IES MASTER · R (4) Conventional Class Test-1 (CPM + Soil) 28th Feb 2018 Mob. : E-mail: 8010009955, 9711853908 [email protected], [email protected] Regd. office : Phone : F-126,

Sl. No. 1 ABCDEF(M)Civil EngineeringConventional Class Test -1

28th Feb 2018

Time Allowed : Three Hours Maximum Marks : 300

Institute for Engineers (IES/GATE/PSUs)

IES MASTER

SECTION-A

1. (a) The atterberg limits of a clay are : LL = 60%, PL = 45% and SL = 25%. The specific gravityof soil solids is 2.70 and the natural moisture content is 50%

(i) What is its state of consistency in nature?(ii) Calculate the volume to be expected in the sample when moisture content is reduced

by evaporation to 20%. Its volume at liquid limit is 10 cm3. [10 Marks]Sol.1. (a)

wl = 60%, wp = 45%, ws = 25%w = 50% and G = 2.70

(i) Consistency index, Ic =

lw w 100Ip and Ip = wl -- wp

Ic =

60 50 100 66.67%60 45

It is a medium (firm) clay

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(2) Conventional Class Test-1 (CPM + Soil) 28th Feb 2018

8010009955, 9711853908 [email protected], [email protected]. : E-mail:

Regd. office : Phone : F-126, (Upper Basement), Katwaria Sarai, New Delhi-110016 011-41013406

(ii) vs

10

w = 25%s w = 45%p w = 60%l

Volu

me

(cm

)3

(%)

vp

Let Ms be mass of the solids in the sample.Mass of the water at LL = 0.6 Ms gmVolume of water at LL = 0.6 Ms mlVolume of solids at LL = 10 – 0 0.6 Ms

10 – 0.6 Ms= s

w

MG

10 – 0.6 Ms = ss

M M 10.305 g2.70

Volume of solids =10.305 3.817 ml

2.70

Volume calculation at SL (i.e. at ws = 25%)Mass of water= 0.25 Ms = 0.25 × 10.305 = 2.57625 ml

Volume of solids in the sample = 3.817 ml Volume of sample = 3.817 + 2.57625 = 6.39325 ml

= 6.40 mlCalculation of volume at w = 20%As at the shrinkage limit (25%) volume = 6.40 mlReducing w.c after SL does not change the volume of sample.Volume of sample (at w = 20%) = 6.40 ml

1. (b) A building has to be supported on a RC raft foundation of dimensions 14 m × 21 m. Thesubsoil is clay, which has an average unconfined compressive strength of 15 kN/m2. Thepressure on the soil due to the weight of the building and the loads that it will carry will be140 kN/m2 at the base of the raft. The building has provision for basement floors. At whatdepth should the bottom of the raft be placed to provide a factor of safety of 3 against shearfailure?

gclay = 19 kN/m3. Use Skempton’s approach for bearing capacity calculations.[13 Marks]

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Conventional Class Test-1 (CPM + Soil) 28th Feb 2018 (3)



1. (b) Data Given:Average unconfined strength, qu = 15 kN/m2

Df

Basement

c =2uq 15 7.5 kN/m

2 2

Safe bearing pressure on the soil = 140 kN/m2

FOS against shear Failure = 3Let the depth of raft foundation be Df

We have for clay, = 0 and for such a condition, net ultimate bearing capacity as per skemptonqnu = cNc ... (i)

As per skempton the value of Nc is given by

Nc = f0.2D B5.0 1 1 0.2B L

= 5.0 f0.2 D 141 1 0.2

14 21

= 5.67 (1 + 0.0143 Df ) ... (ii)The building has a provision for basement, therefore

2PB

= nuf

q DFOS

... (iii)

Thus,

nuf

qD

FOS = 140

cf

cN DFOS

= 140

7.53

[5.67 (1 + 0.0143 Df )] + 19 Df = 140

D f = 6.55 m ... (iv)Thus the bottom of the raft should be placed at a depth of 6.55 from the ground surface.

1. (c) (A) In an experiment set up as shown in fig., flow is taking place under a constant headthrough the soils 1 and 2, of different hydraulic properties.

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(4) Convent ional Class Test-1 (CPM + Soil) 28th Feb 2018



Overflow

0.40 m

0.25 m

0.30 m

Over flow

Soil 2

Soil 1A

B

Datum

(i) Determine the total head and piezometric head at the point A.(ii) If 40% of the excess hydrostatic pressure is lost in flowing through soil 1, then what

is the total head and piezometric head at point B?(iii) if the permeability of soil 1 is 0.40 mm/s, what quantity of water is flowing through

unit plan area of the soil per second ?(iv) What is the coefficient of permeability of soil 2 ?

1. (c) (A) (a) Available total head = 0.4 mThis head is dissipated in flowing through the soil sample 1 and 2.

piezometric head at A = 0.3 + 0.25 + 0.4= 0.95 m

datum head = –0.25 – 0.30= –0.45 m

total head = 0.95 – 0.45= 0.40 m

(b) hydrostatic pressure lost in flowing throughA = 0.4 × 0.4 = 0.16 m

total head at B = 0.4 – 0.16 = 0.24 mdatum head at B = –0.25 m

piezometric head at B = 0.24 – (–0.25)= 0.49 m

(c) K1 = 0.4 mm/sQ = KiA

where i =head lost in flowing through soil 1

length of soil 1 = 0.160.3

Q = 3 0.160.4 10 10.3

= 4 32.13 10 m / sec

(d) Quantity of water flowing in soil 2 will be same as that in soil 1.Let the coefficient of permeability of soil 2 is K2.

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Q = K2.i2 A

i2 = LH 60% 0.4 0.96L 0.25

2.13 × 10–4 = K2 × 0.96 × 1K2 = 0.22 mm/sec

1. (c) (B) If the void ratios and specific gravities of soil 1 and 2, are as below, then

Soil 1 : e = 0.55,G = 2.65

Soil 2 : e = 0.65,G = 2.70

Determine: (i) the discharge velocity and seepage velocity through each soil, and (ii) thehydraulic head at which instability occurs. [12 Marks]

Sol. 1.(c) (B)

(i) Discharge velocity =4Q 2.13 10 m/sec

A 1

= 2.13 × 10–4 m/sec

Seepage velocity s =n

n =e

1 e

n1 =1

1

e 0.55 0.3551 e 1 0.55

n2 =2

2

e 0.65 0.3941 e 1 0.65

Seepage velocity, 1s =

44

1

2.13 10 6 10 m/secn 0.355

1s = 0.6 mm/sec

2s =4

2

2.13 10n 0.394

= 0.54 mm/sec(ii) Hydraulic gradient in soil 2 is more than that in soil 1.

instability will occur in soil 2 first.Critical hydraulic gradient for soil 2

=G 1 2.7 1 1.031 e 1 0.65

for instability, the hydraulic gradient in soil 2 should be 1.03

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i.e., LHL

= 1.03

LH0.25 = 1.03

HL = 0.26Let the hydraulic head at which instability occur is H.

H × 60% = 0.26H = 0.43 m

1. (d) For the given PERT network, determine

(i) Expected time, Standard deviation and variance of the PROJECT and show thecritical path also.

