GATE CE 2011 Official Paper (Previous Year Paper)

Explanation:

Let Log (P) = (1/2) Log (Q) = (1/3) Log (R) = l

∴ P = 10^k, Q = 10^2k, R = 10^3k

Now,

Option 1:

P²=Q³R²

(10^k)² = (10^2k)³ (10^3k)² 

10^2k ≠ 10^12k

Option 2:

Q² = PR

10^4k = (10^3k)(10^k)

10^4k = 10^4k

∴Q² = PR follows the given relation.

Option 3:

Q² = R³P ⇒ 10^2k ≠ 10^7k

Option 4:

R=P²Q² ⇒ 10^3k ≠ 10^6k

∴ Only option(2) is correct.

The correct answer is option 1 i.e. Incomprehensible.

Explanation-

Inexplicable means incapable of being explained, interpreted or accounted for. 

• Incomprehensible: impossible to comprehend: UNINTELLIGIBLE.
• Indelible: that cannot be removed, washed away, or erased.
• Inextricable: forming a maze or tangle from which it is impossible to get free
• Infallible: incapable of error: UNERRING

Inexplicable means not explicable; that cannot be explained, understood, or accounted for. So the best synonym here is incomprehensible.

The correct answer is option 2 i.e. Split.

Explanation-

Amalgamate means to unite in or as if in an amalgam, let's have a look at all the given options. 

• Merge: to cause to combine, unite, or coalesce.
• Split: to divide lengthwise usually along a grain or seam or by layers
• Collect: to bring together into one body or place
• Separate:  to set or keep apart: DISCONNECT, SEVER

​

Amalgamate means to combine or unite to form one organization or structure. So the best option here is split. Separate on the other hand, although a close synonym, it is too general to be the best antonym in the given question while Merge is the synonym; Collect is not related.

The correct answer is Option 2 i.e Restrained

Explanation:

Reading the above statement we find that:

• The tone of the sentence clearly indicates a word similar to understated and tentative.
• The word should also be an antonym of strong.

Let's look at the meaning of the marked option:

• Restrained: Not excessively showy or ornate; understated and timid.

Hence from the above meaning, we find that restrained is similar in meaning to understated and tentative, and fits perfectly in the give sentence.

Thus, the correct answer is option 2 Restrained.

Additional Information 

Let's look at the meaning of the other given options. 

• Hyperbolic: Delibrately; or exaggerated.
• Argumentative: Given to arguing.
• Indifferent: Having no particular interest; unconcerned.

The correct answer is Option 3 i.e visiting 

Explanation:

Reading the given sentence we find that.

• The word contemplated is a transitive verb.
• A transitive verb always needs to transfer it's action on to something which is an object.
• Hence, here after the transitive verb contemplated we need a noun that will be the object.
• Out of the given options visiting is a gerund. And therefore it grammatically follows the verb contemplated.

Hence the correct answer is Option 3 i.e visiting.

Important Points

• A gerund is an -ing form of verb which acts as a noun.
• In the above case, we need a noun after the verb contemplated.
• And the gerund visiting is the appropriate option as it is a verb acting as a noun.
• Therefore the action of the verb contemplated is transferred to the gerund visiting.

Thus the correct sentence will be: "I contemplated visiting Singapore for my vacation bu decided against it."

Concept:

According to the given information:

Most dangerous microbe ∝ probability that microbe will overcome human immune system.

Most dangerous microbe ∝ area (growth of microbe)

Most dangerous microbe ∝ (quantity required)^-1

\(So, the;most;dangerous;microbe;\propto;\frac{Probability;\times;Area}{Quantity;required}\)

\(So, the;most;dangerous;microbe;=;\frac{K;\times;Probability;\times;Area}{Quantity;required}\)

\(So, the;most;dangerous;microbe;=;\frac{K;\times;Probability;\times;\pi{d^2}}{4;\times;Quantity;required}\)

Calculation:

Given:

For microbe P:

Probability (P) = 0.4, Diameter = 50 mm and Quantity required = 800 milligram of microbe/mass of body in kg.

\(The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.4\times50^2}{800}\Rightarrow\frac{\pi{K}}{4}\times\frac{5}{4}=1.25\frac{\pi{K}}{4}\)

For microbe Q:

Probability (Q) = 0.5, Diameter = 40 mm and Quantity required = 600 milligram of microbe/mass of body in kg.

\(The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.5\times40^2}{600}\Rightarrow\frac{\pi{K}}{4}\times\frac{4}{3}=1.33\frac{\pi{K}}{4}\)

For microbe R:

Probability (R) = 0.4, Diameter = 30 mm and Quantity required = 300 milligram of microbe/mass of body in kg.

\(The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.4\times30^2}{300}\Rightarrow\frac{\pi{K}}{4}\times\frac{6}{5}=1.2\frac{\pi{K}}{4}\)

For microbe S:

Probability (S) = 0.8, Diameter = 20 mm and Quantity required = 200 milligram of microbe/mass of body in kg.

\(The;most;dangerous;microbe;=\frac{\pi{K}}{4}\times\frac{0.8\times20^2}{200}\Rightarrow\frac{\pi{K}}{4}\times\frac{8}{5}=1.6\frac{\pi{K}}{4}\)

Microbe S > Microbe Q > Microbe P > Microbe R.

∴ toxicity is more for microbe S and the company should target it in its first attempt.

The correct answer is Option 3 i.e what the leading causes of death are

Explanation:

Reading the above statement we find that.

• The given sentence starts by stating how to deal with bereavement and grief after a tragedy occurs and not about precautions.
• Therefore, irrespective of the causes of death, a school student rarely gets into details of causes
• Hence it is very clear that the leading causes of death are not included in a unit on bereavement.

Hence, the correct answer is option 3 i.e what the leading causes of death are.

Concept:

\({\bf{Volume}};{\bf{of}};{\bf{liquid}};{\bf{left}} = {\bf{x}}{\left( {\frac{{{\bf{x}} - 1}}{{\bf{x}}}} \right)^{\bf{n}}}\)

where x = Original amount of liquid, n = Number of times replaced.

Calculation:

Given:

x = 10 litre, n = 3

Now, we know that

\({\bf{Volume}};{\bf{of}};{\bf{spirit}};{\bf{left}} = {\bf{x}}{\left( {\frac{{{\bf{x}} - 1}}{{\bf{x}}}} \right)^{\bf{n}}}\)

∴  Volume of spirit left = \(10{\left( {\frac{{10 - 1}}{{10}}} \right)^3} = \frac{{729}}{{100}} = 7.29\)  litres

Concept:

The transporter receives the same number of order each day and he currently has pending orders.

Let 'x' be the number of daily orders and 'y' be the number of pending orders.

Calculation:

Given:

Condition I:

Use of 7 trucks each day finishes all orders in 4 days.

∴ total numbers of trucks used = 7 + 7 + 7 + 7 ⇒ 28.

Total orders received in 4 days = 4x.

Pending orders = y.

∴ 4x + y = 28      eq(1).

Condition II:

Use of 3 trucks each day finishes all orders in 10 days.

∴ total numbers of trucks used = 3 × 10 ⇒ 30.

Total orders received in 10 days = 10x.

Pending orders = y.

∴ 10x + y = 30      eq(2).

