GATE CE 2020 Official Paper: Shift 2 (Previous Year Paper)

Concept:

Impartial is the opposite of partial. In the same way, we have to look for the opposite word of popular**.**

⇒ Opposite of popular is unpopular.

⇒ Synonyms of unpopular are awful, horrible, miserable, poor, rotten.

Concept:

• Positions of the playground, road, lab, and lift are in the fixed position.
• We have to find the places for P, Q, R, S, and HoD according to their satisfaction.

Explanation:

1. P wanted a room adjacent to ‘LAB’.

2) Q wanted to be close to the ‘LIFT’.

3) R wanted the view of the playground

, or 

,or 

4. S wanted the corner office

5. Position of R and HoD is not fixed, so the possible solutions can be

, or 

Concept:

Composite function:

Consider f(x) as one function and g(x) as another function

Then f of g(x) is denoted by f ∘ g(x)

Also g of f(x) is denoted by g ∘ f(x)

Where,

f ∘ g(x) = f(g(x))

f(g(x)) is calculated by substituting g(x) in place of x in function f(x)

g ∘ f(x) = g(f(x))

g(f(x)) is calculated by substituting f(x) in place of x in function g(x)

Calculation:

$f\left( x \right) = {x^2};for;each;X;;\left( { - \infty ,;\infty } \right)$

$f\left( x \right) = {x^2}$

$f(f\left( {f\left( x \right))} \right) = f\left( {f\left( {{x^2}} \right)} \right) = f\left( {{x^4}} \right) = {\left( {{x^4}} \right)^2} = {x^8}$

$\frac{{f(f\left( {f\left( x \right))} \right)}}{{f\left( x \right)}} = \frac{{{x^8}}}{{{x^2}}} = {x^6} = {\left( {{x^2}} \right)^3} = {\left( {f\left( x \right)} \right)^3}$

Concept:

• Real interest rate:

It is the rate of interest on which a borrower pays the amount to the lender. In this rate of interest rate inflation rate is also adjusted.

• The nominal rate of interest:

It is the rate of interest on which the borrower pays the amount to the lender. In this rate of the interest rate of inflation is not considered.

• The relation between the real rate of interest, the nominal rate of interest, and inflation rate:

The real rate of interest = Nominal rate of interest – rate of inflation

Explanation:

Statement 1: under the high inflation, the real rate of interest is low and borrowers get benefited.

• The real rate of interest = Nominal rate of interest – rate of inflation.
• As we can see, if the rate of inflation increases, the real rate of interest decreases, and the rate at which borrowers had to pay the lender before is also less, hence the borrower gets benefited.
• Hence statement 1 is correct.

 

Statement 2: under the low inflation, the real rate of interest is high and borrowers get benefited.

• The real rate of interest = Nominal rate of interest – rate of inflation.
• As we can see, if the rate of inflation decreases, the real rate of interest increases, and the rate at which borrowers had to pay the lender before is high, hence the lender gets benefited, borrowers are in loss.
• Hence statement 2 is wrong.

 

Statement 3: under the high inflation, the real rate of interest is low and lenders get benefited.

• The real rate of interest = Nominal rate of interest – rate of inflation
• As we can see, if the rate of inflation increases, the real rate of interest decreases, and the rate at which borrowers had to pay the lender before is less, hence the borrowers get benefited, lenders are in loss.
• Hence statement 3 is wrong.

 

Statement 4: Under the low inflation, the real rate of interest is low and borrowers get benefited

• The real rate of interest = Nominal rate of interest – rate of inflation
• As we can see, if the rate of inflation decreases, the real rate of interest increases, and the rate at which borrowers had to pay the lender before is high, hence the lenders get benefited, borrowers are in loss.
• Hence statement 4 is wrong.

Concept:

Leap year: A leap year has 366 days and it repeats every 4 years

No. of odd days in a non-leap year = 365/7 = 1 (remainder)  

No. of odd days in a leap year = 366/7 = 2 (remainder)

Where,

365 = no. of days in a year

7 = no. of days in a week.

Calculation:

Hence to find out the same calendar as for the year 2019, the number of odd days from that year to the 2019 year must be zero.

 

Hence the 2011 year calendar is not as same as that of 2019.

 

Hence the 2012 year calendar is also not as same as the 2019 calendar.

 

Hence the 2013 year calendar is the same as that of the 2019 calendar.

Short Tricks:

We can eliminate option 2012 as follows

The 2019 year is not a leap year, hence the leap year 2012 will not be the answer even if the No. of odd days is zero.

