GATE CE 2015 Official Paper: Shift 2 (Previous Year Paper)

‘Respectably’ is inappropriate as per the context of the sentence, it means in a way that achieve a reasonable result.

‘Reputably’ is also grammatically incorrect as the correct word is ‘reputable’ which means a person having a good reputation.

‘Respectively’ means ‘in a particular sequence’ and does not fit the given context.

‘Respectfully’ means ‘with respect’

Only statement (B) is correct as the official must have answered ‘respectfully’.

‘Insured’ means a person or thing that is insured against any untoward occurrence by an insurance company.

‘Ensured’ means to make sure of the certainty of something.

Thus only ‘B’ is used correctly as ‘he’ declared with certainty that the company will not have to bear any loss.

Statement ‘A’ is incorrect as a minister is not interested in ‘insuring’ people. Also he can ensure that everything would be all right.

In ‘C’ the actor cannot promise his own safety to himself, he must have registered with an official organisation in order to ‘insure’ it.

‘D’ the teacher too cannot ‘insure’ the students of good results. He can 'ensure' students good result

Hence the statement with correct use of underlined word is option 2

‘Vernacular’ means the regional language spoken by a native person.

‘Regional’ means ‘relating to a characteristic of region’. Therefore, a vernacular can be said to be a ‘regional language’

‘Indigenous’ is something that can be found locally or in one place. Therefore this is also related to ‘vernacular’.

‘Colloquial’ is a language spoken normally everyday which are informal in nature.

‘Indigent’ is a poor needy person,

So from the above words the word ‘Indigent’ does not talk about any local or regional thing like all the others.

Hence it is not a synonym for the word vernacular.

Given that both angles are 60^Owe can conclude that the triangle is an equilateral triangle, thereby making the sides equal as well, making the distance between them 6 m.

Further, by drawing a line perpendicular from the final point to the base of the above created triangle, we get a right-angled triangle with side 2 m (equal to distance walked south) and base 2 m (6 m-4 m).

Therefore, by Pythagoras theorem, the required distance will be the square root of (2²) + (2)² = \(2\sqrt 2 \) m

Number of kings in a pack of 52 cards = 4

Given, four cards are randomly selected from a pack of 52 cards and the first two cards are kings.

We know that, Probability = No. of favorable outcomes/Total number of outcomes

Probability that the third card is a king = ²C₁/⁵⁰C₁ = 2/50

‘Cheerful’ is ‘chirpy and happy’

‘Dreamy’ is ‘a person who is always dreaming or is lost in good thoughts.’

‘Hard’ is ‘firm’, ‘resistant’

‘Dreary’ means dull 

‘Dismal’ too means ‘dull’, ‘listless’,

Hence it is the right answer.

Given,

Capacity of a solution tank A is 70% of the capacity of tank B.

From statement 1: Tank A is 80% full and tank B is 40% full.

Alone with statement 1, we can’t find the gallons of solution in tank A and tank B.

From statement 2: Tank A if full contains 14,000 gallons of solution.

Alone with statement 2, we can't find the gallons of solution in tank A and tank B

Now, combining statement 1 and 2.

Gallons of solution in tank A = 80% of 14000 = 11200 gallons

We know that Solution in tank A is 70% of the capacity of tank B.

∴ Capacity of tank B = 14000/0.7 = 20000 gallons

Gallons of solution in tank B = 40% of 20000 = 8000 gallons

∴ Both the statements I and II together are sufficient.

At the first place, 9 digits can come excluding 0.

At the second place, 9 digits can come excluding the 1^st digit (as repition is not allowed).

At the third place, 8 digits can come (as repition is not allowed).

At the fourth place, 7 digits can come (as repition is not allowed).

Thus, number of four digit numbers which can be formed with the 10 digits 0, 1, 2,……9

Number of 'four digit number' = 9 × 9 × 8 × 7 = 4536

From the given table, we can see that only in type I, type III and type V there is growth between 2006 and 2012.

Sales of type I battery in 2006 = 75

Sales of type I battery in 2012 = 115

Growth of type I battery \(= \frac{{115 - 75}}{{75}} \times 100% = 53.33% \)

Sales of type III battery in 2006 = 114

Sales of type III battery in 2012 = 160

Growth of type III battery \(= \frac{{\left( {160 - 114} \right)}}{{114}} \times 100% = 40.35% \)

Sales of type V battery in 2006 = 108

Sales of type V battery in 2012 = 145

Growth of type V battery \(= \frac{{145 - 108}}{{108}} \times 100% = 34.26%\)

Hence we can see that the highest growth in sales is of the type I batteries.

Explanation:

Given: There are 11 ES teachers.

Of these, 6 are given to be able to teach only ES.

Therefore, 11 – 6 = 5 teachers have the possibility of fulfilling the given condition.

It is given that 4 can teach ES and TD.

Therefore, the number of persons able to teach ES and EM but not TD = 5 – 4 = 1.

Concept:

The method of finding Maxima and Minima of Y = f(x)

1. Find f’(x) and f”(x) for given function Y = f(x)

2. Equate f’(x) to zero to obtain stationary points x = a

3. calculate f”(x) at each stationary points x = a (i.e f”(a))

4. we obtain following three conditions

(i) If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)

(ii) If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)

(iii) If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a 

Explanation:

Here X_o is the stationary point

f(x) has a local minimum at x = x₀

If f’(x₀) = 0 and f”(x₀) > 0

Hence option 4 is correct

Concept:

According to Newton–Raphson Method:

Newton-Raphson method has quadratic convergence, i.e order of convergence = 2

Newton-Raphson method converges more rapidly than the other methods.

The method can be used for solving algebraic and transcendental equations and it can also be used when the roots are complex.

The iteration formula is,

\({X_{N + 1}} = {X_N} - \frac{{f\left( {{X_N}} \right)}}{{f'\left( {{X_N}} \right)}}\)

Correction to be added to X_N

ΔX = X_N+1 - X_N 

Calculation:

Given,

f(x) = -2 + 6x – 4x² + 0.5x³

xini = 0

∴ f’(x) = 6 – 8x + 1.5x²

f(xini) = f(0) = -2

f’(xini) = f'(0) = 6 

By Newton Raphson Method,

\({X_{N + 1}} = {X_N} - \frac{{f\left( {{X_N}} \right)}}{{f'\left( {{X_N}} \right)}}\)

\({X_{1}} = {0} - \frac{{-2}}{{6}}\)

\({X_{1}} = \frac{{1}}{{3}}\)

ΔX = X1 - Xini

ΔX = 1/3 - 0

∴ ΔX = 1/3

Concept:

Some important relations

1) \({{\rm{e}}^{{\rm{ix}}}} = \cos {\rm{x}} + {\rm{i}}\sin {\rm{x;;}}\)

2) \({{\rm{e}}^{ - {\rm{ix}}}} = \cos {\rm{x}} - {\rm{i}}\sin {\rm{x}}\)

3) \({{\rm{e}}^{{\rm{i\pi }}}} = \cos {\rm{\pi }} + {\rm{i}}\sin {\rm{\pi }} = - 1\)

4) \(\frac{{{\rm{ - 1}}}}{{\rm{i}}}{\rm{ = ;i}}\)

Calculation:

\(\begin{array}{l} I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{\rm{cosx \ +\ isinx}}}}{{{\rm{cosx \ -\ isinx}};}}{\rm{dx}}\ = \mathop \smallint \limits_0^{\frac{\pi }{0}} \frac{{{e^{ix}}}}{{{e^{ - ix}}}}dx =\mathop \smallint \limits_0^{\frac{\pi }{2}} {e^{2ix}}dx\ = \frac{1}{{2i}}\left[ {{e^{2ix}}} \right]_0^{\frac{\pi }{2}} = \frac{1}{{2i}}\left[ {{e^{i\pi }} - 1} \right] \= \frac{1}{{2i}}\left( { - 1 - 1} \right)\ = - \frac{1}{i} \= i \end{array}\)

Conecpt:

L - Hospital's Rule:

\({\rm{If;}}\mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}{\rm{;is;of;the;form}}\frac{0}{0}\left( {{\rm{or}}} \right)\frac{\infty }{\infty }\)

\({\rm{then;}}\mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \mathop {\lim }\limits_{{\rm{X}} \to {\rm{a}}} \frac{{{\rm{f'}}\left( {\rm{X}} \right)}}{{{\rm{g'}}\left( {\rm{X}} \right)}} = \frac{{{\rm{f'}}\left( {\rm{a}} \right)}}{{{\rm{g'}}\left( {\rm{a}} \right)}}\)

When f’(a) = g’(a) ≠ 0

Calculation:

\(\begin{array}{l} y = \mathop {\lim }\limits_{x \to ∞ } {\left( {1 + \frac{1}{x}} \right)^{2x}}\ \log y = \mathop {\lim }\limits_{x \to ∞ } 2x \log\left( {1 + \frac{1}{x}} \right) \end{array}\)

Which is in the form of ∞ × 0

To covert this into \(\frac{0}{0}\) form, we rewrite as

\(\log y = \mathop {\lim }\limits_{x \to ∞ } \frac{{2\log \left( {1 + \frac{1}{x}} \right)}}{{\frac{1}{x}}}\)

Now it is in \(\frac{0}{0}\) form

Using L’ Hospital’s rule

\({\log {\rm{y}} = \mathop {\lim }\limits_{x \to 8} \frac{{\frac{{2 \times - \frac{1}{{{x^2}}}}}{{1 + \frac{1}{x}}}}}{{ - \frac{1}{{{x^2}}}}}}\)

\({{\rm{log}};{\rm{y}} = \mathop {\lim }\limits_{x \to 8} \frac{2}{{1 + \frac{1}{x}}} = 2}\)

∴ y = e²

Shortcut trick:

The standard limit form

\(\mathop {\lim }\limits_{{\rm{X}} \to ∞ } {\left[ {1 + \frac{{\rm{1}}}{{\rm{X}}}} \right]^{\rm{aX}}} = {{\rm{e}}^{\rm{a}}}\)

Concept:

The rank of a Matrix

A non-negative integer ‘r’ is said to be the rank of matrix A, if

1. There exists at least one non-zero minor of order ‘r’

2. All minors of order (r+1) if they exist, are zeros.

Then we write rank of a matrix A = ρ(A) = r

Calculation:

Given,

A = [aij], 1 ≤ i, j ≤ n with n ≥ 3 and aij = i. j.

