answers - ies masteriesmaster.org/public/archive/2016/im-1482669055.pdf25. (c) pile foundations...

1. (b)

2. (d)

3. (b)

4. (c)

5. (c)

6. (b)

7. (b)

8. (c)

9. (a)

10. (d)

11. (b)

12. (c)

13. (c)

14. (b)

15. (b)

16. (d)

17. (c)

18. (a)

19. (a)

20. (b)

21. (c)

22. (b)

23. (a)

24. (b)

25. (c)

26. (c)

27. (b)

28. (b)

29. (a)

30. (c)

31. (a)

32. (a)

ESE-2017 PRELIMS TEST SERIESDate: 25th December, 2016

ANSWERS

33. (c)

34. (b)

35. (c)

36. (a)

37. (c)

38. (d)

39. (c)

40. (c)

41. (d)

42. (d)

43. (a)

44. (d)

45. (b)

46. (d)

47. (a)

48. (a)

49. (a)

50. (a)

51. (a)

52. (a)

53. (c)

54. (b)

55. (b)

56. (a)

57. (a)

58. (c)

59. (a)

60. (a)

61. (b)

62. (c)

63. (c)

64. (c)

65. (b)

66. (b)

67. (b)

68. (b)

69. (d)

70. (b)

71. (b)

72. (b)

73. (b)

74. (a)

75. (d)

76. (d)

77. (c)

78. (a)

79. (b)

80. (d)

81. (b)

82. (d)

83. (b)

84. (a)

85. (d)

86. (a)

87. (c)88. (a)

89. (a)

90. (c)

91. (d)

92. (b)

93. (b)

94. (c)

95. (b)

96. (a)

97. (d)

98. (b)

99. (d)

100. (a)

101. (b)

102. (d)

103. (d)

104. (b)

105. (a)

106. (d)

107. (b)

108. (b)

109. (a)

110. (a)

111. (a)

112. (a)

113. (a)

114. (d)

115. (a)

116. (b)

117. (b)

118. (b)

119. (b)

120. (c)

121. (c)

122. (d)

123. (c)

124. (b)

125. (b)

126. (d)

127. (a)

128. (c)

129. (c)

130. (a)

131. (d)

132. (d)

133. (d)

134. (d)

135. (a)

136. (d)

137. (d)

138. (c)

139. (c)

140. (c)

141. (a)

142. (a)

143. (a)

144. (c)

145. (d)

146. (d)

147. (b)

148. (a)

149. (c)

150. (a)

151. (c)

152. (b)

153 (c)

154. (c)

155. (a)

156. (a)

157. (d)

158. (d)

159. (a)

160. (d)

IES M

ASTER

(2) (Full-length-2)-25 December 2016

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

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

1. (b) Ground moulded & burnt in kilns implies itis 2nd class brick.1st class brick are table moulded (burnt inkilns)3rd class brick are ground moulded (burntin clamps).

2. (d)

3. (b) Shrinkage along longitudinal direction is lessthan that in transverse direction.

4. (c) Rueping process is used for treating timberof mixed species & timber containingsapwood & heartwood.

5. (c) Dry process consumes less fuel than wetprocess.

6. (b) 1:3 mixture of cement sand is made forstrength test of cement.

7. (b) Curing reduce permeability or increaseimpermeability of concrete.

8. (c) Slump value Degree of workability0-25 very low

25-50 low

50-100 medium

100-125 high

9. (a) In excess of 1.5%, carbon does notcombine with iron and is present as freegraphite. If there is no free graphite in thematerial composition, it is said to be steel.On the other hand, presence of free graphiteindicates that the material is cast iron.

10. (d)

Time headway

Optimum speed

flead

way

11. (b) Ductility is the distance in cm that briquettecan be stretched before breaking.

12. (c) For no of storeys exceeding 2, maximumslenderness ratio using lime mortar is 13.

13. (c)

Solution of baryta is a solution of bariumhydroxide Ba(OH)2

Ba(OH)2 + CaSO4 BaSO4 + Ca(OH)2

14. (b) nC tan

Sand specimen C = 0

n 100 kN / m²

2300 tan 45900 2

= 30°

11003

57.74 kN / m²

15. (b) Q = 6350 10

270 =

260K 0.1

100 4

K = 0.0275 cm/sec.

