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CIVIL ENGINEERING - Paper II

F-126, Katwaria Sarai, New Delhi - 110 016 011-41013406, Mobile: 8130909220, 9711853908Ph:

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Mains Exam Solution



1.(a) The velocity vector in an incompressible flow is given by

2 2V i j kxy 2xyz 6tz6xt yz 3t xy

(i) Verify whether the continuity equation is satisfied.(ii) Determine the acceleration in x direction at point A[1, 1, 1] and t = 1.0.

[12 Marks]

Sol. We have,

V = 2 2(6xt yz )i (3t xy )j (xy 2xyz 6tz)k

(i) Continuity equation in 3-dimensional form is

( u) ( v) ( w)t x y t

= 0

For incompressible flow

u v w 0x y z

Here,

ux

=2(6xt yz ) 6t

x

vy =

2(3t xy ) 2xyy

wz

=(xy 2xyz 6tz) 2xy 6t

z

Here,u v wx y z

= 6t + 2xy – 2xy – 6t = 0

So, continuity equation is satisfied.

(ii) ax =u u u uu. v. wx y z t

Here,ut

=2(6xt yz ) 6x

t

ux

=2(6xt yz ) 6t

x

uy =

22(6xt yz ) z

y

uz

=2(6xt yz ) 2yz

z

So, ax = 6x +(6xt + yz2). 6t + (3t + xy2) z2

+ (xy – 2xyz – 6tz) 2yzPutting x = 1, y = 1, z = 1 and t = 1 in ax, we get

ax = 6 × 1 + (6 × 1 ×1 + 1×12)·(6×1) + (3×1 + 1×1)2.12

+ (1×1 – 2 (1×1×1) – 6) · 2×1×1= 38 units




Mains Exam Solution

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1.(b) Three tube wells of 25 cm diameter each are located at the three vertices of an equilateraltriangle of side 100 m. Each tube well penetrates fully in a confined aquifer of thickness25 m. Assume the radius of influence for these wells and the coefficient of permeability ofthe aquifer as 300 m and 40 m/day respectively.

(i) Calculate the discharge when only one well is pumping with a drawdown of 3 m.

(ii) What will be the percent change in discharge of this well if all the three wells wereto pump such that the drawdown is 3 m in all the wells ?

[12 Marks]

Sol.Given :s = 3m, rw = 0.125 m, R = 300 m, K = 40 m/day, b = 25 m(i) When only one well is pumping at steady state.

T = K b = 1000 m2/day

w

RQs Inr2 T

Q = 32421.82 m day

(ii) When all the three wells are pumping at steady state.

3

jijtotal i

i 1

Qrs S F

2 T

Q1 = 2 3Q Q

st = *

3Q RIn2 T r

where, 1/3*11 12 13r r r r

r11 = 0.125 m, r12 = r13 = 100m, and st = 3 mQ = 1875.25 m3/day

Percentage reduction in discharge = 2421.82 1875.25 100 22.57

2421.82

%

1.(c) Draw the schematic diagram of a gravity dam and indicate the major forces acting on it.Draw the diagram of the uplift force when (i) drain is not provided and (ii) drain isprovided.

[12 Marks]

Sol. The various external forces acting on a gravity dam may be(1) Water Pressure (2) Uplift Pressure (3) Pressure due to earthquake force (4) Silt Pressure(5) Wave Pressure (6) Ice Pressure (7) Weight of the dam

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Mains Exam Solution



(1) Water Pressure

Water pressure (p) is the major external force acting on a dam. The horizontal water pressure,exerted by the weight of the water stored on the u/s side on the dam can be estimated from rulesof hydrostatic pressure distribution.

When the u/s face is vertical, the intensity is zero at the water surface and equal to wH at the

base where, w = unit weight of water and H = depth of water..

The resultant force due to this external water, P 2w

1 H2

, acting at H/3 from base.

W.M.L

H/3

wH

2w

1 H2

P =

Gravity dam

H

(2) Uplift Pressure

Water seeping through the pores, cracks and fissures of the foundation material, exert an upliftpressure on the base of the dam.

An uplift force reduces the downward weight of the body of the dam and hence destabilises thedam.

According to USBR recommendations, the uplift pressure intensities at the heel and the toe shouldbe taken equal to their respective hydrostatic pressures and joined by straight line in between asexplained in figure below.

WML

H

WH

H

Uplift diagram when there is no drainage gallery

Pu

WH

WH

TailwaterNo drainage

gallery

WH

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Mains Exam Solution

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(3) Earthquake Forces

The effect of an earthquake is equivalent to imparting an acceleration to the foundations of the damin the direction in which the wave is travelling.

Earthquake wave may move in any direction.• Both horizontal acceleration (ah) and vertical acceleration (av) are induced by an earthquake. The

values of these acceleration are generally expressed as percentage of the acceleration due togravity (g), i.e., a = 0.1 g or 0.2 g, etc.

(4) Silt PressureSilt gets deposited against the u/s face of the dam. If h is the height of silt deposited, then the forceexerted by this silt in addition to external water pressure, can be represented by Rankine’s formula as

: Psilt = 2sub a

1. .h K2 (it acts at

h3 . from base)

where, Ka = 1 sin1 sin = coefficient of active earth pressure of silt

( = angle of internal friction of soil, and cohesion is neglected).

sub = submerged unit weight of silt material ; h = height of silt deposited.

Note : If the u/s face is inclined, the vertical weight of the silt supported on the slope also acts as vertical force.

(5) Wave PressureWaves are generated on the surface of the reservoir due to blowing winds, which causes a pressuretowards the d/s side. Wave pressure depends upon the wave height.(6) Ice PressureIn cold countries, ice gets formed on the water surface of the reservoir. The dam face has to resistthe thrust exerted by the expanding or melting ice. This force acts linearly along the length of thedam and at the reservoir level.(7) Weight of the dam

The weight of the dam body and its foundation is the major resisting force. (ii) Reservoir empty case

Empty reservoir without earthquake forces are computed for determining bending diagrams, etc.for reinforcement design, for grouting studies or other purposes.

Empty reservoir with a horizontal earthquake force produced towards the u/s has to be checkedfor non-development of tension at toe.

Diagram of uplift force when

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Mains Exam Solution



(i) Drain is not providedWML

H

WH

H

Uplift diagram when there is no drainage gallery

Pu

WH

WH

TailwaterNo drainage

gallery

WH

(i) Drain is provided

MWL

H

Heel Toe

wH

ordinate of uplift at drainage gallery

w w w1H ( H H )3

=wH

Drainage gallery

1.(d) A city has the following recorded population :Year 1971 : 60000Year 1991 : 120000Year 2011 : 180000Estimate (i) the saturation population and (ii) expected population in the year 2031 bylogistic curve method.

[12 Marks]




Mains Exam Solution

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Sol. It is given thatP0 = 60000 ; t0 = 0P1 = 120000 ; t1 = 20 yrsP2 = 180000 ; t2 = 40 yrs

(i) The saturation population (PS) can be calculated by using equation

20 1 2 1 0 2

S 20 2 1

2P PP P (P P )P

P P P

= 2

22 60000 120000 180000 (120000) [6000 180000]

60000 180000 (120000)

= 240000

sSaturation population (P ) 240000

(ii) We have, m =s 0

0

P P 240000 60000P 60000

m 3

n =0 s 1

101 1 s 0

P (P P )2.303 logt P (P P )

= 102.303 60000 (240000 120000)log

20 120000 (240000 60000)

n 0.0549

Expected population in year 2031,

P =s

projected 01e

P; t (t t )

1 m log (nt)

= 60

= 1e

2400001 3log ( 0.0549 60)

= 1e

2400001 3log ( 0.054g 60)

= ( 0.0549 60)240000 215960.2945

1 3 e

215960

1.(e) A water contains 110 mg/L carbonate ion and 80 mg/L bicarbonate ion at a pH of 10.Calculate the alkalinity exactly at 25°C. Approximate the alkalinity by ignoring hydroxideand hydrogen ion. What is the percentage error in approximation ?

