1. science
2. mathematics
3. geometry

CIVIL ENGINEERING
Conventional Question Practice Program
Date: 21th March, 2015
ANSWERS
1.
(a)
21.
(c)
41.
(d)
61.
(d)
81.
(b)
101. (b)
2.
(b)
22.
(b)
42.
(b)
62.
(d)
82.
(c)
102. (d)
3.
(b)
23.
(a)
43.
(b)
63.
(d)
83.
(b)
103. (a)
4.
(a)
24.
(d)
44.
(b)
64.
(a)
84.
(b)
104. (b)
5.
(b)
25.
(c)
45.
(d)
65.
(b)
85.
(b)
105. (a)
6.
(d)
26.
(a)
46.
(d)
66.
(c)
86.
(a)
106. (d)
7.
(a)
27.
(d)
47.
(d)
67.
(a)
87.
(c)
107. (a)
8.
(d)
28.
(b)
48.
(a)
68.
(c)
88.
(d)
108. (b)
(a)
109. (a)
9
(a)
29.
(c)
49.
(b)
69.
(c)
89.
10.
(a)
30.
(a)
50.
(a)
70.
(a)
90.
11.
(c)
31.
(b)
51.
(d)
71.
(b)
91.
(c)
111. (b)
12.
(a)
32.
(d)
52.
(a)
72.
(d)
92.
(b)
112. (b)
13.
(a)
33.
(b)
53.
(b)
73.
(b)
93.
(c)
113. (a)
14.
(d)
34.
(a)
54.
(b)
74.
(b)
94.
(a)
114. (a)
15.
(b)
35.
(b)
55.
(d)
75.
(a)
95.
(d)
115. (d)
16.
(b)
36.
(a)
56.
(b)
76.
(b)
96.
(a)
116. (a)
17.
(d)
37.
(c)
57.
(c)
77.
(b)
97.
(c)
117. (b)
18.
(a)
38.
(c)
58.
(b)
78.
(d)
98.
(c)
118. (a)
19.
(a)
39.
(b)
59.
(c)
79.
(b)
99.
(c)
119. (a)
20.
(b)
40.
(d)
60.
(b)
80.
(d)
100. (c)
120. (a)
(d)
110. (a)
(2)
1. (a)
Hoop stress,
h
=
1enat = 0.20 × 2.6
enat = 0.52
Pd
2t
Relative density
Longitudinal stress,
Pd
4t
= 0.8 = 80%
7. (a)
(b) Both ends are pinned, leff = l
2EI
= 150 kN
l2
when column is restrained against lateral
movement at its mid height,
P
l ef = l
2
Buckling load,
Buckling load, P1 =
2
vs
A
M
S
IE
Volume of voids
Volume of solids
 1 1
  
Va  Vw
5 3
 
=
V  (Va  Vw )
 1 1
1   
5 3
= 8/7
T2 > T1
All types of soils carried and deposited by
water are known as alluvial deposits.
Deposits made in lakes are called lacustrine
deposits.
T2
DAD curve
11. (c) Load carrying capacity of a long column is
governed by Eulers' formula.
B
C
l, EI
0
0
3
Pl
3 EI
We know
Se = wG
For sand deposit
e =
Marl: A v ery f ine grained calcium
carbonated soil of marine origin.
P
 Deflection of B 
Solid
10. (a) Loess: A loose deposit of wind blown silt
that has been weakly cemented with
calcium carbonate and montmorillonite.
Area
5. (b)
Water
Void ratio
l/2
T1
A
6. (d)
1 1/3
l/2
(b)
Air
T
l
 
2
= 600 kN
Rainfall
depth
1/5
2EI
 2EI 
= 4 2 
 l 
= 4 × 150
3.
9. (a)
S
P2 =
Evaporation from sea water is less than
that of fresh water due to presence of
impurites in the form of salts.
R
L
1
 0.5
h = 2

2.
=
E
L
emax  enat
0.92  0.52
= e
=

e
0.92  0.42
max
min
Pcr =
2EI
2
leff

2EI
2r 2
This depends on both flexural rigidity and
slenderness ratio.
12. (a) Kern, in general, is a zone (or area) in
which an axial arbitrary load can be applied
without inducing tensile stress in the
section.
(3)
In circular section, kern is D/4 in diameter.
16. (b)
h
D/4
D
radius 
inner most radius
hoop stress distribution
13. (a) Elastic critical stress,
2E
2
1
  2

