GATE CE 2014 Official Paper: Shift 1 (Previous Year Paper)

The word 'comprehension' means 'the ability to understand something'. The meanings of the other words are: 

Understanding →  comprehension

Meaning →   what is intended to be, or actually is, expressed or indicated

Concentration →   focus

Stability →   the state of being stable.

Thus, understanding is the correct answer.

The sentence mentions a positive quality of a person which is ability to forgive someone thus a word conveying a positive meaning should fit here. The meanings of the words are as follows:

Vice → immoral behaviour (the sentence does not mention something negative thus this is incorrect)

Virtues →  behaviour showing high moral standards

Choices → an act of choosing between two or more possibilities.

Strength → power

Thus the word 'virtues' is the apt word here as forgiving someone is goodness of a person. This is not a matter of choice or strength thus options 3 and 4 do not fit here. Option 2 is thus the correct answer.

It is not mentioned that Rajan wanted to work only in a group thus option 1 is invalid.

It is not mentioned that the two of them were grouped together. It is clear that it was a matter of their choice and Sajan had chosen to work individually.Thus 2 is invalid.

It is mentioned that Sajan explained to Rajan that he liked to work independently which makes it clear that he had not given in to his request. Thus 3 is invalid too.

It can be understood that Rajan was unhappy with Sajan working on his own as he had expected that they would work together. His expectation was not met. Thus option 4 is logically valid here.

y = 5x² + 3

The tangent is given as

\(\therefore \frac{{dy}}{{dx}} = 10;x\)

Tangent at x = 0 and y = 3 = 0

Slope is the angle made with x-axis, here \(\frac{{dy}}{{dx}} = 0,\) so the slope is parallel to the x-axis.

Fixed daily cost = 50,000

Variable cost = 800 Q

Cost of production for 100 tonnes = fixed cost + variable cost

= 50000 + 800 × 100

= 50000 + 80000 = 130,000 Rs.

Cost of production per tonn\(= \frac{{130,000}}{{100}} = 1300;Rs.\)

Here in the first letter, 4 is added but in OYEIK the first letter has an addition of 6 (9 + 6 = 15).

Similarly, at the 2^nd place, there is addition of 4 digits but at last there is addition of 5(20 + 5 = 25).

So, the last option is odd.

Given,

There is a six-storeyed building, the ground floor is numbered 1, the floor above it 2 and so on.

Given conditions:

1) Anuj lives on an even-numbered floor → Possible floor numbers are 2, 4 and 6.

2) Bhola does not live on an odd-numbered floor → Possible floor numbers are 2, 4 and 6.

3) Chandan does not live on any of the floors below Faisal’s floor → Chandan's floor number > Faisal's floor number.

4. Dilip does not live on floor number 2

5) Eswar does not live on a floor immediately above or immediately below Bhola → The difference between floor number of Eswar and Bhola > 1

6) Faisal lives three floors above Dilip → The difference between floor number of Faisal and Dilip = 3

All these conditions are satisified in option 2

Thus, respective positions are Anuj - 2, Bhola - 6, Chandan - 5, Dilip - 1, Eswar - 3, Faisal - 4

Let the smallest angle, second largest angle, and the largest angle of a triangle be α, β, and γ respectively

Given

α = (2/3)^rd of the smallest angle of a quadrilateral

γ = 2α

The ratio between the angles of the quadrilateral is 3:4:5:6

We know that,

The sum of angles of quadrilateral  = 360°

∴ 3X + 4X + 5X +6X = 360°

∴ 18X = 360°

∴ X = 20°

∴ The smallest angle of quadrilateral = 3X = 3 × 20° = 60°

∴ The largest angle of quadrilateral = 6X = 6 × 20° = 120°

Now,

α = (2/3)^rd of the smallest angle of a quadrilateral

α = (2/3) × 60° = 40°

γ = 2α =2 × 40° = 80°

We know that,

The sum of angles of triangle = 180°

∴ α + β + γ = 180°

∴ 40° + β + 80° = 180°

∴ β = 60°

The smallest angle, second largest angle and the largest angle of a triangle are 40°, 60°, and 80° respectively.

The second largest angle of the triangle = 60°

The largest angle of the quadrilateral = 120°

The sum of the second largest angle of the triangle and the largest angle of the quadrilateral,

 Sum = 60° + 120° = 180°

Assuming that the total number of people in the country Y be 100.

Given,

There are thrice as many people in country X as in country Y.

∴ The total number of people in the country X = 3 × 100 = 300

1. One percent of the people in country X are taller than 6 ft.

    1% of 300 = 3, i.e 3 persons are taller than 6 ft in country X.

2. Two percent of the people of country Y are taller than 6 ft.

    2% of 100 = 2, i.e 2 persons are taller than 6 ft in country Y.

Total number of persons of country X and country Y = 300 + 100 = 400

Total number of persons of country X and country Y taller than 6 ft = 3 + 2 = 5

Taking both countries together, the percentage of people taller than 6 ft

⇒ \(\frac{5}{{400}} \times 100 = 1.25% \)

In the question, the monthly rainfall chart is given, based on 50 years of rainfall in Agra.

Checking the validity of the given statements.

(i) On average, it rains more in July than in December ⇒ correct.

(ii) Every year, the amount of rainfall in August is more than that in January. ⇒ may not be correct because average rainfall is provided.

(iii) July rainfall can be estimated with better confidence than February rainfall. ⇒ From the chart, it is clear the gap between the 5 percentile and 95 percentile from average is higher in February than that in July ⇒ correct.

(iv) In August, there is at least 500 mm of rainfall ⇒ May not be correct, because it is a 50-year average.

So the correct answer is option (B), i.e only statements (i) and (iii) are correct.

Given:

\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{x + \sin x}}{x}} \right)\)

The given expression \(\left( {\frac{{x + \sin x}}{x}} \right)\) is of \(\frac{\infty }{\infty }\) indeterminate form and to solve this divide the numerator and denominator by ‘x’.

\(\mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{x + \sin x}}{x}}}{{\frac{x}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 + \left( {\frac{{\sin x}}{x}} \right)}}{1}\)

∵ \(\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} ;f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)\)

Using above,

\(\mathop {\lim }\limits_{x \to \infty } \frac{{1 + \left( {\frac{{\sin x}}{x}} \right)}}{1} = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\sin x}}{x}} \right) = 1 + 0 = {\bf{1}}\)

This is simple matrix multiplication:

Given:

 \(J = \left[ {\begin{array}{*{20}{c}} 3&2&1\ 2&4&2\ 1&2&6 \end{array}} \right]\) and \(K = \left[ {\begin{array}{*{20}{c}} 1\ 2\ { - 1} \end{array}} \right]\),

∴ K^T = [1 2 -1]

Now,

\({K^T};J;K = \left[ {1;2; - 1} \right]\left[ {\begin{array}{*{20}{c}} 3&2&1\ 2&4&2\ 1&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\ 2\ { - 1} \end{array}} \right] = \left[ {6;8; - 1} \right]\left[ {\begin{array}{*{20}{c}} 1\ 2\ { - 1} \end{array}} \right] = \left[ {23} \right]\)

∴ K^T J K = 23.

Concept:

Let f(x) be any probability density function where ‘x’ is random variable.

The probability that x lies between x₁ and x­₂ is given by,

\(P\left( {{x_1} \le x \le {x_2}} \right) = \mathop \smallint \limits_{{x_1}}^{{x_2}} f\left( x \right)dx\)

In this case, x is evaporation and probability density f (x).