[12 Marks]

Sol. Concentration of carbonate ion ( 23CO ) = 110 mg/lt

=110 3.67 meq / lt602

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Mains Exam Solution



3.67 meq/lt of 23CO = 3.67 meq/lt of CaCO3

Concentration of bicarbonate ion ( 3HCO ) = 80 mg/lt

=80 meq / lt611

= 1.31 meq/lt

1.31 meq/lt of 3HCO = 1.31 meq/lt of CaCO3

pH = 10 pOH 4

–log [H+] = 10

[H+] = 1 × 10–10 mol/lt

= 1 × 10–10 × 1 gm/lt= 1 × 10–7 meq/lit

1 × 10–7 meq/lt of H+ = 1 × 10–7 meq/lt of CaCO3.Also, –log10[OH–] = 4

[OH–] = 10–4 mogl/lt

10–4 mol/lt of OH– = 4 1710 gm/eq/lt17

= 0.1 meq/lt

0.1 meq/lt of OH– = 0.1 meq/lt of CaCO3.Now, Approximate alkalinity (ignoring H+ and OH–)

Expressed in meq/lt = 3 3[HCO ] [CO ]

Where quantities in paranthesis are concentration in meq/lt.= [3.67] + (1.31) = 4.98 meq/lt

Approximate alkalinity = 4.98 × 50 = 249 mg/l as CaCO3

By exact calculation:

Alkalinity (in meq/lt) = 23 3[HCO ] [CO ] [OH ] [H ]

= 4(1.31 3.67 0.1 1 10 )meq / lt

= 5.0799 meq/ltExact alkalinity = 5.0799 × 50 = 253.995 mg/l as CaCO3

Percentage error in approximation

=253.995 249 100

253.995

= 1.966% Ans.

2.(a) A trapezoidal channel is to be designed to convey a discharge of 50 m3/sec at a velocity of2 m/sec. The bed width to depth ratio is 0.8. The side slopes are 1 H : 1 V. Calculate thebed width, depth of flow and bed slope of the channel. Assume Manning’s coefficient,n = 0.02.

[20 Marks]

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Mains Exam Solution

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Sol. GivenQ = 50 m3/s

B

1x

y

V = 2 m/s

By = 0.8

Side slope = 1H : 1Vn = 0.02

A = 2Q 25 mv

A = (B + xy)y25 = (0.8y + 1y)y25 = 1.8y2

y = 3.727 mB = 0.8 × 3.727 = 2.982 m

P = 2B 2y 1 x 13.524 m

v = 2/3 1/21R Sn

2 =2/3

1/21 25 .S0.02 13.524

S = 0.00071 B = 2.982 m, y = 3.727m, S = 0.00071

2.(b) Define a unit hydrograph. Explain two basic assumptions made in the derivation of unithydrograph. Following are the ordinates of a 4-hr unit hydrograph. Using this, derive theordinates of a 12-hr unit hydrograph (do not plot the graph) :

Time (hr ) 0 4 8 12 16 20 24 28 32 36 40 44Ordinate of 4 hr UH 0 20 80 130 150 130 90 52 27 15 05 0

What are the uses and limitations of unit hydrograph ?

[20 Marks]

Sol. A unit hydrograph is defined as the hydrograph of direct runoff resulting from one unit depth (1 cm)of rainfall resulting from one unit depth (1 cm) of rainfall excess occuring uniformly over the basin andat a uniform rate for a specified duration (D hours).The terms unit here refers to a unit depth of rainfall excess which is usually taken as 1 cm.Assumptions of unit hydrograph theory :(1) Time Invariance : The direct run-off hydrograph for a given effective rainfall for a catchment is

always the same irrespective of when it occurs. Hence, any previous rainfall event is not consideredto effect the new rainfall.

(2) Linear Response : The ordinates of DRH are directly proportional to the effective rainfall hyetographordinates.For example : If a 6-hr U.H due to 1 cm rainfall is given, then 6-hr hydrograph due to 2 cm rainfallwould just mean doubling the U.H ordinates.(The assumption of linear response enables the use of principle of superposition).

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Mains Exam Solution



(3) Effective rainfall is constant over the catchment during the unit time, i.e. intensity is constant.(4) Effective rainfall should be uniformly distributed over the basin.The calculations are performed in tabular form in given table. In this

Column 3 = ordinates of 4-h UH lagged by 4-hColumn 4 = ordinates of 4-h UH lagged by 8-hColumn 5 = ordinates of DRH representing 3 cm ER in 12-hColumn 6 = ordinate of 12-h UH = (Column 5)/3

Time(h) A

BLagged by

4-h

CLagged by

8-h

DRHof 3 cm in 12-h

(m /s)(Col. 2+3+4)

3

Ordinate of12-h UH

(m /s)(Col. 5)/3

3

Ordinates of 4-h UH (m /s)3

1 2 3 4 5 6

048

1216202428323640444852

02080

1301501309052271550

–0

2080

1301501309052271550

––0

20801301501309052271550

020

10023036041037027216994472050

06.7

33.376.7

120.0136.7123.390.756.331.315.76.7

1.700

Uses:

Unit hydrograph is used for:

(i) Development of flood hydrograph for extreme rainfall magnitudes for use in the design of hydraulicstructure.

(ii) Extension of flood flow records based on rainfall records.(iii) Development of flood forecasting and warning system based on rainfall.Limitations of Unit hydrograph theory :(1) U.H cannot be used for catchment area > 5000 km2

(2) U.H cannot be used for catchment area < 2 km2

(3) Precipitation must be from rainfall only. Snow melt run-off cannot be satisfactorily represented byU.H.

(4) The catchment should not have unusually large storage in terms of ponds, tanks etc which affectsthe linear relationship between storage and discharge.

(5) U.H are not expected to give good results if the precipitation is decidedly non-uniform.




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2.(c) (i) How will you estimate the total storage capacity of a distribution reservoir ? Supportyour answer with suitable sketches and formulae.

[15 Marks]

Sol. Distribution reservoir, also called service reservoirs, are the storage reservoirs, which store the treated waterfor supply during emergencies (such as during fires, repair, etc.) and also help in absorbing the hourlyfluctuations in the normal water demand.

Storage Capacity of distribution Reservoirs:The total storage capacity of a distribution reservoir is the summation of:

1. Balancing storage: The quanity of water required to be stored in the reservoir for equlising orbalancing fluctuating demand against constant supply is known as the balancing storage (orequalising or operating storage). The balance storage can be worked out by mass curve method.

2. Breakdown Storage: The breakdown storage or often called emergency storage is the storagepreserved in order to tide over the emergencies posed by the failure of pumps, electricity, or anyother mechanism driving the pumps. A value of about 25% of the total storage capacity of reservoirsor 1.5 to 2 times of the average hourly supply, may be considered as enough provision foraccounting this storage.

3. Fire storage: The third component of the total reservoir storage is the fire storage. This provisiontakes care of the requirements of water for extinguishing fire. A provision of 1 to 4 per person perday is sufficient to meet the requirement.

The total reservoir storage can finally be worked out by adding all the three storages.

Mass Curve Method

A mass curve is the plot of accumulated supply or demand versus time.

The supply is also known as inflow and demand as outflow. First mass curve of supply, knownas supply line is drawn and over this demand curve is superimposed.

The amount of balancing storage is determined by adding the maximum ordinates between thedemand and supply lines.

First hourly demand for all 24 hours from the day of maximum requirement is determined. Cumulativedemand is plotted against time, which is known as mass curve of demand. Next cumulative supplyagainst time is plotted, which is a straight line if the supply is constant. The storage required iscalculated as the sum of the two maximum ordinates between demand and supply lines.