16. (d)

4.5%

d

w/c

In cohesionless soil compaction is mainlydone by vibration which breaks theinterlocking between particles and particlescan go into denser packing. Pure sand canbe compacted to maximum density undercompletely dry condition or completelysaturated condition. Dry density achievedis minimum at about 4-5% moisture content.

17. (c)

18. (a) 2 =

5/2

2 2 23Q 1

2 z 1 r / z

IES M

ASTER

(Full-length-2)-25 December 2016 (3)



= 2

2

3 3 132 2

(2 )

19. (a)

20. (b) 2T U4

for U 60%

22 2

1 1

t Ut U

2

260t 17850

= 256.32

Additional time = 256.32 – 178 = 78.32

21. (c) t =

wG es

1 e

32.65 0.65 0.5 10 18 kN/m1.65

wsat

(G e)1 e

= 20 kN/m3

sub = 20 – 10 kN/m3 = 10 kN/m3

11 = 18 × 4 + 20 + 6 × 10 = 152 kN/m3

22. (b)

23. (a)

24. (b) 4STa

bS 0.7V 6

50.7 30 6 11.8318

T = 6.15 sec

25. (c) Pile foundations undergo settlementUnder-reamed piles can be used for non-expansive soils.

26. (c) The lime stabilized bases or sub-bases forma water resistant barrier which stopspenetration of rain water. For heavy plasticsoils large quantity of asphalt and cement

is required hence stabilization throughbitumen and cement is not preferred.

27. (b)

28. (b) G.C. =

30 R 75

R R

G.C. = 75 75 1.25%R 60

Compensated gradient = 6 – 1.25 = 4.75

29. (a)

30. (c)

31. (a)

32. (a)

33. (c)

34. (b)

7°20´ 45°40´

S

EW

7°20´

M.T T.MN

T· B = 45°40´ + 7°20´T· B = S 53°0´ E

35. (c) f = 20 cm; d = 10 cmi = 4 mm;

C = f + d = 0.2 + 0.1 = 0.3

K = f 20i 0.4 = 50

Staff intercept,

S = 2 × (2.5 – 1) = 3.0 m

Horizontal distance between the staff stationand instrument station,

D = Ks + C

= 50 × 3 + 0.3

= 150.3 m

IES M

ASTER




36. (a) d = rhH

where d = relief displacementH = height of aircraft

h = height of objectr = radial distance

37.(c) Distance ‘d’ between the lighthouse and aship

d = h 100

0.0673 0.0673 = 38.54 km

38. (d)

r

area = A = r2

dA = 2 r dr

eA = 2 rer = 2 × 7 × 0.02

= 0.8792 m2

0.88 m2

39. (c) Permissible angle misclosure,

E = K N

= 20" 9

= 60"Where K = for ordinary theodolite traverseK = 20"

40. (c) It is best suitable when from one setting,elevation of several points are to bedetermined

41. (d)

42. (d)

43. (a)

Correction to MSL =LhR

=300 2.25

6370

= 0.106m 106mm

44. (d)

45. (b) In resource levelling the uniform demand ofresource is achieved and project durationcan increased but not reduced.

46. (d) Total float of 20 – 30 activity is= LS – ES – tij= 23 – 10 – 8 = 5

47. (a)

No.

of a

ctiv

ities

tE

Duration

48. (a) The limit state of serviceability analysis isdone at service (working) loads by linearelastic theory. So all the assumption of wsmare valid for limit state of serviceability.

49. (a) Maximum clear span

=

2

25bwhich ever

b is minimum100d

(1) = 25 × 0.3 = 7.5m

(2) = 20.3100

0.5 = 18m

So l = 7.5 m

50. (a) Deflection after partition erection andapplication of finishes should not exceed.

span350

which ever or is less20 mm

So9000350

= 25.7mm

or 20mm

So it should not exceed 20mm

[ lC. . 23.2 IS 456 - 2000]

51. (a) There are 9 conditions for design of twoway slabs which are restrained or continuousat the edges. These coefficients are basedon yield line theory. [Johansen’s Theory]

IES M

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52. (a) DL = 0 15 0 30 24 1 08 kN m . . . /

LL = 5 kN/mTotal load = 5 + 1.08 = 6.08 kN/m

38 4150 300I 3 375 10 mm

12.