\(f\left( E \right) = \left{ {\begin{array}{*{20}{c}} {\frac{1}{2};0 \le E \le 5;mm/day}\ {0;;;otherwise} \end{array}} \right.\)

Calculation:

The probability E lies between 2 and 4 mm/day is P(2 ≤ E ≤ 4)

\(P\left( {2 \le E \le 4} \right) = \mathop \smallint \limits_2^4 f\left( E \right)dE = ;\mathop \smallint \limits_2^4 \left( {\frac{1}{5}} \right)dE = \left. {\frac{1}{5}} \right|_2^4 = \frac{2}{5} = 0.4\)

Concept:

The following properties should be memorized for eigen values:

1. Sum of eigen values = Trace of matrix and Trace of matrix is the sum of its diagonal elements.

2. Product of eigen values = Determinate of matrix

3. No. of eigen values = Size of the matrix.

Calculation:

Given:

\(M = \left[ {\begin{array}{*{20}{c}} {215}&{650}&{795}\ {655}&{150}&{835}\ {485}&{355}&{550} \end{array}} \right]\)

Its Diagonal elements are: 215, 150, 550.

Sum of eigen values = Trace of matrix = 215 + 150 + 550 = 915

Mistake Point:

Do not try to evaluate its eigen values by using standard procedure as calculation becomes very tough and it is time consuming process.

Calculation:

Plot the given co-ordinates on cartesian system as shown:

\(a = \sqrt 5 .b = \sqrt {18} ,c = \sqrt 5 \)

Area of triangle, \(ABC = \frac{1}{2} \times \) (Product of any two adjacent sides) × (sine of angle between them)

Let’s take sides AB and AC and angle b/w them is ‘θ’.

As per cosine rule,

\(\cos \theta = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{18 + 5 - 5}}{{2 \times \sqrt {18} \times \sqrt 5 }} = \sqrt {\frac{9}{{10}}} \)

\(\sin \theta = \sqrt {1 - {{\cos }^2}\theta } = 1 - \frac{9}{{10}} = \sqrt {\frac{1}{{10}}} \)

Area \(= \frac{1}{2} \times AB \times AC \times \sin \theta = \frac{1}{2} \times \sqrt 5 \times \sqrt 8 \times \frac{1}{{\sqrt {10} }} = \frac{3}{2}\) = 1.5

Kinematical mechanism method of ultimate load analysis:

It states that collapse is formed by assuming a mechanism will always be greater than or equal to true collapse load. i.e**. P ≥ P_u.**

It satisfies the equilibrium and mechanism conditions.

It is called Upper bound theorem as it gives upper limit of the true collapse load.

Static method of ultimate load analysis:

It states that collapse load found on the basis of any collapse bending moment diagram will be less than or equal to true collapse load. i.e**. P ≤ P_u.**

It satisfies the Equilibrium and Yield conditions.

It is called Lower bound theorem as it gives lower limit of the true collapse load.

Load factor:

It is the factor used to increase the working load to account the effect of various uncertainties in working load calculation. It is given as

Design load = Load factor × Working load

Factor of safety:

It is the factor used to reduce the ultimate strength of material so that more safety may be ensured while designing a structure.

Design strength = Ultimate strength/FOS

Shear Centre is defined as the point through which the load must be applied to produce zero twisting moment on the section. If load passes through the shear centre, then there will be only bending stress in the cross section and no twisting.

The following points should be remembered for shear center:

1. If a beam has two axes of symmetry, then the shear centre will lie at their point of intersection.
2. For a section having one axis of symmetry, the shear centre lies on that axis of symmetry.

The shear flow in channel section causes clockwise twisting as shown.

To balance that clockwise twisting, external resultant load must cause anticlockwise twisting and it can be possible if load lies an LHS of the channel section.

Further, since channel section has X-X as of symmetry and Y-Y axis is not the symmetric axis so, ‘R’ cannot be shear centre.

‘S’, and ‘θ’ cannot be shear centre because they cause external loading twisting in clockwise rather than anticlockwise.

So, only possible location of shear centre is ‘P’.

The following procedure to be adopted to find the collapse load.

1. Find the all possible locations of plastic Hinge. In this case plastic hinges will be at A, B only. There will no plastic hinge at C because at C bending moment is zero due to hinge connection.
2. No of plastic hinge required to form collapse mechanism, N = Degree of static indeterminacy + 1

Degree of static indeterminacy for the given beam = 1

⇒ N = 2 i.e. two plastic hinge will be needed to form mechanism and that will be point A and B.

3. Use principle of virtual work i.e.

External work done + Internal work done = 0

Calculation:

Possible hinge 0 elastic hinge.

External work done \(= P{\rm{\Delta }} = \left( {\frac{{PL\theta }}{2}} \right)\)

Internal work done: \(\begin{array}{*{20}{c}} { - \left( {{M_P}\theta } \right)}& + &{{M_P}\left( { - \theta - \theta } \right)}& = &{ - 3{M_P}\theta }\ \downarrow &{}& \downarrow &{}&{}\ {at;A}&{}&{at;B}&{}&{} \end{array}\)

By virtual work:

W_ext + W_int = 0

\(\Rightarrow \frac{{PL\theta }}{2} + \left( { - 3{M_P}\theta } \right) = 0 \Rightarrow P = \frac{{6{M_P}}}{L}\)

Concept

As per IS 456: 2000, the bearing strength of concrete (f_br) is given by:

 f_br = 0.45 f_ck (For limit state)

Calculation

 Given, 

 Grade of concrete M20 i.e. f_ck = 20 MPa  

 ∴ f_br = 0.45 × 20 = 9 MPa

Explanation:

As per IS 800 : 2007, CL : 9.2.1,

The high shear case is V > 0.6 V_s and it says that there is no reduction in moment capacity of cross-section as long as it is not subjected to high shear.

Concept

Method 1: Using basic concept

The degree of static indeterminacy means no of unknowns beyond available equation of equilibrium equations.

Segment RS is cable and cable can resist only tensile force, so there will be one unknown at joint R.

Further, Joint P is hinged and therefore, there will be only two reactions in x and y direction as shown in figure:

Total unknown = 1 (at Joint R) + 2 (at Joint P) = 3

For frame PQR, there are 3 Equations of Equilibrium available as shown in above figure, by using these equation these 3 unknown can be easily predicted.

Equilibrium equation:

∑F_x = 0, ∑F_y = 0, ∑M_p = 0

∴ D_s = 3 – 3 = 0

Method 2: Using direct formula:

Degree of Static Indeterminacy (D_s):

D_s = D_se + D_si

D_se = Degree of external static indeterminacy.

D_si = Degree of internal static indeterminacy.

D_se = Support Reactions – Number of Equilibrium equations

D_si = 3  C – R

C = No of closed loops

R = No of releases = m - 1

Calculation:

No of reactions = 2 + 2 = 4 (2 reactions at joint S and P each)

No of equilibrium reactions = 3

D_se = 4 - 3 = 1

No of closed loops, C = 0

Releases

Moment releases at joint R = 2 - 1 = 1

D_si = 3  C – R = 3  – 1 = -1

D_s = D_se + D_si = 1 - 1 = 0

∴ Static Indeterminacy, D­_s = 0

Concept:

As per Muller Breslau Principle, to draw influence line diagram of shear force at any section, cut the beam at that section and lift in such a way that angle at joint A and B are equal, then corresponding deflected shape of the beam is itself an influence line diagram ( do not remove support A and B). The ordinate of influence line diagram at that section will be (a + b)/L, where a and b are the length of the span on LHS and RHS of that section and L is the total length of beam.

