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1 (c)
2 (b)
3 (a)
4 (b)
5 (c)
6 (b)
7 (a)
8 (a)
9 (c)
10 (b)
11 (c)
12 (b)
13 (d)
14 (c)
15 (c)
16 (b)
17 (b)
18 (b)
19 (c)
20 (b)
21 (d)
22 (b)
ESE-2019 PRELIMS TEST SERIESDate: 23rd December, 2018
ANSWERS
23 (b)
24 (b)
25 (d)
26 (c)
27 (b)
28 (d)
29 (b)
30 (c)
31 (a)
32 (c)
33 (b)
34 (d)
35 (a)
36 (c)
37 (d)
38 (c)
39 (b)
40 (d)
41 (c)
42 (d)
43 (a)
44 (b)
45 (a)
46 (c)
47 (b)
48 (c)
49 (a)
50 (b)
51 (d)
52 (b)
53 (c)
54 (c)
55 (c)
56 (a)
57 (c)
58 (d)
59 (a)
60 (a)
61 (c)
62 (c)
63 (c)
64 (d)
65 (d)
66 (d)
67 (c)
68 (c)
69 (a)
70 (d)
71 (b)
72 (d)
73 (a)
74 (b)
75 (b)
76 (a)
77 (c)
78 (b)
79 (a)
80 (b)
81 (a)
82 (a)
83 (a)
84 (d)
85 (d)
86 (c)
86 (c)
88 (c)
89 (c)
90 (b)
91 (a)
92 (a)
93 (b)
94 (b)
95 (b)
96 (b)
97 (c)
98 (b)
99 (c)
100 (b)
101 (b)
102 (d)
103 (d)
104 (d)
105 (c)
106 (d)
107 (a)
108 (d)
109 (b)
110 (a)
111 (c)
112 (b)
113 (d)
114 (a)
115 (b)
116 (a)
117 (b)
118 (b)
119 (a)
120 (c)
121 (b)
122 (b)
123 (b)
124 (a)
125 (a)
126 (c)
127 (c)
128 (b)
129 (b)
130 (b)
131 (b)
132 (a)
133 (d)
134 (b)
135 (b)
136 (d)
137 (d)
138 (b)
139 (c)
140 (a)
141 (a)
142 (c)
143 (a)
144 (b)
145 (a)
146 (b)
147 (a)
148 (a)
149 (d)
150 (d)
IES M
ASTER
(2) Full Syllabus
1. (c)In ladder network
Total project time = 15 25 30
4
= 40 days
Time saved = 15 + 25 + 30 – 40
= 30 days
2. (b)Normal time = 14 day
Normal cost = Rs 1360
Crash time = 12 days
Crash cost = Rs 1600
Cost slope = crash cost – normal costnormal time – crash time
= Rs 1600 Rs 1360
14 12 days
= Rs 120/day
= Rs 5/hour
3. (a)Bar chart does not indicate which activities arecritical.
4. (b)
5. (c)
Correct value = original record × corrected slope/Original slope
=
1157601
18
=1876015
= 912 mm
6. (b)
Surface floats are affected by external factor suchas wind so for exact velocity multiplied by reductionfactor.
Rod float and canister float are improved float whichare not affected by the external factor.
