WHAT IS QUALITY OF A PRODUCT OR SERVICES?

EME-072: QUALITY MANAGEMENT
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|||---- WHAT IS QUALITY?
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Everyone has had experiences of poor quality when dealing with business organizations. These experiences might involve an airline that has lost a passenger’s luggage, a dry cleaner that has left clothes wrinkled or stained, poor course offerings and scheduling at your college, a purchased product that is damaged or broken, or a pizza delivery service that is often late or delivers the wrong order. So, what is the exact definition of Quality.

Although Quality is a vague concept up to some extent, but we can still define it. So, we define "Quality of a Product" as the degree of its excellence and fitness for the purpose.
Although, some of the quality characteristics can be specified in quantitative terms, but no single characteristics can be used to measure the quality of a product on an absolute scale. 

Quality of a product means all those activities which are directed to

  (i) Maintaining and improving such as setting of quality targets,

           (ii) Appraisal of conformance

          (iii) Taking corrective action where any deviation is noticed
          (iv) And planning for improvements in quality.

Quality is a measure of the user satisfaction provided by a product, it includes
            (i) Functional efficiency
           (ii) Appearance
          (iii) Ease of installation and operation
          (iv)  Safety reliability
           (v) Maintainability
          (vi) Running and maintenance cost
         (vii) Continued fault free service/ after-sales service.

There are two elements of quality, namely 

(i) Quality of Design
(ii) Quality of Conformance.

Quality is initially created by the designer in the form of product specifications and manufacturing instructions where as the design provides user satisfaction, the product must be conformed to the design.

Making quality a priority means putting customer needs first. It means meeting and exceeding customer expectations by involving everyone in the organization through an integrated effort. Total quality management (TQM) is an integrated organizational effort designed to improve quality at every level.

So, to be a successful brand a product must possess the best quality. But, how does one build quality into a product?

It is obvious that inspection alone can not build quality into a product unless quality has been designed and manufactured into it.

The quality of a product in a company is determined by the philosophy, commitment, and the quality policy of the top management and the extent to which these policies can be put into actual practice.

TQM is about meeting quality expectations as defined by the customer; this is called customer-defined quality. However, defining quality is not as easy as it may seem, because different people have different ideas of what constitutes high quality. Let’s begin by looking at different ways in which quality can be defined.

► Total quality management (TQM):

"An integrated effort designed to improve quality performance at every level

of the organization."

► Customer-defined Quality:

"The meaning of quality as defined by the customer."

► Conformance to Specifications:

"How well a product or service meets the targets and tolerances determined by its designers."

► Fitness for Use:

"A definition of quality that evaluates how well the product performs for its intended use."

► Value for Price Paid:
"Quality defined in terms of product or service usefulness for the price paid."


Quality Control and User-defined Characteristics of Quality:

The perception of quality is heavily dependent upon the types of processes adopted to maintain the quality of the product during manufacturing and distribution of the product. Those processes are called as Quality Control processes. In modern concept of quality control, mainly TQC or Total Quality Control, Quality Assurance and Quality Management have been termed as "QUALITY CONTROL".

Quality of a product is determined by the combined effects of various departments such as Design, Engineering, Purchase, Production and Inspection.

Quality is perceived differently by different people, but understood by almost everyone. The customer as a user takes the quality of fit, finish, appearance and performance in a manufactured product where as service quality may be evaluated on the basis of the "degree of satisfaction".

As the customer has the final saying about the quality of a product; therefore, the measurable characteristics in a product or service are basically translation of the customer needs.

Once the specifications are developed depending upon the customer satisfaction, next the ways to measure as well as monitor the characteristics should be devised.

This becomes the basis of further improvement or continuous improvement of the product or the service.

The ultimate objective of all the processes is to ensure the customer satisfaction so that they become ready to pay for the product or the service.
 

QUALITY MANAGEMENT : TOTAL QUALITY MANAGEMENT (TQM)

Total Quality Management (TQM) is a management approach that focuses on continuous improvement of the quality of products, services, and processes. The goal of TQM is to enhance customer satisfaction and increase business efficiency by reducing errors, defects, and waste.