2ip 1000N mm/ , 2

sA 100mm

i i sP PA 1000 100N 100 kN

Rotation due to prestress, Pc

PeL2E I

p 8100 50 6000

2 36 3 375 10.

= 0 00123. radian

Rotation due to loads, 3

ec

wl24E I

3 2

e 86 08 6 1000

24 36 3 375 10

.

.

= 0.00450 radian

Net rotation = 0.00450 – 0.00123 = 0.00327 radian

Elongation of cable 2 50 0 00327. = 0.327 mm

Increase in stress = 50 327 2 1 10

6000. .

= 11.445 N/mm2

Percentage increase = 11 445 100 1 1451000

. . %

53. (c) A steel trunked guide leads each wire fromcable to anchorage point along a gentlecurvature. In addition to the guide, a centralblock is also provided to anchor centralwires.

54. (b) Minimum reinforcement

Ast =

y

0.85 bdf

b = 200

d = 300 – 25 = 275

fy = 250

Ast = 0.85 200 275

250Ast = 187 mm2

55. (b) Over all FOS in limit state:

On concrete: 1.5 on loading and 1

0.446 on

material concrete so overall

= 1.5 3.336 3

0.446

On steel: 1.5 1.724 1.780.87

Working stress is more conservative andless economical than limit state method.

56. (a) According to IS 456-2000 – minimum tensilestrain in steel at collapse state should be

y

s m

f.002

E

So, = 5415 .002

1.5 2 10

= 0.00338

57. (a) An activity with the lowest cost slope is tobe considered first for crashing.

58. (c) Gross weight of the tractor = 20000 kg= 20 tHence total rolling resistance offered by theroad = 20 × 45 = 900 kg

Load on driving tyres

= 60 20000 12000 kg

100Maximum possible rimpull prior to slipageof tyres= 0.65 × 12000 = 7800 kg

Maximum ef fective rimpull is= 7800 – 900= 6900 kg

59. (a)

FDB =

1ns

i

C1

c

=1521

64

= 0.5

D1 = Ci × FDB = 64000 × 0.5 = 32000

B1 = 64000 – 32000 = 32000

D2 = 32000 × 0.5 = 16000

B2 = 32000 – 16000 = 16000

D3 = 16000 × 0.5 = 8000

B3 = 16000 – 8000 = 8000 Rs.

IES M

ASTER




60. (a) The output of the power shovel dependson:

(I) Class of material(ii) Depth of cut(III) Angle of swing(IV) Job condition(V) Management condition(VI) Size of the hauling units(VII) Skill of the operator(VIII) Physical condition of shovel

The optimum depth of cut is that depth forwhich output is greatest and dipper comesup with full load without excessivedownward pressure and tension. This depthvaries with the type of soil and size of thedipper. Ideal production of shovel is basedon operating at 90° angle of swing.

61. (b) Since the non coagulated particles will belesser in size and hence to make themsettle, more time is to be needed for thelower velocity they have.On the other hand, to remove these smallerparticles overflow rate needs to be lower asthe settling velocity of these smaller particleswill also be small.

62. (c) Micro strainers are used for removal of algaeand planktons but not useful for removingvery fine parties like clay and colour. It isnot very usefull for running water and functionefficiently in stored water. It is used beforesand filters as it increases the efficiencyupto 50%.

63. (c) In this first stage i.e. of acid fermentation,of sludge digestion, the fresh sewage sludgebegins to act upon by anaerobic andfacultative bacteria called acid formers.These organism solubilize the organic solidsthrough hydrolysis.

64. (c) Chloride content may be found in municipalsewage from kitchen waste, human facesand urinary discharges.Large amount of chlorides may enter fromindustries like ice cream, meat salting etc.Therefore large chloride content indicatesindustrial pollution.Mohr’s test — Chloride contentDeterminationOrthotolidine Test — Chlorine Determination

65. (b) Winklers method is used in general todetermine the D.O content of sewage.

