GATE CE 2020 Official Paper: Shift 1 (Previous Year Paper)

Explanation

The correct answer is option 2- ivory towers.

To live in an ivory tower is an idiom which means having no knowledge or experience of the practical problems of everyday life; be detached from the ground realities of life. 

Note:

The context of this sentence refers to the fact that academicians are so caught up in their academic and theoretical arguments, that they often fail to see the practical and real-life challenges that common people face.

Explanation

The correct answer is option 3- voracious.

A voracious reader is one who shows great eagerness and enthusiasm in reading. 

Here, although the sentence 'He reads indiscriminately' refers to the unbiasedness and all-encompassing quality of the reader, the word 'voracious' is a much better fit for the blank than the word 'all-round'.

The meanings of the other words-

All-round: having a great many abilities or uses; versatile

Precocious: (of behaviour or ability) having developed at an earlier age than is usual or expected

Wise: having or showing experience, knowledge, and good judgment

Explanation

The correct answer is option 1- Usage.

From the first part of the analogy, it becomes clear that a noun is needed after its corresponding verb. 

The noun form of the verb 'use' is 'usage'. 

Hence it is the correct fit in the second part of the given analogy.

Explanation:

 

Thus 45 = ST

Explanation

Let the two numbers be x and y respectively.

Sum = 100 ⇒ x + y = 100

After subtracting 5 from each number, the two numbers becomes (x - 5) and (y - 5).

Product = 0 ⇒ (x - 5) × (y - 5) = 0

⇒ Either x = 5 or y = 5

If x = 5, then y = 95

& If y = 5, then x = 95

So one of the original numbers is 95 and the other is 5.

Concept:

The correct answer is option 4. 

• Option 1 can be negated from the get-go as Howard Gardner had never insisted that the teaching curriculum and evaluation methodology needed to be multi-dimensional. What he had said was that human intelligence can be sub-categorized into multiple kinds.
• Option 2 can be negated in a similar fashion as the educationists were directly involved in the development and assessment of multiple intelligences, not Howard Gardner.
• Option 3 can be negated as the development and assessment of multiple intelligences was not the primary goal of the educationists. The primary goal was to prescribe multi-dimensional curriculums and evaluation parameters. The effect of such a prescription would be seen in the development and assessment of multiple intelligences.

Option 4 follows directly from the second sentence of the passage.

Explanation:

The correct answer is option 1- RSPTQ.

From the conditions given in the question, it is evident that-

• P, Q, and T will not be sleeping adjacent to R
• P and S will not be sleeping adjacent to Q.**

The only arrangement which satisfies the conditions given above is RSPTQ

Explanation

The first number is 2 and the last number = 34

Let the sequence be 2, _, _, _, _, _, _, _, 34.

Let the common difference of the sequence be d.

Total number of terms in sequence = 9

For any A.P.

T_n = a + (n - 1)d.

Where T_n = n^th term of an A.P and a = First Term.

∵ 34 is the 9^th term of the sequence given above.

⇒ 34 = 2 + (9 - 1)d ⇒ d = 4

Hence the sequence becomes 2, 6, 10, 14, 18, 22, …34

Sum of n terms of an A.P is given as:-

${S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$

Sum of seven numbers inserted = sum of A.P – (first term + last term)

$= \frac{9}{2}\left( {2 \times 2 + \left( {9 - 1} \right) \times 4} \right) - \left( {2 + 34} \right) = 126$

Concept:

The concept of cyclicity is used to identify the unit digit in any power.

Various cycles of unit digit:

(1) The cycle of 2ⁿ is (2, 4, 8, 16, 32) and hence the cycle consists of 4 terms i.e 2 repeats itself after 4 terms.

(2) The cycle of 3ⁿ is (3, 9, 27, 81, 243) and hence the cycle consists of 4 terms.

Similarly,

The cycle of 9ⁿ is (9, 81, 729) and hence the cycle consists of 2 terms.

Calculation:

Unit digit of 26591749¹¹⁰⁰¹⁶ = unit digit of 9¹¹⁰⁰¹⁶

Divide 110016 by the number of terms in the cycle, the remainder = 0

Whenever the remainder is 0, the last term of the cycle is considered as a unit digit.

Thus, the unit digit of 9¹¹⁰⁰¹⁶ = unit digit of 26591749¹¹⁰⁰¹⁶ = 1

Explanation:

Let the total expenditure be x.

Money spent on education = 15% = 0.15x

Money spent on transport = 10% = 0.1x

$\Rightarrow \frac{{Money;spent;on;education;}}{{Money;spent;on;Transport}} = \frac{{0.15x}}{{0.1x}} = 1.5$

⇒ Money spent on education = 1.5 × Money spent on Transport

i.e. the money spent on education is 50% more than the money spent on transport.

Concept:

Identification of Non-linear Differential Equation:

 

If any differential equation consists at least one of the above properties, then it is called non-linear differential equation and if any differential equation is free from all the above properties, then it is a linear differential equation.

Order and Degree of the Differential equation:

Order: The highest order derivative occurring in a differential equation is known as its order.

Degree: It is the exponent of the highest order derivative when the differential equation is made free from radicals.

e.g. ${\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2} + 2{\left( {\frac{{dt}}{{dx}}} \right)^3} + 6 = 0$

Order = 2

Degree = 2

Note: Degree ≠ 3, because it is not the exponent of the highest order derivative.

Calculation:

Given:

$D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}}$

Here, the product of the dependent variable D(θ) and its derivative  $\frac{{{\delta ^2}\theta }}{{\delta {z^2}}}$ is present. Hence it is a non-linear differential equation.

Order = 2 and Degree = 1

Thus the above equation is a non-linear differential equation of Order = 2.

Concept:

To solve $\mathop {\lim }\limits_{x \to ∞ } \frac{{f\left( x \right)}}{{g\left( x \right)}}$ of this type, divide both numerator and denominator by the variable with the highest power.

