OCTANE AND CETANE NUMBERS

Self ignition temperature (SIT) of a fuel is the temperature at which the fuel ignites on its own without spark. If large amount of mixture in an engine cylinder auto ignites, there will be a rapid rise in pressure causing direct blow on engine structure accompanied by thudding sound. This causes vibrations in the engine. The phenomenon is called knocking.

If however, a small pocket of fuel-air mixture auto ignites, pressure waves are generated which travel with the speed of sound across the cylinder. These pressure waves are of such small duration that indicator diagram mechanism fails to record them. These waves interact within themselves and with the cylinder walls, creating characteristics ping sound. The phenomenon is called pinking.

The engine runs rough, overheats and loses efficiency due to knocking and pinking.

The processes of knocking and pinking are related to the nature of the fuel and relative merits of the fuel are decided on the basis of their anti-pinking and anti-knock property. The merit is measured by octane number such that a fuel of high octane number will be liable to less pink or knock as compared to a fuel of low octane number in the same engine. It is important to note that the same fuel will show same tendency to pink or knock in all engines.

Commonly used fuel in SI engines is a mixture of iso-octane and n-heptane. Iso-octane has minimum tendency to knock and this fuel is arbitrarily assigned an octane number of 100 (ON = 100) where as n-heptane has maximum knocking tendency with ON = 0. The octane number of a given fuel is percentage of iso-octane in the mixture of iso-octane and n-heptane. Thus a fuel other than mixture of iso-octane and n-heptane if assigned an ON of 80, it means, it will knock under standard operating condition similar to the mixture of 80% iso-octane and 20% n-heptane.

The tendency to knock in an engine increases with the increase in compression ratio. The highest compression upto which no knocking occurs in a given engine is called highest useful compression ratio (HUCR).

Certain chemical compounds when added to the fuel successfully suppress the knocking tendency. Tetra-ethyl lead [Pb(C₂ H₅)₄] also commonly called TEL and tetra-methyl lead [Pb(CH₃)₄] also referred to as TML are effective dopes in the automobile fuel to check knocking. They are called as anti-knocking agents. However, because of lead poisoning effects TEL and TML are not being used now-a-days. In stead, some organic auto knocking agents have been developed to check the undesirable effects like knocking.

In CI engine air alone is compressed to a compression ratio of 15 to 20 (commonly). The fuel is injected under a pressure of 120 to 210 bars about 20° to 35° before TDC. As the fuel in the engine starts to evaporate the pressure in the cylinder drops and it delays the ignition process by a small amount. The time between beginning of injection and the beginning of combustion is known as the delay period which consists of time for atomization, vapourization and mixing along with time of chemical reaction prior to auto-ignition. The combustion of fuel continues in the expansion and is called after burning. Increased delay period causes accumulation of atomized fuel in the combustion chamber and as the pressure and temperature continue to rise at one instant, the bulk of fuel auto-ignites. This would result in high forces on the structure of the engine causing vibration and rough running.

The CI engine fuel rating is based on ignition delay and is measured in terms of cetane number. Cetane fuel [C₁₆ H₃₄] has very low delay period and is arbitrarily assigned a cetane number of 100. Another fuel a α-methyl-napthalene [C₁₁ H₁₀] has poor ignition quality and is assigned zero cetane number. The volume percentage of cetane in a mixture of cetane and a-methyl naphthalene is the cetane number of the fuel that produces same delay period as the mixture under specified test conditions. Additives such as methyl nitrate, ethyl thio-nitrate and amyl nitrate increase cetane number of a fuel respectively by 13.5%, 10% and 9% if added to the extent of 0.5%.

MOCK CLASS TEST: THERMODYNAMICS Sub: Code: EME-303; Mahamaya Technical University

Time: 2 hrs                                                                                                   Maximum Marks: 50 

Attempt all the questions: 

SECTION A: 

1) Attempt the following questions:                                                                        (5 x 2 = 10) 

a) Define system, surroundings and universe. 

b) Distinguish between Heat pump and Refrigerator. 

c) What is Exergy and Anergy? 

d) Explain the law of degradation of energy. 

e) What is triple point of water? 

SECTION B: 

2) Attempt any three questions:                                                                               (3 x 5 = 15) 

a) Distinguish between macroscopic and microscopic approaches of thermodynamics. 

b) Discuss the neccessity of 2nd law of thermodynamics. 

c) 2 kg of a gas at 10 bar expands adiabatically and reversibly till its pressure drops to 5 bar. During the process 120 kJ of non-flow work is done by the system and the temperature drops from 377°C centigrade to 257°C. Calculate the value of the index of expansion and the characteristics gas constants. 

d) Steam at a pressure of 4 bar absolute and having dryness fraction 0.8, is heated at constant volume to a pressure of 8 bar absolute. Find the final temperature of the steam. Also, find the total heat absorbed by 1 kg of steam. 

e) 2 kg of air at NTP is heated at constant volume untill the pressure becomes 6 bar. Find the change of entropy of the system. 