Concept:

The number of terms in the arithmetic progression is calculated by,

b = a + (n – 1) × d

The n^th term in the arithmetic progression is given by,

$n = \frac{{b - a}}{d} + 1;$

The sum of terms in arithmetic progression (A.P) is calculated by,

$\% MathType!MTEF!2!1!+- \% feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbmaeaaaaaa \% aaa8qacaWFZbGaeyypa0ZaaSaaa8aabaWdbiaa-5gaa8aabaWdbiaa \% ikdaaaGaey41aq7aaeWaa8aabaWdbiaa-jgacqGHRaWkcaWFHbaaca \% GLOaGaayzkaaaaaa!407C! s = \frac{n}{2} \times \left( {b + a} \right)$(s = \frac{n}{2} \times \left( {b + a} \right);\)

S = sum of terms in A.P

n = number of terms in the sequence

b = last term in the sequence = n^th term in A.P

a = first term in the sequence

d = common difference

The odd positive integers from 1 to 100 is 1, 3, 5, 7……. 99.

First number = a = 1

Last number = b = 99

Common difference = d = 3-1 = 2

So, the number of terms in arithmetic progression = n

$n = \frac{{b - a}}{d} + 1 = \frac{{99 - 1}}{2} + 1 = \frac{{98}}{2} + 1 = 50$

So, the sum of the arithmetic progression = s

$s = \frac{n}{2} \times \left( {b + a} \right) = \frac{{50}}{2} \times \left( {99 + 1} \right) = 2500;$

The even positive integers from 150 to 200 are 150, 152, 154 …….. 200

First number = a = 150

Last number = b = 200

Common difference = d = 152 - 150 = 2

So, the number of terms in arithmetic progression = n

$n = \frac{{b - a}}{d} + 1 = \frac{{200 - 150}}{2} + 1 = 25 + 1 = 26$

So, the sum of the arithmetic progression = s

$s = \frac{n}{2} \times \left( {b + a} \right) = \frac{{26}}{2} \times \left( {200 + 150} \right) = 4550$

$\frac{{{\bf{The}};{\bf{sum}};{\bf{of}};{\bf{the}};{\bf{odd}};{\bf{positive}};{\bf{integers}};{\bf{from}};1;{\bf{to}};100}}{{{\bf{The}};{\bf{sum}};{\bf{of}};{\bf{the}};{\bf{even}};{\bf{positive}};{\bf{integers}};{\bf{from}};150;{\bf{to}};200}} = \frac{{2500}}{{4550}} = \frac{{50}}{{91}};$

Calculation:

The total number of students in school = 1000

Students who play chess = 300

Students who play football = 600

Students who play both chess and football = 50

Hence the total number of students playing sports = 250 + 50 + 550 = 850

∴ The total number of students not playing any of the sports = 1000 – 850 = 150

Explanation:

LED bulbs sold by

Y in April = (2/5) × 15% = 6%

Y in May = (4/5) × 10% = 8%

Y in June = (11/20) × 10% = 5.5%

∴ Total bulbs sold by Y from April to June = 6 + 8 + 5.5 = 19.5%

Bulbs sold by X and Y firm from April to June = 35% of (total bulbs sold by X and Y from January to June) = 50000

∴ The number of bulbs sold by Y from April to June = (50000/35) × 19.5% = 27857.14

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.

Given:

 $\frac{{{d^2}u}}{{d{x^2}}} - 2{x^2}u + \sin x = 0$

It is free from all the four characteristics described in the table. Hence it is a linear differential equation.

Product of dependent and independent variable i.e. x and u is present. Hence it is a non-homogenous equation.

Explanation:

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

$\therefore \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {9{x^2} + 2020} }}{{x + 7}}$

$= \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\sqrt {{x^2}\left( {9 + 2020/{x^2}} \right)} }}{{x\left( {1 + \frac{7}{x}} \right)}}$

$= \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{x\sqrt {9 + \frac{{2020}}{{{x^2}}}} }}{{x\left( {1 + \frac{7}{x}} \right)}}$

Replace x by infinity,

$= \mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\sqrt {9 + \frac{{2020}}{\infty }} }}{{\left( {1 + \frac{7}{\infty }} \right)}} = 3$

Important Points

Other Indeterminate forms are \(\frac{0}{0},\frac{\infty }{\infty },;0 \times \infty ,;\infty - \infty ,;0^\circ ,;\infty ^\circ ;;& ;{1^\infty }\)

Concept:

General Quadrature Formula (G.Q.F):-

\(I = \mathop \smallint \nolimits_a^b f\left( x \right)dx = \mathop \smallint \nolimits_{{x_0}}^{{x_n}} f\left( x \right)dx\)

$= h\left[ {n{y_0} + \frac{{{n^2}}}{2}{\rm{\Delta }}{y_0} + \left( {\frac{{{n^3}}}{3} - \frac{{{n^2}}}{2}} \right){{\rm{\Delta }}^2}\frac{{{y_0}}}{{2!}} \ldots } \right]$

