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.

CENTROIDS OF LINES

CENTROID OF A STRAIGHT LINE

If we take a straight line of length (L), then its midpoint will be at a distance (L/2) from either end of the line. Let us denote the centroid as the point G(X_g,Y_g).

Suppose we have a straight line AB of length (L) that makes an angle θ with X axis. Let the coordinate of point A is (X_o,Y_o). Let the centroid be at G(X_g,Y_g), then

X_g = X_o + (Lcos θ)/2
Y_g = Y_o + (Lsin θ)/2

                                                                                                                                                             

     Again, suppose the coordinate of B is given instead of point A. Let it is given as B(Xi,Yi). Then, it will be                                                       

Xg = X₁ - (Lcos θ)/2
Yg = Y₁ - (Lsin θ)/2

For Horizontal lines θ = 0° and for Vertical lines θ = 90°

CENTROID OF A CURVED LINE

The steps to derive the centroid of a quarter circular arc of radius R. 

 

Centroid of a curved line can be derived with the help of calculus.

i) Draw the figure in a X-Y coordinate system. Let the curved line has been represented by a function θ

ii) Take an arbitrary point P(X,Y) on the curve. Join the line OP, where O is the origin of the coordinate system. Let OP has a length L and makes an angle (θ) with X axis. Therefore, we can write

                  

                                   X = Rcosθ ----- (a)
                             Y = Rsinθ ----- (b)

iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write

                                   dL = Rdθ ------ (c)

                             X_g = (1/L) ∫(XdL)
                                  = (1/L) ∫ Rcosθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ ------- (d)

                             Y_g = (1/L) ∫ YdL
                                  = (1/L) ∫ Rsinθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ -------- (e)

CENTROID OF A QUARTER CIRCULAR ARC OF RADIUS R

Suppose we have a quarter circular arc in a co-ordinate system as shown in the figure. Total length of the arc AB = (πR)/2 . We take an arbitrarily small length of the arc CD and denote it as dL.

So,

           dL = Rdθ  ------ (iii)    [ as s=Rθ ]

where R = Radius of the quarter circular arc.

Let the co-ordinate of the point D be D(x,y) where

   
               X = Rcosθ -----(iv) and

            Y = Rsinθ -----(v)

Hence   X_g = (1/L)∫x.dL  ;  here  L = (πR)/2  ;        

                                           X = Rcosθ      

                                          dL = Rdθ

    

             X_g = (2/πR)   ₀∫^π/2Rcosθ.Rdθ 

     =    (2/πR) R²  ₀∫^π/2cosθ.dθ

 =      2R/π
   

   Y_g = (2/πR)   ₀∫^π/2Rsinθ.Rdθ 

     

 =      (2/πR) R²  ₀∫^π/2sinθ.dθ

 =      2R/π

Hence, for a quarter circular arc of radius R will be G(2R/π,2R/π)                                                                

                                                                                      

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure.So, a composite line is consisted of several straight or curved lines.


Let a composite line is made of n number of lines, which may straight or curved lines.

STEP-ONE:

Draw the composite line and break it into n number of parts. Let the lengths of these lines are L_1, L_2,  L_{3 ........} L_n. Let the centroids of these lines are G₁(X₁,Y₁),G₂(X₂,Y₂), G₃(X₃,Y₃) ........ G_n(X_n,Y_n).

Calculate length (L_i), and coordinates (X_i,Y_i) for each and every parts.

 

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

X_g = (∑L_iX_i)/(∑L_i) 

    => (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
   

Y_g = (∑L_iY_i)/(∑L_i)

    => (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
   

CENTROID OF AN AREA

 

CENTROID OF AN AREA

Engineering Mechanics EME-102

Geometrical Center of an area (A) is often termed as Centroid or Center of an Area.

Suppose we have an area A in a certain X-Y coordinate system, we divide the area into n parts and named them as A₁, A₂, A₃, .... A_n,. Let the coordinates of those tiny elemental areas are as (X₁,Y₁), (X₂,Y₂), (X₃,Y₃) ..... (X_n,Y_n).

As area can be represented by a vector, hence, Area A can be treated as the resultant of the tiny elemental vectors A₁, A₂, A₃, .... A_n... Let the direction of the resultant vector passes through the point G(X_g,Y_g) on the plane of the area. The point G(X_g,Y_g) is called the CENTROID of the area A. (The direction of any area is along the perpendicular to the area drawn at the centroid of the area).

Like other vectors, an area has a moment about an axis and be represented by the product of the radial distance between the area and the axis and the area itself. So if an elementary area A₁ has a coordinate (X₁,Y₁) it means the area is at a distance X₁ from the Y axis and Y₁ from the X axis. Therefore the moment produced by A1 about Y axis is X₁A₁ and about X axis is Y₁A₁.

 Therefore the summation of all the moments produced by each and every elemental area about Y axis will be ∑A_iX_i and about X axis will be ∑A_iY_i.

Again, the resultant area A passes through the point G(X_g,Y_g). Therefore the moment produced by the area A about Y axis will be AX_g and about X axis will be AY_g.