∴ The matrix obtained

\(\left[ {\begin{array}{*{20}{c}} {1{\rm{;}}}&2&3\ 2&4&6\ {\rm{n}}&{2{\rm{n;}}}&{3{\rm{n}}} \end{array}} \right]\)

R₂ – 2R₁, R_n – nR₁

\(\left[ {\begin{array}{*{20}{c}} {1{\rm{;}}}&2&3\ 0&0&0\ 0&{0{\rm{;}}}&0 \end{array}} \right]\)

The number of non zero row = 1

All the minors of order greater than 1 are zero.

∴ Rank of A = 1

Reaction at B

Δ_B = 0 (Compatibility condition)

\(\begin{array}{l} \frac{{10 \times {{\left( {0.75} \right)}^3}}}{{3EI}} + \frac{{2.5 \times {{0.75}^2}}}{{2EI}} - \frac{{{R_B} \times {{0.75}^3}}}{{3EI}} = 0\ \frac{{9{R_B}}}{{64EI}} = \frac{{135}}{{64EI}} \end{array}\)

∴ R_B = 15 kN

BM at a distance x from free end

BM_x = 0

(10 × x) – (15 × (x – 0.25)) = 0

10x = 15x - 3.75

5x = 3.75

∴ x = 0.75 m

∴ From end A, distance = 1 - x = 0.25 m

Concept:

Principle stress equation is given by,

\({{\rm{σ }}_1},;{{\rm{σ }}_2} = \frac{{{{\rm{σ }}_x} + ;{{\rm{σ }}_y}}}{2} \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_x} - ;{{\rm{σ }}_y}}}{2}} \right)}^2} + {{\left( {{τ _{xy}}} \right)}^2}} \)

Where,

σ₁ = Maximum principal stress

σ₂ = Minimum principal stress

σ_x = Stress in x direction

σ_y = Stress in Y direction

τ_xy = Shear stress in xy plane

Maximum shear stress is given by

\({{\rm{τ }}_{{\rm{max}}}} = \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_{\bf{x}}} - ;{{\rm{σ }}_{\bf{y}}}}}{2}} \right)}^2} + {{\left( {{{\bf{τ }}_{{\bf{xy}}}}} \right)}^2}} \)

Calculation:

σ_x = σ_y = 50 MPa, τ_xy = 0 MPa  

\({{\rm{τ }}_{{\rm{max}}}} = \pm \sqrt {{{\left( {\frac{{{{\rm{σ }}_{\bf{x}}} - ;{{\rm{σ }}_{\bf{y}}}}}{2}} \right)}^2} + {{\left( {{{\bf{τ }}_{{\bf{xy}}}}} \right)}^2}} \)

\({{\rm{\tau }}_{{\rm{max}}}} = \pm \sqrt {{{\left( {\frac{{50 - ;50}}{2}} \right)}^2} + {{\left( 0 \right)}^2}} = 0;{\bf{MPa}}\)

Hence option 4 is correct

Important point:

Under hydrostatic loading condition, stresses at a point in all directions are equal and hence no shear stress. Mohr’s circle reduces to a point.

Guided roller:

At guided roller support there are two reactions 

1. One horizontal reaction (R_x)

2. One moment reaction (M_z)

The vertical guided roller has two reactions i.e it will have infinite stiffness in that direction

∴ K_x = ∞, K_θ = ∞ 

But the vertical guided roller support is free to move in vertical direction i.e it has zero stiffness in that direction

∴ K_y = 0 

Hence, Kx = ∞, Kθ = ∞ and Ky = 0

Concept:

Minimum eccentricities

\({{\rm{e}}_{{\rm{x;min}}}} = \max \left{ {\begin{array}{*{20}{c}} {\frac{{{{\rm{l}}_{\rm{x}}}}}{{500}} + \frac{{\rm{D}}}{{30}}}\ {20;mm} \end{array}} \right.\)

\({{\rm{e}}_{{\rm{y;min}}}} = \max \left{ {\begin{array}{*{20}{c}} {\frac{{{{\rm{l}}_{\rm{x}}}}}{{500}} + \frac{{\rm{b}}}{{30}}}\ {20;mm} \end{array}} \right.\)

where l_x = l_y = Unsupported length

Calculation:

Given,

l = 3 m = 3000 mm,

b = 450 mm, d = 600 mm

\({e_y} = \frac{{3000}}{{500}} + \frac{{450}}{{30}} = 21;mm\)

\({e_x} = \frac{l}{{500}} + \frac{d}{{30}} = \frac{{3000}}{{500}} + \frac{{600}}{{30}} = 26;mm\)

The maximum eccentricity with respect to major axis = e_x = 26 mm

The maximum eccentricity with respect to minor axis = e_y = 21 mm

Prying forces:

The flexibility of connected parts can lead to deformations that increases the tension applied to the bolts. The additional tension is called prying force.

The prying forces can be kept small  by using a thick plate or by limiting the distance between the bolt and plate edge.

The corresponding prying force can be calculated as follows

\({\rm{Q}} = \frac{{{{\rm{M}}_{\rm{p}}}}}{{\rm{n}}}\)

Where,

n = Number of bolts

M_p = Plastic moment

\({{\rm{M}}_{\rm{p}}} = \frac{{{{\rm{f}}_{\rm{y}}}}}{{1.15}}\frac{{{\rm{W}}{{\rm{t}}^2}}}{4}\)

t = Thickness of end plate

Concept:

According to IS 800 : 2007

Maximum Slenderness Ratio(λ):

It is defined as the ratio of the effective length of a member to the radius of gyration of the cross-section about the axis under consideration

The maximum effective slenderness ratio \(\frac{{{\bf{KL}}}}{{\bf{r}}}\) , values of a beam, strut or tension member shall not exceed those given in the table below.

\(\lambda =\frac{{{\bf{KL}}}}{{\bf{r}}}\)

Where,

KL = Effective length of the member

r = Appropriate radius of gyration based on the effective section

Table: Maximum values of effective slenderness ratio 

Explanation:

As you can see in the table, 

For member M, the maximum permissible slenderness ratio is 180

For member N the maximum, permissible slenderness ratio is 350.

Hence the value of M is less than value of N

Concept: 

Voids ratio:

It is defined as the ratio of total volume of voids to the volume of solids in given soil mass.

\({\rm{e}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{{{\rm{V}}_{\rm{s}}}{\rm{;}}}}\)

e > 0, voids ratio has no upper limit

Water content (W):

Water content also called as moisture content is defined as the ratio of weight of water to the weight of soil solids in the soil mass

\({\rm{W}} = \frac{{{{\rm{W}}_{\rm{w}}}}}{{{{\rm{W}}_{\rm{s}}}}} × 100\)({\rm{W}} = \frac{{{{\rm{W}}{\rm{w}}}}}{{{{\rm{W}}{\rm{s}}}}} × 100\)

For dy soils, W = 0

For moist soils it is around, W = 60 %

For a Saturated soils, W > 0

Degree of Saturation (S):

Degree of saturation of a soil is defined as the ratio of volume of water to the volume of voids in the soil mass

\({\rm{S}} = \frac{{{{\rm{V}}_{\rm{w}}}}}{{{{\rm{V}}_{\rm{v}}}}} × 100\)

For dy soils, S = 0

For a Saturated soils, S = 100 %

For partially saturated soils it ranges as 0 % ≤ S ≥ 100 %

The relation between the degree of saturation (S), voids ratio (e), moisture content (W), and specific gravity (G) is given by,

e × S = W × G 

Calculation:

Given W% = 100%, 

For fully saturated soil mass, we know that 

S = 100%

e × s= w × G

In this case, 1 × e = 1 × G

e = G

Hence void ratio is equal to the specific gravity of the soil.

Friction circle method:

This method is based on total stress analysis, but it enables the angle of shearing resistance to be taken into account.

It is convenient approach for both graphical and mathematical solutions.

It should be noted that some soils, such as saturated silts and unsaturated clays, do exhibit a Φ value under un-drained condition.

The friction circle method assumes a circular slip surface; it is in the form of a circular arc of radius R with its centre at ‘O’ and radius of slip circle is Rsinϕ as shown in the figure below.

Ultimate bearing capacity (q_u):

The ultimate bearing capacity is the gross pressure at the base of the foundation at which soil fails in shear.

Net ultimate bearing capacity (q_nu):

It is the net increase in the pressure at the base of the foundation that causes shear failure of soil. It is equal to the gross pressure minus overburden pressure.

q_nu = q_u - γ D_f

Where,

γ = Unit weight of foundation soil

D_f = Depth of the foundation

According to Terzaghi's theory

The ultimate bearing capacity of the footing is given by,

q_u = C N_c + γ D_f N_q + 0.5 γ B N_γ 

Where,

C = Cohesion of soil, N_c, N_q and N_γ are Terzaghi's bearing capacity factor

For clays, Nc = 5.7 Nq = 1 and Nγ = 0

∴ For clays, qu = C 5.7 + γ Df 

q_nu = q_u - γ D_f

qnu = C 5.7 + γ Df  - γ Df = 5.7 C

∴ q_nu = C N_c

∴ Net ultimate bearing capacity of the foundation in the clay stratum is independent of width and depth of the footing.

