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.

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. 

Private Engineering Colleges in Ghaziabad: Will They Survive?

There are some very good Engineering colleges in and around Ghaziabad. These colleges not only topped the annual ranks of formerly UPTU or its later avatars GBTU and MTU, but during these periods they have curved a niche for themselves.

There are colleges like Ajay Kumar Garg Engineering College or AKGEC, ABES, KIET, RKGIT and IMSEC in Ghaziabad which are doing good in imparting Technical Education and already established a brand name in this arena. They draw fair numbers of students every year but there are other colleges which are practically starving due to the lack of students as well as quality students.

The second rung colleges in Ghaziabad:

All the engineering colleges in Western UP (including NCRs ie. Ghaziabad, Noida and Greater Noida) are affiliated to the Mahamaya Technical University, Noida. There are several good colleges in Ghaziabad like Ideal Institute of Technology in Govindpuram, VIET in Dadri, BBDIT in Meerat Road, Sunderdeep Engineering College in Dasna are as good as the private colleges of Karnataka. Then there are VITS, SGIT, LKEC near Jindal Nagar, SIET and RKGEC in Pilakhuwa.

The last rung colleges are the newly established colleges like Bhagwati Institute of Technology in Masuri, Aryan Institute of Technology, Jindal Nagar, Bhagwant Institute of Technology, MAIT in near Jindal Nagar, Satyam, ICE in Pilakhuwa. The problem they are suffering is the lack of students. Last year many seats remained vacant, even the concerned colleges offered more than 15% in commission, still number of students getting admission was very low.

Last year the scenario was very grim, many colleges were finding tough to pay the salaries to their employees. Moreover, as the number of quality students dwindled over the passage of time, pass rate also plunged dramatically.

Just imagine the predicament of the colleges here, in one side the students of the subsequent batches are coming more dull and blunt where as the syllabus has been being modified every third year and every new syllabus is tougher than its previous versions. So, can you guess the outcome? Yes, rapidly falling over all pass rate and the fall of the ranks of these poor colleges. The cascading effects of these events are the sharp fall of the revenue earned by these colleges which in turn makes them unable to pay good salary to its employees which again becomes the cause of mass exodus of the good teachers to the cash rich colleges of Greater Noida and as a result the survival of these colleges gradually becomes tougher. It's a vicious trap and none of the colleges know how to deal with the situation.

Introduction To the Combustion of Fuels

Combustion:

Principle of Combustion:

Combustion is the conversion of a substance called a fuel into chemical compounds
known as products of combustion by combination with an oxidizer. The combustion
process is an exothermic chemical reaction, i.e., a reaction that releases energy as it
occurs.

Thus combustion may be represented symbolically by:
Fuel + Oxidizer = Products of combustion + Energy

Here the fuel and the oxidizer are reactants, i.e., the substances present before the
reaction takes place. This relation indicates that the reactants produce combustion
products and energy. Either the chemical energy released is transferred to the
surroundings as it is produced, or it remains in the combustion products in the form of
elevated internal energy (temperature), or some combination thereof.

Fuels are evaluated, in part, based on the amount of energy or heat that they
release per unit mass or per mole during combustion of the fuel. Such a quantity is
known as the fuel’s heat of reaction or heating value.

Heats of reaction may be measured in a calorimeter, a device in which chemical
energy release is determined by transferring the released heat to a surrounding fluid.
The amount of heat transferred to the fluid in returning the products of combustion to
their initial temperature yields the heat of reaction.

In combustion processes the oxidizer is usually air but could be pure oxygen, an
oxygen mixture, or a substance involving some other oxidizing element such as
fluorine. Here we will limit our attention to combustion of a fuel with air or pure
oxygen.

Chemical fuels exist in gaseous, liquid, or solid form. Natural gas, gasoline, and
coal, perhaps the most widely used examples of these three forms, are each a complex
mixture of reacting and inert compounds. We will consider each more closely later in
the chapter. First let’s review some important fundamentals of mixtures of gases, such
as those involved in combustion reactions.

Therefore, combustion refers to the rapid oxidation of fuel accompanied by the production of heat, or heat and light. Complete combustion of a fuel is possible only in the presence of an adequate supply of oxygen.

Oxygen (O₂) is one of the most common elements on earth making up 20.9% of our air. Rapid fuel oxidation results in large amounts of heat. Solid or liquid fuels must be changed to a gas before they will burn. Usually heat is required to change liquids or solids into gases. Fuel gases will burn in their normal state if enough air is present.