Calculation:

Given:

$\mathop {\lim }\limits_{x - ∞ } \frac{{{x^2};-;5x;+;4}}{{4{x^2};+;2x}}$

Replacing x with ∞ in above expression, it is an indeterminate form $\left( {\frac{∞ }{∞ }} \right).$

Take out common the highest degree term from numerator and denominator.

$\therefore ;\mathop {\lim }\limits_{x \to ∞ } \left( {\frac{{{x^2}\left[ {1;-;\frac{5}{x};+;\frac{4}{{{x^2}}}} \right]}}{{{x^2}\left[ {4;+;\frac{2}{x}} \right]}}} \right)$

$\because\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0$

$\therefore;\mathop {{\rm{lim}}}\limits_{x \to ∞ } \left( {\frac{{\left[ {1;-;0;+;0} \right]}}{{\left[ {4;+;0} \right]}}} \right)$

$\therefore\frac{1}{4}$

Explanation

Let f(x) = In (x)

 

Let the approximate value of In (2) be x.

By using linear interpolation,

$\frac{{1.792 - 0}}{{6 - 1}} = \frac{{x - 0}}{{2 - 1}}$

⇒ x = 0.358

Thus the percentage error $= \frac{{True;;value - Approx;;value}}{{True;value}}$ $= \left( {\frac{{0.69 - 0.358}}{{0.69}}} \right) \times 100$

= 48.11% ≈ 48%

Important Points

• Because of the log scale involved in actual value it is noted that the error is extremely large under linear scale which makes sense.

Concept:

Ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Length major axis of ellipse = 2a

Length of minor axis of ellipse = 2b

Area $= \mathop \int!!!\int \limits_R^{} 1 \cdot dy \cdot dx$

Calculation:

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \Rightarrow y = \sqrt {b\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)}$.

For the first quadrant, take a vertical strip as shown. Here, y coordinate varies from 0 to $\sqrt {b\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)} = \frac{b}{a}\sqrt {{a^2} - {x^2}}$.

Also, the x-coordinate varies from 0 to a

∴ Area \(= \mathop \smallint \limits_0^a \mathop \smallint \limits_0^{\frac{b}{a}\sqrt {{a^2} - {x^2}} } \left( 1 \right)dy;dx\)

\(= \mathop \smallint \limits_0^a \left( {\frac{b}{a}\sqrt {{a^2} - {x^2}} } \right)dx\)

$= \frac{b}{a}\left[ {\frac{{x\sqrt {{a^2} - {x^2}} }}{2} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{a}} \right)} \right]_0^a = \frac{b}{a} \times \frac{{{a^2}}}{2} \times \frac{\pi }{2} = \frac{{\pi ab}}{4}$

∴ The total area of ellipse $= 4 \times \frac{{\pi ab}}{4}$ = πab units

Important Points

1. x² + y² = a² ; Represents a circle centred at (0, 0) and radius ‘a’ units.

2. $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$; Represents Ellipse with major axis length 2a and minor axis length 2b and vertex at (0,0)

3. $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$; Equation of Hyperbola .

4. $\frac{{x^2}}{{{a^2}}} - \frac{{{y^2}}}{{{a^2}}} = 1$; Equation of Rectangular Hyperbola.

Concept:

Planar Truss: When all the members and nodes of a truss lie within a 2-dimensional plane it is called a planar truss.

Members of a truss carry only axial force. If the member of a truss does not carry any force under some specific load conditions, those members are called zero-force members.

Identification of zero-force members:

1. If at any joint two members are non-collinear and no load is acting at the joint, then both the members will be zero-force members.

e.g.

In the truss shown above, there are two non-collinear members at joint D and there is no load acting at this joint, hence both the members BD and CD are zero force members.

2. If at a truss joint, 3 members are meeting and two of them are collinear and no load is acting at the joint, then the non-collinear member will be a zero-force member.

e.g.

Calculation:

A 2D truss as shown below:

Using property 1 explained above, at joint D members CD and EC are zero force members. Similarly, EF and EC are zero force members, and at joint C, CF and BC are zero force members,

At joint F;

∵ ∑ F_y = 0 ∴ F_FG = F_EF

∵ ∑ F_X = 0   ∴ F_BF = P

Thus the truss reduces to

This is an indeterminate truss (indeterminate to 1 degree) . Let the redundant be the vertical reaction at joint G.

θ = 45°

At joint B ; ∑ F_y = 0 ⇒ F_BG sin θ + P = 0

$\Rightarrow {F_{BG}} = - P\sqrt 2$

∑ F_X = 0 ⇒ F_BA + F_BG cos θ = 0

⇒ F_BA = P

Negative sign in F_BG indicates compression.

At joint A, ∵ ∑ F_Y = 0 ⇒ F_AG = P

∵ ∑ F_X = 0 ⇒ R_AH = 0

For stability of truss, ∑ F_x = 0

Thus horizontal reaction at G = P.             

Apply a unit load at G and in the direction of redundant assumed.

By using property 1, explained above F_AB = F_BG = 0

By using the unit load method redundant R is given as

$R = \frac{{ - \sum {u_i}\left( {\frac{{{P_i}{L_i}}}{{{A_i}{E_i}}}} \right)}}{{\sum \left( {\frac{{u_i^2{L_i}}}{{{A_i}{E_i}}}} \right)}}$

Where,

u_i = Member force due to the unit load applied.

P_i = Member force due to external load.

$\Rightarrow R = - \frac{{\left( {1 \times P \times \ell } \right)}}{{{1^2} \times \ell }}$

⇒ R = -P

∴ Force in GA = P + (-P) × ℓ = 0

Thus we have a total of 8 zero-force members.

Key Points

Force in member GA can also be found using the fact that truss members carry the only axial force and if the axial deflection of any member is zero that member will be a zero force member.

In the given truss, due to hinge at A and G no axial deformation of the member AG is possible.

Hence GA is a zero force member.

Concept:

The design bond stress is defined as the shear force per unit nominal surface area of reinforcing bar. The grip of reinforcement and concrete due to adhesion, shrinkage gripping and bearing on account of ribs on the deformed bar is termed as bond.

