BPSC AE Paper 4 (General Engineering Science) 2019 Official Paper (Previous Year Paper)

Explanation:

Cantilever beam with uniformly distributed load (w = W/L):

Shear force:

SFxx = wx

Bending Moment:

$B{M_{xx}} = \frac{{{wx^2}}}{2}$

$B{M_{max}} = B{M_{x=l}}=BM_{fixed~end}=\frac{{{wl^2}}}{2}$------(1)

Additional Information

Simply supported beam of length (L) loaded with uniform loading intensity (w = W/L):

Bending Moment (Mx) at a section x-x is given by:

${{\rm{M}}_{{\rm{x}} - {\rm{x}}}} = \frac{{{{w}}L}}{2}x - \frac{{{{w}} {x^2}}}{2}$

When plotted against x, bending moment gives rise to a parabolic curve and the shear force and bending moment can be drawn in the following way will appear as follows:

The maximum bending moment for a simply supported beam with a uniformly distributed load $= {wL^2\over 8}$   -------(2)

By dividing equation (1) and (2) we get: 

The ratio of their maximum bending moment = ${{wL^2\over 2}\times{8\over wL^2}} = 4$

Concepts:

Chain line: The line joining the main stations are called main survey lines or chain lines.

Tie line: The line joining tie stations is termed a tie line. It usually fulfills a dual purpose i.e., it checks the accuracy of the framework and enables the surveyor to locate the interior details which are far away from the main chain line.

Check line: The line joining the apex point of a triangle to some fixed points on its base is known as the Check line.  It is also called a proof line. It is taken to check the accuracy of the triangle.

Baseline: The line on which the framework of the survey is built is known as the Baseline. It is the most important line of the survey as it fixes up the direction of all the above other lines.

Generally, the longest of the main survey lines is considered the baseline. It should be laid on level ground as much as possible.

Concept:

An optical square is used for setting out right angles. It consist of a small circular metal box of diameter 5 cm and depth 1.25 cm. It has a metal cover which slides round the box to cover the slits.

Principle of optical square:

According to the principle of reflecting surfaces, the angle between the first incident ray and the last reflected ray is twice the angle between the mirrors. In this case, the angle between the mirrors is fixed at 45˚. So, the angle between the horizon sight and index sight will be 90˚.

Construction of Optical Square:

A horizon glass H is fixed at the bottom of the metal box. The lower half of the glass is unsilvered and the upper half is silvered.

An index glass “I” is also fixed at the bottom of the box which is completely silvered.

The angle between the index glass and horizon glass is maintained at 45˚.

Concepts:

Important Definitions in Theodolite:

1. Axis of Bubble tube:  It is the imaginary line tangential to the longitudinal curve of the bubble tube at its midpoint.
2. A-Line of Sight is a line that passes through the optical center of the objective and the intersection of crosshairs and the line of sight when horizontal is called the line of collimation.
3. Axis of the telescope: It is the imaginary line passing through the optical center of the objective lens and the optical center of the eye-piece lens of a telescope.
4. Vertical Axis: It is the axis of rotation of the telescope in the horizontal plane.
5. Horizontal Axis: It is the axis of rotation of the telescope in the vertical plane.

Concepts:

Method of repetition:

• When the horizontal angle is desired to be more than the least count of the instrument or to be measured more accurately than repetition method is used.
• In this method, the desired angle is measured several times, and the correct value of the angle is obtained by dividing the accumulated reading by the no. of repetitions.
• The nos. of repetitions made usually in this method is six- three with the face left and three with the face right. In this way, angles can be measured more accurately than a Vernier.

Reiteration method/ Direction method/ Method of series:

• It is adopted for measuring the horizontal angles when many angles of well-distributed points are to be measured from the same instrument station with high precision.
• In this method, angles are measured successively starting from a point termed an initial station and the angle between the terminating station and the initial station is the last observation.
• This process of measuring the angles is to obtain a check on their sum being equal to 360° and is called closing the horizon. When the horizon is closed, the final reading of the vernier should be the same as its initial reading if there is no discrepancy.

Explanation:

Collimation Method:

• This method is generally adopted if there are lot many numbers of intermediate stations and also if the number of shifting instruments is less
• It does not provide a check on Intermediate sights
• The height of the instrument is calculated at each set up of the instruments
• Generally used in contour surveys, in small areas

Profile levelling:

• It is the method of determining the level of a ground surface along a predetermined line which may be the centre-line of a road, canal, railway or pipeline. The predetermined line may be a single straight line or a series of connected straight lines. The method is also known as longitudinal levelling or sectioning.

Differential levelling: 

• It is the method of levelling to determine the elevation of points located at some distance apart or to determine the elevation difference between two points or to establish benchmarks.
• The height of collimation is generally used in differential levelling.
• The height of Collimation(benchmark + staff reading) is the height of the horizontal plane through the telescope.
• The method is used in order to find the difference in elevations between two points:
• If they are far apart
• The difference in elevation between the two points is too great
• If there are obstacles intervening

 Check Levelling:

• It is conducted to check a series of levels, which have previously been fixed.
• At the end of each day's work, a line of levels starting from the point and returning to the starting point of that day is run with the object of checking the work done on that day.