7. (a)
Index of wetness = Rainfall in yearAnnual avg. rainfall
=80 66.67%
120
So deficiency = 100 – 66.67 = 33.33%
If rainfall deficiency is 30 – 45% large deficiency
If rainfall deficiency is 45 – 60 serious deficiency
If rainfall deficiency is > 60 disastrous deficiency
8. (a)
Q = 1
2
h2.303 log A st h
0.085 =
210
2.303 3log D 3100 60 2 4
D 23.10m
9. (c)
10. (b)
B1 = 4 m, B2 = 1m
V1 = 16 4 m/s
(4 1)
E1 = 1 + 24 1.8 m
2 10
q2 = 16 161
m3/s/m
1 12 3 3 13
c 216 256(y ) (25.6) 2.94 m10 10
(EC)2 = 1.5 × 2.94 = 4.42
E1 = (EC)2 + z
z = E1 – E2
= 1.8 – 4.42 = –2.62 m (depression)
11. (c)
11
1
VFgy
124F 6 5
9.81 1.6
Lj = 6.1 y2
221
1
y 1 1 1 8Fy 2
22 1
1y 1 1 8F2
y2 = 21
1 1 1 8 6 y2
2 1y 0.5 ( 1 289)y
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-04) (3)
y2 = 18yLj = 6.1 × 8 × 1.61 = 78 m
12. (b)
13. (d)
14. (c)
24T L LU U , J d
2GJ J
4LUd
15. (c)
x
d2d
L
0
TdxGJ
xdx xD 2d d 2L L
L
40x
T 32dxG D
=
4L L
4 4 40 0
32T dx 32T L dxG dGd x 2L x2
L
16. (b)3
464PR n
Gd
17. (b)
Strain Energy stored, U = 2 3
432P R n
Gd
2 3
43
32 100 50 10U 500 Nmm
80 10 10
18. (b)
B
A B4 2
20kN
40 kNm
A B
20kN
80 kNm+ B 20
40C
B
3 2
c B20 2 40 2
23EI 2EI
= 80 4 160 160 10402 2mm
3EI 3EI 2EI 3EI
19. (c)
D B
B
circular squareArea Area
22D B
4
....(1)
Also,
33
cC3 3
Sq sq
DZM 3 D32
BM Z 16 B6
2C
3Sq
M 3 B D 3 D 3 2B 3M 4 B 4 B 4 B 2
20. (b)W
L/2 L/2A B
Mid, spanWLM2
3
WL D2 2f
DB12
and
2 3WLDBf
3WLDBf
IES M
ASTER
(4) Full Syllabus
21. (d)
For plane stress condition
1 2 x y
2 y x 1 50 100 50 MPa
2 y 50MPa
22. (b)
At P, shear force, V = 0
and Blending moment, M = 2
8l
for point P,,2
2
Bending 3 2
d8 2My 3
I bd 4bd12
ll
2
23 64 10000
100MPa4 300 400
and P 0 .
23. (b)
Since no flexural cracks appeared.
1 CK0.7 f
CK3.4 0.7 f
CKf 23.59MPa
Also diagonal compression failure occurred.
max CK0.631 f
1 2CK0.631 f
2
CK
3.4 3.000.631 f
2
CKf 25.71MPa
Therefore, out of given options, grade of concretemust be 25 MPa.
24. (b)
Mohr circle is :
(, 0)
–
0 (, 0)
25. (d)
m = 8, r = 10, j = 9, m' = 3
Ds = 3m + r – 3j – (m 1)
= 24 + 10 – 27 – 2 = 5
Alternatively,
No. of cuts req. = C = 3
R' = No. of restraint req. = 1 + 1 + (3 – 1) = 4
Ds = 3C – R'
= 3 × 3 – 4 = 5
26. (c)
A
B
C
PLP
D
Slope deflection equations :
DA D4EIML
DB D3EIML
DC D3EIML
Equilibrium equation :MDA + MDB + MDC = PL
2
DPL10EI
2
AD2EI PL PLML 10EI 5
27. (b)
1kN
1kN
1kN
1kN
1kN
1kN
0kN
1kN
2 kN
A
C
B
D
Member F (kN)AC 1BC 2
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-04) (5)
D1kN. T Lu
5D 10 50 4 2 4 2
D 2 mm
28. (d)
Since B is on the line of symmetry, B 0
AB
6EIM25
Alternatively,
A C0, 0
as no external loading is applied
Fixed end moment = 0
BA B A2EI 3.M 25 5
BC B C2EI 3.M 25 5
At joint B : Equilibrium equation, MBA + MBC = 0
B 0
CB2EI 3. 6EIM5 5 25
29. (b)
6m
1 kN/m
1 kN/m
18m 18mRA RB
A BHA HB
C
Taking moment about B
Ra × 36 + 6 × 3 = 18 × 9
Ra = 4
Taking moment about C
Ra × 18 = Ha × 6 + 6 × 3
a4 18 18 H
6
Ha = 9 kN
30. (c)
Column analogy method is force method in whichunknowns are forces.
31. (a)
Column analogy method is a force method ofanalysis.
