answers - ies master · begining of test (t = 0) u = excess pore pressure after the time t. so, u i...
TRANSCRIPT
1. (c)
2. (a)
3. (a)
4. (a)
5. (a)
6. (a)
7. (a)
8. (d)
9. (c)
10. (d)
11. (c)
12. (b)
13. (d)
14. (c)
15. (d)
16. (b)
17. (d)
18. (b)
19. (b)
20. (a)
21. (d)
22. (d)
ESE-2019 PRELIMS TEST SERIESDate: 09th December, 2018
23. (a)
24. (d)
25. (b)
26. (a)
27. (b)
28. (b)
29. (b)
30. (d)
31. (d)
32. (b)
33. (c)
34. (a)
35. (c)
36. (c)
37. (c)
38. (d)
39. (c)
40. (c)
41. (b)
42. (c)
43. (b)
44. (c)
45. (a)
46. (c)
47. (d)
48. (d)
49. (c)
50. (a)
51. (a)
52. (c)
53. (a)
54. (c)
55. (a)
56. (b)
57. (d)
58. (c)
59. (c)
60. (b)
61. (b)
62. (a)
63. (b)
64. (d)
65. (c)
66. (a)
ANSWERS
67. (a)
68. (d)
69. (d)
70. (b)
71. (d)
72. (a)
73. (c)
74. (c)
75. (c)
76. (b)
77. (d)
78. (d)
79. (a)
80. (b)
81. (d)
82. (c)
83. (c)
84. (c)
85. (d)
86. (b)
87. (b)
88. (d)
89. (c)
90. (c)
91. (a)
92. (c)
93. (c)
94. (c)
95. (c)
96. (c)
97. (c)
98. (b)
99. (b)
100. (b)
101. (b)
102. (d)
103. (b)
104. (d)
105. (c)
106. (c)
107. (c)
108. (b)
109. (b)
110. (b)
111. (a)
112. (c)
113. (d)
114. (a)
115. (c)
116. (b)
117. (c)
118. (d)
119. (a)
120. (a)
121. (a)
122. (d)
123. (b)
124. (d)
125. (a)
126. (c)
127. (a)
128. (c)
129. (b)
130. (a)
131. (a)
132. (c)
133. (c)
134. (b)
135. (a)
136. (b)
137. (b)
138. (b)
139. (a)
140. (a)
141. (d)
142. (a)
143. (d)
144. (d)
145. (a)
146. (b)
147. (a)
148. (c)
149. (a)
150. (d)
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (2)
1. (c)
2. (a)
xdu 3dx
yzdw dv 0 0 0dy dz
3. (a)
Modulus of elasticity = Yield stress
strain
= 2240 N / mm
0.0012 = 20 × 104 N/mm2
= 2 × 105 MPa
Modulus of elasticity in compression =Modulus of elasticity of tension.
4. (a)
5. (a)
P MA Z
Comp,max 4
P eP P PedA Ad Ad 2864
P e1A d 8
ten, maxP e1A d 8
comp, max
ten,max
1 e d 8 131 e d 8 3
de5
6. (a)
Cr E
1 1 1 1 11P P P 600 400 240
P = 240 kN
7. (a)
8. (d)
9. (c)
Due to Poisson’s effect, longitudinal strainleads to lateral strain which will lead toanticlastic curvature for an originallyrectangular cross-section.
For a doubly symmetric section, shear
centre always coincide with the centroid ofthe section.
10. (d)
From Mohr circle :
(120, 0)
( , 50)n
(20, 0) (70, 0)
Radius of Mohr Circle = 50 MPa
20MPa
Maximum shear stress = 120 20
2
= 50 MPa
max 50 20 30 MPa
11. (c)
1 2 1 2max, abs max , ,
2 2 2
For thin spherical cells, 1 2and are same equal
to pd4t .
max, abspd8t
12. (b)
Maximum permissible stress Circumferentialstress
pd802t
2t 80p
d
2 50 80p 3.2 MPa
2500
Therefore max internal pressure allowed = 3.2MPa.