Where, h = Step size

n = Number of strips

1. Trapezoidal rule: Taking n = 1 strip at a time and neglecting second and higher-order differences in G.Q.F

\(\mathop \smallint \nolimits_{{x_0}}^{{x_1}} f\left( x \right)dx = h\left[ {1.{y_0} + \left( {\frac{1}{2}} \right){\rm{\Delta }}{y_0} + Neglect} \right]\)

$= h\left[ {{y_0} + \frac{1}{2}\left( {{y_1} - {y_0}} \right)} \right]$

$= \frac{h}{2}\left( {{y_0} + {y_1}} \right)$

Again,

\(\mathop \smallint \nolimits_{{x_1}}^{{x_2}} f\left( x \right)dx = h\left[ {1.{y_1} + \frac{1}{2}{\rm{\Delta }}{y_1} + Neglect} \right]\)

$= h\left[ {{y_1} + \left( {\frac{1}{2}} \right)\left( {{y_2} - {y_1}} \right)} \right]$

$= \frac{h}{2}\left( {{y_1} + {y_2}} \right)$

Similarly, \(\mathop \smallint \nolimits_{{x_2}}^{{x_3}} f\left( x \right)dx = \frac{h}{2}\left( {{y_2} + {y_3}} \right)\) 

\(\mathop \smallint \nolimits_{{x_{n - 1}}}^{{x_n}} f\left( x \right)dx = \frac{h}{2}\left( {{y_{n - 1}} + {y_n}} \right)\)

\(I = \mathop \smallint \nolimits_a^b f\left( x \right)dx = \frac{h}{2}\left[ {{y_0} + {y_n} + 2\left( {{y_1} + {y_2} + {y_3} + \ldots {y_{n - 1}}} \right)} \right]\)

$\Rightarrow I = \frac{h}{2}\left[ {{y_0} + {y_n} + 2\left( {{y_1} + {y_2} + {y_3} \ldots + {y_{n - 1}}} \right)} \right]$

2. Simpson’s 1/3^rd Rule: If we take n = 2 strip at a time and neglect 3^rd and the higher-order difference in G.Q.F

$I = \frac{h}{3}\left[ {{y_0} + {y_n} + 4\left( {{y_1} + {y_3} + {y_5} \ldots } \right) + 2\left( {{y_2} + {y_4} \ldots } \right)} \right]$

3) Simpson’s 3/8^th Rule: If we take n = 3 strips at a time and neglecting 4^th and higher-order difference in G.Q.F.

\(I = \mathop \smallint \nolimits_a^b f\left( x \right)dx = \mathop \smallint \nolimits_{{x_0}}^{{x_3}} f\left( x \right)dx + \mathop \smallint \nolimits_{{x_3}}^{xb} f\left( x \right)dx \ldots \mathop \smallint \nolimits_{{x_{n - 3}}}^{{x_n}} f\left( x \right)dx\)

$I = \frac{3}{8}h\left[ {{y_0} + {y_n} + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} \ldots } \right) + 2\left( {{y_3} + {y_6} + {y_9}} \right)} \right]$

4) Rectangle Rule

In the Rectangle rule, we approximate f_|a,b| using a single interpolation point ‘a’. Our polynomial interpolant will thus be a constant polynomial p(t) = f(a), as shown in figure  and we can calculate its area I_R using:

I_R = f(a) ⋅ (b - a)

Thus,

 1) Trapezoidal Rule gives the exact result for a polynomial of degree 1 because we have neglected 2^nd order difference in G.Q.F while the result exceeds from exact value for higher degree polynomials.

2. Simpson’s 1/3^rd Rule gives the exact result for a polynomial of degree 2, while the result exceeds from exact value for higher degree polynomials.

3. Simpson’s 3/8^th Rule gives the exact result for a cubic polynomial.

4. Rectangle Rule gives the exact result for a constant function.

Calculation:

Given:

$\frac{{\partial v}}{{\partial y}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}};$ y ≥ 0; x₁ ≤ x ≤ x₂

∵ y ≥ 0 ⇒ It can be replaced with ‘t’.

$\therefore \frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}$

This is a 1-D Heat equation. It measures temperature distribution in a uniform rod.

The general solution is u = f(x, t)

u = (c₁ cos px + c₂ sin px) $\left( {{c_3}{e^{ - {c^2}{p^2}t}}} \right)$

Auxiliary solutions include both initial and boundary conditions.

1. Number of initial conditions = Highest order of time derivative in partial differential = 1

2. The number of boundary conditions:

$\frac{{\partial v}}{{\partial t}} = \frac{{{\partial ^2}v}}{{\partial {x^2}}}$ ; To solve this partial differential equation, it needs to be integrated twice that will introduce two arbitrary constants.

