COURSE FILE: VITS GHAZIABAD

COURSE FILE: ©subhankar_karmakar

COURSE FILE:

INSTITUTE NAME : VIVEKANAND INSTITUTE OF TECHNOLOGY AND SCIENCE
SUBJECT NAME : ENGINEERING MECHANICS
SUBJECT CODE : EME-102
FACULTY NAME : SUBHANKAR KARMAKAR
DEPARTMENT : MECHANICAL ENGINEERING DEPARTMENT
YEAR  : FIRST YEAR

INDEX:

1 : LESSON PLAN
2 : TIMETABLE
3 : COURSE PLAN
4 : ASSIGNMENT
5 : LECTURE NOTES

LESSON PLAN : EME-102 / 202 : ENGINEERING MECHANICS

UNIT	TOPICS SYNOPSIS	No. of Lectures	DATE
UNIT I	Two Dimensional Force Systems: Basic concepts, Laws of motion, Principle of Transmissibility of forces, Transfer of a force to parallel position , Resultant of a force system, Simplest Resultant of Two dimensional concurrent and Non-concurrent Force systems, Distributed force system, Free body diagrams, Equilibrium and Equations of Equilibrium, Applications.	5	date
UNIT I	Friction: Introduction, Laws of Coulomb Friction, Equilibrium of Bodies involving Dry-friction, Belt friction, Application.	3	date
UNIT II	Beam: Introduction, Shear force and Bending Moment, Differential Equations for Equilibrium, Shear force and Bending Moment Diagrams for Statically Determinate Beams.	5	date
UNIT II	Trusses: Introduction, Simple Truss and Solution of Simple truss, Method f Joints and Method of Sections.	3	date
UNIT III	Centroid and Moment of Inertia: Centroid of plane, curve, area, volume and composite bodies, Moment of inertia of plane area, Parallel Axes Theorem, Perpendicular axes theorems, Principal Moment Inertia, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their Axis of Symmetry.	6	date
UNIT IV	Kinematics of Rigid Body: Introduction, Plane Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity.	4	date
UNIT IV	Kinetics of Rigid Body: Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D’Alembert’s Principles and Dynamic Equilibrium.	4	date
UNIT V	Simple Stress and Strain: Introduction, Normal and Shear stresses, Stress- Strain Diagrams for ductile and brittle material, Elastic Constants, One Dimensional Loading of members of varying cross-sections, Strain energy.	3	date
UNIT V	Pure Bending of Beams: Introduction, Simple Bending Theory, Stress in beams of different cross sections.	3	date
UNIT V	Torsion: Introduction, Torsion of shafts of circular section, torque and twist, shear stress due to torque.	3	date

Text books:

1. Engineering Mechanics by Irving H. Shames, Prentice-Hall

2. Mechanics of Solids by Abdul Mubeen, Pearson Education Asia.

3. Mechanics of Materials by E.P.Popov, Prentice Hall of India Private Limited.

TIME TABLE:

DAY	9.30	10.20	11.10	12.00	LUNCH	1.40	2.30	3.20	4.10
MON	9.30	ME-D	11.10	ME-F	LUNCH	1.40	ME-D	3.20	ME-D
TUE	9.30	ME-D	11.10	12.00	LUNCH	ME-F	2.30	ME-D	4.10
WED	ME-F	10.20	ME-D	12.00	LUNCH	1.40	2.30	3.20	4.10
THU	ME-D	10.20	11.10	ME-F	LUNCH	1.40	2.30	3.20	4.10
FRI	9.30	ME-F	11.10	ME-D	LUNCH	1.40	2.30	3.20	4.10
SAT	ME-D	10.20	ME-F	12.00	LUNCH	ME-F	2.30	3.20	4.10

COURSE PLAN:

SUBJECT NAME: ENGINEERING MECHANICS

SUBJECT CODE : EME-102

SCOPE : The course aims to provide deeper knowledge, a wider scope and improved understanding of the study of motion and the basic principles of mechanics and strength of materials. It is a concept based subject and it needs the application capabilities of the concepts on the part of the students.

SESSIONAL EVALUATION SCHEME:

PARTICULAR	WEIGHTAGE	MARKS
TWO SESSIONALS	60%	30
ATTENDANCE	20%	10
TEACHER'S ASSESSMENT(TA)*	WEIGHTAGE	MARKS

*TA will be based on the Assignments given, Unit test Performances and Attendance in the class for a particular student.

Lecture Schedule of Unit – 1

Total Number of Lectures: 8

• Lecture Details & Synopsis :



• Lecture- 1: Introduction, mass, particle, rigid body, position vector, change of position, velocity, momentum, change of momentum, force acceleration, Newton’s law of motion, conservation of momentum, conservation of energy.