13. (d)
Analyse only half beam,100 kNm
B
100 = B4EI5
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (3)
B 4100 5
4 2.5 10
= 0.005 radian
14. (c)
M = MOA + MOB + MOC
oOA
4EIM
L
oOB
3EIM
L
oOC
3EIM
L
o10EIM
L
oOC
3EI 3M ML 10
OCc
M L 3M L ML6EI 10 6EI 20EI
15. (d)
Influence line diagram for reaction at ‘C’
100
38
5 1318 8
3 m 8 m 5 m
1 m
120
DCBA
Maximum reaction at support C
= 13 100 120 1 282.5 kN8
When 100 kN load is at A
Reaction at C = 3 100 37.5 kN8
Case II : Reaction at C when 120 kN load is atA
1.5120
100
2/8
3/8
= 3 2120 100 20 kN8 8
16. (b)
No. of members, m = 8
No. of reactions, r = 7
No. of joints, j = 8
Additional DOF = 4 in internal hinge + (3 – 1) injoint hinge = 6
Dk = (3j – r) + additional DOF – No. of inextensiblemembers
= (3 × 8 – 7) + 6 – 9
= 14
17. (d)
Members, AK, BJ, DI, DE, EF, HG and GF arezero force members.
18. (b)
Flemish Bond gives better appearance thanEnglish Bond.
• Construction of Flemish bond requiresgreater skill in comparison to English Bond.
19. (b)
The inner annual ring surrounding the pithconstitute the heart wood. It indicates deadportion of tree and doesn’t actively take part ingrowth of tree. It imparts rigidity to the tree andhence it provides strong and durable timber forvarious engineering purposes.
20. (a)
Initial setting time time limit for handlingthe cement mortar.
Final setting Time beginning of developmentof mechanical strength.
21. (d)
22. (d)
23. (a)
Absorption of FA = 1 846
100 = 8.46 litres
Absorption of CA = 0.5 1152 5.76 litres100
Total amount of water to be used = 150 + 8.46+ 5.76 = 164.2 Kg/m3
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (4)
FA = 846 – 8.46 = 837.5 Kg/m3
CA = 1152 – 5.76 = 1146.2 Kg/m3
24. (d)
All of above parameters effects swelling pressureof soil.
25. (b)
26. (a)
Max. dry unit weight obtained by compaction
maxw
dG 2.7 10y
wG .30 2.71 1s .80
= 313.42 kN/m
27. (b)
In electrical analogy method an analogy existsbetween the darcy law and ohm’sn law:
Analogous quantities
Flow of water Flow of current
1.hLaw k AL
E.I k AL
2. Discharge ( ) current (I)
3. Head (h) voltage (E)
4. Length (L) Length (L)
5. Area (A) Area (A)
6. Permeability (k) Conductivity (k)
28. (b)
degree of consolidation given as
i
i
U U%U 100U
where, Ui = initial excess pore pressure at thebegining of test (t = 0)
U = excess pore pressure after thetime t.
So, Ui = 200 kN/m2
U = 60 kN/m2
200 60%U 100 70%200
29. (b)
In U.U test, no drainage is permitted duringthe consolidation stage. The drainage isalso not permited in the shear stage.
30. (d)
It approximates the curved failure envelopeby a straight line, which may not give correctresults.
For some clayey soils, there is no fixedrelationship between the normal and shearstresses on the plane of failure. The theorycannot be used for such soils.
31. (d)
The following assumptions were taken in Tezaghi’sbearing capacity theory
(i) The base of footing is rough.
(ii) The footing is laid at a shallow depth. i.e.
fD B .
(iii) The shear strength of soil above the baseof footing is neglected. The soil above thebase is replaced by a uniform suchargerDf.
(iv) The load on the footing is vertical and isuniformly distributed.
(v) Fooling is long ie. L/B is infinite. where Lis length of footing and B is the width.
32. (b)
= wG 2.7 10 19.31 e 1 0.4
kN/m3
Coefficient of earth pressure = 1 sin at rest(K0) = 0.5
v at 2.7 m = 19.3 × 2.7 kN/m2
= 19.3 = 52.1 kN/m2
H = 0.5v = 26 kN/m2
q = v H1 13.052
kN/m2
P = v H1 39.052
kN/m2.
