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ESE 2020 Preliminary Examination

Detailed Solutions of

Civil Engineering (Set-A)

Click here for detailed solutions

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ESE 2020 Preliminary Examination Civil Engineering | Set-A

Expected Cutoff of ESE 2020 Prelims (Out of 500 Marks) Branch

Gen

OBC

SC

Actual Cutoff of ESE 2019 Prelims (Out of 500 Marks)

ST

Branch

Gen

OBC

SC

ST

CE

210-220 205-215 170-180 170-180

CE

188

185

143

159

ME

245-255 245-255 210-220 210-220

ME

187

187

166

169

EE

225-235 215-225 195-205 195-205

EE

221

211

191

172

E&T

235-245 225-235 185-195 185-195

E&T

226

221

176

165

Civil Engineering Paper Analysis : ESE 2020 Prelims Exam Sl.

Subjects

No. of Qs.

1

Building Materials

15

2

Strength of Materials

18

3

Structural Analysis

06

4

Design of Steel Structures

13

5

RCC & Prestress Concrete

10

6

Construction Practice, Planning & Management

12

7

Fluid & Hydraulic Machines + OCF

12

8

Engineering Hydrology

02

9

Irrigation Engineering

08

10

Environmental Engineering

13

11

Geo-technical Engineering

14

12

Surveying and Geology

13

13

Transportation Engineering

06

14

Railway Engineering

08

UPSC ESE/IES Prelims 2020 Civil Engineering Analysis & Expected Cut-off by MADE EASY Faculties

https://youtu.be/YKX-lrNxvZE

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Page 2

ESE 2020 Preliminary Examination Civil Engineering | Set-A 1.

When the deposit of efflorescence is more than 10% but less than 50% of the exposed area of the brick, the presence of efflorescence is (a) Moderate (b) Slight (c) Heavy (d) Serious

Ans.

(a) NIL negligible SLIGHT ≤ 10% MODERATE 10 – 50% HEAVY/HIGH > 50% SERIOUS > 50% and accompanied by powdering or flacking of exposed surface. End of Solution

2.

Mohs scale is used for stones to determine (a) Flakiness index (b) Durability (c) Strength (d) Hardness

Ans.

(d) End of Solution

3.

Which of the following conditions are recommended for using sulphate resisting cement? 1. Concrete to be used in foundation and basement, where soil is not infested with sulphates. 2. Concrete used for fabrication of pipes which are likely to be buried in marshy region or sulphate bearing soils. 3. Concrete to be used in the construction of sewage treatment works. (a) 2 and 3 only (b) 1 and 2 only (c) 1 and 3 only (d) 1, 2 and 3

Ans.

(a) End of Solution

4.

Which one of the following cements is a deliquescent? (a) Quick setting Portland cement (b) White and Coloured cement (c) Calcium Chloride cement (d) Water Repellent cement

Ans.

(c) End of Solution

5.

Consider the following data for concrete with mild exposure Water-cement ratio = 0.50 Water = 191.6 litre The required cement content will be (a) 561 kg/m3 (b) 472 kg/m3 3 (c) 383 kg/m (d) 294 kg/m3

Ans.

(c)

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Page 3

ESE 2020 Preliminary Examination Civil Engineering | Set-A W (By mass) = 0.5 C W 0.5 Volume of water = 191.6 litre = 0.1916 m3 Mass of water = 0.1916 × 103 = 191.6 kg

C=

C=

191.6 = 383.2 kg 0.5 End of Solution

6.

The strength of a fully matured concrete sample is 500 kg/cm2. When cured at an average temperature of 20°C in day, 10°C in night, datum temperature T0 is –11°C. If Plowman constants A is 32 and B is 54, the strength of identical concrete at 7 days will be nearly (a) 333 kg/cm2 (b) 312 kg/cm2 (c) 272 kg/cm2 (d) 243 kg/cm2

Ans.