Solving eq (1) and eq (2).

x = 0.33 ⇒ Daily orders

y = 26.66 ⇒ Pending orders.

Condition III:

Use of 'n' trucks each day finishes all orders in 5 days.

∴ total numbers of trucks used = 5 × n ⇒ 5n.

Total orders received in 5 days = 5x.

Pending orders = y.

∴ 5x + y = 5n   

∴ (5 × 0.33)  + (26.66) = 5n

∴ n = 5.66 ≈ 6

∴ minimum 6 number of trucks required so that there will be no pending order at the end of the 5^th day.

Concept:

Total cost = Fixed cost + Variable cost

To find maxima and minima of a function y = f(x), follow these steps.

Step 1

\(Find;\frac{{dy}}{{dx}},;and;put\frac{{dy}}{{dx}} = 0.\)

Find the value of x and this value is said to be the stationary point, this is a necessary condition to find the extremum value of a function.

Step 2

\(Find;\frac{{{d^2}y}}{{d{x^2}}};\)

Check the value at the stationary point obtained in Step 1.

A function f(x) has a maxima at x = a if f’(a) = 0 and f”(a) < 0

A function f(x) has a minima at x = a if f’(a) = 0 and f”(a) > 0

A function f(x) has no maxima and minima at x = a if f’(a) = 0 and f”(a) = 0.

Calculation:

Given:

F = 100/q, V = 4q

Total cost (TC) = Fixed cost + Variable cost

\(∴;TC=4q;+;\frac{100}{q}\)

Step 1:

\(\frac{{d(TC)}}{{dq}} = 0\)

\(\frac{d(TC)}{dq}=4;-;\frac{100}{q^2}\)

∴ q = ± 5

\(\frac{{d^2(TC)}}{{dq^2}} =+ve;for;minima\)

\(\frac{{d^2(TC)}}{{dq^2}} =-ve;for;maxima\)

\(\frac{d^2(TC)}{dq^2}=\frac{200}{q^3}\)

At q = 5

\(\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow1.6;;(+ve;,;∴;minima)\)

At q = -5

\(\frac{d^2(TC)}{dq^2}=\frac{200}{5^3}\Rightarrow-1.6;;(-ve;,;∴;maxima)\)

∴ at q = 5, Total cost (TC) will be minimum.

Explanation:

Given,

[A] is a square matrix and it is neither symmetric nor skew-symmetric

Also given,

[A^T]^T = [A]

[S] = [A] + [A]T , [D] = [A] - [A]T

Lets take  Case (i)

[S]^T = ( [A] + [A]T )^T = [A]T + A = [S]

So [S] is a Symmetric Matrix

Now Case (ii)

[D]^T = ([A] - [A]T)^T = [A]^T - [A^T]T = [A]T - [A] = - ([A] - [A]T) = - [D]

So [D] is skew Symmetric Matrix

Explanation:

Given

Now,

f(x) = x2 – N = 0

Differentiating,

f’(x) = 2x

Now,

Using Newton-Raphson Method,

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

\({x_{i + 1}} = {x_i} - \frac{{\left[ {x_i^2 - N} \right]}}{{2x_i}}\)

\({x_{i + 1}} = \frac{{2x_i^2 - x_i^2 + N}}{{2x_i}}\)

\({x_{i + 1}} = \frac{{{x_i}}}{2} + \frac{N}{{2x_i}}\)

xi+1 = (xi + N/xi)/2

Explanation:

Given,

There are 2 containers,

Container 1 has ( 4 Red and 3 Green Balls)

Probability of Red = 4/7

Container 2 has ( 3 Blue and 4 Green Balls)

Probability of being blue = 3/7

One Ball is Drawn from each container Randomly

Probability (One Red & another is Blue) = Probability (first is red & Second is Blue) 

= 4/7 x 3/7

= 12/49

Concept:

The value of Effective throat thickness either depends upon K values which depend on the angle between fusion faces or depend upon the size of the weld

Effective throat thickness (tt) = K × Size of the weld (s)

The angle between fusion faces is 99°, so K = 0.65

Throat Thickness (t_t) = 0.65 × s

Working axial load

\(\begin{array}{l} = \frac{{Factgored;axial;load}}{{1.5}}\ = \frac{{2000}}{{1.5}} = 1333.33;kN \end{array}\)

The allowable bearing pressure on concrete may be given as

\(\begin{array}{l} {\sigma _{all}} = \frac{{Direct;load}}{{Area;of;base;palte}}\ = \frac{{1333.33 \times {{10}^3}}}{{650 \times 420}} = 4.88;N/m{m^2} \end{array}\)

The permissible stress in direct compression in various grades of concrete as per IS 456 : 2000 are tabulated below:

The permissible stress in concrete should be more than the allowable bearing pressure. Thus the minimum grade of concrete which should be used is M 20.

Concept:

Corrosion of reinforcing steel and other embedded metals is the leading cause of deterioration in concrete. When steel corrodes, the resulting rust occupies a greater volume than the steel. This expansion creates tensile stresses in the concrete, which can eventually cause cracking, delamination, and spalling

Corrosion in steel occupies a volume several times greater than the volume of reinforcing steel hence exerts radial pressure on adjoining concrete layer. Cracks can be seen developing in the radial direction from the reinforcing steel.

Concept:

Sieve grain analysis is done to classify the coarse-grained soils 

Fineness modulus is basically an Index that defines the coarseness and fineness of the aggregate. It is obtained by adding a cumulative percentage of aggregates retained on each of the standard sieves.

Fineness modulus of Q can be taken as the geometric mean of the fineness modulus of P and R.

\(F{M_Q} = \sqrt {F{M_P} \times F{M_R}}\)

Concept:

• The cross section of a thermo mechanically treated (TMT) reinforcing bar has a soft ferrite-pearlite core with a hard martensitic rim.
• The sudden quenching and drastic change in temperature toughen the outer layer of the steel bar, thus making it super tough and durable.
• The TMT bars are then subjected to atmospheric cooling to equalise the temperature difference between the inner core and the exterior. Once the TMT bar cools down, it slowly turns into a ferrite-pearlite mass.
• The inner core remains soft giving the TMT bar tensile strength and ductility.

Explanation:

Given

We know the bending stress varies linearly.

Also Along the neutral axis, Bending stress is Zero.

Here as Mentioned Point (P) is at the Centre of  the span and also along the neutral axis

So, it is evident that @ P there is neither Shear force nor Bending stress

As shear force is Zero. There exists no shear stress at Point P

The state of stress can be shown as 

@ Point Q

As mentioned Point Q is on the bottom of the member which is subjected to Tension

As we know when the beam is subjected to downward loading the top portion of fibers (above NA) is subjected to Compression and the bottom fibers (Below NA) are subjected to tension.

At Point Q the Bending stress is maximum (Tension side). Shear Force at Pont Q is also Zero.

So State of stress at point Q can be shown as

Concept:

Critical hydraulic gradient (icr): 

The quick condition occurs at a critical upward hydraulic gradient ic when the seepage force just balances the buoyant weight of an element of soil.

The critical hydraulic gradient is typically around 1.0 for many soils.

\({i_{cr}} = \frac{{{G_s} - 1}}{{1 + e}} \)

G_s - Specific gravity of soil solids, e - Void ratio

At the critical hydraulic gradient. the effective stress is equal to zero.

