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QUESTION: 1

Select the word that fits the analogy:

Partial : Impartial : : Popular : ________

Solution:

Partial and Impartial are opposite, in the same way popular and unpopular are opposite words.

QUESTION: 2

The monthly distribution of 9 Watt LED bulbs sold by two firms X and Y from January to June 2018 is shown in the pie-chart and the corresponding table. If the total number of LED bulbs sold by two firms during April-June 2018 is 50000, then the number of LED bulbs sold by the firm Y during April-June 2018 is _________.

Solution:

LED bulbs sold

By ...(A)

LED bulb sold by Y during April-June

A + B + C = (6 + 8 + 5.5)% = 19.5%

Total LED bulb sold by X and Y 35% this value = 50000

So, LED bulb sold by Y (April to June)

[As 35% of total = 5000

Y (April- June) = 19.5% of total 0.195 × 1442857.14 = 2785.142]

No option is matching.

Mistake in paper was that examiner intended to give total as 50000 and the options were place accordingly as 19.5% of 50000 = 9750 which is (d).

But this will be wrong as total is NOT 50000 as per language. So correct answer is 27857.142.

Which matches with none of options.

QUESTION: 3

After the inauguration of the new building, the head of department (HOD) collated faculty preferences for office space. P wanted a room adjacent to the lab. Q wanted to be close to the lift. R wanted a view of the playground and S wanted a corner office.

Assuming that everyone was satisfied, which among the following shows a possible allocation?

Solution:

QUESTION: 4

In a school of 1000 students, 300 students play chess and 600 students play football. If 50 students play both chess and football, the number of students who play neither is ________.

Solution:

Total number of students playing sports = 850

Total number of students not playing sports = 1000 – 850 = 150

QUESTION: 5

Select the most appropriate word that can replace the underlined word without changing the meaning of the sentence:

Now-a-days, most children have a tendency to __belittle__ the legitimate concerns of their parents.

Solution:

Belittle means to undervalue/ underestimate some as unimportant. Disparage fits in the most appropriate manner.

QUESTION: 6

For the year 2019, which of the previous year’s calendar can be used?

Solution:

Number of odd days in 2019 = 1, so 2013 calendar is same as 2019.

QUESTION: 7

If f (x) = x^{2} for each then is equal to _________.

Solution:

f(x)= x^{2}

f(f(f(x))) = f (f (x^{2})) = f (x^{4}) = (x^{4})^{2} = x^{8}

So,

QUESTION: 8

Rescue teams deployed _________ disaster hit areas combat ________ a lot of difficulties to save the people.

Solution:

QUESTION: 9

Nominal interest rate is defined as the amount paid by the borrower to the lender for using the borrowed amount for a specific period of time. Real interest rate calculated on the basis of actual value (inflation-adjusted), is approximately equal to the difference between nominal rate and expected rate of inflation in the economy.

Which of the following assertions is best supported by the above information?

Solution:

QUESTION: 10

The ratio of ‘the sum of the odd positive integers from 1 to 100’ to ‘the sum of the even positive integers from 150 to 200’ is _________.

Solution:

Ratio = 50 : 91

From 1 to 100 = 50 odd number

From 150 to 200 = 26 even number

*Answer can only contain numeric values

QUESTION: 11

For an axle load of 15 tonne on a road, the Vehicle Damage Factor (round off to two decimal places), in terms of the standard axle load of 8 tonne, is _________.

Solution:

Axle load = 15 T

Standard axle load = 8 T

VDF = [15/8]^{4} = 12.35

QUESTION: 12

Muskingum method is used in

Solution:

QUESTION: 13

The traffic starts discharging from an approach at an intersection with the signal turning green. The constant headway considered from the fourth or fifth headway position is referred to as

Solution:

QUESTION: 14

Soil deposit formed due to transportation by wind is termed as

Solution:

Soil deposited by wind is Aeolian soil.

QUESTION: 15

The relationship between oxygen consumption and equivalent biodegradable organic removal (i.e. BOD) in a closed container with respect to time is shown in the figure.

