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Answer Key

1. (d) 2. (d) 3. (b) 4. (b) 5. (d) 6. (a)

7. (b) 8. (d) 9. (b) 10. (c) 11. (a) 12. (b)

13. (d) 14. (c) 15. (c) 16. (c) 17. (a) 18. (a)

19. (b) 20. (d) 21. (a) 22. (c) 23. (d) 24. (c)

25. (b) 26. (b) 27. (a) 28. (c) 29. (c) 30. (a)

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1. (d)

Explanation:

The reaction provided from the water will be udl.

Therefore the situation can be analysed as:

Let the reaction has w udl intensity, then

W + W = wL

w =

Therefore, Bending Moment at center

= W ×

– w ×

×

=

-

×

=

-

= 0.

2. (d)

Explanation:

Strain Energy stored per unit volume, U =

As σ =

U =

= U P2

U = kP2 (k = Constant) Therefore, U1 = kp1

2 and U2 = kP22

Therefore, U = k(P1 + P2)2 U = kP1

2 + kP22 + 2kP1P2

U = U1 + U2 + 2kP1P2 U – 2kP1P2 = U1 + U2

⇒ U > U1 + U2.

3. (b)

Explanation:

Major Principal Stress,

=

√( )

Hence, σx = 8 MPa, σy = 6MPa and

τxy =4 MPa

=

√( )

= 7 √

= 7 √ = 11.123 MPa.

4. (b)

Explanation:

Bending stresses are given by

=

σ =

× y

Now, σ at N.A = 0 and σ will be maximum on the extreme fibre of shaded region.

Therefore, σmax =

× ymax

=

= 6 N/mm2

Therefore, σavg =

=

= 3 N/mm2

Force = σavg × Area = 3×40×80 = 9.6 kN.

5. (d)

Explanation:

The individual free body diagrams are:

l/4 l/2 l/4

W W

0.8 m

50 kN 50 kN

1.2 m

35 kN 35 kN

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Then the total elongation = Δ1 + Δ2 + Δ3

=

+

+

=

[

] 106 mm

=

= 1.27 mm.

6. (a) Explanation:

Material A is more brittle than material B, as

the strain or deformation observed in material B is more for same stress. But the ultimate

strength of A is more than that of B, as it can bear more stress.

7. (b)

Explanation:

Cut the frame in such a way that each part of

the frame looks like a stable cantilever frame. DS = 3 × (Number of cuts required for making stable cantilever) – (Number of reactions

required for making stable cantilever).

Number of cuts required = 4 So, DS = 3 × 4 – 0 = 12.

8. (d)

Explanation:

The ILD of force in member U1 L1 is, when

load is exactly at L1. U1 L1 will be having zero force because all the force will be taken by the support.

If the applied unit load is moving from L2 to L6. We can consider the equilibrium of the left

part.

So, the ILD of force in U1 L1 is same as that of

the vertical reactions V1

So, the full ILD of the member U1 L1.

So, ordinate at L2 = 0.8.

1.0 m

45 kN 45 kN

1

2

3

4

20 m 5 m

V2 V1

FV = 0

1

V1

FU L

FU L = v1

=

= 0.8

FU L



9. (b) Explanation:

Number of cuts required = 4

Number of reactions required = 3 Number of additional equations because of hinges = 2 (at A) + 3 (at B) + 1 (at C)

= 6

Since two members joined at a point by an internal hinge produce one additional

equation, similarly three members joined at a point by internal hinge produce two additional

equations. Also four members at a joint by an internal hinge produce three additional equations.

Thus, DS = 3 × 4 – 3 – 6 = 3.

10. (c) Explanation:

ILD for B.M. at a distance of 4 m from left

support is:

Now, maximum B.M. will be, when load is

distributed in such a way on both side of the

cross section that W

= W

(l1 – x)

⇒

(4 – x) =

(16 – 5 + x)

⇒

=

⇒ 12 – 3x = 2 + 2x

⇒ 10 = 5x

⇒ x = 2 m. So, max value of B.M. will be when:

Total area = Area of ABCD + Area of CDEF

= ½ [2.4 + 1.2] × 2 + ½ [2.4 + 1.2] × 3

= 3.6 + 5.4

= 9.0 m2. Thus, B.M. = 10 kN/m × 9 m2 = 90 kNm.

11. (a)

Explanation:

At joint B, two non – collinear member meet. So, force in each member should be zero. So, FAB = 0. Also, FCB = 0.

12. (b)

Explanation:

At joint D, member BD and CD is collinear. So, force in member AD should be zero.

A

B

C

b a

= 2.4

l1

=

x 6 m

(5-x)

4 m

F

A

B

C

E

1.2

2.4

6 m

4 m

2 m 3 m



13. (d)

Explanation:

For calculating depth of neutral axis xu, we will equate:

Total compressive force = Total Tensile Force. Area of stress diagram × width = Tensile Force.

*

+ × 0.67 fck × 300 = 0.87 fy Ast

*

+ × 0.67 fck × 300 = 0.87 × fy × 4 ×

× 20

2

xu × 0.67 × 20 × 300 = 0.87 × 415 × 4 ×

× 20

2

xu = 158 mm.

14. (c)

Explanation:

Calculation of xulim: xulim = 0.48 × 450 = 216 mm

Calculation of xu: 0.36 × fck b xu = 0.87 fy Ast

0.36 × 25 × 230 × xu = 0.87 × 415 × 5 ×

× 16

2

Xu = 175.346 mm

Since xu < xulim, the beam is under reinforced.

Mu = 0.36 fck b Xu (d – 0.42 Xu) × 10-6 Mu = 136.60 kNm.