33. (c)
Side of cube a = 10 mm = 0.01 m
Total surface area of cube A = 6a2 = 6 × 0.012
= 6 × 10–4 m2
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (5)
Volume of cube V = (0.01)3 = 10–6 m3
Specific surface area of cube
= 4
16
6 10 600 m10
34. (a)
Available shearing resistance,
2f subc zcos tan
= 15 + (20 – 10) × 4 cos245° tan30°
= 26.54 kN/m2
Mobilised shearing resistance,
= satzcos sin
= 20 × 4 × cos45° × sin45° = 40
F.O.S. = 26.54 0.66
40
35. (c)
36. (c)
S = 37t100(1 (0.63) )
= 100 [1 – (0.63)1.5] = 50%
37. (c)
For trickling filter
Surface area = 0 eQ QHydraulic Loading Rate(including recirculation)
= 375 2 10
20
= 7500 m2
Surface area excluding recirculation
=
3
275 10 6250m12
Surface area = 750 m2
38. (d)
All the three are possible reasons for increase inacidity of sludge.
39. (c)
Li = 14 (Qs)1/3
= 14 × (27)1/3 = 42 m
40. (c)
– Water pipes are not designed for self
cleansing velocity.
– Sewer pipes are laid on continuous gradientto permit gravity flow.
41. (b)
Air required = air 2
w% of O
=
3
1200 kg / day1.15 kg / m 0.21
= 4968.94 m3/day
O2 transfer efficiency of aeration equipment =10%
Air needed per minute =
4968.940.1 60 24
= 34.50 m3/min
42. (c)
G = PV
= 3
410 10
10.03 10 10.8
= 960.81 sec–1
43. (b)
Vs = d
dt
td = s
dV = 4
34.2 10
td = 1.98 hr
44. (c)
Assume total solids in undigested sludge
= 100 kg
Volatile solids = 60 kg
Fixed solids = 40 kg
Fixed solid remain unchanged after digestionassume volatile solid after digestion = x kg
Given, x = 0.3 (x + 40)
x = 17.15 kg
% volatile reduced to gases
=
60 17.15 100 71.4
60
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (6)
45. (a)
50
50w
5
5w
w w w15 50 53
Seepage :Pressure at Drainage gallery
= w w w1H' [ h H']3
= w w w15 [50 5 ]3
= w20 .
46. (c)
Silt excluder up stream of head regulator
Silt ejector down stream of head regulator..
47. (d)
48. (d)
T.R. =
Total mass transpired by the plant during its full growth
Mass of dry matter produced
Mass of dry matter produced is generally takenas entire mass of plants including its roots.
49. (c)
50. (a)
The integration of the hydrograph gives the flowmass curve and mathematically it is expressedby
V(t) = t
0
Q t dt
51. (a)
AI = PET AET 100
PET
= 150 120 100 20%
150
52. (c)
Probability of exceedence of flood = 1 0.02
50
Probability of occurring at least once in 100 yrs
= 1 – (1 – 0.02)100
= 1 – (0.98)100
= 0.87
53. (a)
54. (c)
Relative height (jump)
= 2 1j
1
y yh (height of jump)
E
Relative initial depth = 1
1
yE
Relative sequent depth = 2
1
yE
55. (a)
1
1
v 81 9.03 9gy 9.81 8.2
2
1
y 1 1 1 8 81y 2
2
1
y0.5 1 649
y
2
1
y 12.23y
y2 = t8.2 12.23 100.35 m y
Free repelled jump.
56. (b)
Using direct step method
0 f
ExS S
2 1
1 2
E Ex
10 (sf sf )2
5 2 3x 30 m1 0.1(0.15 0.5)2
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (7)
57. (d)
q = 35 5m /s/m1
v = 5 2.5 m/s2
2 2v 2.5 0.3125 m2g 2 10
v 2.5 2.5Fgy 10 2 20
= 0.56 < subcritical
E1 = 0.312 + 2 = 2.31 m
E2 = 2.31 – 0.10 = 2.21 m
yc =
1 12 23 3 1
3q 5 (2.5) 1.35g 10
Ec = 1.5 × 1.35 = 2.025 m
Since E2 > EC
Upstream depth will remain unchanged.