Where  =slenderness ratio
s =
R
Unconfinedcompressive
strength of undisturbedsoil
Sensitivity =
Unconfinedcompressive
strength of remouldedsoil
14. (d)
A
p
S
n
IE
pd
Hoop stress =
in thin spherical shell
pt
C
K
J
G
No. of joints, J = 11
No. of inextensible members, n = 11
kinematic indeterminancy, Dc = 3J – R – n
20. (b)
=
h h

E
E
=
h
pd
(1  ) 
(1   )
E
4tE
C2
(  l  h )
=

A
I
H
= 3 × 11– 7 – 11 = 15


Hoop strain (h) = h   l
E
E
Hoop strain, h =
E
D
No. of reactions, R = 7


F
B
M
n
qu  undisturbed
qu  remoulded
18. (a)
19. (a)
15. (a)
h
St =
S
Effective Length Effective length is the
length of the equivalent pinned-end member,
or the distance between points of inflection
in the deflection curve. It depends on the
boundary condition at the end of member.
E
Effective length
Radius of gyration
T
 =
17. (d)
pd
(1  )
4tE
Hinge C1
C4
C3
C5
Hinge
d chanage in diameter

d
original diameter
Change in diameter,
pd2
(1  )
d =
4tE
No. of restraint required to make the hinged joint
rigid = m – 1
(4)
where m is no. of members meeting at the hinge
R = (4 – 1)+ (2 – 1) = 4
No. of cuts required to make it tree like structure
C = 5
Degree of static indeterminancy DS = 3C – R
and direction of flow as in tidal rivers
(4) Cost of installation is independent of size
of river.
Accuracy of this measurement is limited by
(a) air entrainment
= 3 × 5 – 4 = 11
(b) salinity and temperature changes
21. (c) Total degree of static indeterminancy in
truss
DS = m + r – 2j
where m = No. of members = 20
r = No. of reactions = 6
(c) high load of suspended solid
(d) Fluctuating weed growth
28. (b)
29. (c)
j = No. of joints = 10
= 6
R
30. (a)
DS = 20 + 6 – 2 × 10
22. (b) The 'Saturated unit weight' is defined as
the bulk unit weight of the soil mass in the
saturated condition. 'Submerged' or
'Buoyant' unit weight of a soil is its unit
weight in the submerged condition.
... (i)
32. (d)
33. (b) At pin support there will be 2 reactions
(horizontal reaction and vertical reaction)
34. (a) In slope deflection method, number of
unknown displacement rotations are equal
to kinematic indeterminacy (Dk)
S
 sub =  sat   w
E
31. (b)
T

A
If may be noted that a submerged soil is
invariably saturated, while a saturated soil
need not be submerged.
M
Equation (i) is a direct application of
Archimedes principle which states that the
apparent loss of weight of a substance when
weighed in water is equal to the weight of
water displaced by it.
Dk = 3J – R – n
where, J = Number of joints = 4
R = No. of reactions = 5(3 at A and 2 at D)
n = Number of inextensible members = 3
 Dk = 3 × 4 – 5 – 3 = 4
S
Thus,  = sat – w
IE
23. (a) Consistency represents the relative ease
with which a soil can be deformed. This
term is mainly used for clayey soil and is
related to water content i.e., how with
change in water content the consistency of
soil changes.
P
35. (b) A
B
This determinate structure may not have zero
deflection at it’s ends.
24. (d)
36. (a)
25. (c)
37. (c) Kinematic indeterminancy,
DK = 3j – R – n
26. (a)
27. (d) Specific advantages are:
(1) Rapid and high accuracy
(2) Surface for automatic recording of data
(3) Can handle rapid changes in the magnitude
J = 9 ; R = 5 ; n = 10 ;
DK = 3×9 – 5 – 10 = 12
(5)
Alternatively,
Hence degree of kinematic indeterminacy = 9
3,  x3
40. (d) For equilibrium
4, x4
9
FH = 0
6, x6
7, x7
MAB  MBA MCD  MDC