Active pressure of C-ϕ soils

Where,

ZC = Depth of tension zone or depth of tension crack

HC = Critical height or maximum depth of the unsupported vertical trench

Ka = Coefficient of active earth pressure

Pa = Active earth pressure

\({{\rm{P}}_{\rm{a}}}{\rm{ \ =\ }}{{\rm{K}}_{\rm{a}}}{{\rm{\sigma }}_{\rm{V}}}{\rm{\ -\ 2C}}\sqrt {{{\rm{K}}_{\rm{a}}}} \)

In the figure, at the top of a trench

σV = 0,

\({{\rm{P}}_{\rm{a}}}{\rm{\ = }};{\rm{ - 2C}}\sqrt {{{\rm{K}}_{\rm{a}}}} \)  

\(\begin{array}{*{20}{l}} {{{\rm{K}}_{\rm{a}}}{\rm{q = 2c}}\sqrt {{{\rm{K}}_{\rm{a}}}} }\ {{\rm{q = }}\frac{{{\rm{2c}}}}{{\sqrt {{{\rm{K}}_{\rm{a}}}} }}} \end{array}\)

Important point:

The Depth of the tension zone 'Z_c' is given by

\({{\rm{Z}}_{\rm{c}}}{\rm{ = }}\frac{{{\rm{2C}}}}{{{\rm{\gamma }};\sqrt {{{\rm{K}}_{\rm{a}}}} }}\)

The critical depth of unsupported vertical trench 'H_c' is given by,

H_c = 2 × Z_c

\({{\rm{H}}_{\rm{C}}}{\rm{ = }}\frac{{{\rm{4C}}}}{{{\rm{\gamma }};\sqrt {{{\rm{K}}_{\rm{a}}}} }}\)

Concept:

Froude number is given by:

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} {\rm{;}}}}\)

Where,

D = Y = Hydraulic depth

Geometrical similarity:

Geometrical similarity exists between the model and the prototype.

If the ratios of corresponding length dimensions in the model and the prototype are equal, such a ratio is called length scale ratio.

Length scale ratio: 

\({{\rm{L}}_{\rm{r}}} = \frac{{{{\rm{L}}_{\rm{m}}}}}{{{{\rm{L}}_{\rm{p}}}}} = \frac{{{{\rm{b}}_{\rm{m}}}}}{{{{\rm{b}}_{\rm{p}}}}} = \frac{{{{\rm{d}}_{\rm{m}}}}}{{{{\rm{d}}_{\rm{p}}}}}\)

Area scale ratio:

\({{\rm{A}}_{\rm{r}}} = \frac{{{{\rm{A}}_{\rm{m}}}}}{{{{\rm{A}}_{\rm{p}}}}} = \frac{{\left( {{{\rm{L}}_{\rm{m}}} \times {{\rm{b}}_{\rm{m}}}} \right)}}{{\left( {{{\rm{L}}_{\rm{m}}} \times {{\rm{b}}_{\rm{p}}}} \right)}} = {\rm{L}}_{\rm{r}}^2{\rm{;}}\)

Explanation

Froude's law of similarity

(F_r)_m = (F_r)_p

\({\left( {\frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} }}} \right)_{\rm{m}}} = {\left( {\frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} }}} \right)_{\rm{p}}}\)

\({{\rm{V}}_{\rm{r}}} = \frac{{{{\rm{V}}_{\rm{m}}}}}{{{{\rm{V}}_{\rm{p}}}}} = \sqrt {\frac{{{{\rm{g}}_{\rm{m}}}}}{{{{\rm{g}}_{\rm{p}}}}}} \times \sqrt {\frac{{{{\rm{D}}_{\rm{m}}}}}{{{{\rm{D}}_{\rm{p}}}}}} \)

\({{\rm{V}}_{\rm{r}}} = \sqrt {{{\rm{g}}_{\rm{r}}}} \times \sqrt {{{\rm{D}}_{\rm{r}}}} \)

\({{\rm{V}}_{\rm{r}}} = \sqrt {{{\rm{L}}_{\rm{r}}}} \)

Concept:

The total accelration is given by,

\({{\rm{a}}_{\rm{n}}} = \frac{{{\rm{d}}{{\rm{V}}_{\rm{n}}}}}{{{\rm{dt}}}} = \left[ {\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{t}}}}} \right] + \left[ {{\rm{U}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{x}}}} + {\rm{V}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{y}}}} + {\rm{W}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{z}}}}} \right]\)

Where,

\(\left[ {\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{t}}}}} \right] =\) Local accelration or temporal acceleration

\(\left[ {{\rm{U}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{x}}}} + {\rm{V}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{y}}}} + {\rm{W}}\frac{{∂ {{\rm{V}}_{\rm{n}}}}}{{∂ {\rm{z}}}}} \right]=\) Convective acceleration

Local acceleration:

Acceleration calculated due to variation of velocity with respect to time.

Convective Acceleration:

Accleration calculated due to variation of velocity with respect to space is known as convective accelration.

Convective acceleration in x direction is given as

\({\rm{u}}\frac{{∂ {\rm{u}}}}{{∂ {\rm{x}}}} + {\rm{v}}\frac{{∂ {\rm{u}}}}{{∂ {\rm{y}}}} + {\rm{w}}\frac{{∂ {\rm{u}}}}{{∂ {\rm{x}}}}\)

Calculation:

\(\begin{array}{l} {V_{avg}}_1 = 1.5;m/s\ {V_{av{g_2}}} = 15;m/s \end{array}\)

∂x = 0.375 m

Convective acceleration \( = {\rm{u}}\frac{{∂ {\rm{u}}}}{{∂ {\rm{x}}}}\)

\( {\rm{u}}\frac{{\partial {\rm{u}}}}{{\partial {\rm{x}}}} = \frac{{15\left( {15 - 1.5} \right)}}{{0.375}} = 540;m/{s^2}\)

Explanation 

A hydraulic jump occurs when flow changes from a supercritical flow (unstable) to a sub-critical flow (stable).

There is a sudden rise in water level at the point where the hydraulic jump occurs.

Rollers (eddies) of turbulent water form at this point. These rollers cause dissipation of energy.

A hydraulic jump occurs in practice at the toe of a dam or below a sluice gate where the velocity is very high.

For supercritical flow, depth is less ( y_p < y_c ), Similarly for subcritical flow depth is more ( y_s > y_c)

 In general post Jump has to be y_a but due to losses, depth changes to y_s

 ∴ Pre jump depth is lower than alternate depth. Sequent depth is also lower than alternate depth but greater than pre-jump depth.

i.e. y_p < y_s < y_a

Following are the properties of aquifer

1) Porosity (n):

Porosity is measure of storage property of aquifer

\({\rm{n}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{\rm{V}}}\)

n > 25 % → Sufficient amount of groundwater present

5 % < n > 25 % → Moderate amount of groundwater

n < 5 % → Less amount of groundwater

2) Specific yield (S_y)

It is the ratio of the volume of water that is extracted from a saturated rock (due to gravity) to the total volume of the rock.

\({{\rm{S}}_{\rm{y}}} = \frac{{{\rm{Volume;of;water;extracted}}}}{{{\rm{Volume;of;aquifer}}}}\)

3) Specific Retention (S_r)

Specific retention of a rock or soil is the ratio of the volume of water a rock can retain against gravity to the total volume of the rock.

The relation between porosity (n), specific yield (S_y) and surface retention (S_r)

n = S_y + S_r

Concept:

Total Dissolved Solids (TDS):

Total solids mean all solids i.e suspended or dissolved solids. It is measured by evaporating (at a temperature of 104°C ) the sample and measuring the residue. It can also be measured by the electrical conductivity of water.

TDS = Suspended solids + Dissolved solids

The drinking water standard for TDS is less than 500 mg/liter

Calculation:

TDS (in%) in the sample

\({\rm{TDS}} = \frac{{500 \times {{10}^{ - 3}}g}}{{1000g}} \times 100 = 0.05% \)

Important Points

If TDS is given in ppm, i.e 500 ppm then also we get the same answer.

\({\rm{TDS}} = \frac{{500}}{{{{10}^6}}} \times 100 = 0.05{\rm{;% }}\)

Sulfur dioxide (SO₂):

Sulfur dioxide affects human health when it is breathed in. It irritates the nose, throat, and airways to cause coughing, wheezing, shortness of breath, or a tight feeling around the chest. The effects of sulfur dioxide are felt very quickly and most people would feel the worst symptoms in 10 or 15 minutes after breathing it in. Those most at risk of developing problems if they are exposed to sulfur dioxide are people with asthma or similar conditions.

Carbon monoxide (CO):

Carbon monoxide mainly causes adverse effects by combining with hemoglobin to form carboxyhemoglobin (HbCO) in the blood. This prevents hemoglobin from carrying oxygen to the tissues. Additionally, myoglobin and mitochondrial cytochrome oxidase are thought to be adversely affected. Carboxyhemoglobin can revert to hemoglobin, but the recovery takes time as the HbCO complex is fairly stable.

Important point:

Other harmful gases are,

1)Oxides of nitrogen (NO_x):

It causes asphyxiation and respiratory discomfort.

2) Hydrocarons (HC):

It cause irregular heart rhythms, or damage to the kidney or liver.

3) Hydrogen sulphide (H₂S):

It cause nausea, headache, Skin and eye irritation.

4) Hydrogen fluoride (HF).

It is more harmful to vegetation and animals

Concept:

Superelevation (‘Cant’ or ’ Banking’):

It is the transverse slope to counteract the centrifugal force and to reduce the tendency of the vehicle to overturn or skid.

Rise of the outer edge of the pavement with respect to inner edge on a horizontal curve thus providing a transverse slope troughout the length of the horizontal curve.

\({\rm{Superelevation;}} = {\rm{;e}} = \tan {\rm{θ }} = \frac{{\rm{E}}}{{\rm{B}}}\)

Where,

E = Width of pavement

B = Total super elevated height of outer edge.