58. (c)
59. (c)
60. (b)
Correction = 30.1 30 100 3.33 m
30
Corrected length = 1000 + 3.33 = 1003.33
61. (b)
62. (a)
Quartzite is a non-foliated metamorphic rock.
Parent rock can be any of three types ofrock igneous, sedimentary or metamorphic.
63. (b)
Hypocentre: it is a location of earthquake in acrustal plate. It is also known as focus.
Epicentre: The projection on the surface of earthdirectly above the focus.
64. (d)Limitation of bar chart:
1. It doesn’t indicate the critical activities ofthe project.
2. The financial aspect involved in the projectis unknown from the bar chart. Becausecritical path can’t be identified, there is noquestion of crashing activites.
65. (c)Calculation of expected time,
tE = 0 m pt 4t t
6
EActivity t1 2 52 3 83 4 104 5 53 5 7
1 2 3
4 5
5 8
10
5
7
TE = 28 days
As, Z =
s ET T
=30 282
= 1
P = 85%
66. (a)
Free float, Ff = j i ijEET T t
67. (a)
As intefering float is equal to slack of head eventfor this we have to plot TE & TL in network diagram.
0 0 2
2
77
1010 12 12
9 955
6 8FA
2
2
B3
D5
E5
H 1
I2
G3
C4
ST for F = 10 – 10 = 0
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (8)
68. (d)
T(time) C(cost)
Q(quality)
AB
D
E
at T, Q and C, time quality and cost respectivelyare the only individual dominant factors.
at pt B, time and quality are controlling factorsonly.
at pt. D, time and cost are controlling factorsonly.
at pt. E, quality and cost are controlling factorsonly.
at pt A, a balance is strunk b/w the 3 constraintsof construction project management i.e quality,time and cost.
69. (d)70. (b)71. (d)
For the crawler tractor—
Coefficient of traction is high as 0.9
Speed is low
Most efficient in muddy area and in flatarea it is not efficient.
High pulling affort.
Capacity is high.
72. (a)
Years purchase is defined as the capital sumrequired to be interested in order to receive a netannual income of 1` at a certain rate of interest.
Y.P. = c
100i i (when sinking fund is also recovered)
=
100 12.55 3
73. (c)
Assuming t = 2.5 secs, f = 0.35 for V = 80 kmph
As there is ascending gradient on one side ofsummit and descending gradient on the otherside, the effect of gradients on the SSD isassumed to get compensated and hence ignoredin the calculations.
SSD = 0.278 Vt + 2V
254f
= 0.278 × 80 × 2.5 + 280
254 0.35
= 127.6 m 128 m
N = 0.03 – (– 0.05) = 0.08
Assuming L > SSD
2 2NS .08 128L4.4 4.4
= 297.9 > SSD ok
L = 298 m
74. (c)
75. (c)
Lc = concrete
2f
= 2 16 0.88 m
1.5 24
76. (b)
77. (d)
H.C. = 8 25 0.2 20 tonnes
2
78. (d)
79. (a)
The maximum harbour depth below lowest lowwater is
= Loaded draft + 1.2 m for soft bottom= Loaded draft + 1.8 m for rock bottom
80. (b)
In general, lining of 2.8 cm per 30 cm of borediameter is provided.
Therefore, thickness of lining = 8002.8300
= 74.67 cm 75 cm
81. (d)
Size of shaft is governed by following factors:
(i) Amount of muck to be hoisted.
(ii) Hoisting system to be adopted.
(iii) Size of muck car.
(iv) No of workmen.
(v) Type of equipments to be used.
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (9)
82. (c)
The Buoyancy force due to mercury.
1BF = 1 1gV = 13.6 × 1000 × 9.81 × V1
The Buoyancy force due to water
2BF = 2 2gV = 1000 × 9.81 × V2
Given : 1 2V V V total volume of metallic
body.
From the Archimedes principle of floatation.