Calculation:

Given, e = 0.7, Gs = 2.67

\(\Rightarrow {i_{cr}} = \frac{{{G_s} - 1}}{{1 + e}} = \frac{{2.67 - 1}}{{1 + 0.7}} =0.98\)

 ∴ The critical hydraulic gradient is 0.98.

Explanation:

General shear failure

Occur in medium to dense sand and stiff clays and has the following properties:

a) A well – defined failure pattern

b) Sudden, catastrophic failure accompanied by tilting of foundation

c) Bulging of the ground surface adjacent to the foundation.

So if the intergranular packing of sand decreases then the sand will behave as loose sand. Hence likelihood of general shear failure will decrease

This occurs in soils that show brittle – type stress-strain behavior. In this, all 3 zone of failure fully develops.

Failure plane is one at which angle of obliquity is maximum. 

Failure shear stress is one at which difference of shear stress and shear strength is minimum.

θ_c = Angle of the critical plane to the major principal plane

\({{\rm{2\theta }}_{\rm{c}}} = \frac{\pi }{2} + {{{\beta _{maximum}}}}{}\)

\({{\rm{\theta }}_{\rm{c}}} = \frac{\pi }{4} + \frac{{{\beta _{maximum}}}}{2}\)

where, β_max = Angle between resultant stress and normal stress on critical plane = Friction angle of soil = ϕ

Explanation:

Contact Pressure

Loads from the structure are transferred to the soil through the footing. As a reaction to it, soil exerts upward pressure on the bottom surface of the footing which is termed as contact pressure.

The distribution of contact pressure under different types of footings on different types of soils are given below:

 

For Rigid footing:  Settlement is uniform

For flexible footing:  Contact pressure is uniform.

So For the Given case

In X (Flexible Footing) contact pressure is uniformly distributed.

In Y (Rigid Footing) Contact pressure is not uniformly distributed.

Concept:

The settling velocity of sand particles is given by Stoke's Law,

\({{\rm{V}}_{\rm{s}}} = \frac{{g \times \left( {G - 1} \right) \times {d^2}}}{{18;ν }}\)...... (1)

G -Specific gravity of sand particles

ν - Kinematic viscosity

d - Diameter of sand particles

Calculation:

Given:

Let V₁ and V₂ be settling velocities of larger and smaller grain respectively

V₁ = 0.5 mm/sec

Diameter of larger grain d₁ = 0.075 mm

Diameter of smaller grain d₂ = 0.0075 mm

g is constant, specific gravity identical for both grains, both the particles are falling through water hence ν remains same. Hence equation 1 becomes

⇒ \(\frac{{{V_1}}}{{{V_2}}}; = ;{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\)

⇒ \(\frac{{{0.5}}}{{{V_2}}}; = ;{\left( {\frac{{{0.075}}}{{{0.0075}}}} \right)^2}\)

⇒ V₂ = 0.005 mm/sec

Concept:

Factors affecting Hydrograph:

Shape of catchment: Hydrograph from a catchment with broader end near the outlet has higher and early peak.

(b) Compactness factors: It is defined as the ratio of perimeter of catchment to the perimeter of form circle. Form circle is the circle having the same area as that of the catchment.

 

As the compactness factors decreases, average length of travel to the outlet decreases, hence decreasing infiltration and increasing resultant discharge.

(c) Catchment slope:

In general as the slope of catchment increases, flow velocity increases, thereby decreasing the infiltration and decreasing base period and increasing runoff discharge. For small catchment when overland flow is more dominant, general land slope plays more important role, whereas for large catchment where channel flow is more dominant, main stream slope plays a more important Role.

(d) Drainage density: It is defined as the sum of lengths of all the stream divided by the catchment area.

In general as drainage density increases, flow velocity increases, infiltration decreases, base period decreases and peak discharge increases.

(e) Land use: Urbanisation reduces infiltration by increasing flow velocity, thereby decreasing the base period and increasing the peak of the hydrograph.

The effect of urbanization on storm runoff hydrograph from the watershed is to:

i. increase the volume of runoff-infiltration losses are less

ii. decrease the time to peak discharge-losses are less and runoff rate is more

iii. decrease the time base-time to reach the remote particles is less

iv. increases the peak discharge-losses due to infiltration is less

Concept:

Critical depth of flow yc is given by (rectangular section)

\({{\rm{y}}_{\rm{c}}} = {\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}}\)

​q = discharge per unit width

Hence it can be said that the critical depth depends only on the geometry of the channel and the discharge through the channel.

Concept:

For a completely submerged body, the equilibrium is defined by the relative positions of centre of gravity (G) and centre of buoyancy (B) 

Stable Equilibrium: When the point B is above G, it is said to be in stable equilibrium

Unstable Equilibrium: When the point B is below G, it is said to be in unstable equilibrium

Neutral Equilibrium: When B and G are at same point, the body is said to be in neutral equilibrium.

Concept:



The flow is supercritical, so let us consider the Supercritical region.

As the height of the hump is increased, choked condition is reached. That is critical flow condition (choked condition) is reached at the hump.

E_hump = E_C.

The minimum specific energy is reached at the hump location.

The critical flow condition at the hump cannot be changed and it continues to be critical. If we want to increase the height of the hump, the upstream flow conditions should be changed.

For the increase in hump height Δ z' > Δ z_m, the point E₁ is shifted to E₁^' which is at a lower level than E₁. 

Hence there is a drop in water level at the upstream section of the hump.

Concept:

The sequence of purification of water is given by

1. Screening
2. Plain sedimentation
3. Coagulant aided sedimentation
4. Filtration
5. Disinfection
6. Aeration
7. Softening
8. Miscellaneous treatments like fluoridation, recarbonation, liming, and desalination.

For the water to be treated by secondary sedimentation, water has to be coagulated. Before coagulation Rapid mixing is done.

Rapid mixing is a process by which coagulant is thoroughly mixed with water after which flocculation will occur.

Before water is treated by a rapid sand filter it has to be coagulated. And the filtered water is disinfected with chlorine.The sequence of surface water treatment is given as

Concept:

The removal of total coliform is given by

\(Percentage;removal\left( {% R} \right) = \frac{{{N_0} - {N_t}}}{{{N_t}}} \times 100% \)

Log removal (log R) = log10(No) – log10(Nt)

Where, No = Initial no. of organisms

Nt = Number of organisms at time t.

Calculation:

Given - N₀ =10⁶, N_t = 10²

\(Percentage;removal\left( {% R} \right) = \frac{{{10^6} - {10^2}}}{{{10^6}}} \times 100%=99.99% \)

Log removal (log R) = log₁₀10⁶ – log1010² = 4

Explanation:

 

Note:

Soot is a general term that refers to impure carbon particles resulting from the incomplete combustion of hydrocarbons.

Soot, as an airborne contaminant in the environment, has many different sources but they are all the result of some form of pyrolysis.

Concept:

The Poisson’s model that accounts for nonuniformity of flow which is derived by assuming the random pattern of vehicle arrivals at a specified point is given by

\(P\left( n \right) = \frac{{{{\left( {λ t} \right)}^n} \times {e^{ - λ t}}}}{{n!}}\)..... ( 1 )

where

P(n) = probability of having n vehicles arrive in time t,

n - Number of vehicles in time t

λ = average vehicle flow or arrival rate in vehicles per unit time,

t = time interval

e = base of the natural logarithm (e = 2.718)

Time headway is defined as the time difference between any two successive vehicles when they cross a given point.