Total storage required= A + B

Max.ordinate (B)

m

Max.ordinate = (A)

m

Supply curve(continuous atconstant rate)

Demand curve

Time in hours0

Cum

ulat

ive

dem

and

orsu

pply

in 1

0 li

tres

x

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Mains Exam Solution



(Continuous supply)

Cum

ulat

ive

dem

and

orsu

pply

in 1

0 li

tres

x

A

Supply curve

Pumpingperiod

Total storage required= A + B

Demandcurve

B

0 6 AM 6 PM(Limited Hour Supply)

B

A

A2

1 24 hrs

3

5

4

Watergettingstored

Watergetting

depleted

Watergettingstored

Cumulativedemand orsupply

Time

Demand Supply

From (2–4) Demand rate > supply rate water getting depleted

From (4–5–1–2) Demand rate < supply rate water getting stored

From (4–5) accumulation = BFrom (1–2) accumulation = AFrom (2–3) depletion = AFrom (3–4) depletion = B

Max. water that would be stored = A + B (i.e. From 4–2)

Max. water that would be depleted = A + B (i.e. From 2–4)

2(c) (ii) Compute the average sound pressure level from the following sound pressure readings :(1) 39 dBA(2) 52 dBA(3) 67 dBA(4) 77 dBA

[5 Marks]

Sol. We have individual readings as39 dBA, 52 dBA, 67 dBA, 77 dBA

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Mains Exam Solution

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We know that,

Sound pressure level = LP = 20 × log10 rmsP

20 Pa

Now, 39 dBA = rms10

P20 log

20

Prms = 1782.50Similarly, for 52 dBA

52 = rmsP20 log

20

Prms = 7962.14Prms for 67 dBA

67 = rmsP20 log

20

Prms = 44774.42Prms for 77 dBA

77 = rmsP20 log

20

rmsP = 141589.16

39 dBA + 52 dBA + 67 dBA + 77 dBA 1 in rms

= 2 2 2 2(1782.5) (7962.14) (44774.42) (141589.16)

= 148723.945 Pa

Resultant sound pressure level =148723.9520log

20

= 77.43 dBA

3. (a) Estimate the hydraulic gradient in a 2.2 m diameter smooth concrete pipe carryinga discharge of 3.4 cumecs at 20°C temperature by using (i) Darcy-Weisbach formula,(ii) Manning’s formula and (iii) Hazen-William’s formula. The kinematic viscosity ofwater at 20°C = 1.004 × 10–6 m2/s, Hazen-William’s coefficeint of hydraulic capacity ofthe smooth pipe = 130 and Manning’s coefficient = 0.013.

[20 Marks]

Sol. GivenSmooth concrete pipe

Diameter (d) = 2.2 mDischarge (Q) = 3.4 m3/sec at temperature 20°C

Kinematic viscosity of water 20°C ( ) = 1.004 × 10–6 m2/secHazen William coefficeint (CH) = 130

Manning’s coefficient (n) = 0.013

(i) Assuming flow to be running full.

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Mains Exam Solution



Velocity, (V) =2

Q 3.4 0.894 m/secA (2.2)

4

Reynold’s number, (Re) =Vd

= 60.894 2.2

1.004 10

= 1.96 × 106

This is the case of turbulent flow through smooth boundary.

As per Blasius,

Friction factor, (f) = 1/4e

0.316(R )

= 6 1/40.316

(1.96 10 )

= 0.00844Darcy-Weisbach equation:

Head loss (hL) =2L Vf

d 2g

LhL

= Slope (s) = 2f V

d 2g

Slope (s) =20.00844 (0.894)

2.2 2 9.81

=1

6398.88

1Slope (s)6398.88

Ans.

(ii) As per Manning’s formula,

Velocity, (V) =2/3

1/21 d (s)n 4

2/3

1/21 2.20.894 (s)0.013 4

1Slope (s)3336.22

(iii) According Hazen William’s formula,V = 0.849 × CH × R0.63 × S0.54

V =0.63

0.54d0.849 130 s4

S0.54 = 0.630.894

2.20.849 1304




Mains Exam Solution

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S0.54 = 0.0118

Slope, (S) = 1/0.540.0118

=1

3720.39

1Slope (S)3720.39

3. (b) A wastewater treatment plant consists of primary treatment clarifier followed by anactivated sludge treatment unit. The primary and secondary sludge are mixed,thickened in a gravity thickener and sent to further treatment. Wastewater, treatmentplant and sludge characteristics are as follows: Influent SS = 220 mg/L; primary clarifier diameter = 25 m Influent BOD = 250 mg/L; aerator volume = 3000 m3

Effluent BOD = 30 mg/L; MLSS in aerator = 3000 mg/L Flow = 20000 m3/day; solids in thickner supernatant = neglisgible Primary sludge = 5% solids; secondary sludge = 0.75% solids and thickned sludge

= 4% solids Efficiency of primary clarifier for SS and BOD removal are 58% and 32

respectively Biomass conversion factor in aerator = 0.35

Determine–(i) Solids loading in kg/day to the sludge disposal facilities;

(ii)FM ratio in aerator;

(iii) Percent volume reduction by the thickener.

[20 Marks]

Sol. (i) Calculation of mass of primary solids and volume of primary sludge:The area of primary clarfier is

A =2 2

2d (25) 490.87 m4 4

The overflow rate =3

220000 m /d490.87 m

= 40.74 m/dayEfficiency of primary clarifier is

SS = 58% and BOD = 32%Mass of primary solids removed is found by

MP = 0.58 × 0.22 kg/m3 × 20000 m3/day= 2552 kg/day

Percent solid content is less than 20% so we can approximately take the density of sludge equal tothat of water.

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Mains Exam Solution



The volume of primary sludge is given by

VP = PM1000 S

=2552

1000 0.05

= 51.04 m3/dayCalculation of mass of secondary solids and volume of secondary sludge:Food consumed by aerator is:

BODin = (1 – 0.32) × 250 = 170 mg/ltBODout = Effluent BOD = 30 mg/lt

BOD consumed in aerator = (170 – 30) mg/lt= 140 mg/lt= 0.14 kg/m3

The biomass conversion factor, Y = 0.35 (given)

The above factor accounts for the net quantity of living biomass generated in the reactor. The sameamount has to be wasted to keep the system in equilibrium.

The mass of secondary volatile solids is found out by

Mvs = Q(BOD5 consumed in the aerator) × Y

=3

3kg m0.35 0.14 20000

daym

= 980 kg/day

Mass of total secondary solid wasted, (MS) =

980 980MLVSS 0.8MLSS

= 1225 kg/dayThe volume of secondary sludge (Vs) is

Vs = SM1000 S

=1225

1000 0.0075

= 163.33 m3/day

Total mass of sludge to thikner, MT = P sM M (2552 1225) kg/day

= 3777 kg/day

Sludge loading to the sludge disposal facilities = MT = 3777 kg/day(ii) F/M ratio calculation:Food applied to the aerator = BODin = 0.17 kg/m3 = 20000 m3/d × 0.17 kg/m3 = 3400 kg/day

Assuming MLVSS = 0.8 × MLSS = 0.8 × 3.0 kg/m3 = 2.4 kg/m3

The biomass in the reactor = (MLVSS in aerator × Aerator volume)

=3

3kg2.4 3000 mm

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Mains Exam Solution

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= 7200 kg

The F/M ratio =3400 kg / day

7200 kg

= 0.472 day–1

(iii) Total volume of sludge to the thickner, VT = 3P s(V V ) (51.04 163.33) m /day

= 214.37 m3/dayTotal volume of thickened sludge

Vthick = 33777 kg/day

1000 kg/m 0.04

= 94.425 m3/day

% Volume reduction achieved by thikener =214.37 94.425 100

214.37

= 55.952% Ans.

3. (c) Explain geometric similarity, kinematic similarity and dynamic similarity. Twohomologous pumps are to run at the same speed of 600 r.p.m. Pump A has an impellerof 50 cm diameter and discharges 0.4 m3/sec of water under a head of 50 m. Determinethe size of pump B and its net head if it is to discharge 0.3 m3/sec.

[20 Marks]

Geometrical similarity : For geometrical similarity to exist, the ratio of corresponding length dimensionbetween the model and prototype must be same.