\({\rm{e}} + {\rm{f}} = \frac{{{V^2}}}{{gR}},{\rm{;V;in;m}}/{\rm{sec}}\)

\({\rm{e}} + {\rm{f}} = \frac{{{V^2}}}{{127R}},{\rm{;V;in;kmph}}\)

where,

R = radius of horizontal curve in m

F = coefficient of lateral friction = 0.15

Calculation:

R = 335 m, θ =33° 

V = 320 km/hr

\(e + f = \frac{{{V^2}}}{{127;R}}\)

\(\begin{array}{l} f = \frac{{{V^2}}}{{127;R}} - e\ \therefore f= \frac{{{{320}^2}}}{{127 \times 335}} - \tan 33^\circ \end{array}\)

∴ f = 1.761

Note point:

While deriving the above equation, we consider '1 - ef ≈ 1' but this is not valid in this case because banking provided is very high (tan33° = 0.65). so formula used should be \(\frac{{e + f}}{{1 - ef}} = \frac{{{V^2}}}{{127R}}\) which will result in f = 0.685.

Design length is the average of (i) and (iii) as per Indian Roads Congress (IRC) recommendations for storage lanes.

Concept:

According to H.I method or Collimation method,

H.I = R.L + B.S

R.L = H.I – (B.S/F.S)

The Arithmatic check is given by,

ΣB.S - ΣF.S = Last R.L - First R.L  

Where,

H.I = Height of instrument, R.L = Reduced level

B.S = Backsight, F.S = Foresight

Calculation:

Using H.I method

ΣF.S. > ΣB.S.

∴ ΣF.S. – ΣB.S. = R.L. of first station – R.L. of last station

∴ ΣF.S. – ΣB.S = 5.645 – 3.085 = 2.56 m

R.L. of first station – R.L. of last station = 2.56

R.L. (last station) = R.L. (first station) – 2.56 = 100.00 – 2.56 = 97.44 m

Concept:

Correction due to curvature

\({{\rm{C}}_{\rm{c}}} = \frac{{{{\rm{d}}^2}}}{{2{\rm{R}}}}\)

d = Horizontal distance between two points in kM

R = Radius of curvature of earth in km = 6370 km

After submitting values of d and R in km we will get C_c in m

Correction due to curvature will be always negative

∴ Cc = - 0.0785 d2, d is substituted in km

Corrections due to refraction

\({{\rm{C}}_{\rm{R}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{7}}}{{\rm{C}}_{\rm{c}}}\)

\({{\rm{C}}_{\rm{R}}} = \frac{1}{7}\frac{{{{\rm{d}}^2}}}{{2{\rm{R}}}}\)

Correction for refraction is always positive

∴C _R = 0.01122 d², d is substituted in km

Combined correction:

C = C_c + C_R

C =  - 0.0785 d2 + 0.01122 d2

C = - 0.06735 d2

Calculation:

d = 1 km

C = – 0.0785 d² + 0.0112 d²

C = – 0.0673 d² 

 where d is in km, C is in m

C = 0.0673 × 1² = 0.0673 m

Explanation:

\(\begin{array}{l} f\left( x \right) = \left{ {\begin{array}{*{20}{c}} {\frac{x}{4}\left( {4 - {x^2}} \right),}&{0 \le x \le 2}\ 0&{,otherwise} \end{array}} \right.\ {\mu _x} = \mathop \smallint \limits_0^2 xf\left( x \right)dx \end{array}\)

∴ Mean \(\left( {{\mu _x}} \right) = \mathop \smallint \limits_0^2 x\frac{x}{4}\left( {4 - {x^2}} \right)dx\) 

\(\begin{array}{l} {{\rm{Mean}}}= \mathop \smallint \limits_0^2 \left( {{x^2} - \frac{{{x^4}}}{4}} \right)dx\ {{\rm{Mean}}}= \left[ {\frac{{{x^3}}}{3} - \frac{{{x^5}}}{{20}}} \right] = \frac{8}{3} - \frac{{32}}{{20}} = \frac{{16}}{{15}} \end{array}\)

Mean = 1.066

\(\frac{{{d^2}y}}{{d{x^2}}} = - 12{x^2} + 24x - 20\)

Integrating both sides w.r.t x

\(\frac{{dy}}{{dx}} = - 4{x^3} + 12{x^2} - 20x + {c_1}\)

Again integrating both sides w.r.t x

y = -x⁴ + 4x³ – 10x² + c₁x + c₂                                    ……….(i)

At x = 0 and y = 5

5 = c₂

At x = 2, y =21

y = -x4 + 4x3 – 10x2 + c1x + c2   

21 = – 16 + 32 – 40 + 2c₁ + c₂

2c₁ = 21 + 16 – 32 + 40 – 5

2c₁ = 40

c₁ = 20

y = – x⁴ + 4x³ – 10 x² + 20x + 5  

Put x = 1

y = – 1 + 4 – 10 + 20 + 5

y = 18

Concept:

Characteristics polynomial of a matrix:

Let A be a square matrix of order n and λ be any scalar quantity. Then the polynomial in λ of degree n formed by solving the characteristic equation |A - λI| = 0 is called the characteristics polynomial of the matrix.

Eigenvalues:

The roots of the characteristic equation are called the eigenvalues or characteristic roots of latent roots of the matrix A.

Eigenvectors:

If λ is the eigenvalue of the matrix A then a non-zero vector X which satisfies AX = λX is called the eigenvector of the matrix corresponding to the eigenvalue λ.

Calculation:

\(\left[ {\begin{array}{*{20}{c}} 2&1\ 1&p \end{array}} \right]\)

|A – I λ| = 0

\(\left[ {\begin{array}{*{20}{c}} {2 - {\rm{\lambda }}}&1\ 1&{p - {\rm{\lambda }}} \end{array}} \right] = 0\)

(2 – λ) (p – λ) – 1 = 0

λ² – (p + 2) λ + (2p – 1) = 0

By putting values of p from options

By putting option (d) \(\frac{{14}}{3}\) in above equations gives value \(5,\frac{5}{3}\)

Hence ratio of two eigen values \( = \frac{5}{{\frac{5}{3}}} = 3:1\)

So option (d) is correct.

Concept:

Simpson's Rule:

This method is based on approximating the function f(x) by fitting qudratics through sets of three points

\({\rm{I}} = \frac{{\rm{h}}}{3}\left[ {\left( {{{\rm{Y}}_{\rm{o}}} + {{\rm{Y}}_{\rm{n}}}} \right) + 4\left( {{{\rm{Y}}_1} + {{\rm{Y}}_3} + \ldots {{\rm{Y}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{Y}}_2} + {{\rm{Y}}_4} + \ldots {{\rm{Y}}_{{\rm{n}} - 2}}} \right)} \right]{\rm{;}}\)

In this rule, we have neglected al differences above second, so 'Y' will be a polynomial of second dgree only.

If f(x) is a polynomial function of degree ≤ 3, then Simpson's rule gives exact value of the integral.

Calculation:

a = 0, b = 0.8, Δx = h =0.4

f(x) = 0.2 + 25 x – 200x² + 675x³ – 900x⁴ + 400x⁵

By Simpson’s 1/3 rule

\(\begin{array}{l} y\left( x \right) = \mathop \smallint \limits_0^{0.8} \left( {0.2 + 25x - 200{x^2} + 675{x^3} - 900{x^4} + 400{x^5}} \right)dx\ y\left( x \right)= \frac{4}{3}\left[ {{y_0} + 4{y_1} + {y_2}} \right] \end{array}\)

y₀ = y(0) = 0.2

y₁ = y(0.4) = 2.456

y₂ = y(0.8) = 0.232

\(y\left( n \right) = \frac{{0.4}}{3}\left( {0.2 + 4 \times 2.456 + 0.232} \right) = 1.367\)

Concept:

The stiffness of the spring ‘K’ is given by,

\({\rm{K}} = \frac{{\rm{P}}}{{\rm{Δ }}}\)

Where,

P = Load in N

Δ = Deflection in the spring in cm

Strain energy in springs

\({\rm{Strain\ energy}} = \frac{1}{2} \times {\rm{;K}} \times {{\rm{Δ }}^2}\)

Calculation:

Deflection under load = Δ = 2θ 

work done by force P

 \({\rm{Workdone = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{P}}\left( {{\rm{\theta L}}} \right)\)

Strain energy stored in spring

\(\begin{array}{l} {\rm{Strai\ energy }}= \frac{1}{2}\left[ {k\left( {2\theta } \right)} \right]\left( {2\theta } \right) \end{array}\)

Equating workdone and strain energy

\(\frac{{\rm{1}}}{{\rm{2}}}{\rm{P}}\left( {{\rm{\theta L}}} \right){\rm{ = }}\frac{{{\rm{4k}}{{\rm{\theta }}^{\rm{2}}}}}{{\rm{2}}}\\therefore {\rm{ \theta = }}\frac{{{\rm{PL}}}}{{{\rm{4k}}}}\)

Concept:

Bending stress equation or pure Bending equation:

\(\frac{M}{I} = \frac{{{\sigma _{max}}}}{Y} = \frac{E}{R}\)

Where,

M = Maximum bending moment

I = Moment of inertia of the section about the neutral axis

σ max = Bending stress at a distance Y from N.A

Y = Distance of the layer from the N.A

E = Modulus of elasticity

R = Radius of curvature given by

R = 1/C

C = Uniform curvature of the beam in m^-1

Calculation:

Given,

C = 0.004 m^-1, L = 10 m

\(\begin{array}{l} R = \frac{1}{C} = \frac{1}{0.004} = 250m\ OA = \sqrt {{{250}^2} - {5^2}} \end{array}\)

= 249.95 m

Deflection = AA’ = 250 – 249.95 

Deflection of the beam at the midspan = 0.05 m

Concept:

The stiffness of the spring ‘K’ is given by,

\({\rm{K}} = \frac{{\rm{P}}}{{\rm{\Delta }}}\)

Where,

P = Load in N

Δ = Deflection in the spring in cm

\({\rm{\Delta }} = \frac{{{\rm{P}}{{\rm{L}}^3}}}{{3{\rm{EI}}}}\)

Calculation:

Given,

L = 200 mm, K = 2 N/mm, P = 50 N, E = 200 GPa

Deflection of point B = Deflection of spring

\({{\rm{\Delta }}_{\rm{B}}}{\rm{ = }}\frac{{\left( {{\rm{P \ -\ R}}} \right){{\rm{L}}^{\rm{3}}}}}{{{\rm{3EI}}}}{\rm{ = }}\frac{{\rm{R}}}{{\rm{K}}};\)

Where, R = force in the spring

\(\frac{{\left( {{\rm{50\ - \ R}}} \right){\rm{20}}{{\rm{0}}^{\rm{3}}}}}{{{\rm{3 \times 200 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ \times }}\frac{{{\rm{5 \times 1}}{{\rm{0}}^{\rm{3}}}}}{{{\rm{12}}}}}}{\rm{ = }}\frac{{\rm{R}}}{{\rm{2}}}\)

0.064 (50 – R) = R

3.2 = R + 0.064 R

R = 3.0075 N

Concept:

C.G. of system = 2.5 m from any load

For a maximum bending moment, the system of load should be

Maximum BM will occur below load at C.