Assume that the rate of oxygen consumption is directly proportional to the amount of degradable organic matter and is expressed as where, L_{1} (in mg/litre) is the oxygen equivalent of the organics remaining at time t and k (in d^{–1}) is the degradation rate constant. L_{0} is the oxygen of organic matter at time, t = 0

In the above context, the correct expression is

Solution:

*Answer can only contain numeric values

QUESTION: 16

A one-dimensional consolidation test is carried out on a standard 19 mm thick clay sample. The oedometer’s deflection gauge indicates a reading of 2.1 mm, just before removal of the load, without allowing any swelling. The void ratio is 0.62 at this stage. The initial void ratio (round off to two decimal places) of the standard specimen is ________.

Solution:

Oedometer reading = 2.1 mm

16 mm thick, e = 0.62

e_{0} = 0.82

QUESTION: 17

The velocity components in the x and y directions for an incompressible flow are given as u = (–5 + 6x) and v = –(9 + 6y), respectively. The equation of the streamline is

Solution:

Given: u = -5 + 6x

v = –(9 + 6y)

Equation of streamline

Integrating it,

ln(–5 + 6x)^{1/6} = –ln (9 + 6y)^{1/6} + ln C^{1/6}

Take antilog,

(–5 + 6x)(9 + 6y ) = constant

u.v = constant

QUESTION: 18

A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The peak flow of 60 m^{3}/s occurs at 20 hours from the start of the storm. The area of catchment is 300 km^{2}. The rainfall excess of the storm (in cm), is

Solution:

∴ Rainfall excess = 3.24 cm

*Answer can only contain numeric values

QUESTION: 19

24-h traffic count at a road section was observed to be 1000 vehicles on a Tuesday in the month of July. If daily adjustment factor for Tuesday is 1.121 and monthly adjustment factor for July is 0.913, the Annual Average Daily Traffic (in veh/day, round off to the nearest integer) is _______.

Solution:

T_{24} = 1000 veh (Tuesday)

DAF = 1.121

AADT = ?

MAF = 0.913

T_{week} = T_{24} × DAF

= 1000 × 1.121 = 1121

AADT = MAF × ADT= (0.913 × 1121)

= 1023.473

= 1023 VPD

*Answer can only contain numeric values

QUESTION: 20

Velocity distribution in a boundary layer is given by where u is the velocity at vertical coordinate y, U_{∞} is the free stream velocity and δ is the boundary layer thickness. The values of U_{∞} and δ are 0.3 m/s and 1.0 m, respectively. The velocity gradient (in s^{–1}, round off to two decimal places) at y = 0, is ________.

Solution:

Given :

u_{∞} = 0.3 m/s

δ =1 m

At y = 0 and δ = 1

= 0.47 s^{–1}

QUESTION: 21

A weightless cantilever beam of span L is loaded as shown in the figure. For the entire span of the beam, the material properties are identical and the cross-section is rectangular with constant width.

From the flexure-critical perspective, the most economical longitudinal profile of the beam to carry the given loads amongst the options given below, is

Solution:

(–PL) + (PL) + (–M_{A})= 0

M_{A} =0

For most economical,

Maximum cross-section is given where maximum bending moment occurs.

So, option (a) is correct.

*Answer can only contain numeric values

QUESTION: 22

Two identically sized primary settling tanks receive water for Type-I settling (discrete particles in dilute suspension) under laminar flow conditions. The surface overflow rate (SOR) maintained in the two tanks are 30 m^{3}/m^{2}.d and 15 m^{3}/m^{2}.d. The lowest diameters of the particles, which shall be settled out completely under SORs of 30 m^{3}/m^{2}.d are designated as d_{30} and d_{15} respectively. The ratio d_{30}/d_{15} (round off to two decimal places), is ___.

Solution:

For type-I setting, Stokes law is applicable.

V_{s} ∝ d^{2}

QUESTION: 23

As per IS 456:2000, the pH value of water for concrete mix shall NOT be less than

Solution:

1. Minimum pH value of water for concrete = 6.0

As per IS code provision no. 5.4.2, the pH value of water shall not less than 6.0.