A B

C

D

E

a b

cD

E

ABab

= BC CD DEbc cd de

The geometric parameters are length, width, height, area, volume, diameter.

m

P

LL =

m mr

P P

W H LW H

2m

rP

A (L )A

3mr

P

V =(L )V

Kinematic similarity: Kinematic parameters are velocity, acceleration, discharge.

· At all corresponding points in the model and prototype the ratio of velocity as well as accelerationmust be same (both in magnitude and direction).

· Such similarity can be attained if flownets for the model and prototype are geometrically similar

Velocity ratio =m r

rp r

V LVV T

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Mains Exam Solution



Acceleration ratio =m r

r 2p r

a L= a =a T

Discharge ratio =3

m rr

p r

Q L= Q =Q T

Dynamic similarity: For dynamic similarity to exist between model and prototype, identical type of forces(viscous, pressure, elastic etc) must be parallel and must bear the same ratio at all corresponding setsof points

Dynamic parameters are force and power

Force ratio =4

m r rr r r 2

p r

F ρ L= ×V ×a =F T

Power ratio =5

m r rr r 3

p r

P L=F .V =P T

For kinematic similarity: Geometrical similarity must exist

For dynamic similarity: 1. Geometrical similarity must exist.2. Kinematic similarity must exist but is not the sufficient condition for dynamic

similarity.

For complete similarity, geometrical, kinematic and dynamic similarity must exist.Pump A

DA = 50 cm, QA = 0.4 m3/sec, HA = 50 mNA = NB = 600 rpm

Pump B :DB = ?, HB = ?, QB = 0.3 m3/s

3Q

D N

= 3 B

QD N

A B

B A

Q NQ N

=3

A

B

DD

0.4 10.3

=3

B

50D

DB = 45.428

2 2H

D N

= 2 2H

D N

A

B

HH =

2 2A A

B B

D ND N

=250 1

45.428

B

50H = 1.211




Mains Exam Solution

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BH 41.274m

4. (a) Explain the terms ‘initial regime’ and ‘final regime’ as explained in Lacey’s regimetheory of stable channels. Design a stable channel for carrying a discharge of 30 m3/sec using Lacey’s method assuming a silt factor equal to 1.0

[20 Marks]

Sol. Initial Regime

It is the first stage of regime attained by a channel after it is in service.

If a channel is excavated with smaller width and flatter bed slope, then as the flow takes placein the channel, bed slope of the channel is increased due to deposition of silt on the bed of thechannel to develop an increased flow velocity. Hence, the given discharge is allowed to flow throughthe channel of smaller width.

With increase in the bed slope , the depth may also vary but the width of the channel does’ntchange because the sides of the channel are usually cohesive and hence they resist erosion.So, keeping the discharge, silt grade, silt charge and width fixed and only by varying bed slopeand depth, the channel attains stability. This condition is known as initial regime.

Final Regime

It is the ultimate state of regime attained by a channel when in addition to bed slope and depth,the width of the channel is also adjusted as per requirement.

In this condition, the resistance of the sides of the channel is ultimately over come due tocontinuous action of water.

So, the channel adjust its width, depth and bed slope in order to obtain a stable channel. Thiscondition is known as final regime.

Channel Shapecarrying mediumsized silt

Channel shape carrying coarse silt

Channel shape carrying fine silt

Such a channel in which all variables are equally free to vary, has a tendency to assume a semi-elliptical section. The coarser the silt, the flatter is the semi-ellipse, i.e. greater is the width of thewater-surface. The finer the silt, the more nearly the section attains a semicircle.

Note : (i) The various equations developed by Lacey are applicable to channels which has attained final regime. (ii)Hence, we can conclude that total no. of independent equations that form the Lacey’s regime theory is 3.

Given, Q = 30 m3/secSilt factor, (f) = 1.0

So, for a regime channel using Lacey theory, we have

(i) Velocity =1/6 1/62 2Qf 30 1

140 140

= 0.774 m/sec

(ii) Hydraulic mean depth, (R) =25 V

2 f

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=25 0.774

2 1

= 1.498 m

(iii) Area of the channel, (A) =Q 30V 0.774

= 38.759 m2

(iv) Wetted perimeter, (P) = 4.75 Q

= 4.75 30

= 26.017 m

(v) Bed slope, (S) =5/3

1/6f 1

5887.533340 Q

1

5888Assuming the cross section of channel to be trapezoidal in shape.

B

y m =21

A =yB y2

P = B + 2.236 y

Now, 38.759 = (B 0.5y)y

or, B =38.759 0.5y

y

Also, 26.017 = B+ 0.236 y

or, 26.017 =38.759 0.5y 2.236 y

y

y = 1.68m, 13.31 mAdopting, y = 1.68 m

So, B =38.759 0.5 1.681.68

= 22.23 mFinally

22.23 m

y = 1.68 m2

1

4. (b) (i) Define field capacity, permanent wilting point and average moisture content.

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Explain how these will be useful in deciding the frequency of irrigation. (Aschematic diagram showing less and more frequent irrigation is to be drawn forclarity)

[3+4+3 = 10 Marks]

Sol. Field Capacity

Field capacity is defined as the maximum amount of moisture which can be held by a soil againstgravity. After the gravity water has drained off.

At field capacity the large or non-capillary pores of the soil are filled with air and the small orcapillary pores are filled with water.

Field capacity is the upper limit of the capillary water or the moisture content available to the plantroots.

The soil moisture tension at field capacity ranges between 1/10 to 1/3 atmospheres.Permanent wilting point

The permanent wilting point is usually expressed as the weight of the moisture held by the soilper unit weight of the dry soil when the plants are permanently wilted.

Permanent wilting point is the moisture content at which the films of water around the soil particlesare held so tightly that the plant roots cannot extract enough moisture at sufficiently rapid rate tosatisfy transpiration requirements thus resulting in the wilting of the plants.

The soil moisture tension of a soil at the permanent wilting point ranges from 7 to 32 atmospheresdepending on soil texture, kind and condition of the plants etc.

Average moisture content

Depending upon the depth of irrigation water applied and frequency the average moisture contentin the soil may lie between the lower limit wilting point or upper limit field capacity.

Average moisture content of soil is moisture present in the soil that is avilable for plant.

Concept of frequency of irrigation:

Soil moisture in the root zone varies between field capacity (upper limit) and permanent wilting point (lowerlimit). When the soil moisture is consumed by plants through their roots, the soil moisture depletes. However,it is not allowed to deplete upto the permanent Wilting point, as it would result in considerable fall in cropyield. The optimum level up to which the soil moisture is allowed to deplete is called optimum moisturecontent.

The irrigation water should be supplied as soon as the moisture falls upto this optimum level, and its quantityshould be sufficient enough to bring the m.c upto its field capacity.

Water will be utilised by the plants after the fresh irrigation dose is given, and the soil moisture will start falling.It is then again recouped by a fresh irrigation dose as soon as the soil moisture reaches the optimum level.

The time interval between two successive irrigation dose is termed as frequency of irrigation.

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PWP

Time

OMC

F.C

frequency of irrigation

Moi

stur

e co

nten

t of s

oil Readily available

moisture

Wilting point m.c

Frequency of irrigation depends upon the amount of readily available moisture in the root zone and the rateof consumptive use.

Frequency of irrigation =Depth of water applied during irrigation

consumptive use

Field capacity average moisture content

PWP

Time

(a) More Frequent Irrigation

Soi

l moi

stur

e co

nten

t

Field capacity

Average moisture content

PWP

Soi

l moi

stur

e co

nten

t

Time

(b) Less frequent irrigation

4. (b) (ii) In a hydraulic jump occurring in a horizontal channel, the Froude’s numberbefore the jump is 10.0 and energy loss is 3.2 m. Estimate sequent depths, dischargeintensity and Froude’s number after the jump.