32. (c)
33. (b)
Speed-density relationship is given
V = 48 – 0.24 K …(i)
V = Vf – f
J
VK
K …(ii)
Comparing equation (i) and (ii)
Vf = 48 km/hr
fJ
J
V 480.24 K 200 veh/kmK 0.24
qmax =
f JV / K 48 200 2400 veh/hr4 4
Velocity at capacity flow = fV24 km/hr
2
Density at capacity flow = JK100 veh/km
2
34. (d)
Simultaneous systemIn this system all the signals along a given roadalways show the same indication (green, red etc.)at the same time
Alternate systemIn this system alternate signals or groups of signalsopposite indications in a a route at the same time.
Simple progressive systemA time schedule is made to permit as nearly aspossible a continuous operation of group of vehiclesalong the main road at a resonable speed.
Flexible progressive systemIn this system it is possible to automatically varythe length of cycle, cycle division and the timeschedule at each signalized intersection with thehelp of computer.
35. (a)
It is chemically inactive (inert).
It is thermoplastic i.e. it becomes soft onheating and in the reverse process hard oncooling. It oxidises slowly.
36. (c)
n365A 1 r 1N VDF DF
r
12365 1200 1.08 13 1
0.08
= 24.93 msa
IES M
ASTER
(6) Full Syllabus
37. (d)
Shoving is a form of plastic movement within thelayer resulting in localised bulging of the pavementsurface.
All are correct statements regarding causes ofshoving.
38. (c)
Bird baths : Localised pavement surface areaswith slightly lower surrounding pavements andis due to subgrade failure.
Ravelling : It is the ongoing separation ofremoval of aggregate particles from the surface.
Pot holes : Small bowl shaped depression inthe pavement surface that penetrate all theway through the asphalt layer till the surface.
Subsidence : It is a localised or abrupt loweringof the road surface. It max result from poorlycompacted backfill, poor local drainage.
39. (b)
Rail joint is the weakest link in the track.
The ill effect of the rail joints are :
1. Maintenance effort of track increases.
2. Life of rails, sleepers and fastenings are ad-versely affected.
3. Lot of noise pollution
4. Sabotage chances increases as there is apotential danger of removal of fish plate.
5. Quality of track suffers
7. Increased fuel consumption.
40. (d)
41. (c)
Length of rail on BG track = 12.8
Sleeper density = n + 5
= 12.5 + 5 18 per rail
No. of rails = 1000 78.12 7912.8
No. of sleepers = 18 × 79 = 1422
42. (d)
In the direction of wind, air speed is less thancruising speed (air speed of aircraft is speedrelative to wind)
Capacity of aircraft is the measure of passen-gers, cargo and fuel to be accommodated inthe aircraft.
43. (a)
Corrected length, l' = Basic length + correction
= l + c
Correction due to temperature and elevation
= 300 m ( correction due to elevation is zero)
So, correction due to gradient
= (l 300) 0.3 20
100
So, l' = (l 300) 0.3 20l 300100
2000 = 5.3l 300 5.35 5
l 1586 m
44. (b)
The LCN method is used for design of bothrigid and flexible pavement.
The daily and seasonal variation of tempera-ture cause a considerable effect.
The best direction of runway is along the long-est line of windrose diagram.
45. (a)
D = KS + c
k = 100, C = 0.3 m
D = 100 × (1.555 – 0.900) + 0.3
= 65.8 m
So correct option is (a).
46. (c)
d S 206265R nD
S = 30 3 150
206265
S = 0.0654 m
47. (b)
Gradient is 1 in 25
X
25
1
2 2x 25 1 25.02m
fall in 20 m will be 20 1 0.799
25.02
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-04) (7)
Slope correction
Cg =
2 2h (0.799)2L 2 20
Cg = –0.01597 m
gC 1.597 cm
48. (c)
Let the depth of the tank be ‘d’ m. volume usingtrapezoidal formula
= 3d 24 8 (16d)m2
Volume using prismoidal formula.
= d 1 24 8 4(5 3)2 3
= 346 d m3
We have 4616d d 43
d = 6m
49. (a)
Napier has two rules:
(i) The sine of a part is equal to the product ofthe tangents of two adjacent parts.
(ii) The sine of a part is equal to the product ofthe cosines of the two opposite parts.
50. (b)
In active remote sensing, sensors emits radiationwhich is targeted towards the object to be investigedwhile in passive remote sensing, sun providesenergy for remote sensing.