Calculation:

Given:

Time Headway is greater than or equal to t₁∴ Number of vehicles arriving is zero in time t₁ ( n = 0)

Equation 1 becomes

\(P\left( 0 \right) = \frac{{{{\left( {λ t} \right)}^0} \times {e^{ - λ t}}}}{{0!}}\)

⇒ P(0) = e^-λt₁

Concept:

Transition curve is provided to change the horizontal alignment from straight to circular curve gradually and has a radius that decreases from infinity at the straight end (tangent point) to the desired radius of the circular curve at the other end (curve point).

The rate of change of centrifugal acceleration is called jerk. The length of the transition curve depends on three criteria

1. Rate of change of centrifugal acceleration
2. Rate of change of superelevation
3. Empirical formula's
4. Rate of change of centrifugal acceleration:

\({{\rm{L}}_1} = \frac{{{{\rm{v}}^3}}}{{{\rm{CR}}}}\)

v - Design speed in m/sec;

R - Radius of the horizontal curve, C - Rate of change in centrifugal acceleration or jerk

IRC suggests that \({\rm{C}} = \frac{{80}}{{75 + 3.6{\rm{v}}}}\) [C ranges from 0.8 to 0.5]

Additional Information

Rate of change of super elevation:

\({{\rm{L}}_2} = {\rm{Be'N}}\)

 e’ - Effective superelevation, N - The rate of change in super-elevation  

B - Total width of pavement including the widening if any

When the pavement is rotated about center e = e/2

When the pavement is rotated about the inner edge e = e

IRC Empirical Formula:

\({{\rm{L}}_3} = \left{ {\begin{array}{*{20}{c}} {\frac{{2.7{{\rm{V}}^2}}}{{\rm{R}}},;;For;plain;and;rolling;terrain}\ {\frac{{{{\rm{V}}^2}}}{{\rm{R}}},;;For;hilly;and;steep;terrain} \end{array}} \right.\)  where V is the speed in km/hr and R is the radius of horizontal curve

Concept:

Marshall stability and flow:

Marshall stability of a test specimen is the maximum load required to produce failure when the specimen is preheated to a prescribed temperature placed in a special test head and the load is applied at a constant strain (5 cm per minute). during the stability test, a dial gauge is used to measure the vertical deformation of the specimen. The deformation at the failure point expressed in units of 0.25 mm is called the marshall flow value of the specimen.

Graphical plot:

The average value of each of the above properties is found for each mix with the different bitumen contents. Graphs are plotted with the bitumen content on the x-axis and the following value on the y-axis.

(i) Marshall stability value

(ii) Flow value

(iii) Unit weight

(iv) Percent voids in the total mix

(v) Percent voids filled with bitumen

From the diagram, we can say as the bitumen content increases the flow value increases monotonically.

Explanation:

Curvature:

• The dumpy level provides a horizontal line of sight and the line of sight is not a level line.
• The curvature makes the apparent staff reading more and the object appears lower than its real level on the ground.
• Hence the curvature correction (C_c) is negative and applied to staff reading.

Important Points

Correction due to curvature

Cc = -0.0785 d2

Corrections due to refraction

Cr = +0.0112 d2

∴ Composite correction

C = –0.0785 d2 + 0.0112 d2

C = –0.0673 d2 

d - Horizontal distance from point of observation to the station where staff is kept.

where d is in km, C in m

Note: 

Correction due to refraction = 1/7(Correction due to curvature)

μ = 3x² – 3y²

\(\frac{{\partial \mu }}{{\partial x}} = 6x\ &\ \frac{{\partial \mu }}{{\partial y}} = - 6y\) 

By Cauchy-Riemann equation

\(\begin{array}{l} \frac{{\partial \mu }}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}}\ &\ \frac{{\partial \mu }}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}\ \therefore \frac{{\partial \nu }}{{\partial y}} = 6x\ &\ \frac{{\partial \nu }}{{\partial x}} = 6y \end{array}\) 

We know

\(dv = \frac{{\partial \nu }}{{\partial x}}dx + \frac{{\partial \nu }}{{\partial y}}dy\) 

⇒ dv = 6ydx + 6xdy

On integration

ν = 6xy + k      where k is constant

Concept:

A function f(x) is continuous at x = a if,

Left limit = Right limit = Function value = Real and finite

A function is said to be differentiable at x =a if,

Left derivative = Right derivative = Well defined

Analysis:

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{;f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\frac{{\rm{\pi }}}{2}} \right)\)  for continuity

\(\begin{array}{*{20}{c}} {{\rm{lim}}}\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{;}}\frac{{{\rm{λ }}\cos {\rm{x}}}}{{\frac{{\rm{\pi }}}{2} - {\rm{x}}}} = 1\)

To find the LHS value, we can apply the L-Hospital’s rule

\(\begin{array}{*{20}{c}} {{\rm{lt}}}\ {{\rm{x}} \to \frac{{\rm{\pi }}}{2}} \end{array}{\rm{;}}\frac{{ - {\rm{λ }}\sin {\rm{x}}}}{{ - 1}} = 1\)

∴ λ = 1

\(I = \mathop \smallint \limits_0^a \frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx\)

By property,

 \(\mathop \smallint \limits_0^a f\left( x \right) = \mathop \smallint \limits_0^a f\left( {a - x} \right)dx\)

\(I = \mathop \smallint \limits_0^a \frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}dx\)

Adding both integrals, We get

\(2I = \mathop \smallint \limits_0^a \frac{{\sqrt x }}{{\sqrt x + \sqrt {a - x} }}dx + \mathop \smallint \limits_0^a \frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}dx\)

\(2I = \mathop \smallint \limits_0^a dx \)

2I = a

\(I = \frac{a}{2}\)

Explanation:

Vectors can be multiplied in two ways – scalar product and vector product.

Vector product of two vectors (A̅ and B̅) results in a vector and the resultant vector is perpendicular to both  A̅ and B̅. It is denoted as  A̅ × B̅, and

A̅ × B̅ = |A̅||B̅| sin θ

Where, θ is the angle between the two vectors A̅ and B̅

Scalar product of of two vectors (A̅ and B̅) results in a scalar. It is denoted as  A̅ ⋅ B̅, and

 A̅ × B̅ = |A̅||B̅| cos θ

Where, θ is the angle between the two vectors A̅ and B̅.