TQM involves all employees in the organization, from top management to frontline workers, and requires a commitment to quality from all levels. It is based on a set of principles that include customer focus, continuous improvement, employee involvement, process-centered approach, and data-driven decision making.

The implementation of TQM requires a systematic approach that involves the following steps:

1. Establishing a quality policy and goals
2. Forming a quality improvement team
3. Identifying customer needs and expectations
4. Mapping and analyzing business processes
5. Developing and implementing improvement plans
6. Measuring and evaluating results
7. Recognizing and rewarding success

TQM has been widely adopted by organizations in various industries, including manufacturing, healthcare, and service sectors, as it provides a framework for continuous improvement and ensures customer satisfaction.

1. Establishing a quality policy and goals

Total Quality Control (TQM) is a management approach that emphasizes continuous improvement in all aspects of an organization. Establishing a quality policy and goals is an essential component of TQM, as it provides direction for the organization to achieve its quality objectives.

Here are some steps to establish a quality policy and goals in TQM:

Identify the organization's mission and values: The quality policy and goals should align with the organization's mission and values. Therefore, it's important to clearly define the organization's purpose and beliefs before developing the policy and goals.

Define the quality policy: The quality policy is a statement that communicates the organization's commitment to quality. It should be concise and easily understandable, and should outline the organization's approach to meeting customer needs and expectations.

Develop quality goals: Quality goals are specific, measurable targets that the organization aims to achieve. They should be aligned with the quality policy and the organization's mission and values. Some common quality goals include reducing defects, improving customer satisfaction, and increasing productivity.

Involve employees: To ensure that the quality policy and goals are achievable and relevant, it's important to involve employees in the process. This can be done through brainstorming sessions, focus groups, or surveys. Employees should feel empowered to contribute their ideas and suggestions.

Monitor progress: Once the quality policy and goals are established, it's important to monitor progress towards achieving them. This can be done through regular performance reviews, customer feedback, and quality audits. If progress is not being made, it may be necessary to revise the policy and goals.

Continuously improve: TQM is based on the principle of continuous improvement. Therefore, the quality policy and goals should be reviewed and updated regularly to ensure that they remain relevant and effective.

In summary, establishing a quality policy and goals in TQM requires a thorough understanding of the organization's mission and values, involvement of employees, monitoring progress, and a commitment to continuous improvement.

ENGINEERING MECHANICS LECTURE NOTES:

Topic: Introduction of the Concept of Force.


CHANGE IN POSITION:


To know force well, first we have to understand what do we mean by Change. What does it mean when we say the position of the body has been changed? Whenever we find the state of object becomes different than that of the same object before some time say Δt, then we say that there exists a change in the state of the object. Suppose the change occurs in the position of the body. But to find the initial position of a body, we need a co-ordinate system.


THE CAUSE OF CHANGE:


It has been seen that to induced a change or to make a change in the position of an object we must have to change the energy possess by the body. To transfer energy into the object we shall have to apply FORCE on the body. Therefore Force is the agency that makes a change in position of a body.


THE CONCLUSION: GALILEO'S LAW OF INERTIA OR NEWTON'S FIRST LAW OF MOTION.


So, if there is no force on an object the position of the object won't change with respect to time. It means if a body at rest would remain at rest and a body at uniform motion would remain in a steady motion. This law is known as Galileo's Law of Inertia or Newton's first law of motion.






Topic: FORCE SYSTEM


Q: WHAT IS A FORCE SYSTEM? CLASSIFY THEM WITH EXAMPLES.


ANSWER:                                                                                                            


                                           A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces.


Different types of force system:


(i) COPLANAR FORCES:

 If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.




(ii) CONCURRENT FORCES:

 A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. Con-current forces may or may not be coplanar.






(iii) LIKE FORCES:

  A parallel coplanar force system consists of two or more forces whose lines of action are all parallel to one another. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.




(iv) UNLIKE FORCES:

 If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".
















For more notes on force system click here


(v) NON COPLANAR FORCES:

The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.