Azide modification: In this method,interference caused by nitrate is removedefficiently.Permanganate modification: If thesample contains iron (Fe+ ion), addition of1 ml of potassium fluoride and azideadopted to suppress the interference of Fe+3

66. (b) In super adiabatic lapse rate, theenvironment lapse rate is more thanadiabatic lapse rate and thus the risingparcel of air continues to rise.

67. (b) Maximum upper limit of BOD will be thetheoretical oxygen demand (ThOD) i.e.when entire glucose will get oxidized.

6 12 6 2 2 2192mg180gm

C H O 6O 6CO 6H Ol

180 mg/l exerts a ThoD of 192 mg/l

250 mg/l exerts a ThoD of

= 192 250 266.67 mg L180

68. (b) The oxidation of organic matter present insewage starts as soon as the sewage fallsinto the river.This starts consuming dissolved oxygenfrom the river. The process of oxidationcontinues till all organic mater has beenoxidized and thus considerably reducing theBOD. This serves as most important partof self purification.

69. (d)The chlorine content does not increases withtime. They can be applied to water either indry form or as solution. They increase the pHof water on account their lime content. Theycontain a very low chlorine content.

70. (b) K = 0.75

A = 8 km2 = 8 × 106 m2 = 800 hactare

Tc = 30 min

ic = c

100T 20

for 20 < To < 100 min

IES M

ASTER




= 10030 20

= 2 cm/hr

Q = 1 0.75 800 2

36

= 31200 33.33 m /s36

71. (b)

(1) The grit cannot be used for preparationof concrete as it contains sufficientorganic matter.

(2) The water used for washing the grits maycontain several organic matter which maybe sticking to the grits, thus this watershould be sent to treatment plant forfurther treatment.

(3) The length of channel is increased beyondtheoretical value by 10 – 50% to accountfor non idealities in flow & settlingparticles.

72. (b) The septic conditions are most likely tooccur when the top layer of sand is cloggeddue to non-oxidation of some suspendedsolids which are transformed into humus.This occurs in the layer just below the toplayer as the organic matter cannot penetratedeep inside the sand filters due to the sizeof contact media.

73. (b)1. The flow scheme of an extended aeration

process and its mixing regime are similarto the complete mix process.

2. The process permits low food/mass ratioand low organic loading, high MLSSconcentration. This leads to high BODremoval upto 95-98%.

3. The air or oxygen requirement is quitehigh, which increase the running cost ofplant considerably.

4. No sludge digestor is separately requiredbecause the solids undergo considerableendogenous respiration and get wellstabilized over long detention time.

74. (a) Sludge volume index =1000 40ml / gm

25

RQQ =

t6

T

X10 – XSVI

= 6

5000 0.2510 – 500040

75. (d)Cylinder is rotated clock wise which has atangential velocity of r at point C in oppositedirection to the flow.

So stagnation can occur at cSpeed r = 33

= 33

rpm = 33 60 315 rpm

2

76. (d)

77. (c) h = 2 2 2

2 1w r r

2g

15 =

22 2w 0.30 0.10

2g

w =

2 22 15 9.810.3 0.1

= 60.65 rad/s

N =

60w2

=

60 60.652 3.14

= 579.48 = 580 rpm

78. (a)In an open channel whenever flow will takeplace, losses will occur so total energy willdecrease in the direction of flow hence energygradient line will not be horizontal.

79. (b)

P = 0 QH = 0.9 × 9.81 × 300 × 35

= 92704.5 KW

For a specific speed of 400, power producedper machine (P1) is given by

IES M

ASTER




400 = 1

54

150 P

(35)

P1 = 5 22 400(35)

150

= 51536 KW

No. of turbines reqd. =

92704.5 251536

80. (d)

NDL

CDLS2

MildSteeper

Steep

S3

CDL

NDL

M2

NDL

81. (b) Assume that pressure at top = P

So, Pressure at bottom = wP 5 250

P = 200 kN/m2

5m

200 kN/m²

250 kN/m²

inside width=3m

Pressure prism

Total force = area of prism×inside width

= 1 5 (250 200) 32

= 3375 kN

= 3.375 MN

82. (d) Pressure of water at the base of piezometer

= 2 × 0.625 × w

So h × w = 2 × 0.625 w

h = 1.25m above the base of piezometer

So H = 1.25 + 0.75 = 2m above the bottomof tank

83. (b) Total volume = 1000 m3

Total Buoyant force = 1000 × 1000kg= 1000 tonne

Additional required weight = 1000–500

= 500 tonne

So Volume of water which gives 500 tonneweight

V × 1000 = 500 × 1000

V = 500 m3

84. (a)