The stress acts on the interface between bars and surrounding concrete and along the direction parallel to the bars.

The local bond stress varies along the length of the reinforcement while the average bond stress gives the average value throughout its development length. This average bond stress is still used in the working stress method. However, in the limit state method of design, the average bond stress has been designated as design bond stress τ_bd and the values are given in cl. 26.2.1.1. The same is given below as a ready reference.

The basic requirement of RCC is that steel and surrounding concrete act together and there should be no slippage of the steel bar relative to the surrounding concrete.

The actual bond stress distribution is:

Important Points

As per IS 456: 2000 bond stresses are assumed to be uniform along the length of reinforcement.

Uniform bond stress – Distribution.

Bond stress value for plain bars in tension is given for various grades of concrete as:

 

• τ_bd  = 0.16 (f_ck)^2/3
• For deformed bars the value of bond stress is increased by 60% and for bars in compression, it is increased by 25%.

Concept:

For the state of pure shear, normal stresses on the plane must be equal to zero.

Also, the principal stresses are equal to shear stress.

Mohr’s circle for the state of pure shear is

Calculation:

Given:

 \(\sigma = \left[ {\begin{array}{*{20}{c}} {{\sigma _{xx}}}&{{\sigma _{xy}}}\ {{\tau _{xy}}}&{{\sigma _{yy}}} \end{array}} \right]\)

Principal stress $= \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{22}} - {\sigma _{yy}}}}{1}} \right)}^2} + \tau _{xy}^2}$

Shear stress $= \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + {{\left( {{\tau _{xy}}} \right)}^2}}$

For the pure shear state of stress –

Principal stress = shear stress

$\Rightarrow \frac{{{\sigma _{xx}} + {\sigma _{yy}}}}{2} \pm \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2} = \sqrt {{{\left( {\frac{{{\sigma _{xx}} - {\sigma _{yy}}}}{2}} \right)}^2} + \tau _{xy}^2}$

⇒ σ_xx  + σ_yy = 0

Important Points

In the case of pure shear:

• The magnitude of principal stresses are equal and they are opposite in nature.
• Mohr’s circle is centered at origin and radius = τ
• τ_max in-plane (τ) which is also the absolute maximum shear stress.

Concept:

The four major compounds which are constituents of cement are:

a) Tricalcium silicate  (C₃S): 3CaO.SiO₂

b) Dicalcium silicate (C₂S): 2CaO.SiO₂

c) Tricalcium Aluminate (C₃A): 3CaO.Al₂O₃

d) Tetra-calcium Alumino Ferrite (C₄AF): 4CaO.Al₂.Fe₂O₃

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These compounds are also known as Bogue Compounds.

Hydration of cement:

When water is added to cement a chemical reaction between water and cement takes place leading to the evolution of heat. This is called hydration of cement. The product formed is C-S-H gel.

The heat liberated during this reaction is called the heat of hydration.

• C₃S readily reacts with water, produces more heat of hydration and is responsible for the early strength of concrete.
• C₂S hydrates more slowly and produce less heat of hydration and are responsible for later strength of concrete.

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1. Tricalcium Silicate C₃S – (25 – 50%) – Normally 40%

• It is considered as the best cementing material and is well-burnt cement.
• It hydrates rapidly generating high heat and develops an early hardness and strength.
• Raising of C₃S content beyond the specified limits increases heat of hydration and solubility of cement in water.
• The heat of hydration is 500 J/g.

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2. Dicalcium Silicate (C₂S) - (25 – 40%) - (Normally 32%)

• It hydrates and hardens slowly and takes a long time to add to the strength (after a year or more) i.e. it is responsible for ultimate strength.
• It imparts resistance to chemical attack.
• Raising of C₂S content renders clinkers harder to grind, reduces early strength, decreases resistance to freezing and thawing at an early age and decreases heat of hydration.
• At an early age, less than a month, C₂S has little influence on strength and hardness. While after one year, its contribution to the strength and hardness is proportionately almost equal to C₃S.
• The heat of hydration is 260 J/g.

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3. Tricalcium Aluminate (C₃A) - (5 – 11%) -(Normally 10.5%).

• It rapidly reacts with water and is responsible for the flash set of finely grounded clinkers.
• The rapidity of action is regulated by the addition of 2-3% of gypsum at the time of grinding the cement.
• It is most responsible for the initial setting, the high heat of hydration and has a greater tendency to volume changes causing cracking.
• Raising the C₃A content reduces the setting time, weakens resistance to sulphate attack and lowers the ultimate strength, heat of hydration and contraction during air hardening.
• The heat of hydration of 865 J/g.

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4. Tetracalcium Alumino Ferrite - (C₄AF 8 – 14%) (Normally 9%)

• It is responsible for the flash set but generates less heat.
• It has the poorest cementing value. Raising C₄AF content reduces the strength slightly.
• The heat of hydration 420 J/g.

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The rate of heat evolution of the principal compound if equal amount of each is considered will be in following descending order: C₃A (865 J/g) > C₃S (500 J/g) > C₄ AF (420 J/g) > C₂S (260 J/g) .

Thus by increasing the C₂S content the heat of hydration decreases.

Note:

The development of strength of the four principal compounds of cement with age:

The rate of hydration is increased by an increase in the fineness of the cement. However total heat evolved is the same. The rate of hydration of the principal compounds is shown in the figure and will be in the following descending order: C₄AF > C₃A > C₃S > C₂S

Important Points

Hydration products of C₂S are considered better than that of C₃S. It is because of the lesser time formation of lime when C₂S hydrates than those in hydration of C₃S.

2C₃S + 6H → C₃S₂H₃ + 3 Ca(OH)₂

2C₂S + 4H → C₃S₂H₃ + Ca(OH)₂

Concept:

Los angeles abrasion test is a preferred test for examination of hardness property of the aggregates and has been standardized in India (IS 2386 : Part IV). Abrasion resistance is the property of aggregates by virtue of which they resist surface wear and scratches caused due to rubbing action of another surface or material.