Explanation:

Bar chart:

• The bar chart is the technique of project management. It consists of two co-ordinate axes in which the time required for the completion of the activity is marked on the X-axis and actively involved in its execution of the project is marked on Y-axis in form of bars.
• A bar chart consists of a number of vertically or horizontally placed bars and the length of the bars defines the relative magnitude of the values. It uses rectangular bars with lengths proportional to the values that they represent. A Gantt chart is a horizontal bar chart used for project management

Characteristics of a bar chart:

• The start and endpoints of the bar indicate the time of the finish and start time of the activity.
• The length of the bar represents the time required for the completion of an activity.
• The width of the bar does not signify anything i.e. it can be of any size.

Gantt Chart:

• Gantt charts are mainly used to allocate resources to activities.
• The resources allocated to activities include staff, hardware, and software. Gantt charts are useful for resource planning.
• A Gantt chart is a special type of bar chart where each bar represents an activity.
• This chart lists the tasks to be performed on the vertical axis, and time intervals on the horizontal axis.
• The bars are drawn along a timeline. The length of each bar is proportional to the duration of time planned for the corresponding activity.
• With the use of this chart, we can check the progress of the project with
• A visual display of the whole project,
• Timelines and deadlines of all tasks,
• Relationships and dependencies between the various activities,
• Project phases

Explanation:

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.

To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.

PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.

• PERT (Project Evaluation and Review Technique) approach takes account of the uncertainties. In this approach, 3-time values are associated which each activity. So it is probabilistic.

• CPM (Critical Path Method) involves the critical path which is the largest path in the network from starting to ending event and defines the minimum time required to complete the project.So it is deterministic.

Difference between PERT and CPM (Critical Path Method)

Explanation:

A project may be defined as a combination of interrelated activities which must be executed in a certain order before the entire task can be completed.

The aim of planning is to develop a sequence of activities of the project so that the project completion time and cost are properly balanced.

To meet the objective of systematic planning, the management has evolved several techniques applying network strategy.

PERT (Programme Evaluation and Review Technique) and CPM (Critical Path Method) are network techniques which have been widely used for planning, scheduling and controlling the large and complex projects.

• PERT (Project Evaluation and Review Technique) approach takes account of the uncertainties. In this approach, 3-time values are associated which each activity. So it is probabilistic.

• CPM (Critical Path Method) involves the critical path which is the largest path in the network from starting to ending event and defines the minimum time required to complete the project.So it is deterministic.

Difference between PERT and CPM (Critical Path Method)

Concept:

Crash time is the minimum activity duration to which an activity can be compressed by increasing the resources and by increasing the direct cost.

The slope of the line gives the amount of increase in the direct cost per unit time for crashing an activity.

Concept:

Cost Slope: The direct cost curve is a curve that can be approximated by a straight line, depending upon the flatness of the curve. The slope of this straight is the cost slope. It is very helpful in the project cost analysis.

${{\rm{C}}_{\rm{s}}} = \frac{{{\rm{Crash;Cost}} - {\rm{Normal;Cost}}}}{{{\rm{Normal;Time}} - {\rm{Crash;Time}}}}$

The activity with minimum cost slope should be crashed first, as it results in optimum crashing of the project.

Explanation:

Co-efficient of Discharge (Cd) is defined as the ratio of the actual discharge from an orifice to the theoretical discharge from the orifice.

$C_d={Q\over Q_{th}}={Actual~ velocity × Actual~ area\over Theoretical~ velocity × Theoretical ~area}$

\(C_d={Actualvelocity\over Theoretical velocity}×{Actual ~area\over Theoretical ~area}=C_v× C_c\)

so co-efficient of contraction in terms of Cd and Cv is given as

$C_c={C_d\over C_v}$ i.e. Cd = Cc × Cv

Additional Information

Coefficient of Velocity  (Cv) is defined as the ratio between the actual velocity of a jet of liquid at vena-contracta and the theoretical velocity of the jet.

${C_v} = \frac{V}{{\sqrt {2gH} }}$

Co-efficient of contraction (Cc) is defined as the ratio of the area of the jet at vena-contracta to the area of the orifice.

${C_c} = \frac{{{a_c}}}{a}$

Concept:

Minor losses caused by the disruption of the flow due to the installation of appurtenances, such as valves, bends, and other fittings.

The various losses are given by 

1). Sudden expansion loss:

${\left( {{h_L}} \right)_{exp}} = \frac{{{{\left( {{v_1} - {v_2}} \right)}^2}}}{{2g}}$

2). Exit loss:

${\left( {{h_L}} \right)_{exit}} = \frac{{v^2}}{{2g}}$

3). Entrance loss:

${\left( {{h_L}} \right)_{ent}} = \frac{{0.5 ;×; v^2}}{{2g}}$

Concept:

$kinematic~ viscosity= \frac {dynamic ~ viscocity}{density ~of ~fluid}$

$ν = \frac {μ}{ρ }$

density = specific gravity × 1000

Calculation:

Given:

μ = 0.011 poise = 0.0011 kg/ms, specific gravity = 0.95, 

∴ ρ = 0.95 × 1000 = 950 kg/m3

$ν = \frac {μ}{ρ }$

$ν = \frac {0.0011}{950}=1.16 \times 10^{-6}=0.0116$ stokes

Concept:

Forces on submerged bodies:

• When a fluid flowing over a stationary body, a force is exerted by the fluid on the body. Similarly, when body is moving in a stationary fluid, a force is exerted by the fluid on the body.
• The total force exerted by the fluid on the body is perpendicular to the surface of the body. Thus total force is inclined to the direction of motion.
• This total force can be resolved into two components, one in direction of motion and other perpendicular to direction of motion.

