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.

NEW SYLLABUS FOR ENGINEERING MECHANICS: FIRST YEAR OF MTU FOR 2012-13

ENGINEERING MECHANICS

L T P

3 1 2

UNIT I

Two Dimensional Concurrent Force Systems: Basic concepts, Units, Force systems, Laws of motion, Moment and Couple, Vectors - Vector representation of forces and moments - Vector operations. Principle of Transmissibility of forces, Resultant of a force system, Equilibrium and Equations of equilibrium, Equilibrium conditions, Free body diagrams, Determination of reaction, Resultant of two dimensional concurrent forces, Applications of concurrent forces.                                                                     8

UNIT II
Two Dimensional Non-Concurrent Force Systems: Basic Concept, Varignon’s theorem, Transfer of a Force to Parallel Position, Distributed force system, Types of Supports and their Reactions, Converting force into couple and vise versa.                                                                                                                   3
Friction: Introduction, Laws of Coulomb Friction, Equilibrium of bodies involving dry-friction, Belt friction, Ladder Friction, Screw jack                                                                                                         3
Structure: Plane truss, Perfect and Imperfect Truss, Assumption in the Truss Analysis, Analysis of Perfect Plane Trusses by the Method of Joints, Method of Section.                                                           4


UNIT III
Centroid and Moment of Inertia: Centroid of plane, curve, area, volume and composite bodies,
Moment of inertia of plane area, Parallel Axes Theorem, Perpendicular axes theorems, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their, Axis of Symmetry. Pappus-theorems, Polar moment of inertia.                                                                                                               8

UNIT IV
Kinematics of Rigid Body: Introduction, Plane Rectilinear Motion of Rigid Body, Plane Curvilinear Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity. 8


UNIT (V)
Kinetics of Rigid Body: Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D’Alembert’s Principles and Dynamic Equilibrium, Friction in moving bodies.              8

Text books:
1. Engineering Mechanics Statics , J.L Meriam , Wiley
2. Engineering Mechanics Dynamics , J.L Meriam , Wiley
3. Engineering Mechanics – Statics & Dynamics by A Nelson, McGraw Hill
4. Engineering Mechanics : Statics and Dynamics, R. C. Hibbler
5. Mechanics of Solids by Abdul Mubeen, Pearson Education Asia.
6. Mechanics of Materials by E.P.Popov, Prentice Hall of India Private Limited.



ENGINEERING MECHANICS- LAB

(Any 10 experiments of the following or such experiments suitably designed)

1. Polygon law of Co-planer forces (concurrent)
2. Bell crank lever -Jib crane
3. Support reaction for beam
4. Collision of elastic bodies(Law of conservation of momentum
5. Moment of inertia of fly wheel.
6. Screw fiction by using screw jack
7. To study the slider-crank mechanism etc. of 2-stroke & 4-stroke I.C. Engine models.
8. Friction experiment(s) on inclined plane and/or on screw-jack.
9. Simple & compound gear-train experiment.
10. Worm & worm-wheel experiment for load lifting.
11. Belt-Pulley experiment. .
12. Experiment on Trusses.
13. Statics experiment on equilibrium
14. Dynamics experiment on momentum conservation
15. Dynamics experiment on collision for determining coefficient of restitution.
16. Simple/compound pendulum

Steam Turbine and Power Plants

Steam turbine

A steam turbine is a device that extracts thermal energy from pressurized steam and uses it to do mechanical work on a rotating output shaft. Its modern manifestation was invented by Sir Charles Parsons in 1884.

Because the turbine generates rotary motion, it is particularly suited to be used to drive an electrical generator – about 90% of all electricity generation in the United States (1996) is by use of steam turbines. The steam turbine is a form of heat engine that derives much of its improvement in thermodynamic efficiency through the use of multiple stages in the expansion of the steam, which results in a closer approach to the ideal reversible process.

Back Pressure Steam Turbine

Steam turbines are the prime movers in generating electricity. Back pressure steam turbines are a type of steam turbine that is used in connection with industrial processes where there is a need for low or medium pressure steam.

The high pressure steam enters the back pressure steam turbine and while the steam expands – part of its thermal energy is converted into mechanical energy. The mechanical energy is used to run an electric generator or mechanical equipment, such as pumps, fans, compressors etc.

The outlet steam leaves the back pressure steam turbine at “overpressure” and then the steam returns to the plant for process steam application such as heating or drying purposes.

Steam Turbine Power Plants:

Steam turbine power plants operate on a Rankine cycle. The steam is created by a boiler, where pure water passes through a series of tubes to capture heat from the firebox and then boils under high pressure to become superheated steam. The heat in the firebox is normally provided by burning fossil fuel (e.g. coal, fuel oil or natural gas). However, the heat can also be provided by biomass, solar energy or nuclear fuel. The superheated steam leaving the boiler then enters the steam turbine throttle, where it powers the turbine and connected generator to make electricity. After the steam expands through the turbine, it exits the back end of the turbine, where it is cooled and condensed back to water in the surface condenser. This condensate is then returned to the boiler through high-pressure feedpumps for reuse. Heat from the condensing steam is normally rejected from the condenser to a body of water, such as a river or cooling tower.

Steam turbine plants generally have a history of achieving up to 95% availability and can operate for more than a year between shutdowns for maintenance and inspections. Their unplanned or forced outage rates are typically less than 2% or less than one week per year.

Modern large steam turbine plants (over 500 MW) have efficiencies approaching 40-45%. These plants have installed costs between $800 and$2000/kW, depending on environmental permitting requirements.