The hardness test is of the following type:

1. Los Angeles abrasion test 2) Devel abrasion test  3) Dorry abrasion test

Los Angeles Abrasion Test :

The principle of the Los Angeles Abrasion Test is to find the percentage wear due to relative rubbing action between the aggregate and steel balls used as an abrasive charge. Los Angeles machine consists of circular drum of internal diameter 700 mm and length 520 mm mounted on a horizontal axis enabling it to be rotated. An abrasive charge consisting of cast iron spherical balls of 48 mm diameters and weight 340-445g is placed in the cylinder along with the aggregates.  

The cylinder is then locked and rotated at the speed of 30-33 rpm for a total of 500-1000 revolutions depending upon the gradation of aggregates.

After specified revolutions, the material is sieved through 1.7 mm sieve and passed fraction is expressed as percentage total Weight of the sample.

This value is called the Los Angeles abrasion value. A maximum Value of 40 percent is allowed for WBM base course m Indian conditions. For Bituminous concrete, a maximum value of 35 percent is specified.

Important Points

Some of the important aggregate tests are –

1. Crushing test: Resistance to crushing under gradually applied crushing load.

2. Impact test: To determine the toughness of aggregate.

3. Soundness test: To determine the resistance of aggregates to weathering action.

4. Abrasion test: To determine the hardness of aggregates

5. Shape test: Flaky Index, Elongation index and Angularity Number are determined.

6. Smith Test: To find soluble matter in stone aggregates.

7. Specific Gravity and Water Adsorption test.

Concept:

(1) Liquidity Index: It is given as -

${I_L} = \frac{{{\omega _n} - {\omega _p}}}{{{\omega _L} - {\omega _p}}}$

When ${I_L} > I \Rightarrow {\omega _n} - {\omega _p} > {\omega _L} - {\omega _p}$

∴ ω_n > ω_L

Hence the soil is in a state of liquid consistency.

Where, ω_n = Natural water content.

ω_L = Liquid limit water content.

ω_p = Plastic limit water content.

(2) Triaxial Test is the most widely used shear strength test suitable for all types of soil. It is performed in 2 stages on the basis of which, there are 3 types of triaxial test –

(a) Consolidated Drained Test: Drainage is allowed in both stages.

(b**) Consolidated undrained Test**: Drainage is allowed in stage 1 and not allowed in stage 2.

(c**) Unconsolidated Undrained Test:** Drainage is not allowed in any of the stages.

A normally consolidated clay has zero cohesion when tested in CU test because a completely saturated clay under σ̅ = 0 behaves as a slurry.

(3) The ultimate bearing capacity of sandy soil as obtained from the plate load test increases with

an increase in the size of the footing.

$\frac{{{q_{uf}}}}{{{q_{up}}}} = \frac{{{B_f}}}{{{B_p}}}$

But for clayey soil, the ultimate bearing capacity is independent of the size of the footing.

(4) Boussinesq’s equation:

${\sigma _z} = \frac{3}{{2\pi }}\frac{Q}{{{z^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{r}{z}} \right)}^2}}}} \right)^{5/2}}$

Westergaard’s equation:

${\sigma _z} = \frac{1}{\pi }\frac{Q}{{{z^2}}};{\left( {\frac{1}{{1 + 2{{\left( {\frac{r}{z}} \right)}^2}}}} \right)^{3/2}}$

Vertical stress directly below a point load i.e. when r = 0

 

Hence the stress obtained using Boussinesq’s equation is 1.5 times higher than that of Westergaard’s equation.

Important Points

• Consistency Index $\left( {{I_C}} \right) = \frac{{{\omega _L} - \omega }}{{{\omega _L} - {\omega _p}}}$
• I_L + I_C = 1

Concept:

Bearing Capacity can be calculated based on the following field test:

(a) Standard Penetration Test (S.P.T)

(b) Plate Load test

(c) Static cone penetration test/Static Cone resistance test.

Standard Penetration Test (S.P.T)-

S.P.T. Test procedure

• Suitable for Granular soil.
• Split spoon samples are used in the borehole.
• The borehole is advanced to a depth at which N-value is to be calculated.
• The split-spoon sampler is allowed to penetrate the soil by applying an impact load of 65 Kg having a free fall of 75 cm.
• The sampler is allowed to penetrate for 150 mm depth, but reading is not noted i.e.(No. of blows required for 150 mm penetration is not Noted).
• Then the sampler is allowed to penetrate, further for 300 mm and No. of blows required to penetrate the sampler to this 300 mm is the SPT N-value
• The next test is carried out at level 750 mm below the previous test reference level.
• If borehole depth is large, then the interval of the next test is taken at a depth of 2.5 to 2 m or the change of strata.

 

Calculation of S.P.T N value:

• Bearing capacity of Granular soil can be calculated by using corrected S.P.T N-values.
• The values of N – is determined at Number of Selected boreholes and the average value of corrected (N) is calculated.
• While calculating the average value of (N), any value greater than 50% of the average value is discarded.

 

e.g. if $\left[ {\frac{{{N_1} + {N_2} + {N_3} + {N_4}}}{4}} \right] = 30$ and if N₃ = 40

Then (N₃) is accepted as [30 × 1.5 = 45] and (40 < 45) and if N₂ = 48 then (N₂) is discarded as 48 is > 45

• New average is taken as $= \left[ {\frac{{{N_1} + {N_3} + {N_4}}}{3}} \right]$
• Minimum of all the avg. values of various boreholes are used in the design of foundations.

Calculation:

Given:

The number of blows for each 150 mm penetration is 8, 6, and 9.

The number of blows for the first 150 mm penetration is not considered.

Hence the correct SPT-N value = 6 + 9 = 15

Important Points

The following corrections are applied to the N-values obtained above:-

1. Overburden Correction

• It is necessary because the N-value will affect it due to the confinement of soil at various depth.
• Two granular soils possessing the same relative density but having different confining pressures are tested, the one with higher confining pressure will give higher N value.
• Since the confining pressure increases with depth, the N values at shallow depths are underestimated and the N values at larger depths are overestimated.
• If N₀ = observed S.P.T. value

 

${N_1} = {N_0} \times \frac{{350}}{{\left( {\bar \sigma + 70} \right)}}$

Where, σ̅ = Effective stress at the level of the test (KN/m²)

N₁ = Corrected N-value of overburden.