(a)

M = (20 – (–11)) × 12 × 7 + (10 – (–11)) × 12 × 7 = 2604 + 1764 = 4368°C – Hrs Strength, f = a + b log10 (M × 10–3) = 32 + 54 log10 (4368 × 10–3) = 66.57% Strength of concrete at 7 days = 0.6657 × 500 = 332.85 kg/cm2 = 333 kg/cm2 End of Solution

7.

A sample of concrete is prepared by using 500 g of cement with water cement ratio of 0.55 and 240 N/mm2 intrinsic strength of gel. The theoretical strength of concrete on full hydration will be nearly (a) 148 N/mm2 (b) 126 N/mm2 (c) 104 N/mm2 (d) 82 N/mm2

Ans.

(c) Weight of cement = 500 gm

W = 0.55 C Weight of water = 500 × 0.55 = 275 gm Volume of water = 275 ml 0.657C 0.657 ×500 328.5 = = 0.756 = 0.319 C + W 0.319 × 500 + 275 434.5 Theoretical strength of concrete = 240 (0.756)3 = 103.71 N/mm2

Gel space ratio =

End of Solution

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Page 4

ESE 2020 Preliminary Examination Civil Engineering | Set-A 8.

The cement and water slurry coming on the top and setting on the surface is called (a) Crazing (b) Efflorescence (c) Sulphate deterioration (d) Laitance

Ans.

(d) End of Solution

9.

Polymer concrete is most suitable for (a) Sewage disposal works (b) Mass concreting works (c) Insulating exterior walls of an air-conditioned building (d) Road repair works

Ans.

(a) End of Solution

10.

Which one of the following limes will be used for finishing coat in plastering and white washing? (a) Semi Hydraulic lime (b) Kankar lime (c) Magnesium/Dolomitic lime (d) Eminently Hydraulic lime

Ans.

(c) End of Solution

11.

Which one of the following light weight element will be added to enhance the protective properties for X-ray shielding mortars? (a) Sodium (b) Potassium (c) Lithium (d) Calcium

Ans.

(c) End of Solution

12.

Which one of the following stone is produced by moulding a mixture of iron slag and Portland cement? (a) Imperial stone (b) Garlic stone (c) Ransom stone (d) Victoria stone

Ans.

(b) End of Solution

13.

When a round bar material with diameter of 37.5 mm, length of 2.4 m, Young’s modulus of 110 GN/m2 and shear modulus of 42 GN/m2 is stretched for 2.5 mm, its Bulk modulus will be nearly (a) 104 GN/m2 (b) 96 GN/m2 (c) 84 GN/m2 (d) 76 GN/m2

Ans.

(b)

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Page 5

ESE 2020 Preliminary Examination Civil Engineering | Set-A 9KG 3K + G 3 KE + EG = 9 KG 9 KG – 3 KE = EG

E=

K=

EG 110 × 42 = = 96.25 GN/m2 9G − 3E 9 × 42 − 3 × 110 End of Solution

14.

A punch of 20 mm diameter is used to punch a hole in 8 mm thick plate. If the force required to create a hole is 110 kN, the average shear stress in the plate will be nearly (a) 410 MPa (b) 320 MPa (c) 220 MPa (d) 140 MPa

Ans.

(c) Shear stress, τ = =

P P = A πdt

110 × 103 = 219 MPa ≃ 220 MPa π × 20 × 8 End of Solution

15.

A reinforced concrete circular section of 50,000 mm2 cross-sectional area carries 6 reinforcing bars whose total area is 500 mm2. If the concrete is not to be stressed more than 3.5 MPa and modular ratio for steel and concrete is 18, the safe load the column can carry will be nearly (a) 225 kN (b) 205 kN (c) 180 kN (d) 160 kN

Ans.