Ordinate of ILD at C \( = \frac{{\left( {12 - 1.25} \right) \times \left( {12 + 1.25} \right)}}{{24}}\)

Ordinate of ILD at C = 5.935

Ordinate of ILD at D \( = \frac{{5.935 \times 8.25}}{{13.25}}\)

Ordinate of ILD at D = 3.695

Maximum BM = (5.935 + 3.695) × 3 = 28.89 kN-m

Maximum bending stress

\(\sigma =\frac{M}{Z} = \frac{{28.89 \times {{10}^3}}}{{16.2 \times {{10}^{ - 6}} \times {{10}^9}}} = 1.78;GPa\)

According to IS 456 : 2000

Limit state method concept:

1. In this concept, the structure shall be designed to withstand safely all loads liable to act on it throughout its life, it shall also satisfy the serviceability requirements, such as limitations on deflection and cracking.

2. The acceptable limit for the safety and serviciability requirements before failure occurs is called a limit state.

According to clause 6.1.1

The characteristic strength is defined as the strength of material below which not more than 5 percent of the test results are expected to fall.

∴ Chances of no failure = 100 – 5 = 95 % = 0.095

∴ Probability of failure = [1 – P²] = [1 – (0.95)²] = 0.0975

Concept:

The stress at the top and bottom fibre of the beam is given by

\({{\rm{f}}_{{\rm{top}}}} = \frac{{\rm{P}}}{{\rm{A}}} - \frac{{{\rm{Pe}}}}{{\rm{Z}}}\)

\({{\rm{f}}_{{\rm{bottom}}}} = \frac{{\rm{P}}}{{\rm{A}}} + \frac{{{\rm{Pe}}}}{{\rm{Z}}}\)

Where,

P = Prestressing force in kN, A = Cross - sectional area of the beam

e = Eccentricity of the tendon, Z = Sectional modulus

Calculation:

Final pre-stressing force after loss = 750 kN

Net loss = 10%

Hence, initial prestressing force \( = \frac{{750}}{{0.9}} = 833.33\ kN\)

Initial stress at top and bottom \( = \frac{P}{A} \mp \frac{{{Pe}}}{Z}\)

\( = \frac{{833.33 \times {{10}^3}}}{{250 \times 400}} \pm \frac{{833.33 \times {{10}^3} \times 100 \times 6}}{{250 \times {{400}^2}}}\)

Initial stress at top and bottom = 8.33 ± 12.50

At top = 8.33 – 12.50 = - 4.17 N/mm² (T)

At bottom = 8.33 + 12.50 = 20.83 N/mm2 (C)

No of plastic hinges formed at collapse = r + 1 = 3

There can be two collapse mechanisms

Case (I)

\(\begin{array}{l}θ = 2\alpha\ {\rm{\Delta }} = \frac{L}{3} × {\rm{θ }} = \frac{{2{\rm{L}}}}{3} × a; \end{array}\)

Internal work done = External work done

⇒ 2M_pθ + 2M_pθ + 2M_pα + M_pα = W × (L/3)θ 

\(\begin{array}{l} 4{M_p}{\rm{θ }} + 3{{\rm{m}}_{\rm{p}}}\frac{{\rm{θ }}}{2} = \frac{{{\rm{WL}}}}{3}{\rm{θ ;}}\ \frac{{11}}{2}{M_p}{\rm{θ }} = \frac{{{\rm{WL}}}}{3}{\rm{θ }}\ \therefore W = \frac{{16.5}}{L}{M_p} \end{array}\)

Case (II)

2M_pθ + M_pθ + M_pθ + M_pθ = W × (L/3)θ 

\(\begin{array}{l} \therefore W = \frac{{15{M_p}}}{L} \end{array}\)

Hence, minimum collapse load is \(\frac{{15{M_p}}}{L}\)

Concept:

According to the three-phase system of soil mass,

V = V_V + V_S­

Where,

V_V = Total volume of voids = V_a + V_w

V_a = Volume of air, V_w = Volume of water

V = Total volume of soil, V_S = Volume of soil solids

Unit weight of soil

It is defined as the ratio of the total weight of soil to the total volume of soil mass

\({{\rm{γ }}_{\rm{s}}} = \frac{{\rm{W}}}{{\rm{V}}}\)

The specific gravity of soil

It is defined as the ratio of unit weight of a given volume of solids to the unit weight of an equivalent volume of water.

\({\rm{G}} = \frac{{{{\rm{γ }}_{\rm{s}}}}}{{{{\rm{γ }}_{\rm{w}}}}}\)

Voids ratio:

It is defined as the ratio of total volume of voids to the volume of solids in given soil mass.

\({\rm{e}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{{{\rm{V}}_{\rm{s}}}{\rm{;}}}}\)

e > 0, voids ratio has no upper limit

Calculation:

Given,

W = 1010 gm, Ws = 918 gm

V = 588 cm³, G_s = 2.67

 γ_w = 1000 kg/m³ = 1 gm/cm³ 

\(\begin{array}{l} {G_s} = \frac{{{W_s}}}{{{V_s}{Y_w}}}\ 2.67 = \frac{{918}}{{{V_s} \times 1}} \end{array}\)

V_s = 343.82 cm³

\(\begin{array}{l} {V_v} = V - {V_s} = 588 - 343.82 = 244.18{m^3}\ e = \frac{{{V_v}}}{{{V_s}}} = \frac{{244.18}}{{343.82}} = 0.71 \end{array}\)

Concept:

Compression index

It is equal to the slope of the linear portion of the void ratio versus log σ̅  plot, thus

\({{\rm{C}}_{\rm{c}}} = \frac{{{\rm{\Delta e}}}}{{{{\log }_{10}}\left( {\frac{{{\rm{\bar \sigma }}}}{{{{{\rm{\bar \sigma }}}_o}}}} \right)}}\)

Where,

σ̅ = Final effective stress = σ̅_o + Δσ̅   

Δe = Change in void ratio, σ̅_o = Initial effective stress   

Let e_o be the initial void ratio. Let the volume of solids be unity. Therefore the initial volume V_o is equal to 1 + e_o. If Δe is the change in void ratio due to change in volume ΔV, we have ΔV = Δe. Thus,

\(\frac{{{\rm{\Delta V}}}}{{{{\rm{V}}_{\rm{o}}}}} = \frac{{{\rm{\Delta e}}}}{{1 + {{\rm{e}}_{\rm{o}}}}}\)

The change in volume is also propotional to change in height

\(\frac{{{\rm{;\Delta V}}}}{{{{\rm{V}}_{\rm{o}}}}} = \frac{{{\rm{\Delta H}}}}{{{{\rm{H}}_{\rm{o}}}}}\)

Where ΔH = Consolidation settlement, H_o = Initial height

\(\therefore ;\frac{{{\rm{\Delta H}}}}{{{{\rm{H}}_{\rm{o}}}}} = ;\frac{{{\rm{\Delta e}}}}{{1 + {{\rm{e}}_{\rm{o}}}}}\)

Now the relation between Consolidation settlement and compression index is given by,

\({\rm{\Delta H}} = \frac{{{{\rm{H}}_{\rm{o}}}{\rm{}}{{\rm{C}}_{\rm{c}}}}}{{1 + {{\rm{e}}_{\rm{o}}}}}{\log _{10}}\left( {\frac{{{{{\rm{\bar \sigma }}}_{\rm{o}}} + {\rm{\Delta \bar \sigma }}}}{{{{\rm{\sigma }}_{\rm{o}}}}}} \right)\)

Calculation:

H_o = 4m, e_o = 1.30

C_c = 0.6, C_v = 1 m²/year

 \(\begin{array}{*{20}{l}} {{\rm{\Delta H}} = \frac{{0.6\times 4}}{{2.3}}\log \left( {\frac{{{{\bar \sigma }_i} + {{\bar \sigma }_i}}}{{{{\bar \sigma }_i}}}} \right)}\ {{\rm{\Delta H}} = 0.3141;{\rm{m}}}\ {{\rm{\Delta H}} = 314.1;{\rm{mm}}} \end{array}\)

Concept:

Calculating ultimate bearing capacity of a pile by λ - method

This method is used for pile embedded in clay only

Q_U = A_b C N_c + A_s λ (2C_a + σ_Va) 

Ab C Nc = End bearing component

 As λ (2Ca + σVa) = Surface bearing component

A_b = Base area of pile = π/4 × d²

C = Cohesion of soil, N_c = Terzaghi's bearing capacity factor = 9 (for clays)

A_s = Surface area of pile = π d l

λ = Coeffeceint depending on soil type, σ_Va = Average vertical effective stress along pile length 

C_a = Weighted average cohesion along pile length 

Calculation:

Given,

Length = 25 m, λ = 0.15,

d = 0.4 m, γ = 18 kN/m³ 

Neglecting the end bearing component

∴ Q_U = λ (σ_v + 2C_u)A_s

C_u (average value) \( = \frac{{40 + 50 + 60 + 70 + 80}}{5}\) = 60 kPa

σ_v = 18 × 12.5 = 225 kPa

\({{\rm{Q}}_{\rm{U}}}{\rm{ = 0}}{\rm{.15}}\left[ {{\rm{225 + 60 \times 2}}} \right]{\rm{\pi}}{{\rm{\times0.4}}} {\rm{ \times 25}}\) 

Q_U = 1625.77 kN

\(q = 10\sqrt 3 + 0.5p\) ------- (1)

\(q = \frac{{{\sigma _1} - {\sigma _3}}}{2}\)

\(p = \frac{{{\sigma _1} + {\sigma _3}}}{2}\)

\(\frac{{{\sigma _1} - {\sigma _3}}}{2} = c\cos \phi + \frac{{{\sigma _1} + {\sigma _3}}}{2}\sin \phi \) -------------- (2)

Comparing equation (1) and (2)

 \(\begin{array}{l} \sin \phi = 0.5\ \phi = {\sin ^{ - 1}}\left( {0.5} \right) = 30^\circ \ C\times cos\phi = 10\sqrt 3 \ c = \frac{{10\sqrt 3 }}{{\left( {\frac{{\sqrt 3 }}{2}} \right)}} = 20 \end{array}\)

Concept:

Bell’s equation giving active earth pressure of C-ϕ soils is,

\({{\rm{P}}_{\rm{a}}}{\rm{ = (}}{{\rm{K}}_{\rm{a}}}{\rm{ \times }}{{\rm{\sigma }}_{\rm{V}}}{\rm{) - }}\left( {{\rm{2C}}\sqrt {{{\rm{K}}_{\rm{a}}}} } \right)\)

Where,

σV = Vertical earth pressure

C = Cohesion of soil

Ka = Coefficient of active earth pressure

\({{\rm{K}}_{\rm{a}}}{\rm{ = ta}}{{\rm{n}}^{\rm{2}}}\left( {{\rm{4}}{{\rm{5}}^{\rm{^\circ }}}{\rm{ - }}\frac{{\rm{\phi}}}{{\rm{2}}}} \right)\)

ϕ = Angle of internal friction or angle of shearing resistance

Calculation:

γ = 18 kN/m³

 \(\begin{array}{l}{{\rm{K}}_{\rm{a_1}}} = {\tan ^2}\left( {45 - \frac{\phi }{2}} \right) = {\tan ^2}\left( {45 - \frac{{30}}{2}} \right) = \frac{1}{3}\ {{\rm{K}}_{\rm{a_2}}} = {\tan ^2}\left( {45 - \frac{\phi }{2}} \right) = {\tan ^2}\left( {45 - 0} \right) = 1 \end{array}\)

P_a at A:

σv = 0

P_a = 0

P_a at B due to layer 1:

σ_v = 18 × 3 = 54 kN/m³

\(P_a = K{a_1}{\sigma _v} = \frac{1}{3} \times 54 = 18kN/{m^2}\)

P_a at B due to layer 2:

 \(\begin{array}{*{20}{l}} {{{\rm{\sigma }}_{\rm{v}}}{\rm{ = 54\ kN/}}{{\rm{m}}^{\rm{2}}}}\ {{{\rm{p}}_{\rm{a}}}{\rm{ = K}}{{\rm{a}}_{\rm{2}}}{{\rm{\sigma }}_{\rm{v}}}{\rm{ - 2c}}\sqrt {{\rm{k}}{{\rm{a}}_{\rm{2}}}} }\ {{\rm{ p_a= 1 \times 54 - 2 \times 20}}\sqrt {\rm{1}} {\rm{ \ = 14 \ kN/}}{{\rm{m}}^{\rm{2}}}} \end{array}\) 

P_a at C due to layer 3:

 \(\begin{array}{*{20}{l}} {{{\rm{\sigma }}_{\rm{v}}}{\rm{ = 18 \times 3 + 18 \times 3 = 108}};{\rm{kN/}}{{\rm{m}}^{\rm{2}}}}\ {{{\rm{p}}_{\rm{a}}}{\rm{ = 1 \times 108 - 2 \times 20}}\sqrt {\rm{1}} {\rm{\ = 68}};{\rm{kN/}}{{\rm{m}}^{\rm{2}}}} \end{array}\) 

Active earth pressure diagram

 \(\begin{array}{*{20}{l}} {{{\rm{P}}_{{{\rm{a}}_{\rm{1}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times 3 \times 18 = 27}};{\rm{kN/m}}}\ {{{\rm{P}}_{{{\rm{a}}_{\rm{2}}}}}{\rm{ =14 \times 3 = 42}};{\rm{kN/m}}}\ {{{\rm{P}}_{{{\rm{a}}_{\rm{3}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times }}\left( {{\rm{68 - 14}}} \right){\rm{ \times 3 = 81}};{\rm{kN/m}}} \end{array}\) 

Total active earth pressure = 27 + 42 + 81 = 150 kN/m

Concept:

Gross command area (GCA):

It is the total area that can be irrigated by a canal so that an unlimited quantity of water is available. A canal is usually aligned along with the watershed between two drainage valley.

In other words, GCA is the total area lying between the drainage boundaries that can be irrigated by the canal system. It is measured in hectares.

GCA = CCA + UCA

Where,

CCA = Clturable command area, UCA = Unculturable command area

Duty of canal water (D)

Duty is defined as the area of the land irrigated by a unit discharge of water flowing continuously for the duration of the base period of the crop.

\({\rm{D}} = \frac{{{\rm{Area;of;land;being;irrigated}}}}{{{\rm{Rate;of;water;supply}}}} = \frac{A}{Q}\)

Duty is measured in hectares/cumec (ha/cumec)

Delta of a crop (Δ):

It is defined as the total thickness of water required for complete maturity of the crop within its base period. It is measured in cm

The sugarcane crop has maximum delta, i.e Δ = 120 cm. Other crops have delta less than 120 cm

Duty and delta can be related by

\({\bf{D}} = \frac{{8.64 \times {\bf{B}}}}{{\bf{\Delta }}}\)

Where, D is in ha/cumec, B is base period or kor period in days, and Delta (Δ) is in meters

Calculation:

Given,

CCA = 2000 hectares

Gram and wheat are Rabi crops,

For Gram,

Δ_g = 12 cm = 0.12 m, B_g = 18 days

Discharge required for gram

\({Q_g} = \frac{{{A_g}}}{{{D_g}}} = \frac{{2000 \times 0.3}}{{8.64 \times \frac{{18}}{{0.12}}}} = 0.4630\frac{{{m^3}}}{s}\)

For wheat,

Δ_w = 15 cm = 0.15 m, B_w = 18 days

Discharge required for wheat,

\({Q_w} = \frac{{{A_w}}}{{{D_w}}} = \frac{{2000 \times 0.5}}{{8.64 \times \frac{{18}}{{0.15}}}} = 0.9645\frac{{{m^3}}}{s}\)

So both are grown in the same period

Total discharge = Q_g + Q_w = 1.4275 m³/s

Concept:

Hydrostatic force or total pressure

The net force acting on one side of a gate due to the hydrostatic pressure intensity when the gate is supporting the fluid is given by,

F = γ × A × h̅ 

Where,

γ = Unit weight of the fluid which the gate supports = ρ × g

A = Cross-section area of gate

h̅ = Height of fluid above the C.G of gate

Calculation:

Given,

S.G of oil = 0.8

∴ ρ_oil = S.G of oil × ρ_water = 0.8 × 1000 = 800 kg/m³

F = ρgAh̅ 

\(\begin{array}{*{20}{l}} {{\rm{\bar h = 1}}{\rm{.5 + }}\frac{{\rm{2}}}{{\rm{3}}}{\rm{ × 1}}{\rm{.5 = 2}}{\rm{.5}};{\rm{m,}}\{\rm{A = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ × 1}}{\rm{.5 × 2 = 1}}{\rm{.5}}{{\rm{m}}^{\rm{2}}};}\ {{\rm{F = 800 × 9}}{\rm{.81 × 1}}{\rm{.5 × 2}}{\rm{.5 × 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ = 29}}{\rm{.43}};{\rm{kN}}} \end{array}\)

Concept:

Continutiy equation in Three-Dimension

\(\frac{{\partial {\rm{\rho }}}}{{\partial {\rm{t}}}} + \frac{{\partial \left( {{\rm{\rho U}}} \right)}}{{\partial {\rm{X}}}} + \frac{{\partial \left( {{\rm{\rho V}}} \right)}}{{\partial {\rm{Y}}}} + \frac{{\partial \left( {{\rm{\rho W}}} \right)}}{{\partial {\rm{Z}}}} = 0{\rm{;}}\)

Where,

U, V, and W are components of velocity in X, Y and Z direction respectively

When the flow is steady\(,;;\frac{{\partial {\bf{\rho }}}}{{\partial {\bf{t}}}} = 0\)

\(\therefore \frac{{{\rm{\partial }}\left( {{\rm{\rho U}}} \right)}}{{{\rm{\partial X}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho V}}} \right)}}{{{\rm{\partial Y}}}} + \frac{{{\rm{\partial }}\left( {{\rm{\rho W}}} \right)}}{{{\rm{\partial Z}}}} = 0\)

When flow is steady and incompressible, ρ = constant

\(\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} + \frac{{\partial {\rm{W}}}}{{\partial {\rm{Z}}}} = 0{\rm{;}}\)

When the flow is steady, incompressible and 2-D, \(\frac{{\partial {\bf{W}}}}{{\partial {\bf{Z}}}} = 0\)

\(\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\)

For flow being irrotational

\( \frac{{\partial {\rm{V}}}}{{\partial {\rm{X}}}} - \frac{{\partial {\rm{U}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\)

Calculation:

Given:

\(\begin{array}{l} U = \frac{{{y^3}}}{3} + 2x - {x^2}y\ V = x{y^2} - 2y - \frac{{{x^3}}}{3} \end{array}\)

As per continuity equation, for flow to occur

\(\therefore \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\)

\(\begin{array}{l} \frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} = 2 - 2xy\ \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 2xy - 2 \end{array}\)

2 - 2xy + 2xy - 2 = 0

Since, \(\frac{{\partial {\rm{U}}}}{{\partial {\rm{X}}}} + \frac{{\partial {\rm{V}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\), then flow is incompressible.