QUESTION: 24

Superpassage is a canal cross-drainage structure in which

Solution:

*Answer can only contain numeric values

QUESTION: 25

A soil has dry weight of 15.5 kN/m^{3}, specific gravity of 2.65 and degree of saturation of 72%. Considering the unit weight of water as 10 kN/m^{3}, the water content of the soil (in %, round off to two decimal places) is ________.

Solution:

γ_{d} = 15.5 kN/m^{3}, G = 2.65, S = 72%

e = 0.7096

w = 19.28%

*Answer can only contain numeric values

QUESTION: 26

The maximum applied load on a cylindrical concrete specimen of diameter 150 mm and length 300 mm tested as per the split tensile strength test guidelines of IS 5816 : 1999 is 157 kN. The split tensile strength (in MPa, round off to one decimal place) of the specimen is _______.

Solution:

P = 157 kN

D = 150 mm

L = 300 mm

In split tensile strength test, split tensile

strength of concrete

= 2.22 N/mm^{2}

QUESTION: 27

The state of stress represented by Mohr’s circle shown in the figure is

Solution:

In pure shear condition, Mohr’s circle has its center at origin.

QUESTION: 28

The ratio of the plastic moment capacity of a beam section to its yield moment capacity is termed as

Solution:

Ratio of M_{p}/M_{y} = Shape factor

QUESTION: 29

A sample of 500 g dry sand, when poured into a 2 litre capacity cylinder which is partially filled with water, displaces 188 cm^{3} of water. The density of water is 1 g/cm^{3}. The specific gravity of the sand is

Solution:

W_{s} = 500 gm

V_{s} = 188 cc

QUESTION: 30

The value of is

Solution:

QUESTION: 31

The ordinary differential equation is

Solution:

Its solution is of the type u = f(x), i.e., dependent variable is u.

Hence, given equation is Linear & Non-Homogeneous.

QUESTION: 32

A gas contains two types of suspended particle having average sizes of 2 μm and 50 μm. Amongst the options given below, the most suitable pollution control strategy for removal of these particles is

Solution:

QUESTION: 33

The integral

is estimated numerically using three alternative methods namely the rectangular, trapezoidal and Simpson’s rules with a common step size. In this context, which one of the following statement is TRUE?

Solution:

Because integral is a polynomial of 3rd degree so Simpson’s rule will give error free answer.

*Answer can only contain numeric values

QUESTION: 34

A fair (unbiased) coin is tossed 15 times. The probability of getting exactly 8 Heads (round off to three decimal places), is ________.

Solution:

P(H) = 1/2

P(T) = 1/2

Probability of getting exactly 8 heads out of 15 trial =

= 0.196

QUESTION: 35

The following partial differential equation is defined for u : u ( x , y )

The set of auxiliary conditions necessary to solve the equation uniquely, is

Solution:

Given:

∵ y is given as ≥ 0 so we take it as time.

Hence, above equation is nothing but one-D heat equation which requires one initial condition and two boundary condition.

*Answer can only contain numeric values

QUESTION: 36

A hydraulic jump occurs, in a triangular (V-shaped) channel with side slopes 1:1 (vertical to horizontal). The sequent depths are 0.5 m and 1.5 m. The flow rate (in m^{3}/s, round off to two decimal places) in the channel is _________.

Solution:

For a horizontal and frictionless channel

Specific Force (F) = Constant

⇒

⇒

If Y_{1} and Y_{2} are conjugate depths

⇒

⇒

Q = 1.728 m^{3}/sec

*Answer can only contain numeric values

QUESTION: 37

A concrete beam of span 15 m, 150 mm wide and 350 mm deep is prestressed with a parabolic cable as shown in the figure (not drawn to the scale). Coefficient of friction for the cable is 0.35, and coefficient of wave effect is 0.0015 per metre.

If the cable is tensioned from one end only, the percentage loss (round off to one decimal place) in the cable force due to friction, is ________.