5mm

V

= 10 = 0.1

f = 0.4Shear stress on lower plate

= vy

Considering a unit area 1 m2

Force = vy

weight of unit area liquid

1 × y × = w

and friction force = w × f

So,

2v fyy f vy

=

20.4 (0.005) 100000.1

v = 1msec

85. (d)u = 2x – 3y and v = – 3x – 2y

u =

x

x = – (2x – 3y)

= – 2x + 3y

= – x2 + 3xy + f(y) ..... (1)

Again,

y = – v = 3x + 2y

= 3xy + y2 + f(x) ..... (2)

From equation (1) and (2)

= – x2 + 3xy + y2

86. (a)

87. (c) Basic capacity

= 100024 960 veh / hour

(19 6)

IES M

ASTER




88. (a) Since pressure drop in flow is directlyproportional to friction loss

P hf and hf = 2fL V×

D 2gFor laminar flow, the friction factor

f =

64 64=Re VD

h f =2

264 V×

2gVD

h f = 264 × V

2 gD

h f V

For laminar flow P V..Hence flow is laminar.

89. (a)

Bvmax

y

shear stressdistribution

0velocitydistribution

= p B y

2x

here B = 0.03 m

Q = 600 cm3/s per cm width of plates

p x = 2

12 VB

= 312 Q

B 0.01

= 6

312 1.2 600 10

(0.03) 0.01

= 32,000 pa/m

at y = 0.01 m

= 32,000 × 0.03 0.01

2

= 160 pa

90. (c) 25t

2 2l

5 2510 252 2l

5L 10 mm

L 100 m

91. (d)The model test results can be utilized to obtainin advance the useful information about theperformance of the prototype only if there exista complete similarity between the model andprototype.There are in general three types of similaritiesto be established for complete similarity toexist between the model and its prototype.(i) Geometric Similarity : Geometric

similarity exists between the model andthe prototype if the ratio of correspondinglength dimensions in the model and theprototype are equal.

(ii) Kinematic Similarity : Kinematicsimilarity exists between the model andthe prototype if (a) the paths of thehomologous mov ing particles aregeometrically similar, and (b) if the ratiosof the velocities as well as acceleration ofthe homologous particles are equal.Homologous point means correspondingpoint in the model and prototype.

(iii) Dynamic Similarity : Dynamic similarityexists between the model and theprototype which are geometrically andkinematically similar if the ratio of all theforces acting at homologous points in thetwo systems viz., the model and theprototype are equal. Thus for flows to bedynamically similar, the ratios of thevarious forces acting on the fluid particlesin one flow system should be equal to theratios of similar forces at correspondingpoints in the other flow system.

In compressible flow system, Machnumber should be same in the twosystem.

92. (b)Often a compound pipe consisting of severalpipes of varying diameters, and length is tobe replaced by a pipe of uniform diameterwhich is known as equivalent pipe. If D isthe diameter and L is the length of theequivalent pipe then it would carry the samedischarge Q if the head loss due to frictionin the equivalent pipe is same as that in thecompound pipe.

h f =2 2fLv 16v

2gd 2g

IES M

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or,2

2fLQ

2gd.A =2

216Q2gA

or,fL

2gd =162g

or,fLd = 16

L =16d 16 0.15 80m

f 0.03

93. (b) Ultimate load = 1.2 × ( 12) 100 / 1000

= 4.52 kN Safe load = Ultimate load / load factor

= 4.52/1.5 = 3 kN.

94. (c)As the web begins to buckle, the web loosesits ability to resist the diagonal compression.The diagonal compression is then transferredto transverse stiffeners and the flanges. Thevertical component of the diagonal compressionis supported by the stiffeners and flanges resistthe horizontal component. The web resists onlythe diagonal tension, and this behaviour of thewebs is called tension field action.