\({\theta _A} = \frac{{A'B'}}{{AB'}} = \frac{{\frac{a}{L}}}{{\frac{L}{4}}} = \frac{1}{L}\)

\({\theta _B} = \frac{{B'C'}}{{B'B}} = \frac{{\frac{b}{L}}}{{3\frac{L}{4}}} = \frac{1}{L}\)

The degree of disturbance in sampling depends on the design parameter of sampler such as inside clearance, outside clearance and area ratio.

Inside Clearance: This helps in reducing the friction between soil sample and sampling tube when soil enters in sampling tube. It lies in between 1 to 3%.

Outside Clearance: It helps in reducing friction between sampling tube and soil when sampler is driven and withdrawn. It lies in between 0 to 2%.

Area ratio: It should be kept less as possible as such lesser area ratio implies sharp edge. It should be less than 10 % but in no case it should exceed 20%.

Inside clearance, \({C_i} = \left( {\frac{{{D_3} - {D_i}}}{{{D_i}}}} \right) \times 100;;{C_i} \to \left( {1 - 3} \right)% \)

Outside clearance, \({C_0} = \left( {\frac{{{D_0} - {D_4}}}{{{D_4}}}} \right) \times 100;;{C_0} = \left( {0 - 2} \right)% \)

Area Ratio, \({A_r} = \left( {\frac{{D_0^2 - D_i^2}}{{D_i^2}}} \right) \times 100\ \)

Its value should not be greater than about 20% for stiff formations while for soft sensitive clays should not be greater than 10%.

For a saturated cohesive soil, a triaxial test yields the angle of internal friction (φ) as zero. The conducted test will be Unconsolidated Undrained (UU) test.

Unconsolidated Un-drained (UU) Test: In this test expulsion of pore water is not permitted in both the stages. It is used for clays in short term analysis for clays under un-drained conditions at fast loading rate.

Consolidated Drained (CD) Test: In this test expulsion of pore water is permitted in both the stages. It is used for short term and long term stability analysis in saturated sands and long term stability analysis in clays.

Consolidated Un-drained (CU) Test: In this test, expulsion of pore water is permitted in 1^st stage but not in second stage. It is used for investigation of safety of earthen dam which may occur due to sudden drawdown of water table.

Unconfined Compression (UC) Test: In this test, cell pressure is kept zero and specimen is loaded vertically till it fail in shear. This is performed generally for highly cohesive clays having ϕ = 0.

There is friction between pile and surrounding soil and this friction always acts in opposite direction of relative motion.

Now, if surrounding soil settles more than the pile, then relative motion of soil with respect to pile is in downward direction and hence, friction will act in upward direction on soil and on pile it will act in down ward direction as per Newton’s third law of Motion i.e. action and reaction are equal and opposite). This is called negative skin friction.

This negative skin friction reduces the bearing capacity of pile because the depth or potion of pile up-to which negative skin friction acts becomes the part of applied load on pile.

Important point:

Causes of Negative Skin friction

1. Sudden lowering of Water Table
2. Soft soil strata followed by stiff strata
3. Application of surcharge over surrounding soil

Concept:

γ_sat = 18 kN/m³

γ_w = 10 kN/m³

ϕ’ = 34°

C’ = 0

FOS (Factor of safety) for infinite slope of cohesionless soil with seepage is given as:

\(FOS = \left( {1 - \frac{{{\gamma _w}h}}{{{\gamma _{avg}}z}}} \right)\frac{{\tan \phi }}{{\tan \beta }}\)

Where h is the depth of water table.

\({\gamma _{avg}} = \frac{{\gamma \left( {z - h} \right) + {\gamma _{sat}}h}}{{\left( {z - 4} \right) + \left( h \right)}}\)

Where (z - h) depth of soil has unit weight ‘γ’ and soil below water table is fully saturated (γ_sat).

If seepage takes place at ground i.e. (h = 2)

Then \(FOS = \left( {\frac{{{\gamma _{sat}}}}{{{\gamma _{sat}}}}} \right)\left( {\frac{{\tan \phi }}{{\tan \beta }}} \right)\)

Calculation:

Given:

γ_sat = 18 kN/m³,

γ_w = 10 kN/m³,

ϕ = 34°

\(FOS = \left( {\frac{{18 - 10}}{{18}}} \right)\left( {\frac{{\tan 34^\circ }}{{\tan \beta }}} \right) = 1.5\)

\(\Rightarrow \tan \beta = 0.2 \Rightarrow \beta = 11.3^\circ \)

Concept:

For an incompressible homogeneous fluid, mass of the fluid is conserved i.e. equation of continuity is valid.

⇒ A₁V₁ = A₂V₂

Where A₁, A₂ are the cross-sectional areas at section 1 and 2 respectively.

V₁ and V₂ are the velocity of flow at section 1 and 2 respectively.

A₁V₁ = A₂V₂

\(\Rightarrow \frac{\pi }{4}{\left( {{D_1}} \right)^2}{V_1} = \frac{\pi }{4}{\left( {{D_2}} \right)^2}{V_2}\)

\(\Rightarrow \frac{\pi }{4}{\left( {.15} \right)^2} \times 2.5 = \frac{\pi }{4}{\left( {0.05} \right)^2} \times {V_2}\)

⇒ V₂ = 22.5 m/s

Concept:

A flow duration curve is a plot of flow and percent of time that a particular discharge was equalled or exceeded. The area under the flow duration curve gives the average daily flow for a given period for which flow duration curve is plotted.

Application:

1. To find out the maximum and minimum flow for a hydroelectric power plant.
2. The shape of a flow-duration curve is used to evaluate the stream and catchment characteristics.
3. A very steep curve means high flow for very short period of time and it would cause flood. On the other hand, flat curve means low flow region and it can sustain the flow.

When outflow from a reservoir is uncontrolled as in freely operating spillways, the peak in outflow hydrograph will occur at the point of intersection of the inflow and out flow or at the recession limb of the inflow hydrograph.

Kinematic viscosity, \(v = \frac{{Dynamic;viscosity;\left( \mu \right)}}{{Mass;density;\left( \rho \right)}}\)

Mass Density (ρ):

\(\rho = \frac{{Mass}}{{Volume}}\)

Dimension of mass is M

Dimension of volume, = L³

So, dimension of \(\rho = \frac{M}{{{L^3}}} = M{L^{ - 3}}\)

Dynamic viscosity (μ): 

We know that:

\(I = \mu \frac{{du}}{{dy}};and;I = \frac{F}{A}\)

\(\Rightarrow \mu = \frac{I}{{\frac{{du}}{{dy}}}} = \frac{{\frac{F}{A}}}{{\left( {\frac{{du}}{{dy}}} \right)}}\)

\(\mu = \frac{F}{{A\left( {\frac{{du}}{{dy}}} \right)}}\)

Dimension of ‘F’ is MLT^-2

Dimension of ‘A’ is L²

Dimension of \(\frac{{du}}{{dy}}:is:\left( {\frac{1}{T}} \right)\)

⇒ Dimension of μ is \(\frac{{ML{T^{ - 2}}}}{{L\frac{1}{T}}} = {M^{ - 1}}L{T^{ - 1}}\)

Dim of \(v = \frac{\mu }{\rho } = \frac{{M{L^{ - 1}}{T^{ - 1}}}}{{M{L^{ - 3}}}} = {L^2}{T^{ - 1}}\)

The metals found in natural water are classified in toxic and non-toxic metals.

