 0.2778 m3/sec

Wet weather flow

360

AIRQWWF

(md) machine direction ribs

(cm

d) c

ross

mac

hine

dir

ectio

n ri

bs

Aperture

Junction



R = Rainfall intensity in mm/hr

Time of concentration tc = 40 + 10 = 50 min

as t > 20 min, a = 40, b = 20

Rainfall R =bt

a4.25

c

2050

404.25

= 14.51 mm/hr

360

51.145.050QWWF

= 1 m3/sec

Combined sewer discharge

Q = QDWF + QWWF

= 0.2778 + 1 = 1.2778 m3/sec

We know 2/13/2 SRn

1AQ

4

dR

1.2778 = 2/13/2

2 S4

d

015.0

1d

8

Sewere is running half A = 2d8

Assume manning’s constant n = 0.015

d8/3 = 3.889

d = 1.6641 m

Diameter of the sewer pipe is 1.6641 m

(d) (i)

Sol:

Advantages of the Tilting of Rails:

1) The tilting of rails maintains the gauge

properly

2) the wear of the head of the rail is

uniform due to tilting of rails

3) The tilting of rails increases the life of

sleepers as well as rails.

Disadvantages of the coning of wheels:

1) The smooth riding is produced by the

coning of wheels. But the pressure of

the horizontal component near the

inner edge of the rail has a tendency to

wear the rail quickly.

2) The horizontal component tends to turn

the rail outwardly and hence the gauge

is sometimes widened.

3) If no base-plates are provide, the

sleepers under the outer edge of the rail

are damaged.

(ii)

Sol:

Length of rail for B.G track = 12.80 m ≃ 13

Total length of track = 26 m

Total number of rail is13

26= 2 rails

Sleeper density = (n + 6)

= (13 + 6)

= 19 sleepers

Hence, total number of sleepers

= 2 19

= 38 sleepers

: 5 : Test – 14




(e)

Sol:

Given Horton’s equation

f = 5 + 10e–2t

Generally Horton’s equation

ft = fc + ktcc eff

o

Where fc = 5, cc ffo = 10

hr/mm15foc

at t = 0, ft = 15 mm/hr

at t = 1, ft = 5 + 10 e–2 = 6.3533 mm/hr

at t = 2, ft = 5 + 10 e–4 = 5.1831 mm/hr

at t = 3, ft = 5 + 10 e–6 = 5.02478 mm/hr

at t = 4, ft = 5 + 10 e–8 = 5.0033 mm/hr

at t = 5, ft = 5 mm/hr

From graph 6 = 5 + 10e–2 t

t = 1.1513 hr

Infiltration loss

4

1513.1

t2 dte105I

= 14.3 mm/hr

Surface runoff = Rainfall – infiltration (we

take in i > f values only)

= (1 – 0.1513) 6 + 10 + 7 – 14.3

= 7.7922 mm = 7.7922 10–3 m

Volume of runoff V = A 7.7922 10–3

= 36

36

m.M10

107922.7101000

= 7.7922 M.m3

02.

(a)

Apply Bernoulli’s equation between

(S) & (D)

D

2DD

s

2s Z

g2

VPZ

g2

VpS

SDDs ZZ

PP

78.081.9

101

1

0 1 2 3 4 51.1513

23456789101112131415

t (hrs)

i (mm/hr) 1.5 m

4 m

1.5 m

A B

S

D

C

: 7 : Test – 14


Ps = 6.65 kPa


(C) & (D)

D

2DD

C

2CC Z

g2

VPZ

g2

VP

DC

2D ZZg2

V

5.581.92VD

VD = 10.38 m/s


(A) & (C)

C

2CC

A

2AA Z

g2

VPZ

g2

VP

5.1086.120PA

PA = 14.37 9.81 0.8

kPa772.112PA


(A) & (B)

B

2BB

A

2AA Z

g2

VPZ

g2

VP

g2

VPP 2BAB

5.5PP AB

PB = 69.608 kPa

Assuming vapour pressure 2.5 kPa,

atmospheric pressure 101 kPa

Ps = 2.5 kPa


(S) & (D)

D

2DD

S

2SS Z

g2

VPZ

g2

VP

8.081.9

101Z

8.081.9

5.2S

ZS = 12.55 from D

Maximum height of summit

= 12.55 – 5.5

= 7.05 from water surface

(iv) maximum vertical depth of the pipe

= 12.55 m

(b)

Sol:

Rock

3 m

2 m

4 m

O

2 m

4 m

8 m

12 m

14 m

17 m

FictitiousFooting

Silt

Clay

Clay

Clay-I

Clay-II

Clay-III

12

m693

2L

3

2



Assume that the total load 2500 kN acts at a

depth of3

2L from the bottom of the pile cap

on a fictions footing as shown in the above

figure. The three clay layers (I, II & III)

shown above under go consolidation

settlement. The consolidation settlement is

determined by the following equation for each

clay layer.

o

o10

o

cf log

e1

CHS

o = effective in-situ a stress at centre of

each clay layer

= Increase in effective stress at centre of

each clay layer due to fictious footing

load.