51. (d)
Spectral resolution: Ability of a sensor to definefine wavelength intervals. It refers to number of bandsin which sensor can take measurement.
Temporal resolution: It refers time gap betweentwo successive image of same area.
Spatial resolution: It refers to amount of detailthat can be detected by a sensor/smallest unit-area measured.
Radiometric resolution: It is the ability of sensorto discriminate very slight differences in energy.
52. (b)
Ideal fluids are assumed to be both incompressibleand inviscid.
Water is incompressible, so it is more nearto an ideal fluid than air.
There is norelative motion b/w the particlesand every particle having same velocity.
53. (c)
Laminar boundary layer thickness over a flat plateis
ex
5x R
2
4
2 10 54 V 4
102
22 10 v 2 20
10
v 25 m/s
54. (c)
u 2xx
v 2y
For steady, incompressible fluid flow :
u v 0x y z
2x + 2 + 0z
w = – (2x + 2) z + f(x, y).
55. (c)
Circulation = vorticity × Area.
Vorticity = zv u2W 8 4 4x y
Area A1 = 2 × 1 = 2
(2,4)
(2,2)
(3,4)
(3,4)
y=4
y=2
x=2 x=3
A1
Circulation = 4 × 2 = 8.
56. (a)
If cross sectional area of throat is reduced muchthen pressure may fall below vapour pressure andcavitation may occur.
IES M
ASTER
(8) Full Syllabus
57. (c)
2 2 2
L0.5V fLV V200 180 H
2g 2gD 2g
2V 0.04 20020 1.52g 1
220 2 10 V9.5
20 220 m sec199.5
/
58. (d)
For dynamic similarity
(Re)m = (Re)p
m p
VD VD
air m
air p
µ1
Temperature = Constant so P = constant
pP
m m
P 1P 10
p
m
V 1 110V 20 2
Vm = 800 × 2 = 1600 kmph
59. (a)
If boundary layer is turbulent throughout
1/50.376
x Re x
For 5 × 105 < Re < 107
1/55
0.376 0.3761 1 2 1032 10
= 0.0188m = 18.8 mm
60. (a)
Laminar boundary layer on smooth plate.
5 xx Rex
and 0
020
0.664 1V / 2 Re x x
Turbulent boundary layer on smooth plate.
4/5
1/50.376 x
x Re x
0 1/51
x .
61. (c)
Euler’s model law :When apart from intertial forces, only pressure forceis dominant, this law is applicable. In this case :
um pEu E
r
r
r
V 1P
This law is applicable
• In pipe flow at high pressure [viscous effectsare neglected]
• In case of cavitation
• Pressure due to sudden closure of valve.
62. (c)
N H
1 1
2 2
N HN H
2
250 200N 400
2N 250 2 = 353.5 rpm.
63. (c)
For irrigation purposes we required more dischargehence axial flow pumps is more reliable.
64. (d)
Fluid slip : Under certain circumstances, the angleat which the fluid leaves the impeller may not bethe same as the actual blade angle is called fluidslip.
65. (d)
Aluminium manufacture is accomplished in twophases; the bayer process of refining the bauxiteare to obtain aluminium oxide, and the Hall-Heroultprocess of smelting the aluminium oxide to releasepure aluminium.
66. (d)
Slaking is also known as hydration of lime.
Hydration2 2CaO H O Ca OH Heat
The amount of water used is 2 or even 3 timesgreater than theoretical water depending uponcomposition, degree of burning and slakingmethods, because a part of the water used isvaporized by the released heat.
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-04) (9)
67. (c)
The proportion of various ingredients of bricks areas follows :
Silica 50 - 60%
Alumina 20 - 30%
Lime 10%
Magnesia 1%
68. (c)
AsCu is a preservative composed of
2 5 2
4 2
2 2 2 2 7
As O 2H O 1 Part by weightCuSO 5H O 3 Part by weightK Cr / Na Cr O 4 Part by weight
69. (a)
A concrete mix with excess amount of coarseaggregate will lack sufficient mortar to fill the voidsystem resulting in loss of cohesion and mobilityand mix formed will be harsh in nature.
70. (d)
Total load = 20 + 10 = 30 kN
Eccentricity = 20 50 33.33 mm20 10
71. (b)
Low alkali content in cement requires high dosageof air extraining admixture.