Toxic metals are: Cadmium (Cd), lead (Pb), Mercury (Hg), Silver (Ag), Arsenic (As) and Barium (Ba).

Non-Toxic are:

Sodium (Na) - Permissible value – 50 -250 mg/l

Copper (Cu) - Permissible value – 0.05 to 1.5 mg/l

Calcium (Ca) - Permissible value up to 60 mg/l

Zinc (Zn) – Permissible value - 5-15 mg/l

Manganese (Mn) - Permissible value – 0.05 – 0.5 mg/l

Magnesium (Mg) - Permissible value - up to 40 mg/l

Iron (Fe) - Permissible value- 0.1-1 mg/l

Molecular wt. of CH₄ is 16 gm

Molecular wt. of CO₂ is 44 gm

When CH₄ reacts with O₂, leads to formation of CO₂ and H₂O. The reaction involved is:

\(C{H_4} + 2{O_2};\mathop \to \limits^{{\rm{;\Delta }};} C{O_2} + 2{H_2}O\)

1 mole of CH₄ gets completely oxidized and formed 1 mole of CO₂

⇒ 16 gm of CH₄ → formation of 44 gm CO₂

1 gm of CH₄ → formation of \(\left( {\frac{{44}}{{16}}} \right)gm;C{O_2}\)

1000 gm of CH₄ → formation of \(\frac{{44}}{{16}} \times 1000 = 2750;C{O_2}\)

∴ CO₂ formed is 2.75 kg

Explanation:

The peak hour factor (PHF) is defined as.

\({\left( {PHF} \right)_x} = \frac{{Hourly;Traffic;volume}}{{\left( {\frac{{60}}{x}} \right) \times Maximum;'x'minute;Traffic}}\)

Less PHF means that more congestion.

\(\frac{1}{{\left( {\frac{{60}}{x}} \right)}} \le {\left( {PHF} \right)_x} \le 1\)

The minimum value of PHF is  and Maximum value is 1.

Given:

x = 15 minute.

\(\therefore {\left( {PHF} \right)_{15;min}} = \frac{x}{{60}} = \frac{{15}}{{60}} = \frac{1}{4} = 0.25\)

Temperature stresses are developed in rigid pavements due to variations in temperature. This is caused by following two factors:

1. Seasonal variation in temperature and
2. Daily Variation in temperature

Daily variation in temperature:

It causes warping stress due to temperature gradient across the slab thickness.

During day, top surface will try to expand but weight of slab will resist and hence, compression will be developed in top surface and tension will develop at bottom surface.

During night, the bottom surface will be at higher temperature and hence, has tendency to expand, but weight of slab will resist it, generating the compression at bottom surface and tension at top surface.

Note:

1. Thermal stress is developed only when there is resistant against thermal strain. When slab is weight less, slab is free to expand as there is no resistant and hence, temperature stresses will not be induced in slab.
2. Seasonal Variation in temperature:

It causes frictional stresses due to overall change in temperature of the slab. During summer season as the mean temperature of the slab increases, the concrete pavement expands towards the expansion joints and in winter season opposite phenomena occurred.

QS = 51.87 – 49.6 = 2.27 m

PS = 20 m

Let PT is x then TS = (20 - x)

RT = 51 – 49.6 = 1.4 m

In Δ PRT and PQS, tan θ is taken

\(\tan \theta = \frac{{RT}}{{PT}} = \frac{{QS}}{{PS}} = \frac{{1.4}}{{PT}} = \frac{{2.27}}{{20}}\)

⇒ PT = 12.33 m

\(\frac{{{d^2}y}}{{d{x^2}}} = - \frac{P}{{EI}}.y = \begin{array}{*{20}{c}} -m^2y \end{array}\)

Solution of above equation is:

y = a sin mx + b cos nx

at x = 0, y = 0

0 = a × 0 + b × 1

∴ b = 0

at x = L, y = 0

0 = a sin mL

∴ mL = nπ 

\(\therefore m = \frac{{n\pi }}{L}\)

\(y = a\sin \left(\frac{{nπ }}{L}\right)x\)

Calculation:

ΣF_Horizontal = 0

90 cos30° = 40 cos45° + F cosθ 

77.94 = 28.28 + F cosθ 

or F cosθ = 49.66 kN ........(i)

ΣF_Vertical = 100

90 sin30° + 40 sin45° + F sinθ = W

45 + 28.28 + F sinθ = 100

F sinθ = 26.72 kN ...............(ii)

dividing equation (ii) by equation (i), we get

tanθ = 26.72/49.66 = 0.5380

θ = 28.28° 

Substituting θ = 28.28° in equation (i), we get

F cos28.28° = 49.66

F = 56.39 kN

x = t³ + 2t

\(\therefore {V_x} = \frac{{dx}}{{dt}} = 3{t^2} + 2\)

\({a_x} = \frac{{{d^2}x}}{{d{t^2}}} = 6t\)

y = -3e^-2t

v_y = +6e^-2t

\({a_y} = \frac{{{d^2}s}}{{d{t^2}}} = - 12{e^{ - 2t}}\)

Z = 2 sin (5t)

∴ V_z = 10 cos 5t

a_z = -50 sin 5t

at, t = 0

a_x = 0

a_y = -12

a_z = 0

\(\therefore Acceleration = \sqrt {a_x^2 + a_y^2 + a_z^2} \)

\(= \sqrt {0 + {{\left( { - 12} \right)}^2} + 0} = 12\)

Concept:

Poission[Arrival rate evolutionary process]

It is defined as:

\(P\left( {X:λ > 0} \right) = p\left( x \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{{e^{ - λ }}{λ ^x}}}{{x!}};;λ > 0;and;0 \le x < ∞ }\ {0;;;;;;;;otherwise} \end{array}} \right}\)

• It measures the signal arriving rate of the Receiver.
• Measures the defect item probability.
• it is time-dependent so it is also called "evolutionary process"

Condition:

1. observations are infinitely large ( n → ∞ )

2. Probability of success is small ( p → 0)

Calculation:

Given,

Mean, λ = 5

P(x < 4) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)

= \(\frac{{{e^{ - 5}}{5^0}}}{{0!}} + \frac{{{e^{ - 5}}{5^1}}}{{1!}} + \frac{{{e^{ - 5}}{5^2}}}{{2!}} + \frac{{{e^{ - 5}}{5^3}}}{{3!}}\)

= \({e^{ - 5}}\left[ {1 + 5 + \frac{{25}}{2} + \frac{{125}}{6}} \right]\)

= \({e^{ - 5}}\left[ {\frac{{118}}{3}} \right]\)

= 0.265

Calculation:

V_Q + V_R = 80 ........(i)

ΣM_R = 0

80 × 3 = V_Q × 2

∴ V_Q = 120 kN

From equation (i), V_R = -40 kN

From fig, tanϕ = 1/4

⇒ \(\sin \phi = \frac{1}{{\sqrt {17} }}\), \(\cos \phi = \frac{4}{{\sqrt {17} }}\)

∴ tanθ = 3/4

⇒ cosθ = 4/5

Considering joint Q,

α = 104.03° - 90° = 14.03° 

ΣF_y = 0,

F_PQ cosα  + V_Q = 0

F_PQ cos14.03 = -120

F_PQ = -123.6897 kN

ΣF_x = 0,

F_PQ sinα = FQR

F_QR = - 29.986 kN

F_QR ≈ -30 kN =  30 kN (Compressive)

Concept

Maxwell-Betti Theorem:

Maxwell-Betti's law of real work is a basic theorem in structural analysis.