(ii) Probability of completion of project in 34 days.(iii) Time duration that will provide 90% probability of its completion in time.The three time estimates of each activity are mentioned on the network. [15 Marks]

2-5-8 4-7-16 7-10-134-9-20 3-7-17

2-3-10

2-4-63-5-13

8-11-20

3

1 2 4 6

5 7 8

Z –0.7 –0.8 –0.5 –0.9 –

P 24.20 21.19 30.85 18.41 10.00

1.28

Sol. 1(d)Act t t t t

1–2 2 5 8 5 1 12

0 m p e

–3 4 9 20 10 2.67 7.132–4 4 7 16 8 2 42–5 8 11 20 12 2 44–5 0 0 0 0 0 03–6 3 7 17 8 2.33 5.434–6 7 10 13 10 1 15–7 3 5 13 6 1.67 2.786–7 2 3 10 4 1.33 1.777–8 2 4 6 4 0.67 0.45

2

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21

5 7 8

4 613

S=0B

t = 6e t = 4e

t = 8e t = 10e

t = 8e

t = 12e

t = 5e

t = 0e

t = 10e

t = 4e

33

13

Project completion time = 31 days.

Critical paths are(a) 1–2–3–6–7–8(b) 1–2–4–6–7–8As there are two critical paths, hence to determine most critical path we have to find out variance.

For path (a) = 1 7.13 5.43 1.77 0.45 3.97

For path (b) = 1 4 1 1.77 0.45 2.87

Hence most critical path is (a) i.e. 1–2–3–6–7–8Probability corresponding 34 days

Z = S ET T 34 31 0.7563.97

p =78.81 75.8075.80 0.056 77.486%

0.1

sT for p = 90%

Z = 1.28 for p = 90%

1.28 = ST 313.97

ST 36.082 days

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1 (e) The delivery price of an equipment is Rs. 1.10 lakh and its installation charge is Rs. 10000,Life of the equipment is 6 years, the rate of interest (i) for sinking fund is 8%. Calculate thedepreciation for the 4th year and book value at the end of 4th year by the following methods.

(a) Sink fund method(b) Sum of digits of years method [10 Marks]

Sol. 1. (e) iC 1,10,000 , sC 10,000 n = 6 yr, i = 8%

Sinking fund method.

D = i s

nC C i

n (1 i) 1

D =6

110000 10000 0.086 (1.08) 1

D = 2271.92

Depreciation at the 4th year.

4D = 4 12271.92(1.08) = 2861.96 2862

Sum of digits of year method

mD = i sn m 1 (C C )n(n 1)

2

4D =6 4 11,00,000 76

2

= 14285.71 14286

2. (a) A building was to be constructed on a clay stratum. Preliminary analysis indicated a settlementof 50 mm in 6 years and an ultimate settlement of 250 mm. The average increase of pressurein the clay stratum was 24 kN/m2.

The following variations occurred from the assumption used in the preliminary analysis:

(i) The loading period was 3 years, which was not considered in the preliminary analysis.(ii) Borings indicated 20% more thickness for the clay stratum than originally assumed.(iii) During construction the water table got lowered permanently by 1 meter.Estimate :

(1) The ultimate settlement(2) The settlement at the end of the loading period(3) The settlement 2 years after completion of building

[13 Marks]

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Sol. 2 (a) (1) Since water table got lowered by 1m, there will be increase in effective stress.

increase in effective stress = 1

= 9.81 KN/m2

Actual increase of pressure in the clay stratum= 24 + 9.81= 33.81 KN/m2

initial thickness = H1

actual thickness = 1.2 H1

We know, H = m p H

1

2

HH

=

1

1

m 24 Hm 33.81 1.2H

2

250H =

2433.81 1.2

2H = 422.62 mm

(2) loading period = 3 years

Settlement at the end of loading period will be due to settlement caused after half of loading periodi.e., 1.5 years.

ultimate settlement = 250 mm

degree of settlement in 6 years (U1) =50 0.2

250

t1 = 6 yearst2 = 1.5 years

U2 = ?

2122

UU =

1

2

tt

222

0.2

U=

61.5

U2 = 0.1

Settlement at the end of loading period= 0.1 × 422.6 = 42.26 mm

(3) t3 = 1.5 + 2 = 3.5 years2322

UU =

3

2

tt

23

2U

0.1 =3.51.5

U3 = 0.153

Settlement 2 years after completion of building = 0.153 × 422.62 = 64.56 mm

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2. (b) A 20 m high earthen bund has the following dimensions:

Free board = 2 m; Top width = 3 m

U/S slope = 12 :1; D/S slope = 2 :1.2 Both (H : V)

A horizontal filter extends to a distance of 15 m from the d/s toe. The material of the bundshas a vertical permability of1 × 10–4 cm/sec, and a horizontal permeability of 9 times that of vertical permeability.Determine the seepage loss per m length of the bund. [12 Marks]

Sol. 2 (b)

2m L

0.3L

3m

1H=18m

15mS

d

O D

12

D'

52

P

K y = 1 × 10–4 cm/secKx = 9 × 10–4 cm/sec

From the figure, L =518 45 m2

H = 20 – 2 = 18 m

d = 50.3 45 2 3 20 2 152

d = 46.5 m

d =4

y4

x

K 1 10d 46.5K 9 10

d = 15.5 m

From the property of parabola, we have

d S = 2 2d H

15.5 S = 2 215.5 18

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S = 8.25 m

Now Keq = 4 4x yK K 9 10 1 10

Keq = 3 × 10–4 cm/sec

Seepage discharge per unit length of dam

q = eqK S

= 3 × 10–4 × 10–2 × 8.25= 2.476 × 10–5 m3/sec/m

2. (c) An alternative, A requires an initial investment of Rs. 500,000 and an annual expense of Rs.250,000 for the next 10 years. Alternative B, on the other hand, requires an initial investmentof Rs. 750,000 and an annual expense of Rs. 200,000 for the next 10 years. Whichalternative would you prefer if interest rate were 10 per cent? [15 Marks]

Sol. 2. (c) The cash-flow diagram corresponding to alternative A is given below

5.00

2.50 2.50 2.50 2.50 2.50

+veincoming

–veoutgoing

10543210

2.50

Present cost of alternative A = 500000 + 250000 (P/A, 10%, 10)

= 500000 + 250000 (6.1446)

= Rs. 2036150.00

The cash-flow diagram corresponding to alternative B is given below

7.50

2.00 2.00 2.00 2.00 2.00

+veincoming

–veoutgoing

10543210

2.00

Cash Flow diagram for Alternative B

year

Present cost of alternative B = 750000 + 200000 (P/A, 10%, 10)

= 750000 + 200000(6.1446)

= Rs. 1978920.00

It can be noticed from the two cash-flow diagrams that the cost data alone are provided. Thus,alternative with the lowest cost at present time would be the most preferable. In this case, sincethe present cost for alternative B is less than that of alternative A, it is preferable to choose

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alternative B.