• Lecture- 2: Definition of force, characteristics of force, Force as a vector, Force addition, triangle’s & parallelogram laws of force addition, magnitudes & direction of resultant force, negative force, resolution of force, oblique and orthogonal resolutions, component of a vector along a line, classification of a force system, force system in one dimension, like & unlike forces, two dimensional force system, co-planar force system, non coplanar force system, concurrent force system, coplanar concurrent force system, coplanar parallel force system



• Lecture- 3: The concepts of rigid body, principle of transmissibility of forces, resultant of coplanar concurrent force system, equilibrium of forces, conditions of static equilibrium for concurrent force system, actions & reactions in case of equilibrium in (i) spherical balls in a channel, (ii) blocks of mass in an inclined plane, (iii) reactions in strings, wires & ropes. Types of force (i) tension (ii) compression. Concepts of free body diagrams. Lami’s theorem.



• Lecture- 4: Applications of the conditions of static equilibrium in case of concurrent forces in the analysis of a concurrent force system & numericals based on this. Numericals based on the resultant of a force system. Numericals based on Lami’s theorem.



•Lecture- 5: Normal reactions, concepts of friction, angle of friction, coefficient of friction, angle of repose, laws of coulomb friction, limiting friction, coefficient of static friction & kinematics friction, Equilibrium of bodies involving dry friction. Use of Friction, Friction as a necessary EVIL.




• Lecture- 6: Numericals based on static friction, ladder friction, friction in inclined plane, numericals on ladder friction & friction in inclined plane. Objective type questions in friction.




• Lecture- 7: Theory of Belt Friction, Slack & tight side of a belt, Concepts of Included angle, power delivered by belt drive, Numericals on Belt friction & objective type Questions.



• Lecture- 8: Doubt clearing Sessions on Unit- 1, (Static Equilibrium Analysis, Resultant Forces, Resolution of Forces, Lami’s Theorem, Concepts of Dry & belt Friction.)





•Reference books:
•(i)Engineering Mechanics by Timoshenko & Young
(ii)Engineering Mechanics by R. K. Rajput
(iii) Engineering Mechanics by Irving H. Shames

Lecture Schedule of Unit – 2

Total Number of Lectures: 8

• Lecture Details & Synopsis:



• Lecture- 9: Concepts of Beam, Classification of Beams, simply supported beam, cantilever beam, over hanging beam, continuous beam,Types of Support Reactions, Pin/hinged joints, Roller joints, fixed joints, determination of support reactions in beam, types of loading in beams, concentrated load, distributed load on the beam, uniformly distributed load (UDL), uniformly varying load (UVL), pure moment loading.



• Lecture- 10: Concepts of Shear Force, sign convention for shear force, determination of shear force at each point of the beam over the complete length of the beam, shear force diagrams (SFD), differential equations for equilibrium, concepts of bending moments, sign conventions for bending moments, determination of bending moments at each point of the beam over the complete length of the beam, bending moment diagrams (BMD), maximum bending moment, point of contra-flexure and its importance.



• Lecture- 11: SFD & BMD in case of (i) simply supported beam, (ii) cantilever beam, (iii) over-hanging beam with (a) concentrated loading, (b) uniformly distributed loading, (c) uniformly varying loading.



• Lecture- 12: Numericals on SFD & BMD for all types of beam.



• Lecture- 13: Numericals on SFD & BMD for all types of beam and to find point of contra-flexure.



• Lecture- 14: Concepts of Truss, Linkages, and Joints, Classification of Trusses, Perfect Truss, Deficient Truss, Redundant Truss, Simple Truss, Analysis of a Truss by (i) Method of Joints (ii) Method of Sections.



• Lecture- 15: Numericals on Truss analysis by method of joints & method of Sections.



• Lecture- 16: Numericals on Truss analysis by method of sections.

Lecture Schedule of Unit – 3

Total Number of Lectures: 6

• Lecture Details & Synopsis:



• Lecture- 17: Concepts of geometrical Centroid, Center of Mass & Center of Gravity, Centroid of Plane, Curve, Area, & Volume, determination of centroid of composite bodies.



• Lecture- 18: Numericals on determination of Centroid of composite bodies.



• Lecture- 19: Concepts of Rotation & Moment of Inertia, concepts of area moment of inertia & mass moment of inertia, Determination of moment of Inertia with the help of calculus, Parallel axis theorem & Perpendicular axis theorem of Moment of Inertia.



• Lecture- 20: Concepts of Principal Moment of Inertia, determination of Mass Moment of Inertia of (i) Circular Ring, (ii) Disc, (iii) Cylinder, (iv) Sphere & (v) Cone about their axis of symmetry



• Lecture- 21: Numericals on determination of Moment of Inertia of different objects.



• Lecture- 22: Numericals on determination of M.O.I of different objects.