72. (d)73. (a)
Critical section will be at face of support SF atface of support
300mm
3m
800mm
Vu = 20 × 3 × 1.5 + 50 × 3 = 240 kN
Vu = factored S.F at support)
Mu = 20 3 3 31.5 50 3 360 kN-m
2 2
v =
3u
3uM 360 10 1V tan 240 10d 0.8 6bd 800 300 800 300
= 0.6875 N/mm2
74. (b)
Development length Ld = s
db4
Since deformed bars are used in compression, thevalue of design bond stress will be increased by60% and 25% for use of deformed bars and acompression respectively.
d
16 0.87 415L4 1.6 1.25 1.4
= 515.8 mm
75. (b)For short column emin < 0.5 × lateral dimension
Puz = 0.4 fck Ac + 0.67 fy Asc
Ac = Ag – Asc
Ac = 400 × 400 – 24 254
= 158036.50 mm2
Puz = 0.40 × 20 × [158036.5] + 0.67 × 415 ×
24 254
= 1810 kN
76. (a)For waist slab spanning transversly, the tangentialcomponent causes waist slab to bend in its ownplane.
77. (c)
In post-tensioning the transfer of prestress isindependent of transmission length.
In post-tensioning large anchorage devicesare used while in pre-tensioning largeanchorage device is not required.
78. (b)Correct sequence of post-tensioning option is
�1 – 4 – 6 – 3 – 5 – 2
79. (a)
Ka = 1 sin30 11 sin30 3
A
B
= 16 kN/m3
= 30°µ = 0.45
pA = 21 15 5kN / m3
pB = 21 15 4 16 26.33 N/m3
IES M
ASTER
(10) Full Syllabus
Thrust on wall, P = 21 5 26.33 4 62.67 N/m2
Facor of safety against sliding = WP
= 0.45 100
62.67
= 0.72
80. (b)
81. (a)
3 3
v v
s s
v s v s3 3
v v3 3
s s
Sample ‘A’ Sample ‘B’e 1 e 2
v 1m v 1mV V
e e 2V V
V V V 2V
V 0.5m V 0.666m
V 0.5m V 0.333m
As both the soil samples are mixed andcompacted, therefore Vs of sample = VSA +VSB = 0.50 + 0.333 = 0.833 m3
As the samples is compacted to volume of1m3.
Vv = sample = 1–0.833 = 0.167 m3
porosity = vV 0.167 0.167 0.17V 1
n 0.17 100 17% Ans.
82. (a)
L d
d
V VV = 36%
p d
d
V VV = 24%
VL – Vd = .36 Vd
VL = 1.36 Vd
Vp = 1.24 Vd
S.R. =
L P
d
L P
V VV
W W
=
d d
d
1.36 V 1.24 VV
10044 32
S.R. = 12 112
.
83. (a)
Using Kozeny-Carman formula3eK
1 e
1
2
KK =
31
13 32
2
e 11 e 2
e 1.52.51 e
1
2
Kk = 0.37
K2 = 0.012 0.0324cm sec0.37
84. (d)sand
6m = H day
sandGiven, Cv = 5 × 10–4 cm2/sec
V = 50% = 0.50
Tv = 20.50 0.1963 0.197
4For double drainage condition,
d = H 6 3m 300cm2 2
Time required for 50% consolidation
T50 =
2 2v,50
4v
T d 0.197 300C 5 10
= 410.41 days
85. (d)
1. Soil structure on dry side of optimum isfloculated due to predominance of attractiveforce but on wet side of optimum, repulsiveforce is dominant hence structure is dispersed.
2. Permeability depends upon the size of voids.On the dry side of optimum, soil structure isflocculated and have more and larger voidshence more permeability occurs.
3. Pore water pressure is more on wet side ofoptimum because there is excess water onthis side and there is deficiency of water ondry side of optimum and having less porewater pressure.
4. Soils compacted on dry side of optimumshrinks less on drying than wet side becuase
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-04) (11)
of small water content as compared to wetside.
86. (c)
Compactive effort per unit volume
= 6 325 5 4.9 0.45 9.81 J.
1000 10 m
= 2703.88 kJ/m3.
87. (c)
Newmark’s charts are based on Boussinesq theorywhile Fenske charts are based on Westergaard’stheory.