SECTION C: 

Attempt part (a) or part (b) of the following questions                                                 (5 x 5 = 25) 

3) (a) Explain the thermodynamic equilibrium and quasi-static process. 

(b) Prove the equivalence of Kelvin-Planck statement and Clausius statement. 

4) (a) A steam turbine developing 110 kW is supplied steam at 17.5 bar with an internal energy of 2600 kJ/min, specific volume = 15.5 m³/kg and velocity of 275 m/s. Heat loss from the steam turbine  37.6 kJ/kg. Neglecting the changes in potential energy, determine the steam flow rate in kg/hr. 

(b) A reversible engine takes 2400 kJ/min from a reservoir at 750 K develops 400 kJ/min of work during cycle. The engine rejects heat at two reservoir at 650 K and 550 K. Find the heat rejected to each sink. 

5) (a) Explain the causes of internal and external irreversibility. 

(b) Explain the importance of Gibb's function and Gibb's free energy. 

6) (a) 5 kg steam at pressure 8 bar and temperature 300°C is adiabatically mixed with 4 kg steam at 6 bar and 250°C. Find the final condition of the mixture. Also find the change in entropy. 

(b) Hot steam is flowing through a perfectly adiabatic pipe. At point A the temperature of the steam is 250°C and pressure is 4 bar, while at the point B, its temperature is 275°C and pressure is 3.5 bar. Find the direction of the flow. 

7) (a) 5 kg of Oxygen is enclosed within a vessel of 0.05 m³ at a temperature 200°C, is being supplied 120 kJ of energy through heating. Find the final pressure and temperature. 

(b) One kg of an ideal gas is heated from 18.3°C to 93.4°C. Assuming R = 287 J/kg-K and  γ  = 1.18 for the gas. Find out (i) specific heats, (ii) change in internal energy, and (iii) change in enthalpy and entropy.

BEAMS, JOINTS AND SUPPORTS

KINEMATICS ANALYSIS : MOVING BODIES......

KINEMATICS ANALYSIS: MOVING BODIES......
Mechanics, physics, mechanical engineering

What makes a body moving?

OBSERVATION ONE:

"From our perception, we can say, a body moves if we apply either a push or a pull on the body. There are several  instances, when after applying push or pull, the body still doesn't move. What is the exact reason behind this?"

OBSERVATION TWO:

"Almost to apply a push or pull on a body, a physical contact is needed, with out any physical contact it is not possible to exert push/pull on a body, although there are exceptions too.

(i) We know on every material body existing on earth experience a downward pull towards the center of Earth and to exert this pull, the object and the earth don't need any physical contact. This phenomena is aptly named as the 'Force of Gravity.'

(ii) The second option is Magnet. A magnet can pull as well as push another magnet from a distance with out any physical contact.

(iii) When we place a charged particle near another charged particle, we watch the particles can exert push as well as pull without any physical contact. Like a proton repels another proton, ie they exert push on each other. But a proton attracts an electron by pulling each other to come close.

So there are two types of motion. Type one is the example of a Cricket ball going to boundary after being hit by the bat. So, here impact is the driver of the motion.

But an apple falling from a tree is also in motion, but for this no impact is there. In fact the attraction between the apple and the earth is responsible for the motion. By attraction, it means the tendency of going closer in two objects. Here motion of apple occurs without being hit.


LINEAR MOTION:

We know that Force is a kind of physical quantity having both magnitude as well as direction. So, force is a vector quantity. We also know that applying triangle's law or parallelogram law of forces addition, we can add two forces to get an equivalent force which is known Resultant force.

So, application of force on an object brings a change in position of the object. This change is position is called displacement. An object in a coordinate system has a position vector to define the position vector. Any change of position will bring change in its position vector.

Suppose in a coordinate system we have an object at a position vector (r). Let after a time interval of (dt), the object changes its position by an amount (dr). Hence, the rate of change of position will be (dr)/(dt). Rate of change of position

means the change of position in unit time. Rate of change of position is called velocity of the object. It is denoted by (v).

Hence, (v) = (dr)/(dt).
The unit of velocity is m/s in SI units and cm/s in cgs system. The most popular unit is km/hr.