P = 0
4
4
As D and A are hinges, so MAB = 0
8
Hinge
Roller
x2 = x3  x 4  x a
x1 = x5  x6  xb
 Unknowns are
1, 2 , 3 ... 3 , 8 , 9 =
xa , xb = 2
x7 = 1

9


 12 unknowns


J6
38. (c)
HA + HD + P = 0

MBA MCD

P
4
4
41. (d)
= 0
MBC = MFBC +
=
2(2EI)
(2B  C )
L
R
5, x5
1, x1
4EI
(2B  C )
6
[  MFBC= 0]
E
 2 x 2
42. (b)
5
4
L EI
J4
S
7 J7 8
13 14
J5
T
B
9
12
J
3
9
J8 15
6
10
11 8
2
1
J1
J3
J2
M
RB
No. of external reactions, R = 3
No. of equilibrium equations = 3
(M = 0, Fx = 0, SFV = 0)
S
 External indeterminacy, Dse = R – 3 = 0
IE
Total degree of indeterminacy, Ds = m + r – 2j

m = 15 ; r = 3 ; J = 9
Ds = 15 + 3 – 9 × 2 = 0
Internal degree of indeterminancy
Dsi = Ds –Dse = 0 – 0 = 0
39. (b)
A
B
C
D
E
F
QA
QBA
 xA  0
 yA  0
QBC
 yB
 xB  0
QC
QD
 xC  0  xD  0
L D
EI
A
RA
A
 xE  0
M
L, EI
L EI
C
At joint D, moment PL is acting which will be
distributed in the ratio of stiffness of members
DA, DB and DC
4EI
Stiffness
KDA =
L
4EI
KDC =
L
4EI
KDB =
L
4EI / L
Distribution factor for DA, =
4EI 4EI 4EI


L
L
L
1
=
3
M

MDA = D.F × M =
3
As end A is fixed, so carry over factor for
QED
QEF
 yE
M
M
AD is

1
2
MAD =
1
M
 MDA =
2
6
(6)
43. (b) Due to symmetry of structure and loading,
there will not be any bending in column
DC.
Hence moment, MCD = 0
B
MBA
C
MCD

A
R
D
BC is rigid so it will not bend
MBA = MAB =

6EI
L2
49. (b) Ordinate of IUH is given by
E
44. (b)

48. (a) By using the polygon of forces together
with the funicular polygon, the three possible
cases which may occur in the case of a
system of forces in a plane can be
investigated entirely by graphical methods.
If the polygon of forces is not closed, the
given system of forces reduces to a
resultant force. If the polygon of forces is
closed but the first and last sides of the
funicular polygon which are parallel do not
coincide the system reduces to a resultant
couple. If the polygon of forces is closed
and the first and last sides of the funicular
polygon coincide i.e. the funicular polygon
is closed also, the system of forces is in
equilibrium.
6EI
L2
3EI 3 EI

=
(2L)2 4 L2
3 ( 60)

=
= – 7.5 kNm
4
6
1 ds
... (i)
i dt
i = intensity of rainfall
s = ordinate of S-curve
– 60 =
Side sway  is same in both the columns
For column AB, end A is hinged
3EI
MBA =
L2
column, end D is hinged
3EI
12EI

MCD =
2
(L / 2)
L2
MBA
3EI / L2
=
MCD
12EI / L2
M


S
In
Thus, the ordinate of IUH is the slope of
S - curve of intensity 1 cm/hr.
= 1 : 4
IE
46. (d)
A
50. (a)
A
45. (d)

T
MCD
S
and
U(t) =
EI = 10000 kN-m
L = 10 m
Here B = 0.001 radians
A = 0
Using slope deflection method,
2EI
MAB = 0 
2A  B 
L
2  10000
 0.001
= 0
10
= 2 kN-m
= Probability of occurance once in next 5 year
1
=
1 
 1  
C1   1 

 10   10 
5 1
 0.328
(b) Probability of 20 year flood occuring in next
year
1
11
1 
 1  
= 1C1 
 1 