Drag Force:

• The component of total force in direction of motion is called “Drag”. Thus drag is the force exerted by the fluid in direction of motion.

Lift Force: 

• The component of total force in direction perpendicular to motion is called “Lift”. Thus lift force is the force exerted by the fluid in direction perpendicular to the motion.

Examples of immersed bodies having drag and/or lift forces:

1. A tall chimney exposed to wind;
2. Flow of water past a bridge pier;
3. Flow of fluids past blades in fans, blowers, compressors, turbines etc.;
4. Motion of aeroplanes, submarines, torpedoes etc.

Examples of bodies where both drag and lift forces are produced:

1. Propeller blades;
2. Aerofoils;
3. Hydrofiles;
4. Rotating cylindrical bodies;
5. Kites etc.

Following points are worth noting:

1. In contrast to drag, the lift forces may exist even in ideal fluids by the presence of circulation.
2. Real fluids also require vortices or circulation around the body for producing lift.
3. In motion of arifoils with finite spans, there is another kind of drag force associated with the lift force, called the "induced drag".

Explanation:

Darcy Weisbach Equation for friction losses in circular pipe:

${h_f} = \frac{{4f \times L \times {V^2}}}{{2 \times g \times D}}$

where

L = length of the pipe,

D = diameter of the circular pipe,

V = mean velocity of the flow,

f = Darcy’s friction factor = 4 × Cf 

C_f = coefficient of friction

hf = head loss due to friction

For Laminar Flow

Friction Factor $f = \frac{{64}}{{{R_e}}};and;C_f = \frac{1}{4} \times \frac{{64}}{{{R_e}}} = \frac{{16}}{{{R_e}}}$

We know,

f = 4C_f

⇒ Cf = 16 / R_e

Where C_f is the coefficient of friction and f is the friction factor.

For Turbulent flow

$f = \frac{{0.316}}{{R_e^{\frac{1}{4}}}}$

Explanation:

Flow process:

• The processes occurring in the open system which permit the transfer of mass to and from the system are known as the Flow process.
• Flow work is the amount of work that is required to displace an infinitely small fluid element to flow into or out of the control volume.

​

• Control mass is known as a closed system and control volume is known as an open system.
• The boundary of the control volume is known as a control surface.

Non-Flow Processes

These are compression and expansion processes on gases in a cylinder with complete leakproof. In these, there is only energy transfer with zero mass transfer.

These nonflow processes can be the followings:

• constant pressure process
• constant volume process
• constant temperature process
• reversible adiabatic process
• poly-tropic process
• constant internal energy process

Adiabatic Process:

• Adiabatic process: When a thermodynamic system undergoes a change in such a way that no exchange of heat takes place between Systems and surroundings, the process is known as an adiabatic process.
• During the adiabatic process, no exchange of heat takes place between the system and the surroundings. i.e. ΔQ = 0.
• Change in pressure (P), volume (V), and temperature (T) takes place.

Concept:

Mayer’s Equation for an ideal gas:

Cp – Cv = R

The gas constant (R) is equal to the difference of two specific heats.

The ratio of Specific heat is known as an adiabatic index.

$\gamma = \frac{{{C_p}}}{{{C_v}}}$

Where Cp = Specific heat of the gas at constant pressure and Cv = Specific heat of the gas at constant volume

∴ The ratio of the specific heat of a gas at constant pressure and the specific heat of the gas at constant volume is always constant.

Explanation:

The work done is given by the area under the P - V curve.

W = ∫ v dP

As clearly seen the Larger the value of n, the smaller is the area and so smaller is work obtained. In other words

W = ∫ v dP

$W = \frac{{{P_1}{V_1} - {P_2}{V_2}}}{{n - 1}}$

So, W is maximum, when n = 0 i.e. constant pressure process

W is always zero for n = ∞ i.e. constant volume process.

The heat is given by the area under the T - S curve.

dQ = ∫ T dS

or, Q = T(s2 - s1), represents the area under T-S diagram.

Hence, the Area under the process line in P - V (Pressure-Volume) and T - S (Temperature - Entropy) diagrams give work and heat interaction (Heat absorbed or Heat rejected) respectively.

Explanation:

Carnot cycle:

The Carnot cycle consists of 4 processes

• 1-2 isothermal heat addition
• 2-3 reversible adiabatic expansion
• 3-4 isothermal heat rejection
• 4-1 reversible adiabatic compression

A cycle is said to be reversible only when each process in a cycle is reversible.