A velocity may change, it may change in direction or it may change in magnitude. Suppose, we have an object moving with a velocity (v) at any instant. Suppose after an interval of time (dt), its new velocity becomes (v + dv), where (dv) is the change in velocity. Hence (dv)/(dt) will be the rate of change of velocity and it indicates the change of velocity per unit time. This physical quantity is called acceleration. Negative acceleration which is rate of decrease in velocity is called retardation or deacceleration too.

Acceleration may be changed; the rate of change of acceleration is called impulse. Like when a bat touches a moving ball, it has an impact and this changes its acceleration due to this magnitude and direction of the ball changes. If a large magnitude of force act on a body for a very short period of time, it is called impulse.

Free falling under the forces of gravity is a case of constant acceleration and that is denoted by (g) and it is equal to g = 9.81  m/s ² .

THREE EQUATIONS OF MOTION

These three equations are valid only in the case of constant acceleration. Every particle falling under the gravity will satisfy these equations.

We know (dv)/(dt) = a,
hence, (dv) = a(dt) or by integrating both side from initial state t = 0; v = u to final t = t; v = v,
we get,
v - u = a(t - 0) or v = u + at

Again, a = (dv)/(dt)
a = {(dv)/(dx)}.{(dx)/(dt)}
a = v.(dv)/(dx)
hence, v.dv = a.dx
Integrating both sides from initial v= u, x = 0 to final v = v, x = s
we get,
v² - u² = 2 a (s - 0)
v² - u² = 2as

From the defination of velocity, we get
v = (dx)/(dt)
but, v = u + at
hence,
u + at = (dx)/(dt)
dx = (u +at). dt
Integrating both sides of the equation from initial condition t = 0, x = 0 and final condition t = t, x = s, we get

(s - 0) = u(t - 0) + (a.t²)/2
s = ut + (at²)/2

Average Velocity = (u + v)/2
= (u + u + at)/2 = u + at/2

Total distance, s = Average Velocity x time

s = (u + at/2) x t
s = ut + (at²)/2

Uses of these equations:

1) Suppose we have a particle travelling with 5 m/s and an acceleration of 1 m/s^2 is applied on the body. What will be the

velocity after a time of 10 s?

Ans: Here, u = 5 m/s, a = 1 m/s ²  and t = 10 s, then

velocity after 10 s,
v = u + at
v = 5 + 1 x 10 = 15 m/s

2) A car moving with a velocity 60 km/hr suddenly applies the brake. As a result, the car comes to a halt after running 50 m

after applying brake. What will the value of retardation? Find the time it needs to come to rest after the application of brake.

Ans: Here, initial velocity u = 60 km/hr = (60 x 1000)/(60 x 60) m/s = 16.67 m/s
final veloity, v = 0 m/s
Total distance travelled s = 50 m
a = (v² - u²)/2s
a = (0 - 16.67²)/(2 x 50)
a = - 2.78 m/s²

again time, t = (v - u)/a
t = (0 - 16.67)/(-2.78)
t = 5.99 s

The Concept of Relative Velocity:

Suppose in a road two car is moving. The faster car at 40 km/hr and the slower car at 30 km/hr in the same direction. Now, if anyone watches the faster car from the slower car, he won't be see it running at 40 km/hr, in stead he will see the velocity at 15 km/hr. When we watch from a moving body or better we say moving reference frame, the velocity of other bodies seem be reduced. Again if we take the same two cars running in opposite directions to each other, the velocity of the each car will be at 55 km/hr as seen from the other. This is due to relative velocity. The relative velocity of a body is the velocity of the body relative to an observer.

Suppose, a car is moving with a velocity V_c and a train is moving with a velocity V_t. Then velocity of car with respect to a person sitting in the train will be V_ct = V_c – V_t and velocity of the train to a person in the car will be  V_tc = V_t – V_c  .

3) A train is moving towards east with a velocity 120 km/hr and wind is blowing towards west with a velocity 20 km/hr. What will the velocity of the wind to an observer in the train?

Ans: We shall take towards east direction as positive and towards west direction as negative.
Let the velocity of train is (V_t) and velocity of wind is (V_w)

So,  V_t  = 120 km/hr and
V_w = - 20 km/hr

V_wt =  V_w  - V_t
V_wt = - 20 - 120 = - 140 m/s and it means wind is flowing from west.

4) A car is moving on a horizontal road at 40 km/hr. Suddenly rain started to pour down at a velocity 30 km/hr. Find the

velocity of the rain drops with rest to an observer in the car. Also find the angle with which rain drops would appear to strike the car.