 20  
20 
 0.05
(c) Probability of non occurance of 20 year flood
in next 20 year
1 

1 


20 
20
 0.358
(d) Exceedance Probability of 20 year flood
occuring 2 times in next 10 year
= Probability of occurance once in next one
year
=
47. (d)
5
= Probability of occurance once in next one
year
B
2
(a) Probability of 10 year flood occuring in next
5 year
10
C2 p2 q10 2
2
8
1 
 1  
= 45  
 1 
  0.0763
 20  
20 
(7)
51. (d)
57. (c)
Discharge
A
D
C
ERH
58. (b)
B
C
Rainfall excess – Direct runoff hydrograph
Unit hydrographs
D
Rainfall intensity = Hyetograph (intensity
versus time curve)
Rainfall averaging – Isohytes (Isohytes is
the line of having equal rainfall depth and
used to measure the average rainfall over
an catchment)
A
B
Mass curve – Cumulative Rainfall
Time
Hydrograph
Duration
UH
D = 4 hr
IUH
D  0
60. (b)
R
61. (d) Average rainfall over a catchment area can
be estimated using following methods-
E
1. Arithmetic mean method: In this
method average rainfall is taken to be
the mean of rainfall datas collected at
various raingauge stations of the
catchment area i.e., every rain gauge
station is giv en equal weightage
regardless of its location. This can be
used when the rainfall is almost uniformly
distributed over the whole catchment area
which rarely occurs in nature. The
method is fast but does not give accurate
results.
S
Figure shows a typical variation of the
shape of unit hydrographs for different value
of D. As D is reduced, the intensity of rainfall
excess being equal to 1/D increases and
unit Hydrograph becomes more skewed.
Therefore the peak increases.
59. (c)
T
The shape of these different unit hydrographs
depends upon the value of duration of unit
hydrograph (D hr).
A
Hence, peak discharge of IUH is greater
than that of 4 hr UH.
IE
S
M
53. (b) Standard project flood (SPF) is the greatest
storm that may reasonably be expected
without modifying the rainfall data where
Probable Maximum Precipitation (PMP) is
the greatest or extreme rainfall of a given
duration that is physically possible over a
stration or basin. Hence, SPF is always
less than PMP.
54. (b) Isochrone is a line joining points having
equal time of travel of surface runoff to the
catchments outlet.
55. (d)
Evapotranspiration  Lysimeter
Capillary potential  Tensiometer
56. (b) A cyclone is a more or less circular area of
low atmospheric pressure in which winds
blow spirally inward in counter clockwise
direction in the Northern Hemisphere & in
clockwise direction in the Southern
Hemisphere.
2. Theissen polygon method: This method
gives weightage to the various rainfall
datas based on area close to the rain
gauge station called theissen polygon
areas. The method is fast when once
the weights are known. But it does not
take care of the variability in rainfall due
to elevation difference. (i.e., topographical
influence are not taken care of). New
polygon is required to be drawn when,
due to addition or deletion of raingauges
to the network, weight of each station
changes.
3. Isohyetal method: This is the most
accurate method. It utilizes all relevant
datas and properly interpret them.
However the method is very slow and
laborious.
4. Normal ratio method is used to
estimate missing rainfall data.
So correct option is (d).
(8)
62. (d)
h
ln
63. (d)
A
 e 0
A0
h
A
 e 0 
A0
64. (a) There are three methods to get the average
depth of rainfall over an area1. Arithmetic mean method
67. (a) Inconsistency of rainfall data can be
checked by double mass curve method.
2. Theissen polygon method
Normal ratio method is used for estimation
of missing rainfall data.
3. Isohyetal method
Mass curv e is a plot between the
accumulated rainfall against time in
chronological order.
Mass curve of rainfall is a plot of the
accumulated precipitation against time and
gives information about duration, magnitude
and intensity of rainfall.
R
Hyetograph is the curve between rainfall
intensity and time interval.
Rainfall depth-duration-frequency (DDF)
curves describe rainfall depth as a function
of duration for given return period and are
important for the design of hydraulic
structures.
T
65. (b)
68. (c) Rain gauge is used to measure the rainfall
depth at any station. It is also known as
Ombrometer, Pluviometer and Hyetometer.
So, the correct option is (c).
h
where  = density of bar
M
      = stress at a section
69. (c) Optimum number of raingauge stations (N)
is given by
A