2. (d) The two machines A and B have the following costs with the money worth as 8% per year;

A B

First cost Rs 10,000 Rs 25,000

Salvage value Rs 1,100 Rs 1,500Uniform expenditure at end of year Rs 3,000 Rs 2,000Irregular expenses at end of 1st year Rs 1,000 –Irregular expenses at end of 3rd year – Rs 2,500Benefit from quality control – Rs 600(at end of year)Life 2 years 5 years

Compare the machines for suitablity of selection on the following bases;

(i) Present worth(ii) Equivalent Annual cost worth(iii) Capitalized cost worth [20 Marks]

Sol. 2.(d) The fig shows the cash flow diagram for life of 2 years.

10,000 3000

1000

m/c (A)1100

3000

20 1

PA(2) = Present worth of m/c (A) at 0 time of its life period (2 years)

=

P P P10000 3000 ,8%,2 1100 ,8%,2 1,000 ,8%,1A F F

=

2

2 21.08 1 1 110000 3000 1100 1000

1.080.08 1.08 1.08

= Rs 15,333

PB(5) = Present worth of m/c (B) at 0 time of its own life period ( 5 years)

0

600

1

2000

600

2

2000

600

3

2000

600

4

2000

600

5

2000

1500

2500

25000

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= P P P25000 2000 600 ,8%,5 2500 ,8%,3 1500 ,8%,5A F F

=

5

5 3 51 0.08 1 1 125000 1400 2500 1500

0.08 1 0.08 1 0.08 1.08

= Rs 31,553

Since the lives of machines are different. The common life period will be L.C.M. of their livesi.e. L.C.M. of 2 and 5 i.e. 10 years.

0 2 4 6 8 10

15333 15333 15333 15333 15333

PA(10) = overall present worth of machine A on the basis of 10 yrs period

=

P P15333 15333 ,8%,2 15333 ,8%,4F F

P P15333 ,8%,6 15333 ,8%,8F F

= 2 4 6 8

1 1 1 115333 153331.08 1.08 1.08 1.08

= Rs 57,695

PB(10) = Overall present worth of machine is for common life period of 10 yrs.0 5 10

31553 31553

= P31553 31553 ,8%,5F

= 5

131553 315531.08

= 53,027

Since PB(10) < PA(10) Machine B is selected(ii) Equivalent Annual worth method

A(A) = Equivalent Annual worth of machine A

= AAP 2 ,i,nP =

2

20.08 1 0.0815333

1 0.08 1

= Rs 8598

Similarly,A(B) = Equivalent annual worth of machine (B)

= BAP 5 ,8%,5P =

5

50.08 1 0.0831553

1 0.08 1

= Rs 7903

Since A(B) < A(A)

Machine B is selected

(iii) Capitalized Cost MethodC(A) = Capitalized worth of machine A

= Equivalent annual worth of machine A

i = 85980.08 = 107,475

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CB = 79030.08 = 98787

Since CB < CA

Machine B is selected.

3. (a) A series of tests on a sample of silty clay indicated the following index properties :

L.L. = 53.9%P.L. = 23.4%Natural w.c. = 51.2%D60 = 0.0050 mmD10 = 0.0007 mm

Unconfined compressive strength (undisturbed) = 2180 kN / m .

Unconfined compressive strength (remoulded) = 285 kN / mCompute (i) the uniformity coefficient, (ii) the plasticity index (iii) the liquidity index, and (iv)the sensitivity of this soil. (v) In what region can the soil be placed on the I.S. plasticity chart?

[10 Marks]

3. (a) (i)

Cu =60

10

D 0.0050 7.14D 0.0007

(ii) Plasticity index = L.L. – P.L.= 53.9 – 23.4= 30.5%

(iii) Liquidity index = IL = p

PI

= 0.9114

(iv) Sensitivity =unconfined compressive strength (undisturbed)

unconfined compressive strength (disturbed)

=180 2.1285

(v) IP = 0.73 L 20

= 0.73 (53.9 – 20)= 24.747

Soil lies in high compressive zone and lies above A-line. hence called CH.

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3. (b) In order to determine the average permeability of a bed of sand 14 metres thick, overlyingon impermeable stratum, a well was shunk through the sand and a pumping test was carriedout. After a certain interval the discharge was 12.4 litres per second and drawdowns in theobservations wells at 16 m and 33 metres, from the pumping well were found to be 1.787m and 1.495 m respectively. If G.W.L was originally 2.14 below ground level, find

(i) the permeability of the sand layer(ii) approximate value of the effective grain size. [10 Marks]

3. (b) (i)

2e

12 22 1

rQlogrK

(h h )

r1 = 16 m, r2 = 33 m,

h1 = 14 – (2.14 + 1.787) = 10.073 m

h2 = 14 – (2.14 + 1.495) = 10.365 m

e

2 2

12.4 33log1000 16K 0.479 mm/sec

(10.365 10.073 )

(ii) 210K 100D

210

0.479D100

D10 = 0.0692 mm

3. (c) A layer of clay 2 m thick is subjected to a loading of 20.5kg / cm . One year after loading,the average consolidation is 50%. The layer has double drainage. (i) What is the coefficientof consolidation ? (ii) If the coefficient of permeability is3 mm/year, what is the settlement after one year, and (iii) how much time will the layer taketo reach 90% consolidation. [10 Marks]

3. (c) thickness of clay layer = 2 m

Max. length of drainage path, H =2 1 m2

U = 50%

T = 20.5 0.1964

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= 0.5 kg/cm2

= 0.5 × 104 Kg/m2

(i) T = 2C tH

t = 1 yr.

0.196 =C 1

1

C = 0.196 m2/yr

(ii) K = C m

3 m3 10yr

= 2 30.196 m / yr m 1000 kg/m

m = 1.53 × 10–5 m2/kg

H = 0m H

= 1.53 × 10–5 × 0.5 × 104 × 2= 0.153 m= 153 mm

(iii) U = 90%

T = 1.781 – 0.933 log (100 – 90)

= 0.848

1T = 0.196

t1 = 1 yr, t2 = ?

1

2

TT

=

1

2

tt

0.1960.848 =

2

1t

t2 = 4.33 yr.

3. (d) What is the present equivalent value of Rs. 50000, five years from now at 14% per annumcompounded semi-annually? [10 Marks]

Sol. 3 (d)

Nominal intrest rat is 14% per annum

If compunded half yearly nominal intrest rate is 7%

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PF

= n1

(1 i)

P = 1050000

(1 0.07)

P = Rs. 25417

3. (e) Determine the gradability of a crawler tractor pulling a high pressure rubber tyred self loadingscraper and its load. The following information is available:Tractor HP = 200Weight of tractor = 18.5 tWeight of loaded scraper = 31.5 tRated drawbar pull of tractor = 14.8 tRolling resistance for the tractor = 90 kg/tRolling resistance for the scraper = 105 kg/tTake pull required per tonne for the unit per 1% slope = 10 kgAssume available drawbar pull as 0.85 times the rated drawbar pull of tractor.

[20 Marks]

Sol. 3 (e) Applying suitable factor of safety on rated drawbar pull.