Solution:

Jacking from one end

x = L = 15 m

Wobble correction factor, K = 0.0015

Coefficient of friction = 0.35 = μ

P = Not given

p_{0} = Unknown

Change of gradient,

% loss of stress in steel due to friction

= (0.0015 × 15 + 0.35 × 0.064) × 100

= 4.49%

*Answer can only contain numeric values

QUESTION: 38

The Fourier series to represent x – x^{2} for −π ≤ x ≤ π is given by

The value of a_{0 }(round off to two decimal places), is _______.

Solution:

*Answer can only contain numeric values

QUESTION: 39

The diameter and height of a right circular cylinder are 3 cm and 4 cm, respectively. The absolute error in each of these two measurements is 0.2 cm. The absolute error in the computed volume (in cm^{3}, round off to three decimal places), is _______.

Solution:

Let diameter, x = 3 and height = y = 4 and error = ± 0.2

∴ V = f (x, y )

So,

i.e.,

= 1.65 × 3.14 = 5.18 (approx)

i.e., absolute error = | 5.18 | = 5.18

*Answer can only contain numeric values

QUESTION: 40

The ion product of water (pK_{w}) is 14. If a rain water sample has a pH of 5.6, the concentration of OH^{–} in the sample (in 10^{–9} mol/litre, round off to one decimal place), is ________.

Solution:

pH + pOH = 14

pOH = 14 – 5.6 = 8.4

–log [OH^{–}] = 8.4

[OH^{–}]= 10^{–8.4} moles/lt

=10^{–8.4+9} × 10^{–9} moles/lt

= 3.98 × 10^{–9} moles/lt

*Answer can only contain numeric values

QUESTION: 41

A concrete dam holds 10 m of static water as shown in the figure (not drawn to the scale). The uplift assumed to vary linearly from full hydrostatic head at the heel, to zero at the toe of dam. The coefficient of friction between the dam and foundation soil is 0.45.

Specific weights of concrete and water are 24 kN/m^{3} and 9.81 kN/m^{3}, respectively.

For NO sliding condition, the required minimum base width B (in m, round off to two decimal places) is __________.

Solution:

μ = 0.45

γ_{conc}. = 24 kN/m^{3}

= 15.873 m

QUESTION: 42

Permeability tests were carried out on the samples collected from two different layers as shown in the figure (not drawn to the scale). The relevant horizontal (k_{h}) and vertical (k_{v}) coefficients of permeability are indicated for each layer.

The ratio of the equivalent horizontal to vertical coefficients of permeability, is

Solution:

*Answer can only contain numeric values

QUESTION: 43

A sample of water contain an organic compound C_{8}H_{16}O_{8} at a concentration of 10^{–3 }mol/litre. Given that the atomic weight of C = 12 g/mol, H = 1 g/mol, and O = 16 g/mol, the theoretical oxygen demand of water (in g of O_{2} per litre, round off to two decimal places), is ___________.

Solution:

C_{8}H_{16}O_{8} of conc. 10^{–3} moles/lt required O_{2} (in gm/lt)

C_{8}H_{16}O_{8 }+ 8O_{2} → 8CO_{2} + 8H_{2}O

1 mole 8 mole

1 mole of C_{8}H_{16}O_{2} requires 8 moles of O_{2} for its decomposition

240 gm = 128 gm

or 10^{–3} moles = 8 × 10^{–3} moles

= 8 × 10^{–3} × 32

= 0.256 gm/lt

*Answer can only contain numeric values

QUESTION: 44

A constant head permeability test was conducted on a soil specimen under a hydraulic gradient of 2.5. The soil specimen has specific gravity of 2.65 and saturated water content of 20%. If the coefficient of permeability of the soil is 0.1 cm/s, the seepage velocity (in cm/s, round off to two decimal places) through the soil specimen is ________.