• Overburden correction will not be applied if σ̅ > 280 KN/m²

 

2. Dilatancy Correction

• It is applied to the already corrected N-values for overburden pressure. Dilatancy correction is required only if N₁ > 15 in saturated fine sand and silt
• (N₁ > 15) represents the Dense sand which will tend to dilate under rapid loading (undrained condition) and (-VA) pore water pressure will develop. Hence, observed (N) value will be more because shear resistance will increase.

 

Corrected N value after Dilatancy Correction $:{N_2} = 15 + \frac{1}{2}\left( {{N_1} - 15} \right)$

Concept:

Darcy’s Law:

The flow of free water through soil is governed by Darcy’s law. As per this law, for laminar flow through homogenous soil, the velocity of flow (v) is given as

ν = ki

where, k = coefficient of permeability, i = hydraulic gradient

ν = discharge velocity or superficial velocity.

Validity of Darcy’s law:

It is valid when the flow through the soil is laminar. This assumption was based on the fact that there is no head loss associated when the water flows through soil due to the formation of turbulent rollers.

For the flow of water through soil, the flow is laminar if Reynold’s number is less than 1.

• In fine-grained soil, flow is laminar.
• In coarse-grained soil, the flow becomes turbulent when the size of particles is very large i.e. in gravel range.

Thus the assumed soil mass must be homogenous and fully saturated for the validity of Darcy’s law.

Hence Darcy’s law applies to laminar flow in porous media.

Important Points

Coefficient of Permeability (k):

It is defined as the velocity of flow which would occur under a unit hydraulic gradient. The unit of k is m/sec or cm/sec. It depends upon the particle size and many other factors.

Seepage velocity is different from discharge velocity. Darcy’s law assumes that the flow takes place through the complete cross-section, but in reality, it flows only through the voids between the soil particles.

Seepage velocity calculations:

Seepage velocity (V_s) = q/A_v ⇒ q = A_v × V_s

Discharge velocity (V) $= \frac{q}{A} \Rightarrow q = A \times V$

∵ A_v × V_s = A × V

$\Rightarrow {V_s} = \frac{V}{{\left( {\frac{{{A_v}}}{A}} \right)}} = \frac{V}{n}$ ; n → porosity.

Thus the seepage velocity is always greater than the actual velocity.

Concept:

Metacenter:

It is the point of intersection of the line of action of buoyant force before and after rotation.

1. The basic floating position is calculated Aand the body’s center of mass G and center of buoyancy B are computed.
2. The body is tilted a small angle Δθ, and a new water line is established for the body to float at this angle. The new position B’ of the center of buoyancy is calculated. A vertical line is drawn upward from B’ that intersects the previous line of buoyancy at a point M, called the metacenter, which is independent of Δθ for small angles.
3. If point M is above G, that is, if the metacentric height $\overline {MG}$ is positive, a restoring value indicates the stability of the body.

Thus, a floating body is said to be in a state of stable equilibrium if the metacenter of this lies above the center of gravity. In other words, the metacentric height is positive.

Important Points

Metacentric Height: It is the distance between the center of gravity and the metacenter.

$\overline {MG} = BM - BG$

Stability of submerged body:

It is determined based on the location of the center of buoyancy B and the center of gravity G with respect to each other.

1) 

2) 

Concept:

Velocity Potential:

It is defined as a scalar function of space and time such that its partial negative derivative with respect to any direction gives the velocity of flow in that direction.

If ϕ = f(x, y, z, t)

$\Rightarrow \frac{{ - \partial \phi }}{{\partial x}} = u;;;\frac{{ - \partial \phi }}{{\partial y}} = \nu ;;;\frac{{ - \partial \phi }}{{\partial z}} = \omega$

If ϕ in a particular direction is constant, the flow will have no velocity component in that direction and hence flow will not occur in that direction.

If Velocity potential satisfies the Laplace equation, the Continuity equation gets satisfied and flow is possible.

$\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = 0$

Calculation:

For the given velocity field,

Velocity in x-direction = u sin θ

Velocity in y-direction = u cos θ

From the definition of velocity potential,

$\frac{{ - \partial \phi }}{{\partial x}} = u\sin \theta$

Integrating on both sides,

ϕ = -u sin θ x + f(y)     ---(i)

Also, $\frac{{\partial \phi }}{{\partial y}} = - u\cos \theta$

$\Rightarrow \frac{{\partial ( - u\sin \theta \cdot x + f\left( y \right)}}{{\partial y}} = - u\cos \theta$

⇒ 0 + f’(y) = u cos θ

⇒ f(y) = -(u cos θ) ⋅ y + c      ---(ii)

Thus using equation (i) & (ii),

ϕ = -u sin θ ⋅ x + (-u cos θ) ⋅ y

⇒ ϕ = ± U (x sin θ + y cos θ)

Important Points

The assumption of Irrotational flow leads to the existence of the Velocity potential and velocity potential concept is used for integration of Euler’s equation to yield Bernoulli's equation.

Explanation:

Given,

Pan evaporation = 100 mm ; effective rainfall = 20 mm ;

Crop effective = 0.4 ; Irrigation Efficiency = 0.5

Total water required by crop = crop coefficient × pan evaporation = 0.4 × 100 = 40 mm

Effective Rainfall = 20 mm

∴ Additional water required by crop = (40 - 20) = 20 mm

∵ The efficiency of irrigation = $\frac{{Amount;of;water;available;for;use}}{{Amount;of;water;applied;in;the;field}}$

⇒ Amount of water applied $= \frac{{20}}{{0.5}} = 40;mm$

Important Points

Part of the rainwater percolates below the root zone of the plants and part of the rainwater flows away over the soil surface as run-off. This deep percolation water and run-off water cannot be used by the plants**.** The remaining part is stored in the root zone and can be used by the plants. This remaining part is called effective rainfall.