Carnot cycle consists of 2 isothermal and 2 Reversible adiabatic (Isentropic) processes. An isothermal process is a very slow process and the adiabatic process is a very fast process and the combination of a slow process and fast process are very difficult.

So a Carnot cycle is also known as an impractical cycle. It is used only to compare other actual cycles.

Hence, Carnot and Rankine cycle differs in the heat addition process.

Explanation:

Otto cycle:

The air-standard-Otto cycle is the idealized cycle for the spark-ignition internal combustion engines.

Otto cycle is the one which has two constant volume heat transfer processes and two adiabatic work transfer processes.

The Otto cycle 1-2-3-4 consists of the following four processes:

• Process 1-2: Reversible adiabatic compression of air

• Process 2-3: Heat addition at constant volume

• Process 3-4: Reversible adiabatic expansion of air

• Process 4-1: Heat rejection at constant volume

• During the constant volume process heat addition and heat rejection take place and no work transfer, because of this it is also known as a constant volume cycle.

• During the adiabatic processes [compressions/expansion] only work transfer taken place but no heat transfer occurs.

Important Points 

Explanation:

The formula for the efficiency of the Otto and Diesel cycle are as follows:

${\eta _{otto}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}$

${\eta _{diesel}} = 1 - \frac{1}{{{r^{\gamma - 1}}}}\frac{{{\alpha ^\gamma } - 1}}{{\gamma \left( {\alpha - 1} \right)}}$

From the Diesel cycle efficiency formula,

• In Otto and Diesel cycle, compression ratio part is common in the efficiency formula.
• For the practical cutoff ratio, the extra term in the efficiency formula of the Diesel cycle is always greater than one.
• The extra term in the efficiency formula of the Diesel cycle will approach unity when the cutoff ratio approaches unity, which is quite impractical.

When the cut-off ratio (α) is equal to zero then the efficiency of the Diesel cycle approaches Otto cycle efficiency.​

Concept:

Coefficient of Coupling (k):

The coefficient of coupling (k) between two coils is defined as the fraction of magnetic flux produced by the current in one coil that links the other.

Two coils have self-inductance L1 and L2, then mutual inductance M between them then Coefficient of Coupling (k) is given by 

$k=\frac{M}{\sqrt {L_1L_2}}$

Where,

$M=\frac{N_1N_2A}{\mu_o \mu_r l}$

$L_1=\frac{\mu_o \mu_rN_1^2A}{ l}$

$L_2=\frac{\mu_o \mu_rN_2^2A}{ l}$

N1 and N2 is the number of turns in coil 1 and coil 2 respectively

A is the cross-section area

l is the length

Calculation:

Given,

L1 = 75 mH

L2 = 105 mH

k = 0.45 

From the above concept,

$k=\frac{M}{\sqrt {L_1L_2}}$

$M = 0.45 ×\sqrt{75 \ mH\times105 \ mH}$

M = 39.93 mH

Concept:

The maximum flux in a transformer core is given by the emf equation,

$E = 4.44 f N \Phi_{max}$

For a transformer designed for a fixed voltage, the maximum core flux is inversely proportional to frequency.

Explanation:

When a transformer designed to operate at 50 Hz is connected to a 5 Hz supply of the same voltage, the frequency is reduced to one-tenth of its rated value.

Since voltage is constant, the core flux increases as frequency decreases:

$\Phi_{max} \propto \frac{1}{f}$

At 5 Hz, the flux becomes very high, pushing the core deep into saturation. Due to core saturation:

• Magnetizing current increases sharply
• Core losses rise abnormally
• Heavy heating occurs in the core and windings

This excessive heating can cause insulation damage, overheating, and visible smoke from the transformer.

Correct Answer:

Transformer may start to smoke.

Concept:

• The rating of the Transformer is always expressed in kVA.
• kVA is the unit of apparent power.
• kVA rating is given by:

$S = {\sqrt{3} V_LI_L}$

where S = apparent power

V_L = line voltage

I_L = line current

Calculation:

Given V_L = 415V

I_L = 139.1A

$S = {\sqrt{3} \times 415\times 139.1}$

S = 99985.23 VA

S = 99.98 kVA ≈ 100kVA

Calculation of Thevenin's Resistance:

• The terminal from where Thevenin resistance is to be calculated has to be open-circuited.
• All independent voltage sources along with their internal resistances have to be short-circuited.
• All independent current sources along with their internal resistances have to be open-circuited.
• All dependent voltage and current sources have to remain in the same state.

Additional Information Thevenin's Theorem states that “In any linear, bidirectional circuit having more than one independent source, having more number of active and passive elements, it can be replaced by a single equivalent circuit consisting of equivalent voltage source VTh and series with equivalent circuit resistance RTh“.

Where

VTh = Thevenin Voltage or open circuit voltage

RTh= Thevenin Resistance 

For RTh All independent sources replaced by their internal resistance (ideal voltage source replaced by a short circuit and ideal current source replaced by an open circuit.

Vth is open circuit voltage

MMF:

We know that, Ohm’s law for magnetic circuit states that the MMF is directly proportional to the magnetic flux where reluctance is the constant of proportionality.

MMF = flux × reluctance

flux ∝ MMF  

↓ MMF → ↓ flux

So, MMF decreases with decrease in magnetic flux.