Solution:

Void ratio,

Porosity,

Seepage velocity,

= 0.72 cm/sec

*Answer can only contain numeric values

QUESTION: 45

A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle of elevation to the top of a 4 m long staff, held vertical at station B, is 7°. The horizontal distance between station A and B is 400 m. Neglecting the errors due to curvature of earth and refraction, the RL (in m, round off to three decimal places) of station B is __________.

Solution:

V = 400 tan 7° = 49.113

x = (49.113 – 4) = 45.113

RL_{B} = 212.25 + 45.113 = 257.363 m

*Answer can only contain numeric values

QUESTION: 46

The plane truss has hinge supports at P and W and is subjected to the horizontal forces as shown in the figure (not drawn to the scale).

Representing the tensile force with ‘+’ sign and the compressive force with ‘–’ sign, the force in member XW (in kN, round off to the nearest integer), is _________.

Solution:

Force in PQ

Considering the section above (1) – (1)

Taking moment about ‘R’

ΣM_{RL} =0

(10 × 4) + (10 × 8) + F_{PQ} × 4 = 0

F_{PQ }= -120/4 = -30kN = 30kN (Comp.)

QUESTION: 47

A 4 × 4 matrix [P ] is given below

The eigen values of [P] are

Solution:

|P | = 70 and Trace (P ) = 15

So, only option, i.e., (c) (1, 2, 5, 7) satisfies.

QUESTION: 48

The flow-density relationship of traffic on a headway is shown in the figure

The correct representation of speed-density relationship of the traffic on this highway is

Solution:

QUESTION: 49

Alkalinity of water, in equivalent/litre (eq/litre), is given by

where, { } represents concentration in mol/litre. For a water sample, the concentration of HCO^{-}_{3 }= 2 x 10^{-3} mol/litre, CO^{2-}_{3} = 3.04 × 10^{–4} mol/litre and the pH of water = 9.0. The atomic weights are : Ca = 40; C = 12; and O = 16. If the concentration of OH^{–} and H^{+} are NEGLECTED, the alkalinity of the water sample (in mg/litre as CaCO_{3}), is

Solution:

Alkalinity of water sample is due to presence of

[HCO_{3}^{–}] and [CO_{3}^{2–}]

Total alkalinity = 1 mole of [HCO_{3}^{–}] + 2 mole of [CO_{3}^{2–}] in terms of CaCO_{3}

= (2 × 10^{–3} × 50 + 2 × 3.04 × 10^{–4} × 50) × 10^{3} mg/l

= 130.4 mg/l as CaCO_{3}

QUESTION: 50

Group-I gives a list of test methods for evaluating properties of aggregates. Group-II gives the list of properties to be evaluated.

__Group-I : Test Methods__

P. Soundness test

Q. Crushing test

R. Los Angeles abrasion test

S. Stripping value test

__Group-II : Properties__

1. Strength

2. Resistance to weathering

3. Adhesion

4. Hardness

The correct match of test methods under Group-I to properties under Group-II, is

Solution:

*Answer can only contain numeric values

QUESTION: 51

A waste to energy plant burns dry solid waste of composition : Carbon = 35%, Oxygen = 26%, Hydrogen = 10%, Sulphur = 6%, Nitrogen = 3% and Inerts = 20%.

Burning rate is 1000 tonnes/d. Oxygen in air by weight is 23%. Assume complete conversion of Carbon to CO_{2}. Hydrogen to H_{2}O, Sulphur to SO_{2} and Nitrogen to NO_{2}.

Given Atomic weighs : H = 1, C = 12, N = 14, O = 16, S = 32.

The stoichiometric (theoretical) amount of air (in tonnes/d, round off to the nearest integer) required for complete burning of this waste, is __________.