S
For a bar of uniform strength
A = A e
0
E
DAD curves are always a falling curve
because as the area increases the average
depth over the area decreases.
So the correct option is (a).
66. (c)
2
A0 = c/s area at the bottom section
where,
and
S
IE
d
unit weight = 
h
d0
0
(A  dA)0  A0  A . dh
0 dA  Adh
 Cv 
N = 

 
C v = coefficient of variation = 40%
 = admissible error = 10%
2

dh
h
d2  d02 e 0

 40 
N =    16
 10 
N  16
 Correct option is (c)
70. (a) -index is the average infiltration rate above
which the total rainfall volume is equal to
the runoff volume. It is the average infiltration
rate during the period of rainfall excess.
Following are the methods to measure the
evapotranspiration (consumptive use)1. Using lysimeter
2. Using field plots
3. Using water budget equation
dA

 A  0
In A 
 dh

h c
0
at h = 0, A = A0  C = ln A0
4. Using empirical equations Penman’s equation
 Blanney-criddle’s equation
Dilution technique is used f or f low
measurement of small and turbulent
streams.
(9)
Snyder gave the synthetic unit hydrograph
theory in which he related v arious
hydrograph parameters with the basin
characteristics.
Q
30 m3 /s
71. (b) The penman’s equation for determining
evapo-transpiration is based on both the
energy balance method and the mass
transfer method. So correct option is (b).
= 2.88 cm
Rainfall = 0.5 × 12 = 6.00 cm
In order to reduce evaporation from the water
bodies certain chemicals such as cetyl
alcohol (hexadecanol) and stearyl alcohol
(octadecanol) are used. So, the correct
option is (d).
R
79. (b)
80. (d)
81. (b)
82. (c)
83. (b)
= 0.2 × 0.1 × 300 × 104
= 60000m3
Runoff
Q = Time of flow
M
Average runoff,
78. (d)
84. (b)
85. (b)
86. (a)
87. (c) The equilibrium discharge for an S–curve
hydrograph is given by,
A 3
m / sec
D
where, A is area of catchment is km2 and D
is duration is hrs.
2.778 300
 420m3 /s
Q =

2
Correct option is (c).
Q =
S
60000 3
m /hr
=
10
= 100 m3/minute
IE
75. (a) Kirpich equation is an empirical equation
used for the estimation of the time of
concentration.
76. (b)
77.
3.12
= 0.26 cm/h
12
S
Runoff = k  p  A

  index =
A
74. (b) Catchment area, A = 300 hectare
= 300 ×104 m2
Rainfall, p = 10 cm = 0.1 m
Runoff
, k = 0.2
Rainfall
= 3.12 cm in 12 hours
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73. (b) Peak ordinate decreases with increase in
the duration of unit hydrograph for the same
basin. So, a 2-hour unit hydrograph will have
peak ordinate greater than that of a 4-hour
unit hydrograph. So the correct option is (b).
Loss = 6.00 – 2.88
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72. (d)
t
64 h
Duration of the rainstorm would not affect
the maximum possible discharge from a
small catchment corresponding to particular
rainfall intensity.

2.778
88. (d)
Q
tr
Unit duration
Pnet = 1 cm
tp
(b) Direct-runoff volume
1
 30  64  60  60
=
2
= 3.456 × 106 m2/s
=
3.456  106
120  106
= 0.0288 m
UH
Qp
Base width T
Time
Basin lag, tp = 30 hr
Time of rainfall excess, tr = 2 hr
Time when the peak discharge occurs from
the start of the storm
(10)
94. (a) Discharge m3 /s
tr
2
 tp =
 30  31hr
2
2
t pk =
36 m3/s
89. (a) The 2hr unit hydrograph is obtained by
dividing the ordinates of storm hydrograph
by the direct runoff depth which is 4 units.
So correct option is (a).
Time
36 hr hour
We know that,
90. (d) Here,
Area of UH = area of catchment × 1 cm
4 1
 0.75 cm / hr
 -index =
4
Now, runoff due to 10 cm of rainfall in 4 hr if index due to 10 cm of rainfall in 4 hr remains at
the same value equal to 0.75 cm/hr then,
Hence option (a) is correct.
95. (d)
 Correct option is (d)
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
A = 233.28 km2  233 km2
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10  runoff
4
Runoff = 7 cm
0.75 =
1
1
 36  3600  36 = A 
100
2
1
1
 144  Q  60  60 = 596 × 106 ×
2
100
96. (a)
91. (c) Given,
T
25.92 × 104 Q = 596 × 104
T = 8 years
1 1