[10 Marks]

Sol. Given,Hydraulic jump in a horizontal channel.Froude number before jump, (F1) = 10.0

Energy loss (FL) = 3.2 mWe know that,

2

1

yy = 2

11 1 8F 12 [Assuming rectangular channel]

or2

1

yy = 21 1 8 10 1

2

or2

1

yy = 13.65

or y2 = 13.65 y1

or EL =3

2 1

1 2

(y y )4y y




Mains Exam Solution

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or 3.2 =3

1 1

1 1

(13.65 y y )4 13.65 y y

or 3.2 × 4× 13.65 =3 3

121

12.65 yy

y1 = 33.2 4 13.65 0.086 m

12.65

So, y2 = 13.65 × 0.086= 1.17 m

hence, final sequent depths are 0.086 m, 1.17 m.

Now, using2

1 2 1 22qy y (y y )g

or 0.086 × 1.17 (0.086 + 1.17) =22 q

9.81

q = 0.787 m3/sec.mCalculation of Froude’s number after jump (F2)

22F =

21

321

8F

1 8F 1

or 22F =

2

2 3

8 10

( 1 8 10 1)

2F 0.198

4. (c) (i) Explain in detail the various process parameters required to control the aerobiccomposting of solid waste. Discuss the relevance of each parameter also.

[10 Marks]

Sol. Compositing is a controlled aerobic, biological conversion of organic wastes into a complex, stable finalproduct having a number of beneficial uses, most commonly for agriculture and landscaping. In other words,composting of refuse is a biological method of decomposing solid wastes.

Factors affecting the composting process :(i) Nutrient levels (macro - C, H and O and micro-N, P, K, Mg, S, Fe, Ca, Mn, Zn, Cu, Co and Mo)(ii) Nutrient balance (C/N ratio) C/N ratio of the input material in the compost heap is an important

factor for the bacterial activity to continue, since the bacteria use nitrogen for building their cellstructures (as proteins) and carbon for food (as energy).

(iii) Aeration When acids accumulate during the early stages of composting, aeration can be doneto return the compost pH to an acceptance range.

(iv) Moisture and temperature(v) pH (between 5.5 and 8.5)(vi) Particle size of the feedstock material.

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4. (c) (ii) Discuss the isokinetic sampling process of flue gas stack sampling and explainwith the help of diagram, how the results will be affected if the sampling is notdone isokinetically.

[10 Marks]

Sol. Isokinetic sampling processIn Figure a flowsheet for isokinetic sampling is presented, the systems can be applied for (complex) organiccomponents which are either in gaseous, aerosol, liquid or solid state at the sampling point. The samplingcan be performed either by “main stream (a)” or “side stream (b)” (also called “partial stream” sampling).

11

107

386

54

9 9a9 9

(a) Main-stream sampling

13

2

(b) Side-stream sampling

31

32

9 9a9 9

11

107

8

6

54

7

118

4

10

Key1 Nozzle 2 Probe3 Filter (either behind or in front of the probe)4 Dryer5 Main-stream valve6 By-pass valve

Key7 Pump8 Gas volume meter9 absorber9a Safety bottle (optional)10 Gas flow meter11 Temperature and pressure measurement

Figure: Isokinetic sampling system

A particle filter (3) is placed either in front or behind the sampling probe. This renders necessary heatingof the filter in order to fully evaporate the sampled gas passing through the filter. The sample gas streamfrom the stack is then either passed through absorbent filled gas wash bottles (9; multiple impingers inseries) to remove gaseous components or aerosols, the gas stream is dried (4) and the dried gas streamregistered (10). The combined concentration of a component is determined through analysis of the filter andthe wash bottle content. Alternatively in the side stream method, a smaller stream is split from the mainstream and passes through a series of absorbent filled wash bottles. Pre-requisite for a split of the gasstream is that the sample gas stream is “gas only” (no droplets, particles) The choice which of the twomethods are preferred is often dependent on the gas conditions (e.g. wet gas would dilute the absorbent;therefore = side stream) or on the combination of component concentration and limit of detection (lowconcentration = Main stream). In addition, a side stream setup is necessary in order to fulfil the optimalgas flow requirements for impingers and adsorption tubes.

All flue gas sampling has to be performed isokinetically to ensure that representative samples are collected.

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Mains Exam Solution

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If sampling is done other than at isokinetic velocity, then it will cause errors for two reasons. First samplingat greater or less than isokinetic rates tends to cause respectively larger or a smaller volume to be withdrawfrom the flue gases than accounted for by the cross-section area of the probe. Secondly, particles greaterthan 3.5 size have sufficeint inertia so that particle motion may deviate significantly from the gas flowstream line pattern. Thus, if sampling is not done isokinetically, then we will be unable to obtain representativesample.

(a) Isokinetic sampling

(b) Flow rate less than Isokinetic rate

(c) Flow rate greater than Isokinetic rate

Isokinetic sampling

Section—B

5. (a) A pit of 6.4 m deep is to be excavated in a fine sand stratum completely saturated upto the ground surface. The saturated unit weight of the sand was obtained as 20.3kN/m3. To stabilize the bottom of the excavation (prevent boiling), it was decided todrive steel sheet piles to act as cutoff walls that encircle the excavation. Determine thetotal length of sheet pile wall to provide a factor of safety of 1.5 against sand boiling.Assume specific gravity of soil, Gs = 2.7 and unit weight of water, = 9.81/kN/m3.

[12 Marks]

Sol.

AF

6.4 m

Sheet Pile

BH

D

E

CThere will be an upward flow of water through the soil mass.

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The differential head which will cause this flow, h = 6.4 mAgain thickness of the soil mass through which the flow occur, L = BC = ED = H

Hydraulic gradient, i =h 6.4L H

Saturated Unit weight S = wG e1 e

20.3 =2.7 e 9.811 e

e = 0.59

Critical hydraulic gradient iC =G 11 e

iC =2.7 1

1 0.59

iC = 1.0693

Give Factor of safety = Ci1.5i

1.5 =1.0693

6.4H

H = 8.98 m

So, total length of sheet pile = 6.4 + 8.98= 15.38 m

5. (b) An unsupported cut as shown in the figure below was made at a site for which unitweight of soil, s = 18.2 kN/m3, cohesion, C = 25 kN/m2 and angle of internal friction, 10 . Determine the lateral stress at—

(i) the top of the excavation;

(ii) the bottom of the excavation;

(iii) The maximum depth of potential tension crack for the excavation.

What is the maximum depth up to which the excavation can be carried out safelywithout any support?

4.2 m

s = 18.2 kN/mC = 25 kN/m

= 10°

3

2

[12 Marks]




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Sol. Given data

sat = 18.2 kN/m3

C = 25 kN/m2

= 10°

Ka =1 sin 1 sin10 0.7041 sin 1 sin10

Active earth pressure is given as

pa = a a aK q K z 3C K

Here, q = 0(i) Active earth pressure at top of excavation

Z = 0

pa = a2C K

pa = 2 25 0.704

pa = -41.95 KN/m2

(ii) Active earth pressure at bottom of excavationZ = 4.2 m

pa = a S aK Z 2C K

pa = 0.704 18.2 4.2 2 25 0.704

pa = 11.859 KN/m2

(iii) Maximum depth of potential tension crack for the excavation.where, pa = 0

a t aK Z C K = 0

Zt =a

2CK

Zt =2 25

18.2 0.704

Z = 3.27 m [From the top of soil]Maximum depths of excavation without any support

= 2 Zt

= 2 3.27

= 6.55 m [From the top of soil]

5. (c) How are runways oriented? Explain the term ‘wind coverage’ and ‘crosswindcomponent’.

[12 Marks]

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Sol.

Runway is so oriented that landing and takeoff is done by heading into the wind i.e., opposite to winddirection.

The wind data of direction, duration and intensity should be collected from wind rose diagram overa minimum period of 5 years.

The site should be located to windward direction of the city, so that minimum smoke from the cityis blown over the site.

Cross Wind Component It is not possible to obtain the direction of wind along the direction of the central line of runway through

out the year.