Available drawbar pull = 0.85 × 14.8 × 1000 = 12580 kg

Since the design of crawler tractor as per manufacturer’s specifications is based on the rollingresistance of 55 kg/t,

Rolling Resistance to be overcome by the tractor = 18.5 × (90 – 55) = 647.5 kg

Rolling Resistance to be overcome by the scraper = 31.5 × 105 = 3307.5 kg

Combined Rolling Resistance to be overcome by the unit

= 647.5 + 3307.5 = 3955 kg

Net pull available to negotiate the grade = (12,580 – 3955) kg = 8625 kg

Combined weight of the unit = 18.5 + 31.5 = 50.0 t

The pull required per tonne for the unit per 1% slope = 10 kg

The pull required for 50 tonne of the unit per 1% slope = 10 × 50 = 500 kg

500 kg pull is required for 1% slope

8625 kg pull is required for 8625500 = 17.25% Ans.

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4. (a) A retaining wall 6 m high retains sand with 30 deg and unit weight 324 kN / m uptoa depth of 3m from top. From 3 m to 6m, the material is a cohesive soil with

2c 20 kN / m and 20 deg. Unit weight of cohesive soil is 318kN / m . A uniform

surcharge of 2100 kN / m acts on the top of soil. Determine the total lateral pressureacting on the wall and its point of application.

[15 Marks]

Sol 4.(a)

3m

3m

= 30, = 24 kN/mt3Z

C = 20 kN/m , = 202

t = 18 kN/m3

100 kN/m2

For Sand : Ka= 1 sin1 sin =

1 sin30 11 sin30 3

pa = a ak ( z q) 2C k

at Z = 0, a1p (0 100) 2 0 1/ 33

= 33.33 kN/m2

at Z = 3, a1p [24 3 100] 2 0 1/ 33

= 57.33 kN/m2

for clay :

ka =1 sin20 0.491 sin20

equivalent surcharge q = 100 + 24 × 3= 172 kN/m2

at Z = 3, a a ap k q 2C k

= 0.49 × 172 – 2 × 20 0.49

= 56.28 kN/m2

at Z = 6, a a ap k (172 18 3) 2 C k

= 0.49 × 226 – 2 × 20 0.49

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= 82.74 kN/m2

3m

3m

P1

P2

24 57.33 kN/m2

56.28 kN/m2

P3

P426.46 kN/m2

82.74 kN/m2

33.33 KN/m2

Total lateral pressure = P1 + P2 + P3 + P4

=1 133.33 3 24 3 56.28 3 26.46 32 2

= 100 + 36 + 168.84 + 39.69P = 344.52 kN/m

Point of application x from bottom

x =1 1 2 2 3 3 4 4

1 2 3 4

P x P X P X P XP P P P

=100 4.5 36 4 168.84 1.5 39.69 1

344.52

= 2.57 m

4. (b) A 12 m long, 300 mm diameter concrete pile is driven in a uniform deposit of sand (= 40º).The water table is very much down and is not likely to rise in future. The average dry unitweight of sand is 16 kN/m3. Using Nq = 137, calculate the safe load capacity of the pilewith a factor of safety of 2.5. Assume the critical length of pile as 15 times of the diameter.K for dense sand = 2.0. [15 Marks]

4. (b) Critical depth=15 × 0.3 = 4.5 m

V = Hcritical = 16 × 4.5 = 72 KN/m2 = q

12 m

4.5 m

7.5 m

V = 72 kN/m2

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Bearing capacity of pile = qNqqub = 72 × 137

qub = 9864 kN/m2

For concrete pile

Skin friction capacity = av sK tan A = 3 '4 = 30°

Qsu = av sK tan A As = Surface area of pile

=

1 2av

s v sK tan A K tan S2

= 2 tan 30° × 722 × × 0.3 × 4.5 + 2 × tan 30°

× 72 × × 0.3 × 7.5= 763.972 kN

Qu = Qbu + Qsu

= Qub × Ab + Qsu

= (9864 × 0.7854 × 0.32 + 763.972) kNQu = 1461.21 kN

Qsafe = uQF.O.S.

Qsafe = 1461.212.5

= 584.49 kN

Qsafe = 584.49 kN

4. (c) Construct a bar chart for a project with the activity data as noted below. Also compute thetotal duration of the project.

Activities Duration (days) Preceded byA 4 NoneB 3 AC 5 BD 4 NoneE 7 B, DF 6 DG 8 EH 5 FI 4 HJ 3 G, I [15 Marks]

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Sol. 4(c)

I

H

G

F

E

D

C

B

A

4 7 10 12 14 15 19 22 25

Time

Activity

J

Total duration project is 25 days

4. (d) What is Break even Analysis? Write functions and limitations of break even analysis[15 Marks]

Sol. 4.(d) Break even Analysis

Break even Analysis is a graphical representation of the relationship between cost and revenue forall possible volume of output.

Break even analysis is basically done to find out the point at which total revenue equals total cost& profit potential under varying conditions of out put and cost.

Break Even point is therefore at which neither a profit nor a loss is incurred.

Sometimes it is also called as Cost–Volume–profit studies.

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Profit Margin

Profit

Angle ofincidence Variable cost

Cost/Expenses

Reve

nue

Fixed cost

Loss

No. of Output Units

Cost

/Rev

enue

Functions1. Suitable for business firm to study cost revenue relationship.

2. Useful in making engineering decisions.

3. Useful in selection of favourable option of business.

4. Possibility of profit is determined for any rate of production.

5. It shows weather business is good or bad by angle of incidence. Greater the angle of incidencemeans more profit margin and, we know that profit margin should be more for good business.

Limitations

1. Break even analysis is a nice tool for small business.

2. It provides a static picture where as business processes are dynamic in nature because themarket conditions do not remains constant.

3. Revenue line may not be always a straight line.

4. Analysis of break even becomes difficult when company produces different/variety of products.

5. Cost and revenue are related only with number of units produced. They have no relation withthe time.

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SECTION-B

5. (a) In a proctor’s compaction test the maximum dry density was found to be 1.8 gm/c/c andO.M.C 15.2%. If the specific gravity of the soil grains is 2.65, calculate degree of saturationand void ratio. [8 Marks]

Sol 5. (a) Vdmax = 1.8 gm/ccO.M.C. w = 0.152G = 2.65

wd

G1 e

2.65 11.81 e

e = 0.47

es = wG0.47 × s = 0.152 × 2.65

s = 0.857 = 85.7%

5. (b) A foundation trench is to be excavated in a stratum of stiff clay, 8 m thick, underlain by abed of sand. In a trial bore hole, the ground water is observed to rise to an elevation 2mbelow the ground surface. (a) Find the depth to which excavation can be safely carried outwithout the danger of the bottom becoming unstable under uplift pressure of ground water.The specific gravity of clay particles is 2.72, and void ratio of 0.72 (b) If excavation is tobe carried safely to a depth 6m, how much should the water-table be lowered in thevicinity of the trench? [12 Marks]

Sol 5.(b) Let us assume that x m depth of excavation can be safely done. Then,

8m

x

G = 2.75e = 0.72

2m

6m

For stability, total downward weight = total upward pore water pressure

sat clay =G e 2.72 0.721 e 1 0.72

= 2

upward pore water pressure = 6 6

total downward weight = 8 x 2

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8 x 2 = 6

x = 5 m(b) If excavation is carried out upto 6m depth. Hence upward pressure has to be reduced

sat2 = 6 y

2 2 = 6 y

y = 2mW.T. will be lowered by 2m.