Using this theorem, it can be established that the flexibility coefficients in compatibility equations, formulated to solve indeterminate structures by the flexibility method, form a symmetric matrix and this will reduce the number of deflection computations.

The Maxwell-Betti law also has applications in the construction of influence lines diagrams for statically indeterminate structures.

The Maxwell-Betti law applies to any stable elastic structure (a beam, truss, or frame, for example) on unyielding supports and at a constant temperature, states.

Let’s suppose two systems of forces

According to Maxwell-Betti theorem,

∴ Pi × y1 = Pj × y2

Where, y = Deflection, and P is the force

Calculation:

P₁ × Δ₁₂ = P₂ × Δ₂₁ 

10 × 2 = 20 × Δ21

⇒ Δ21 = 1 mm

PQ is short by 3 mm

We have to find out vertical displacement of joint R in mm

Δ_R = ∑u(λ) 

Let in apply unit load at R as shown below

u_PR sin θ = \(\frac{1}{2}\)

uPQ + uPR cos θ = 0

uPQ = -u_PR cos θ = \(\rm -\frac{1}{2\sin\theta}\cdot \cos\theta \) 

u_PQ = \(\rm -\frac{1}{2}\space cot\theta = \frac{-1×4/3}{2}=\frac{-2}{3}\)

Δ_R = u_PQ × λ_PQ = \(\rm \frac{-2}{3} (-3) = 2\) mm upwards

Concept:

Nominal shear reinforcement,

\({{τ }_{v}}=\frac{{{V}_{u}}}{bd}\)

Minimum shear reinforcement is given by

\(\frac{{{A}_{sv}}}{b{{S}_{v}}}=\frac{0.4}{0.87~{{f}_{y}}}\)

Calculation:

τ_uc = 0.36 N/mm² [For M20] for  \(\frac{{{A}_{st}}}{bd}\times 100=0.25\)

τuc = 0.48 N/mm2 [For M20] for  \(\frac{{{A}_{st}}}{bd}\times 100=0.5\)

Factored SF (V_u) = 45 kN 

We have to calculate the dia of Fe500 two-legged stirrup to be used at a spacing of 325 mm c/c 

\({{τ }_{v}}=\frac{{{45\times1000}}}{230 \times 450} =0.434; N/mm^2\)

\(Percentage ; tensile~stress=\frac{4\frac{\pi ~}{4}\times {{12}^{2}}}{230\times 450}\times 100=0.437~%\)

\({{τ }_{c}}=0.36+\frac{0.12}{0.25}\times \left{ 0.437-0.25 \right} = 0.45 ; N/mm^2\)

∴ τ_v = τ_c  

Minimum shear reinforcement is 

\(\frac{{{A}_{sv}}}{230 \times {{375}}}=\frac{0.4}{0.87~{{\times 415}}} =95.55;mm^2\)

\({{A}_{sv}}=2\times \frac{\pi }{4}\left( {{ϕ }^{2}} \right)= 95.55\)

ϕ = 7.79 ≈ 8 mm

Explanation:

Given, P_u = 250 kN and No. of bolts(n) = 6

Tan θ = \(\frac{3}{4}\)

Cos θ = \(\frac{4}{5}\)

Sin θ = \(\frac{3}{5}\)

Total tension force = P_u cos θ = \(\frac{4}{5}P_u\)

Total shear force = Pu sin θ = \(\frac{3}{5}P_u\)

Tension in each bolt = \(\frac{Total \ tension\ force}{No.\ of\ bolts}\) = \(\frac{4\ \times \ P_u}{5\ \times 6}\)

Tension in each bolt = \(\frac{4\ \times \ 250}{5\ \times ;6}\)

Tension in each bolt = 33.33 kN

Shear force in each bolt = \(\frac{Total \ shear \ forcee}{No.\ of\ bolts}\) = \(\frac{3\ \times \ P_u}{5\ \times ;6}\)

Shear force in each bolt = \(\frac{3\ \times \ 250}{5\ \times; 6}\)

Shear force in each bolt = 25 kN

Concept:

Minimum tension reinforcement:

\(\frac{{{A_{st\ min}}}}{{b; \times; d}} = \frac{{0.85}}{{{f_y}}}\)

Where,

Ast min = Minimum tension reinforcement, b = width of the beam, d = depth of the beam, fy = yield strength of the steel

Calculation:

Given, b = 230 mm, d = 500 mm, Dia of bars = 12 mm, f_y = 500 N/mm²

\(\frac{{{A_{st\ min}}}}{{b; \times; d}} = \frac{{0.85}}{{{f_y}}}\)

\({{{A_{st\ min}}}} = \frac{{0.85;\times; {{b ;\times; d}}}}{{{f_y}}}\)

\({{{A_{st\ min}}}} = \frac{{0.85;\times; {{230; \times; 500}}}}{{{500}}}\)

\({{{A_{st\ min}}}} = 195.5 \ mm^2\)

Number of rebars = \(\frac{195.5}{\pi(12^2/4)}\)

Number of rebars = 1.729 = 2

Modulus of elasticity of concrete

\(E = 5000\sqrt {25}\)

= 25000 N/mm²

\(\begin{array}{l} \therefore \frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\ \Rightarrow \frac{1}{R} = \frac{\sigma }{{Ey}} = \frac{{5.8}}{{E{X_u}}}\ = \frac{{5.8}}{{58 \times 25000}} \end{array}\)

\(= 4 \times {10^{ - 6}}\) per mm

⇒ Curvature = 4 × 10^-6 per mm

Explanation:

Load settlement curve for the sandy soil is given below:

• General shear failure occurs is shallow foundation when placed on medium to dense sand. At the time of failure most of the soil within stress zone reaches in plastic static except central portion. Foundation gets tilled and heaving /buldging will seen at sides. If load settlement curve is plotted, then clear failure is found,
• Local shear failure occurs in loose sands and soft clays in shallow foundation. Before failure, large elastic settlements are recorded. The stress zone does not extend up to ground level. Therefore little or no builging occurs. If load settlement curve is plotted, clear failure is not observed.
• Punching shear failure occurs in deep footing/piles in loose sands and soft clays.

Concept:

Relative density of a soil sample having natural void ratio (eo) , maximum Void ratio(emax) and minimum void ratio (emin) is given by:

\({I_D} = \left( {\frac{{{e_{max}} - {e_o}}}{{{e_{max}} - {e_{min}}}}} \right) \times 100\)

Calculation:

Given,

emax = 0.85, emin = 0.50

eo = ?