Here asked \(\left| {\begin{array}{*{20}{c}} {{\rm{\vec a}} \times }{{\rm{\vec b}}} \end{array}} \right|{;^2}\)

It is given magnitudes of vectors are a and b

as  A̅ × B̅ = |A̅||B̅| sin θ

\(\left| {\begin{array}{*{20}{c}} {{\rm{\vec a}} \times }{{\rm{\vec b}}} \end{array}} \right|{;^2}\) = (absinθ)² = a²b²sin² θ

As we know sin²θ = 1 − cos² θ

So a2b2sin2 θ is written as 

a2b2sin2 θ = a2b2 (1 − cos2 θ)

We know A̅ × B̅ = |A̅||B̅| cos θ

So this becomes a2b2  − a2b2cos2 θ  = a2b2 − ( \({\rm{\vec a}}\).\({\rm{\vec b}}\))2

Given:

\(\frac{{dy}}{{dx}} + \frac{y}{x} = x\)

Comparing with

\(\frac{{dy}}{{dx}} + py = q\)

We get,

\(p = \frac{1}{x};q = x\)

\(TF = {e^{\smallint pdx}} = {e^{\smallint \frac{1}{x}dx}} = {e^{\ln x}} = x\)

Now,

General solution can be written as,

Y (IF) = ∫ q (IF) dxt C

Y (x) = ∫ x (x) dxt C

xy = ∫ x² dx + C

\(xy = \frac{{{x^3}}}{3} + C\)

At, x = 1 & y  =1

\(\left( 1 \right)\left( 1 \right) = \frac{{\left( 1 \right)}}{3} + C\)

\(\therefore C = \frac{2}{3}\)

\(\therefore xy = \frac{{{x^3}}}{3} + \frac{2}{3}\)

\(\therefore y = \frac{{{x^2}}}{3} + \frac{2}{{3x}}\)

Concept:

Possible locations of Plastic Hinges:

1. At the fixed end of a beam
2. At the intermediate joint of a continuous beam
3. At the rigid joint of a frame
4. At the point of maximum bending moment (also at greatest curvature)
5. At the point where the section of beam changes
6. Under the external load

Number of independent mechanisms-

The number of independent mechanisms (i) is given as

i = N - r

N = Number of possible plastic hinges 

r = static indeterminacy

Calculation:

Given:

Number of the possible plastic hinges (N) = 2 (Under the load, at the fixed end)

r = 4 - 3 = 1

i = 2 - 1 = 1

From Δ ABD,

\({\rm{θ }} = \frac{{\rm{Δ }}}{7}\)

∴ Δ = 7 × θ     ...(1)

From Δ CBD

\(ϕ = \frac{{\rm{Δ }}}{3}\)

∴ Δ = 3 × ϕ     ....(2)

From 1 and 2

7 × θ = 3 × ϕ 

External work done = W × Δ = W × 7 × θ (∵ Δ = 7× θ)

Internal work done = M_p × θ + M_p × (θ + ϕ)

\({\rm{Internal;work;done}} = {{\rm{M}}_{\rm{p}}} × {\rm{θ }} + {{\rm{M}}_{\rm{p}}} × \left( {{\rm{θ }} + \frac{{7 × {\rm{θ }}}}{3}} \right)\)     ...(∵ \(\phi = \frac{{7 × θ }}{3}\))

By the principle of virtual work, Equating external and internal work done

\({\rm{W × 7× θ }} = {{\rm{M}}_{\rm{p}}} × {\rm{θ }} + {{\rm{M}}_{\rm{p}}} × \left( {{\rm{θ }} + \frac{{7 × {\rm{θ }}}}{3}} \right)\)

\(\bf W = \frac{{13 \times {M_p}}}{{21}}\)

 

Plastic hinges:

If the cross-section of the beam is subjected to yield stresses throughout the depth, it cannot take further loads. If any additional load is applied, then the beam rotates at that section. Since the beam rotates due to moment, we call it a plastic moment, we say that a plastic hinge is formed at that section.

Plastic hinge means stress distribution becoming rectangular i.e. stress at every point is equal to yield stress.

Explanation:

Given

Bending Moment at a distance x from P, M_x = Wx

So in the PQ part Bending Moment (BM) is varying linearly with x.

BM @ Q = WL & There is no twisting Moment in member PQ

Now the moment (WL) @ Q acts as a twisting moment (WL) for member QR (Not varying Twisting Moment)

Hence Option B is wrong

And because of Load W @ Q the bending moment in the member QR vary linearly. 

BM @ R = W(2L)

Now Moment 2WL acts as twisting Moment for member RS. and because of Load W Bending moment in the section RS varies linearly.

Concept

We know,

In an Anchorage bond

Maximum force resisted by bond strength = (π × D × L× σ _b)

Maximum tensile force resisted by bar = \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)

Where,

 L is development length

D is dia of bar,

σ_b is bond strength (between the bar and concrete)

σ_st is the tensile strength of the bar

P is pull out Force.

Maximum Pull should be minimum of both resistance for the safety of the bar. So it should not fail in any of the above case.

∴ Maximum value of P = Minimum ( (π × D × L× σ _b) or \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) )

Concept:

As per IS 456:2000

Nominal shear stress = \({\tau _v} = \frac{V}{{bd}}\) 

Where, V = Ultimate shear force, b = width of beam, d = effective depth of beam

Nominal shear stress should never be greater than maximum shear stress with shear reinforcement.

Calculation:

Two Beams P & Q of B = 400 mm

d = 7.50 mm

τ_{c, max} = 2.0 MPa

τ_c = 0.75 Mpa

Design shear force for P = 400 kN

Design shear force for Q = 750 kN

For beam P,

V_u = 400 kN

\({\tau _V} = \frac{V}{{bd}} = \frac{{400 \times {{10}^3}}}{{400 \times 750}} = 1.33;N/m{m^3} < {\tau _{c,;max}} = 2.0\) N/mm²

⇒ V_us = (1.33 – 0.75) × 400 × 750 = 175 kN

For Beam Q,

V_u = 750 kN

\({\tau _v} = \frac{{750 \times {{10}^3}}}{{400 \times 750}} = 2.5;N/m{m^2}; > {\tau _{c,;max}}\)

So, section for Beam Q should be revised.

Concept:

 

Intermediate Transverse stiffeners:

1. These are also called vertical stiffeners.
2. Vertical stiffener is not subjected to any load and is selected to provide lateral stiffness.
3. These stiffeners increase the buckling resistance of the web caused by shear. So, these stiffeners should be sufficiently strong to withstand shear transmitted by the web and to prevent local buckling of the web.

Intermediate Longitudinal stiffeners:

1. These are also called Horizontal stiffeners.
2. Longitudinal stiffeners increase the buckling resistance considerably as compared to transverse stiffeners when the web is subjected to bending.
3. These are provided in the compression zone of the web to reduce the chances of web buckling due to bending in compression.
4. The first horizontal stiffener is provided at one-fifth of the distance from the compression flange to the tension flange.
5. Longitudinal stiffeners are not continuous and are provided between vertical stiffeners.

Important Points

• In plate girder, the flange is designed to resist bending moment

• The web is designed to resist the shear resulting from loads

Explanation:

The Tensile resistance of a pile is given by,

Qt = Qeb + Qsf + W

Q_t = qbAb + qsAs

Where,

qb = unit end bearing resistance = cN_c

Ab = Base area of the pile

\({A_b} = \frac{\pi }{4}({D^2 -d^2})\)

d = Diameter of the pile

qs = Unit adhesion of soil with pile on surface area = αC̅

α = Adhesion factor between clay and pile

c̅ = Average unit cohesion of clay over the depth of pile.