(b) Asc Ac

Ag = Asc = Ac = P=

50000 mm2 500 mm2 50000 – 500 = 49500 mm2 Ac · σcc + Asc · σsc

= Ac ·σcc + Asc ·m·σcc = 49500 × 3.5 + 500 × 18 × 3.5 = 204750 N = 204.75 kN = 205 kN (say) End of Solution

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Page 6

ESE 2020 Preliminary Examination Civil Engineering | Set-A 16.

The strain energy U stored due to bending of the cantilever beam due to point load at the free end will be (a)

W 2l 3 6EI

(b)

W 2l 2 6E I

(c)

W 3l 3 36EI

(d)

W 2l 3 36EI

where :

Ans.

W = Concentrated load l = Length of a cantilever EI = Flexural rigidity

(a) W L

(M x ) 2 d x = L W 2 x2d x = W 2L3 ∫ 2 EI ∫ 2 EI 6 EI L

U=

0

0

End of Solution

17.

A steel bar 2 m long, 20 mm wide and 15 mm thick is subjected to a tensile load of 30 kN. If Poisson’s ratio is 0.25 and Young’s modulus is 200 GPa, an increase in volume will be (a) 160 mm3 (b) 150 mm3 3 (c) 140 mm (d) 130 mm3

Ans.

(b) 3

30× 10 P = 100 MPa = 20 × 15 A σ Volumetric strain, ε v = (1 − 2 µ) E 100 ΔV = (1 − 0.5) × 20 × 15 × 2000 200 × 103 = 150 mm3 Axial stress, σ =

∴

End of Solution

18.

A bolt is under an axial thrust of 9.6 kN together with a transverse force of 4.8 kN. If factor of safety is 3, yield strength of bolt material is 270 N/mm2 and Poisson’s ratio is 0.3, its diameter as per maximum principal stress theory will be nearly (a) 13 mm (b) 15 mm (c) 17 mm (d) 19 mm

Ans.

(a) Axial stress, σ =

9.6 × 103 12.23 × 103 N/mm2 = 2 π 2 d d 4

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Page 7

ESE 2020 Preliminary Examination Civil Engineering | Set-A

Shear stress, τ =

4.8 × 103 6.115 × 103 N/mm2 = 2 π 2 d d 4

Maximum principal stress, σ1 =

σ

2

2

+

⎛σ ⎞ 2 ⎜⎝ 2 ⎟⎠ + τ 2

= =

6115 ⎛ 6115⎞ ⎛ 6115⎞ + ⎜ +⎜ ⎝ d2 ⎟⎠ ⎝ d2 ⎟⎠ d2 1

d2

2

[6115 + 8647.916]

14762.9

N/mm2 d2 According to maximum principal stress theorem, =

σ1 ≤ 14762.9

d2 ∴

σy

FOS

270 ≤ 3

d ≥ 12.81 mm End of Solution

19.

In a material the principal stresses are 60 MN/m2, 48 MN/m2 and –36 MN/m2. When 1 the values of E = 200 GN/m2 and = 0.3, the total strain energy per unit volume will m be nearly (a) 43.5 kNm/m3 (b) 35.5 kNm/m3 3 (c) 27.5 kNm/m (d) 19.5 kNm/m3

Ans.

(d) Strain energy per unit volume =

1 2 [σ + σ22 + σ32 − 2µ (σ1σ2 + σ2 σ3 + σ3 σ1 )] 2E 1

=

1 2 2 2 12 [60 + 48 + 36 − 0.6(60 × 48 − 48 × 36 − 60 × 36)] ×10 2 × 200 × 109

=

1012 [7200 − 0.6(2880 − 1728 − 2160)] 11 4 × 10

10 (7200 + 604.8) 4 = 19.51 kNm/m3

=

End of Solution

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Page 8

ESE 2020 Preliminary Examination Civil Engineering | Set-A 20.

Ans.