Concept:

Disinfection is defined as the method of inactivation of harmful microorganisms in water either by the physical or chemical process.

Chlorination is one of the major methods of disinfection.

When chlorine is added to water, the reaction takes place between them.

At pH < 5, chlorine does not react with water and remains as free chlorine or unreacted chlorine.

Reaction of chlorine with water

Out of all these forms of chlorine, HOCl^- is the most destructive. It is 80% more effective than OCl^- ion.

Hence pH is maintained just below 7 for most effective chlorination.

Important Points

• Chlorine immediately reacts with ammonia present in water to form chloramines. Out of all chloramines formed dichloramine is predominant. They are less effective than free chlorine but they are stable and remain in the water for a greater duration and hence protect from future contamination.
• Different forms in which chlorine is added:

<br>

(i) As free chlorine

(ii) Hypochlorites (Bleaching powder)

(iii) Chloramines

(iv) Chlorine dioxide

Concept:

pH:

It is a measure of the acidity or alkalinity of a solution. It measures the presence of hydrogen ion concentration relative to that of a standard solution. Mathematically, pH is the negative logarithm of the hydrogen ion concentration [H+] present in the solution expressed in molarity.

pH = -log₁₀ [H⁺]

Molarity:

It is defined as the number of moles of solute per litre of solution.

i.e. ${\rm{Molarity}} = \frac{{Moles;of;solute}}{{Volume;of;solution;in;\ell }}$

Calculation:

When HCl is added to water, it dissociates into H⁺ and Cl^-

HCl + H₂O ⇌ H⁺ + Cl^- + H₂O

Molecular weight of HCl = ( 1 +35.5) = 36.5 gm

Thus, 36.5 gm of HCl dissociates to 1 gm of H⁺

⇒ 1 gm of HCl dissociates to $\frac{1}{{36.5}};gm$ of H⁺

⇒ 35.67 mg of HCl dissociates to $\left( {\frac{1}{{35.5}} \times 35.67 \times {{10}^{ - 3}}} \right);gm$ of H⁺

$= \frac{{\left( {\frac{1}{{36.5}} \times 35.67 \times {{10}^{ - 3}}} \right)}}{1};moles;of;[{H^ + }]$

∴ The concentration of H⁺ in molarity $= \frac{{\frac{1}{{36.5}} \times 35.67 \times {{10}^{ - 3}}}}{{1;litre}} = 0.977 \times {10^{ - 3}}\frac{{moles}}{{litre}}$

∴ pH = -log₁₀ [H⁺] = -log₁₀ [0.977 × 10^-3] = 3.01

Important Points

pH > 7 ⇒ Basic solution

pH < 7 ⇒ Acidic solution

pH + p(OH) = 14

pH is measured using a potentiometer in which the potential exerted by [H⁺] ions is measured.

 

• mmole/ℓ =millimoles/ℓ = 10^-3 moles/ℓ

Explanation:

$\frac{{{d^2}x}}{{d{t^2}}} - 5\frac{{dx}}{{dt}} + 6x = 0$

Above given equation is a linear differential equation of order = 2. Such equations are solved using CF + PI method.

Let $D = \frac{{dx}}{{dt}}$

$\Rightarrow \frac{{{d^2}x}}{{d{t^2}}} - 5\frac{{dx}}{{dt}} + 6x = 0$

⇒ D²x – 5Dx + 6x = 0

⇒ (D² – 5D + 6)x = 0

Thus auxiliary equation (obtained by replacing D with m) is m² – 5D + 6 = 0.

Roots of above obtained auxiliary equation are-

m² – 5D + 6 = 0

⇒ (m – 2) (m - 3) = 0

⇒ m₁ = 2 or m₂ = 3

When both roots of auxiliary equation are real and distinct

$C.F = {C_1}{e^{{m_1}t}} + {C_2}{e^{{m_2}t}}$

Thus general solution of above D.E is

$x = {C_1}{e^{2t}} + {C_2}{e^{3t}}$

Applying boundary conditions, x(0) = 0

⇒ O = C₁ + C₂

⇒ C₁ = -C₂     ---(i)

Using initial condition $\frac{{dx}}{{dt}}\left( 0 \right) = 10$

$\frac{{dx}}{{dt}} = 2G{e^{2t}} + 3{C_2}{e^{3t}}$

⇒ 10 = 2C₁ + 3C₂       ---(ii)

Solving equation (i) & (ii) simultaneously,

C₁ = -10 and C₂ = 10

Thus, x = -10e^2t + 10e^3t

Explanation:

By using Taylor’s series expansion,

$f\left( {x + h} \right) = f\left( x \right) + hf'\left( x \right) + \frac{{{h^2}f''\left( x \right)}}{{2!}} + \frac{{{h^3}f'''\left( x \right)}}{{3!}} + \frac{{{h^4}{f^4}\left( x \right)}}{{4!}} + \frac{{{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{{h^6}}}{{6!}}{f^6}\left( x \right) + \frac{{{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$f\left( {x - h} \right) = f\left( x \right) - hf'\left( x \right) + \frac{{{h^2}f''\left( x \right)}}{{2!}} - \frac{{{h^3}f'''\left( x \right)}}{{3!}} + \frac{{{h^4}{f^4}\left( x \right)}}{{4!}} - \frac{{{h^5}}}{{5!}}{f^5}\left( x \right) + \frac{{{h^6}}}{{6!}}{f^6}\left( x \right) - \frac{{{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$f\left( {x + 2h} \right) = f\left( x \right) + 2hf'\left( x \right) + \frac{{4{h^2}f''\left( x \right)}}{{2!}} + \frac{{8{h^3}f'''\left( x \right)}}{{3!}} + \frac{{16{h^4}{f^4}\left( x \right)}}{{4!}} + \frac{{32{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{64{h^6}{f^6}\left( x \right)}}{{6!}} + \frac{{128{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$f\left( {x - 2h} \right) = f\left( x \right) - 2hf'\left( x \right) + \frac{{4{h^2}f''\left( x \right)}}{{2!}} - \frac{{8{h^3}f'''\left( x \right)}}{{3!}} + \frac{{16{h^4}{f^4}\left( x \right)}}{{4!}} - \frac{{32{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{64{h^6}{f^6}\left( x \right)}}{{6!}} - \frac{{128{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$f\left( {x + 3h} \right) = f\left( x \right) + 3hf'\left( x \right) + \frac{{9{h^2}f''\left( x \right)}}{{2!}} + \frac{{27{h^3}f'''\left( x \right)}}{{3!}} + \frac{{81{h^4}{f^4}\left( x \right)}}{{4!}} + \frac{{243{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{729{h^6}{f^6}\left( x \right)}}{{6!}} + \frac{{2187{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$f\left( {x - 3h} \right) = f\left( x \right) - 3hf'\left( x \right) + \frac{{9{h^2}f''\left( x \right)}}{{2!}} - \frac{{27{h^3}f'''\left( x \right)}}{{3!}} + \frac{{81{h^4}{f^4}\left( x \right)}}{{4!}} - \frac{{243{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{729{h^6}{f^6}\left( x \right)}}{{6!}} - \frac{{2187{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