5. (c) Discuss the I.S. classification based on grain size and classify the following soils :

4.75 mm 0.075 mm 0.002 mm L P 1. 95 93 12 20 10 2. 98 8 — Non Plastic

UniformlyGraded

3. 93.2 72

S.N. % Passing Atterberg LimitLIMIT LIMIT

— Non-Plastic 4. 96 94 24 40 18

[10 Marks]

Sol 5.(c)To classify the given soilsSoil at S.No. 1

%Passing

4.75 mm95

0.075 mm93

0.002 mm12

Liquid Limit20

Plastic Limit10

Plastic Index IP20 – 10 = 10

Atterberg Limit

Since > 50% passes through 0.075 mm sieve hence the given soil is fine grained soilLiquid limit WL = 20 and IP = 10Plotting the point on the plasticity chart.The point comes above the A line represented by IP = 0.73 (WL – 20) and since the WL < 35.Hence the given soil is clay of low plasticity i.e. CL.Soil at S.No. 2

%Passing

4.75 mm98

0.075 mm8

0.002 mm–

WL

Non Plastic

WP

Uniform Graded

IP–

The given soil is a coarse grained soil with greater percentage of coarse fraction passes through4.75 mm sieve but retained on 75 µ sieve with only 8% passing through 75 µ sieve (i.e. between 5%and 12%).

Hence the given soil shall be given dual symbol representing that its a poor graded silty sand i.e.SP–SM

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Soil at S.No. 3

%Passing

4.75 mm93.2

0.075 mm72

0.002 mm–

WL

Non Plastic

WP

–

IP–

The given soil is a fine grained soil with greater percentage passing through 0.075 mm sievemoreover the given soil is non plastic in nature. Hence the given soil is non plastic silt i.e. (ML

Soil at S.No. 4

%Passing

4.75 mm96

0.075 mm94

0.002 mm24

WL

40

WP

18

IP22

The given soil is a fine grained soil with WL = 40 and IP = 22. Plotting the point on the plasticitychart. The point lies above the A-line and since 35 < WL < 50. Hence the soil can be classified asclay of intermediate plasticity i.e. CI.

5. (d) The following table gives the data for the duration and costs of each activity of a projectnetwork shown in figure. The indirect cost of the project is Rs. 3000/week. Determine theoptimum duration of project and the corresponding minimum cost. Draw the time scaledversion of the network.

1

2

4

3

6(3)5(3)

5(3)

4(1)

8(5)

1 2 6 7000 3 14,5001 3 8 4000 5 8,5002 3 4 6000 1 9,0002 4 5 8000 3 15,0003 4 5 5000 3 11,000

Activity Normal Normal Cost Crash Crash CostDuration Rs. Duration Rs.(week) (week)

Fig. (a): Network

[15 Marks]

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(26) Convent ional Class Test-1 (CPM + Soil) 28th Feb 2018



Sol. 5(c)

1

2

4

3

6(3)5(3)

5(3)

4(1)

8(5)

10,10

15,15

6,6

0,0

Critical path is 1 – 2 – 3 – 4Project duration = 15 weeksTotal project cost = Direct cost + Indirect costDirect cost = 7000 + 4000 + 6000 + 8000 + 5000 = 30,000Indirect cost = 3000 × 15 = 45,000T.P.C = 30,000 + 45,000 = 75,000Calculation of cost slopes

1 2 14,500 7,000 7,500 6 3 3 2,5001 3 8,500 4,000 4,500 8 5 3 1,5002 3 9,000 6,000 3,000 4 1 3 1,0002 4 15,000 8,000 7,000 5 3 2 3,5003 4 11,000 5,000 6,000 5 3 2 3,000

Activity ΔC(Rs.) Δt [weeks] Cost slope[Rs. / week]

Time Scaled version of network

41 326(3) 4(1) 5(3)

5(3)

8(3)

Ist Stage CrashingActivity 2-63 has least cost slope among critical activites. Crash this activity by 2 weaks.New project duration = 13 weeks.T.P.C. 75,000 ~ 2 × 3000 + 2 × 10000 = 71,000

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41 326(3) 2(1) 5(3)

5(3)

8(5)

13 weeks

Stage-1: Duration 13 weeks

IInd Stage CrashingActivity 2-3 and 1-3 has least combined cost slope i.e 2500. We will crash them simultaneously by1 week.New project duration = 12 weeksT.P.C = 71000 + 1× 3000 + 1 × 2500 = 70500

41 326(3) 1(1) 5(3)

5(3)

7(5)

12 weeks


IIIrd Stage CrashingNow activity 3-4 has least cost slope i.e. 3000, we will crash it by 1 week.New project duration = 11 weeks.T.P.C = 70500 – 1 × 3000 + 1 × 3000 = 70500

41 326(3) 1(1) 4(3)

5(3)

7(5)

11 weeks


IVth Stage CrashingActivity 1-3 and 1-2 have least combined cost slope i.e. 4000, we will crash it by 2 weeks.New project duration = 9 weeks.T.P.C. = 70500 – 2 ×.3000 + 2 × 4000 = 725000

41 324(3) 1(1) 4(3)

5(3)

5(5)

9 weeks


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Vth Stage CrashingActivity 3-4 and 2-4 can be crashed simultaneously by 1 weeks.New project duration = 8 weeks.T.P.C. = 72500 – 1 × 3000 + 1 × 6500 = 76000

41 324(3) 1(1) 3(3)

4(3)

5(5)

8 weeks


Hence, the optimum project duration = 11 weeks.

The corresponding least cost = Rs. 70500.

5. (e) For the network shown in Fig. below assume that, after working 15 days on the project, thefollowing conditions exist:

(a) Activities 1-2, 1-3 and 1-4 are complete as originally planned(b) Activity 2-4 is in process and will be completed in 3 more days(c) Activity 3-6 is in process and will need 18 more days for completion(d) Activity 6-7 appears to present some problem and its new estimated time of completion

is 12 days.(e) Activity 6-8 can be completed in 5 days instead of the originally planned 7 days.(f) No other activity have been started and the estimates of original time remains same.

1

2

3

4 7

6 8

5

10

6

79

18

8

6

7

7

12

20

5

Update the network based on the assessment at the end of 15 days. Include all activities inthe new project. [15 Marks]

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Sol. 5(e)

Before Updating

41

3 6 8

7

11

0

29

36

42

289

16

9

6

10 5

12

18

7 8

20

76

7

3 32 5

10

0

22

36

42

279

16

Critical path is 1–2–4–7–8 after updating.

41

3 6 8

7

15

15

33

45

3719

25

0

0

0

18

18

3 8

20

126

5

3 32 5

15

15

33

45

3315

20

5151

5

Now critical path will be 1–3–6–7–8.