= 0.10
T 10
S
p =
97. (c)
A
q = 1–p
= 1 – 0.10 = 0.90
M
=
nC
r
2C
1
pr qn–r
(0.10)1
(0.9)2–1
92. (b)
S
= 2 × 0.1 × 0.9 = 0.18
Probability of occurrence
1
1

T 30
Probability of non-occurrence
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p =
1
29

q = 1 – p = 1
30 30
 Risk that the structure will be over topped
in the second year exactly
= q× p
29 1

=
30 30
= 0.03222 = 3.22 %
Indian standard
equivalent of
proctor test
Light
Heavy
Standard Modified
proctor
proctor compaction compaction
test
test
test
test
Probability of flood magnitude will be
exceeded once during the next 2 years.
=
Q = 23 m3/s
Weight of
hammer
P (kg)
2.495
4.54
2.6
4.9
Fall of
hammer
Q (mm)
304.8
457.2
310
450
Number of
blows per
layer R
25
25
25
25
Number of
layers S
3
5
3
5
944
944
1000
1000
Volume of
mould (CC)
In India, the Indian standard specification are
followed in laboratory testing.
98. (c) Density index or relative density
emax  e
Dr = e
max  emin
0.9  0.6
0.6 = 0.9  e
min
Void ratio in densest state,
emin = 0.4
(11)
99. (c) Macaulay method (the double integration
method) is a technique used in structural
analysis to determine the deflection of EulerBernoulli beams. Use of macaulay’s
technique in very convenient for cases of
discontinuous and/or discrete loading.
Typically partial uniformly distributed loads
(udl) and uniformly varying load (uvl) over
the span and a number of concentrated
loads are conveniently handled using this
technique. The starting point for macaulay’s
method is the relation between bending
moment and curvature from Euler-Bernoulli
beam theory.
Analysis of trusses : Method of sections:
the method of joints is good if we have to
find the internal forces in all the truss
members. In situations where we need to
find the internal forces only in a few specific
members of a truss, the method of section
is more appropriate.
Method of sections:

Imagine a cut through the members of
interest.

Try to cut the least number of members
(preferably 3)

Draw FBD of the 2 different parts of the
truss

Enforce equilibrium to find the forces in
the 3 members that are cut.
d y
= M
dx2
This equation can be integrated twice to find
y if the value of M as a function of x is
known.
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2
T
This method can be used in determinate truss
only.
100. (c) Let rainfall excess be ER (in cm)
 Volume of surface runoff
1
ER
6
=  60  3600  30 = 300  10 
2
100
 ER = 1.08 cm = 10.8 mm
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M
A
S
Column analog method : This method of
structure analysis is used to analyse the
indeterminate structures, specially the fixed
beams, arches etc. It is based on the analog
between the fixed end moments induced in
a fixed beam to the pressure induced at the
base of a column subjected to loading of the
bending moment diagram of the basic
determinate structure of the given fixed beam.
It means that first we have to make the fixed
beams a determinate structure and then we
have to find out the bending moment diagram
for that structure and that diagram is used
as a loading on the top of a column having
cross-section as that of the beam, means,
length equal to the span of the beam and
width is kept as one. The pressures induced
at the two ends of the column base are
calculated and they are equal to the fixed
end moments induced at the ends of the
fixed beams. They are superimposed with
the simple supported bending moments and
the final bending moment diagram is got.
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EI
Kani’s method : The Kani’s method is self
correcting, that is, the error, if any, in a cycle
is corrected automatically in the subsequent
cycle. The checking is easier as only the
last cycle is required to bechecked. The end
moments are calculated by the application
of rotation contribution method i.e. kani’s
method for the analysis of portal frame.
101. (b) As per Euler’s buckling stress for a long
column,
critical buckling stress, pcr =
2E
2
Where,
leff
          = slenderness ratio = r
min

or,
pcr 
1
2
1
pcr  2
leff
Description
Critical
buckling
stress
Effective
Length (leff)
2E
l
 l