W = 1 2B BF F
gV = 13.6 × 1000 × 9.81 V1 + 1000 × 9.81× V2.
4.5 × 1000 × 9.81 × V = 13.6 × 1000 × 9.81 ×V1 + 1000 × 9.81 × (V – V1).
4.5 V = 13.6 V1 + V – V1
3.5 V = 12.6 V1
V1 = 27.77% of V
V2 = 72.22% of V.
83. (c)
Electrical analogy method : This method is basedon the fact that the flow of fluids and flow ofelectricity through a conductor are analogous.These two systems are similar in the respectthat
(i) Electric potential is analogous to the velocitypotential
(ii) The electric current is analogous to thevelocity of flow, and
(iii) The homogeneous conductor is analogousto the homogeneous fluid.
84. (c)
Applying Bernoulli's equation when valve is closed.
V1 = V2 = 02 2
1 1 2 11 2
P V P VZ Z2g 2g
rh
o
2 11 2
P PZ Z
3
1 270 10z z 7m
10 1000
Now when valve is open2 2
1 1 2 21 2 L
P V P VZ Z h2g 2g
1 21 2 L
P PZ Z h
1 2V V
23 + 7 = hL Lh 30m .
85. (d)
Dimensionless parameter to specify instability2 du
dy
this is zero at near the wall (y = 0)
and at the centre of pipe du 0dy
. This
instability is maximum somewhere between thesetwo.
86. (b)
Thickness of Hydrodynamically smooth boundary
kinematic wiscosity 1Flow velocity
.
87. (b)
Oil (G = 0.6)
water (G = 1)o
A
B
B
C
D
Downward force on AB
=
20.8 0.8 0.8 1 (10 0.6)
4= 0.82 kN
Upward force on BCD
= 21.6 0.8 1 0.6 10 (1.6) 10 1
8` = 17.73 kN
Net force (vertical) = 17.73–0.82 = 16.9 kN
88. (d)
u = 2y, v = 8x + 2
Equation of streamline, dy vdx u
dydx =
8x 22y
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (10)
2y dy = (8x + 2) dx
Integrating both sides, we get
y2 = 4x2 + 2x + c
y2 = 2x (2x + 1) + c
89. (c)
Displacement thickness, * =
00
u1 dyu
= 0
y1 dy
=
2
0
yy2
=2
Momentum thickness, =
0
0 0
u u1 dyu u
= 0
y y1 dy
=
2 3
20
y y2 3
= 6
shape factor =
* / 2 3
/ 6
Alternatively; for
m
o
u y * m 2u m
Hence shape factor =
1 2 3
1
90. (c)
For maximum efficiency, H = 3hf
H = 2
53f l Q12.1d
100 =
2
53 0.03 50 Q
12.1 (0.5)Q = 2.9 cumecs
= 2900 l/sec.
91. (a)
Head loss in pipes connected in parallelwill be same irrespective of their sizes andlengths.
Water hammer pressure is more in rigidpipes than flexible pipes.
92. (c)
Rankine efficiency of hydraulic ram = q(H h)
Qh
=2.5(30 3) 100
40 3
=2.5 27 100
120
= 56.25 %.
93. (c)
Slip = Qth – Qactual
% Slip = th act
th
Q Q100
Q
When the slip becomes 'negative' it is callednegative slip, Qactual > Qtheoritical.
Negative slip occurs when suction pipeis long, delivery pipe is short and speed of rotationis high. In this case the delivery valve opensbefore the suction stroke is completed due toinertial pressure in suction pipe leading to Qactual> Qtheoritical.
94. (c)
The instantaneous velocity Vd in the delivery pipemay be obtained as :
Vd =d
A r sina
(Vd)mean = d2
00
A r sin da
d
= d
A ra
(Vd)max = d d
A A rr sin 90ºa a
d mean
d max
(V ) 1 .(V )
95. (c)
To reduce the acceleration head we needto reduce the length of suction pipe (ls) andlength of delivery pipe (ld) in which fluctuationof velocity occurs. This is done by the fitting
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (11)
air vessels closed to cylinder as possible.Thus between air vessel and cylinder onlyfluctuating velocity will occur. Below airvessel on suction side and above air vesselon delivery side velocity will be constant.