6. (a) A concrete dam of 20 m base width stores 6 m depth of water. A sheet pile cut off is providedat the upstream end of the base of the dam up to a depth of 10 m. The base of the damis 1.0 m below the ground surface, and the pervious foundation extends to a depth of 16.0m. below which an impervious straturm exists. The flow net for this structure is drawn, asshown in the Fig. below.

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(i) Compute the seepage flow below the dam in cumecs per meter length of dam, if thecoeffcient of permeability (K) for the foundation soil is 31.5 10 cm / s .

(ii) Compute the exit gradient and F.S. against boiling, if the elementary square at the toeof the dam has been measured to have dimensions of 2.4 m.

Use unit wt. of water = 310 kN / m . [10 Marks]

6. (a)

(i) seepage flow q = f

d

NKh

N

h = 6mK = 1.5 × 10–3 cm/sec

= 1.5 × 10–5 m/secN f = 3, Nd = 10

q = 5 31.5 10 610

q = 2.7 × 10–5 m3/sec/m run

(ii) exit gradient =head loss per drop

length of flow in end field square

=

610 0.252.4

If the critical hydraulic gradient is taken approximately equal to 1 the FS against boiling is

= 40.25

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6. (b) The soil profile at a site for a proposed office building consists of a layer of fine sand 10.4mthick above a clay layer 2m thick. Below the clay there is a deposit of coarse sand. Theground water table was observed at 3m below ground level. The void ratio of the sandis 0.76 and specific gravity is 2.7. The building will impose a vertical stress increase of 140kPa at the middle of clay layer. The clay is over consolidated with an OCR = 2.5, w =38%, Cr = 0.05, Cc = 0.3 and Gs = 2.7. Assume the soil above the water table to besaturated. Determine the primary consolidation settlement of the clay. [10 Marks]

Sol. 6.(b)

Fine sand

Coarse Sand

Clay

e = 0.76G = 2.7

10.4m

2m

3m

7.4m

AA = 38%

Clay, e0 = G 0.38 2.7 1.026

sat sand =G e1 e

= 32.7 0.76 9.81 19.28 KN/m1 0.76

sub sand =G 11 e

= 32.7 1 9.81 9.47 KN/m1 0.76

sub clay =2.7 1 9.81

1 1.026

= 8.23 KN/m3

0A = sat sand sub sand sub clay3 7.4 1

= 19.28 × 3 + 9.47 × 7.4 + 8.23 × 1= 136.15 KN/m2

A = 140 KN/m2

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A1 = 136.15 + 140 = 276.148 KN/m2

OCR = C

where C = preconsolidation stress

= existing effective stress

2.5 = C

136.15

C = 340.375 KN/m2

C > 1A

H =1

0

Ar o

0 A

C Hlog

1 e

=0.05 2 276.148log1 1.026 136.15

H = 15.16 mm

6. (c) A footing, 2 m square, rests on a soft clay soil, with its base at a depth of 1.5 m from groundsurface. The clay stratum is 3.5 m thick and is underlain by a firm sand stratum. The claysoil has L.L. = 30%, G= 2.7, water content at saturation = 40%, cohesion = 0.5 kg/cm2

(f = 0). It is known that the clay stratum is normally consolidated. Compute the settlementthat would result if the load intensity equal to safe bearing capacity of soil were allowedto act on the footing. Natural water table is quite close to the ground surface. For givenconditions, bearing capacity factor (Nc) is obtained as 6.9. Take factor of safety as 3.Assume load spread of 2 (vertical) to 1 (horizontal). [15 Marks]

Sol. 6. (c)

2m

1.5 m

A2 m

1

21.Q2

1.Q2

0 2 m

A1

1.Q2

1.Q2

L.L = 30%G = 2.7W = 40%c = 0.5 kg/cm

s2

N = 6.9 C

1 m

Firm Sand Stratum

B

Calculation for net safe bearing capacity of soil for square footing in clayey soil, qns

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As per Terzaghi, for square footing on clayey soil net ultimate bearing capacityqnu = 1.3c NC

qnu = 1.3 × 0.5 × 6.9= 4.485 kg/cm2

= 44.85 t/m2

Net safe bearing capacity of soil,

qns = nuqFOS

= 244.85 14.95t / m3

B = 14.95 t/m2

A =214.95 2 2 6.64 t / m

(2 1)(2 1)

initial effective stress at A, A = sub × 2.5

sub = w2.7 1G 1 1

1 0.4 2.71 e

sub = 0.8173 t/m2

A = 0.8173 × 2.5 = 2.043 t/m2

Cc = 0.009(LL – 10%) = 0.009(30 – 10) = 0.18

H =Ac A

A

C Hlog

1 e

H =0.18 2 2.043 6.64log

1 0.4 2.7 2.043

H = 10.88 cm

6. (d) Draw the network for the the following project and number the events according to fulkerson’srule. [12 Marks]

Activity Immediate Activity ImmediatePr edecessor Pr edecessor

A G CB A H C&DC A I E & FD AE B J G & HF C K I & J

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Sol. 6. (d)B

A F

E I

C

G

JH

KD

3 6

4

5 7

8 9

21

6. (e) Discuss in brief types, basic parts and operations of power shovels.[13 Marks]

Sol. 6. (e)Power Shovels Basically a shovel is a tool for digging, lifting, and moving bulk materials, such as soil, coal,

gravel, snow, sand, or ore.

Shovels are extremely common tools that are used extensively in agriculture, construction,and gardening.

When a shovel is mounted on a Power vehicle it is called as Power Shovel

Power shovels are used mainly to excavate earth and load into trucks or tractor-drawnwagons.

Power shovels can excavate all types of earth except solid rock without prior loosening.Types of Power Shovels1. Crawler mounted Shovels

2. Rubber Tyre mounted Shovels

Crawler mounted Shovels Crawler mounted Shovels are mounted on Crawler tracks.

Crawler mounted Shovels have very low travel speed.

Crawler mounted Shovels exert low pressure on the soil and hence suited for muddy and softground surface.

Rubber Tyre mounted Shovels Rubber Tyre mounted Shovels may be mounted on Rubber-tyres.

Rubber-tyre-mounted shovels have higher travel speeds are useful for small jobs whereconsiderable travelling is involved.

Rubber-tyre-mounted shovels exert considerable pressure on the soil surface hence suitedwhere the road and the firm ground surfaces.

The basic parts of a power shovel include Mounting

Cab

Boom

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Dipper stick

Dipper

Other attachments to the shovel include hoe, dragline, clamshell and crane.

Operations of Shovels Positioning the shovel near the face of the earth to be excavated.

The dipper is lowered to the floor of the pit, with the teeth pointing into the face.

A penetration force is applied through the dipper shaft and at the same time tension isapplied to the hoisting line to pull the dipper up the face of the pit.

If the depth of the face (called the depth of cut) is just right, the dipper will be filled as itreaches the top of the face.

If the depth is shallow it will not be possible to fill the dipper completely without excessivecrowding and hoisting tension.

If the depth of cut is more than is required to fill the dipper, the depth of penetration of thedipper into the face must be reduced if the full face is to be excavated or to start theexcavation above the floor of the pit.