Ans: Let the velocity of the car be V_c = 40 km/hr and rain drop velocity is  V_r = - 30 km/hr. The angle between them is θ = 90°.
(V_rc) = (V_r) - (V_c)
(V_rc)² = (V_r)² + (V_c)² - 2(V_r)(V_c) cos θ
= 30² + 40² + 0
= 2500
(V_rc) = 50 m/s

ANGULAR MOTION:

Suppose a line AB displaces side wise such that A remains at same point, but the other end B comes to new position C. This is an angular displacement. Angles are measured in radian. 1 radian is the angle formed by an arc equals to magnitude of radius (r). A full circle produces 2π.

Suppose, a line of length (r) makes an angular displacement of (dθ) in time (dt). Then the rate of change of angular displacement is given by (dθ)/(dt) and it is called angular velocity, (w). Hence, (w) = (dθ)/(dt).

If we apply torque or moment in the line, the angular velocity will be changed. Let during the time interval, (dt), the change in angular velocity be (dω). The rate of change of angular velocity will (dω)/(dt) and it is called as angular acceleration and denoted as α.

α = (dω)/(dt)
(dω) = α.(dt)
Taking initial value ω = ωₒ, t = 0, to final value ω = ω, t = t and integrating both side, we get,
ω - ωₒ = α.(t - 0)
ω = ωₒ + αt

angular velocity ω = (dθ)/(dt)
ωₒ + αt = (dθ)/(dt)
(ωₒ + αt)(dt) = dθ
Taking initial value θ = 0, t = 0, to final value, θ = θ, t = t, and integrating both side we get,

ωₒ.(t - 0) + (α/2)(t² - 0) = θ - 0

θ = ωₒt + (αt²/2)

α = (dω)/(dt)
= {(dω)/(dθ)}.{(dθ)/(dt)}
α = ω.(dω)/(dθ)
ω.(dω) = α.(dθ)
taking the initial value ω = ωₒ, θ = 0 and final value ω =ω, θ = θ and integrating both sides,
(ω²)/2 - (ωₒ²)/2 = α.(θ - 0)
(ω²) = (ωₒ²) + 2.α.θ



MOTION OF A RIGID BODY

Plane Motion:

A rigid body is said to be in plane motion when all parts of the body move in parallel planes. The plane motion of a rigid body may be classified into several categories like :

1) Translation
2) Rotation
3) General plane motion.

Translation:

ENGINEERING MECHANICS CLASS TEST: ONE

CLASS TEST: ONE

Time: 1 hr 30 min                                                                                        Max. Marks: 30

                                                                                                                         

1) Answer the following question in brief                                                              2 x 6 = 12   

                                                                 

a)      Distinguish clearly between composition of forces and resolution of forces.

b)      Show that the algebraic sum of the resolved part of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction.

c)      Differentiate between coplanar forces and concurrent forces clearly.

d)     State and explain the laws of transmissibility of forces.

e)      Explain Newton’s third law of motion.

f)       What is a couple? Explain its characteristics.

2) Answer any two of the following questions                                                     6 x 3 = 18

a)      A smooth circular cylinder of radius 1500 mm is lying in a triangular groove, the right side of which makes 20° angle and left side makes 40° angle with horizontal. Find the reaction at each contact if there is no friction and the cylinder weight is 400 N.

                                                                                             

b)      A right circular cylinder of weight 5 kN rests on a smooth inclined plane and is held in position by a cord AC as shown in the figure. Find the tension in the cord if there is a horizontal force of magnitude 1 kN acting at C.

c)      Four forces of magnitude 10 kN, 18 kN, 15 kN and 12 kN are acting along the diagonals and sides of a regular pentagon as shown in the figure. Find the resultant force of the given force system.

d)     Prove that if a body is at equilibrium under three forces, then the forces are concurrent forces.

SMALL ENGINEERING COLLEGES OF GHAZIABAD: A BLEAK FUTURE

Economics says "When Supply is more than the Demand of a product, the Price falls." This is particularly true in the case of Technical Education in U.P. and perhaps upto some extent in the country itself.

Over the past few years the supply is outstripping the demand for Engineering and Management seats in the country. Just take the example of Uttar Pradesh, the most populous state of India is home to about 333 Engineering colleges which cumulatively offer a total seats of 1,15,379 in Engineering Education where as according to University datas the total number of students took admission in various engineering colleges after qualifying SEE amounts to mere 25,903. So, what happens to the vacant seats? And this year the figures are not going to be improved it seems.