Due to movement of piston, when there isexcess velocity above average b/w suctionside air vessel and cylinder, water issupplied from air vessel.
96. (c)
97. (c)
98. (b)
1000critical s/c
200
1 m
The critical section for maximum moment islocated halfway between the centreline andedge of the wall in this case
Given qu = 200 kN/m2
= 1000 200
2 4 = 450 mm
For 1 m width inside, the design bending moment
is given as, Mu = 2
u0.45W 1
2
= 0.2025200 1
2 = 20.25 kNm/m
99. (b)
As maximum hoop tension will develop at bottomand is given by
= wH D
2=
10 3 22
= 30 kN/m
100. (b)101. (b)102. (d)
Loss of stress due to elastic deformationdepends upon modular ratio and it isindependent of relexation of steel.
Loss of stress due to shrinkage is timedependent.
In pretensioned there is only one type ofimmediate loss i.e. elastic shortering.
103. (b)
Initial stress in wires
=
3
2
300 10 1590 MPa2 60
4Initial stress in concrete
=
3300 10 4.8 MPa250 250
Losses:
6 3
210Elastic deformation= 4.8 33.6 MPa30
Shrinkage loss 200 10 210 10 42 MPa8other loss= 1590 127.2 MPa
100202.8 MPa
Effective prestress
= 1590 – 202.8 = 1387.2 N/mm2
104. (d)
Moment due to prestress, M = Pe
Pe Pe
Upward deflection at mid span = 2ML
8EI
= 2PeL
8EI
= 2
3PeL
bd8E12
= 2
33PeL2bd E
105. (c)
MOR = 0.36 fck bx [d – .416 x]
For balance section x = xu = kd = .48 × 500 =240 mm.
IES M
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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (12)
MOR = .36 × 20 × 300 × 240 × [500 – .416 × 240]
= 207.44 kN-m 210 kN-m.
106. (c)
y SV
V
0.87f AS = v C b
u
vV
b dVu = 1.5 × V
v =
300 1000 2MPa500 300
VS =
0.87 415 2 1004 378.1mm
3002 1.5
300 mm and 0.75 × d = 0.75 × 500 = 375 mm
107. (c)
Li = 2
i iB 2
i i
WhVWh …(A)
VB = Ah × Total load
Total load = 1700 kN
Load at each floor level = 1700 425 kN
4
2i iWh = 425 × 82 + 425 × 62 + 425 × 42 + 425
× 22 = 51000 kN
at storey level 2, storey force =
2425 4850
51000= 113.33 kN
108. (b)
Shear wall are very efficient in terms ofconstruction cost and effectiveness in minimizingearthquake damage.
109. (b)
aS 1.36, 0.55 T 4g T
= 1.36 2.2670.6
110. (b)
Damping ratio, 1
j 1
u1 ln2 j u
= 1 50ln
2 5 25
= 0.022
= 2.2%
111. (a)
For springs connected in parallel,
ek k 2k 3k
Equivalent stiffness of entire system,
e
1 1 1 1k k 3k 4k
e
1 (12 4 3)k 12k
e12kk19
112. (c)
Elasticity Idealised Curve
PlasticityStress fy
Strain(A) It will increase the margin of safety
(B) In case of tension members, fractures mayexists.
(C) We Generally neglect the strain hardeningeffect in Plastic analysis.
113. (d)
(i) Members supporting lifting or rolling loads.
(ii) Members subjected to repeated cycles ofstress from vibrating machine.
(iii) Members subjected to wind inducedoscillation for a large numbers of cycles inlife.
(iv) Members subjected to crowd inducedoscillation of a large numbers of cycles inlife.
114. (a)
Bolted connection allow greater slip betweencomponents than rivetted connection.
115. (c)
Structural members subjected to bendingaccompanied by large axial comprensive loadsat the same time are known as beam column.
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (13)
A Beam column differs from column only bypresence of eccentricity of load application, endmoment, transverse load.