95. (b)The actual Merchant Rankine formula is givenby

n n nc cr y

1 1 1

f f f

or1/n

yc

ny

cr

ff

f1 f

The 1984 IS 800 adopts a FOS of 1.67

1/ny

Cn

y

cr

0.6ff

f1

f

For a zero slenderness ratio, the column failsthrough yielding and compressive stress

C yf 0.6 f with the multiplication factor

becoming unity. Thus in this formula thebuckling effect of the column is considered by

the multiplying factor of

1/nny

cr

1

f1

f

96. (a)Safe load that can be resisted per mm

length of weld = u

mw

f0.7 S3

[s = 6 mm ; mw = 1.25 ; fu = 410]

=

0.7 6 410 795.36 N/mm3 1.25

Length of the weld = d = × 150

= 471.24 mm Greatest twisting moment

= 795.36 × 471.24 × 6150 102

= 28.11 kNm.

97. (d) The larger diameter bolts are particularlyfavourable in connections where sheargoverns because shear capacity varies assquare of the bolt diameter.

98. (b) Gross dia of rivets

= 20 + 1.5 = 21.5 mm

For most critical section

Anct = t × (b – n × d)

Anct = 1.2 × (15 – 2 × 2.15) = 12.84 cm2

Maximum tension in the flat

= 2150 12.84 101000

= 192.6 kN

99. (d) Shear lag reduces the effectiveness of thecomponent plate of the tension member thatare note connected directly to a gussetplate, the outstanding legs are kept shorterin length. For this reason unequal angleswith long legs connected are preferred.Shear lag effect can be reduced byincreasing the length of the end connection.

100. (a)

IES M

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300

50

200

600

200 200

200

t >

2 2

y mo

2.5w (a 0.3b )f /

w =

31.5 P 1.5 700 10 5.83 MPaA 600 300

a = 200 mm

b = 50 mm

t >

2 22.5 5.83 1.1 (200 0.3 50 )250

= 50.17 mm

101. (b)

102. (d)

103. (d)

104.(b)Standard temperature at elevation h = 15 –0.0065 × h

= 15 – 0.0065 × 800 = 9.8°C

105.(a)At full saturation,

Vw = Vv

Vv = eV

1 ee = wG = 0.2 × 2.65 = 0.53

Vv = 53

1.53 = 34.64m3

106. (d) IP = 56 – 25 = 31 and soil is fine grained

Ip = 0.73 (WL – 20)

= 0.73 × (56 – 20) = 26.28

107.(b) Pa = a aK H 2c K

Ka = 1

Z = 2c

=

2 9.81 119.62

Pa at 10m = 19.62 × 10 – 2 × 9.81= 176.58 kN/m2

So,Total pressure = 1 (10 1) 176.582

= 794.61 kN/m

108.(b)

FOS = 2

t

t

c H cos tanH cos sin

sin (19.47°) =13

tan (19.47°) =1

2 2

fos =

tantan

=

tan30 1.63tan19.47

109.(a)The local thickening of flat slab at the junctionwith column is known as drop panel.

110. (a)

32

v52.5 10 0.496 N/mm230 460

For m20, c max a a2.81MP 0.496MP (o.k.).

pt =

25 164 0.95%

230 460c = 0.61 N/mm2.c > v .

So, minimum shear reinforcement is provided.

v

v y

As 0.4bs 0.87f

Sv <

22 6 0.87 2504

0.4 230Sv < 133.62 mm.

Also,Sv > 0.75d = 0.75 × 460 = 345 mm.> 300 mm

Sv = 130 mm.

111.(a)

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d = 2

cq 1 sin1 sin

= 2

300 1 sin 3018 1 sin 30

= 1.852 m

112.(a)Duty of water increases from head of canal toits tail. For tank irrigation, duty is usuallyexpressed in terms of ha/Mm3 (ha-million m3).

113. (a) Q is same in both canals

fM > fN

Also, s = 5/3

1/61 f

3340 Qfor same discharge

5/3s f

SM > SN

Channel M will have steeper longitudinalslope.

114. (d)An aggrading river is a silting river. Such ariver increases its bed slope, called buildingup of slope.