Computation of o at centre of each clay

layer:

For clay layer-I

o = 2 + 2 + (4 + 2)

= 2 16 + 2 19.2 + 6 (19.2 – 9.81)

= 126.74 kPa

(w is taken as 9.81 kN/m3)

For clay layer II

o = 2 16 + 2 19.2 + 8 (19.2 – 9.81)

+ 1(18.24 – 9.81)

= 153.95 kPa

For clay layer –III

81.9205.181.924.18195.153o

= 177.67 kPa

Computation of at centre of each clay

layer:

ZLZB

Q

oo for 2V to 1H load

dispersion

Q = 2500 kN

Bo = width of pile block = (3S +d)

= (3 0.9 + 0.3)

= 3 m

Lo = length of pile block = (4S + d)

= 4 0.9 + 0.3 = 3.9 m

Z = depth to the centre of each clay layer,

from the fictious footing.

For layer- I; (Z = 2 m)

kPa75.8429.323

2500

For layer- II; (Z = 5 m)

s

sLo

Bo

: 9 : Test – 14


kPa11.3559.353

2500

For layer- III; (Z = 7.5 m)

88.205.79.35.73

2500

kPa

Computation of Settlements:

Layer-I :

o

o10

o

C1 log

e1

CHS

74.126

75.8474.126log

8.01

23.04 10

= 0.114 m

Layer-II:

95.153

11.3595.153log

08.11

34.02S 102

= 0.029 m

Layer-III:

67.177

88.2067.177log

7.01

2.03S 103

= 0.017 m

Total consolidation settlement,

S = S1 + S2 + S3

S = 0.114 + 0.029 + 0.017

= 0.16 m

= 16 cm

Total settlement, S = 16 cm

(c)

Sol:

The time required for the pedestrians to

cross the road = walk time +pV

W

sec67.262.1

2010

The green time Given for concurrent

vehicles to cross the road

= Given interval + intergreen time

= 14 + 6 = 20 s

Which is less than pedestrian crossing time

of 26.67. Therefore, pedestrian crossing time

is insufficient.

v = 50 18

5 = 13.89 m/s

gf2

vvtSSD

2

5.081.92

89.13189.13

2

= 33.56 m

Time required to cross the intersection

v

WSSD

89.13

20656.33 = s28.4

89.13

56.59

The intergreen time (IG) = Amber + all red

time

= 2 + 2 = 4 sec

IG < 4.28 sec

Dilemma occurs at the junction

To clear the dilemma the IG should be

increased to 4.28 sec

: 11 : Test – 14


03. (a)

Sol:

Risk = 1 - safety

= 1 - 0.8 = 0.2

Risk = 1 – qn

0.2 = 1 – q20

q20 = 0.8

(1 – p)20 = 0.8

p = 0.0110

90p

1T years

For return period of T = 90 years

p1nny t

= – 011.01nn

yt = 4.5043

Frequency factor KT =2825.1

5778.0y t

2825.1

5778.05043.4

= 3.0616

Magnitude of flood at 90 years

XT = TKX

= 1200 + 3.0616 650

= 3190 m3/sec

To increase the life of the structure 40 years

Risk = 1 – (1 – p)n

0.2 = 1 – (1 – p)40

(1 – p)40 = 0.8

p = 5.5630 10–3

Return period T = 180p

1 years

For return period T = 180 years

yt = p1nn

yt = 5.188

Frequency factor2825.1

5778.0yK t

T

2825.1

5778.0188.5

= 3.6

Magnitude of flood at 180 years

Tkxx TT

= 1200 + 3.6 650

= 3537 m3/sec

(b)

Sol:

Assuming the each pipe as following flow

Consider a loop ABCDA

40B

70

C 20

k = 2

80

30

A60

k = 2

D

30

50

k = 1

k = 1



Hardly cross procedure for 1st correction.

Pipe Assumed flow Qak(given)

HL = k 2aQ

a

L

Q

H Corrected Q after first correctionQ = Qa + 1

AB 30 2 1800 60 9.23

BC 70 1 4900 70 49.23

CD 50 2 5000 100 29.23

DA –30 1 –900 30 –50.77

HL = 10,800a

L

Q

H260

Where, 77.202602

800,10

Q

Hx

H

a

L

L1

Hardy cross procedure for 2nd correction

Pipe Assumed flow Qak(given)

HL = k 2aQ

a

L

Q

H Corrected Q after first correctionQ = Qa + 1

AB 9.23 2 170.38 18.46 4.3544

BC 49.23 1 2423.6 49.23 44.3544

CD 29.23 2 1708.78 58.46 24.3544

DA –50.7 1 –2577.6 50.77 –55.6455

HL = +1725.16a

L

Q

H= 176.92

92.1762

16.1725

Q

Hx

H

a

L

L2

= – 4.8755

The final flow rate in each pipe

40

44.3544

24.3544

80

55.6455

A60

4.3544B

C

D

20

: 13 : Test – 14




(c)

Sol:

(i) To find seepage loss rate, Q:

Nf = 6, Nd = 12, H = 12 m

K = 1.333 10–6 m/sec

= 1.333 10–6 60 60 24

= 0.1152 m/day

Seepage loss, Q =d

f

N

NKH

12

6121152.0

= 0.691 m3/day.m

(ii) To find factor of safety against heave on the

D/S of sheet pile:

