Saturday, 27 August 2011

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 Fx = Fcosθ and
component along Y-axis is Fy = 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 Fx = Fcosθ, and that along Y-axis is Fy = Fsinθ.


STEP 2:


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


STEP 3:


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


STEP 4:

DIRECTION OF THE RESULTANT
(α)



α equals to the tan inverse of (ΣFx/ΣFy).


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, ΣFi = 0;
(ii) Net moment on the body must be zero ie, ΣMo= 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) ΣFx= 0; (ii) ΣFy = 0 
and for coplanar non concurrent force system, the equilibrium conditions are as follows
                          (i) ΣFx= 0; (ii) ΣFy = 0; (iii) ΣMi= 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, .....An. Let's take an elemental area (Ai) is at (Xi) distance from Y axis, and at (Yi) from X axis.


Hence, the moment of the area Ai about X-axis is AiYi and about Y-axis the moment will be AiXi. 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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Wednesday, 24 August 2011

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(Xg,Yg).

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 (Xo,Yo). Let the centroid be at G(Xg,Yg), then

Xg = Xo + (Lcos θ)/2
Yg = Yo + (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 (). So we can write

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

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

                             Yg = (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   Xg = (1/L)∫x.dL  ;  here  L = (πR)/2  ;        
                                           X = Rcosθ      
                                          dL = Rdθ
    

             Xg = (2/πR)   0π/2Rcosθ.Rdθ 

     =    (2/πR) R2  0π/2cosθ.dθ

 =      2R/π
   
   Yg = (2/πR)   0π/2Rsinθ.Rdθ 
     
 =      (2/πR) R2  0π/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 L1, L2,  L3 ........ Ln. Let the centroids of these lines are G1(X1,Y1),G2(X2,Y2), G3(X3,Y3) ........ Gn(Xn,Yn).

Calculate length (Li), and coordinates (Xi,Yi) for each and every parts.
 
Now, if the centroid of the composite line be G(Xg,Yg)

Xg = (∑LiXi)/(∑Li


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

Yg = (∑LiYi)/(∑Li)

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

Friday, 19 August 2011

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  (Xg,Yg)

Xg = (∑LiXi)/(∑Li



    = (L1X1 + L2X2 + L3X3)/(L1 + L2 + L3)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Yg = (∑LiYi)/(∑Li



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



Wednesday, 17 August 2011

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, .... An,. Let the coordinates of those tiny elemental areas are as (X1,Y1), (X2,Y2), (X3,Y3) ..... (Xn,Yn).



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, .... An... Let the direction of the resultant vector passes through the point G(Xg,Yg) on the plane of the area. The point G(Xg,Yg) 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 ∑AiXi and about X axis will be ∑AiYi.

Again, the resultant area A passes through the point G(Xg,Yg). Therefore the moment produced by the area A about Y axis will be AXg and about X axis will be AYg.

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,
AXg = A1X1 + A2X2 + A3X3 + ...... + AnXn

and
AYg = A1Y1 + A2Y2 + A3Y3 + ...... + AnYn


           

                 For an area, a centroid G(Xg,Yg) can be defined using calculus by the equations,
Xg = (1/A)x.dA   ------ (i)
Where dA = elemental area and A= total area.
Yg = (1/A)y.dA   ------ (ii)

HOW TO DERIVE THE VALUES OF Xg and Yg FOR BASIC GEOMETRIC FIGURE:

STEPS TO FIND Xg



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) Xg = (∫ 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) Yg = (∫ 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, .... An. Let the centroids are G1(X1,Y1), G2(X2,Y2), G3(X3,Y3), ...... Gn(Xn,Yn).

(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  (Xg,Yg)
Xg = (∑AiXi)/(∑Ai

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

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