Engineering Mechanics

ENGINEERING MECHANICS

(In S.I. Units)

For

B.E./B. Tech. 1st YEAR

[ANNA UNIVERSITY, TAMILNADU]

By Dr. R.K. BANSAL B.Sc. Engg. (Mech.), M. Tech., Hons. (I.I.T., Delhi), Ph.D., M.I.E. (India) Formerly, Professor in Mechanical Engineering Department of Mechanical Engineering Delhi College of Engineering Delhi Presently, Dean (Academics) Northern India Engineering College, New Delhi AND SANJAY BANSAL B.E. (Computer)

LAXMI PUBLICATIONS (P) LTD

BANGALORE N CHENNAI N COCHIN N GUWAHATI N HYDERABAD JALANDHAR N KOLKATA N LUCKNOW N MUMBAI N PATNA RANCHI N NEW DELHI Compiled by : Smt. Nirmal Bansal

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Price : ` 350.00 Only. First Edition : 2005 ; Second Edition : 2007 ; Reprint : 2008, 2009, 2010, 2011 ; Third Edition : 2012

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EEM-0638-350-ENGG MECHANICS (TN)–BAN C—2179/010/09 Typeset at : Excellent Graphics, Delhi. Printed at : Ajit Printers Contents

Chapters Pages 1. Basics and Statics of Particles 1—95 1.1. Introduction ... 1 1.2. Units and Dimensions ... 5 1.3. Laws of Mechanics ... 14 1.4. Lame’s Theorem ... 17 1.5. Parallelogram and Triangular Law of Forces ... 17 1.6. Vectors ... 17 1.7. Vector Operations ... 20 1.8. Resolution and Composition of a Force ... 25 1.9. Coplanar Forces ... 31 1.10. Resultant of Coplanar Forces ... 32 1.11. Equilibrium of a Particle ... 57 1.12. Equilibrium of a Rigid Body ... 58 1.13. Forces in Space ... 69 1.14. Equilibrium of a Particle in Space ... 85 1.15. Equivalent Systems of Forces ... 86 1.16. Principle of Transmissibility ... 88 1.17. Single Equivalent Force ... 88 Highlights ... 88 Exercise-1 ... 90 2. Equilibrium of Rigid Bodies 96—180 2.1. Introduction ... 96 2.2. Free Body Diagram ... 97 2.3. Types of Supports and Their Reactions ... 114 2.4. Requirements of Stable Equilibrium ... 115 2.5. Moments and Couples ... 116 2.6. Moment of a Force about a Point and about an Axis ... 116 2.7. Vectorial Representation of Moments and Couples ... 117 2.8. Varignon’s Theorem (or Principle of Moments) ... 119 2.9. Equilibrium of Rigid Bodies in Two Dimensions and in Three Dimensions ... 121 2.10. Methods for Finding Out the Reactions of a Beam ... 122 2.11. Problems for Equilibrium of Rigid Bodies in Two-Dimension and Three Dimensions ... 123 2.12. Analysis of Pin-Jointed Plane Trusses (or Frames) ... 139 Highlights ... 172 Exercise-2 ... 173 3. Properties of Surfaces and Solids 181—245 3.1. Introduction ... 181 3.2. Determination of First Moment of Area and Centroid of Sections—Rectangle, Circle, Triangle by Integration ... 183 ( v ) ( vi )

Chapters Pages 3.3. Centroid of T-section, I-section, Angle-section, Hollow-section etc. ... 198 3.4. Centroid of Volume ... 203 3.5. Second Moment of Area (or Area Moment of Inertia) ... 206 3.6. Theorem of the Perpendicular Axis ... 207 3.7. Theorem of Parallel Axis ... 208 3.8. Determination of Second Moment of Area (or Area Moment of Inertia) of Plane Area like Rectangle, Triangle, Circle etc. from Integration ... 209 3.9. Moment of Inertia of T-section, I-section, Angle-section, Hollow-section etc. By using Standard Formula ... 216 3.10. Polar Moment of Inertia ... 223 3.11. Product of Inertia ... 224 3.12. Principal Axes ... 225 3.13. Principal Moments of Inertia ... 227 3.14. Mass Moment of Inertia ... 234 3.15. Derivation of Mass Moment of Inertia for Rectangle Section, Prism, Sphere etc. From First Principal ... 235 Highlights ... 241 Exercise-3 ... 242 4. Dynamics of Particles 246—418 4.1. Introduction ... 246 4.2. Velocity ... 246 4.3. Acceleration ... 246 4.4. Relationship of Velocity, Acceleration and Displacement ... 247 4.5. Relative Motion ... 276 4.6. Curvilinear Motion ... 280 4.7. Equations of Motions Along a Circular Path ... 281 4.8. Newton’s Laws ... 288 4.9. Momentum and Angular Momentum (or Moment of Momentum) ... 328 4.10. Laws for Rotary Motion ... 334 4.11. Work Energy Equation of Particles ... 354 4.12. Impulse and Momentum ... 383 4.13. Impact of Elastic Bodies ... 384 Highlights ... 405 Exercise-4 ... 409 5. Friction and Elements of Rigid Body Dynamics 419—512 5.1. Frictional Force ... 419 5.2. Limiting Force of Friction and Definitions of Certain Terms ... 419 5.3. Laws of Coulomb Friction ... 421 5.4. Simple Contact Friction ... 421 5.5. Rolling Resistance ... 455 5.6. Belt Friction ... 457 5.7. Transmission of Power Through Belts ... 459 5.8. Translation, Rotation and General Plane Motion ... 473 ( vii )