F.O.S safety against heave,uw h

zF

Z = depth of sheet pile below bed level

hu = uplift pressure head at bottom of sheet

pile

Z = 6 m, hu = 3.5 m

= sat – w = 19.5 – 9.81 = 9.69 kN/m3

w = 9.81 kN/m3

uw h

zF

5.381.9

669.9

= 1.69

12 m

6 m15 m

z

w hu

: 15 : Test – 14


(d)

Sol:

Assume

Kv = 0.98, speed ratio, Ku = 0.46

gH2KV v

30081.9298.0

= 75.19 m/sec

gH2Ku u

30081.9246.0

= 35.29 m/sec

QH

Po

300Q9810

10600085.0

3

Q = 2.399 m3/sec

60

DNu

60

550D = 35.29 m/sec

D = 1.225 m

10

1

D

d

10

225.1d

= 0.1225 m = 122.5 mm

Area of jet

2)1225.0(4

a = 0.0118 m2

Total jet area requiredV

Q

19.75

399.2

= 0.0319 m2

No. of jets required0118.0

0319.0

= 2.7 ≃ 3

Thus three jets are required each having a

diameter

2dd

A

2d4

0319.0

d = 116.4 mm

Check:

5.10

1

225.1

1164.0

D

d

04. (a)

Sol:

Discharge Q = 9000 m3/day

hr/it24

109000 3

= 375000 lit / hr

Rate of filtration = 3000 lit/hr/m2

To find the size & Number of filter beds

Total area of filter =filtrationofRate

eargDisch



3000

375000

= 125 m2

Assume size of each filter = 10 5

Number of filters =50

125 = 2.5 ≃ 3

(ii) Design of under–drainage system.

Manifold– lateral under drainage system.

Let us assume that the area of the perforation

is 0.2% of the total filter area.

Total area of perforation = 0.2% Filter

area

50100

2.0 = 0.1 m2

Assuming the area of each lateral

= 2 times the area of perforation

= 2 0.1

= 0.2 m2

Assuming the area of the manifold to be about

twice the area of laterals.

Area of Manifold = 2 0.2 = 0.4 m2

Diameter of manifold (d)

4.0d4

2

d = 0.7136 m

d = 71.36 cm

length of each lateral =2

713.05

= 2.14 m

Assuming spacing between lateral

= 20 cm c/c

Number of lateral on each side

lateraleachofspacing

length =

2.0

10 = 50

Total Number of lateral = 2 50 = 100

lateralofNumber

lateralofareaTotallateraleachofArea

100

2.0 = 0.2 10–2

2102.04

lateralofDiameter =0.05 m

= 5 cm

Assuming the diameter of perforation = 8 mm

231084

nperforatioeachofArea

= 5.026 10–5 m2

Number of perforation required

nperforatioeachofArea

nperforatioofareaTotal

510026.5

1.0

= 1989.65

Pipe

Laterals

0.713 m

0.2m

5 m

10 m

Central Manifold

: 17 : Test – 14


Perforation on each side

lateralofNumber

nperforatioofnumberTotal

100

65.1989 = 19.96 ≃ 20

Spacing of perforation

lateraleachpernperforatioofNumber

lateraleachoflength

20

14.2 = 107 10–3 m

= 10. 7 cm c/c

Check:

05.0

14.2

lateralofDiameter

lateraloflength = 42.8 ≯ 60

Hence ok.

Rate of backwash = 10, 000 lit/hr/m2

Volumes of back wash water

= ROB DOB area of each filter

= 5060

30000,10

= 250 m3

% water used for back washing

=filter

filter

ADOFROB

ADOBROB

100

30)6024(3000

3010000

100

= 7.09 %

(b)

Sol:

Spillway is a high weir, Cd = 2.2

To calculate approximate value of He,

Q =Cd Le2/3

eH Le = 8 12

= 96 m

10000 = 2.2 (96) 2/3eH

m08.13962.2

10000H

3/2

e

Given

Height of spillway above the river bed

h = 210 110 = 100 m

33.164.708.13

100

H

h

d

It is a high spillway, velocity head can be

neglected,

7.108.13

10008.13

H

hH

e

e

Therefore, there is no effect of tailwater

condition.

Up stream slop: Upstream face of dam and

spillways is kept vertical. In the lower part a

batter of 1:10 may be provided Cd remains

unchanged.

Effective length of spillway:

Le = L 2 [ NKP + Ka] He

With rounded nose having 90 cut,



Kp = 0.01 and Ka = 0.1, N = 7

Le = 96 –2 [ 7(0.01) + 0.1] 14.5

= 91.07 (He = assumed 14.5)

Q = CLe2/3

eH

10000 = 2.2 (91.07) 2/3eH

He = 14.44 m 14.5 m ok

(assumed)

Therefore crest profile will be designed for Hd

= 14.44 m.