This year a total 1,60,561 candidates had registered for the State Entrance Examination, out of whom, 1,29,924 have qualified. But, there are approximately 1.33 lakh B.Tech seats in the Engineering colleges affiliated to GBTU and MTU.

There are clearly a huge gap between the supply and demand of Engineering seats. In this situation, it has been believed that many small colleges will be bankrupt due to the lack of students. Many colleges have defered the salaries of the teachers and other employees due to the revenue crunch. It seems a grim scenario ahead for those colleges.

Due to the revenue crunch, promoters of small colleges are taking the refuge of cost cutting, and as a part of that they are trying to trim their faculty strength. Surely, this will affect the quality of the education as the each teacher will be over burdened and perhaps have to take five classes per day, where as the AICTE limits the load at best 18 classes per week. Also, most of the colleges don't follow the exact teacher students ratio of 1:20 prescribed by the apex body.

Many promoters are planning to opt out by selling their stakes in the colleges. The causes of their exits are the facts that running colleges in western U.P. is no longer a profitable business. They have cited that due to lack of students in taking admission, the colleges are no longer the chickens that lay gold eggs, which were in fact so just three years ago. So, why these colleges suddenly loss their values? What are the reasons behind these failures?

There are several reasons for the fall in numbers of students opting B.Tech courses. The most vital reason is the very high tution fees in colleges under MTU and GBTU compared to colleges in other states like Karnataka, Punjab and Rajasthan. Most of the colleges here charge more than 90,000.00 in the first year B.Tech where as colleges in other states charges below 60,000.00, even colleges in Punjab and West Bengal charge below 50 thousand and this is going to be a major factor.

The 2nd factor is the placement after the completion of the degree. Although many colleges claim tall, citing a long list of companies taking interest to place their students in very good packages, but reality always bites hard. The negative publicity by the ex students are also eating the pie here and there is no solution other than boosting the placement record by making a good relation with the HR of these companies by the respective college authorities.

The third most important factor is the sagging quality of the available faculty members. Many teachers although possess M.Tech degrees are not competent to impart quality education due to lack of depth of required knowledge as well as the essential communicating power required to be a good teacher. In some cases, due to over burdened schedule, a good teacher becomes unable to teach in the class. Just imagine the mental fatigue a teacher experienced while taking 5th or 6th class in a day when each class is of 55 min. duration.

A college has to show money to run the next three years during the visits from AICTE. So all the colleges have to show enough balance to pay the salaries of the employees for atleast three years, otherwise they won't get the permission to run the colleges, still some of them couldn't pay the salaries of the teachers and staffs. Why? Becouse they must have showed enough balances to acquire the clearences during the AICTE visits. Where does the money go? Vanished! Or siphoned off? There are several "skips" of the rules and regulations these college authorities used to practise.

QUESTIONS BANK 8 : FORCE AND FORCE SYSTEM

CONCEPT OF FRICTION

CONCEPT OF FRICTION

FRICTION IS ALL PERVADING, FRICTION IS OMNIPRESENT
FRICTION IS AN UNIVERSAL PHENOMENA

FRICTION: Friction may be defined as the resistive force acting in opposite direction in which the body tends to move or it moves. Frictional force always acts tangentially at points of contact.


Friction may be classified into two categories.


• Static Friction and
• Kinetic Friction


Static Friction is the friction experienced by a body when it is at rest under the action of external friction.


Kinetic Friction is the friction experienced by a body when it moves. Kinetic friction may be classified as Sliding Friction and Rolling Friction.


As a body can move in two ways, one is sliding and the other is rolling, there are two types of kinetic friction


• Sliding Friction and
• Rolling Friction


Sliding Friction: When a body slides over a surface without having any rotational tendency about a horizontal axis, it experiences a sliding friction. Sliding frictional force always try to dampen the movement as quickly as possible.


Rolling Friction: It is the friction experienced by a body when it rolls over the surface with an angular velocity as well as linear velocity. Rolling is a combination of translation as well as rotation about a horizontal axis passing through the centre of gravity.

Suppose we have a block of weight (W) lying on the ground as shown in the figure. The block will be at equilibrium as the weight will be neutralised due to the normal reaction provided by the ground.


Now let a horizontal (P) force is gradually applied to the block. Initially, when the force P is small, the block will not move, but if we increase the magnitude of P, then one moment will come when the block will start to move along the direction of applied force.


But from Newton's 2nd law, we know that whether the magnitude of force P is small or large, in all the cases the block should start to move with an acceleration F/m. But, practically the block starts to move when the magnitude of the applied forces reaches a definite value. Why does the block behave so? What does it mean?