Lecture Schedule of Unit – 4

Total Number of Lectures: 8

• Lecture Details & Synopsis:



• Lecture- 23: Introduction of rigid body, Motion of Rigid Body, Velocity & Acceleration under Translational Motion, Equation of motion due to gravity, concepts of Relative Velocity, Problems on Projectile Motion.



• Lecture- 24: Concepts of Rotational Motion, Angular Displacement, Angular Velocity, Laws of Motion for Rotation, Concepts of Moment, Torque & Couple, Angular Acceleration, Relations between angular velocity & linear velocity, Relation between angular acceleration & linear acceleration, concepts of centripetal acceleration, concepts of Pseudo Force ie. Centrifugal acceleration.



• Lecture- 25: Motion on Level road, Banking of road & Super elevation of rails, Analysis of Slider-crank mechanism (Four bar mechanism) & numericals on rotational motion.



• Lecture- 26: Numericals on Rotational motion & its application.



• Lecture- 27: Concepts of Force, Newton’s Laws of Motion, Definition of Mass, Gravitational Mass & Inertial Mass, Concepts of Work & Energy, Conservation of Mass Principle, Principle of Conservation of Momentum.



• Lecture- 28: Principle of Conservation of Energy, Work- Energy Theorem, Concepts of Conservative Force & Potential Energy. Collision of two bodies, Elastic & Inelastic Collision, Impulse & Impulsive Force, Impulse & change of Momentum. Power.



• Lecture- 29: Concepts of Dynamic Equilibrium, Inertial Mass & D’ Alembert’s Principle of Dynamic Equilibrium, Motion on an Inclined Plane, Analysis of Lift Motion, Analysis of Motion of Connected Bodies (i) System of Pulleys (ii) Two Bodies connected by a string.



• Lecture- 30: Numericals on Dynamic Equilibrium & System of Pulleys.

Lecture Schedule of Unit – 5

Total Number of Lectures: 10

• Lecture Details & Synopsis:



• Lecture- 31: Deformation of Rigid Bodies under the action of External Force, Resistance against deformation & induction of internal resistive force, Unit deformation & strain, internal force & stress, linear deformation and normal stress, Hooke’s Law & Modulus of Elasticity ( E, Young’s modulus), angular deformation, Shear Strain, & shear Stress, Modulus of Rigidity ( G ), Complimentary Shear Stress.




• Lecture-32: Simple Stress-Strain Diagrams for (i) Ductile Materials, (ii) Brittle Materials, One Dimensional Loading of members of Varying Cross Sections (i) Circular Bar of Uniform Taper, (ii) Bar of Uniform Strength (iii) Bar of (a) Uniform & (b) Taper Cross section due to Self Weight, (iv) Composite Bar, Impact loading (i) Gradually Applied load, (ii) Suddenly Applied Load.



• Lecture- 33: Concepts of Strain Energy & Resilience, Concepts of (i) Longitudinal & (ii) Lateral Strain, Poisson’s Ratio, Hydro-static Compression & Volumetric Strain, Bulk Modulus (K), relation between (i) E, G, & K, (ii) E, K, m (iii) E, G, m. simple numericals on stress & strain.



• Lecture- 34: Numericals on Simple Stress & Simple Strain, Shear Strain & Shear Stress, numericals on composite bars.



• Lecture- 35: Concepts of Pure Bending, Assumptions in simple theory of bending, Concepts of Bending Stress, Neutral Layer & Neutral Axis, Bending Stress Diagrams, Difference between Simple Stress & Bending Stress, Derivation of Bending Equation, Section Modulus (Z), Relation between max. Tensile & max. Compressive Stress,



• Lecture- 36: Stress in Beams of different cross sections, Numericals on Bending Stresses.



• Lecture- 37: Doubt clearance class on (i) stress, strain (ii) pure bending



• Lecture- 38: Introduction of Shaft & Torsion, concept of pure torsion, Polar Moment of Inertia ( J ), Section Modulus (Z), Polar Modulus ( Zp), Assumption for Deriving the Torsional Formulas, Torsional Equation,



• Lecture- 39: Torsional Rigidity or Torsional Stiffness ( K ), Comparison of strength of (i) Solid & (ii) Hollow Circular Shaft (Tmax), Power Transmission by a Shaft, Importance of Angle of Twist, numerical based on Torsion in Shaft.



• Lecture- 40: Doubt clearing classes on Torsion.





• Reference Books:

• Engineering Mechanics by R. S. Khurmi

• Engineering Mechanics by Bhavikatti

• Engineering Mechanics by D. S. Kumar.

• Engineering Mechanics by Timoshenko & Young.

ENGINEERING MECHANICS: ENGINEERING MECHANICS: THEORY OF FRICTION & FRICTIONAL FORCES

HOW TO FIND THE RESULTANT OF A FORCE SYSTEM?