 2
r
 min





2
leff =
l
2
(12)
 Mx = 0
2E
2
leff =
l
2
 My = 0
 Mz = 0
l
 2l 
r 
 min 
2
leff = 2l
2
leff  l
2E
l
 l 
r 
 min 
102. (d)
 Slope deflection method is a stiffness
method in which unknown joint displacement
are found out by applying the equilibrium
condition at end .
107. (a) DAD curves are always a falling curve
because as the area increases the average
depth over the area decreases.
108. (b) Double mass curve is a plot, used for
checking the consistency of the rainfall
data. In this method data of the annual
rainfall of the station x and also the average
rainfall of the group of base stations covering
a long period is arranged in the reverse
chronological order accumulated. A graph
is plotted between accumulated annual
rainfall of a station X vs. accumulated rainfall
of a group of stations, this graph is known
as double-mass-curve.
A
S
 In slope deflection equation we use the
principle of superposition by considering
separately the moments developed at each
support due to each of the displacement
 A ,  B ,  and then the loads, so
displacement at joint are independent.
106. (d) Theissen polygon method: This method
gives weightage to the various rainfall datas
based on area close to the rain gauge
station called theissen polygon areas. The
method is fast when once the weights are
known. But it does not take care of the
variability in rainfall due to elevation
difference. (i.e., topographical influence are
not taken care of). New polygon is required
to be drawn when, due to addition or
deletion of raingauges to the network, weight
of each station changes.
R
2E
E
l / 2
r 
 min 
T
l
103. (a)
M
104. (b)
105. (a) Equilibrium of rigid bodies -2D
For a rigid body in static equilibrium,
the external forces and moments are
balanced and will impart no translational
or rotational motion to the body.

The necessary and sufficient condition
for the static equilibrium of a body are
that the resultant force and couple from
all external f orces f orm a system
equivalent to zero.

F = 0
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

 M0 =   r  F  = 0

Resolving each force and moment into
its rectangular components leads to 6
scalar equation which also express the
conditions for static equilibrium.
 Fx = 0
109. (a)
1. The normal annual rainfall of a station is
obtained as the arithmetic average of the
successive annual rainfall in the last 30
years.
2. The standard deviation computed for the
rainfall of the same 30 years of rainfall
data is taken as a measure of the variability
of the rainfall during the same set of years.
110. (a) W-index : The W-index is a refined version
of  -index. It excludes the depression
storage and interception from the total
losses. It is the average infiltration rate
during the time rainfall intensity exceeds
the capacity rate. That is
 Fy = 0
W =
 Fz = 0
F P  Q  S

