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 COMPLEX GEOMETRIC FIGURES:

So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a₁, a₂, a₃, ..... a_n.

Let the elemental areas are at a distance x₁, x₂, x₃, ..... x_n, from Y axis and y₁, y₂, y₃, ...y_n from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 

Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (X_g,Y_g)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of X_gA about Y axis and Y_gA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a₁x₁+ a₂x₂+ + +a_nx_n) = AX_g
we can use summation sign ∑ to represent these equations,
∑a_ix_i = (∑a_i)X_g
=> X_g = (∑a_ix_i)/((∑a_i)

Sum(a₁y₁+ a₂y₂+ + +a_ny_n) = AY_g
∑a_iy_i = (∑a_i)Y_g
=> Y_g = (∑a_iy_i)/((∑a_i)

Algorithm to find out the Centroid G(X_g, Y_g) of a Complex Geometric Figure.

Step1:
Take a complex 2D figure like an Area or Lamina.

Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.

Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.

Step4:
Compute the area (a_i), coordinates of their own centroid G_i (x_i, y_i) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.

Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.

Step6:
If the Centroid of the complex figure be G(X_g,Y_g)then,

=> X_g = (∑a_ix_i)/((∑a_i)

=> Y_g = (∑a_iy_i)/((∑a_i)

Here G₁ is the centroid of the part one where G₂ is the centroid of the circular area that has to be removed where as G₃ is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.

ENGINEERING. MECHANICS:

FORCE AND FORCE SYSTEM

Topic: FORCE SYSTEM

1) What is a FORCE SYSTEM? Classify them with examples and diagrams.

Ans: 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. concurrent 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. 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".


(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.

_________________________________________________________________________________
N.B. 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. It is important to note that for any given system of forces, there is only one resultant.


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: F_x = Fcosθ, Fy = Fsinθ. There are a few geometric relationships that seem to common in general building practice in North America. These relationships relate to roof pitches, stair pitches, and common slopes or relationships between truss members. Some of these are triangles with sides of ratios of 3-4-5, 1-2-sqrt3, 1-1-sqrt2, 5-12-13 or 8-15-17. Committing the first three to memory will simplify the determination of vector magnitudes when resolving more difficult problems.


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.


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.
__________________________________________________________________________________


2) What is STATIC EQUILIBRIUM? 
    What are the conditions of static equilibrium for
            (i) concurrent force system
            (ii) 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_i = 0.

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

For concurrent force system total moment about the concurrent point is always zero as all the forces pass through the point, and we know the moment of a force passing through the point about which we shall take moment is always zero. Hence, the conditions of equilibrium for concurrent forces will be  

Net force on the body must be zero ie, ΣF_i = 0; and we can resolve it along X axis and along Y axis, ie.  (i) ΣF_x = 0; and  (ii) ΣF_y = 0.

for coplanar non concurrent force system, the equilibrium conditions are

(i) ΣF_x = 0; and  (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 ie. each of the force passes through a common point. Hence, about that common point total moment of all the forces will be zero.

3) What are different types of joint? 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.


PIN 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 a 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.

ROLLER JOINTS:


 

MULTIPLE CHOICE QUESTIONS: sub: engg. mechanics. Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech

Q.1) The example of Statically indeterminate structures are,
a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.2) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3, 

c. m > 2j - 3,
d. m > 2j + 3

Q.3) The property of a material to withstand a sudden impact or shock is called,
a. hardness 
b. ductility, 
c. toughness, 
d. elasticity of the material

Q.4) The stress generated by a dynamic loading is approximately _____ times of the stress developed by the gradually applying the same load.

Q.5) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus

Q.6) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above

Q.7) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces

Q.8) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body

Q.9) The variation of shear force due to a triangular load on simply supported beam is
a. uniform 
b. linear 
c. parabolic 
d. cubic

Q.10) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be

a. 1/√3

b. √3

c. 1

d. 0

11. A force F of 10 N is applied on a mass of 2 kg. What is the acceleration of the mass?

A. 2 m/s²

B. 5 m/s²

C. 10 m/s²

D. 20 m/s²

Answer: B

12. What is the moment of a force of 50 N applied at a distance of 2 meters from a fixed point?

A. 25 Nm

B. 50 Nm

C. 100 Nm

D. 200 Nm

Answer: C

13. A 2000 kg car traveling at 20 m/s collides with a 500 kg car traveling at 10 m/s in the opposite direction. What is the velocity of the cars after the collision?