88. (c)
1 260 20 240kPa
3 100 20 80kPa
From, 21 3 tan 45 2c tan 45
2 2
2240 80tan 452
3 tan 452
60 = 45 + 2
30
3cos2
89. (c)
90. (b)
Given; B = 3 m
L = 5m
Df = 2 m
As per skempton
Net ultimate bearing capacity, qnu = cNc
Nc =
fD B5 1 0.2 1 0.2B L
=
2 35 1 0.2 1 0.23 5
= 6.347
qnu = 50 × 6.347
= 317.33 kN/m2.
91. (a)
Due to densification of sand caused during drivingthe piles, point resistance increases which isapproximately equal to 2 to 3 times of pointresistance is in case of bored piles.
92. (a)
Ar = 2 20 i
2i
d d100
d
12.5 =
2 20
2d 75
10075
do = 79.55 mm
Wall thickness = 79.55 75 2.27 mm
2
93. (b)
NCH = TH – Alkalinity, (If –ve take zero)
Total hardness = 50 50160 4020 12
= 567 mg/L as CaCO3
TH = CH + NCH
NCH = 567 – 200
= 367 mg/L as CaCO3
94. (b)
Dosage of Cl2 applied
= 6
38.5 10 (mg / d) 0.34 mg / l25000 10 (l / d)
Cl2 demand = 0.34 – 0.2 = = 0.14 mg/l
95. (b)
filtration is a process for separating suspended andcolloidal impurities from water by passage througha porous medium or porous media.
Removal of turbidity is essential not only for therequirement of aesthetic acceptability but also forefficient disinfection which is difficult in the presenceof suspended and colloidal impurities.
96. (b)
The amount of the suspended solids that volatilizesat 500 ± 50°C is taken to be a measure of activebiomass concentration. The presence of nonlivingorganic particles in the influent wastewater willcause some error (usually small) in the use ofvolatile suspended solids as a measure of biomass.
Mixed liquor volatile suspended solids is a measureof the active biological mass in the aeration tank.
97. (c)
Sludge age = w u
VXX
w uX = 2000 kg/day
IES M
ASTER
(12) Full Syllabus
V = 3
3m 124000 4hr 4000mday 24hr/day
X = MLSS = 2500 mg/l
3
64000 10 2500 mg/Sludge age 5days
2000 10 mg/dayl l
98. (b)
Stabilization is the process of converting the organicsolids to more refractory (inert) forms so that theycan be handled or used as soil conditioners withoutcausing a nuisance or health hazard throughprocesses referred to as digestion. These arebiochemical oxidation processes.
99. (c)
Higher the F/m ratio, lower will be the BODremoval.
Increase in sludge wasting rate reduces sludgeage which reduces nitrification.
Filamentous bacteria growth is promoted byhigher sludge age, lower f/m ratio and highertemperature.
100. (b)
101. (b)
Kavg = 0.3 0.9 0.4 0.3 0.3 0.15
1 = 0.435
Q = 1 KiA
36
2 = 1 0.435 7 A
36A = 23.645 ha
102. (d)
103. (d)
104. (d)
Concentration of OH– ion = 10–5.8 mmol/litre
= 10–5.8 × 10–3 mmol/litre
10–8.8 mol/litre
pOH = – log [OH–]
= – log 10–8.8
= 8.8
pH = 14 – 8.8
pH = 5.2
105. (c)
Maximum slenderness ratio
1. A tension member in which reversal of directstress occurs due to load other than wind orearth quake load = 180
2. A member subject to compressive forcesresulting only from combination of wind/earthquake actions, provided the deformation ofsuch a member does not adversely effectsstress = 250
3. A tension member in which, reversal of directstress is due to wind or earth quake = 350
106. (d)
Floor Beam - A major beam of a floor systemusually supporting joists in a building, a transversebeam in a bridge floor.
Girder - In Building, girders are the same as floorbeam, also a major beam in any structure. Floorbeams are also reffered as Girders.
Girt - A horizontal member fastened to and spanningbetween columns of an industrial building, used tosupport wall cladding such as corrugated metalsheeting.
Joists - A beam supporting floor construction butnot a major beam.