Well, it means that there must exist an opposite resistance force that acts opposite to the applied force, having the same magnitude as that of applied force. But, this resistance force has a limit. When the resistance force reaches a maximum value, then the further increase in applied force can not be neutralized and as a result the body starts to move. When the value of the resistance force becomes maximum, any further increase of applied force set the movement in the body. The condition just before a body starts to move is called as "Limiting Conditions." The resistance force is called Frictional force and it becomes maximum, when the body attains the limiting condition. At this position the maximum frictional force is called "Limiting Friction."


Now just look at the experiment again, if we applied an infinitesimal force, the block doesn't move, which means the frictional force thus generated must be equal to the infinitesimal applied, else the body will experience a net force and the body will start to move. Now, if we increase force by some amount, the body will still be static, until the applied force reached the value of limiting friction. So, what does this indicate?

It indicates that frictional forces are variable. As we increase applied load, the frictional force also increases from zero to the highest value of limiting friction.


When the horizontal applied force is zero, the frictional force must be zero else the body will experience a net force. It means that when there is no applied force, frictional force just vanishes. Hence, it is known as pseudo force as its existence is dependent upon the applied force. (Remember there is another force centrifugal force which depends upon the centripetal force in rotational motion, hence, it is also a pseudo force.)


So, as the applied force starts to increase, the frictional force also increases thus maintaining the equilibrium conditions. But, the frictional force can not neutralize the applied force P, when its magnitude crosses the maximum value frictional force that can be generated due to friction on the contact surfaces.

Therefore when P > (fs)max, where (fs)max is the maximum magnitude of frictional force, the body will be in motion. When P = (fs)max, we call it as the case of impending motion or limiting condition. 

So, on what factor does the maximum frictional force on a surface depend upon? Certainly, it doesn't depend upon the applied force, although frictional force depends upon the applied force, but how much will be the maximum value of friction entirely depends upon the surface and the geometry of that surface and the Normal reaction the surface produces to counter balance the weight of the body. Here, one should remember normal reaction depends upon the mass of the body and the inclination of the surface with horizontal. 

COULOMB'S LAW OF DRY FRICTION: 

So, Coulomb's Law of dry friction states that, when there is a body resting on a surface is subjected to an applied force P, the maximum frictional force that would be generated directly depends upon the value of Normal reaction experienced by the body. 

(f_s)_max ∝ N 

N = mg cos θ, where mg is the weight of the body and θ is the inclination angle of the plane with horizonal. For a flat plane N = mg.

(f_s)_max = µN 

where µ is the proportionality constant and N is the normal reaction the body experiences from the surface. 

COEFFICIENT OF FRICTION: 

Here the constant (µ) plays a vital role. On what factor does the constant mu depend upon? It has been observed that the value of (µ) is greatly affected by the roughness of the surface upon which the body rests. It's value is a combined property of the contact surface as well as the surface roughness of the body itself. If we replace the body with another body of same mass but different material the value of (µ) changes. Also, if we place the body upon a different surface then also the value of mu changes. So, the value of mu is such a property that defines the characteristics of friction between the body and the contact surface. Hence, it is aptly named as the coefficient of friction. 

There are basically two types of Co-efficient of Friction. 

• Co-efficient of Static Friction 
• Co-efficient of Kinetic Friction 

ANGLE OF FRICTION (φ) 

When a block of mass is at rest on a surface and a horizontal force P is applied on the body to move it, a frictional force will be there to oppose any movement of the body. This force will act on the contact surface. Normal reaction is also acting upward on the contact surface. So total force on the contact surface will be resultant of normal reaction and frictional force. The angle made by this resultant force with normal reaction is called the angle of friction. 

FRICTION ON AN INCLINED PLANE: 

The direction of a frictional force depends upon the tendency of movement. 

Suppose we get two identical block of weight W in identical planes at angle α with horizontal as shown in the figure. 

Due to the force component W.sin α acting downwards along the plane, the body will have a tendency to move downwards along the plane. 

As the body would try to move downwards, a frictional force will be generated at the contact surface which would try to oppose the tendency to move downwards of the body, i.e., it would try to resist the downwards movement of the body. So, it will act upwards along the plane. 

Normal reaction produced by the inclined surface at the contact point or area. The normal reaction will be equal and opposite the force component of the weight of the body at a perpendicular direction to the inclined plane hence, N = W cos α, where N is the normal reaction. 

Now suppose we plane adjust the inclination of the plane, it means we can either increase or decrease the inclination of the plane. When the inclination is very small, the downward force component W sin α will be small and an equal magnitude frictional force will be produced and neutralize the downward force. Hence, the body will be at rest. 