t
t
(13)
where F is the total infiltration, t is time during
which rainfall intensity exceeds infiltration
capacity, P is total precipitation corresponding
to t, Q is the total storm runoff and S is the
volume of depression storage and interception.
in discharge (or runoff) at the watershed
outlet. During the initial periods of the storm,
the increase in runoff is rather gradual as
the falling precipitation has to meet the initial
losses in the form of high infiltration,
depression storage and gradual building up
of storage in channels and over the
waershed surface. As the storm continues,
losses decrease with time and more and
more rainfall excess from distant parts of
the watershed reaches the watershed outlet.
The runoff, then, increases rapidly with time.
When the runoff from all parts of the
watershed reaches the watershed outlet.
simultaneously, the runoff attains the peak
(i.e., maximum) value. This peak flow is
represented by the crest segment of the
hydrograph. The recession limb of the
hydrograph starts at the point of inflectio
(i.e., the end of the crest segment) and
continues till the commencement of the
natural ground water flow.
111. (b) Perennial stream: one which always
carries some flow. Ground water contributes
throughout the year. Glaciers also contribute
the flow in the river.
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112. (b) Flow duration curve of a stream is a plot of
discharge against the percent of time the
flow was equalled or exceeded some of the
important uses of flow duration curve are
1. In evaluating various dependable flows in
planning of water resources projects.
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2. In flood control studies.
3. In the design of drainage systems
117. (b) Preparing unit hydrographs for a given
catchment and covering wide range of
durations is usually not possible due to
lack of data. Therefore, one needs to convert
an existing or derived unit hydrograph for
one storm duration (say, D hours) to another
(say, nD hours) that could be either shorter
(to better cope with spatial and intensity
variations) or longer (to reduce the
computational work and also recognizing
the coarseness of the available data).
M
A
S
113. (a) The point of inflection on the recession limn
is commonly assumed to mark the time at
which surface inflow to the channel system
or the overland flow ceases. Thereafter the
recession curve represents withdrawal of
water from storage within the channel
system. Thefore, it is more or loss
independent of variations in rainfall and
infiltration.
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4. In evaluating the hydropower potential of
a river.
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114. (a) In very large catchment basins, storms may
not meet the conditions of constant intensity
within effective storm duration and uniform
areal distribution. Therefore, each storm may
give different direct runoff hydrograph under,
otherwise, identical conditions. Therefore,
unit hydrograph is considered applicable for
catchments having area less than about
5000 km 2. Very large catchments are
usually divided into smaller sub-basins and
the hydrographs of these sub-basins are
proceesed to obtain composite hydrograph
at the basin outlet.
115. (d) Flow mass curve in always a rising curve
or horizontal when there is no inflow or runoff
added into the stream.
116. (a) A single-peaked hydrograph, figure consists
of (i) a rising limb, (ii) the crest segment,
and (iii) the recessio of failling limb. The
rising limb (or concentration curve) of a
hydrograph represents continuous increase
118. (a)
119. (a) Permeability on wet side of optimum is less
than the dry side of optimum.
Project
Compaction
water
content
Reason
Core of an
earth dam.
Wet of
optimum
To reduce permeability
and prevent cracking in
core.
Homogenous
earth dam
Dry of
optimum
To have a stronger soil
& to prevent build up of
high pore water pressure.
Sub-grade
of pavement
Wet of
optimum
To limit volume change.
120. (a) Kani’s method
Gaspar kani’s method of structural analysis is
similar to cross moment distribution in that
both these methods use Gauss-seidell iteration
(14)
The above advantages make Kani’s method
one of the most powerful techniques applicable
to all types of continuous beams and frames.
Moment distribution method.
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This method consists of solving slope deflection
equations by successive approximation that
may be carried out to any desired degree of
accuracy. Essentially, the method begins by
assuming each joint of a structure is fixed.
Then by unlocking and locking each joint in
succession, the internal moments at the joints
are distributed and balanced until the joints
have rotated to their final or nearly final
positions. This method of analysis is both
repetitive and easy to apply. Manual analysis
of gable frames mostly uses the moment
distribution or slope deflection methods. These
methods are usually lengthy and have no builtin-error elimination capability.
M
A
S
 It has a built-in error elimination so that
computational errors automatically disappear
in subsequent operations. This also makes
possible the introduction of any changes in
loads or member lengths that may become
necessary during calculations without
necessitating a new analysis. Such changes
are inserted in the computational scheme
wherever required and the analysis simply
continued.
for highly irregular frames with multiple side
sway. Compared to most other methods,
Kani’s method involves substantially less labour
and time in the analysis of such frames.
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procedure to solve the slope deflection equation
without explicitly writting them down. However,
whereas the moment distribution method
obtains the unknowns (i.e., the end moments
of the structural members) by iterating their
increments, Kan’s method iterates these
unknowns themselves. This method essentially
consists of a single, simple numerical operation
performed repeatedly by the joints of a structure
in a chosen sequence. Results of any desired
accuracy may be obtained by peforming this
operation a sufficient number of times using
the required number of significant digits. Kani’s
method is specially useful for the analysis of
multistorey frames. It has the advantages of
simplicity, speed, economy of time, labour and
space and of accuracy. However, perhaps, the
two most attractive features of this method are
as follows:
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 It requires only one table of calculations even