As = Surface area of pile = πDL

W is weight of pile, 

Given, 

α = 0.3, c = 50, W = 20 , D = 750 mm = 0.75 m,

d = 350 mm = 0.35 m, Nc = 9 

Q_t = αC̅A_s + W + cN_cA_b 

= 0.3 × 50 × π×(0.35)×8 + 20 + 50 × 9 × \( \frac{\pi }{4}({0.75^2 -0.35^2})\)

= 307.408 kN

Concept:

The time required for any degree of consolidation is given by Taylor's formula

\(\left( {{{\rm{T}}_{\rm{v}}}} \right) = \frac{{{{\rm{C}}_{\rm{v}}}{\rm{t}}}}{{{{\rm{d}}^2}}}\)

T_v - Time factor

C_v - Coefficient of consolidation

t - Time taken for consolidation

d - Drainage path

Drainage path represents the maximum distance the water particles have to travel to reach the free drainage layer.

Calculation:

Given:

Let the subscript X denote parameters related to Site X and Y denote parameters related to Site Y

Site X

t_X = 36 months

d_X= 10 m

Site Y

d_Y = 5 m ( Sand layer is the free drainage layer)

Time factor is a function of degree of consolidation

Since identical surcharges are placed at both sites, both sites undergo the same consolidation, the time factor is the same in both the sites

∴ (T_v)  _X = (T_v) _Y 

Coefficient of consolidation is a function of the properties of soil and the magnitude of the load

Since the silt clay layer is identical in both the sites and identical surcharges are used at both the sites, Coefficient of consolidation is the same in both cases.

∴ (C_v) _X = (T_v) _Y

Since (Tv) _X  = (Tv) _Y

_{\(\frac{{{{\left( {{C_v}} \right)}_X} \times {t_X}}}{{d_X^2}} = \frac{{{{\left( {{C_v}} \right)}_Y} \times {t_Y}}}{{d_Y^2}}\)}

\(\frac{{36}}{{{{10}^2}}} = \frac{{{t_Y}}}{{{5^2}}}\)

t_Y = 9 months

 

Double drainage condition refers to the situation when both faces are permeable and pore water can dissipate through both faces but in single drainage only one face is permeable and hence it takes a longer time in single drainage.

Single drainage d = H

Double drainage d = H/2

Concept:

If both ends take part in shearing:

\(\text{T}=\text{ }!!\pi!!\text{ }{{\text{d}}^{2}}{{\text{ }!!τ!!\text{ }}_{\text{f}}}\left[ \frac{\text{H}}{2}+\frac{\text{d}}{6} \right]\)

Where

T = Torque (maximum moment), d = diameter of vane

H = height/length of vane, τf = shear strength of clay

Calculation:

Given:

T = 4.6 Nm, d = 50 mm = 0.05 m

H = 100 mm = 0.1 m

For saturated silty clay

τf = shear strength of clay

Since the instrument was inserted completely both ends take part in shearing

\(\text{4.6}=\text{ }!!\pi\times!!\text{ }{{\text{(0.05)}}^{2}}{{\times\text{ }!!τ!!\text{ }}_{\text{f}}}\times\left[\frac{\text{0.1}}{2}+\frac{\text{0.05}}{6} \right]\)

τ_f = 10 kPa

Additional Information

For vane shear test with one end in shearing

\(\text{T}=\text{ }!!\pi!!\text{ }{{\text{d}}^{2}}{{\text{ }!!τ!!\text{ }}_{\text{f}}}\left[ \frac{\text{H}}{2}+\frac{\text{d}}{12} \right]\)

Explanation:

We know,

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

Also can be written as,

\({\rm{Head;loss}} = {{\rm{h}}_{\rm{f}}} = \frac{{{\rm{8fL}}{{\rm{Q}}^2}}}{{{\pi^2\rm{gD^5}}}}\)

Case 1

Given, 

Length = 1500 m, Dia of Pipe = 0.6 m, Head loss = 20 m

\({\rm{Head;loss}} = 20 = \frac{{{\rm{8f(1500)}}{{\rm{Q}}^2}}}{{{\pi^2\rm{g0.6^5}}}}\)

Q_intial = 0.112/√f   --- (1)

Case 2:

Pipe is seperated into 2 pipes for 25% more discharge.

Let assume each pipe carries a discharge of Q

So now the equation become

2Q₂ = 1.25×Q_initial

Q2 = 0.625×Q_initial

We know 

\({\rm{Head;loss}} = 20 = \frac{{{\rm{8f(1500)}}{{\rm{0.625Q}}^2}}}{{{\pi^2\rm{gd^5}}}}\)

0.413 = fQ²_initial/d⁵ 

Q²_initial = 0.413d⁵/√f  

Q_initial = 0.642d⁵/√f   ---- (2)

Solving (1) and (2)

We get D = 49.7 cm ≈ 50 cm

Concept:

Hydraulic Jump**:** 

In an open channel when a rapidly flowing stream abruptly changes into a slowly flowing stream, a distinct rise or jump in elevation of fluid surface happens, this phenomenon is known as Hydraulic Jump.

For a hydraulic jump, the height of the slow stream or subcritical stream is (y₂) = \(- \frac{y_1}{2} + \sqrt{\frac{y_1^2}{4}\ + \frac{q^2}{gy_1}}\)

y₁ = Height of the rapidly flowing stream or supercritical stream

g = Acceleration due to gravity

q = Discharge per metre width in m³/sec

Calculation:

Given:

y1 = height of horizontal apron = 46 cm = 0.46 m

q = 9 m3/sec

∴ y₂ = \(- \frac{0.46}{2} + \sqrt{\frac{(0.46)^2}{4}\ + \frac{9^2}{9.81\times 0.46}}\)

y2 = 5.77 m.

Concept:

Specific yield is the ratio of volume of water that drains from sample under gravity to the total volume of sample. Specific yield is the measure of storage change in an aquifer. 

Change in storage = Drop in water level x Area x specific yield

= 150 x 3 x 0.25 = 112.5 ha.m

Concept:

\(Average;TSP;concentration = \frac{{Final;wt.;of;filter;paper - Initial;wt.;of;filter;paper}}{{Volume;of;air;sampled}}\)

Calculation:

Initial air flow rate = 1.5 m³/min

Final air flow rate = 1.4 m³/min

The decline in air flow is linear.

So average air flow rate \(= \frac{{1.5 + 1.4}}{2} = 1.45;{m^3}/min\)

\(\therefore TSP;conceutration = \frac{{\left( {10.283 - 9.787} \right) \times {{10}^6}}}{{1.45 \times 60 \times 24}} = 237.5;\mu g/{m^3}\)

Concept:

Chlorination is one of the major methods of disinfection, When chlorine is added to water, the reaction takes place between them.