Most of the 79% of air (that is not oxygen) is nitrogen, with traces of other elements. Nitrogen is considered to be a temperature reducing dilutant that must be present to obtain the oxygen required for combustion.

Nitrogen reduces combustion efficiency by absorbing heat from the combustion of fuels and diluting the flue gases. This reduces the heat available for transfer through the heat exchange surfaces. It also increases the volume of combustion by-products, which then have to travel through the heat exchanger and up the stack faster to allow the introduction of additional fuel air mixture.

This nitrogen also can combine with oxygen (particularly at high flame temperatures) to produce oxides of nitrogen (NO_x), which are toxic pollutants.

Carbon, hydrogen and sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and sulphur dioxide, releasing 8084 kcals, 28922 kcals & 2224 kcals of heat respectively.

Under certain conditions, Carbon may also combine with Oxygen to form Carbon Monoxide, which results in the release of a smaller quantity of heat (2430 kcals/kg of carbon) Carbon burned to CO2 will produce more heat per pound of fuel than when CO or smoke are produced.

C + O₂ → CO₂ + 8084 kCals/kg of Carbon

2C + O₂ → 2 CO + 2430 kCals/kg of Carbon

2H₂ + O₂ → 2H₂O + 28,922 kCals/kg of Hydrogen

S + O₂ → SO₂ + 2,224 kCals/kg of Sulphur

3 T’s of Combustion:

The objective of good combustion is to release all of the heat in the fuel. This is accomplished by controlling the "three T's" of combustion which are

1. Temperature high enough to ignite and maintain ignition of the fuel,
2. Turbulence or intimate mixing of the fuel and oxygen, and
3. Time sufficient for complete combustion.

Commonly used fuels like natural gas and propane generally consist of carbon and hydrogen. Water vapor is a by-product of burning hydrogen. This robs heat from the flue gases, which would otherwise be available for more heat transfer.

Natural gas contains more hydrogen and less carbon per kg than fuel oils and as such produces more water vapor. Consequently, more heat will be carried away by exhaust while firing natural gas.

Too much, or too little fuel with the available combustion air may potentially result in unburned fuel and carbon monoxide generation. A very specific amount of O2 is needed for perfect combustion and some additional (excess) air is required for ensuring complete combustion. However, too much excess air will result in heat and efficiency losses.

Not all of the heat in the fuel are converted to heat and absorbed by the steam generation equipment. Usually all of the hydrogen in the fuel is burned and most boiler fuels, allowable with today's air pollution standards, contain little or no sulfur. So the main challenge in combustion efficiency is directed toward unburned carbon (in the ash or incompletely burned gas), which forms CO instead of CO₂.

                                                                                                                             Subhankar Karmakar

MOCK QUESTION PAPER: APPLIED THERMODYNAMICS (2 units only)

                                                                   Paper Code: EME-402

B.  Tech - ME

(SEM.IV) Sessional Examination, 2011 – 12

Applied Thermodynamics

Time:   3hrs                        Total Marks:  100

    Note:   (1)           Attempt all questions.

         (2)  Be precise in your answer.

SECTION-A:

Q.1: Answer the following questions as per the instructions.           

2X10=30

 (i) What is the importance of feed pump in steam engine?

(ii) What is reversible adiabatic process?

(iii) Explain the term isothermal compressibility?

(iv) What is missing quantity?

(v) What is Work Ratio in Carnot vapour cycle?

(vi) Explain the term “Specific steam consumption.”

(vii) What is thermal efficiency of a steam engine?

(viii) What is indicated power?

(ix) What is mean effective pressure of a steam engine?

(x) What is inversion temperature?

SECTION-B:

Q.2: Answer any three parts of the followings:     

                                                                                                               3X10=30

a) Derive the Tds equations.

b) Derive the expressions of mass discharge of steam through a Nozzle.

c) A single cylinder double acting steam engine is supplied with dry and saturated steam at 11.5 bar and exhaust occur at 1.1 bar. The cut-off occurs at 40% of the stroke. If the stroke equals 1.25 times the cylinder bore and engine develops 60 kW at 90 rev/min. Determine the bore and the stroke of the engine. (Assume hyperbolic expansion and diagram factor of 0.79.)

Also calculate the theoretical steam consumption

d) Dry saturated steam enters a steam nozzle at a pressure of 12 bar and is discharged at a pressure of 1.5 bar. If the dryness factor of the discharged steam is 0.95, what would be the final velocity of the steam? Neglect initial velocity of steam.

If 12% heat drop is lost in friction, find the % reduction in the final velocity.