To find the fourth order approximation of third derivative, we need to eliminate all the lower order derivatives up to 6^th.

$f\left( {x + h} \right) - f\left( {x - h} \right) = 2hf'\left( x \right) + \frac{{2{h^3}f'''\left( x \right)}}{{3!}} + \frac{{2{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{2{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$ …1)

$f\left( {x + 2h} \right) - f\left( {x - 2h} \right) = 4hf'\left( x \right) + \frac{{16{h^3}f'''\left( x \right)}}{{3!}} + \frac{{64{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{256{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$ …2)

$f\left( {x + 3h} \right) - f\left( {x - 3h} \right) = 6hf'\left( x \right) + \frac{{54{h^3}f'''\left( x \right)}}{{3!}} + \frac{{486{h^5}{f^5}\left( x \right)}}{{5!}} + \frac{{4374{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$ …3)

By performing the operation, [2(equation 1) – equation 2], we get

2[f(x + h) – f(x - h)] – [f(x + 2h) - f(x – 2h)]

$= - \frac{{12{h^3}f'''\left( x \right)}}{{3!}} - \frac{{60{h^5}{f^5}\left( x \right)}}{{5!}} - \frac{{252{h^7}{f^7}\left( x \right)}}{{7!}} + \ldots$

$= - 2{h^3}f'''\left( x \right) - \frac{1}{2}{h^5}{f^5}\left( x \right) - 0.05{h^7}{f^7}\left( x \right) + \ldots$ …4)

By performing the operation [3(equation 1) – equation 3], we get

3[f(x + h) – f(x - h)] – [f(x + 3h) - f(x – 3h)]

$= - \frac{{48{h^3}}}{{3!}}f'''\left( x \right) - \frac{{480{h^5}{f^5}\left( x \right)}}{{5!}} - \frac{{4368}}{{7!}}{h^7}{f^7}\left( x \right)$

= -8h³ f’’’(x) – 4h⁵ f⁵(x) – 0.758 h⁷ f⁷(x) …5)

By performing the operation [equation 5 – 8 (equation 4)], we get

3[f(x + h) – f(x - h)] – [f(x + 3h) – f(x – 3h)] – 8[2[f(x + h) – f(x - h)] – [f(x + 2h) – f(x – 2h)]] = 8h³ f’’’(x) – 0.358 h⁷ f⁷(x)

⇒ 3f(x + h) – 3f(x - h) – f(x + 3h) + f(x – 3h) – 16f(x + h) + 16f(x - h) – 8f(x + 2h) + 8f(x - 2h) = 8h³ f’’’(x) – 0.358h⁷ f⁷(x)

⇒ 8h³ f’’’(x) – 0.358 h⁷ f⁷(x) = -f(x + 3h) + 8f(x + 2h) – 13f(x + h) + 13f(x - h) -8f(x - 2h) + f(x – 3h)

$\Rightarrow f'''\left( x \right) - \left[ {\frac{{f\left( {x + 3h} \right) + 8f\left( {x + 2h} \right) - 13f\left( {x + h} \right) + 13f\left( {x - h} \right) - 8f\left( {x - 2h} \right) + f\left( {x - 3h} \right)}}{{8{h^3}}}} \right] = 0.358{h^4}{f^7}\left( x \right)$

The above equation gives the fourth order (h⁴) approximation of third derivative.

Therefore, the fourth-order central difference approximation of the third derivative is

$f'''\left( x \right) = \frac{{ - f\left( {x + 3h} \right) + 8f\left( {x + 2h} \right) - 13f\left( {x + h} \right) + 13f\left( {x - h} \right) - 8f\left( {x - 2h} \right) + f\left( {x - 3h} \right)}}{{8{h^3}}}$

$f'''\left( {{x_i}} \right) = \frac{{ - f\left( {{x_{i + 3}}} \right) + 8f\left( {{x_{i + 2}}} \right) - 13f\left( {{x_{i + 1}}} \right) + 13f\left( {{x_{i - 1}}} \right) - 8f\left( {{x_{i - 2}}} \right) + f\left( {{x_{i - 3}}} \right)}}{{8{h^3}}}$

Concept:

The value of any stress function can be found by drawing a qualitative ILD using the Muller-Breslau principle.

The Muller Breslau Principle for determinate structure states that the ordinate value of an influence line for any stress function on any structure is proportional to the ordinates of the deflected shape that is obtained by removing the restraint offered to correspond to the stress function from the structure and introducing a force that causes a unit displacement in the positive direction.

Given:

This is a determinate structure over which a UDL of 50 kN/m moves. The maximum negative moment is found by drawing ILD for the moment at R.