Hence 2 boundary conditions and 1 initial condition are required to solve this Partial differential equation.

Concept:

Shape factor: It is defined as the ratio of plastic moment capacity of a beam section to its yield moment capacity.

Mathematically, $S.F = \frac{{{M_p}}}{{{M_y}}} = \frac{{{f_y}{Z_p}}}{{{f_y}Z}} = \frac{{{Z_p}}}{Z}$

Where, f_y = Yield stress.

Z_p = Plastic modulus of a section.

Z = Elastic modulus of a section.

Z_p is calculated about the equal area axis and Z is calculated about the centroidal axis.

• Shape factor shows the reserve of the strength of one section beyond the yield point. It doesn't show the reserve of the strength of the complete structure.
• It depends upon the geometry of the cross-section.

<br>

Aspect Ratio: It is the ratio of width to the height of any section.

Slenderness Ratio:  It is the ratio of the effective length of a column to the least radius of gyration.

Mathematically, \(\lambda = \frac{{{\ell {eff}}}}{{{r{min}}}}\)

Where, ℓ_eff = Effective length of column which depends on end conditions.

r_min = Least radius of Gyration.

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.

 

Uniaxial tension: Tensile stress acts along one axis only.

Mohr’s circle:

 

Biaxial Tension of equal magnitude:'

Mohr’s circle: It is a point on the normal stress axis.

Hydrostatic stress:

Mohr’s circle: It is a point on the normal stress axis.

Mistake Points

The Mohr’s circle for Biaxial tension of equal magnitude and Hydrostatic stress represent a point on σ-axis but in case of hydrostatic stress, the point is on the negative side and on biaxial tension the point is on the positive side of the Mohr’s circle

Concept:

The most economical profile for the cantilever will be the one for which depth of beam will vary along the length as per the shape of the Bending Moment Diagram.

Bending Moment Diagram represents the variation of bending moment along the length of the beam due to applied external load. Bending moment at any section is the summation of moments due to all transverse loads and the couple either to the left or to the right of the section.

Given:

 

Let the reactions be as shown above.

∵ ∑ F_y = 0 ⇒ R_A = P

∵ ∑ M_A = 0 ⇒ M_A + P.L – PL = 0 ⇒ M_A = 0

The SFD and BMD for the beam is drawn

 

From the BMD shown, the maximum bending moment occurs at the free end and the minimum bending moment occurs at the fixed end. Hence the most economical profile will be one having maximum depth at the free end and minimum depth at the fixed end.

Among the 4 options given, option (3) satisfies the above condition.

Important Points

• The shear force diagram does not get any kink due to the presence of a concentrated couple anywhere in the beam.
• Relationship between Bending moment, Shear force (V) and load applied (ω):

1. $\frac{dv}{dx}=\omega ;$ Rate of change of shear force at any section is equal to load intensity at that section.
2. $\frac{dM}{dx}=V;$ Rate of change of bending moment at any section is equal to shear force acting at that section.

Concept:

Important properties of water for concrete:

1. The pH value must not be less than 6.

2. Presence of salts of manganese, tin, zinc, copper and lead causes a reduction in strength.

3. Presence of sugar causes retardation in the setting.

The permissible limit for solids as per IS 456 of 2000:

Concept:

Time Headway:- It is the time interval between the passes of the rear bumper of successive vehicles at a point.

• When the signal at an intersection turns green, the vehicles in the queue will start crossing the intersection. The first driver in the queue needs to observe and react to the signal change at the start of green time and the second driver does the same but only when he see the first vehicle moving.
• This results in a shorter headway than the first because the driver had an extra vehicle length to accelerate. This process carries through with all following vehicles where each vehicle’s headway will be slightly shorter than the preceding vehicle.
• This continues until a certain number of vehicles have crossed the intersection and start-up reaction and acceleration no longer affect the headways.
• The headway will remain constant beyond this point and this constant headway is known as the saturation headway. It generally occurs after the fourth or fifth vehicle passes the intersection.

Important Points

• Distance between the rear bumpers of successive vehicles at any specific time is called space headway.
• The reciprocal of density gives the space headway between vehicles at that time.

Concept:

Soil is also classified based on the mode of deposition or mode of transportation.

Soil deposit formed due to transportation by wind is called Aeolian deposit e.g. Loess

Important Points

According to transporting agencies, the soil is classified as:

(i) Alluvial Deposit → Deposited by river water.

 (ii) Lacustrine Deposit → Deposited by still water lakes.

(iii) Marine Deposit → Deposited by seawater.

(iv) Colluvial Deposit → Deposited by gravity e.g. Talus

(v) Glacial Deposit → Deposited by ice e.g. Till

Concept:

Specific Gravity: It is defined as the ratio of unit weight of solids to the unit weight of water. It is denoted by ‘G’. It is a unitless quantity.