Concepts:

The compressive strength of good Portland cement and standard sand mortar cement at the end of 3 days should not be less than 11.50 N/mm² and that at the end of 7 days should not be less than 17.50 N/mm².

Note:

1. Standard sand is known as Ennore sand and it is obtained from Tamil Nadu. It is used for the preparation of cement mortar. For the preparation of cement mortar, the cement-sand ratio is 1:3 in which one part is cement and 3 part is sand is used.
2. The compressive strength of cement is determined by cube test on cement mortar cubes compacted by means of a vibrating machine with vibration frequency should be 12000 ± 400 vibrations per minute and vibration time is of 2 minutes.

Concepts:

Hydraulic lime contains lime with silica or alumina and can set under water.  It contains 30 percent of silica and 5 percent of alumina along with iron oxide. It is also called Water Lime.  It is off-white in color.

They are classified on basis of strength: feebly, moderately and eminently hydraulic lime.

1. Feebly hydraulic lime contains 5-10% of silica and alumina, slakes in minutes, and sets in about three weeks. It is used for less expensive work and in mild climates.
2. Moderately hydraulic lime contains 15 – 25 % of silica and alumina, slakes in one to two hours, and sets in approximately one week. It is used for better quality work and exterior walls in freezing climates.
3. Eminently hydraulic lime has a chemical composition similar to Ordinary Portland cement (OPC). This consists of 25 to 30 % of silica and alumina. It slakes very slowly, and sets in approximately a day. It is used in harsh environments such as damp locations and near saltwater.

Explanation:

Bulking of sand:

• Bulking of fine aggregate or sand is the phenomenon of an increase in sand volume due to the increase of moisture content. Bulking test on fine aggregates has to be performed before using it in construction.
• The moisture content in the sand makes thin films around sand particles. Hence, each particle exerts pressure. Thus they move away from each other causing increasing in volume.
• The bulking of the aggregates is dependent on two factors:

a) The fineness of the aggregates

b) Percentage moisture content

As shown in the figure below, the bulking of the sand increases with the increase in moisture content. This happens up to a limit beyond which any moisture addition will decrease the volume.

A fully saturated fine aggregate does not show any bulking. The rate of bulking is inversely proportional to the size of the aggregates. Hence, fine aggregates bulk more compared to coarse aggregates.

Explanation:

Vehicle in paints:

• It is also known as a binder
• The vehicle is an oil to which the base is mixed.
• It holds the constituents of paint in suspension and helps spread it over the surface to be painted and imparts
  

durability, toughness, and waterproofness to the paint film and resistance to weathering and

gloss to the painted surface and forms the body of the paint.

• Examples are natural drying oils such as linseed oil, nut oil, poppy oil, and tung oil

Bronze paints:

• These paints are prepared by disbrushing aluminum bronze and copper bronze in nitrocellulose lacquer as a vehicle
• They produce a very reflective type surface and are hence very useful for being applied to radiators
• these paints are equally effective for painting interior or exterior metallic surfaces

Anticorrosive paints:

• These paints are used to protect metallic structural steelworks
• There are several paints that exhibit these properties i;e oil paints
• These paints essentially consist of linseed oil as a vehicle and red lead, zinc oxide, iron oxide, etc. as a base

Concept:

Header bond is also known as heading bond. In header bonds, all bricks in each course are placed as of the headers on the face’s walls.

Stretcher bond is also called a running bond, is created when bricks are laid with only their stretchers showing, overlapping midway with the courses of bricks below and above.

$\therefore \frac{{Length;of;a;stretcher}}{{;Length;of;a;header;}} \approx 2$

∴ The number of vertical joints in the stretcher bond is half that of vertical joints in the header bond in a course.

The number of vertical joints in a stretcher course is 1/2 times the number of joints in the header course.

Explanation:

As per IS-1904(1966): 

• The maximum permissible settlement for isolated foundations is 40mm on sandy soil and 65mm in clayey soils.
• The permissible settlement for the raft foundation on clay soil is 65-100 mm and for sandy soil, it is 65 mm.

Additional Information 

Differential settlement:

Foundation on clay soil = 40 mm.

The foundation on Sandy soil = 25 mm.

Explanation:

Killed steels: In killed steel, the Oxygen is removed completely. The solidification of such steels does not give gas porosity (blowholes).

• So Killed steels are those steels Which are deoxidized in the ladle with silicon and aluminium
• Killed steel is a type of steel from which there is practically no evolution of gas during solidification of the ingot after pouring, because of the complete deoxidation, and formation of pipe in the upper central portion of the ingot, which is later cut off and discarded.
• All alloy steels, most low-alloy steels, and many carbon steel are usually killed.
• The continuous casting billets are also killed. The essential quality criterion is soundness.
• Killed steel is characterized by a homogeneous structure and even distribution of chemical compositions and properties.
• Killed steel is produced by the use of a deoxidizer such as Al and a ferroalloy of Mn or Si
• However, calcium silicide and other special deoxidizers are sometimes used.