If the direction of wind is at an angle to runway centre line, its component are shown below

C RunwayLTaking offDirection

vsin vWind

Vcos

– vcos – the component along the direction of runway

– vsin – the normal to the central line of runways

The normal component is known as cross wind component, it interrupt safe landing and take-offoperations.

Type of Aircraft Permissible limit of cross wind component (KmpH)

Small aircraftMixed trafficBig traffic

152535

Wind Coverage : The percentage of time in a year during which cross wind component with in the permissible limit

as specified above, is called wind coverage.

According to FAA, runway landing mixed air traffic should be planned that wind coverage is 95% andpermissible crosswind component should not exceed 25 KmpH.

For busy airport, wind coverage may be increased to 98–100%.

5. (d) Calculate equilibrium cant on MG curve of 6 degree for an average speed of 50 km/hr. Also find out the maximum permissible speed after allowing maximum cantdeficiency.

[12 Marks]

Sol. Given data

Degree of curve = 6°Average speed = 50 Km/hr

For meter gauge, G = 1.00 m

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Mains Exam Solution

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Radius of curve =1750

6

= 291.67 m(i) Actual cant provided

eact =2GV

127R

= 21.00 50127 291.67

= 6.75 cm < 10 cmSo actual cant provided = 6.75 cm

(ii) Maximum permissible speed is given by eth i.e. eth = eact + cant deficiencyCant deficiency is permitted for MG = 5.1 cmSo, eth = 6.75 + 5.10

= 11.85 cm

So, eth =2maxGV

127R

0.1185 =2max1.000 V

127 291.67

Vmax = 66.25 KmphCheck maximum speed as per Martin’s formula.For V < 100 Kmph and MG

Vmax = 4.35 R 67

Vmax = 4.35 291.67 67

Vmax = 65.2 Kmph

So, maximum permissible speed 65.2 Kmph

5. (e)(i) What is repetition method in theodolite surveying? What are different instrumentalerrors which can be eliminated by the repetition method?

[6 Marks]

Sol. The method of repetition is used to measure a horizontal angle to a finer degree of accuracy than thatobtainable with the least count of the vernier. By this method, an angle is measured two or more timesby allowing the vernier to remain clamped each time at the end of each measurement instead of settingit back at zero when sighting at the previous station. Thus an angle reading is mechanically added severaltimes depending upon the number of repetitions. The average horizontal angle is then obtained by dividingthe final reading by the number of repetitions.

A higher degree precision is achieved during method of repetition because the following errors are eliminated.1. By reading both the verniers, errors due to eccentricity of verniers and centres are eliminated.2. By taking both the faces reading, errors due to the line to sight and the trunnion axis being out of

adjustment are eliminated.

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3. By taking the reading on the different part of circle, errors due to inaccurate graduations on the mainscale are eliminated.

4. But taking more number of observations, errors due to inaccurate bisection of the signal are eliminated.5. By dividing the cumulative angle value with the number of observations, other errors are also minimized.

5.(e) (ii) A level was set up between two stations A and B. The distance of level from stationA was 520 m and the reading on the staff held at A was 1.620 m. The distance of levelfrom station B was 780 m and the reading on the staff held at B was 2.120 m. Thereduced level (RL) of point A was 100.000 m. What is the RL of point B? Assume thatthere is no error in the instrument.

[6 Marks]

Sol.

1.620 m 2.120 m

520780 m

A

B

Correct staff reading at station A = 21.620 0.0673 1.602m0.52

Correct Staff Reading at station B = 22.120 0.0673 2.079m0.780

ABh = 2.079 1.602 0.477m

R.L. of B = ABR.L. of A h 100 0.477 99.523m

6. (a) Liquid limit (LL) and plastic limit (PL) tests were carried out on a soil sample as perIndian Standard method. The values were 60% and 36% respectively for LL and PL.What is the type of soil based on the above test data as per Indian StandardClassification System? Justify your answer.

[10 Marks]

Sol. Given,

Liquid Limit WL = 60%Plastic Limit WP = 36%

Plastic Index IP = L PW W

IP = 60 – 36

IP = 24%Plastic index as per A line.

IP = L0.73 W 20

IP = 0.73 60 20




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IP = 29.2%

IP = 29.2% > Actual Plastic Index

So, soil is silt.Because WL > 50%, soil is highly compressible.So, give soil should be MH or OH.

MH or OH

10 20 35 50 60

29.2

24

74

CL

CH I = 0.73 (W – 20)

P

L

CI

IPPlasticity

Index(W – W )L P

MI or OIMLOL

6. (b) Two square footings with equal contact pressure of 250 kPa are at 5 m apart (centre-to-centre). The size of the one footing (A) is 2 m × 2 m and the other one (B) is 2.5 m × 2.5 m.Determine the vertical stress at 2 m vertically below (i) the footing (A), (ii) the footing(B) and (iii) the midpoint between the footings. Use Boussinesq’s point load formula.

[10 Marks]

Sol. Given data

q = 250 KPaZ = 2 m

A2 m

Q = 250×2×2=1000 KNA

2 m

B2.5 m

Q = 250×2.5×2.5 =1562.5 KNB

2 m

Z = 2 mB

5 mZ = 2 mA

(i) Vertical stress below footing (A) as per Boussinesq equation

A =

5/2 5/2

A B2 22 2A B

1 1Q 3Q3

r r1 12 Z 2 ZZ Z

Here, QA = 1000 KN, QB = 1562.5 KN, rA = 0, rB = 5 m, Z = 2m

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So, A =

5/2 5/2

2 22 2

1 13 1000 3 1562.5

0 52 2 1 2 2 12 2

A = 119.366 + 1.318

A = 120.64 KN/m2

(ii) Vertical strees below footing B.

B =

5/2 5/2

A B2 22 2A B

1 13Q 3Q

r r1 12 Z 2 ZZ Z

Here, QA = 1000 kN, QB = 1562.5 KN, rA = 5 m, rB = 0, Z = 2m

B =

5/2 5/2

2 22 2

1 13 1000 3 1562.5

5 02 2 1 2 2 12 2

B = 0.8434 + 186.510

B = 187.35 KN/m2

(iii) Stress at mid point between the footing.QA = 1000 KN, QB = 1562.5 KN, rA= 2.5 m, rB = 2.5 m, Z = 2 m

So, C =

5/2 5/2

2 22 2

1 13 1000 3 1562.5

2.5 2.52 2 1 2 2 12 2

= 11.356 + 17.744

C = 29.10 KN/m2

6. (c) A plate bearing test with a 0.3 m diameter plate was carried out on a thick depositof sand. The shearing failure of the plate was occurred when a load of 3.5 kN wasapplied. The unit weight of the sand was 19.2 kN/m3 and water table was found to beat a depth of 1.0 m below the ground surface. If a square foundation of size 1.5 m ×1.5 m is planned in the same sand deposit but placed at a depth of 1.0 m below theground surface, what will be the allowable bearing capacity of the footing? Assumesaturated unit weight of sand also as 19.2 kN/m3 and unit weight of water as 9.81kN/m3. The chart given below may be used:




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C qN N N0 5.70 1.00 0.002 6.30 1.22 0.184 6.97 1.49 0.386 7.73 1.81 0.628 8.60 2.21 0.91

10 9.61 2.69 1.2512 10.76 3.29 1.7014 12.11 4.02 2.2316 13.68 4.92 2.9418 15.52 6.04 3.8720 17.69 7.44 4.9722 20.27 9.19 6.9124 23.36 11.40 8.5826 27.09

14.21 11.3528 31.61 17.81 15.1530 37.16 22.46 19.7332 44.04 28.52 27.4934 52.64 36.51 36.9636 63.53 47.16 51.7038 77.50 61.55 73.4740 95.67 81.27 100.3942 119.67 108.75 165.6944 151.95 147.74 248.2946 196.22 204.20 426.9648 258.29 287.86 742.6150 347.52 415.16 1153.15

[20 Marks]

Sol. The plate load test is to be assumed at the foundation level.