Velocity of approach

44.141003796

10000Va = 0.75 m/s

m03.0g2

Vh

2a

a , negligible

Down stream profile:

For a vertical up stream face,

yH2x 85.0d

85.1

85.0

85.1

85.0d

85.1

44.142

x

H2

xy

349.19

xy

85.1

Tangent point is given by

0.25.0

1

dx

dy

2349.19

x85.1

dx

dy 85.0

x = 35.7728 m

y = 38.6736 m

x y

1 0.5

2 0.186

3 0.394

4

5

:

:

:

34 35.2

35.7728 38.673

According to US army corps, the upstream

curve of ogee spillway having vertical

upstream face should have the following

equation.

d85.0

d

85.1d H126.0

H

H27.0x724.0y

–0.4315 625.0d

35.0d H27.0xH

The upstream profile extends up to

x = –0.27 Hd = –0.27 14.44 = –3.89 m

Hd = 14.44 m

At

x =0, y = 8.0436 m

x = –0.50 y =6.2578 m

x = –1.0 y = 4.7213 m

x = –2.0 y = 2.436 m

x = –3.0 y = 1.315 m

x = –3.8988 m y = 1.8194 m

: 19 : Test – 14


(c)

Sol:

Stations A and B are free from local

attraction because the F.B and B.B

difference of line AB is 180o.

Stations C & D are affected by local

attraction

(i) Calculation of included angles

A = FB of AB – FB of AD

= (124o30 – 17o45) = 106o45

B = FB of BC – FB of BA

= (68o15 – 304o30) + 360o = 123o45

C = FB of CD – FB of CB

= (310o30 – 246o) = 64o30

D = FB of DA – FB of DC

= 200o15 – 135o15 = 65o

Arithmetic check: Sum of included angles

= 360o = (2n – 4) 90o

No correction for angles

(ii) Corrected bearings of lines:

Corrected FB of BC = 68o15 as station ‘B’

is free from local attraction

Corrected BB of BC = 248o15

Corrected FB of CD = corrected FB of CB

+ C

= 248o15 + 64o30 = 312o45

Corrected BB of CD = 132o45

Corrected FB of DA = corrected FB of DC

+ D

= 132o45 + 65o = 197o45

Corrected BB of DA = 17o45 (which is

correct)

(iii)Corrected TB of all the lines:

Corrected T.B of AB = 124o30 – 2o20

= 122o10

Corrected TB of BC = 68o15 – 2o20

= 65o55

Corrected TB of CD = 312o45 – 2o21

= 310o25

Corrected TB of DA = 197o45 – 2o20

= 195o25

(d)

Sol:

Discharge Q = AV

20 = by V

V =by

20 (i)

The critical depth for a rectangular channel

section is by equation.

3/1

2

2

c gb

Qy

3/1

2

2

b81.9

20

3/2b

442.3



Using Manning’s formula

2/13/2 SRn

1V

From equation (i)

2/1

3/2

c

c

c

)0064.0(y2b

by

015.0

1

by

20

3/2c

3

21

c

)y2b(

)by(75.3

3/21

3/5c

)y2b(

)by(

3/2

3/2

3/5

3/2

b

442.32b

b

442.3b

75.3

3/2c b

442.3y

092.2bb

884.6b

3/5

3

21

3/2

3/2

092.2bb

804.6b 5

/2

3/2

6/53/2

b026.3b

884.6b

b = 2.41 m

05. (a)

Sol:

By Manning’s formula, we have

2/13/2 SRn

1AQ

2/13/2

3000

1

225.7

25.7

02.0

125.7Q

= 16.35 m3/sec

We know the gradually varied flow equation

3

2fo

gA

TQ1

SS

dx

dy

……… (i)

3/4

22

f R

VnS

y = 2 + 0.75 = 2.75 m

A = 7.5 2.75 = 20.625

P = 7.5 + 2 2.75 = 13

m5865.113

625.20

P

AR

7927.0625.20

35.16

A

QV m/sec

3/4

22

f )5865.1(

)7927.0()02.0(S

= 1.3584 10-4

3

2

3

2

)625.20(81.9

5.7)35.16(

gA

TQ

= 0.0233

From equation (i)

)0233.01(

10)3584.1333.3(

dx

dy 4

= 2.021 10–4

The water surface slope Sw with respect to

horizontal is given by

dx

dySS ow

= (3.333 – 2.021) 10–4

= 1.312 10–4

: 21 : Test – 14


(b)

Ans:

1. Soil Stabilization with Cement:

The soil stabilized with cement is known as

soil cement. The cementing action is believed

to be the result of chemical reactions of

cement with siliceous soil during hydration

reaction. The important factors affecting the

soil-cement are nature of soil content,

conditions of mixing, compaction, curing and

admixtures used. This technique can be used

for both cohesive and non-cohesive soils.

2. Soil Stabilization using Lime:

Slaked lime is very effective in treating heavy

plastic clayey soil. Lime has been mainly

used for stabilizing the road bases and the

subgrade. Lime decreases the liquid limit,

plasticity index and expansiveness of clay.

3. Soil Stabilization with Bitumen:

Asphalts and tars are bituminous materials

which are used for stabilization of soil,

generally for pavement construction.

Bitminous materials when added to a soil, it

imparts both cohesion and reduced water

absorption. Depending upon the above actions

and the nature of soils, bitumen stabilization

is classified in following four types:

Sand bitumen stabilization

Soil bitumen stabilization

Water proofed mechanical

stabilization, and

Oiled earth

4. Chemical Stabilization of Soil:

Calcium chloride being hygroscopic and

deliquescent is used as a water retentive

additive in mechanically stabilized soil bases

and surfacing. The vapor pressure gets

lowered, surface tension increases and rate of

evaporation decreases. The freezing point of

pure water gets lowered and it results in

prevention or reduction of frost heave

5. Electrical Stabilization of Clayey Soils:

Electrical stabilization of clayey soils is done

by method known as electroosmosis. This is

an expensive method of soil stabilization and

is mainly used for drainage of cohesive soils.