Soil is compacted to density 1800 kg/m³ @ water content 8%

mass density of water = 1000 kg/m³, G_s = 2.7 

w = 8%, ρb = 1800 kg/m3, ρw = 1000 kg/m3 

we need to determine natural void ratio by the equation 

\({{\rm{\rho }}_{\rm{b}}} = \frac{{{\rm{G}} + {\rm{wG}}}}{{1 + {\rm{e}}}} \times {{{\rho }}_{\rm{w}}}\)

⇒ \(1800 = \frac{{2.7 + 0.08 \times 2.7}}{{1 + {\rm{e}}}} \times 1000\)

e = 0.62

Relative density, \(\left( {{{\rm{I}}_{\rm{D}}}} \right) = \frac{{{{\rm{e}}_{{\rm{max}}}} - {\rm{e}}}}{{{{\rm{e}}_{{\rm{max}}}} - {{\rm{e}}_{{\rm{min}}}}}} = {\rm{;}}\frac{{0.85 - 0.62}}{{0.85 - 0.5}} =65.71%\)

The relative density (in %) of the soil is 65.71%

Concept:

Coefficient of volume compressibility is given by

\({m_v} = \frac{{ - \left( {\frac{{{\rm{\Delta }}e}}{{{\rm{\Delta }}\bar \sigma }}} \right)}}{{1 + {e_o}}}\)

Δe = Change in void ratio of the soil

eo = initial void ratio

\({\rm{\Delta }}\bar \sigma = initial;effective;stress\)

Calculation:

Given,

σo ' = 175 kPa ; eo = 1.1 ; σo ' + Δσo ' = 300 kPa ; e = 0.9

Δe = 1.1 – 0.90 = 0.20

\({\rm{\Delta }}\bar \sigma = 175 - 300 = - 125;kPa\)

eo = 1.1

The coefficient of volume compressibility is:

\({m_v} = \frac{{ - \left( {\frac{{0.20}}{{ - 125}}} \right)}}{{1 + 1.10}} =7.619 \times {10^{ -4}}\) m²/kN

Concept:

Quantity of flow is:

\(Q = kh\left( {\frac{{{N_f}}}{{{N_d}}}} \right) \)

Where,

K = permeability

H = head loss due to seepage

Nf = Number of flow channels, Nd = Number of equipotential drops

Nf = Number of flow lines - 1, Nd = Number of equipotential lines -1

Calculation:

Given,

k = 3.8 × 10-6 m/sec

From the figure

Number of flow channels = 3

Number of equipotential lines = 10

Head under which flow is taking place = 6.3 m.

\(Q = {10^{ - 6}} \times 3.8 \times 6.3 \times \frac{{3 \times 1}}{{10}} = 7.182 \times {10^{ - 6}}{m^3}/s\)

Q = 7.182 cm3/s

Concept:

The force exerted by a horizontal jet normally on a vertical fixed plate is given by

In X-direction

FX = ρ A V2, where A and V is the area and velocity of jet respectively.

In Y-direction

FY = 0

Calculation:

Given,

V = 5 m/sec, Area = 0.0028 m2

ρ = 1000 kg/m3 (Water)

Force on a fixed plate

FX = ρ A V2

FX = 1000 × (0.0028) × (5)2

FX = 70 N

Additional InformationThe force exerted by a horizontal jet normally on a vertical moving plate is given by

In X-direction

FX = ρ A (V – u)2, where A and V is the area and velocity of jet respectively, and u is the velocity of the plate.

In Y-direction

FY = 0

Concept:

Venturimeter is a device that is used to measure the rate of flow of fluid through a pipe. This device is based on the principle of Bernoulli's Equation

.

The rate of flow is calculated by 

\({Q_{theoretical}} = {C_d} \times \frac{{{A_1} \times {A_2}}}{{\sqrt {A_2^2 - A_1^2} }} \times \sqrt {2 \times g \times h} ;;;;;\left( {{m^3}/sec} \right)\)

Where Cd is coefficient of discharge of venturimeter.

A1 is cross sectional area at throat

A2 is the cross sectional area at enlarged section

h is the difference of pressure head between throat and enlarged section 

\({C_d} = \frac{{{Q_{act}}}}{{{Q_{theoretical}}}}\)

From Bernoulli’s Equation 

Where,

\(\frac{{{P_1}}}{\gamma };and\frac{{{P_2}}}{\gamma };are;preesure;heads;at;respective;sections\)

V1 and V2 are velocities at respective sections

Z1 and Z2 are datum heads at respective sections

hL is the head loss between the section 

Calculation:

Given, diameter at throat (d₁) = 7.5 cm.

Diameter at enlarged end (d₂) = 15 cm.

\({A_1} = \pi \times \frac{{d_1^2}}{4} = ; = \pi \times \frac{{{{7.5}^2}}}{4};c{m^2} = 44.178;c{m^2}\)

\({A_2} = ;\pi \times \frac{{d_2^2}}{4} = \pi \times \frac{{{{15}^2}}}{4};c{m^2} = 176.714;c{m^2}\)

Acceleration due to gravity = 9.81 m/sec²

Flow rate (Q_act) = 30 lit / sec

= 30 × 10 ^-3 m³/sec    {**∵ 1 litre = 10 ^-3 m³/sec}

= 30 × 10³ cm³ /sec.   

Pressure head difference between the throat and enlarged section = 2.45 m 

\(;\frac{{{P_1}}}{\gamma } - \frac{{{P_2}}}{\gamma } = 2.45;m;\) ----- (a)

​\(\frac{{{P_1}}}{\gamma } + \frac{{V_1^2}}{{2 \times g}} + ;{Z_1} = ;\frac{{{P_2}}}{\gamma } + \frac{{V_2^2}}{{2 \times g}} + ;{Z_2} + {h_L}\)​

Between section 1 and section 2

Z1 = Z2  

and neglecting head loss we get

\(\frac{{{P_1}}}{\gamma } + \frac{{V_1^2}}{{2 \times g}} + ; = ;\frac{{{P_2}}}{\gamma } + \frac{{V_2^2}}{{2 \times g}}\)

\(\frac{{{P_1}}}{\gamma } - \frac{{{P_2}}}{\gamma } = \frac{{V_1^2}}{{2g}} - \frac{{V_2^2}}{{2g}};;\)---- (b)

From Equation of continuity

Q = A × V

Note: C.G.S units are used  

\({V_1} = \frac{{{Q_{th}}}}{{{A_1}}} = \frac{{{Q_{th}}}}{{44.178}};:;{V_2} = \frac{{{Q_{th}}}}{{{A_2}}} = \frac{{{Q_{th}}}}{{176.714}}\)

From (a) and (b)

\(245 = \frac{{{{\left( {\frac{{{Q_{theoretical;}}}}{{44.178}}} \right)}^2} - {{\left( {\frac{{{Q_{theoritical}}}}{{176.714}}} \right)}^2}}}{{2 \times 981}}\)

Q_theoretical = 31.5 × 10³ cm³ /sec.

\({C_d} = \frac{{30 \times {{10}^3}}}{{31.5 \times {{10}^3}}}\)

∴ Cd = 0.95

Note:

By using the Q theoretical formula if we solve we get 0.74 as Cd which is wrong (Common mistake we do)

Concept:

Discharge through the rectangular channel according to manning’s is given by:

\(Q = \frac{1}{n}A{R^{\frac{2}{3}}}S_0^{\frac{1}{2}}\)

Where,

n = Manning’s coefficient, R = Hydraulic mean depth

A = Cross-sectional area of the channel, So = Bed slope of the channel

When,

the channel is contracted to its minimum width, specific energy at the contracted section will become:

\({E_{min}} = {E_c} = {y_c} + \frac{{V_c^2}}{{2g}}\)

\({E_c} = \frac{3}{2}{\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}} = \frac{3}{2}{\left( {\frac{{{{\left( {\frac{Q}{{{B}}}} \right)}^2}}}{g}} \right)^{\frac{1}{3}}} \)       ---(i)

Also, initial specific energy (E1) = Ec

Hence minimum contracted with can be obtained.