Combustion (Gas) Turbines:

Combustion turbine plants operate on the Brayton cycle. They use a compressor to compress the inlet air upstream of a combustion chamber. Then the fuel is introduced and ignited to produce a high temperature, high-pressure gas that enters and expands through the turbine section. The turbine section powers both the generator and compressor. Combustion turbines are also able to burn a wide range of liquid and gaseous fuels from crude oil to natural gas.

The combustion turbine’s energy conversion typically ranges between 25% to 35% efficiency as a simple cycle. The simple cycle efficiency can be increased by installing a recuperator or waste heat boiler onto the turbine’s exhaust. A recuperator captures waste heat in the turbine exhaust stream to preheat the compressor discharge air before it enters the combustion chamber. A waste heat boiler generates steam by capturing heat form the turbine exhaust. These boilers are known as heat recovery steam generators (HRSG). They can provide steam for heating or industrial processes, which is called cogeneration. High-pressure steam from these boilers can also generate power with steam turbines, which is called a combined cycle (steam and combustion turbine operation). Recuperators and HRSGs can increase the combustion turbine’s overall energy cycle efficiency up to 80%.



Combustion (natural gas) turbine development increased in the 1930’s as a means of jet aircraft propulsion. In the early 1980’s, the efficiency and reliability of gas turbines had progressed sufficiently to be widely adopted for stationary power applications. Gas turbines range in size from 30 kW (micro-turbines) to 250 MW (industrial frames). Industrial gas turbines have efficiencies approaching 40% and 60% for simple and combined cycles respectively.

The gas turbine share of the world power generation market has climbed from 20 % to 40 % of capacity additions over the past 20 years with this technology seeing increased use for base load power generation. Much of this growth can be accredited to large (>500 MW) combined cycle power plants that exhibit low capital cost (less than $550/kW) and high thermal efficiency.

STRESS, STRAIN AND YOUNG'S MODULUS

STRESS:

When a material is subjected to an external force, it will either totally comply with that force and be pushed away, like a liquid or powder, or it will set up internal forces to oppose those applied from outside. Solid materials generally act rather like a spring – when stretched or compressed, the internal forces come into play, as is easily seen when the spring is released.

A material subjected to external forces that tend to stretch it is said to be in tension, whereas forces which squeeze the material put it in compression.

An important aspect is not so much the size of the force, as how much force is applied per unit of cross-sectional area. The term ‘stress’, symbol σ (Greek letter sigma), is used for the force per unit area, and has the units of pascals (Pa) with 1Pa being one newton per square metre.

Because the reference area is so large, it is normally necessary to use high multiples such as the megapascal (MPa = 10⁶ Pa) and gigapascal (GPa = 10⁹ Pa). However, when we bear in mind that, in electronics, the area over which forces are applied is generally very much smaller, it is useful to keep in mind that one MPa is equivalent to a force of 1 newton applied on a square millimetre of area.



STRAIN:

A material in tension or compression changes in length, and the change in length compared to the original length is referred to as the ‘strain’, symbol ε (Greek letter epsilon). Since strain is a ratio of two lengths it has no units and is frequently expressed as a percentage: a strain of 0.005 corresponds to a ½% change of the original length.



HOOKE'S LAW:

As you know from a spring, if you gradually stretch it, the force needed increases, but the material springs back to its original shape when the force is released. Materials which react in the same way as a spring are said to be ‘elastic’. Typically if we measure the extension of different forces and plot the graph of this, we will find that the extension is proportional to the force applied. Materials that obey Hooke’s Law exhibit a linear relationship between the strain and the applied stress (Figure 1).

Figure 1: Stress-strain graph for an elastic solid

Many metals follow Hooke’s Law until a certain level of stress has been applied, after which the material will distort more severely. The point at which straight line behaviour ceases is called the limit of proportionality: beyond this the material will not spring back to its original shape, and is said to exhibit some plastic behaviour (Figure 2). The stress at which the material starts to exhibit permanent deformation is called the elastic limit or yield point.

Figure 2: Stress-strain graph for a typical metal

 
As Figure 2 shows, if the stress is increased beyond the yield point the sample will eventually break. The term (ultimate) tensile strength is used for the maximum value of tensile stress that a material can withstand without breaking, and is calculated at the maximum tensile force divided by the original cross-sectional area.

Note that there may be substantial differences between the stress at the yield point and on breaking – for example, one source quotes the ‘ultimate tensile strength’ for AISI304 stainless steel as 505 MPa, and the ‘yield tensile strength’ as 215 MPa. For most engineering purposes, metals are regarded as having failed once they have yielded, and are normally loaded at well below the yield point.

With some materials, including mild steel, the stress/strain graph shows a noticeable dip beyond the elastic limit, where the strain (the effect of the load) increases without any need to increase the load. The material is said to have ‘yielded’, and the point at which this occurs is the yield point. Materials such as aluminium alloys on the other hand don’t show a noticeable yield point, and it is usual to specify a ‘proof’ test. As shown in Figure 3, the 0.2% proof strength is obtained by drawing a line parallel to the straight line part of the graph, but starting at a strain of 0.2%.

Figure 3: Stress-strain graph for an aluminium alloy

YOUNG'S MODULUS:

As you will appreciate from the shapes of Figure 2 and Figure 3, the slope of the stress/strain graph varies with stress, so we generally take only the slope of the initial straight-line portion. The stress/strain ratio is referred to as the modulus of elasticity or Young’s Modulus. The units are those of stress, since strain has no units. Engineering materials frequently have a modulus of the order of 10⁹Pa, which is usually expressed as GPa.