6. Soil Stabilization by Grouting:

In this method, stablizers are introduced by

injection into the soil. This method is not

useful for clayey soils because of their low

permeability. This is a costly method for soil

stabilization.



7. Soil Stabilization by Geotextiles and

Fabrics:

Geotextiles are porous fabric made of

synthetic materials such as polyethylene,

polyester, nylons and polyvinyl chloride.

Woven, non-woven and gird form varieties of

geotextiles are available. Geotextiles have a

high strength. When properly embedded in

soil, it contributes to its stability. It is used in

the construction of unpaved roads over soft

soils.

(c)

Sol:

Quantity of sewage produced per day

= 200 150 0.8

= 24,000 l/day

Quantity of sewage produced during detention

time of 24 hrs

2424

000,24 lit

= 24,000 lit

= 24 m3

Now sludge deposition in the tank @ 40

lit/person/year assume cleaning period 6

months

Sludge volume =12

620040

= 4000 lit = 4 m3

Total required capacity of the tank

= Capacity of sewage + capacity of sludge

= 24 + 4 = 28 m3

Assume 1.5 m depth, length : width = 2 : 1

Surface area 667.185.1

28 m2

L B = 18.667

2B2 = 18.667

B = 3.055 3 ,

L = 6 m

Hence, use 6 m 3 m (1.5 + 0.3) m sized

septic tank

Diameter of soak-pit:

Area required for the soak pit =2500

24000

= 9.6 m2

Area of soak-well required = 9.6 m2

6.9d4

2

d = 3.5 m

Diameter of soak pit = 3.5 m

: 23 : Test – 14


(d) (i)

Sol:

B.S I.S F.S Rise Fall R.L R.L

0.405 69.980 A Starting point

1.035 0.630 62.350 B

1.930 0.895 61.455 C

2.895 0.965 60.490 D

3.805 0.910 59.580 E

0.715 4.760 0.955 58.625 F. (Change point)

2.060 1.345 57.280 G

3.160 1.100 56.180 H

4.415 1.255 54.925 I End point

Back sight

= 1.120

Fore sight =

9.175

Rise

= 0

Fall

=8.055

BS – FS = 1.120 – 9.175 = - 8.055 m

Rise – Fall = 0 – 8.055 = – 8.055 m

Last R.L– First R.L = 54.925– 62.980 = – 8.055 m

Distance between the starting point A and the end point B

= 8 25 = 200 m

Fall from A to B = 8.055 m

Gradient of the straight line AB83.24

1

200

055.8

Falling gradient = 1 vertical to 24.83 horizontal



(ii)

Ans:

The fundamental axes of a theodolite are

1. Vertical axis

2. Trunnion axis

3. Line of collimation

4. Altitude level axis

5. Axis of plate level

A theodolite is said to be in proper condition

if the following conditions are satisfied:

1) The axis of the plate is perpendicular to the

vertical axis

2) The trunnion axis is perpendicular to the

vertical axis

3. The line of collimation is perpendicular to

the trunnion (horizontal) axis.

4. The axis of the altitude level is parallel to

the line of collimation.90o

Horizontal Axis

Line of Collimation

Altitude level Axis

Plate level Axis

Plate level Axis

90o90o

Vertical Axis

Fundamental Axes of a Theodolite

: 25 : Test – 14


(e) (i)

Ans:

Classification of River Training works:

1. High water training: This is also called

training for discharge. The river is trained to

provide sufficient and efficient cross-sectional

area for the expeditious passage of maximum

flood. It concerns mainly with alignment and

height of embankment for a given flood

discharge.

2. Low water training: In this case the river is

trained to provide sufficient depth for

navigation during low stage of river. This is

also called training for depth and is usually

achieved by contraction of the width of the

channel.

3. Mean water training: In this case the river is

trained to correct the configuration of river

bed for the efficient transport of sediment

load in order to keep the channel in good

shape. It can be called training for sediment.

Components of river training works:

Marginal embankment (dyke or levee):

The marginal embankment or dyke is an

earthen embankment of trapezoidal section

constructed approximately parallel to the bank

of the river to confine the flood water within a

section between the embankments.

Guide bank: Guide banks must be

constructed on both the approaches to protect

the structure from erosion. It is an earthen

embankment with curved heads on both the

ends.

As the guide bank was first designed by Bell,

it is sometimes known as Bell’s Bund. The

Guide bank serves the following purposes.

a) Protects the barrage from the effect of

scouring and erosion

Bridge Road

Bank protection

S

S

G

Dyke

Guide bank Guide bank

S = SpurG=Groyne

G

G



b) Provides a straight approach towards the

barrage

c) Controls the tendency of changing the

course of the river

Spurs: These are temporary structures

permeable in nature provided on the curve of

a river to protect the river bank from erosion.

These are projected from the river bank

towards the bed making angles 60 to 75 with

the bank of the river.

The function of the spurs is to break the

velocity of flow and to form a water pocket

on the upstream side where the sediments get

deposited.