Q.) WHAT IS FORCE ADDITION? WHAT ARE THE OPERATIONS OF FORCE?


A.) Force is a vector quantity. It has magnitude and as well as direction. Like other vectors two forces can be added, or a force can be substituted from another force, or may be a force can be multiplied by scalars as well as another vector. Unlike scalar quantities, two vector can't be added arithmatically, they must be geometrically added. Suppose we have a force 10 kN acting on a particle towards east, and suppose another force of 10 kN is acting towards north. We know that 10 kg mass +10 kg mass = 20 kg mass, but here forces of 10 kN towards east and 10 kN towards north, when added produces a resultant of magnitude =10*sqrt(2)=14.14 kN.


To add two forces acting on a plane we use
          (i) Triangle's Law and
         (ii) Paralellogram Law.


In case of more than two forces exist, then we use force resolution method to find the resultant.

click on the question to get the answer


QUESTION: WHAT IS A RESULTANT OF A FORCE SYSTEM?


ANSWER: We have already discussed about addition of two forces on a plane by either
                          (i) Triangle's Law or
                         (ii) Paralellogram Law.


For more than two vectors we use


                        (iii) Polygon Law of Force Addition.
                        (iv) Force Resolution Method.


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 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. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. One can progressively resolve pairs or small groups of forces into resultants. Then another resultant of the resultants can be found and so on until all of the forces have been combined into one force. This is one way to save time with the tedious "bookkeeping" involved with a large number of individual forces. Resultants can be determined both graphically and algebraically.


The Parallelogram Method and the Triangle Method are used to find the resultant of a force system. It is important to note that for any given system of forces, there is only one resultant.






Q: WHAT IS A COMPONENT OF A FORCE? WHAT IS FORCE RESOLUTION?


ANSWER: It is often convenient to decompose a single force into two distinct forces. These forces, when acting together, have the same external effect on a body as the original force. They are known as components. Finding the components of a force can be viewed as the converse of finding a resultant. There are an infinite number of components to any single force. And, the correct choice of the pair to represent a force depends upon the most convenient geometry. For simplicity, the most convenient is often the coordinate axis of a structure.


A force can be represented as a pair of components that correspond with the X and Y axis. These are known as the rectangular components of a force. Rectangular components can be thought of as the two sides of a right angle which are at ninety degrees to each other. The resultant of these components is the hypotenuse of the triangle. The rectangular components for any force can be found with trigonometrical relationships:


component of a force F along X-axis is F_x = Fcosθ and
component along Y-axis is F_y = Fsinθ.


numericals


EXTRA NOTE: When forces are being represented as vectors, it is important to should show a clear distinction between a resultant and its components. The resultant could be shown with color or as a dashed line and the components as solid lines, or vice versa. NEVER represent the resultant in the same graphic way as its components.


The Steps to find a Resultant of a force system: (for con-current forces)


STEP 1:


RESOLVE ALL THE COMPONENT FORCES ALONG X-AXIS AND Y-AXIS.


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.


STEP 2:


Add all the X-components or Horizontal components and it is denoted by ΣF_x . Add all the Y-components and denote it as ΣF_y .


STEP 3:


MAGNITUDE OF THE RESULTANT R will be equal to the square root of the sum of square of  ΣF_x and  ΣF_y .


STEP 4:

DIRECTION OF THE RESULTANT (α)


α equals to the tan inverse of (ΣF_x/ΣF_y).


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. A non-concurrent or a parallel force system can actually be in equilibrium with respect to all of the forces, but not be in equilibrium with respect to moments.


[EXTRA NOTE: Graphic Statics and graphical methods of force resolution were developed before the turn of the century by Karl Culmann. They were the only methods of structural analysis for many years. These methods can help to develop an intuitive understanding of the action of the forces. Today, the Algebraic Method is considered to be more applicable to structural design. Despite this, graphical methods are a very easy way to get a quick answer for a structural design problem and can aid in the determination of structural form.]