$G = \frac{{{\gamma _s}}}{{{\gamma _w}}} = \frac{{{\rho _s}}}{{{\rho _w}}}$

Calculation:

Given:

Weight of dry sand = 500 gm

Water displaced by sand = 188 cm³

∴ Volume of sand = 188 cm³

∴ Density of sand $= \frac{{Mass}}{{Volume}} = \frac{{500}}{{188}}$

⇒ ρ_s = 2.659 gm/cm³

∴ specific gravity $= \frac{{{\gamma _s}}}{{{\gamma _w}}} = \frac{{{\rho _s} \times g}}{{{\rho _w} \times g}} = \frac{{2.659}}{1}$

= 2.659 ≈ 2.66

Explanation:

• A common method of Hydrologic channel routing is Muskingum method, in which the channel storage in a reach is expressed as a function of both inflow and outflow discharge.

S = k [x I^m – (1 - x)Q^m]

Where, s = Storage in a channel reach

I = Inflow Discharge

Q = Outflow Discharge

k → Coefficient known as storage time coefficient. It was unit of time, it is approx. equal to time of travel of flood wave through the channel reach.

x → Weightage factor. Its range series from 0 to 0.5

• For naturally occurring channels m = 1, Hence the equation becomes

S = k [x I – (1 - x)Q]

• For x = 0, storage is only a function of outflow discharge. Such reservoirs are also known as Linear Reservoir.
  

Important Points

Flood routing is the technique of determining the flood hydrograph at a section of a river by utilizing the data of flood flow at one or more upstream sections. There are two broad categories of routing:-

1. Reservoir routing:-

• Level pool Routing is assumed i.e. the water level is assumed parallel to the channel bed.
• In level pool routing,
  

(1) Storage is a function of height of water level

      i.e. S = f(h)

(2) Outflow discharge is a function of water surface elevation.

                i.e. Q = f(h)

 

• Methods of Reservoir routing:
  

(a) Modified pul’s method  (Graphical method)

(b) Goodrich metod  (Graphical method)

(c) Standard 4^th order Runga – Kutta method (Universal method)

2. Channel routing:-

• If the flow were uniform throughout, the water surface line would wave been parallel to channel bed. However because of non-uniform condition due to flood, there will be water level which will not be parallel to channel bed.
• The volume of water stored in a channel reach is comprised of:

1. Prism storage
2. Wedge storage
   

​A variety of routing methods are available and they can be broadly classified into two categories:

(i) Hydrologic routing (ii) Hydraulic routing

Types of flood routing:

(1) Hydrologic/lumped routing:

• Discharge is taken as a function of time.
• Continuity equation is used.

 

(2) Hydraulic Routing/Distributed Routing:

• Discharge is taken as a function of time and space.
• Both continuity and momentum equation are used.

Concept:

A cross drainage work is a structure constructed for carrying a canal across a natural drain or river intercepting the canal to dispose drainage water without mixing with the continuous canal supplies.

Types of Cross Drainage Works:-

Based on the relative bed levels, water levels of the canal and the drain and their relative discharge the Cross Drainage works are of the following types:-

(1**) Cross drainage** works carrying the Canal over the Natural Drain:

(i) Aqueduct

• An aqueduct is a hydraulic structure which carries a canal (through a trough) across and above the drainage similar to a bridge in which instead of the road or a railway, a canal is carried over a natural drain.

• In the case of an aqueduct, HFL (highest flood level) of the drainage should remain lower than the level of the underside of the canal trough.
• The drain flows at atmospheric pressure under the work.
• Generally, an inspection road is provided along with the trough

(ii) Syphon aqueduct

• A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the streambed is depressed locally where it passes under the trough of the canal and the barrels discharges the streamflow under pressure.
• A syphon aqueduct is constructed where the water surface level of the drain at high flood is higher than the canal bed.
• The horizontal floor of the barrels is provided with slopes at its ends to join the drain bed on either side.
• The drain water flows under pressure through the barrels which act as inverted syphons and hence this cross drainage work is known as syphon aqueduct.

 

2. Cross drainage works carrying the natural drain over the canal-

(i) Super passage

A super passage is also similar to a bridge in which the natural drain is carried over the canal.

• A super passage is reverse of an aqueduct.
• A super passage is constructed where the bed of the drain is well above the canal F.S.L
• The canal flows at atmospheric pressure under the work.

• In this case it is not possible to provide an inspection road along the canal.

(ii) Syphon

• A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon the canal water is carried through the barrels under the drain.
• A syphon is constructed where the full supply level of the canal is higher than the bed of the drain.
• The barrels in this case also act as inverted syphons through which the canal water flows under pressure.