At a point in a two dimensional stress system, the normal stress on two mutually perpendicular planes are σxx and σ yy and shear stress is τx y. One of the principal stresses will become zero when the value of shear stress τx y is (a) ±(σxxσyy)

(b) ± σ xx − σ yy

(c) ± σ xx + σyy

(d) ± σ xx σ yy

(d) σ1 = ∴

σ xx + σyy 2

2

⎛ σ − σyy ⎞ 2 + ⎜ xx ⎟ + τ xy = +ve ≠ 0 2 ⎝ ⎠

σ2 = 0

σ xx + σ yy 2

2

⎛ σ xx − σ yy ⎞ 2 − ⎜ ⎟ + τ xy = 0 2 ⎝ ⎠ τx2y = σxx ⋅ σyy

τ xy = ± σ xx ⋅ σyy

∴

End of Solution

21.

The deflection δ of the closed coil helical spring is (a) (c)

WR 2n 8Cd3

(b)

128WR 3n

(d)

64WR 3n

Cd 4 64WR 2n Cd 2

Cd 2 where : W is the axial load R is the radius of the coil n is the number of turns of coil C is the modulus of rigidity d is the diameter of the wire of the coil Ans.

(b) Strain energy in spring =

=

=

T 2L 2GJ P 2R 2 (2 πRn ) π 4 d 2G 32 32P 2 R 3 n

Gd

4

By Castigliano’s theorem, δ=

∂U ∂ 32P 2R 3n = = ∂P ∂P Gd 4

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Page 9

ESE 2020 Preliminary Examination Civil Engineering | Set-A 64PR 3n Gd 4 G = C = Modulus of rigidity

δ=

End of Solution

22.

A closely-coiled helical spring of round steel wire 5 mm in diameter having 12 complete coils of 50 mm mean diameter is subjected to an axial load of 100 N. If modulus of rigidity is 80 GPa, the deflection of the spring will be (a) 36 mm (b) 32 mm (c) 28 mm (d) 24 mm

Ans.

(d) Δ=

64PR 3n 4

Gd Δ = 24 mm

=

64 × 100 × 25 3 × 12 80 × 10 3 × 54 End of Solution

23.

A hollow shaft of external and internal diameters as 100 mm and 40 mm respectively is transmitting power at 120 rpm. If the shearing stress is not to exceed 50 MPa, the power the shaft can transmit will be (a) 100 kW (b) 120 kW (c) 140 kW (d) 160 kW

Ans.

(b) τmax =

50 =

16T ⎡ ⎛ d ⎞4 ⎤ 3 ⎢ πd0 1 − ⎜ i ⎟ ⎥ ⎢ ⎝ do ⎠ ⎥ ⎣ ⎦

16T ⎡ 40 ⎞ 4 ⎤ π × 100 3 ⎢1 − ⎛⎜ ⎥ ⎝ 100⎟⎠ ⎥ ⎣⎢ ⎦

T = 9.56 × 106 Nmm ∴

P= =

2π NT 60

2π × 120× 9.56× 103 60

= 120 kW End of Solution

24.

A circular beam of 100 mm diameter is subjected to a shear force of 30 kN. The maximum shear stress will be nearly (a) 5.1 MPa (b) 6.3 MPa (c) 7.5 MPa (d) 8.7 MPa

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Page 10

ESE 2020 Preliminary Examination Civil Engineering | Set-A Ans.

(a) τmax =

4 30 ×10 3 4 F = × = 5.08 MPa ⋅ 3 π 3 A (100) 2 4 End of Solution

25.

A cantilever beam AB as shown in figure is subjected to a point load of 12 kN over a span of 6 m with E = 2 × 105 N/mm2 and I XX = 6 × 107 mm4. The deflection at the free end will be W A

B l

(a) 80 mm (c) 64 mm Ans.

(b) 72 mm (d) 56 mm

(b) Δ=

3 12 × 10 3 × 6000 3 PL = = 72 mm 3E I 3 × 2 × 10 5 × 6 × 10 7

End of Solution

26.