Additional Information

1. Alloy Steel:  When other elements comprising metals and non-metals are added to carbon steel, alloy steel is formed.

• These alloy steels display various environmental, chemical, and physical properties that can vary with the elements used to alloy.
• Alloy steels are made by combining carbon steel with one or several alloying elements, such as manganese, silicon, nickel, titanium, copper, chromium, and aluminum.

1. Carbon Steel: Carbon steel is an iron-carbon alloy, which contains up to 2.1 wt.% carbon.

• Carbon steel can be classified into three categories according to its carbon content
• low-carbon steel (or mild-carbon steel),
• medium-carbon steel
• high-carbon steel.

1. Silicon Steel: Silicon steels are ferritic alloys of iron and silicon that have magnetic properties which make them useful in motors and transformers.

• The silicon additions improve magnetic softness and increase electrical resistivity.
• They also have the undesirable effects of decreasing the Curie temperature, reducing the saturation magnetization, and embrittling the alloy when the silicon additions exceed about 2 wt.%.

Concept:

Steel is an alloy of iron and carbon. Composition up to 2%C is steel and beyond this, it is considered as cast iron.

Eutectoid/Pearlite steel:

A 0.84% carbon steel or eutectoid steel is known as PEARLITE steel. This is much stronger than ferrite or cementite.

Hypo-eutectoid Steel

Plain carbon steels in which carbon percentage is less than 0.8% are called hypo-eutectoid steel.

Hypereutectoid Steel

Plain carbon steels in which carbon percentage is more than 0.8% are called hypereutectoid steel.

Concept:

Distemper:

• Distemper is an early form of whitewash. As a decorative paint, it is easily marked and cannot get wet. It has been used since antiquity for painting walls and other types of house decoration. Because it's not waterproof, it has most often been used on interior surfaces.
• Distemper is a decorative paint and a historical medium for painting pictures. Soft distemper is not abrasion resistant and may include binders such as chalk, ground pigments, and animal glue.
• Normally, 500 to 700 ml of water is required for 1 kg of distemper. Hence we can take 0.6 litres of water for 1 kg of distemper.

Concepts:

Flat Latex is used to make paint water-resistant and paints based on flat latex material are generally adopted for ceilings.

Walls or ceiling painted with paint containing the celluloid transparent sheets is used to draw objects on them or painted for traditional or hand-drawn animations.

Acrylic compounds act as binders in paint and they form a film after the water has evaporated and ensure good adhesion of paint with the wall.

Paints consist of Volatile Organic Compounds (VOCs). VOCs are released into the air as gases from solids or liquids that contain organic chemicals and exert odor.

Pillowy resins are added to paint to make them odorless to some extent.

Explanation:

Turbidity:

• It is the measure of the degree to which water losses its transparency due to the presence of suspended particles
• Its measurement could be used to provide an estimation of TSS (total suspended solids)

Various unit measure turbidity of water are:

NTU:

The nephelometric Turbidity Unit signifies that the instrument is measuring scattered light from the sample at a 90-degree angle from the incident light. 

JTU:

Jackson Turbidity Unit is a historical unit used when measurements were made visually using a Jackson Candle Turbidimeter. Water was poured into a tube until a flame underneath the tube could no longer be distinguished. 

FTU:

Formazin Turbidity unit and define the unit of measurement after the acceptance of formazine as the primary reference standard for turbidity.

From the above ATU is not a unit to measure the turbidity of water.

Concepts:

The minimum velocity at which suspended material in sewerage is not silted up is known as self-cleansing velocity.

Mathematically, it is given as:

$V_s = √{\frac{8kgd(G-1)}{f}}$

K = Constant, for clean inorganic solids = 0.04 and for organic solids = 0.06.

f = Darcy Weisbach friction factor (for sewers = 0. 03).

G = Specific gravity of sediments

g = gravity acceleration

d = Diameter of grain, m.

 Calculation:

Case 1: For sand particles

d = 1 mm; G = 2.65; f = 0.03 K = 0.04 (Sand is inorganic)

Self-cleansing velocity is calculated as:

$V_s = √{\frac{8 \times 0.04 \times 10 \times 0.001(2.65-1)}{0.03}}$

V_s = 0.42 m/s

Case 2 : For organic solids

d = 5 mm; G = 1.2; f = 0.03 K = 0.06 (For organic)

Self-cleansing velocity is calculated as:

$V_s = √{\frac{8 \times 0.06 \times 10 \times 0.005(1.2-1)}{0.03}}$

V_s = 0.4 m/s

The self-cleansing velocity would be a minimum of the above two values i.e., 0.4 m/s.

Concept:

Biological Oxygen Demand(BOD):

Biological Oxygen Demand or Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by microorganisms to break down organic matter in water.

A low BOD is an indicator of good quality water, while a high BOD indicates polluted water.

BOD = ( DO_i - DF_f ) × Dilution factor

Where,

BOD = Biochemical oxygen demand in ppm or mg/lit

DO_i = Initial dissolved oxygen in mg/lit.

DO_f =Final dissolved oxygen in mg/lit.