Groynes: The function of groynes is similar

to that of spur. But these are impervious

permanent structures constructed on the curve

of a river to protect the river bank from

erosion. They extend from the bank towards

the bed by making an angle of 60 to 75 with

the bank.

Classification according to the function it

serves:

(a) Attracting groyne.

(b) Deflecting groyne.

(c) Repelling groyne.

(d) Sedimenting groyne.

Special types of groynes, such as:

(a) Denehy’s T-headed groyne.

(b) Hockey type groyne.

(c) Burma type groyne.

(ii)

Ans:

Canal fall or drop:

It is a structure constructed across a channel

to lower down the water level and dissipate

surplus energy. It is required when the natural

slope of the ground is greater than the

designed bed slope of the channel. The

difference in slopes is adjusted by

constructing a vertical drop.

Ogee falls, rapids and stepped falls, notch

falls, vertical falls and glacis type falls.

1. Ogee fall:

U/S BED

D/S BED

Ogee fall

: 27 : Test – 14


(i) There was considerable draw down

effect on the u/s resulting is bed

erosion.

(ii) Due to smooth transition, the kinetic

energy was preserved till sufficient

depth was scoured out below the fall

to ensure the formation of the

hydraulic jump.

2. Rapid fall:

Such a fall consists of a glacis sloping at 1

vertical to 10 to 20 horizontal. The long glacis

assured the formation of hydraulic jump. The

gentle slope admitted timber traffic. Hence,

the fall worked admirably. However, there

was very high cost of construction.

3. Notch fall:

The notches were designed to maintain the

normal water depth in the u/s channel at any

two discharge values. The depth discharge

relation was thus maintained with close

approximation. As the channel approached

the fall, there were neither drawdown nor

heading up of water.

4. Vertical Drop fall:

In the vertical drop fall, the nappe impinges

clear into the water cushion below.

5. Glacis type fall:

The glacis type of fall utilized the standing

wave phenomenon for dissipation of energy.

Generally glacis type fall is suitable as a

meter. The vertical drop fall is not suitable as

a meter due to the formation of partial

vacuum under the nappe.

Escapes: Structure constructed on an

irrigation channel for the disposal of surplus

water from the channel (It is called Surplus

Escape).

Sometimes, escapes are provided in the head

reaches of canal to scour out bed silt

deposited in the canal. They are called Canal

Scouring Escapes.

Some times, at the tail end of the channel

when it meets a drain, an escape is provided

to maintain the required FSL in the canal.

Such an escape is called Tail escape.

U/S

D/S1 in 15

Rapid Fall

U/S F.S.L Notch pier Notch Side wall

Sill of Notch Foundation wall



06. (a)

Sol:

(i) The discharge Q delivered by the pump is

given by

Q = kB1D11f

V

Given

B1 = 15 mm = 0.015 m

D1 = 400 mm = 0.400 m

1fV = 3 m/s

Assuming k = 1, we get

Q = 1 0.015 0.400 3

Q = 0.0565 m3/s

= 3390 litres/minute

(ii) Manimetric efficiency is given by

1w

mmano uV

gH

1

mano

m1w H

g

uV1

Hm = 30 m; mano = 75% = 0.75

75.0

30

g

uV 1w1 40 m ………. (i)

i.e., total head developed by the pump = 40 m

The pressure rise through the impeller

PP1 is 65% of the total head developed

by the pump. Thus

m26)4065.0(PP1

Applying Bernoulli’s equation between the

inlet and outlet tips of the impeller and

neglecting the head loss in the impeller,

g

uV

g2

Vp

g2

Vp 1w2

112

1

g2

V

g

uV

g2

Vpp 211w

21 1

V = Vf = 3 m/s

Thus by substitution, we get

g2

V40

81.92

)3(26

21

2

Or V1 = 16.84 m/s

But 2f

2w1 11

VVV

2f

21w 11

VVV

1f

V 3 m/s

Thus by substitution, we get

22w )3()84.16(V

1

= 16.57 m/s

By substituting in equation (i), we get

4081.9

u57.16 1

Or57.16

81.940u1

= 23.68 m/s

But60

NDu 1

1

: 29 : Test – 14


Or60

N400.068.23

Or400.0

6068.23N

= 1131 rpm

i.e., speed of the pump = 1131 rpm

(iii) From outlet velocity triangle, we have

1

1

w1

f

Vu

Vtan

Or 4219.057.1668.23

3tan

= 22 52

i.e., blade angle at outlet = 2252

(iv) From inlet velocity triangle

u

Vtan f

Or cotVu f

= 60

u = 3 cot 60 = 1.732 m/s

Also60

DNu

Or60

1131D732.1

Or1131

60732.1D

Or D = 0.029 m = 29 mm

i.e., the diameter of the impeller at inlet

= 29 mm

(b)

Sol:

Q yi= Population Per capita BOD

20 yi = 50000 80 10–3

/mg20020

108050000y

3

i

WR

WWRRmix QQ

yQyQy

315.05

y2315.0051 W

ye = 22.6 mg/l

100y

yy

i

ei

100200

6.22200

= 88.7%

ymix = 1 mg/l = Lo 5k1e1

Lo = 1.2872 mg/l

2315.05

095DO min

= 8.60 mg/l

Do = (DO)sat – (DO)min = 9 – 8.6 = 0.4 mg/l

Time taken to travel 60 km

60602412.0

1060 3

= 5.78 days ≃ 5.8 days

tko

tktk

22

o18.5

221 eDeekk

LkD



8.58.08.58.08.53.0 e4.0ee3.08.0

2870.13.0

= 0.131 mg/l

DO at 60 km (DO)sat – D5.8

= 9 – 0.131

= 8.868 mg/l

o1

12o

1

2

12c Lk

kkD1

k

kn

kk

1t

= 0.5023 days

Point at which it occurs

= velocity Time

= 0.5023 24 60 0.12

= 5.20 km

(c)

Sol: Base flow B = 10 m3/sec

TimeOrdinates 3h

FHG

Ordinates 3

hours

DRH

FHG – B

3 hr UHG

R

DRH

3 hr DRH 1st

rain UHG

3 cm

(i)

3 hr DRH 2nd

rain UHG

1.5 cm

(ii)

DRH

= (i) + (ii)

FHG = DRH

+ B

0 10 0 0 0 - 0 0

3 30 20 10 30 0 30 40

6 70 60 30 90 15 105 115

9 160 150 75 225 45 270 280

12 130 120 60 180 112.5 292.5 302.5

15 100 90 45 135 90 225 235

18 76 66 33 99 67.5 166.5 176.5

21 60 50 25 75 49.5 124.5 134.5

24 42 32 60 48 37.5 85.5 95.5

27 30 20 10 30 24 54 64

30 20 10 5 15 15 30 40

33 10 0 0 0 7.5 7.5 17.5

0 0 0

: 31 : Test – 14


Volume of the runoff using trapezoidal rule

= Area of Direct rainfall hydrograph

....yyy2yy2

h321no

10203250

66901201506020200

2

36003

= 6,674,400 m3

Runoff depth =areaCatchement

DRHofArea

= 10001072.333

400,674,66

mm

= 20 mm = 2 cm

07. (a)

Sol:

Given data:

Sandy soil, = 35o,

d = 17 kN/m3

Gs = 2.7, take w = 9.81 kN/m3

e1

G. swd

e1

7.281.917

e = 0.558

e1

eG swsat

558.01

558.07.281.9 = 20.51 kN/m3

At a depth of 4 m,

= 2.5 d + 1.5

= 2.5 d + 1.5 [sat – w]

= 2.5 17 + 1.5 [20.51 – 9.81]

= 58.55 kPa

(i) On the horizontal plane, the shear

strength of soil,

S = tan (C = 0 for sand)

= 58.55 tan 35o

= 41 kPa (Ans)

(ii) If W.T rises to G.L

At 4 m depth, = 4

= 4[20.51 – 9.81]

= 42.8 kPa

Shear strength S = tan

= 42.8 tan 35

= 29.97 kPa

Change in shear strength due to rise of WT

is = 41 - 29.97

= 11.03 kPa (decreases)

Shear strength decreases by 11.03 kPa

Ssat1.5

2.5 m4 m

d



(b)

Sol:

By Kerman momentum integral equation

xU 2

o

Where = Momentum thickness:

U

U1

U

U

o

dy

y

, dy = d and the limits of are

0 and 1

U

U= f() = 3

2

1

2

3

d

2

1

2

31

2

1

2

3f 3

1

0

3

1

0

6442

4

1

4

3

2

3

4

3

4

9

2

3d

1

0

754532

2854

3

854

3

34

9

22

3

1

0

75453

2854

3

854

3

34

9

4

3

28

1

20

3

8

1

20

3

12

9

4

3

= 0.1392

dx

dU1392.0 2

o

From the boundary conditions for a laminar

boundary layer

0

d

dfU

dy

dUT

0y

o

f() = 3

2

1

2

3

2

3

2

3

2

3

d

df

0n

2

0n

o =

U

2

3

Equating the two expressions for o,

0.1392

U

2

3

dx

dU 2

dxU

77.10d

Integrating on both sides

CxU

77.102

2

At x = 0, = 0 C = 0

xU

x55.21

22

xU

x65.4

xRe

x65.4

xU

Re x

: 33 : Test – 14


(c)

Sol:

Adopt Btigh’s creep theory

L = Total creep length

= 10+13+14+15+18 = 70 m

H = 7 m

10H

LC

at A:

lA = 10+13+14 = 37 M

m7.310

37h '

A

m3.37.37h A

7.3

3.3

A

hi A

(ii) Uplift pressure at B:

m5215141310B

m2.5C

h B'B

m8.12.57h B

Uplift pressure head at B

hB = 1.8 m

Uplift pressure at B

pB = WhB

= 18 kPa

(iii) Thickness of floor at C

lc = 10 + 13 = 23 m

Balanced head

m3.210

23

C

c1h '

c

Unbalanced head

3.27hHh 1cc = 4.7 m

Thickness of floor at c

1s

h

3

4t

c

cc

m5124.2

7.4

3

4

At C: Theoretical thickness of impervious

apron = 5 m

(d)

Sol:

Given: design speed V = 80 kmph,

gradients n1 = + 3.0% and n2 = – 5.0%

(a) Determination of safe stopping sight

distance, SSD

As there is ascending gradient on one side of

the summit and descending gradient on the

other side, the effect of gradients on the SSD

is assumed to get compensated and hence

ignored in the calculations.