A floor has to carry a load of 12 kN/m2. The floor is supported on rectangular joists each 100 mm wide, 300 mm deep and 5 m long. If maximum stress in the joists should not exceed 8 MN/m2, the centre to centre distance of joists will be (a) 430 mm (b) 400 mm (c) 360 mm (d) 320 mm

Ans.

(d) w/ m l

Let h is the distance between two joists. ∴

Now

12 × l × h =w l w = 12h kN/m

σmax = 8=

8=

M max z

wl 2 bd 2 8× 6 6w l 2 8bd 2

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Page 11

ESE 2020 Preliminary Examination Civil Engineering | Set-A 64 × 100 × 300 2 =w 6 × (5000) 2 ∴

w= 12h = h= =

3.84 N/mm 3.84 0.32 m 320 mm End of Solution

27.

A simply supported wooden beam 100 mm wide, 250 mm deep and 3 m long is carrying a uniformly distributed load of 40 kN/m. The maximum shear stress will be (a) 2.4 MPa (b) 2.8 MPa (c) 3.2 MPa (d) 3.6 MPa

Ans.

(d)

S=

Maximum S.F.,

wL 40 × 3 = 60 kN = 2 2

τmax = 1.5 τavg = 1.5×

60× 103 = 3.6 MPa 100× 250 End of Solution

28.

A simply supported beam of span 8 m carries a uniformly distributed load of 24 kN/ m run over the whole span. The beam is propped at the middle of the span. The values of E = 200 × 106 kN/m2 and I = 20 × 10–5 m4. The amount by which the prop should yield in order to make all three reactions equal will be nearly (a) 20 mm (b) 15 mm (c) 10 mm (d) 5 mm

Ans.

(b) w/m

B

A

C

R

R

R

3R = 24 × 8 R = 64 kN ΔB = =

5 wL4 RL3 − 384 EI 48EI 1 EI

⎡ 5 64 × 8 3 ⎤ × 24 × 8 4 − ⎢ ⎥ 48 ⎦ ⎣384

=

1

200 × 109 × 20 × 10 −5 = 14.9 mm

[1280 − 682.666]

End of Solution

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Page 12

ESE 2020 Preliminary Examination Civil Engineering | Set-A 29.

A cantilever beam ACB has end A fixed and subjected to a point load P at free end B. The point C is mid-point of AB and the moment of inertia of AC is twice that of CB. the deflection at the free end will be P

A

Ans.

2I

I

l 2

l 2

(a)

Pl3 3 EI

(b)

Pl 3 48EI

(c)

5Pl 3 96EI

(d)

9Pl 3 48EI

B

(d) P A

C

P

B Δc θc

L θ c· 2

A

PL 2 C

Δ′B

2 PL ⎛ L ⎞ ⎛ L⎞ ·⎜ ⎟ P ⎜ ⎟ ⎝ ⎠ ⎝ 2⎠ θC = 2 2 + 2E I 4E I

=

PL2 PL2 3PL2 + = 8EI 16EI 16EI

PL ⎛ L ⎞ 2 L 3 P ⎛⎜ ⎞⎟ ·⎜ ⎟ ⎝ 2⎠ 2 ⎝ 2⎠ ΔC = + 4 EI 6 EI = ∴

5PL3 PL3 PL3 + = 32EI 48EI 96EI

ΔB = ΔC + θC ·

L + Δ′B 2

=

5PL3 3PL3 PL3 + + 96EI 32EI 24EI

=

9PL3 48EI End of Solution

30.

A beam of uniform cross-section simply supported at ends carries a concentrated load W at midspan. If the ends of the beam are fixed and only load P is applied at the midspan such that the deflection at the centre remains the same, the value of the load P will be

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Page 13

ESE 2020 Preliminary Examination Civil Engineering | Set-A (a) 6 W (c) 2 W Ans.

(b) 4 W (d) W

(...