Dilution factor $= \frac{{Volume\ of\ the\ diluted\ sample}}{{{\rm{Volume \ of\ the\ undiluted\ sewage\ sample}}}}$

Calculation:

Given:

( DOi - DFf ) = 5

Volume of diluted sample = 100

Volume of the undiluted sewage sample = 2

Dilution factor = $\frac{100}{2}$ = 50

BOD5 = 5 × 50 = 250 ppm

∴ BOD of the sewage is 250 ppm.

The velocity of flow in sewer is not given, so Question cannot be solved due to insufficient data.

Concepts:

The detention period means how much time sewage remains inside the tank. It is calculated as:

$t_d = \frac{V}{Q}$

Where

V = Volume of tank

Q = rate of flow of sewage

t_d = detention time

Calculation:

Given:  Detention time, t_d = 2 hrs.

Dimension of tank:  Length, L = 24 m, Width, W = 6 m and Depth, D = 3 m.

Volume of tank = L × B × D

V = 24 × 6 × 3 = 432 m³

The rate of flow of sewage is calculated as:

$2 = \frac{432}{Q}$

Or

Q = 216 m³/hr.

Concept:

Flywheel

 A flywheel is used to control the variation in speed during each cycle of the engine.

The kinetic energy of flywheel can be calculated by:

K.E $= \frac{1}{2}I{w^2}$

where I = moment of inertia of the flywheel, ω = angular velocity of the flywheel

Calculation:

Given:

Mass of flywheel (m) = 20 kg, radius (r) = 200 mm = 0.2 m, N = 600 rpm

$I = \frac{{m{R^2}}}{2} = \frac{{20 ~\times ~{{0.2}^2}}}{2} = 0.4;kg - {m^2}$

\(w = \frac{{2\pi N}}{{60}} = \frac{{2~ \times~ \pi \times 600}}{{60}} = 62.83;rad/s\)

∴ ${\rm{K}}.{\rm{E}} = \frac{1}{2}{\rm{I}}{{\rm{\omega }}^2} = \frac{1}{2} \times 0.4 \times {\left( {62.83} \right)^2} = 789.5{\rm{;J}} \simeq 790{\rm{;J}}$

Concept:

Rolling Resistance: It is the minimum force required at center of roller parallel to the contacting surface so that rolling starts. In rolling resistance there will not be point or line contact there is surface contact.

The length of contact is known as coefficient of rolling resistance.

The force necessary to roll the wheel is given by \({{F}{R}}=\frac{{{\mu }{R}}W}{R}\)

Where μR is coefficient of rolling resistance

W = Load acting on the wheel, R = Radius of the wheel

Calculation:

Given that D = 600 mm, $R=\frac{D}{2}=300~\text{mm}=0.3~m$, μR = 0.3, W = 500 N

\({{F}{R}}=\frac{{{\mu }{R}}W}{R}=\frac{0.3 \times 500}{300}=0.5~N\)

Explanation:

• Momentum is conserved in all collisions.
• In elastic collision, kinetic energy is also conserved.
• In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.
• Hence, In inelastic collision of two particles total linear momentum only is conserved.

Perfectly elastic collision:

If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If law of conservation of momentum holds good during collision while that of kinetic energy is not.

Coefficient of restitution (e)

$e = \frac{{Relative;velocity;after;collision}}{{Relative;velocity;before;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}$

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Concept:

The disc is rolling without slipping so the point of contact (P) of the annular disc with the ground will be in the state of rest. So about p[oint P there will be pure rotational motion.

So, point P can be considered as the instantaneous center of rotation.

The velocity of point P will be the same as the velocity of the center of mass of the disc.

⇒ v = 2Rω 

$\Rightarrow ω =\frac{V}{2R}{{;;;;;}} \ldots \left( 1 \right)$

Where 

The kinetic energy of the disc,

$\Rightarrow KE = \frac{1}{2}{J_p}{ω ^2}{{;;;;;}} \ldots \left( 2 \right)$

Where,

.J_P = The mass  moment of Inertia of the annular disc about point P, ω = The angular velocity of the annular disc 

Calculation:

The mass  moment of Inertia of the annular disc about point P,

$\Rightarrow {I_c} = \frac{1}{2}m{\left( {2R} \right)^2} + \frac{1}{2}m{R^2} = \frac{5}{2}m{R^2}$

Using parallel axis theorem,

$\Rightarrow {I_p} = {I_c} + m{\left( {2R} \right)^2} = \frac{{13}}{2}m{R^2}{{;;;;;}} \ldots \left( 3 \right)$

Using equation (1), (2), and (3),

$\Rightarrow KE= \frac{1}{2} \times \frac{{13}}{2}m{R^2} \times {\left( {\frac{v}{{2R}}} \right)^2}$

$\Rightarrow KE= \frac{{13}}{{16}}m{v^2}$

Concept:

Linear velocity V is given by:

$V=r×ω$

where ω is angular velocity.

Tangential acceleration in the no-slip condition is

$\frac{{dv}}{{dt}} = r\frac{{dω }}{{dt}} = r×α$

at = r × α

Centripetal acceleration is given by:

ac = r × ω2

The instantaneous velocity of the point of contact is zero.

So at the point of contact, Instantaneous tangential acceleration is also zero.

∴ Only centripetal acceleration is there at the point of contact.