Using, the Muller-Breslau Principle, ILD is as shown:

θ₁ + θ₂ = 1 units

for small θ, tan θ ≈ sin θ ≈ θ

In ΔQRR’ and SRR’,

$\frac{\beta }{3}+\frac{\beta }{2}=1\Rightarrow \beta =1.2~units$

Again, ΔPQP’ and ΔQRR’ are similar

$\therefore \frac{\beta }{3}=\frac{\alpha }{1.5}\Rightarrow \alpha =0.6~units$

ΔSTT’ and ΔSRR’ are similar

$\therefore \frac{\gamma }{1.5}=\frac{\beta }{2}\Rightarrow \gamma =0.9~units$

For the maximum negative moment, the UDL should lie from P to Q and from S to T.

Maximum negative bending moment =

$50\times \left[ \frac{1}{2}\times 0.6\times 1.5 \right]+50~\left[ \frac{1}{2}\times 0.9\times 1.5 \right]~$ = 56.25 kN/m

Important Points

To find out the value of stress function due to point loads we draw the ILD for that stress function and find out the ordinate of ILD at the location of the points load and the value of stress function is obtained by multiplying the ordinate of ILD with the value of point load and summed up.

The value of the stress function for UDL is obtained by multiplying the area under the ILD for stress function within the range of UDL with the load intensity (w).

Explanation:

Drawing FBD of the given frame –

$\sin \theta = \frac{3}{5};;and;\cos \theta = \frac{4}{5}$

For vertical equilibrium of the frame,

∑ F_y = 0 ⇒ R₀ sin θ – 90 = 0

$\Rightarrow {R_0} = \frac{{90}}{3} \times 5 = 150;kN$

At point O, the horizontal component of reaction,

${H_0} = {R_0}\cos \theta = 150 \times \frac{4}{5} = 120;kN$

Important Points

• The equilibrium of the structure is decided based on external force only. The internal member forces do not impact the equilibrium of the structure.

Concept:

Beam or frame only bends in the region where bending moment acts.

In the given beam RQ will not bend as there is no bending moment in this zone and hence it will not show any curvature.

$\delta = \frac{{P{\ell ^3}}}{{3EI}},;;\theta = \frac{{P{\ell ^2}}}{{2EI}}$

Calculation:

The deflected curve of the beam can be

δ_Q = δ₁ + δ₂

Using standard results,

${\delta _1} = \frac{{M{{\left( {\frac{L}{2}} \right)}^2}}}{{2EI}} = \frac{{M{L^2}}}{{8EI}}$

${\delta _2} = {\theta _R} \cdot \frac{L}{2} = \frac{{M\left( {\frac{L}{2}} \right)}}{{EI}} \times \frac{L}{2} = \frac{{M{L^2}}}{{4EI}}$

$\Rightarrow {\delta _Q} = {\delta _1} + {\delta _2} = \frac{{M{L^2}}}{{8EI}} + \frac{{M{L^2}}}{{4EI}} = \frac{{3M{L^2}}}{{8EI}}$

Important Points

Deflection calculations using standard results:

Concept:

Various type of joints provided in the cement concrete pavement are:-

1. Expansion joints 2) Contraction joints 3) Construction joints

Contraction joints:

It is provided to control crack due to shrinkage and moisture variation. To regulate the cracks or to determine that the cracks form at the predetermined location, the slab is weakened at certain intervals. These locations are called contraction joints.

• Dowel bars are provided as load transfer devices across transverse joints and they tend to keep the two pavement slabs at same right. They are bonded on one side and free on the other side.
• Various stresses that develop in a dowel bar are:-

 

(a) Shearing stress (b) Bending stress (c) Bearing stress

The contraction cracks appear near the centre of the dowel bars. Hence the most correct description is shown in option.

Important Points

Expansion joint: These are provided to allow the expansion of pavement due to a rise in temperature with respect to construction temperature.

The maximum spacing between the expansion joint is 140 m.

 

Generally, tie bars are provided at the longitudinal joints. Tie bars are not designed as load transfer devices but ensure that the two slabs remain firmly together. The load is transferred to the adjacent slab by the aggregate interlock.

 

Dowel bars: They are the load-bearing devices provided at transverse joints.

Tie Bars: They are not load-bearing devices and provided at longitudinal joints.

Concept:

The phenomenon of backing and queueing up on a highway due to sudden reduction in the capacity of the highway due to accident, reduction in the number of lanes due to bridgework, etc is known as Bottleneck condition.

When such a condition exists, and the normal flow is relatively large the speed of the vehicles must be reduced while passing the bottleneck which is approximately noted by turning on of the brake lights of the vehicle which is observed to move upstream as traffic continues to reach this vicinity of the bottleneck. This is called Shockwave in the traffic stream. 

Here the decreasing flow from q₂ to q₆ depicts the Shockwave condition.

Concept:

Clearway:

• An area beyond the end of the runway, centrally located along the extended centreline of the runway and not less than 150 m in width is known as clearway.
• Provided as a precautionary measure against engine failure.
• Clearway length < 0.5 (1.15 × Take off distance - 1.15 × Lift off distance)

 

The basic runway length is affected by the following factors:

1. Aircraft characteristics: The most critical aircraft is identified and its take-off and landing weight is noted to find the runway length.

2. Maximum of the three conditions -

(a) Normal take-off case

(b) Normal landing case

(c) Emergency stop case

3. Elevation of Airport and Airport Reference Temperature.

(4) The gradient of the Airport.

a) Normal Takeoff case:-

• Runway length + Clearway = Take-off distance
• Take off Distance ≥ 115% of Runway length.
• Width of clear way must be > 150 m.

b) Normal Landing case: The landing case requires that aircraft should come to stop within 60% of the landing distance. The runway of full-strength pavement is provided for the entire landing distance.

Calculation:

As per the considerations given above for the normal take-off case -

Clearway ≤ $\frac{1}{2} \times \left( {1.15 \times {\rm{takeoff;distance}} - 1.15 \times {\rm{liftoff;distance}}} \right)$ 

$\le \frac{1}{2} \times \left( {1.15 \times 1625 - 1.15 \times 875} \right)$ ≤ 431.25 m