Now if we increase the inclination of the plane, the downward force component W sin α will increases too, and frictional force will also be increased. Gradually, a condition will arrive when the downward weight component becomes equal to the maximum frictional force generated on the contact surface. This is limiting condition and also known as impending motion. If we increase the inclination angle α by a small amount, the body will start to move downwards. The angle of the plane when the body is at limiting condition is known as angle of repose. 

ANGLE OF REPOSE (α) 

We can define angle of repose as the angle of the inclination of a plane when a body on the plane is at limiting condition of impending motion due to its self weight component along the inclined plane. 

It is numerically equal to the angle of friction. It is denoted by (α). 

CONE OF FRICTION: 

It is an imaginary cone generated by revolving resultant reaction R about the normal reaction N. R is the resultant of the frictional force and normal reaction. 

Properties of Cone of Friction: 

• The radius of this cone represents the frictional force (fs)max. 
• The semi apex angle of the cone represents the angle of friction. 
• For co-planar forces, in order for motion not to occur the reaction R must be within the cone of friction. 

QUESTIONS BANK 7 : FORCE AND FORCE SYSTEM

QUESTIONS BANK 6 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 5 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

QUESTIONS BANK 4 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTIONS BANK 3 : FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

 

NEW SYLLABUS FOR MANUFACTURING SCIENCE: FIRST YEAR OF MTU FOR 2012-13

MANUFACTURING SCIENCE (EME-101/EME-201)

Unit-I 
Properties, Inspection and Testing of materials
Introduction to stress & strain
Mechanical Properties: Strength, Elasticity, Stiffness, Malleability, Ductility, Brittleness, Resilience, Toughness and Hardness.
Elementary ideas of Creep, Fatigue & Fracture
Testing of metals: Destructive testing, tensile testing, Compression test, Hardness tests, Impact test.

Unit-II
Basic Metals & Alloys: Properties and Applications
Ferrous Materials: Carbon steels, its classification based on % carbon as low, mild, medium & high carbon steel, its properties & applications.
Wrought iron, Cast iron, Alloy steels: stainless steel, tool steel.
Elementary introduction to Heat- treatment of carbon steels: Annealing, Normalizing, Quenching Tempering & case-hardening.
Non-Ferrous metals & alloys: Common uses of various non-ferrous metals & alloys and its composition such as Cu-alloys: Brass, Bronze, Al-alloys such as Duralumin.

Unit III 
Introduction to Metal Forming & Casting Process and its applications
Metal Forming: Basic metal forming operations & uses of such as: Forging, Rolling, Wire & Tube-drawing/making and Extrusion, and its products/applications, Press-work, die & punch assembly, cutting and forming, its applications, Hot-working versus cold-working.
Casting: Pattern & allowances, Moulding sands and its desirable properties, Mould making with the use of a core, Gating system, Casting defects & remedies, Cupola Furnace, Die-casting and its uses.

Unit-IV 
Introduction to Machining & Welding and its applications
Machining: Basic principles of Lathe-machine and operations performed on it, Basic description of machines and operations of Shaper, Planer, Drilling, and Milling & Grinding.
Welding: Importance & basic concepts of welding, Classification of welding processes. Gas-welding, types of flames, Electric-Arc welding, Resistance welding, Soldering & Brazing and its uses.

Unit-V 
Miscellaneous Topics
Manufacturing: Importance of Materials & Manufacturing towards Technological & Socio- Economic developments, Plant location, Plant layout – its types, Types of Production, Production versus productivity.  Miscellaneous Processes: Powder-metallurgy process & its applications, Plastic-products manufacturing, Galvanizing and Electroplating.

QUESTIONS BANK 2: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

1)      Explain the principle of Super-position.

Ans: The principle of superposition states that “The effect of a force on a body does not change and remains same if we add or subtract any system which is in equilibrium.”

In the fig 4 a, a force P is applied at point A in a beam, where as in the fig 4 b, force P is applied at point A and a force system in equilibrium which is added at point B. Principle of super position says that both will produce the same effect.

2)      What is “Force-Couple system?”

Ans: When a force is required to transfer from a point A to point B, we can transfer the force directly without changing its magnitude and direction but along with the moment of force about point B.

As a result of parallel transfer a system is obtained which is always a combination of a force and a moment or couple. This system consists of a force and a couple at a point is known as Force-Couple system.