∴ Net acceleration of the point of contact is

ac = r × ω2

Concept:

Since axial strain is given by $\frac{{{\sigma _a}}}{E} - μ \frac{{{\sigma _c}}}{E}$

Calculation:

Given:

E = 100 GPa, μ = 0.3

Pressure at 1 m depth = 1000 × 10 × 1 = 10000 N/m2

${\sigma _a} = \frac{{pd}}{{4t}} = \frac{{10000 \times 2}}{{4 \times 0.001}}$

= 5 × 106 N/m2  = 5 MPa

${\sigma _c} = \frac{{pd}}{{2t}} = \frac{{10000 \times 2}}{{2 \times 0.001}} = 10 \times {10^6};\frac{N}{{{m^2}}} = 10;MPa$

Axial strain at mid depth

$= \frac{{\left[ {5 - \left( {0.3 \times 10} \right)} \right] \times {{10}^6}}}{{100 \times {{10}^9}}} = 2 \times {10^{ - 5}}$

Concept:

Hooke’s law: As per Hooke’s law, stress is proportional to strain

i.e. σ ∝ ϵ or, σ = ϵ × E

Where, E =  Modulus of Elasticity

For Hooke’s law to be valid:

(a) The material should be homogenous.

(b) The material should be isotropic.

(c) The material should behave in a linearly elastic manner.

Thus for a plane bar with Area ‘A’, Length ‘L’ and Modulus of Elasticity ‘E’,

${\rm{\sigma = }}\frac{{\rm{P}}}{{\rm{A}}};;{\rm{and}};;{\rm{ϵ = }}\frac{{{\rm{δ L}}}}{L}$

As per Hooke’s law,

∵ σ = ϵ × E

$\frac{{\rm{P}}}{{\rm{A}}}{\rm{ = }}\frac{{{\rm{δ L}}}}{{\rm{L}}}{\rm{ × E}}$

${\rm{δ L = }}\frac{{{\rm{PL}}}}{{{\rm{AE}}}}$

Calculation:

Given,

Two bars of unit length, i.e L1 = L2

Elongation ratio is 3 : 5, i.e δL1 : δL2 = 2 : 5

Both subjected to same tensile load, i.e P1 = P2

Both have the same size, i.e A1 = A2

Elongation of the first bar:

${\rm{δ L_1 = }}\frac{{{\rm{P_1L_1}}}}{{{\rm{A_1E_1}}}}$

Elongation of the second bar:

${\rm{δ L_2 = }}\frac{{{\rm{P_2L_2}}}}{{{\rm{A_2E_2}}}}$

Taking ratio,

$\frac{{{\rm{\delta }}{{\rm{L}}_1}}}{{{\rm{\delta }}{{\rm{L}}_2}}} = \frac{{{{\rm{P}}_1}{{\rm{L}}_1}}}{{{{\rm{A}}_1}{{\rm{E}}_1}}} \times \frac{{{{\rm{A}}_2}{{\rm{E}}_2}}}{{{{\rm{P}}_2}{{\rm{L}}_2}}}$

$\frac{{{\rm{\delta }}{{\rm{L}}_1}}}{{{\rm{\delta }}{{\rm{L}}_2}}} = \frac{{{{\rm{E}}_2}}}{{{{\rm{E}}_1}}}$

$\therefore \frac{{{{\rm{E}}_1}}}{{{{\rm{E}}_2}}} = \frac{{{\rm{\delta }}{{\rm{L}}_2}}}{{{\rm{\delta }}{{\rm{L}}_1}}} = \frac{5}{2}$

Hence the ratio of modulus of elasticity of the two materials will be 5 : 2

Concept :

The relation between E, K and μ

E = 2G (1 + μ)

E = 3K (1 - 2μ)

$\text{E}=\frac{9\text{KG}}{3\text{K}+\text{G}}$

Where,

E = Young's Modulus of Rigidity = Stress / strain

G = Shear Modulus or Modulus of rigidity = Shear stress / Shear strain

μ = Poisson’s ratio = - lateral strain / longitudinal strain

K = Bulk Modulus of elasticity = Volumetric stress / Volumetric strain

Calculation:

Given:

μ = 0.25, K/E = ?

E = 3K (1 - 2μ)

$\frac{K}{E}=\frac{1}{3(1~-~2\mu)}$

$\frac{K}{E}=\frac{1}{3(1~-~2\times 0.25)}=\frac{1}{3\times 0.5}=\frac{2}{3}$

Explanation:

Strain Energy: 

• Strain Energy of the member is defined as the internal work done in deforming the body by the action of externally applied forces.
• Strain energy is given by:
• $Strain;Energy = ;\frac{1}{2} \times Stress \times Strain \times Volume$

Resilience:

• Resilience is the total strain energy stored in a given volume of material within the elastic limit.
• On removal of load, this energy is released.
• In other words, it is the area under the load-deflection curve within the elastic limit.

Proof resilience:

• The maximum energy which can be stored in a body up to the elastic limit is called proof resilience.

Modulus of resilience:

• Modulus of resilience is defined as proof resilience per unit volume. It is the area under the stress-strain curve up to the elastic limit.