7. (a) (i) Derive the expression for coefficient of earth pressure at rest.(ii) A smooth vertical wall 4m high retains cohesive soil backfill with 2c 10kN / m ,

0 and 318 kN / m . Determine (1) the depth at which active earth pressureis zero, (2) depth at which total active earth pressure is zero, (3) depth of tensioncrack, (4) active earth pressure at base of the wall, (5) total active earth pressure andits point of application from base of the wall; and (6) total passive earth pressure onthe wall. [15 Marks]

Sol. 7.(a) (i)If the soil mass is considered to be a semi-infinite, homogeneous, elastic and isotropic material, itis possible to evaluate the lateral pressure using the theory of elasticity, since there is no displacementat all.

Let x be the strain in the horizontal direction at depth Z on an element of soil and let the poisson's

ratio and the elastic modulus be and E respectively. For the plane strain condition, x is given by

x = x z x1 ( )E

The earth pressure at rest corresponds to a condition of zero lateral strain or x = 0

x = z1

x = Po = Lateral earth pressure at rest

Po = z1

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Coefficient of lateral earth pressure at rest (Ko) = 1

o zP ko

Where ko = 1

(ii) a1 sin0k 11 sin0

Depth of tension crack

oa

2c 2 10Zk 18 1

Zo = 1.11m(ii) Total active earth pressure will be zero at

= 2 × 1.11 = 2.22m(iii) Depth of tension crack = 1.11m

20kN/m2

1.11m

1.11m 4m

D E

A

B C

(iv) Active earth pressure at the base of wall = B A ak 2C k

= 218 4 1 2 10 11 52 kN/m

(v) Total earth pressure (without crack)

= 1 152 2.89 1– 20 1.112 2

= 64.04kN/m

x =1 1 2 2

1 2

A x A xA A

=

1 20.5 52 2.89 2.89 – 0.5 20 1.11 1.11 2.893 3

0.5 52 2.89 0.5 20 1.11x = 0.5011 m = From base of the wall.

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Total earth pressure (crack)

= 1 52 2.89 12

= 75.14kN/mPosition of the pressure

=1 2.893

x 0.9633m from base of the wall.

(vi) P z P Pp k 2C k

20kN/m2

92kN/m2

At top, top2

Pp 0 2 10 1 20 kN/m

At bottom,bottom

2Pp 18 4 1 2 10 1 92 kN/m

Total passive pressure = 1 20 92 4 12

= 224 kN/m

7. (b) A direct shear box test performed on a remoulded sand sample yielded the followingobservations at the time of failure :Normal load = 0.36 kN ; Shear load = 0.18 kNThe sample area was 236 cm ; Determine :

(i) The angle of internal friction ;(ii) The magnitude and direction of the principal stresses in the zone of failure; and(iii) The magnitude of maximum deviator stress if a sample of the same sand with the

same void ratio were tested in a triaxial test with an allround pressure of 260 kN / m .Assume c= 0.

[15 Marks]

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Sol. 7. (b)

f =shear load

c / s area of sample

=2

40.18 50 kN / m

36 10

=Normal load

C / s area of sample

=2

40.36 100 kN / m

36 10

From Mohr’s circle diagram

O1

2

P

Horizontal

(a) tan = f 50100

= 26.56

(b) in CP P,

cos = f

PC

PC = 50/cos 26.56 = 55.9 kN/m2

tan =CP

CP = 25 kN/m2

1 = OP CP CD

= 100 + 25 + 55.9 = 181 kN/m2

2 = OP P C CB

= 100 + 25 – 55.9= 69.1 KN/m2

= 90 = 116.56°

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31.72

minor principle plane

58.28

major principle plane

(c) 3 = 60 kN/m2

d = ?

1 = 60 + d

3 C 1

P

O

in OPC

sin =1 3

1 3

sin 26.56 =d

d

60 6060 60

d = 97.05 kN/m2

7. (c) A, B, C and D are the activities. Their normal and crash durations and associated costs aregiven in the table below:

Activity Normal duration in

days

Normal cost Rs.

Crash duration in days

Crash cost Rs.

A B C D

8 4 10 6

6,000 2,000 4,000 4,000

4 2 4 4

12,000 14,000 8,000 8,000

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For the entire project the indirect cost is Rs. 1000 per day. A and B are starting activities;C follows B; D follows A and C; D is the finishing activity. Draw CPM Network. Calculatepoints for PTC graph and plot the same. Determine the optimum cost and optimum durationfor the project. PTC is Project-Time-Cost-Trade -Off graph. [20 Marks]

Sol. 7. (c) Network Diagram

1

2

3 4T = 0T = 0

E

L

T = 4T = 4

E

L

T = 20T = 20

E

L

T = 14T = 14

E

L

A D

BC

4(2)

10(4

)

8(4) 6(4)

Critical path = 1 – 2 – 3 – 4

Cost slope

Activity Normal Crash Cost Slope

time cost (Rs.) time C t C/t

A 8 6000 4 12000

B 4 2000 2 14000

C 10 4000 4 8000

D 6 4000 4 8000

cost (Rs.)

Normal Cost at normal project durationNormal project duration = 20 daysDirect cost = 6000 + 2000 + 4000 + 4000 = Rs. 16000Indirect cost = 20 × 1000 = Rs. 20000Total cost = 16000 + 20000 = Rs. 36000

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First stage crashingWe can observe that among critical activities, activity C hasminimum cost slope i.e. 666.67 and has crashing potential of 6days. It can be crashed by 6 day without affecting other parallelactivities.New project duration = 14 days.Direct cost = 16000 + 6 × 666.67 = Rs. 20000Indirect cost = 14 × 1000 = Rs. 14000Total cost = 20000 + 14000 = Rs. 34000

1

2

3 4T = 0T = 0

E

L

T = 4T = 4

E

L

T = 14T = 14

E

LT = 8T = 8

E

L

A D

B C4(4)

8(4) 6(4)

4(2)

Second stage crashingWe can observe that, now we have 2 critical path A –D and B–C–D. Therefore we have to checkvarious alternatives of combinations of cost slope(i) C/S of B + C/S of A = 6000 + 1500 = 7500(ii) C/S of D = 2000 Cost slope of activity D is minimum, therefore it can be crashed for its complete crashing

potential that is 2 days.New project duration = 12 days.Direct cost = 20000 + 2 × 2000 = Rs. 24000Indirect cost = 12 × 1000 = Rs. 12000Total cost = 24000 + 12000 = Rs. 36000

1

2

3 4T = 0T = 0

E

L

T = 4T = 4

E

L

D

B C4(4)

8(4) 4(4)

4(2)T = 8T = 8

E

L T = 12T = 12

E

L

Third Stage CrashingNow we can crash activity A and B simultaneously by 2 days. As these are parallel activitiesand crashing potential of activity B gets expired.New project duration = 10 days.Direct cost = 24000 + 6000 × 2 = Rs. 36000Indirect cost = 10 × 1000 = Rs. 10000Total cost = 36000 + 10000 = Rs. 46000

1

2

3 4T = 0T = 0

E

L

T = 2T = 2

E

L

T = 6T = 6

E

L

T = 10T = 10

E

L

D

B C4(4)

6(4) 4(4)

2(2)

A

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Total Cost Curve:

010 11 12 13 14 15 16 17 18 19 20

34000

36000

38000

40000

42000

44000

46000

48000

50000

Time (days)

Cost

Optimum project duration = 14 days

Optimum project cost = Rs. 34000

7. (d) Determine the output of a bulldozer for the following situations;

(a) Material handled sandy loam top soil having swell = 25%(b) Haul distance = 30 meters(c) Rated mould board capacity = 3 cum loose volume(d) Actual operating time per hour = 45 minutes(e) Forward speed 2.4 km per hour(f) Reverse speed 6.0 km per hour. [10 Marks]

Sol. 7.(d)

If D = Haul distance (m), R = Reverse Speed (km/h), F = Forward speed (km/h), G = Timerequired for gear shifting (minutes).