116. (b)
Length of Purlin, L = Spacing of truss = 5m
Maximum moment along principal axis of purlin
= PL (5 6) 5 15 kN-m10 10
117. (c)
Maximum bending moment,
M = PL 80 5 100 kN-m4 4
Since the beam is laterally supported throughout.Hence, required plastic section modulus
= 0m 6
y
1.1M 100 10f 250
= 4.4 × 105 mm3
= 440 cm3
118. (d)
119. (a)
Purlins may be designed as simplysupported, cantilever or continuous beam.
IS: 800 recommends to design purlin ascontinuous beams. Sag rods are designedas tension members.
Shear centre of purlin should coincide withnode points of the truss.
120. (a)
121. (a)
Power Tw 150 2150060
7.5 kW [ Assertion is also correct]
122. (d)1000 mm
10 mm
200 y
100
NA NA Tansformed sec tionI I
combined Transformedy y
Trans
200 100 100 1000 10 205y
200 100 1000 10
= 135 mm
depth of NA from top most fibre
= (200 + 10) – 135 = 75 mm
123. (b)
124. (d)
Support slip causes decrease in the span of thecable resulting in marginal increase in the sag,resulting in corresponding decrease in the cabletension.
125. (a)
A well seasoned timber is a good timber.Seasoning of timber has a variety of advantages.
1. Reduction of shrinkage and warping afterplacement in structure.
2. Increment of strengths, durability andworkability
3. Suitable for painting
4. Reduction in weight
5. Reduction in tendency to split and decay.
126. (c)
Modulus of elasticity of matrix must be lowerthan that of fibre for effective stress transfer.
Use of fibres of low elastic modulus impartgreater degree of toughness and resistanceto impact while fibres of higher elasticmodulus impart strength and stiffness.
127. (a)
128. (c)
Double tangenet method is used in case of plateload test.
129. (b)
Westergaard’s solution is suitable for sedimentarydeposits as it does not consider soil to beisotropic.
130. (a)
131. (a)
IES M
ASTER
[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (14)
Since no direct relationship exist between theintensity of light scattered at 90° and jacksoncandle turbidity meter, calibration of nephelometerin terms of candle or silica unit is not valid.
132. (c)
Sea water normally contain 20% less oxygenthan that contained in fresh water and containslarge amount of dissolved matter as such thecapacity of sea water to absorb sewage solids isnot as high as that of fresh water of stream.
133. (c)
pH 8pH 7
HOCl H OCl
HOCl is 80% more effective than OCl–. Thus, pHof water should be maintained slightly below 7.
134. (b)
Majority of biological treatment unit hastemperature range of 20° to 40° which is suitablefor mesophilic bacteria.
135. (a)
136. (b)
137. (b)
138. (b)T/2 T/2
yy
= 60°
B
m
139. (a)Assertion & reason both are correct and reasonis correct as and engineering equipment loomtheir value at constant rate.
140. (a)Reason is statement of central limit theorem thatcorrectly explains the assertion.
141. (d)
Traffic capacity is the ability of a roadway toaccommodate traffic volume.
Basic capacity is the maximum number ofpassenger cars that can pass a given point on
a lane or roadway during one hour under themost ideal roadway and traffic conditions.
142. (a)
If the water table is very near to the subgrade ofthe road, it will ultimately cause cracking of theroad surface, because the consistency of thesoil will change from plastic to liquid state leadingto its volumetric decrease.
143. (d)
Flow always taken place in the direction fromhigh TEL to low TEL.
144. (d)
Pressure gradient dpdx
is positive due to
decrease in velocity in the region of boundarylayer separation.
145. (a)Large number of smaller diameter bars, welldistributed in tension zone, reduces thecrackwidth.
So smaller dia bar is used in slab.
146. (b)147. (a)
In PSC, a change in external moments in elasticrange results in a shift of the pressure line ratherthan in an increase in the resultant force in thebeam.
148. (c)
In moment resisting frame. Members and jointsare capable of resisting force primarily by flexure.
149. (a)Beta distribution model for PERT
Gives
x = tE = 0 m pt 4t t
6Weightage of to & tp is equal.
150. (d)
Crop rotation means changing the crops to begrown every year in the same field or differencegroups in rotations in the same field.