Like other vectors, it will obey the Moment Theorem which states the total moment produced by individual vectors will be exactly equal to the moment produced by the resultant vector about a certain axis.

Therefore,
AX_g = A₁X₁ + A₂X₂ + A₃X₃ + ...... + A_nX_n

and
AY_g = A₁Y₁ + A₂Y₂ + A₃Y₃ + ...... + A_nY_n

           

                 For an area, a centroid G(X_g,Y_g) can be defined using calculus by the equations,

X_g = (1/A)∫x.dA   ------ (i)

Where dA = elemental area and A= total area.

Y_g = (1/A)∫y.dA   ------ (ii)

HOW TO DERIVE THE VALUES OF X_g and Y_g FOR BASIC GEOMETRIC FIGURE:

STEPS TO FIND X_g


i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dx) parallel to Y axis and at a distance (x) from Y axis.

iii) Find the height of the strip. Either the height will be constant or the height is a function of (x), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = hdx

v) integrate the expression ∫xdA, but dA = hdx. Therefore, we shall integrate  ∫hxdx over the total area.

vi) X_g = (∫ xdA)/A = (∫ hxdx)/A ; where A = total area = ∫dA = ∫hdx


STEPS TO FIND Yg

i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dy) parallel to X axis and at a distance (y) from X axis.

iii) Find the length (b) of the strip. Either the length will be constant or the length is a function of (y), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = bdy

v) integrate the expression ∫ydA, but dA = bdy. Therefore, we shall integrate ∫bydy over the total area.

vi) Y_g = (∫ ydA)/A = (∫bydy)/A





CENTROID OF A COMPOSITE AREA:




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

INTELLIGENT OBJECTIVE QUESTIONS IN MECHANICS

1) A cantilever beam of square cross-section (100 mm X 100 mm) and length 2 m carries a concentrated load of 5 kN at its free end. What is the maximum normal bending stress at its mid-length cross-section?

(a) 10 N/mm²
(b) 20 N/mm²
(c) 30 N/mm²
(d) 40 N/mm²

2) A hollow shaft of outside diameter 40 mm and inside diameter 20 mm is to replaced by a solid shaft of 30 mm diameter. If the maximum shear stresses induced in the two shafts are to be equal, what is the ratio of the maximum resistible torque in the hollow to that of solid shaft?

(a) 10/9
(b) 20/9
(c) 30/9
(d) 40/9

(3) A cannonball is fired from a tower 80 m above the ground with a horizontal velocity of 100 m/s. Determine the horizontal distance at which the ball will hit the ground. (take g=10 m/s²)

(a) 400 m,
(b) 280 m,
(c) 200 m,
(d) 100 m.

(4) Water drops from a tap at the rate of four droplets per second. Determine the vertical separation between two consecutive drops after the lower drop attained a velocity of 4 m/s. Take g=10 m/s².

(a) 0.49 m
(b) 0.31 m
(c) 0.50 m
(d) 0.30 m

NEED OF QUALITY ASSURANCE

NEED OF QUALITY ASSURANCE

Quality assurance can be a confusing realm for those who don't have any prior experience in this field. Many commonly asked questions by first timers include wanting to know exactly what quality assurance is and why they require such a service. Read on to find out the answers.

How Can We Define Quality Assurance?

Quality assurance is the process of ascertaining, through a systematic set of procedures, whether or not a product or service satisfies the customers' requirements. This is the simplest and most basic definition for quality assurance.

Why Do We Need Quality Assurance?

If your company has manufactured a certain product, it is necessary to get that product checked to verify that it conforms to the expectations and requirements of the customer. This process of checking and verifying a product's quality is known as quality assurance.

On the other hand, if you are the customer, you would definitely want to ascertain that the product that you are purchasing satisfies your requirements. If the product fails in some way to meet your expectations, you can provide feedback to the company who will then try to improve their quality standards for that particular product in order to improve their product performance.

Thus quality assurance works both ways, ensuring satisfied manufacturers as well as customers.

The mass industrialization period saw the widespread introduction of mass production and piecework, which created problems as workmen could now earn more money by the production of extra products, which in turn led to bad workmanship being passed on to the assembly lines. To counter bad workmanship, full time inspectors were introduced into the factory to identify, quarantine and ideally correct product quality failures. Quality control by inspection in the 1920s and 1930s led to the growth of quality inspection functions, separately organised from production and big enough to be headed by superintendents.

The systematic approach to quality started in industrial manufacture during the 1930s, mostly in the USA, when some attention was given to the cost of scrap and rework. With the impact of mass production, which was required during the Second World War, it became necessary to introduce a more appropriate form of quality control which can be identified as Statistical Quality Control, or SQC. Some of the initial work for SQC is credited to Walter A. Shewhart of Bell Labs, starting with his famous one-page memorandum of 1924.

SQC came about with the realization that quality cannot be fully inspected into an important batch of items. By extending the inspection phase and making inspection organizations more efficient, it provides inspectors with control tools such as sampling and control charts, even where 100 per cent inspection is not practicable. Standard statistical techniques allow the producer to sample and test a certain proportion of the products for quality to achieve the desired level of confidence in the quality of the entire batch or production run.


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