​

3. Cross drainage works admitting the drain water into the canal-

In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other. This may be achieved by the following two types of cross drainage works:-

(i) Level crossing

(ii) Inlet and outlet

Concept:

Direct Runoff Hydrograph: It is a plot of discharge vs time past a specific point due to the direct runoff coming from the catchment without including any contribution from Baseflow.

Area of Direct Runoff Hydrograph = Rainfall excess × Area of catchment

Rainfall excess is the part of precipitation that reaches the stream soon after the rainfall.

Calculation:

Given:

Using the property explained above,

$\frac{1}{2} \times 60 \times 90 \times 3600 = \left( {R.E} \right) \times {A_{catchment}}$

⇒ 30 × 90 × 3600 = (R.E) × (300 × 10⁶)

⇒ R.E = 0.0324 m

= 3.24 cm

Important Points

Since it is a direct runoff hydrograph, the discharge at time t = 0 is 0. For a hydrograph that has baseflow contribution, discharge at t = 0 is not zero.

Concept:

A line drawn in the flow field such that velocity at every point of it is tangential to the line is called streamline.

Equation of streamline:

$\tan \theta = \frac{v}{u} \Rightarrow \frac{{dy}}{{dx}} = \frac{v}{u}$ or $\frac{{dy}}{v} = \frac{{dx}}{u}$

• There can be no component of velocity perpendicular to the streamline. Hence flow can never cross a streamline
• Streamlines never intersect each other.

Calculation:

Given,

u = (-5 + 6x) and v = -(9 + 6y)

$\because \frac{dy}{v}=\frac{dx}{u}\Rightarrow \frac{dy}{-\left( 9+6y \right)}=\frac{dx}{-5+6x}$

Integrating both sides, we get –

$\frac{\ln \left( -5+6x \right)}{6}=\frac{-\ln \left( 9+6y \right)}{6}+C$

⇒ ln (-5 + 6x) + ln (9 + 6y) = constant

⇒ ln {(-5 + 6x) × (9 + 6y)} = constant     ...(∵ ln (x) + ln (y) = ln(xy))

⇒ (-5 + 6x) × (9 + 6y) = constant

Important Points

• Equation of streamline in 3D –

$\frac{dx}{u}=\frac{dy}{v}=\frac{dz}{w}$

• Under a steady flow situation, streamline, streakline and pathline are all the same.

Concept:

Biochemical Oxygen Demand: It is defined as the quantity of oxygen required for oxidation of bio-degradable organic matter present in wastewater by aerobic biochemical action.

The aerobic microorganisms present in wastewater utilize the oxygen present which is directly proportional to the amount of organic matter present and the amount of oxygen required at any instant is the BOD exerted at that instant.

Calculation:

Given:

\(\frac{d{{L}{t}}}{dt}=-k{{L}{t}}\Rightarrow \frac{d{{L}{t}}}{{{L}{t}}}=-kdt\)

\(\Rightarrow \mathop{\int }{{{L}{o}}}^{{{L}{t}}}\frac{d{{L}{t}}}{{{L}{t}}}=\mathop{\int }{o}^{t}-kdt\)

\(\Rightarrow \ln \left( \frac{{{L}{t}}}{{{L}{o}}} \right)=-kt\Rightarrow {{L}{t}}={{L}{o}}{{e}^{-kt}}\)

Where, L_t = The total oxygen equivalent of the organics at any time ‘t’.

The difference between the oxygen equivalent of organics at time t = 0 and at other time t gives the BOD remaining after that time.

∴ BOD_t = L_o – L_t = L_o – L_oe^-kt = L_o (1 – e^-kt)

Concept:

Large-sized particles are settled first in the gravitational setting chamber and then smaller particles are separated either with bag filter. Bag filter has the highest collection efficiency among all fillers, for particles less than 10 μ.

Important Points

1. Gravitational Settling Chambers:

• Settling chambers in air-pollution control systems provide enlarged areas to minimize horizontal velocities and allow time for the vertical velocity to carry the particle to the floor.
• The emitted smokes, when made to pass through a settling chamber, drop some of their larger sized particles in the chamber as per Stoke’s Law. The largest size particle (d) that can be removed with 100% efficiency in such a chamber of length L and height H is given by

​\(d=C.\sqrt{\frac{18\mu .{{v}{h}}.H}{g.L.{{\rho }{p}}}}\)

Where, v_h = Horizontal velocity of gas passing through the chamber

• Simple to design and maintain, and low-pressure flow.
• It requires larger space for installation and has low collection efficiency for small-sized particles.
• Only larger sized particles are separated.
• Although theoretically they should be able to remove particulates down to 5 or 10 μm, in actual, they are not practical for the removal of particles much less than 50 μm in size.