SECTION C:
Q.3: Answer any two parts of the following: 

                                                                                         5X2=10

a) Explain the term “Joule-Thomson coefficient.”

b) With proper diagrams explain the term nozzle efficiency.

c) Explain the Clausius Clapeyron equation. Also write their field of application.

Q.4: Answer any one part of the following:   

                                                                                           1X10=10

a) Explain the effect of velocity and pressure in the flow of a nozzle. What is a choked flow? Also explain the concept of critical pressure in isentropic flow through nozzle.

b) Steam at a pressure of 20 bar, 250°C expands in a convergent-divergent nozzle up to the exit pressure of 2 bar. Assuming a nozzle efficiency of 0.94 for supersaturated flow up to the throat and nozzle efficiency as 90%, find (i) velocity at throat, (ii) mass flow rate if the throat diameter is 1 cm and (iii) velocity and diameter of the nozzle.

Q.5: Answer any three questions: 

                                                                         3X10=30

a) Derive the Maxwell’s Equations

b) Prove that C_p - C_V = -T(∂V/∂T)_p²(∂p/∂V)_T.

c) Steam at a pressure of 10 bar, dry saturated enters the nozzle when exit pressure is 0.3 bar. The nozzle efficiency for the convergent position is 96% and that of the divergent portion is 92%. The throat diameter for each nozzle is 6 mm. Find the mass flow rate of steam and the exit diameter required.

d) Air enters a nozzle at 5 bar, 350°C and comes out at 0.95 bar. The efficiency of expansion through the nozzle is 92%. If the mass flow rate of air is 1.5 kg/s, determine the exit diameter of the nozzle and velocity of air at exit.

BASIC WELDING TERMS

What is Arc Welding?
Arc welding is a method of joining two pieces of metal into one solid piece. To do this, the heat of an electric arc is concentrated on the edges of two pieces of metal to be joined. The metal melts, while the edges are still molten, additional melted metal is added. This molten mass then cools and solidifies into one solid piece.

Welding Consumables

Stick Electrode A short stick of welding filler metal consisting of a core of bare electrode covered by chemical or metallic materials that provide shielding of the welding arc against the surrounding air. It also completes the electrical circuit, thereby creating the arc. (Also known as SMAW, or Stick Metal Arc Welding.)
MIG Wire  Like a stick electrode, MIG wire completes the electrical circuit creating the arc, but it is continually fed through a welding gun from a spool or drum. MIG wire is a solid, non-coated wire and receives shielding from a mixture of gases. (Process is also known as GMAW, or Gas Metal Arc Welding.)
Cored Wire (Flux-Cored Wire)  Cored wire is similar to MIG wire in that it is spooled filler metal for continuous welding. However, Cored wire is not solid, but contains flux internally (chemical & metallic materials) that provides shielding. Gas is often not required for shielding. (Process is also known as FCAW, or Flux-Cored Arc Welding.)
Submerged Arc  A bare metal wire is used in conjunction with a separate flux. Flux is a granular composition of chemical and metallic materials that shields the arc. The actual point of metal fusion, and the arc, is submerged within the flux. (Process is also known as SAW, or Submerged Arc Welding.)
Stainless Steel Stainless steel electrodes and wire are used for welding applications where corrosion resistance is required. Stainless steel consumables are designed to match the composition of stainless steel base metals.
Hardfacing A stick of electrode or cored wire that is designed not to fuse two pieces of metal together, but to add a layer of surface metal to a work-piece in order to reduce wear. An example of this is the shovel on an excavator.

Welding Equipment

Stick Welders Heating the coated stick electrode and the base metal with an arc creates fusion of metals. An AC and/or DC electrical current is produced by this machine to create the heat needed. An electrode holder handles stick electrodes and a ground clamp completes the circuit.
TIG Welders  A less intense current produces a finer, more aesthetically pleasing weld appearance. A tungsten electrode (non-consumable) is used to carry the arc to the workpiece. Filler metals are sometimes supplied with a separate electrode. Gas is used for shielding. (Process is also known as GTAW, or Gas Tungsten Arc Welding.)
MIG Welders and Multi-Process Welders Constant Voltage and Constant Current welders are used for MIG welding and are a semi-automated process when used in conjunction with a wire feeder. Wire is fed through a gun to the weld-joint as long as the trigger is depressed. This process is easier to operate than stick welding and provides higher productivity levels. CC/CV welders operate similarily to CC (MIG) welders except that they possess multi-process capabilities - meaning that they are capable of performing flux-cored, stick and even TIG processes as well as MIG.
Engine Driven Welders Large stick or multi-process welders are able to operate independent of input power and are powered by a gasoline, diesel, or LPG engine instead. Ideal for construction sites and places where power is unavailable.
Wire Feeder / Welders For MIG welding or Flux-Cored wire welding, wire feeder welders are usually complete and portable welding kits. A small built in wire feeder guides wire through the gun to the piece.
Semiautomatic Wire Feeders For MIG welding or Flux-Cored welding, semiautomatic wire feeders are connected to a welding power source and are used to feed a spool of wire through the welding gun. Wire is only fed when the trigger is depressed. These units are portable.
Automatic Wire Feeders For MIG, Flux-Cored, or submerged arc welding, automatic wire feeders feed a spool of wire at a constant rate to the weld joint. They are usually mounted onto a fixture in a factory/industrial setting and are used in conjunction with a separate power source.
Magnum Guns / Torches MIG welding guns and TIG welding torches are hand-held welding application tools connected to both the wire feeder and power source. They direct the welding wire to the weld joint and control the wire feed with the use of a trigger mechanism.