QUESTION: WHAT IS STATIC EQUILIBRIUM?
                    What are the conditions of static equilibrium for
                       (i) con-current force system
                      (ìi) coplanar non concurrent force system?


Ans: A body is said to be in 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 be zero. Therefore the general condition of any system to be in static equilibrium we have to satisfy two conditions


(i) Net force on the body must be zero ie, ΣF_i = 0;
(ii) Net moment on the body must be zero ie, ΣM_o= 0.


Now we can apply these general conditions to different types of Force System.


So, for Concurrent force system, the equilibrium conditions are as follows
                          (i) ΣF_x= 0; (ii) ΣF_y = 0 
and for coplanar non concurrent force system, the equilibrium conditions are as follows
                          (i) ΣF_x= 0; (ii) ΣF_y = 0; (iii) ΣM_i= 0


MOMENT ON A PLANE:


For a force system the total resultant moment about any arbitrary point due to the individual forces are equal to the moment produced by the resultant about the same point. Now if the system is at equilibrium condition, then the resultant force would be zero. Hence, the moment produced by the resultant about any arbitrary point is zero. In case of coplanar & concurrent force system, as the forces are concurrent i.e. each of the force passes through a common point, hence moments produce by these three forces about the con-current point would be zero. But in case of non concurrent forces the total moments would be zero only when the body is in equilibrium.


What is Momentum of a Mass?


Momentum is a physical quantity. It is associated with motion. In fact any object in motion is said to have a momentum. A mass at a static equilibrium does have a zero value. As it has a direction as well as magnitude, it is a vector quantity. The magnitude of momentum of an object is equal to the mass of the object multiplied by the magnitude of the object's velocity. It is denoted by (P). Hence, P = mV
where m = mass of the object and V = velocity of the object. Therefore, the direction of momentum will be the direction of velocity itself.


3) WHAT ARE DIFFERENT TYPES OF JOINTS? Discuss them in details.


Answer: The Concepts of Joints. In Engineering terminology any force carrying linear member is called as links. Links can be attached to each other by the fasteners or joints. Hence, we can say to prevent the relative motion between two links completely or partially we use fasteners or joints.


Basically there are three types of joints which we shall discuss and they are named as,
             (i) pin/ hinged joints, (ii) roller joints and (iii) fixed joints.


They are classified according to the degrees of freedom of the links they would allow. Like a pin or hinge joint is consisted of two links joined by the insertion of a pin at the pivot hole. A pin joint doesn't allow vertical or horizontal relative velocities between the two links.


For better understanding of the mechanism of pin joint we would like to make a simplest type of pin joints. Suppose we would take two links and make holes at one of the ends of each link. Now if we insert a bolt through the holes of both the links, then what we get is an example of pin/hinge joints.


A pin joint although restricts any kind of horizontal or vertical displacement but they can not restrict rotation about an axis passing through the hole, in clockwise or anti clockwise direction. Hence it provides two reactions one vertical and one horizontal to restrict any kind of movement along that direction.


4) WHAT IS MOMENT? Differentiate between moment, torque and couple. Also state and prove VARIGNON's theorem of moments.


Whenever we apply force on a rigid body at a point other than its Center of Mass, the body exhibit a rotational motion about the center of mass other than a translational motion. Where as the translational motion is there due to the application of force on the object, the rotational motion is there due to eccentric application of force at a point away from the center of mass. But if we apply force at the center of the mass, then no rotational tendency has been observed. It is also observed that the longer is the distance between the center of mass and the point of application of force, larger is the magnitude of rotation. The physical quantity that is responsible for motion is termed as Moment. It is denoted by M and M is a vector quantity as it has direction, either clockwise or anti-clockwise. So, if a force F act at a point d distance from the center of mass, then the total moment produced about the center of mass is the vector cross product of the position vector of the force, and the force itself.
          
                                 M=d X F ------ ()


Therefore, same force would produce different magnitudes of revolution rate and as rate of revolution depends upon Moment, it should produce different magnitudes of Moments. The axis about which a force tries to rotate an object, is called center of rotation as well as it is the axis of moment also. To find the moment produced by a force about a point (O) in the plane, we need to multiply the magnitude of the force and the perpendicular distance of the line of action of the force from the point (O).