Time required per trip in minutes (or cycle time) = D D GF R

=30 30 0.32.4 1000 6.0 100060 60

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= 0.75 + 0.30 + 0.30 = 1.35 minutes

Output of bulldozer =3 45 cum / hr

1.25 1.35

= 80 cum/hr.

8. (a) (i) Determine the factor of safety with respect to cohesion only for a submergedembankment 25 m high whose upstream face has an inclination of 45°. The soil has

the following properties; c = 40 kN/m2, = 10°, sat = 18 kN/m3. The relevant

stability number is equal to 0.108.(ii) What is the factor of safety if the embankment experiences the effect of sudden

drawdown? For m = 4.5° and i = 45°, the value to the stability number is 0.136.

[10 Marks]8. (a) (i)

Using submerged unit weight,

Sn =c

CF ' H

' = 3sat w 18 9.81 8.19 kN/m

0.108 = C

40F 8.19 25

Fc =40 1.81

0.108 8.19 25

(ii)

wtan =sat

' 1 tan 'F

First trial, taking F = 1, w

8.19tan tan10º18

w = 4.59º

For m = 4.5º and i = 45º, sn = 0.136

0.136 = cc

40 F 0.653F 18 25

Fc = 0.653 < 1 the slope in not safe

8. (b) A sample of dry soil having specific gravity of 2.74 and having a mass of 133.7 gm isuniformly dispersed in water to form 1000 cc of suspension.

(i) Determine the density of suspension immediately after it is prepared.(ii) A 10 cc of the suspension was removed from the depth of 21 cm beneath the top

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surface after the suspension was allowed to stand for 2 min 30 sec. The dry mass ofthe soil in the sample drawn was found to be 0.406 gm. Determine one point on thegrain-size distribution curve corresponding to this observation.

Temperature of suspension = 20°CViscosity of water at 20°C = 0.0102 poise. [15 Marks]

Sol. 8. (b)G = 2.74

Mass of dry soil = 133.7 gmvolume of suspension = 1000 cm3

Volume of soil = 3133.7 48.8 cm2.74 1

Volume of water = 1000 – 48.8= 951.2 cm3

Mass of water = 951.2 gm

density of suspension =mass of soil + mass of water

total volume = 133.7 951.2

1000

= 1.0 849 gm/cc = 1084.9 kg/m3

(b) Volume of sample = 10 cm3

Hs = 21 cm = 0.21 mt = 2 min 30 sec = 150 sec.µ = 0.0102 poise

= 0.0102 × 10–4 KN-s/m2 = 1.02 × 10–6 KN-s/m2

we know,

Hst =

2

sd

18µ

0.21150

=2

6d

18 1.02 10 [2.74 – 1] 9.81

d = 3.88 × 10–5 md = 0.0388 mm

% of particles finer than d,dry mass of soil in the sample = 0.406 gm

% finer than size 0.0388 =concentration of sample collected at time t from height Hs

Concentration of soil from original soil suspension

=

0.40610 100

133.71000

= 30.36%

IES M

ASTER




8. (c) For the given A-O-A network shown in fig, draw A-O-N network and find.(i) EST, EFT, LST & LFT of each activity(ii) TF of each activity(iii) Critical path and project duration [20 Marks]

B

C

D

I

G

H3

7

5

9

5

10

5J

A5

E5

F3

Sol. 8.(c)

A-O-N Network Diagram

A5

C7

B3

E5

D5 H10

I9 F0

F3

G5 J5

00

55

56

89

810

1315

89

1314

1314

2324

2424

2424

1515

1719

1719

2224

1214

1212

1212

1515

55

Calculation of EST, EFT, LST, LFT and FT are done in table below:

Activity tij EST EFT LST LFT TF RemarksA 5 0 5 0 5 0 CriticalB 3 5 8 6 9 1C 7 5 12 5 12 0 CriticalD 5 8 13 9 14 1E 5 8 13 10 15 2F 3 12 15 12 15 0 CriticalG 5 12 17 14 19 2H 10 13 23 14 24 1I 9 15 24 15 24 0 CriticalJ 5 17 22 19 22 2

Fo 0 24 24 24 24 –

Critical path along A – C – F – I.

The project duration is 24 units.

IES M

ASTER




8. (d) Define batching of materials. How are batchings of cement and aggregate done?[15 Marks]

Sol. 8.(d) Batching Materials Batching is the process of measurement of cement, fine aggregate (sand), coarse aggregates

(stone chips) and water for each batch of concrete mixing. It is essential that quantity of each material that goes into a batch must be exact so as to

produce concrete for the desired properties. Batching can be done in following ways.

(i) By volume(ii) By weight

IS-code recommends batching of materials by weight only, as weight batching is quite reliableand dependable in producing the concrete of uniform concrete.

Batching of cement

Generally cement is measured by counting the number of bags because each has a fixed quantity(50 kg).

When the size of the project is large enough to justify the installation of facilities to handle bulkcement, it is usually stored in a suitable silo or fully enclosed bin

A weighing hopper is generally suspended beneath the storage bin to measure the correct cementamount.

Weighing is facilitated by means of a beam scale or springless dial scale, the latter being moreexpensive but more dependable.

Batching of aggregates Batches of fine and coarse aggregates are based on one bag of cement or its multiples unless

bulk cement is used and weighed separately. This type of batching no correction is needed to allow for the bulking of sand, but an allowance

should be made for the weight of water contained in the wet aggregates. For smaller jobs, weigh batching may be done by(i) Simple spring balances(ii) Platform weighing machines(iii) Automatic weighing machines. For very large project, large weigh batching plants have automatic weighing equipment. Water-cement ratio is the most important parameter to control the strength of concrete. Lower

is the W/C ratio, greater is the strength. Therefore quantity of water as determined by concrete-mix design must be added accurately to

get the desired strength without affecting the workability. Batching of water is done either by weight or by volume, both of them gives same results. As it is easier to measure the volume of water, it is done with the help of calibrated tank and

cylinders.

These tanks are designed to discharge a predetermined volume of water into mixers.

Documents

IES MASTER · R (4) Conventional Class Test-1 (CPM + Soil) 28th Feb 2018 Mob. : E-mail: 8010009955, 9711853908 [email protected], [email protected] Regd. office : Phone : F-126,