A degrading type river constantly scour toreduce available excess land slope.

115. (a)Water is supplied to the soil when the moisturelevel falls from FC to OMC (> PWP).

Moisture level is not allowed to fall below OMC,which is greater than PWP. This is because,if moisture level reaches PWP, crop yieldreduces. Hence, this is always avoided.

116. (b)From virtual work principle

A ABP = B BAM

BA =50 0.1

100

= 0.05

117. (b) DS = R - R1

R = removed reactionR’ = applied restrains.As BC Part has no loadingso at a time only one Cable givesreactionSo R = 1 and applied restrain atA = 1 to make it fix DS = 1 – 1 = 0and structure is Stable in all loadings.

118. (b) Take R = Reaction given by lower beam of Both beam at centre will be same

3 3

AB CD

(P R)l Rl48E I 48E I

P D

B

A

C

P – R = 2 R

R = P3

So RA =

PP P32 3

CR PR2 6

A

C

R 2R

119. (b)(1) Only Sway

No fixed end moment due loadingso no non sway analysis

(2) Both analysis has to be done [M is appliedat a Joint so this will be distributed in nonsway analysis]

(3) Only sway(4) All forces goes to the Column so if only

the Column is considered axially deformablethen sway analysis is to be done otherwise No analysis required.

120. (c) Virtual work method can be applied innon elastic Conditions also [eg- Plasticanalysis]Net Virtual work = 0 in the equilibriumconditions only.

121. (c)Removel of AD will not make the structure

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unstable so can be chosen as redundant.If we take HD as redundant and apply unit loadthen no force will develop in the 1, 2, 3 and 4members. So no force will develop due totemperature.If temperature change occurs in the 5 memberand we take HD as redundant only unit forcewill develop in the fifth member so force willdevelop only in the 5 member.

122. (d)

1 8 7

65

432

DK = 1 2 2 3 3 34 4 5 54 56 8 8 6, ,x , ,y ,x , , ,y ,y , ,x ,

DK = 13

123. (c)K21 = Deflection in the direction of (2) due tounit load at 1

1

L

L

1

L

L

21× L

2 = 2 2 3

21m L L L K2EI 2EI 2EI

l

124. (b)

1) ILD for MB

1 = MBBA C

for max. MB load should be on C2) For max. reaction at B

AC B

RB

load can be any where b/w C and B

3) For max. reaction at A

A C B

RA

load can be on A

4) For max. shear at C

1B

load should be just left to C

125. (b) Buckling strength ISolid steel column dia = DHollow steel column dia = d external

= d2

internal

A1 = A2

D = 3 d2

solid

hollow =

4

44

Ddd2

= 35

126. (d)

x

10-x

1600

4001600 400

x 10 x

x = 8 cm

127. (a) Loading diagram obtained from a SFDwill always satisfy force equilibrium, butmay or may not satisfy momentequilibrium.

128.(c) 2

2max

40 40 30 50 MPa2

129. (c)TJ =

GL

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1J

41

d

0.1 =

44

d 21.6 radian

d

130. (a)A crack parallel to the direction of length of hubsuggests that failure was due to tensile hoopstress alone.

131. (d)132. (d)

Change in ground water storage = 90 × 106

90 × 106 = 100 × 106 × 5 × SY

SY = 0.18

n = SY + SR

= 0.18 + 0.12

= 0.30

133. (d)Area under DRH

= 1 (10 70) 10 (70 90)10 10 102 2 2

90 40 4020 40 36002 2

= 12.06 × 106 m3

Rainfall excess = 6

612.06 10300 10

= 0.0402 m = 4.02 cm

134. (d)

= 15 8.7

8

= 0.7875 cm/hr

but 0.6 cm/hr and 0.75 cm/hr < 0.7875 cm/hr

They will not contribute in runoff

= (1.35 2.25 3.45 2.7 2.41 1.5) 8.7

6

= 0.825 cm/hr

135. (a)The number of optimum stations can be

N = 2 2

vC 29.5410

= 8.73 9 stations

According to WMO, atleast 10% of the raingauge stations should be recording type.