15. (c)

Explanation:

Water content (w) =

w =

=

For soil A,

w =

= 0.8

= 0.8

---------- (I)

+

= 1 ---------- (II)

From equation (I) and (II)

= 0.56 kg, = 0.44 kg

For soil B,

w =

= 0.6

= 0.6

---------- (III)

+ = 1 ---------- (IV)

From equation (I) and (II)

= 0.625kg, = 0.375kg

For mixed soil, Mw = 0.44 + 0.375 = 0.815 kg

Md = 0.56 + 0.625 = 1.185 kg

Therefore, w =

= 0.6877 ≈ 68.8%

16. (c)

Explanation:

Void ratio, e =

For soil X,

0.4 =

= 0.4

+ = 1.4 (Given)

Therefore, = 1m3, = 0.4m3

For soil Y,

w = 0.8 =

Stress

Variation

Strain

Variation

300 mm

Xu

Xu

0.67 fck

0.0020

0.0035

500 mm

4 bars of diameter 20 mm




= 0.8

+ = 1.8 (Given)

Therefore, = 1 m3, = 0.8 m3

For soil Z,

e =

=

= 0.6

n =

= 0.375

17. (a)

Explanation:

From figure, Total stress at x-x = (6 × 9.81) + (6 × 22)

= 190.86 kN/m2

Pore water pressure at x-x = 9.81 × 12 = 117.72 kN/m2

Effective stress at x-x = 190.86 – 117.72 = 73.14 kN/m2.

18. (a)

Explanation:

Present Population = 25000

Design Population = 45000 Increase in Population = 45000-25000 = 20000

Increase per year =

= 1000

Now, increase in discharge required to reach design discharge = 5800 – 4200 = 1600 m3/d. At present per head water consumption is =

= 0.168 m3/d

Increase in water consumption = 0.168 × 1000

= 168 m3/d Therefore, number of years required to reach

design discharge =

= 9.5 years.

19. (b)

Explanation:

Treatment of water volume = 4.2 million litres

= 4.2×106×10-3 m3 = 4.2×103 m3

Treatment discharge =

m3/s

= 0.04861 m3/s

Settling velocity, Vs = 0.27 mm/s = 0.27×10-3 m3/s

Surface area of tank required =

m2

= 180 m2.

20. (d)

Explanation:

Given speed V = 60 kmph =

m/s

Coefficient of friction, f = 0.35

Braking distance =

= (

)

= 40.45 m.

21. (a)

Explanation:

For A, RF =

1 cm on map = 5000 cm on ground.

For D, RF =

Decreasing order of scale,

=

>

>

>

.

X X

6m

6m Sand Bed (γsat = 22 kN/m3)

River (γw = 9.81 kN/m3)



22. (c)

Explanation Shrunk Scale (SS) = Shrinkage Factor (SF) ×

Original Scale (OS).

Now, SF =

=

L

SS = 9.6 =

L × 10

(On map) OL =

= 10.41 cm

OL on ground = 10.41 × 150 = 1561.5 cm

(Since 1 cm = 150 cm scale is given)

OL (ground) = 15.615 m.

23. (d)

Explanation:

SS = SF × OS =

×

(Since SF =

L)

=

SS =

.

24. (c)

Explanation: PB – 0.7 × 0.88 × 9.81 + 0.2 × 0.88 × 9.81 + 0.8 × 13.6 × 9.81 – 0.4 × 13.6 × 9.81 – 0.3 × 9.81 = PA

PA – PB = 46.107 kPa ≈ 46.11 kPa.

25. (a)

Explanation:

Equation of streamline is given by:

=

=

ln x = - ln y + ln c, where c is constant. xy = c

y =

At x = 3, y = 1

1 =

c = 3 Therefore, xy = c

xy = 3.

26. (b)

Explanation

ax =

+

+

= u + v + w u = 2y2, v = 3x2, w = 0

ax = 2y2 × 0 + 3x2 × 4y ax = 12x2y

Acceleration at (x, y) (1, 1)

ax = 12 (1) (1) = 12 m/s2.

27. (a)

Explanation CD = CV × CC

CC =

= Area of Vena – Contracta.

= Area of Orifice.

CC = (

)

CD = (

)

× 0.97 = 0.35.

Result of Scholarship Test (2nd Year PEC Civil Engg.) held on 22.03.2018

S. No. Name Father’s Name Marks Rank Scholarship

1 Divyanshu Gupta Mukesh Gupta 14 1 100 %

2 Hardik Gupta Anil Gupta 14 1 100 %

3 Priti Kumari Santosh Kumar 13 3 100%

4 Pawan Kumar Deshwal

Subhash Chand 8 4 100 %

5 Ankit Kaushal Jagde Kaushal 7 5 75%

6 Piyush Jadon Sarman Singh 6 6 75 %

7 Abhishek Kumar Satish Kumar 6 6 75 %

8 Nancy Jaswinder Singh 4 8 75 %

9 Parth Goyal Bhupinder Goyal 1 9 60 %

10 Shivom Sharma Dharam Sharma -1 10 NIL

11 Kriti Salwan Dinesh Salwan -13 11 NIL

12 Priyanka Aamrjit Pal -14 12 NIL

13 Suvidha Nayyar Kapil Nayyar -14 12 NIL

14 Shubham Sharma Umesh Kumar -16 14 NIL

15 Ashutosh Sharma N.N. Sharma -18 15 NIL

16 Ridham Sharma Ashwani Kumar Bhardwaj -18 15 NIL

17 Vishwas Parmar Amit Pal Singh -20 17 NIL

18 Jyoti Singh Bhanu Pratap Singh -24 18 NIL

19 Anubhav Ojha Prem Ojha -28 19 NIL

20 Rajiv Kumar Paramjit -30 20 NIL

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