2. Venturi Scrubbers:-

• Venturi Scrubbers are most efficient for removing particulate matter in the size range of 0.5 to 5 μm, which makes them especially effective for the removal of submicron particulates associated with smoke and fumes.
• At velocities from 60 to 180 m/s, the contaminated gas passes through a duct that has a venture shaped throat section. A coarse water spray is injected into the throat, where it is atomized by the high gas velocities. The liquid droplets collide with the particles in the gas stream, and the water and particles fall for later removal.
• Venturi scrubbers can efficiently remove gaseous as well as particulate contaminants.

3. Electrostatic Precipitators:-

• In electrostatic precipitators, the emitted gas flue gas is passed through a highly ionized atmosphere high-voltage field; and in that zone particulates get electrically charged and get separated from the gaseous stream with the help of electrostatic forces.
• Four basic steps required in the operation of a high-voltage single-stage electrostatic precipitator:

a) electrical charging of the particulates

b) collection of charged particles on a grounded surface

c) neutralization of the charge at the collector

d) removal of the particulate for disposal.

4. Fabric Filters

• In fabric filters, the gas stream laden with particles is passed through a woven or felted fabric that filters out particulate matter, allowing the gaseous matter to flow on.
• Small, particles are initially retained on the fabric by direct interception, inertial impaction, diffusion, electrostatic attraction, and gravitational settling.
• The collection of sub-micron particles is accomplished by sieving after a mat of dust gets found on the fabric.
• Filter bags, are capable of removing most particles as small as 0.5 μm and will remove substantial quantities of particles as small as 0.1 μm with an efficiency greater than 99%.

 

Explanation:

$6\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} - y = 0$

Let $D = \frac{d}{{dx}}$

∴ 6 D² y + Dy – y = 0

⇒ (6D² + D – 1) y = 0

∴ The auxiliary equation obtained by replacing D with m is (6 m² + m - 1) = 0

Roots of A.e :

6m² + m – 1 = 0

⇒ Roots $\frac{{ - 1\pm\sqrt {{1^2} - 4 \times 6 \times \left( { - 1} \right)} }}{{2 \times 6}}=\frac{-1\pm\sqrt{25}}{12}$

$\therefore {m_1} = - \frac{1}{2};;and;;{m_2} = \frac{1}{3}$

∴ General solution of the above equation is

$y = {C_1}{e^{{m_1}x}} + {C_2};{e^{{m_2}x}}$

$\Rightarrow y = {C_1};{e^{ \frac{1}{3}x}} + {C_2}{e^{- \frac{1}{2}x}};$

Explanation:

Given:

$\left[ P \right]=\left[ \begin{matrix} 0 & 1 & 3 & 0 \ -2 & 3 & 0 & 4 \ 0 & 0 & 6 & 1 \ 0 & 0 & 1 & 6 \ \end{matrix} \right]$

Sum of eigen values = Trace of [P] = 0 + 3 + 6 + 6

⇒ λ₁ + λ₂ + λ₃ + λ₄ = 15

Product of eigen values = Determinant of [P]

Expanding about 1^st column

$\left| P \right|=0\left[ \begin{matrix} 3 & 0 & 4 \ 0 & 6 & 1 \ 0 & 1 & 6 \ \end{matrix} \right]-\left( 2 \right)\left[ \begin{matrix} 1 & 3 & 0 \ 0 & 6 & 1 \ 0 & 1 & 6 \ \end{matrix} \right]+0$

⇒ |P| = 70

Option 4 satisfies both the conditions.

i.e. λ₁ + λ₂ + λ₃ + λ₄ = 15

& λ₁ × λ₂ × λ₃ × λ₄ = 1 × 2 × 5 × 7 = 70

Additional Information

Properties of Eigen Values:

Consider a matrix A_n×n & let λ₁, λ₂, λ₃ … λ_n are the eigenvalues. Then,

1. Sum of the eigenvalues = Trace (A)
2. Product of the eigenvalues = |A|
3. Zero is an eigenvalue of a matrix if the matrix is singular i.e. both way relation holds.
4. If λ is an eigenvalue of Mat A, then the eigenvalue of any polynomial of A can be calculated just by replacing ‘A’ by ‘λ’.

Concept:

Strain Energy: The energy stored in a body on account of deformation produced is called as strain energy. It consists of two parts:

(a) Elastic strain energy

(b) Inelastic strain energy

Strain energy stored due to axial force ‘P’ is given as

$U = \frac{{{P^2}L}}{{2AE}}$

Where, P = Axial force

L = length

A = Area of cross-section

E = Modulus of elasticity

The bar is prismatic i.e. having a constant cross-sectional area.