For a force system i.e. a system of several forces acting on an object, it is possible to get the same effect on the object by the force system replacing it by a single force, that will be equivalent to the summation of the component forces acting on the object. The single force that will produce exactly the same effect on the object in stead of the force system is called Resultant of the force system.

We know that two forces acting on an object lying on a plane can be added together by

• (i) Triangle's Law or
• (ii) Parallelogram 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 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.

THE STEPS TO FIND A RESULTANT OF A CON-CURRENT FORCE SYSTEM:

STEP 1:

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

If a force F acts on an object at an angle ß with the positive X-axis, then its component along X-axis is F cosß, and that along Y-axis is F sinß.


STEP 2:

ADD ALL THE X-COMPONENTS OR HORIZONTAL COMPONENTS AND IT IS DENOTED BY ΣF_x AND

ADD ALL THE Y-COMPONENTS OR VERTICAL COMPONENTS AND IT IS DENOTED BY ΣF_y.

STEP 3:

FIND THE MAGNITUDE OF THE RESULTANT R

We know from Geometry that

R = √{(ΣF_x)² + (ΣF_y)²}

STEP 4:

FIND THE DIRECTION (α) OF THE RESULTANT FORCE (R)

We know that

tan α = (ΣF_y/ΣF_x)

hence,

α = tan^-1(ΣF_y/ΣF_x)

TWO DIMENSIONAL FORCE SYSTEM

Q: WHAT DO YOU UNDERSTAND BY THE TERM "FORCE"? WHAT IS THE EFFECT OF FORCE ON A PARTICLE AND A RIGID BODY? EXPLAIN WITH SUITABLE EXAMPLES.

Answer:

FORMAL DEFINITION:

A FORCE is that which can cause an object with mass to ACCELERATE. Force has both MAGNITUDE and DIRECTION, making it a vector quantity. According to Newton's second law, an object with constant mass will accelerate in proportion to the net force acting upon it and in INVERSE PROPORTION TO ITS MASS (M). An equivalent formulation is that the net force on an object is equal to the RATE OF CHANGE OF MOMENTUM it experiences. Forces acting on three-dimensional objects may also cause them to rotate or deform, or result in a change in pressure. The tendency of a force to cause angular acceleration about an axis is called TORQUE. Deformation and pressure are the result of stress forces within an object.


EXPLANATION OF MECHANICAL FORCE AND IT'S EFFECT ON A PARTICLE:

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.

• 2 DIMENSIONAL FORCE

In physics, force is a vector quantity that is used to describe the interaction between two objects. In a two-dimensional system, forces can act in two different directions, which are typically labeled as the x-axis and the y-axis.

When dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object. The direction of the net force is determined by the angle of the resultant force vector.

To calculate the net force in two dimensions, we must first break down each force into its x and y components. The x-component of a force is the amount of force acting in the x-direction, and the y-component is the amount of force acting in the y-direction. Once we have the x and y components for each force, we can add them together to find the net force.

The magnitude of the net force can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the magnitude of the net force, and the other two sides are the x and y components of the net force.

In summary, when dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. To calculate the net force, we must first break down each force into its x and y components and then add them together. The magnitude and direction of the net force can be determined using trigonometry.

• ORTHOGONAL RESOLUTION OF A FORCE

Orthogonal resolution of a force is a technique used in physics to break down a force vector into its components along two orthogonal axes, typically the x and y axes. This technique is useful in analyzing the motion of an object under the influence of a force and can help determine the net force acting on an object.

To perform orthogonal resolution of a force, we first need to identify the angle that the force vector makes with respect to one of the axes, usually the x-axis. We can then use trigonometry to determine the components of the force vector along the x and y axes.

If the angle between the force vector and the x-axis is θ, the x-component of the force can be found using the equation F_x = F cos(θ), where F is the magnitude of the force. Similarly, the y-component of the force can be found using the equation F_y = F sin(θ).

Once we have the x and y components of the force, we can use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object and can be found by adding the x and y components of each force separately.

Orthogonal resolution of a force is a powerful technique that is used in many areas of physics, including mechanics, electromagnetism, and fluid dynamics. By breaking down a force vector into its components, we can better understand the forces acting on an object and predict its motion under different conditions.

WHAT IS A FORCE SYSTEM? CAN WE CLASSIFY FORCE SYSTEMS?

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


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.

on 20th November, 2010: ©subhankar

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.

INTELLIGENT OBJECTIVE QUESTIONS IN MECHANICS

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

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

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

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

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

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

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

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

NEED OF QUALITY ASSURANCE

NEED OF QUALITY ASSURANCE

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

How Can We Define Quality Assurance?

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

Why Do We Need Quality Assurance?

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

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

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

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

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

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


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