Recording type raingauge = 10 9

100 = 0.9

1

Non-recording type rain gauge = 9 – 1 = 8

136. (d) Acid resistant brick resist deterioration byall acid, except hydrofluoric & hydrochloricacid.

137. (d) Rediset cement release a lot of heat &hence it cannot be used for massconcreting.

138.(c) Increase in super elevation will causetoppling of slow moving vehicles. So superelevation shouldn’t be increased beyond alimit. So A is correct. When super elevationis limited, the fast moving vehicles will havenet centrifugal force and driver will feeldiscomfort. But R is not correct as reasonfor discompfort is given as increase in sidefriction.

139. (c) Weight of a quantity is the trust worthinessof that quantity. The weight of a quantityindicates relative precision of that quantitywithin the set of observations.

Therefore weight are assigned dependingupon the degree of precision.

Weight 1

Variance

140.(c) Aplanation is always is absence of sphericalaberration.

141. (a) Very large cover leads to large crack widthleading to increase of moisture & chemicalattack to concrete resulting in corrosion toreinforcement & deterioration of concrete.

142. (a) Primary torsion is associated with twistingmoments that are developed in a structuralmembers to maintain static equilibrium andare independent of torsional stiffness ofmember.

143. (a) In Symmetrically reinforced cross section,

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shrinkage does not result in any curvatureof member except in statically indeterminateframe elements.

144. (c) Pessimistic time estimate does not includethe possible effects of major catastrophessuch as floods, earthquake etc.

145. (d) The value of asset when its utility isconsidered to be junk is called scrap value.

The value of an asset at the end of itsutility period is called salvage value.

146. (d) In relative stability test, Methylene bluesolution is decolorized by sewage sample.The cause of decolorization is caused byenzymes produced by anaerobic bacteria.

The sooner the decolorization takes place,earlier the anaerobic conditions developsand lesser availability of oxygen and moreis instability of sewage.

147. (b) The grease and oil being lighter in weight,float on the top surface of the sewage.Hence it an outlet draws sewage from lowerlevel, grease and oil get excluded. Basedon this principle, the grease and oil trapchambers are designed in such a way thatthe outlet level is located near the bottomof the chamber, & hence inlet is kept neartop.

148. (a)

U

u

U

At outer layer of boundary layer, the velocitygradient is zero

du = 0dy

i.e. at outer edge, u = USo shear stress from Newton's law of viscosity.

=duu = 0dy

149. (c)As the shear load is increased on a stiffenedweb panel, the web panel buckles. This loaddoes not indicate the maximum shear capacity

of the web. The load can be still increasedand the web panel continues to carry furtherload relying on the tension field action. Part ofthe buckled web takes load in tension. Thistension member action is across the web panelin an inclined direction to web panel diagonal.

150. (a)151. (c)

In lumpsum contract the contractor is verycareful in working out the details of the items,the quantities and the rates, as any mistakewould add to his cost therefore the contractorin this system of bidding where quantities arenot given, generally keeps a margin in theoffer for unforeseen circumstances/items. Thismight lead to increase in the cost.

152. (b)

153 (c) In a catchment for a given duration, rainfall depth

decreases with increase in area.

154. (c) A is true but R is falseAlthough uplift pressure on the upstream flooris quite high but there is weight of water onthe floor which exerts a downward force whichcounterbalances the upward force due to uplift.That’s why thickness of u/s floor is less thanthat of downstream floor.

155. (a)

l2 l2

W

Fixed momentdiagram

W8

l

W4

lFree momentdiagram

Area of fixed moment diagram = –2W

8l

Area of free moment diagram

= 21 W W

2 4 8

l ll

A B– = Area under MEI

diagram between AA

and B = 0Hence answer is (a)

156. (a)

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As depth increases, the over burden pressureincreases due to which N value increases.

157. (d)

Assertion is false as clays have more waterholding capacity and when used as back fillmaterial they increase the pressure due topore water on retaining wall and havedestabilizing effect.

158. (d)Assertion is false reason is correct as it is nearerto zero air void line on wet side, less voids arepresent and less increase in density.

159. (a)

160. (d)

answers - ies masteriesmaster.org/public/archive/2016/im-1482669055.pdf25. (c) pile foundations...

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