A. 6.7 m/s

B. 10 m/s

C. 13.3 m/s

D. 16.7 m/s

Answer: A

14. A 500 N force is applied to a 100 kg object on a flat surface. What is the coefficient of static friction if the object is just about to move?

A. 0.5

B. 0.7

C. 0.8

D. 1.0

Answer: D

15. A beam of length 4 m and moment of inertia of 1000 kg/m² is supported at each end. What is the maximum load that the beam can support if it is uniformly loaded?

A. 500 N

B. 1000 N

C. 2000 N

D. 4000 N

Answer: C

16. A block of mass 2 kg is hanging from a string. What is the tension in the string if the block is stationary?

A. 19.6 N

B. 20 N

C. 29.4 N

D. 30 N

Answer: B

17. A roller coaster car of mass 500 kg is traveling at 20 m/s at the bottom of a  loop-the-loop. What is the minimum radius of the loop required for the car to remain in contact with the track?

A. 40 m

B. 50 m

C. 60 m

D. 70 m

Answer: D

18. A body of mass 10 kg is moving with a velocity of 5 m/s. What is the kinetic energy of the body?

A. 50 J

B. 100 J

C. 125 J

D. 250 J

Answer: B

19. A body of mass 5 kg is placed on an inclined plane which makes an angle of 30° with the horizontal. What is the force acting on the body parallel to the plane?

A. 4.9 N

B. 7.5 N

C. 8.7 N

D. 10 N

Answer: B

20. A force of 100 N is applied on a body of mass 20 kg. What is the work done by the force in moving the body through a distance of 5 meters?

A. 250 J

B. 500 J

C. 1000 J

D. 2000 J

Answer: B

21. What is the principle of moments?

A. The sum of the moments about any point of a system in equilibrium is zero.

B. The sum of the forces acting on a system in equilibrium is zero.

C. The sum of the torques acting on a system in equilibrium is zero.

D. The sum of the accelerations of a system in equilibrium is zero.

Answer: A

22. What is the difference between static and dynamic equilibrium?

A. In static equilibrium, there is no motion, while in dynamic equilibrium, there is motion.

B. In static equilibrium, the forces are balanced, while in dynamic equilibrium, the forces are unbalanced.

C. In static equilibrium, the sum of the forces and moments is zero, while in dynamic equilibrium, the sum of the forces and moments is not zero.

D. In static equilibrium, the sum of the forces and moments is not zero, while in dynamic equilibrium, the sum of the forces and moments is zero.

Answer: C

23. What is the moment of inertia?

A. The resistance of an object to angular acceleration.

B. The force required to rotate an object.

C. The distance between the center of mass and the axis of rotation.

D. The angular velocity of an object.

Answer: A

24.What is the difference between stress and strain?

A. Stress is the deformation per unit length, while strain is the force per unit area.

B. Stress is the force per unit area, while strain is the deformation per unit length.

C. Stress is the force applied to an object, while strain is the resulting deformation.

D. Stress is the resistance of an object to deformation, while strain is the resistance of an object to stress.

Answer: B

25. What is Hooke's Law?

A. The stress applied to an elastic material is proportional to the strain produced.

B. The strain produced in an elastic material is proportional to the stress applied.

C. The deformation produced in an elastic material is proportional to the force applied.

D. The force applied to an elastic material is proportional to the deformation produced.

Answer: A

26.What is the difference between a beam and a truss?

A. A beam is a one-dimensional structure, while a truss is a two-dimensional structure.

B. A beam is made up of several members connected at their ends, while a truss is made up of several members connected at their joints.

C. A beam is used to support loads that are perpendicular to its axis, while a truss is used to support loads that are parallel to its axis.