For irrational flow

\(\begin{array}{l} \frac{{\partial {\rm{V}}}}{{\partial {\rm{X}}}} = {y^2} - {x^2}\ \frac{{\partial {\rm{U}}}}{{\partial {\rm{Y}}}} = {y^2} - {x^2} \end{array}\)

\( \frac{{\partial {\rm{V}}}}{{\partial {\rm{X}}}} - \frac{{\partial {\rm{U}}}}{{\partial {\rm{Y}}}} = 0{\rm{;}}\)

So, flow is irrotational also.

Concept:

Water budget method:

By applying mass balance around water body and by writing water budget equation, evaporation is estimated.

Water budget equation around water body is

Σinflow - Σoutflow = ± ΔS

Σinflow includes rainfall and runoff

Σoutflow includes water used, evaporation and seepage

ΔS = change in storage 

ΔS is positive when water level rises.

ΔS is negative when water level falls.    

Calculation:

Let ‘x’ cm evaporation takes place in month of June.

Total inflow in cm

\( = \frac{{10 \times 30 \times 24 \times 60 \times 60}}{{20 \times {{10}^6}}} \times 100 + 10\)

= 139.6 cm

Total outflow in cm

\( = \frac{{15 \times 30 \times 24 \times 60 \times 60}}{{20 \times {{10}^6}}} \times 100 + 1.8 + x\)

= (196.2 + x) cm

As total outflow is more than total inflow, therefore depression in storage takes place.

Depression in storage (ΔS)

\({\rm{\Delta S = }}\frac{{{\rm{ - 16 \times 1}}{{\rm{0}}^{\rm{6}}}}}{{{\rm{20 \times 1}}{{\rm{0}}^{\rm{6}}}}}{\rm{ \times 100 = - 80}};{\rm{cm}}\)

Σinflow - Σoutflow = ± ΔS

139.6 - (196.2 + x) = - 80

– x = - 80 + 56.6

∴ Evaporation in month of June = x = 23.4 cm

Concept:

According to Darcy’s Weisbach equation:

\({\rm{Head;loss}} = {{\rm{h}}_{\rm{f}}} = \frac{{{\rm{fL}}{{\rm{V}}^2}}}{{2{\rm{gD}}}}\)

Where,

f = Friction factor of the pipe, L = Length of the pipe, V = Velocity of water through the pipe

g = Gravitational acceleration, D = Diameter of the pipe

Losses in the pipe at bends and fittings

\({\rm{h}} = {\rm{k}} × \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}}\)

where k =bend or fitting coefficient, depends on the angle of bend, type of bend, type of fitting and type of threading

Calculation:

Given,

f =  0.024, L_AB = 6km

d = 0.7 m

\(\begin{array}{*{20}{l}} {{{\rm{h}}_{\rm{L}}}{\rm{ = }}{{\rm{h}}_{\rm{L}}}_{\rm{1}}{\rm{ + }}{{\rm{h}}_{\rm{L}}}_{\rm{2}}}\ {30 = \frac{{{\rm{f }}{{\rm{L}}_{\rm{1}}}{\rm{V}}_{\rm{1}}^{\rm{2}}}}{{{\rm{2gd}}}}{\rm{ + }}\frac{{{\rm{f }}{{\rm{L}}_{\rm{2}}}{\rm{V}}_{\rm{2}}^{\rm{2}}}}{{{\rm{2gd}}}}}\ {30 = \frac{{\left( {{\rm{0}}{\rm{.024}}} \right)\left( {{\rm{3000}}} \right){{\rm{Q}}^{\rm{2}}}}}{{{\rm{2 \times 9}}{\rm{.81 \times }}{{\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right)}^{\rm{2}}}{\rm{ \times }}{{\left( {{\rm{0}}{\rm{.7}}} \right)}^{\rm{5}}}}}{\rm{ + }}\frac{{\left( {{\rm{0}}{\rm{.024}}} \right)\left( {{\rm{3000}}} \right){{\left( {{\rm{Q - 0}}{\rm{.15}}} \right)}^{\rm{2}}}}}{{{\rm{2 \times 9}}{\rm{.81 \times }}{{\left( {\frac{{\rm{\pi }}}{{\rm{4}}}} \right)}^{\rm{2}}}{\rm{ \times }}{{\left( {{\rm{0}}{\rm{.7}}} \right)}^{\rm{5}}}}}}\ {30 = 3{\rm{5}}{\rm{.40}}\left[ {{{\rm{Q}}^{\rm{2}}}{\rm{ + }}{{\left( {{\rm{Q - 0}}{\rm{.15}}} \right)}^{\rm{2}}}} \right]} \end{array}\)

0.847 = 2Q² – 0.3Q + 0.0225

Q = 0.721 m³/sec

Q_B = Q – 0.15 = 0.721 – 0.15 = 0.571 m³/sec

The discharge into the reservoir B = Q_B = 0.571 m3/sec

Concept:

\({\rm{CompactionRatio = }}\frac{{{\rm{Volume}};{\rm{before}};{\rm{compaction}}}}{{{\rm{Volume}};{\rm{after}};{\rm{compaction}}}}\)

Total Volume = SW + Cover

Calculation:

Solid waste generated per day= 2 × 200000 = 400000 kg

Total solid waste generated in 25 years = 400000 × 365 × 25 = 3.65 × 10⁹ kg

Compaction ratio,

\(4 = \frac{{{\rm{Volume;before;compaction}}}}{{{\rm{Volume;after;compaction}}}}\)

Volume of waste before compaction,

\({\rm{V}} = \frac{{3.65 \times {{10}^9}}}{{100}} = 3.65 \times {10^7}\ {{\rm{m}}^3}\)

Volume of waste after compaction,

\(\begin{array}{l} {\rm{V'}} = \frac{{3.65 \times {{10}^9}}}{{100\times 4}} = 9.125 \times {10^6}\ {{\rm{m}}^3}\ \frac{{{\rm{SW}} + {\rm{cover}}}}{{{\rm{SW}}}} = \frac{{{\rm{SW}}}}{{{\rm{SW}}}} + \frac{{{\rm{cover}}}}{{{\rm{SW}}}} = 1.5\ \frac{{{\rm{Cover}}}}{{{\rm{SW}}}} = 0.5 \end{array}\)

Cover = 0.5 × 9.125 × 10⁶

Cover = 4.5625 × 10⁶ m³

Total Volume = SW + Cover

Total Volume = (9.125 + 4.5625) × 10⁶ m³

Total Volume = 13.6875 million m³

Concept:

1. Surface Loading Rate (SLR) or Surface Overflow Rate (SOR) 

Volume of water applied in unit time per unit surface area of settling tank is known as Surface Loading Rate. It is also denoted by Vo.

\({\rm{SOR;or;}}{{\rm{V}}_{\rm{o}}} = \frac{{\frac{{{\rm{Volume;of;water;}}}}{{{\rm{Time}}}}}}{{{\rm{Surface;area}}}} = \frac{{{\rm{Discharge;}}}}{{{\rm{Surface;area}}}}\)

∴ \({\rm{Surface\ area\ of\ filter}} = \frac{{{\rm{Discharge;}}}}{{{\rm{SOR}}}}\)

It is expressed in m3/d/m2

2) The number of filters required.

\({\rm{n}} = \frac{{{\rm{Total;surface;area}}}}{{{\rm{Area;of;each;filter}}}}\)

Calculation:

Total water filtered in a day= 60 × 60 × 24 = 86400 m³/day

Total surface area of filter required= \(\frac{{86400}}{120} = 720;{m^2}\)

Area of one filter = 6 × 10 = 60 m²

Total no. of filter = 720/60 = 12 filters

As two filters are out of service, number of filters working = 12 – 2 = 10

Total surface area of filter = 60 × 10 = 600 m²

The new loading rate for 10 filters in use.

\(V_o = \frac{{86400;{m^3}/day}}{{600{m^2}}} = 144\frac{{{m^3}}}{{day.{m^2}}}\)

Concept:

Biochemical oxygen demand (B.O.D):

B.O.D is defined as the amount of oxygen demand by the micro-organisms C₅H₇NO₂ (bacteria) present in it to decompose biodegredable organic matter in waste water under aerobic conditions. B.O.D is measure of strength of waste water.

The amount of organic matter removed (Y_t) is calculated by

Y_t = L_o (1 - e^-Kt)

where,

t = Time in days

L_o = Initial organic matter in waste water at time t = 0

K = B.O.D rate constant depends on temperature

K_T = K₂₀ (1.047)^{T - 20}

T = Temperature in degrees

Percentage of organic matter remaining is given by

\({\rm{% ;exerted}} = \frac{{{\rm{B}}.{\rm{O}}.{{\rm{D}}_{\rm{t}}}}}{{{\rm{B}}.{\rm{O}}.{{\rm{D}}_{\rm{u}}}}} × 100\)

% remaining = 100 - %Exerted

Calculation:

B.O.D remaining after 7 days

B.O.D₇ = B.O.D_u (1- e^-Kt)

B.O.D₇ = 20 × (1- e^{-0.15 × 7})

B.O.D₇ = 20 × 0.65 =13 

% exerted or removed \( = \frac{{13 × 100}}{{20}} = 65% \)

B.O.D remaining after 7 days = (100 – 65) = 35%

Mistake point:

K = B.O.D rate constant 

Normal log means log_e, K_T has following relation between the (base e) value and (base 10) value.