13 m 15 m5 m 7 m

9 m

7 m

C A B



f254

VVt278.0SSD

2

Assuming t = 2.5 sec and f = 0.35 for V = 80

kmph

35.0254

805.280278.0SSD

2

= 55.6 + 72.0 = 127.6 ≃ 128 m

(b) Determination of length of summit curve:

Deviation angle N = 0.03 – (–0.05) = 0.08

Assuming L > SSD,

4.4

12808.0

4.4

NSL

22 = 297.9 m ≃ 298 m

This value of summit curve length L is

greater than SSD of 128 m as per the

assumption and therefore the calculated

length may be accepted for design

Length of summit curve, L = 298 m

08. (a)

Sol:

Given data:

C = 15 kPa, = 30o

n = 40%

Gs = 2.67

For = 30o , Nc = 37.2

Nq = 22.5

N = 19.7

e =n1

n

667.04.01

4.0

Assume w = 9.81 kN / m3

wsat

= (19.64 – 9.81)

= 9.83 kN/m3

For a general shear failure, Terzaghi’s

equation. To find gross allowable bearing

capacity, qs is given below.

DqF

1q nus

D

2

B2.01NB5.0

1NDCNL

B3.01

F

1q

qc

s

As W.T is at G.L, is used in the above

equation

L = Length of footing in metres

283.9

L

32.017.19383.95.0

15.22283.92.3715L

33.01

3

1q s

66.19L

6.0148.29070.422558

L

9.01

3

1q s

D = 2 m

B = 3 m

: 35 : Test – 14


66.19L

1.5883.969.140

L

4.167186

= 443.39 +L

3.109

The given external gross allowable bearing

pressure, q = 455 kPa

Equating, qs = q

455L

3.10939.443

L = 9.40 m

Length required, L = 9.40 m

(b)

Sol:

Correction of Elevation:

This basic length is to be increased at the

rate of 7% per 300 m elevation above mean

sea level.

Correction for elevation =300

100

100

7

= 14 m

Length of runway after correction for

elevation = (600 + 14) = 614 m

Correction for Temperature:

Standard atmospheric temperature at mean

sea-level = 15oC

Taking the temperature gradient as equal to

6.5oC per 1000 m rise in elevation, the

standard temperature at the airport site will

be:

Temperature at R.L 100

1000

1005.615

= 14.35oC

Difference between airport reference

temperature and standard atmospheric

temperature = (28 – 14.35) = 13.65oC

Applying correction at the rate of 1% for

every 1oC

Correction for temperature:

65.13614100

1

= 83.81 say 84 m

Corrected runway length = (614 + 84)

= 698 m

Correction for Gradient:

Effective gradient600

2.95L.R2.98L.R

600

3 or 0.5%

Applying correction for the effective

gradient at the rate of 20% for each 1%

effective gradient

Correction for gradient1

5.0698

100

20

= 69.8 ≃ 70 m

Actual length of runway = (698 + 70) = 768



Check:

Total correction for elevation and

temperature = (14 + 84) = 98 m

Percentage increase 100600

98 = 16.33%

According to ICAO this should not be more

than 35%

(c) (i)

Ans:

In Movable stadia hair method stadia

intercept is variable but staff intercept is

kept constant. A leveling staff on which line

of sight will bisect is called as base. If base

is kept horizontal it is a horizontal subtese

bar. In India, the subtense bars are usually

3.5 m long. It is mounted on a tripod. A

small spirit level is provided to level it. The

alidade provides a line of sight

perpendicular to the bar, which is thereby set

normal to the line of measurement. The

targets are set apart at a known distance and

the horizontal angle between them is read by

a theodolite. The horizontal distance can be

calculated as follows

D = S/β (206265) if β is in seconds

(ii)

Sol:

H

f

cetandisGround

cetandisMapScale

Where

H = height of camera above the selected

datum

Let the average ground be the selected

datum

50.8

0.25020Hor

mH

cm20

m0.250

cm50.8

= 588.2 m

Height of the tower above its base is given

by

m80

2.5885

r

dHh

= 36.7625 m

(iii)

Sol:

At upper culmination (transit)

Declination of the star = = (90 – α) – θ

Where

α = altitude; = latitude

= declination of star

= (90 – 28o) – 50o = 12o

Since latitude () > altitude (α) at upper

culmination, declination of star is towards

north (N)

= 12o N

β/2

β/2 S/2

S/2

D

: 37 : Test – 14


(d)

Sol:

Thus 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

tt

dt

d

tt k

dt

d

Where kT = de oxygenation constant

Integrate on both sides

1n lt = – kt + c

At t = 0, lt = Lo 1n Lo = C

1n ktLo

t

lt = Loe–kt

If yt represents the total amount of organic

matter oxidised in t days.

yt = Lo –lt

yt = Lo – Lo e–kt = Lo (1 – e–kt)

CTt

o

y = (DOInitial – DOFinal) Dilution factor

it/mg1502

10026y C20

5

o

tko

cTt

To

e1Ly

150 = Lo (1 – e–0.23 5)

Lo = 219.50 mg/lit

Lo

Amount oforganic matter

t=0 Time in days t

lt

ace engineering academy · ace engineering academy hyderabad|delhi|bhopal|pune|bhubaneswar|...

Documents