Solution:

Oxygen required for 350 tonne/day

(32/12) x 350 = 933.33

Oxygen required for 100 tonne/day

(32/4) x 100 = 800

Oxygen required for 60 tonne/day

(32/32) x 60 = 60

Oxygen required for 60 tonne/day

(32/14) x 30 = 68.57

Total O_{2 }= 1861.9 tonne/day

Available O_{2} in waste = 260 tonne/day

Required = 1861.9 – 260 = 1601.9 tonne/day

Amount of air required = 1601.9/0.23 = 6964.78 tonne/day ≌ 6965 tonne/day

*Answer can only contain numeric values

QUESTION: 52

A cast iron pipe of diameter 600 mm and length 400 m carries water from a tank and discharges freely into air at a point 4.5 m below the water surface in the tank. The friction factor of the pipe is 0.018. Consider acceleration due to gravity as 9.81 m/s^{2}. The velocity of the flow in pipe (in m/s, round off to two decimal places) is __________.

Solution:

Apply energy equation between (1) and (2)

V^{2} = 6.54

V = 2.557 m/s ≌ 2.56 m/s

QUESTION: 53

Joints I, J, K, L, Q and M of the frame shown in the figure (not drawn to the scale) are pins. Continuous members IQ and IJ are connected through a pin at N. Continuous members JM and KQ are connected through a pin at P. The frame has hinge supports at joints R and S. The loads acting at joints I, J and K are along the negative Y direction and the loads acting at joints I, M are along the positive X direction.

The magnitude of the horizontal component of reaction (in kN) at S, is

Solution:

Remove hinge at support S and replace it with roller support as shown in the figure.

I^{st }Step: Find coordinates of all the points where forces are acting.

II^{nd} step: Find virtual displacements of all the points.

III^{rd} Step: Use principle of virtual work to find unknown horizontal force H_{S}

⇒δU =0

Substituting, θ = 45°, H_{s} = 90/6 = 15 kN

Note : Sign conventions

If a force acts along positive x or positive y-axis, take it as positive.

If a force acts along negative x or negative y-axis, take it as negative.

*Answer can only contain numeric values

QUESTION: 54

The design speed of a two-lane two-way road is 60 km/h and the longitudinal coefficient of friction is 0.36. The reaction time of a driver is 2.5 seconds. Consider acceleration due to gravity as 9.8 m/s^{2}. The intermediate sight distance (in m, round off to the nearest integer) required for the load is ________.

Solution:

Given : f = 0.36; v = 60 km; g = 9.8 m/s^{2}; t_{R} = 2.5s

^{
}

= 41.7 + 39.37 = 81 m

ISD = 2 × SSD = 81 × 2 = 162 m

QUESTION: 55

A prismatic linearly elastic bar of length, L, cross-sectional area A, and made up of a material with Young’s modulus E, is subjected to axial tensile force as shown in the figures. When the bar is subjected to axial tensile force P_{1} and P_{2}, the strain energies stored in the bar are U_{1} and U_{2}, respectively.

If U is the strain energy stored in the same bar when subjected to an axial tensile force (P_{1} + P_{2}), the correct relationship is

Solution:

(P_{1} +P_{2} )^{2} > P_{1}^{2 }+P_{2}^{2}

U > U_{1} + U_{2}

*Answer can only contain numeric values

QUESTION: 56

Two steel plates are lap jointed in a workshop using 6 mm thick fillet weld as shown in the figure (not drawn to the scale). The ultimate strength of the weld is 410 MPa.

As per Limit State Design is IS 800 : 2007, the design capacity (in kN, round off to three decimal places) of the welded connection, is _______.

Solution:

Design capacity of welded connection

P_{s} = f_{b} × l_{eff} × t_{t}

= 413.586 kN

QUESTION: 57

For the hottest month of the year at the proposed airport site, the monthly mean of the average daily temperature is 39°C. The monthly mean of the maximum daily temperature is 48°C for the same month of the year. From the given information, the calculated Airport Reference Temperature (in°C), is

Solution:

T_{a} = 39°C

T_{m} = 48°C

QUESTION: 58

A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45°. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is

Solution:

x/20 = tan 45°

x = 20 m

RL of Q = 433.05 + 2.905 – x – 2

= 433.05 + 2.905 – 20 – 2

= 413.955 m

QUESTION: 59

A 10 m high slope of dry clay soil (unit weight = 20 kN/m^{3}), with a slope angle of 45° and the circular slip surface, is shown in the figure (not drawn to the scale). The weight of the slip wedge is denoted by W. The undrained unit cohesion (c_{u}) is 60 kPa.