K_T (base e) = 2.3 × K_T (base 10)

Concept:

Design parameters of Primary sedimentation tank are

1) Surface Loading Rate (SLR) or Surface Overflow Rate (SOR) 

Volume of water applied in unit time per unit surface area of settling tank is known as Surface Loading Rate. It is also denoted by V_o.

\({\rm{SOR;or;}}{{\rm{V}}_{\rm{o}}} = \frac{{\frac{{{\rm{Volume;of;water;}}}}{{{\rm{Time}}}}}}{{{\rm{Surface;area}}}} = \frac{{{\rm{Discharge;}}}}{{{\rm{Surface;area}}}}\)

It is expressed in m3/d/m2

2) Detention Time (DT)

Average theoritical time taken by water to travel from inlet to outlet of settling tank is known as detention time. Detention time should be greater than settling time of particle to capture the particle in the settling tank.

\({\rm{DT}} = \frac{{{\rm{Volume;of;settling;tank}}}}{{{\rm{Discharge}}}} = \frac{{\rm{V}}}{{\rm{Q}}}\)

It is expressed in hours

3) Flow through velocity

Speed with which the water travel from inlet to outlet of settling tank.

\({\rm{V}} = \frac{{{\rm{Discharge;}}}}{{{\rm{Cross}} - {\rm{section;area;of;tank}}}} = \frac{{\rm{Q}}}{{\rm{A}}}\)

It is expressed in m/min

4) Weir Loading Rate (WLR)

Discharge per unit length of weir is known as weir loading rate.

\({\rm{WLR}} = \frac{{{\rm{Discharge}}}}{{{\rm{Length;of;weir}}}} = \frac{{\rm{Q}}}{{\rm{L}}}\)

It is expressed in m3/day/m

Calculation:

Given,

Dimenisons of tank

L = 32.5 m, B = 8 m, D = 2.25 m

V_o = Overflow rate = 32.5 m3/d/m2

Weir length = 75 m

\({V_0} = \frac{Q}{{BL}}\)

Q = 32.5 × 32.5 × 8

= 8450 m³/d

We know that

\({\rm{WLR}} = \frac{{{\rm{Discharge}}}}{{{\rm{Length;of;weir}}}}\)

WLR = 8450/75 = 112.67 m³/d/m

Concept:

Speed (u):

Travel time is the reciprocal of the speed. 

Spot speed is the instantaneous speed of the vehicle at a specified section or location.

Average speed is the average of the spot speeds of all vehicles passing a given point on the highway.

Space mean speed is the harmonic mean of the spot speeds.

Time mean speed represents the speed distribution of a vehicle at a given point and it is the average of the instantaneous speeds of observed vehicles at a spot.

Traffic density (k):

The number of vehicles occupying a unit length of road at a given instant.

It is expressed in vehicles/km

Traffic volume (q):

The number of vehicles occupying moving in a specified direction on a given road during specified.

It is expressed in vehicles/hr or vehicle/day

Traffic capacity (C):

It is the ability of the road to accommodate traffic volume

\({\rm{C}} = \frac{{1000 × {\rm{u}}}}{{\rm{S}}}\)

where S = average center to center spacing of a vehicle

It is expressed in vehicles/hr/road

The relation between traffic volume, traffic density and speed is given by

q = k × u

Calculation:

Given:

u =70 – 0.7 k

Traffic volume,

q = u × k

q = (70 – 0.7 k) k = 70k – 0.7 k²

Maximum possible volume is capacity of road.

For maximum volume

\(\frac{{dq}}{{dk}} = 0\)

70 – 1.4 k = 0

k = 50 Veh/km

Hence, capacity

C = (70 – 0.7 × 50) × 50

C = 1750 Veh/hr

1) Hardness: 

The ability of a material to resist wear, abrasion, scratching or identation (penetration) by harder bodies is called hardness.

Tests for hardness: Brinell, Rockwell, Vickers

2) Strength:

The strength of a material is its ability to sustain loads without getting distort, collapse or crushed. A material should have adquate strength when subjected to tension, compression, shear, bending, or tension as per the intended use. The maximum stress that any material will withstand is called ultimate strength or tenacity.

3) Toughness:

The amount of Impact energy or shock energy absorbed by a specimen before it fractures is called its impact strength or toughness.

Tests for toughness: Izod and Charpy impact test

4) Soundness:

Soundness is the resistance of the material against disintegration by weathering and freeze-thaw cycles.

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:

8) Stripping value test: Adhesion of Bitumen with aggregates..

Concept:

Determination of the specific gravity

\({{\rm{G}}_{\rm{t}}} = \frac{{100}}{{\frac{{{{\rm{W}}_1}}}{{{{\rm{G}}_1}}} + \frac{{{{\rm{W}}_2}}}{{{{\rm{G}}_2}}} + \frac{{{{\rm{W}}_3}}}{{{{\rm{G}}_3}}} + \frac{{{{\rm{W}}_4}}}{{{{\rm{G}}_4}}}}}\)

Where,

W₁, W₂, W₃, and W₄ are the percentage by weights of the coarse aggregate, fine aggregate, fines and bitumen respectively.

G₁, G₂, G₃, and G₄ are the specific gravity of the coarse aggregate, fine aggregate, fines and bitumen respectively.

Effective specific gravity of aggregates (coarse + fine) is given by

\({\rm{G' = }}\frac{{\left( {{{\rm{W}}_{\rm{1}}}{\rm{ \times }}{{\rm{G}}_{\rm{1}}}} \right){\rm{ + }}\left( {{{\rm{W}}_{\rm{2}}}{\rm{ \times }}{{\rm{G}}_{\rm{2}}}} \right)}}{{{{\rm{W}}_{\rm{1}}}{\rm{ + }}{{\rm{W}}_{\rm{2}}}}}\)

Where,

W1, and W2 are the percentage by weights of the coarse aggregate and fine aggregate respectively.

G1, and G2 are the specific gravity of the coarse aggregate and fine aggregate respectively.

Calculation:

Given,

W₁ = 55.0, W₂ = 35.8, W₃ = 3.7,and W₄ = 5.5

G1= 2.6, G2 = 2.7, G3 = 2.65 and G₄ = 1.01

Theoretical specific gravity

\(\begin{array}{*{20}{l}} {{{\rm{G}}_{\rm{t}}} = \frac{{100}}{{\frac{{{{\rm{W}}_1}}}{{{{\rm{G}}_1}}} + \frac{{{{\rm{W}}_2}}}{{{{\rm{G}}_2}}} + \frac{{{{\rm{W}}_3}}}{{{{\rm{G}}_3}}} + \frac{{{{\rm{W}}_4}}}{{{{\rm{G}}_4}}}}}}\ {{{\rm{G}}_{\rm{t}}} = \frac{{100}}{{\frac{{55}}{{2.6}} + \frac{{35.8}}{{2.7}} + \frac{{3.7}}{{2.65}} + \frac{{5.5}}{{1.01}}}} \ {{\rm{G}}_{\rm{t}}}= 2.42} \end{array}\)

Effective specific gravity of aggregates (coarse + fine) is given by

\({\rm{G'}} = \frac{{\left( {55 \times 2.6} \right) + \left( {35.8 \times 2.7} \right)}}{{55 + 35.8}} = 2.639 = 2.64\)

∴ Theoretical specific gravity of the mix = G_t = 2.42 and the effective specific gravity of the aggregates = G' = 2.64

Concept:

Let S₁ S₃ = l₁ and S₂ S₃ = I₂

Northing of S₃:

500 + l₁ cos 45° = 450 + l₂ cos 26° 34’

l₁ cos 45° – l₂ cos 26° 34’ = – 50                   ----------(i)

Easting of S₃:

500 + l₁ sin 45° = 600 + l₂ sin 26°34’

l₁ sin 45° + l₂ sin 26°34’ = 100                            ----------(ii)

Solving eq. (i) and eq. (ii)

l₁ = 70.707

l₂ = 111.802

∴ Easting of S₃ = 500 + 70.707 sin45° = 550 m

Concept:

Reciprocal Levelling

This method is used for determining the difference between elevations of two points which are situated apart, and it is not possible to set up the instrument midway between the points.

Suppose Point A and point B are situated on the opposite sides of the river then,

The True Reduced level (R.L) difference between station A and station B, when the station B is above station A 

\({{\rm{h}}_{{\rm{AB}}}} = \frac{1}{2}\left[ {\left( {{{\rm{a}}_1} - {{\rm{b}}_1}} \right) + \left( {{{\rm{a}}_2} - {{\rm{b}}_2}} \right)} \right]\)

RL of station B = RL of station A + hAB 

The True Reduced level (R.L) difference between station A and station B, when the station B is below station A ​ 

\({{\rm{h}}_{{\rm{AB}}}} = \frac{1}{2}\left[ {\left( {{{\rm{b}}_1} - {{\rm{a}}_1}} \right) + \left( {{{\rm{b}}_2} - {{\rm{a}}_2}} \right)} \right]\)

RL of station B = RL of station A - hAB 

where, 

a₁ and b₁ are the readings taken while the instrument is at station A

a₂ and b₂ are the readings taken while the instrument is at station B

Calculation:

Given,

Staff readings shows that station, B is below station A.

a₁ = 1.350 m

a₂ = 0.550 m

b₁ = 1.550 m

b₂ = 0.750 m

\(\begin{array}{*{20}{l}} {{{\rm{h}}_{{\rm{AB}}}} = \frac{{\left( {{b_1} - {a_1}} \right) + \left( {{b_2} - {a_2}} \right)}}{2}}\ \begin{array}{l} {{\rm{h}}_{{\rm{AB}}}} = \frac{{\left( {1.550 - 1.350} \right) + \left( {0.750 - 0.550} \right)}}{2}\ {{\rm{h}}_{{\rm{AB}}}} = 0.200 \end{array} \end{array}\)

RL of B = RL of A – 0.200

RL of B = 100.200 – 0.200

RL of B = 100.000 m