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

Reaction of chlorine with water

Calculation:

Given:

pH = 7.5

[Free chlorine residual] = 2 mg/l as Cl₂

Atomic Weight of Cl: 35.5

K = 107.5

pH = - log10 [H+]

7.5 = - log10 [H+]

∴ [H+] = 107.5

\({\rm{OC}}{{\rm{l}}^ - } + {{\rm{H}}^ + } \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over {\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {\rm{;HOCL}}\)

\({\rm{K}} = \frac{{{\rm{[HOCL]}}}}{{\left[ {OC{l^ - }} \right] × \left[ {{H^ + }} \right]}}{\rm{;}}\)

\({10^{7.5}} = \frac{{{\rm{[HOCL]}}}}{{\left[ {OC{l^ - }} \right]× {{10}^{ - 7.5}}}}{\rm{;}}\)

\(1 = \frac{{{\rm{[HOCL]}}}}{{\left[ {OC{l^ - }} \right]}}{\rm{;}}\)

∴ [HOCl] = [OCl^-]

The concentration of free chlorine residual is the sum of the concentration of HOCl and OCl^-

[Free chlorine residual] =[HOCl] + [OCl-]

\(\left[ {{\rm{Free;chlorine;residual}}} \right] = \frac{{2 × {{10}^{ - 3}}}}{{2 × 35.5}}{\rm{;}}\)

⇒ 2.817 × 10^-5 moles/l

  2.817 × 10-5 moles/l=[HOCl] + [OCl-]

[OCl^-] + [OCl-]=  2.817 × 10-5 moles/l (∵  [HOCl] = [OCl-])

\(;{\rm{;}};;\left[ {{\rm{OC}}{{\rm{l}}^ - }} \right] = \frac{{2.817 × {{10}^{ - 5}}}}{2}{\rm{;}}\)

[HOCl] = [OCl-] = 1.4084 × 10^-4 moles/l

• Out of all these forms of chlorine, HOCl- is the most destructive. It is 80% more effective than OCl- ion.
• Hence pH is maintained just below 7 for most effective chlorination.

Explanation:

Flow (q) = Density (k) × Velocity (V)

For Greenshields’s Model (Linear Traffic Speed Density Model)

\(V = {V_f} \times \left( {1 - \frac{k}{{{k_j}}}} \right),\)

Where,

V = Velocity at any instant

Vf = Free mean velocity

k = Density of the flow

kj = Jam density of the flow

Maximum capacity (qmax) = Vmax × Kmax

As per Greenshields’s model

\(Vmax; = ;\frac{{{V_f}}}{2};& ;{k_{max}} = \frac{{{k_j}}}{2};\)

Maximum capacity ⇒ \(\left( {{q_{max}}} \right) = \frac{{{V_f}}}{2} \times \frac{{{k_j}}}{2} = \frac{{{V_f}{k_j}}}{4}\) 

Additional Information

Free Mean Speed:

From the speed-density diagram, it can be defined as the maximum speed at which the number of vehicles in a unit length is zero i.e. density is zero.

Density:: It can be defined as the number of vehicles per unit length. The unit of measurement is vehicles/km.

Jam Density: From speed-density relation and flow-density curve, it can also be seen that it is the maximum density at which there is no flow on the road.

      

Traffic Flow: The number of vehicles passing through a particular point in certain time interval is defined as traffic flow. Also, number of vehicles counted in one hour is called traffic flow (q).

Explanation:

Braking distance at downgrade,

\({\left( {{S_b}} \right)_{downgrade}} = \frac{{{V^2}}}{{254\left( {f - G} \right)}}\)

Where V = Design speed in kmph

f = Co-efficient of longitudinal friction, G = Grade

or 

\({\left( {{S_b}} \right)_{downgrade}} = \frac{{{V^2}}}{{2g\left( {f - G} \right)}}\) v in (m/s)

Braking distance at upgrade,

\({\left( {{S_b}} \right)_{upgrade}} = \frac{{{V^2}}}{{254\left( {f + N} \right)}}\)

or 

\({\left( {{S_b}} \right)_{upgrade}} = \frac{{{V^2}}}{{2g\left( {f + G} \right)}}\)  v in (m/s)

Concept:

The average delay is given by

\(Average;Delay = \frac{{\left( {Area;enclosed;by;arrival;curve) - (Area;enclosed;by;arrival;curve} \right)}}{{No;of;vehicles}}\)

Calculation:

Given:

Cycle time = 50 s

Effective red time = 30 s

Effective green time = 20 s

\(Average;Delay = \frac{{\left( {Area;enclosed;by;arrival;curve) - (Area;enclosed;by;arrival;curve} \right)}}{{No;of;vehicles}}\)

\(Area;of;triangle;formed;by;arrival;curve = \frac{1}{2} \times 40 \times 50\)

\(Area;of;triangle;formed;by;departure;curve = \frac{1}{2} \times 40 \times 20\)

\(Average;Delay = \frac{{\left( {1000) - (400} \right)}}{{40}}\)

Average Delay = 15 s

Concept:

Latitude: Projection of a line on N - S direction called latitude.

Departure: Projection of a line on E - W direction is called Departure.

For a closed traverse

Σ L = 0, Σ D = 0

Where, Σ L = sum of all latitudes, Σ D is the sum of all departures.

Calculation:

Taking angles in the Quadrantal system

Sum of latitudes = L × cos 33.7500° + 300 × cos 86.3847° - 354.524 × cos 10.6181° - 450 × cos 63.9003° + 268 × cos 43.5° 

Sum of latitudes = L × cos 33.7500° - 329.9166

Σ L = 0,

L × cos 33.7500° =  329.9166

L = 396.79 m

Concept:

Since the sand is saturated with seawater we need to use the unit weight of seawater for calculation.

\({G_{sea}} = \frac{{{γ _{Sea}}}}{{{γ _W}}}\)

G_{sea -} Specific gravity of seawater, γ_sea - Unit weight of seawater

γ_w - Unit weight of water

Relative density (I_D) is given by

\({I_D} = \left( {\frac{{{{\rm{e}}_{\max }} - ;e}}{{{{\rm{e}}_{\max }} - ;{{\rm{e}}_{{\rm{min}}}}}}} \right) × 100\)

e_max - void ratio at loosest state, e_min - void ratio at densest state

e - void ratio at state

\({{\rm{γ }}_{{\rm{sat}}}} = \left( {\frac{{{\rm{G}} + {\rm{e}}}}{{1 + {\rm{e}}}}} \right) × {{\rm{γ }}_{\rm{sea}}}\)

γsat – Saturated unit weight of soil, G – Specific gravity of soil solids

e – Void ratio of soil, γ – Unit weight of seawater

Calculation:

Given:

Relative density = 40%, Maximum void ratio = 1.0, Minimum void ratio = 0.5

Specific gravity of seawater =1.03, Unit weight of water = 9.81 kN/m3

Specific gravity of soil solids = 2.67

Using the Eqn \({I_D} = \left( {\frac{{{{\rm{e}}_{\max }} - ;e}}{{{{\rm{e}}_{\max }} - ;{{\rm{e}}_{{\rm{min}}}}}}} \right) × 100\)

\(40 = \left( {\frac{{1 - ;e}}{{1 - ;0.5}}} \right) × 100\)

e = 0.8

\({1.03} = \frac{{{γ _{Sea}}}}{{{9.81}}}\)

γ_sea = 10.1043 kN/m³

\({{\rm{γ }}_{{\rm{sat}}}} = \left( {\frac{{{\rm{2.67}} + {\rm{0.8}}}}{{1 + {\rm{0.8}}}}} \right) × {{10.1043}}\)

γ_sat = 19.478 kN/m³

Total stress distribution:

σA = 20 × 10.1043 = 202.086 kN/m²

σ_B = 202.086 + 30 × 19.478 =786.426 kN/m2

Neutral stress distribution:

u_A = 20 × 10.1043 = 202.086 kN/m2

u_B = 202.086 + 30 × 10.1043 =505.215 kN/m2

Effective stress distribution: 

σ^'A = σA - uA = 202.086 - 202.086 = 0

σ^'_B =  σ_B - u_B = 786.426 - 505.215 = 281.211 ≈ 281 kN/m² 

Concept:

The fluctuations in water level do not alter the effective stress because the total stress and neutral stress increase by the same amount hence nullifying the effect of each other. The change in effective stress is zero.