The factor of safety of the slope against slip failure, is

Solution:

Consider unit length of slope

Area of circular arc

Height of wedge = Volume × γ = (Area × 1) × γ

= 28.54 × 1 × 20 = 570.8 kN

= 3.68

*Answer can only contain numeric values

QUESTION: 60

A 5 m high vertical wall has a saturated clay backfill. The saturation unit weight and cohesion of clay are 18 kN/m^{3} and 20 kPa, respectively. The angle of internal friction of clay is zero. In order to prevent development of tension zone, the height of the wall is required to be increased. Dry sand is used as backfill above the clay for the increased portion of the wall. The unit weight and angle of internal friction of sand are 16 kN/m^{3} and 30°, respectively. Assume that the back of the wall is smooth and top of the backfill is horizontal. To prevent the development of tension zone, the minimum height (in m, round off to one decimal place) by which the wall has to be raised, is __________.

Solution:

To prevent tension crack,

q = γ_{d}x = 40

x = 40/16 = 2.5 m

*Answer can only contain numeric values

QUESTION: 61

The cross-section of the reinforced concrete beam having an effective depth of 500 mm is shown in the figure (not drawn to the scale). The grades of concrete and steel used are M35 and Fe550, respectively. The area of tension reinforcement is 400 mm^{2}. It is given that corresponding to 0.2% proof stress, the material safety factor is 1.15 and the yield strain of Fe550 steel is 0.0044.

As per IS 456.2000, the limiting depth (in mm, round off to the nearest integer) of the neutral axis measured from the extreme compression fiber, is ________.

Solution:

For a RCC T-Beam

(For limiting depth of neutral axis)

Considering d = 500 mm

35 × 500 = 35x_{u,lim} + 44x_{u,lim}

=79x_{u,lim}

Limiting depth of neutral axis

x_{0,lim} = 221.52 mm

QUESTION: 62

Crops are grown in a field having soil, which has field capacity of 30% and permanent wilting point of 13%. The effective depth of root zone is 80 cm. Irrigation water is supplied when the average soil moisture drops to 20%. Consider density of the soil as 1500 kg/m^{3} and density of water as 1000 kg/m^{3}. If the daily consumptive use of water for the crops is 2 mm, the frequency of irrigating the crops (in days), is

Solution:

FC = 30%

PWP = 13%

= 12 cm or 120 mm

Consumptive use = 2 mm/day

So, frequency of irrigation = 120/2 = 60 days

QUESTION: 63

The planar structure RST shown in the figure is roller-supported at S and pin-supported at R. Members RS and ST have uniform flexural rigidity (EI) and S is a rigid joint. Consider only bending deformation and neglect effects of self-weight and axial stiffening.

When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is

Solution:

QUESTION: 64

An ordinary differential equation is given below

The general solution of the above equation (with constant C_{1} and C_{2}), is

Solution:

(6D^{2} + D – 1)y =0

6D^{2} + 3D – 2D – 1 = 0

3D(2D + 1) – 1(2D + 1) = 0

(2D + 1)(3D – 1) = 0

*Answer can only contain numeric values

QUESTION: 65

A footing of size 2 m × 2 m transferring a pressure of 200 kN/m^{2}, is placed at a depth of 1.5 m below the ground as shown in the figure (not drawn to the scale). The clay stratum is normally consolidated. The clay has specific gravity of 2.65 and compression index of 0.3

Considering 2 : 1 (vertical to horizontal) method of load distribution and γ_{w} = 10 kN/m^{3}, the primary consolidation settlement (in mm, round off to two decimal places) of the clay stratum is _________.

Solution:

For clay layer,

e_{0} = 1.357

1. H_{0} = 1.5 m

2.

= (2 × 15 + 0.5 × 18 + 0.75 × 17) – 1.25 × 10

= 39.25 kN/m^{2}

3.

4.

= 74.27 mm

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