For Plate load test :

qup =

2

3.5 10.6 N 0.3 19.2 9.8120.34

N = 58.59

From chart, interpolating the value of Nq corresponding to N = 58.59 is

Nq = 58.59 51.7047.16 61.55 47.1673.47 51.70

Nq = 51.71

Now, for foundation :Sandy soil is given, i.e. C = 0

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Size of foundation = B = 1.5 mNq = 51.71

N = 58.59

= 19.2 KN/m3

SandD = 1 mf

1.5×1.5 msat = 19.2 KN/m

3

qnu = qf sub1N 10 D 0.8 B N2

qnu = 21 kN m0 19.2 1 0.8 1.5 58.5951.71 1 19.2 9.812

= 1303.73 kN/m2

Net ultimate bearing capacity of foundation = 1303.73 kN/m2

Assume factor of safety = 2.5

Net safe ultimate bearing capacity = nuq 1303.73FOS 2.5

= 521.49 kN/m2

6. (d) Write down the construction steps for Water Bound Macadam road. Also compareWBM construction with WMM construction.

[20 Marks]

Sol. The construction of W.B.M. road is carried out in the following stages :

1. Preparation of the Foundation :The foundation for receiving the new layer of W.B.M. may be either the subgrade or subbase course.This foundation layer is prepared to the required grade and camber and the dust and other loosematerials are cleaned. On the existing road surface, the depression and potholes are filled and thecorrugations are removed.

2. Provision of Lateral Confinement :This can be done constructing the shoulders to a thickness equal to that of the compacted W.B.M.layer and trimming the inner sides vertically.

3. Spreading of Coarse Aggregates :The coarse aggregates are spread uniformly to the proper profile and even thickness upon theprepared foundation and checked by templates.

4. Rolling :After spreading the coarse aggregates properly, compaction is done by three-wheeled power roller (6to 10 tones) or equivalent vibrating roller.




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The rolling is started from the edges and progressed gradually towards the center of the road untiladequate compaction is achieved. On super-elevated portion, rolling it started from the inner or loweredge and progressed gradually towards the outer or upper edge of the pavement.

5. Application of ScreeningsAfter the coarse aggregates are rolled adequately the dry screenings are applied gradually over thesurface and try rolling is continued as the screenings are being spread. The brooming operation isalso carried out simultaneously.

6. Sprinkling and GroutingAfter the application of screenings, the surface is Sprinkled with water, swept and rolled.

7. Application of Binding MaterialAfter the application of screenings and rolling, binding materials are applied at a uniform and the slowrate at two or more successive thin layers. After each application of binding material the surface isSprinkled with water and wet slurry with brooms to fill the voids.

8. Setting and Drying :After the final compaction, the W.B.M. course is allowed to set overnight. On the next day the Hungryspots are filed with screenings or binding materials, lightly sprinkled with water if necessary androlled. No traffic is allowed until the W.B.M. layers sets and dries out.

Comparison of the WBM and WMM road construction :Although the cost of construction of the WMM is said to be more than that of the WBM sub-base andbases but the advantages given below will compensate for that. Here are the points of difference.1. The WMM roads are said to be more durable.2. The WMM roads gets dry sooner and can be opened for traffic within less time as compare to the

WBM roads which take about one month for getting dry.3. WMM roads are soon ready to be black topped with the Bituminous layers.4. WMM roads are constructed at the fast rate.5. The consumption of the water is less in case of the WMM roads.6. Stone aggregates used in WBM is lager in size which varies from 90 mm to 20 mm depending upon

the grade but in case of the WMM size varies from 4.75 mm to 20 mm.7. In case of WBM, stone aggregates, screenings and binders are laid one after another in layers while

in WMM, aggregates and binders are premixed in the batching plants and then brought to the sidefor overlaying and compacting.

8. Materials used in the WBM are the stone aggregates, screenings and binder material (Stone dust withwater) while in WMM material used are only stone aggregate and binders.

9. Quantity of the WBM is generally measured in cubic meters while that of the WMM in square meters.

7. (a) A rigid pavement of 25 cm thickness of M40 grade of concrete is supported over asubgrade having modulus of subgrade reaction as 8.0 kg/cm3. If dowel bars are placedat centre-to-centre spacing of 30 cm, calculate the maximum load carried by a singledowel which is just below the wheel. Assume wheel load as 4100 kg, participation ofdowel bars in load distribution up to 1.0 × radius of relative stiffness and load to betransferred by joint as 50%. Poisson’s ratio of the concrete may be taken as 0.15.

[20 Marks]

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Sol. Given, for rigid pavement :

Modulus of subgrade reaction (K) = 8 Kg/cm3

M40 grade concreteSlab thickness (h) = 25 cm

E = 5 33 10 Kg cm

= 0.15

Design wheel load = 4100 KgDesign load transfer = 50%

Spacing of dowel bar = 30 cm

Distance over which dowel bars are effective = 1.0 l ; l = radius of relative stiffness

Radius of relative stiffness ( l ) =

1/43

2

Eh12K 1

=

1/435

2

3 10 2512 8 1 0.15

= 84.06 cmAlso, load transfer capacity of dowel system = (4100×0.50)

= 2050 kg.

Distance over which dowel bars are effective = 1 84.06cm l [Given]

Dowel spacing = 30 cm (given)

Available capacity factor =84.06 30 84.06 601

84.06 84.06

= 1.929

Also, capacity factor =Load transfer capacity of dowel system

P

Where P = Load transfer capacity of single dowel bar.

P =2050 1062.727 kg1.929

7. (b) A pile group consists of four friction piles in cohesive soil. The unit weight andunconfined compressive strength of the soil are respectively 20.2 kN/m3 and 200 kPa.The diameter of each pile is 300 mm, length is 12.0 m and centre-to-centre spacingbetween the piles is 750 mm. Assuming an adhesion factor of 0.6, determine (i) loadcapacity of the group based on the individual pile failure, (ii) load capacity of thegroup based on the block failure and (iii) design load capacity of the group. Assumea factor of safety of 2.0 for individual pile failure and 3 for block failure.

[20 Marks]




Mains Exam Solution

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Sol. Given dataD = s + d

s

= 20.2 KN/m3

UCS = 200 kPad = 0.3 mL = 12 m

Spacing = 0.75 m = 0.6

C =UCS 200 100kPa

2 2

(i) Load capacity of the group based on the individual pile failure. (For friction pile, capacity of bearingmay be ignore for calculation of single pile capacity)

Qu = SCA

= 0.6 100 0.3 12

Qu = 678.58 kNSo, Qu = 4 × 678.58

Qu = 2714.34 kN(ii) Load capacity of the group based on the block failure.

D = s d 0.75 0.3

D = 1.05 m

Qug = b S9CA CA 1

Qug = 9 100 1.05 1.05 1 100 4 1.05 12

Qug = 6032.25 KN(ii) Design Load capacity of group.

= ugu QQMin ,FOS FOS

=2714.34 6032.25Min ,

2 3

= Min [1357.17, 2010.75]QUdesign = 1357.17 kN

7. (c) The following internal angles and length of sides are observed for a closed traverseABCDA (in anti-clockwise direction):

Angle Observed value Side Measured length (m)DAB 92.38 AB 27.15ABC 104 33 BC 52.16BCD 70 46 CD 41.96CDA 92 07 DA 46.73

Adjust the internal angles for closing error. Also adjust the traverse by Bowditchmethod and calculate the consecutive coordinates of points A, B, C and D. Assume lineAD in north direction.

[20 Marks]

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Mains Exam Solution



Sol. Since traverse is running anticlockwise, included angle will be interior angles.

A B C D = 360°4'

Theoretical Sum = 90 3602n 4

Error = 360 4 360 4

Correction = –4'

Correction per angle = 1L 14

Correct angle’s

A = 92°37'

B = 104°32'

C = 70°45'

D = 92°6'

AB

C

D

B.B of DA = 0°

F.B. of Next Line = B.B. previous Line + included angle

F.B. of AB = 0 92 37 92 37

F.B. of BC = 104 32 17 992 37 180

F.B. of CD = 70 45 267 5417 9 180

F.B. of DA = 92 6 180267 54 180 O.K.