      In fig 5 a, a force P acts on a bar at point A, now at point B we introduce a system of forces  in equilibrium (fig 5 b), hence according to principle of superposition there is no change in effect of the original system. Now we can reduce the downward force P at point A and upward force P at point B as a couple of magnitude Pxd at point B (fig 5 c).

3) What do you understand by Equivalent force systems?

Ans: Two different force systems will be equivalent if they can be reduced to the same force-couple system at a given point. So, we can say that two force systems acting on the same rigid body will be equivalent if the sums of forces or resultant and sums of the moments about a point are equal.

4)      What is orthogonal or perpendicular resolution of a force?

Ans: The resolution of a force into two components which are mutually perpendicular to each other along X-axis and Y-axis is called orthogonal resolution of a force.

If a force F acts on an object at an angle θ with the positive X-axis, then its component along X-axis is F_x = Fcosθ, and that along Y-axis is F_y = Fsinθ

5) What is oblique or non-perpendicular resolution of a force?

Ans: When a force is required to be resolved in to two directions which are not perpendiculars to each other the resolution is called oblique or Non-perpendicular resolution of a force.

   
       F_OA = (P sin β)/ sin (α +β)

 F_OB = (P sin α)/ sin (α +β)

QUESTION BANK 1: FORCE AND FORCE SYSTEM

(I am going to publish a question bank for EME-102/EME-202 of 1st yr. MTU; Greater Noida. Some pages from the book .......Subhankar Karmakar)

QUESTION BANK: ENGINEERING MECHANICS

by Er. Subhankar Karmakar

Unit: 1 (Force System)

VERY SHORT QUESTIONS (2 marks):

1)      What is force and force system?

Ans: A force is a physical quantity having magnitude as well as direction. Therefore, it is a   vector quantity. It is defined as an "external agency" which produces or tends to produce or destroys or tends to destroy the motion when applied on a body.

Its unit is Newton (N) in S.I. systems and dyne in C.G.S. system.

When two or more forces act on a body or particle, it is called force system. Therefore, a force system is a collection of two or more forces.

2)      What is static equilibrium? What are the different types of static equilibrium?

Ans: A body is said to be in static equilibrium when there is no change in position as well as no rotation exist on the body. So to be in equilibrium process, there must not be any kind of motions ie there must not be any kind of translational motion as well as rotational motion.

We also know that to have a linear translational motion we need a net force acting on the object towards the direction of motion, again to induce an any kind of rotational motion, a net moment must exists acting on the body. Further it can be said that any kind of complex motion can be resolved into a translational motion coupled with a rotating motion.

“Therefore a body subjected to a force system would be at rest if and only if the net force as well as the net moment on the body is zero.”

There are three types of Static Equilibrium

1.      Stable Equilibrium

2.      Unstable Equilibrium

3.      Neutral Equilibrium

3)      What are the characteristics of a force?

Ans: A force has four (4) basic characteristics.

·         Magnitude: It is the value of the force. It is represented by the length of the arrow that we use to represent a force.

·         Direction: A force always acts along a line, which is called as the “line of action”. The arrow head we used to represent a force is the direction of that force.

·         Nature or Sense: The arrow head also represent the nature of a force. A force may be a pull or a push. If a force acts towards a particle it will be a push and if the force acts away from a point it is pull.

·         Point of Application: It is the original location of a point on a body where the force is acting. 

4)      What are the effects of a force acting on a body?

Whenever a force acts on a body or particle, it may produce some external as well as internal effects or changes.

·         A force may change the state or position of a body by inducing motion of the body. (External effect)

·         A force may change the size or shape of an object when applied on it. It may deform the body thus inducing internal effects on the body.

·         A force may induce rotational motion into a body when applied at a point other than its center of gravity.

·         A force can make a moving body into an equilibrium state at rest.

5)      What is composition and resolution of forces?

Ans: Composition of forces: Composition or compounding is the procedure to find out single resultant force of a force system

Resolution of forces: Resolution is the procedure of splitting up a single force into number of components without changing the effect of the same.

6)      What is Resultant and Equilibrant?

Ans: Resultant: The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

Equilibrant: Any concurrent set of forces, not in equilibrium, can be put into a state of equilibrium by a single force. This force is called the Equilibrant. It is equal in magnitude, opposite in sense and co-linear with the resultant. When this force is added to the force system, the sum of all of the forces is equal to zero.

7)      Explain the principle of Transmissibility?

Ans: The principle of transmissibility states “the point of application of a force can be transmitted anywhere along the line of action, but within the body.”

The fig 3 a shows a force F acting at a point of application A and fig 3 b, the same force F acts along the same line of action but at a different point of action at B and both are equivalent to each other.