107. (a)
No. of plastic Hinges = DS + 1 = 2
Case I: Let Hinge formed at C and B
MP
P 3MP
P P PM M 3M P 3a
5 33 5
P P3M 4M P 3a5
P17M P 3a5
P17MP15a
Case 2: Let Hinge formed at B and D
MP
P 3MP
5a 3a
P PM 4M P 3a
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5a 3a
53
PP
20MM 3aP3
P23MP9a
Case III: Hinge form at C and DP Mp2a
MP
3a
P PM 2M P 3a
3a 2a
32
P PM 3M 3a P
= P4M P3a
Collapse load = P17M15a
108 (d)
(1) No. of Plastic Hinge = Ds + 1
So, always more than one Plastic Hinge isrequired for collapse.
(2) Plastic appreach yields thin sections.
(3) Lower bound theorem is more conservative.
109. (b)
Direct vertical shear in each bolt, FDV = 48 12kN4
Here bolt (4) is critical bolt as it is nearest to forceline and farthast from CG of bolt group.
Torsional shear in bolt (4) FTV =
248 175 250 135
r
FTV = 2 2
48 425 1352 75 2 135
FTV = 57.73 kN
Resultant force in bolt (4) R = FDV + FTV = 12 +57.33 R = 69.73 kN
110. (a)
Intermediate vertical stiffeners are provided in plate
girders to eliminate web buckling under inclinedcompressive stress by dividing the web plate intosmall panels.
111. (c)
Earthquake loads are not significant for roof trussbecause of small self weight the following loadcombination can be considered.
1. Dead load + Show load, if applicable
2. Dead load + Partial/full live load
3. Dead load + wind load + Internal positive airpressure
4. Dead load + wind load + Internal suction airpressure.
112. (b)
Standard equation of motion,
mx cx kx 0
Thus m = 3 kg, C = 8 N-sec/m, k = 27 N/m
Critical damping coefficient
Cc = 2 Km
= 2 27 3
= 18 N-sec/m
113. (d)
114. (a)
3Kx
K
K1
2m
1
1 1 1K 2K K
11
1 3 2KKK 2K 3
eq
1 1 12KK 3KK3
eq
1 3 1 9 5K 5K 3K 15K
Keq = 15K14
Natural frequency, eq
nK 15K2m 28 m
115. (b)
Allowable moisture depth depletion = 0.6 × 120 × 0.25
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= 18 mm
Frequency of irrigation = 185
= 3.6 days
116. (a)
Lining of canal increases channel capacityby allowing water to move faster.
Lining of canal prevents water logging.
117. (b)
Uplift pressure head = 8 – 3 = 5m
Let the thickness of the floor required be t
G t 5
t = 5 1.92 m
2.6118. (b)
119. (a)
Setting = outlet index (m)
channel index (n)For trapezoidal channel, n = 5/3
for orifice outlet, m = 1/2 setting = 1/2 3/105/3
For open flume type outlet, m = 3/2 setting
3/2 95/3 10
120. (c)
Load factor
= Average load over a certain peiod
Peak load during that period
= 4,00,0006,00,000
= 0.67
Capacity factor = Average loadPlant capacity
=
4,00,0001000 0.58000
121. (b)
Time period T = I2
WGM
Comfort in ship is more when time period is largethus GM should be less. But small GM decreasesthe stability of ship. For passenger ship GM isless balance between stability and comfort isensured.
122. (b)
In river models, distorted models are used. Ifundistorted model is used the depth will becomevery small and hence.
(a) There will be difficulty in measurement
(b) Depth becomes so small that flow in modalmay not remain turbulent as is there in caseof original river.
123. (b)
Let us consider a fluid having dimensions ofd dy 1 l variation of shear and pressure on varioussides on as shown in the figure for a twodimensional condition.
P.dy×1
dy d 1y
l
PP d dy 1 l
l
–dhdl
For steady uniform flow, acceleration in the directionof flow will be zero.
P dhd dy dy.d d dy 0y d
l l + l
l l
If h = 0 (for horizontal flow) dl dx
dp ddx dy
In equation is valid for both laminar and turbulentflow.
124. (a)
Smaller jet ratio means smaller diameter of thewheel resulting in closer spacing of buckets whichin turn reduces turbine efficiency.