Line Length W.C.B. Calculated CorrectionAB 27.15 92 37 1.239 0.054 1.293 27.122 0.092 27.030BC 52.16 17 9 49.84 0.104 49.736 15.381 0.177 15.204CD 41.96 267°54 1.537 0.083 1.620

41.932 0.143 42.075DA 46.73 180°0 46.73 0.093 46.823 0 0.159 0.159

0.334 0.334 0.571 0.571

Corrected Calculated Correction CorrectedLatitude Departure

Calculation for Closing error Pex = +0.571, ey = +0.334

e = 2 2x ye e 0.662

=x1

y

etan 59.67

e




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W.C.B. of closing error = 59.67° (1st Quadrant)Calculation for Adjustment of closing error by Bowditch rule

L ABe = yAB 27.15e 0.334 0.054P 168

eD – AB = xAB 27.15e 0.571 0.092P 168

Co-ordinates of end point of a line w.r.t. its initial point ar called consecutive co-ordinatesHence, Consecutive co-ordinates of B(x,y) = (27.030, –1.293)

Consecutive co-ordinates of C(x,y) = (15.204, 49.736)Consecutive co-ordinates of D(x,y) = (–42.075, –1.620)Consecutive co-ordinates of A(x,y) = (–0.159, –46.823)

8. (a) (i) What is spectral reflectance curve (spectral signature) in remote sensing? Explainany four applications of remote sensing in civil engineering.

[10 Marks]

Sol.• The word signature in general, refers to a distinguishing character or a feature which is characteristic

of an object.• Spectral response of features such as water, soil, vegetation, etc. are separable, these permit an

assessment of the type and/or condition of the features. These responses are often referred to asspectral signatures.

• In remote sensing, the term signature specifies the basic property, which directly or indirectly leadsto identification of an object. From this, the characteristic expression of the object that distinguishesit from its surroundings can be known.

• The 4 major characteristics of target facilitating discrimination are variations in :– Spectral (variation in reflectance)– Spatial (spatial arrangement of terrain features providing attributes such as size, shape, etc. of

the object).– Temporal (change in the reflectivity or emissivity with time), and– Polarization (change in the polarization of the radiation reflected or emitted by an object).

• Spectral signature is used in a set of values which may be reflectance or radiances from the objectaveraged over different well-defined wavelength intervals.

Application of Remote sensing in Civil Engineering :(i) Regional planning and site investigations(ii) Town planning and Urban Development(iii) Environment and Geology : Whether for irrigation, power generation, drinking, manufacturing, or

recreation, water is one of our most critical resources. Remote sensing can be used in a variety ofways to help monitor quality, quantity of water resources.

(iv) Water Resources Engineering : By analyzing multi-date remote sensing data, it would be possibleto monitor the effects of dam construction.

8(a) (ii) A simple circular curve of radius 30 chain length has been set out to connect twotangents with external deflection angle of 30°. The chainage of point of tangencyis 300 chains. On further inspection, it is proposed to alter the radius of curve to

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Mains Exam Solution



45 chain length. Calculate the chainage of point of curve and point of tangencyfor revised curve. Also calculate the length of long chord for revised curve. (Chainlength = 20 m)

[10 Marks]

Sol. Given,

(PI)I

(P.C)T1T (P.T)2

= 30°

R = 30 chain = 600 mChain length = 20 m

Chainage of T2 = 300 chain = 6000 m

Chainage of T2 = Chainage of T1 + l

l =R 600 30 314.159m

180 180

Chainage of T1 = 6000 314.159 5685.841m

Chainage of I = Chainage of T1 + T

T =30Rtan 600 tan 160.770m

2 2

Chainage of I = 5685.841 160.770 5846.611m

If radius of curve is revised to 45 chain length.

R' = 45 20 900m

T =30R tan 900 tan 241.154m22

Chainage of T1 (P.C.) = 5846.611 241.154 5605.457 280 chain 28 link

Length of Curve, l =R 900 30 471.239m

180 180

Chainage of T2 (P.C.) = 5605.457 471.239

= 6076.696 303 chain 83Link

Length of Long Chord, L =302Rsin 2 900sin 465.874m 23chain 29Link

2 2

8. (b) The figure given below represents time-consolidation relationship for a 30 mm thick




Mains Exam Solution

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clay sample subjected to a given pressure range under double drainage condition.Determine the coefficient of consolidation, Cv, for the clay sample. How long will ittake (in days) to reach 50% consolidation for the same soil if it was 2.5 m thick anddrained in one direction only? Given

2U%T U 60%4 100

T = 1.781 – 0.933 log (100 U%), U > 60%

0102030405060708090100

0 1 2 3 4 5 6 7 8 9

b

1.15b

a

1.15a

Stra

in

Perc

ent c

onso

lida

tion

( )Time min

[15 Marks]

Sol.

RC

ConsolidationCurve

BA

P 90%

time

Straight line abscissaat every point is1.15 times thatof consolidation line

As per practical result, the intersection of line B with the consolidation curve gives a point Pcorresponding to 90% U.

From the given graph at 90% consolidation.

time = 7.5

time, t = 56.25 minuteFor U = 90%

TV = 1.781 0.933log , U 60%100 U%

= 1.781 0.933log 100 90

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Mains Exam Solution



TV = 0.848We know,

TV = V 2tC

d

Here, d =30 15mm 1.5cm2

t = 56.25 minute

0.848 = V 256.25C1.5

CV = 20.03392 cm min

CV = 248.845 cm day

(ii) For 50% consolidation

TV =250

4 100

TV = 0.196

TV = V 2tC

d [Here d = 2.5 m = 250 cm]

0.196 = 2t48.845

250

t = 250.79 days

8. (c) A two-lane, two-way highway is designed for design speed of 80 km/hr. A verticalcurve is to be provided at intersection of downward gradient of 1 in 50 with anotherdownward gradient of 1 in 20. Calculate the length of the vertical curve fulfilling therequirement of stopping sight distance and overtaking sight distance. The coefficientof longitudinal friction and the acceleration may be taken as 0.35 and 3.6 km/hr/secrespectively.

[25 Marks]

Sol. Given data

11N

50

21N

20

V = 80kmph 22.22m sec

N1 =1

50




Mains Exam Solution

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N2 =1

20

N = 1 21 1N N

50 20

N =3

100f = 0.35

a = 23.6kmph/sec 1 m sec

Stopping sight distance calculation :

SSD =2VVt

2gf

Take reaction time = 2.5 sec.

SSD = 222.2222.22 2.52 9.81 0.35

SSD = 127.45 m

OSD Calculation :

Vb = 80 16 64kmph

Vb = 17.778 m/sec

OSD for two lane two way traffic = 1 2 3d d d

d1 = bV t 17.778 2

d1 = 35.55 m

d2 = 2s b

s = b0.7V 6 0.7 17.778 6

s = 18.444 m

T =4s 4 18.444 8.589seca 1

So, b = bV T 17.778 8.589

= 152.7 m

So, d2 = 2 18.444 152.7

d2 = 189.6 m

d3 = V.T. 22.22 8.589 190.85m

So, OSD = 1 2 3d d d

= 35.55 + 189.6 + 190.85OSD = 416 m

(a) Design of vertical curve based on SSD

Case I : Assume SL SSD

So, length of curve,

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Mains Exam Solution



LS =2NS

4.4

=23 127.47

100 4.4

LS = 110.78 m < SSDSo assumption not correct.

Case II : For SL SSD

LS =4.42SN

= 4.42 127.47 3100

LS = 108.27 mSo, length of summit curve based on SSD = 108.27 m (b) Design of vertical curve based on OSD.

Case I : Assume SL OSD

So, length of curve, LS =2NS

9.6

LS = 23 416100 9.6

LS = 540.80 m > OSD

So, assumption is correct.So, length of summit curve based on OSD = 541 m

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