Calculation:

Given,

Range of normal depth ranges between 0.76 m and 0.8 m

Q = 1.0 m3 /s,  Bed slope (S) = 0.0001, B = 3.0 m

q = Q/B = 1/3 = 0.33

n = 0.015

Let the normal depth of flow be 'y'

We know R = A/P 

Area = 3 × y,

Perimeter = y + y + B = B + 2y

R = 3y/(B + 2y)

Calculating the initial depth of flow:

\(Q = \frac{1}{n}A;{R^{\frac{2}{3}}}S_0^{\frac{1}{2}}\)

\(1 = Q = \frac{1}{{0.015}} × \left( {3 × {y}} \right) × {(\frac{3y}{3 + 2y})^{\frac{2}{3}}} × {\left( {0.0001} \right)^{\frac{1}{2}}}\)

y = 0.78 m

Now, Specific Energy \({E = {y} + \frac{{{q^2}}}{{2g;y^2}}}\) \({ = {0.78} + \frac{{{0.33^2}}}{{2\times9.81\times0.78^2}}}\)

= 0.7893 m

Now,

\(E ={E_c} = \frac{3}{2}{\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}} = \frac{3}{2}{\left( {\frac{{{{\left( {\frac{Q}{{{B}}}} \right)}^2}}}{g}} \right)^{\frac{1}{3}}} \)

⇒  \(0.7893 = \frac{3}{2}{\left( {\frac{{{{\left( {\frac{Q}{{{B}}}} \right)}^2}}}{g}} \right)^{\frac{1}{3}}} \)⇒ \(0.7893 = \frac{3}{2}{\left( {\frac{{{{\left( {\frac{1}{{{B\ min}}}} \right)}^2}}}{g}} \right)^{\frac{1}{3}}} \)

⇒ B _min = 0.836 m

Explanation:

Given,

Identical Heights and Base Areas.

i.e. Area = Constant, Height = Constamt

We know Force on base is given by F = γ A H

as Area and height are constant, Force acting in each base is constant.

Therefore, F1 = F2 = F3

Coming to weight.

Weight is given by W = ρ × g × (Volume)

So weight id proportional to Volume

Volume of container in (Bucket 2) > Bucket 1 > Bucket 3

Therefore, W2 > W1 > W3

Schematic:

Concept:

When a pipes line divides into two or more parallel which again join together at downstream as shown in fig then pipe said to be in parallel.

In order to increase the discharge passing through main pipe, the pipes are connected in parallel.

The discharge in the pipe is equal to sum of the discharge in each of the parallel pipe

Q = Q_A + Q_B

When pipes are arranged in parallel, the loss of head in parallel pipes is same.

h_A = h_B

\( ⇒ \frac{{f{L_A}Q_A^2}}{{12.1;d_A^5}} = \frac{{f{L_B}Q_B^2}}{{12.1;d_B^5}}\)

Equal lengths imply

\(\frac{{Q_A^2}}{{d_A^5}} = \frac{{Q_B^2}}{{d_B^5}}\)

Calculation:

Given

d1 = 4d2

Here assuming it as Pipe A and Pipe B

d_A = 4d m, d_B = d m, Q m³/s discharge;

\(\frac{{Q_A^2}}{{{{4d}^5}}} = \frac{{Q_B^2}}{{{{d}^5}}}\)

\(\frac{{{Q_A}}}{{{Q_B}}} = {\left( {\frac{{4}}{{1}}} \right)^{2.5}} = 32\)

We know from the continuity equation

Q = AV

V = Q/A ⇒ V ∝ Q/d²

\(\frac{{{V_A}}}{{{V_B}}} = {\left( {\frac{{Q_A}}{{Q_B}}} \right)}\times{\left( {\frac{{d_B}}{{d_A}}} \right)}^2 ={\left( {\frac{{32 Q_B}}{{Q_B}}} \right)}\times{\left( {\frac{{d_B}}{{4d_B}}} \right)}^2 = 2\)

Concept:

From Chick’s Watson’s Law:

\({{\rm{C}}^{\rm{n}}} \times {\rm{t;}} = {\rm{constant}}\)

Where C is the concentration of disinfectant

We know Concentration C = W/Q

Where W is the weight of disinfectant per day

Q is the discharge per day

‘t’ is the detention time or contact time

‘n’ is the dilution factor

\({\rm{Contact;time;}}\left( {\rm{t}} \right) = \frac{{{\rm{Volume;of;contact;unit}}}}{{{\rm{Flow;rate}}}}\)

\({\rm{t}} = \frac{{\rm{V}}}{{\rm{Q}}}\)

As \({{\rm{C}}^{\rm{n}}} \times {\rm{t}} = {\rm{constant}}\)

\({\rm{C}}_1^{\rm{n}} \times {{\rm{t}}_1} = {\rm{C}}_2^{\rm{n}} \times {{\rm{t}}_2} = {\rm{constant}}\)

\({\rm{C}}_1^{\rm{n}} \times \frac{{\rm{V}}}{{{{\rm{Q}}_1}}} = {\rm{C}}_2^{\rm{n}} \times \frac{{\rm{V}}}{{{{\rm{Q}}_2}}}\)

As volume being constant and n = 1 

 Equation becomes

 \(\frac{{{W_1}}}{{{Q_1}}} \times \frac{1}{{{Q_1}}} = \frac{{{W_2}}}{{{Q_2}}} \times \frac{1}{{{Q_2}}}\)

\(\frac{{{W_1}}}{{Q_1^2}} = \frac{{{W_2}}}{{Q_2^2}}\)

Calculation:

Given,

The weight of chlorine dose initially (W1) is 32 kg/day

Flow rate initially (Q1) = 16 MLD

Dilution factor = 1

(i) When flow rate changes from 16 MLD to 22 MLD

 \(\frac{{{W_1}}}{{Q_1^2}} = \frac{{{W_2}}}{{Q_2^2}}\)

\(\frac{{32\frac{{{\rm{kg}}}}{{{\rm{day}}}}}}{{16^2{\rm{;MLD}}}} = {\rm{;}}\frac{{{{\rm{W}}_2}}}{{22^2{\rm{;MLD}}}}\)

W2 = 60.5 kg/day

∴ The amount of chlorine dosage added should be 60.5 kg/day.

Explanation:

The correct sequence of treatment of typical turbid surface water is: Coagulation, flocculation, sedimentation, filtration, Disinfection, Storage and Supply

1. Coagulation is the neutralization of charged particles in the turbid water which would otherwise repel each other.
2. Flocculation is the process through which smaller individual particles come together to form large particles.
3. Sedimentation is done so that large floccs which are formed in the flocculation settle down.
4. Finally, Filtration is done to remove very small particle which did not settle down during sedimentation.

Additional Information

Most surface waters contain turbidity more than drinking water standards.

Although fast moving streams may carry larger material in suspension, most of the solids will be colloidal in size and will require chemical coagulation for removal. Depending on the geology of the watershed, hardness may or may not be a problem in surface waters.

If low levels of color and other organic material are present, adsorption onto surface active material, a process not significant in natural water systems, may be necessary.

A wide variety of microorganism, some of which may be pathogenic, are also common constituents of surface waters. Treatment systems commonly used in treating surface water.

Explanation:

Given,

Ca2+ = (4 - 0) meq/L = 4 meq/L

Mg2+ = (5 - 4) meq/L = 1 meq/L

HCO₃^- = 3.5 - 0 = 3.5 meq/L

Hardness is due to the multivalent metallic cation. (Ca2+,Mg2+ )

Total hardness = (4 × 50 + 1 × 50) mg/L as CaCO3

= 250 mg/L as CaCO3.

Alkalinity is due to the presence of HCO₃^- only.