Calculation:

a) Total stress distribution:

σA = 22 × 10.1043 = 222.2946 kN/m2

σB = 222.2946 + 30 × 19.48 = 806.6946 kN/m2

b) Neutral stress distribution:

uA = 22 × 10.1043 = 222.2946 kN/m2

uB =  222.2946 + 30 × 10.1043 =525.4236 kN/m2

c) Effective stress distribution: 

σ'A = σA - uA = 222.2946 - 222.2946 = 0

σ'B =  σB - uB = 806.6946 - 525.4236 = 281.271≈ 281 kN/m2

Hence the effective remains the same hence the change in effective stress is 0 kPa

Concept:

1. \({\rm{Area;of;catchment~(A)}} =\rm \dfrac{{Area;of;unit;hydrograph(Runoff~ volume)}}{{Runoff;depth}}\)

In Unit hydrograph the runoff depth is 1cm

Area of the unit hydrograph is found by Trapezoidal rule

Calculation:

Given:

Ordintaes are  0, 3, 8, 6, 3, 2, and 0 

The area of unit hydrograph is given by,

\(\displaystyle\int_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

Here h = 1 hr

 

Area of unit hydrograph \( = \dfrac{1}{2}\left[ {0 + 0 + 2\left( {3 + 8 + 6 + 3 + 2} \right)} \right] \times 60 \times 60\)

⇒ 79200 m³

^{\({\rm{Area;of;catchment}} = \dfrac{{79200}}{{{{10}^{ - 2}}}}\)}

A = 7920000 m²

A = 7.92 km²

Additional Information

In general for any given hydrograph,

1. Run off volume = Area of Direct runoff hydrograph
2. Runoff volume = Area of catchment × runoff depth

Trapezoidal rule is given by:

\(\displaystyle\int_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{;}} \ldots } \right)} \right]\)

\({\rm{Number;of;intervals(n)}} = \dfrac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{;}}\)

where b is the upper limit, a is the lower limit, h is the step size

Concept:

The ordinate of 't0' Unit hydrograph = \({\rm{Δ }}y × \dfrac{{{T_0}}}{{{t_0}}}\)

Δy -  Difference between ordinates of  shifted S curve and S curve constructed for given ordinates

T0 - Duration of known given hydrograph

t0 - Duration of the new hydrograph

Calculation:

GIVEN:

T0  = 2 hr

ϕindex =  2 mm/hr

Rainfall P = 6.6 cm

Duration of storm = 3 hr, t₀ = 3 hr

The duration of the storm is 3 hr, hence we need to construct a  3 hr Unit hydrograph

 

Peak of 3 hr UH = 6 m3/s

\({\phi _{\rm index}} = \dfrac{{P - r}}{t}\)

\(2 = \dfrac{{66 - r}}{3}\)

r = 60 mm = 6 cm

 

Peak runoff of storm = Peak runoff of UH \( × \dfrac{{{\bf{r}}(cm)}}{{1~cm}}\)

Peak runoff of storm = 6 × 6 = 36 m3/s

Peak flow due to storm = Runoff + Base flow = 36 + 5 = 41 m3/s

Concept:

Force acting in spring of stiffness k is

\(k = \frac{P}{δ }\)

P - Force acting on spring

δ - Deflection in the spring

Calculation:

Given:

The free-body diagram showing the given set of forces is

From similar triangles 

\(\frac{L}{{{δ _1}}} = \frac{{2 × L}}{{{δ _2}}}\)

δ₂ = 2× δ₁

Taking moment about hinge,∑M = 0 we get

P × 2 × L - k × δ₂ × 2 × L - k × δ₁ × L = 0 

P × 2 × L - k × 2 × δ₁ × 2 × L - k × δ1 × L = 0 (∵ δ₂ =2 × δ₁)

\({\delta _1} = \frac{{2 \times P}}{{5 \times k}}\)

\({\delta _2} = \frac{{2\times 2 \times P}}{{5 \times k}}\)(∵ δ2 = 2× δ₁)

\({\delta _2} = \frac{{4 \times P}}{{5 \times k}}\)

Concept:

Force acting in spring of stiffness k is

\(k = \frac{P}{δ }\)

P - Force acting on spring

δ - Deflection in the spring

Calculation:

Given:

we know

\({δ _1} = \frac{{2 × P}}{{5 × k}}\)

\({δ _2} = \frac{{4 × P}}{{5 × k}}\)

From the free-body diagram 

Let R₁ and R₂ represent the force in spring 1 and spring 2

In spring 1

\(k = \frac{R_1}{δ_1 }\)

R₁ = k × δ₁

\({R_1} = k \times \frac{{2 \times P}}{{5 \times k}}\)

\({R_1} = \frac{{2 \times 100}}{5}\)

R₁ = 40 kN

In spring 2

R₂= k × δ₂

\({R_2} = k \times \frac{{4\times P}}{{5 \times k}}\)

\({R_2} = \frac{{4\times 100}}{5}\)

R1 = 80 kN

Concept:

2% of solids (by weight) + 98% of water (by weight) = 100 % of sludge (by weight)

 \({\rm Specific~gravity(G_s)} = \frac{{{\rho _{solid}}}}{{{\rho _{water}}}}\)

Equating the volume 

\(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

Calculation:

Given:

Solids content = 2 % by weight, Specific gravity of Sludge solid = 2.2

2 % of solids means there is 98% of water in sludge.

2% of solids + 98% of water = 100 % of sludge

Let ‘x’ be the density of sludge

Equating total volume 

 \(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

\(\frac{{100}}{x} = \frac{{98}}{{1 \times 1000}} + \frac{2}{{2.2 \times 1000}}\)

X = 1011 kg/m³

∴ Density of sludge = 1011 kg/m³

Concept:

2% of solids (by weight) + 98% of water (by weight) = 100 % of sludge (by weight)

\({\rm Specific~ gravity (G_s)} = \frac{{{\rho _{solid}}}}{{{\rho _{water}}}}\)

Equating the volume 

\(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

Calculation:

Given:

Solids content = 2 % by weight, Specific gravity of Sludge solid = 2.2

2 % of solids means there is 98% of water in sludge.

2% of solids + 98% of water = 100 % of sludge

Let V1 = volume of sludge (Initial)

 V₁ = Volume of water + Volume of solids

\( ⇒ \frac{{98}}{{1 \times 1000}} + \frac{2}{{2.2 \times 1000}}\)

⇒ 0.0989 m3

∴  V₁ = 0.0989 m3

After sludge thickening (Volume of water changes)

New volume of sludge (V2) = V1/2

V2 = 0.04945m3

∴ 0.04945 = Vol. of solids + volume of water 

\(0.04945 = \frac{2}{{2.2 \times 1000}} + \frac{x}{{1000}}\)

 x = 48.545 % 

∴ Solids content by weight = \(\frac{{2}}{{48.545; + 2;}} \times 100\)

⇒ 3.96 % (by weight)