WHAT IS CENTROID? EXPLAIN IT IN YOUR WORDS.


The word Centroid is used to denote the center of a certain area. But, then one may question, what does it mean when we say a particular point, say G is the centroid of a specific area. We shall explain the concepts using figures to make the point crystal clear!
Suppose in a 3D coordinate system we take a lamina on X-Y plane.


Let the total area A be divided into (n) numbers of parts and denote them as A₁, A₂, A₃, .....A_n. Let's take an elemental area (A_i) is at (X_i) distance from Y axis, and at (Y_i) from X axis.


Hence, the moment of the area A_i about X-axis is A_iY_i and about Y-axis the moment will be A_iX_i. So, the total moment (M) of the total area (A) will be the total sum of these tiny moments. Now, we shall introduce an abstract idea that if all the area (A) is concentrated at a point (P) whose coordinates are (x,y) so that it produce the same effect on the surroundings here the effects are moments about X-axis and Y-axis.




5) What are beams? Classify them properly. What is support reactions? 


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 categories. 

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. 

SIMPLY 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. 








6) State coulomb's law of dry friction. Explain the following terms
                  (i) coefficients of static and dynamic friction
                 (ii) angle of friction
                (iii) angle of repose
                (iv) limiting friction
 


CONCEPTS OF FRICTION ......:

Whenever two bodies are in contact with each other, they exert a force R to each other. The force R is called as Contact Reaction. There exists a natural phenomenon associated with two bodies in contact. It has been seen that when the two bodies are in static condition relative to each other, everything remains as normal phenomenon as there exist two equal and opposite normal reaction will act at the contact surface. But, whenever we try to make a relative motion between those two bodies, an amazing thing occurs, a "phantom force" suddenly pops up between the bodies acting on the contact surface whose sole purpose of creation is to oppose any relative motion between those two objects. Although the exact reasons behind the generation of this force is not known, but there exists two separate models of origin of friction, none of them are confirmed, although both of them are used to explain the most possible reason behind the sudden generation of this opposite force or we may even call it Resistance Force. But, one thing is surely confirmed from the standard model of particle physics and that is between four types of fundamental forces, only electro-magnetic force is responsible for the generation of frictional force.




7)   (i) What is Shear Force and Bending Moment? 

    (ii) What do you understand by SFD and BMD? 
 
Ans: Vertical forces that act on a horizontal beam is mainly termed as Shear Force. Where as moments produced by those forces on the beam is
state the working procedure to draw them.


8) What is free body diagram(FBD)? Explain with an example.

9) What is point of contraflexure?

10) What is the difference between truss and frames?
     Explain different types of truss with proper illustration.


11) What is simple stress and strain? Compare the stress strain graphs for ductile and brittle materials.


12) What is strain energy and resilience.explain impact loading? Prove that for the same loading, stress induced due to impact loading is twice of the stress induced due to gradually applied load.

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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 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. If the coordinate of point A is (5,5), find the centroid of the composite line.


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








Part -1 : The line AB : Let the centroid of the line be G₁(X₁,Y₁)

length, L₁ = 40 mm;                  


X₁ = 4 + (40*cos 60⁰)/2 = 14  
Y₁ = 4 + (40*sin 60⁰)/2 = 21.32








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


length, L₂ = 15 mm; 


X₂ = 4 + (40*cos 60⁰) + 15/2 = 31.5 
Y₂ = 4 + (40*sin 60⁰) = 38.64




Part -3 : The line CD : Let the centroid of the line be G₃(X₃,Y₃)


length, L₃ = 20 mm; 

X₃ = 4 + (40*cos 60⁰) + 15 = 39 
Y₃ = 4 + (40*sin 60⁰) - 20/2 = 28.64



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)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Y_g = (∑L_iY_i)/(∑L_i) 

    = (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
    = (40 x 21.32 + 15 x 38.64 + 20 x 28.64)/(40 + 15 + 20) 
    = 26.74

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