Alkalinity = (3.5 × 50) mg/L as CaCO3

= 175 mg/L as CaCO3

⇒Carbonate hardness = minimum of {Alkalinity , Hardness}

=  minimum of {250, 175}

= 175 mg/L as CaCO3

∴ Non-carbonate hardness = total hardness – carbonate hardness

= 250 – 175

Non-carbonate hardness = 75 mg/L as CaCO3

Explanation:

Head loss in pipe due to friction is given as:

\(h_f = \frac {f lV^2}{2gD}\)

where f = friction coefficient, V = average velocity of flow,

D = diameter of pipe, l = length of pipe

\(V= \frac QA\)

where Q = flow rate, A = Cross Area of pipe.

Given:

Q = 0.21 m³/sec, Length of pipe = 100 m, V = 0.75 m/s

f = 0.01, g = 9.81 m/s²

\(V= \frac QA\)

⇒ A = Q/V = 0.21/0.75 = 0.28 m²

we know

Area = πd²/4 = 0.28

∴ d = 0.6 m

Head loss in pipe due to friction is given as:

\(h_f = \frac {f lV^2}{2gD}\)

\(h_f = \frac {0.01\times 100\times 0.75^2}{2\times 9.81\times 0.6}=0.048;cm\)

⇒ h_f = 4.8 cm 

⇒ Minimum gradient = h_f/L = 4.8 cm/100 m (as asked in in cm/100 m length)

Concept:

The design traffic in terms of the cumulative number of standard axles to be carried during the design life of the road as per IRC 37:2012 is:

\(N = \frac{{365 \times \left[ {{{\left( {1 + r} \right)}^n} - 1} \right] \times A \times D \times F}}{r}\)

Where,

N = Cumulative number of standard axles to be cater for in the design in (msa)

A = Initial traffic in the year of completion of construction in terms of the number of commercial vehicles per day.

\(A = P{\left( {1 + r} \right)^x}\)

x = number of years between the last count and the year of completion of construction

P = Number of commercial vehicles as per the last count

r = annual growth rate of commercial vehicles in decimal, n = design life of pavement in years

D = Lane distribution factor, F = Vehicle damage factor

Calculation:

Calculation of VDF: -

\(\begin{array}{l} \sum {N_i}; = ;100% \ \therefore ;VDF; = ;\frac{{10 \times 23.21 + 20 \times 8.496 + 35 \times 2.21 + 15 \times 0.906 + 20 \times 0.2866}}{{100}} \end{array}\)

= 4.986

\(N = \frac{{365 \times \left[ {{{\left( {1 + r} \right)}^n} - 1} \right] \times A \times D \times F}}{r}\)

Where r = Rate of growth = 7.5 %

n = design life = 15 yrs

A = Anticipated traffic = 5000 CV/day

D = VDF = 4.986

F = lane distribution factor = 1

L = lane safety factor = 1.3

\(\therefore ;CSA; = ;\frac{{365}}{{0.075}} \times \left[ {{{\left( {1 + 0.075} \right)}^{15}} - 1} \right] \times 5000 \times 4.896 \times 1 \times 1.3\)

= 309 msa

Explanation

We know.

The minimum distance available on a highway at any spot should be of sufficient length to stop a vehicle traveling at design speed safely without collision with any other obstruction is called SSD

Stopping sight distance (SSD) = Lag distance + Braking distance

 SSD =  \(vt + \frac{{{v^2}}}{{2gf}}\)

Where, v =  design speed of vehicle in m/s

f = Coefficient of friction

We know retarding acceleration (a) = f  × g

Given

SSD = 170 m, Design speed V = 90 km/h = 25 m/sec

 Acceleration due to gravity (g) = 9.81 m/s2, f = 0.35

 \(SSD = vt + \frac{{{v^2}}}{{2gf}}\) 

\(170 = 25t + \frac{{{v^2}}}{{2\times 9.81\times\left( {0.35} \right)}}\)

  \(170 = 25 \times t + \frac{{{{25}^2}}}{{2 \times 3.5}}\)

 170 = 25 × t + 91

 t = 3.15

 ∴ Perception time is 3.15 sec

Concept:

The relationship between speed ( u ) and density ( k ) is given by the Greenshield macroscopic model

\({\rm{v}} = {{\rm{v}}_{\rm{f}}} - \left( {\frac{{{{\rm{v}}_{\rm{f}}}}}{{{{\rm{k}}_{\rm{j}}}}}} \right) × {\rm{k}}\)

u - mean speed at density k ( m/s )

k- density of stream ( veh/km ), uf - free mean speed

kj – jam density

q – Flow ( veh/hr )

Calculation:

We know Flow q = k × u

Given u = 70 – 0.7 k

q = k × ( 70 – 0.7 k )

q = 70 k – 0.7 k2

Capacity condition:

The capacity condition denotes maximum flow

For q to be maximum

\(\left( {\frac{{{\rm{dq}}}}{{{\rm{dk}}}}} \right) = 0\)

\(\left( {\frac{{dq}}{{dk}}} \right) = 70 - 0.14;k = 0\)

k = 50 veh/km

∴ Capacity of road (C) = q = 70 k – 0.7 k2 

= 70 × 50 - 0.7 × 502

C = 1750 veh/hr

Concept:

The optimum signal cycle is given by as per Webster’s Method

\({C_o} = \frac{{1.5L + 5}}{{1 - Y}}\)

Where L = total lost time per cycle, in sec

= 2n + R (n = number of lanes and sum of R is all red-time)

Y = Sum of Ratio of normal flow to the saturation flow.

= y1 + y2 + y₃

y1, y2 and y₃ are the ratio of normal flow to saturation flow 

\({y_1} = \frac{{{q_1}}}{s}, {y_2} = \frac{{{q_2}}}{s} :and;{y_2} = \frac{{{q_3}}}{s}\)

Where q1 = Normal flow on each approach

q2 = Normal flow on each approach, q3 = Normal flow on each approach

S = saturation flow per unit time.

Calculation:

Given:

3 phase (n = 3)

Lost time per phase = 4 sec,

For 3 phase L = 3 × 4 = 12 sec

y1 = 0.20 sec, y2 = 0.30 sec, y₃ = 0.25 sec

∴ Y = y1 + y₂ + y₃ = 0.2 + 0.3 + 0.25

Y = 0.75

The optimum cycle length

\({C_o} = \frac{{1.5 × 12 + 5}}{{1 - 0.75}}\)

Co = 92 sec

Explanation:

Given

We know

Height of Instrument (HI) = RL + Back sight (BS)

RL at any point = HI - Fore Sight (FS)

HI at P = 100 + 1.655 = 101.655 m

RL of Q = HI @ P - FS = 101.655 - (-1.500) = 103.155 m

HI @ Q = RL @Q + BS @ Q = 103.155 + (- 0.950) = 102.205

RL of R = HI @Q - FS @ Q = 102.205 - 0.750 = 101.455 m

Explanation:

• An Alidade is an instrument used for sighting the point in plane table surveying. One edge of the ruler is beveled or Fiducial and the other is graduated (used for drawing line of sight).
• Arrow: It is an instrument used in chain surveying. Arrows are used to mark the position of the ends of the chain on the ground. The arrows are made of good quality, hardened and tempered steel wire of 4 mm diameter.
• Bubble tube: It is an instrument used in levelling. The glass tube containing the liquid and bubble in a spirit level. In levelling, it is used to check the accuracy of the horizontal surface.
• Stadia hair: Stadia hair are crosshairs on the reticle of a theodolite or other surveying instrument that allow stadia metric range finding.​