Cutting

Plasma Cutters A constricted cutting arc is created by this machine, which easily slices through metals. A high velocity jet of ionized gas removes molten material from the application.
Oxyfuel Gas Cutting Oxyfuel gas cutting process involves preheating the base metal to a bright cherry red, then introducing a stream of cutting oxygen which will ignite and burn the metal.

Welding Automation / Robotic Welding

Robotic Welding Systems The combination of a robotic arm, a welding power source and a wire feeder produces welds automatically using various programs, welding fixtures and accessories.
Environmental Systems Also known as fume extraction, these systems are often incorporated into a robotic fixture to remove welding fumes natural to the process from the welding environment. Usually a vacuum unit, they can be portable or mounted onto a wall.

LOADING IN BEAMS

BEAMS & CLASSIFICATION OF BEAMS

BEAM: A beam is a structure generally a horizontal structure on rigid supports and it carries mainly vertical loads. Therefore, beams are a kind of load bearing structures.

Depending upon the types of supports beams can be classified into different catagories.

CANTI-LEVER BEAMS: 

A beam can be at stable equilibrium with a single fixed support at one end and the other end remains free, which is called as the free end while the other end is known as fixed end. This kind of beam is known as Canti lever beam. The fixed joint at the fixed end produces a horizontal, a vertical reactions and a reaction moment at the fixed end.

SIMPLE SUPPORTED BEAM: 

A beam supported as just resting freely on the walls or columns at its both ends is known as simply supported beam.

There will be two vertically upward reactions at the ends of a simply supported beam. A simply supported beam can not resist any horizontal load component.

OVER HANGING BEAM: 

A beam having its end portion or both the end portions extended in the form of a canti-lever beyond the support or supports is called as over hanging beam.

Above those beams are statically determinate. It means that those beams can be analysed applying the conditions of equilibrium. We can determine the values of the unknown reactions.

There are beams which can not be analysed applying the conditions of equilibrium of coplanar forces. These beams are also known as Statically indeterminate structures.

Those types of beams can be classified as,

Fixed beams and Continuous beams.

Fixed Beam: A beam having two fixed joints at the both ends is called fixed beam.

Continuous Beam: The beam which is at rest on more than two supports is called as continuous beam.

What are different types of supports? 

There are four types of supports,

• (i) Simple Supports, 
• (ii) Roller Supports, 
• (iii) Hinged Supports 
• (iv) Fixed Supports.

SOLUTION OF EME-102; CENTROID

HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

 

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.

                       In the figure we have a complex geometrical area composed of three basic geometrical areas. A rectangle, a semi circle and a isosceles triangle. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃.

At first, the composite line is divided into three parts.

Part -1 : The semi-circle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (π/2)x(25)² mm² = 981.74 mm²                  
          X₁ = { 25 -  (4x25)/(3xπ)} mm = 14.39 mm
          Y₁ = 25 mm


Part -2 : The Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)


Area, A₂ = 100 x 50  mm² = 5000 mm²             
          X₂ = 25 + (100/2) = 75 mm
          Y₂ = 25 mm
Part -3 : The Triangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)


Area, A₃ = (1/2) x 50 x 50 mm² = 1250 mm²             
          X₃ = 25 + 50 + 25 = 100 mm
          Y₃ = 50 + (50/3) =  66.67 mm






If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (981.74 x 14.39 + 5000 x 75 + 1250 x 100)/( 981.74 + 5000 + 1250)
    = 71.09
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = (981.74 x 25 + 5000 x 25 + 1250 x 66.67)/ ( 981.74 + 5000 + 1250)
    = 32.20