D. A beam is a rigid structure, while a truss is a flexible structure.

Answer: B

27. What is the difference between a force and a moment?

A. A force is a vector quantity, while a moment is a scalar quantity.

B. A force is a scalar quantity, while a moment is a vector quantity.

C. A force is a push or a pull, while a moment is a twist or a turn.

D. A force is a linear motion, while a moment is a rotational motion.

Answer: C

28. What is the center of mass?

A. The point where the weight of an object is concentrated.

B. The point where the forces acting on an object are balanced.

C. The point where the moments acting on an object are balanced.

D. The point where the acceleration of an object is zero.

Answer: A

29. What is the method used to determine the forces in a truss?

A. Method of joints

B. Method of sections

C. Both A and B

D. None of the above

Answer: C

30. In a truss, which members are in tension and which members are in compression?

A. All members are in tension.

B. All members are in compression.

C. Members with angled force vectors are in tension, and members with vertical force vectors are in compression.

D. Members with vertical force vectors are in tension, and members with angled force vectors are in compression.

Answer: C

31. What is the difference between a simple truss and a compound truss?

A. A simple truss is made up of one triangle, while a compound truss is made up of two or more triangles.

B. A simple truss is made up of straight members only, while a compound truss may have curved members.

C. A simple truss is statically determinate, while a compound truss may be statically indeterminate.

D. A simple truss is used for short spans, while a compound truss is used for long spans.

Answer: A

32.How many unknown forces are there in a simple truss?

A. 2

B. 3

C. 4

D. It depends on the number of joints in the truss.

Answer: B

33. What is the method used to analyze a truss with multiple loadings?

A. Superposition method

B. Substitution method

C. Iterative method

D. None of the above

Answer: A

34. What is the maximum number of reactions that can be present in a truss?

A. 1

B. 2

C. 3

D. 4

Answer: B

35. What is the difference between a statically determinate and a statically indeterminate truss?

A. A statically determinate truss has only one solution for the unknown forces, while a statically indeterminate truss may have more than one solution.

B. A statically determinate truss has more unknown forces than the number of equations available to solve them, while a statically indeterminate truss has fewer unknown forces than the number of equations available to solve them.

C. A statically determinate truss is easier to analyze, while a statically indeterminate truss requires more advanced techniques.

D. A statically determinate truss is always more efficient than a statically indeterminate truss.

Answer: C

36. What is the difference between a pinned support and a roller support?

A. A pinned support allows rotation but not translation, while a roller support allows translation but not rotation.

B. A pinned support allows both rotation and translation, while a roller support allows neither.

C. A pinned support is used for horizontal loads, while a roller support is used for vertical loads.

D. A pinned support is always more stable than a roller support.

Answer: A

37. What is the maximum number of members that can be present in a simple truss?

A. 2n-2, where n is the number of joints

B. 2n-3, where n is the number of joints

C. n-1, where n is the number of joints

D. n+1, where n is the number of joints

Answer: B

©subhankar_karmakar

more OBJECTIVE QUESTIONS on ENGINEERING MECHANICS 

MULTIPLE CHOICE QUESTIONS: ENGINEERING MECHANICS

Choose the correct answer:

Q.1) In a simply supported beam of length L, a UDL of w kN/m acts on the entire span of the beam. The maximum bending moment will be
a. (w.L²)/8 b. (w.L³)/8 c. (w.L²)/4 d. (w.L³)/4

Q.2) If two forces are acting on a particle and the particle is in a stable equilibrium, then the forces are,
a. equal to each other
b. equal but opposite in direction
c. they are unequal but the direction is same.
d. none of the above.

Q.3) The example of Statically indeterminate structures are,
a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.4) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3,
c. m > 2j - 3,
d. m > 2j + 3

Q.5) The property of a material to withstand a sudden impact or shock is called,
a. hardness b. ductility,
c. toughness, d. elasticity
of the material.

Q.6) The stress genarated by a dynamic loading is approximately _____ times of the stress developed by the gradually applying the same load.

Q.7) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus.

Q.8) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above.

Q.9) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces.

Q.10) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body.

Q.11) The variation of shear force due to a triangular load on simply supported beam is
a. uniform b. linear
c. parabolic d. cubic.

Q.12) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be
a. (1/3)^½ b. 3^½ c. 1 d. 0