Calculation:

Strain energy $\left( {{U_1}} \right) = \frac{{P_1^2L}}{{2AE}}$      ---(1)

Strain energy $\left( {{U_2}} \right) = \frac{{P_2^2L}}{{2AE}}$     ---(2)

Strain energy $= \frac{{{{\left( {{P_1} + {P_2}} \right)}^2}L}}{{2AE}} = \frac{{P_1^2L}}{{2AE}} + \frac{{P_2^2L}}{{2AE}} + \frac{{2{P_1}{P_2}L}}{{2AE}}$     ---(3)

∴ from equation (1), (2) and (3)

U > U₁ + U₂

Important Points

Strain energy due to:

(1) Bending moment $= \frac{{{M^2}L}}{{2EI}}$

(2) Torsional Moment $= \frac{{{T^2}L}}{{2GJ}}$

(3) Shear force $= \frac{{{V^2}L}}{{2GA}}$

Strain energy per unit volume is equal to the area under the stress-strain curve.

Explanation

Drawing FBD for the beam –

Deflected shape

Using the standard results –

${\theta _R} = \frac{{\left( {\frac{{PL}}{2}} \right) \times L}}{{6;EI}} = \frac{{P{L^2}}}{{12EI}}$

Important Points

Deflection calculations using standard results:

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

$\delta = \frac{{W{\ell ^4}}}{{8EI}},;\theta = \frac{{W{\ell ^3}}}{{6EI}}$

$\delta = \frac{{{\mu _0}{\ell ^2}}}{{2EI}};;;\theta = \frac{{{\mu _{0\ell }}}}{{EI}}$

$\delta = \frac{{{\omega _0}{\ell ^4}}}{{30EI}};;\theta = \frac{{{\omega _0}{\ell ^3}}}{{24;EI}}$

$\delta = \frac{{P{\ell ^3}}}{{48EI}};;\theta = \frac{{P{\ell ^2}}}{{16EI}}$

$\delta = \frac{5}{{384}}\frac{{\omega {\ell ^4}}}{{EI}};;;;;\theta = \frac{{\omega {\ell ^3}}}{{24EI}}$

$\delta = \frac{{{\omega _0}{L^4}}}{{120EI}};;\theta = \frac{5}{{192}}\frac{{{\omega _0}{\ell ^3}}}{{EI}}$

${\theta _1} = \frac{{{\mu _0}\ell }}{{3EI}};;{\theta _2} = \frac{{{\mu _0}\ell }}{{6EI}}$

$\theta = \frac{{{\mu _0}\ell }}{{4EI}}$

$\delta = \frac{1}{4}\left[ {\frac{{P{\ell ^3}}}{{48EI}}} \right]$

$\delta = \frac{1}{5}\left[ {\frac{5}{{384}}\frac{{{\omega _0}{\ell ^4}}}{{EI}}} \right]$

Concept:

Principal of virtual work:

The total virtual work of external forces acting on the body is zero for any virtual displacement of the body. Virtual Displacement is not experienced but only assumed to exist so that various possible equilibrium positions may be compared. The above principle is used to solve the complex frame.

Work is a scalar given by dot product; which involves the product of a force and distance, both measured along the same line of action.

Work = Force × Displacement

Calculation:

Let us consider the Horizontal reaction at S as R_S. The frame can be redrawn considering the horizontal reaction out of the support and adding roller support at S as shown below.

We need the virtual displacement of only those points at which there is any force component. Here at the point I there is a point load in negative y-directionn and hence y coordinate displacement of the point I is noted. Similarly, the table for the member and virtual displacement is shown below:

 

By the principle of virtual work,

∂U= 0

Where U = Virtual work = force along any direction × Displacement along that direction

$⇒ \left( { - 10 × \sqrt 2 \cos θ ;dθ } \right) × 3 + \left( {10 × - \sqrt 2 \sin θ ;;dθ } \right) + \left( {10 × 5\sqrt 2 \sin θ d;θ}\right) + \left( { - {R_s} × - 6\sqrt 2 \sin θ ;dθ } \right) = 0$

Putting θ = 45° and solving the above equation we get,

R_s = 15 kN

Alternate Method 

 

Let us give angular displacement to members RI and RL amount 'θ' as shown.

Due to which the resulting displacements are 

Δ_VI = Δ_VJ = Δ_VK = Δ_VL = ΔVQ = ΔVM = θ 

Δ_HI = Δ_HL = θ; Δ_HN =2θ;  ΔHJ = ΔHQ = 3θ;

ΔHP = 4θ; ΔHK = ΔHM = 5θ; ΔHS = 6θ

From principle of virtual work,

10 × ΔVI + 10 × ΔVJ  +10 × ΔVK + 10 × Δ_HL  + 10

× ΔHM  - R_H × ΔHS = 0

⇒ 10 × θ + 10 × θ + 10 × θ + 10 × θ + 10 × 5θ - RH × 6θ = 0

⇒ RH = $\frac{{90\theta }}{{6\theta }}$

⇒  RH  = 15 kN