125. (a)
126. (c)
Conveyance, K = 2/31 ARn
=
2/31 AAn P
2/31K
P
127. (c)
Gross irrigation requirement is the sum of fieldirrigation requirements and the amount of irrigation
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water lost in conveyance through the canal system(by evaporation and by seepage).
128. (b)
Smoothness of the pavement plays a major factorfor safe and comfortable driving and thus frictionbetween the wheel and the pavement surface is acrucial factor. There is no relation in smoothnessof pavement and unevenness.
Unevenness index is a measure of unevennesswhich is the cumulative measure of verticalundulation of the pavement surface recorded perunit horizontal length of the road. It affect the vehicleoperating cost, speed, riding comfort, safety, wearand tear of tyres etc.
129. (b)
Level of service is a qualitative measure de-scribing operational conditions within a trafficstream and their perception by driver. it gener-ally describes these conditions in terms offactors such as speed, travel time freedom tomove, traffic interruption, comfort and conve-nience safety etc.
Capacity is the maximum volume that can beaccommodated on the road.
If level of service of road is not good then speedof vehicle decreases thus heading towardsmaximum capacity.
130. (b)
The purpose of the wearing course is to give asmooth riding surface that is dense. It resistspressure exerted by tyres and takes up wear andtear due to traffic. Wearing course also offers awater tight layer against the surface waterinfiltration. In flexible pavement normally abituminuous surfacing is used as a wearing course.In rigid pavements, the cement concrete acts likea base course as well as wearing course.
131. (b)
Maximum landing weight is less than take-off weightdue to fuel consumption.
132. (a)
133. (d)
To avoid potential problems with frost heave thebest material for structural fill will be well gradedsoil because well graded soil is least susceptibleto frost heave.
134. (b)
In rapid sand filtration, as the water is applied inupflow mode to a granular medium or media,frictional resistance is offered by the filter grainsdue to skin friction and form drag. The initial effectat low velocities of flow is to result in reorientationof particles to minimize frictional resistance. At lowbackwash velocities, the filter bed does not expandand its porosity does not change. The head lossor pressure drop is a linear function of upward flowvelocity at low velocities. As the water velocity isincreased, the frictional resistance also increasestill it reaches a value equal to the gravitationalforce acting upon the filter grains. Any furtherincrease in the velocity of water fluidizes the filterbed resulting in bed expansion and increasingporosity.
135. (b)
The coliform group of organisms survives in naturalwaters for relatively longer periods of time, but doesnot reproduce effectively in this environment. Thus,the presence of coliforms in water implies fecalcontamination rather than growth of the organismbecause of favorable environmental conditions.These organisms also survive better in water thanmost of the bacterial pathogens.
136. (d)
One of the biggest problems with clay tiles is theirtendency to crack or shatter due to freezing andthawing cycles in cold regions. Because of this,they are generally used only in warmer climates.
137. (d)
Large end of timber is kept pointing upstream sideduring water seasoning.
138. (b)
139. (c)
Angle of twist, x
0
TdxGJ
Twisting moment will be constant for all crosssections. Angle of twist increases as the distanceof cross section from free end increases.
140. (a)
141. (a)MOR = 0.87 fy Ast (d – 0.416 x4)
= 0.87 fy Ast · z
Upto balance point, change in z is marginal withincrease in pt, hence MOR is linearly varying withAst (or pt).
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142. (c)
Minimum shear reinforcement is provided to arrestthe longitudinal cracks on the side faces due toshrinkage and temperature variation.
143. (a)
144. (b)
Soft storey is the one in which lateral stiffness isless than 70% of that in the storey above or lessthan 80% of the average lateral stiffness of thethree storey above.
145. (a)
146. (b)
147. (a)
Compression members have to resist buckling.Moreover imperfection of cross section dimensions,
eccentricity of loads, lateral loads, residual stressesare other factors which contribute to unwantedstresses making compression members morecritical.
148. (a)
If the members of the horizontal grid are assumedto be rigidly connected and to be subjected tovertical loading only Mz, Mx and Fy will remain.
If there is no horizontal force, then moment due tohorizontal force will be zero.
149. (d)Hoes are primarily used to excavate below groundlevel. They are adopted to dig trenches, pits andbasement.
150. (d)Rubber tyred equipment will slip on smooth surfaceand crawler tractors works better on loose or muddysoil.