Chapters Pages 5.9. Velocity and Acceleration for Combined Motion of Translation and Rotation of Rigid Bodies ... 478 5.10 General Plane Motion ... 485 5.11 Kinematics of Plane Motion ... 488 5.12 Equations of Motion for Plane Motion of a Rigid Body ... 502 Highlights ... 506 Exercise-5 ... 508 Question Bank 513—561 Model Question Paper 562—564

Preface to the Third Edition

Authors are glad to present the Third Edition of the book entitled, ‘Engineering Mechanics’ to the engineering students of all disciples of Anna University, Tamilnadu. The course contents are planned in such a way that the book covers the complete course of first year students of Anna University, Tamilnadu according to the revised syllabus. The Third Edition contains the following five units : Unit I : Basics and Statics of Particles. Unit II : Equilibrium of Rigid Bodies. Unit III : Properties of Surfaces and Solids. Unit IV : Dynamics of Particles. Unit V : Friction and Elements of Rigid Body Dynamics. The book is written in a simple and easy-to-follow language, so that even an average student can grasp the subject by self-study. At the end of each chapter highlights, theoretical questions and many unsolved numerical problems with answers are given for the students to solve them. Mrs. Nirmal Bansal deserves special credit as she not only provided an ideal atmos- phere at home for book writing but also gave inspiration and valuable suggestions. Though every care has been taken in checking the manuscript and proofreading, yet claiming perfection is very difficult. We shall be very grateful to the readers and users of this book for pointing any mistake that might have crept in. Suggestions for improvement are most welcome and would be incorporated in the next edition with a view to make the book more useful.

—AUTHORS SYLLABUS (Anna University, Tamilnadu) GE1X04 ENGINEERING MECHANICS 3 1 0 100 (Common to B.E. (Civil), B. Tech (Chemical / Textile / Textile Tech (Textile Chemistry) Polymer Tech / Biotech / Petroleum Engg / Food Tech / Rubber and Plastics Tech))

OBJECTIVE At the end of this course the student should be able to understand the vectorial and scalar representation of forces and moments, static equilibrium of particles and rigid bodies both in two dimensions and also in three dimensions. Further, he should understand the principle of work and energy. He should be able to comprehend the effect of friction on equilibrium. He should be able to understand the laws of motion, the kinematics of motion and the interrelationship. He should also be able to write the dynamic equilibrium equation. All these should be achieved both conceptually and through solved examples.

UNIT I : BASICS AND STATICS OF PARTICLES 12 Introduction—Units and Dimensions—Laws of Mechanics—Lame’s theorem, Parallelogram and triangular Law of forces—Vectors—Vectorial representation of forces and moments—Vector operations : additions, subtractions, dot product, cross product—Coplanar Forces—Resolution and Composition of forces—Equilibrium of a particle—Forces in space—Equilibrium of a particle in space—Equivalent systems of forces—Principle of transmissibility—Single equivalent force.

UNIT II : EQUILIBRIUM OF RIGID BODIES 12 Free body diagram—Types of supports and their reactions—requirements of stable equilibrium—Moments and Couples—Moment of a force about a point and about an axis—Vectorial representation of moments and couples—Scalar components of a moment—Varignon’s theorem— Equilibrium of Rigid bodies in two dimensions—Equilibrium of Rigid bodies in three dimensions— Examples.

UNIT III : PROPERTIES OF SURFACES AND SOLIDS 12 Determination of Areas and Volumes—First moment of area and the Centroid of sections— Rectangle, circle, triangle from integration—T-section, I-section, Angle section, Hollow section by using standard formula—Second and product moments of plane area—Rectangle, triangle, circle from integration—T-section, I-section, Angle section, Hollow section by using standard formula— Parallel axis theorem and perpendicular axis theorem—Polar moment of inertia—Principal moments of inertia of plane areas—Principal axes of inertia—Mass moment of inertia—Derivation of mass moment of inertia for rectangular section, prism, sphere from first principle—Relation to area moments of inertia.

UNIT IV : DYNAMICS OF PARTICLES 12 Displacements, Velocity and acceleration, their relationship—Relative motion—Curvilinear motion—Newton’s law—Work Energy Equation of particles—Impulse and Momentum—Impact of elastic bodies. UNIT V : FRICTION AND ELEMENTS OF RIGID BODY DYNAMICS 12 Frictional force—Laws of Coulomb friction—Simple contact friction—Rolling resistance— Belt friction. Translation and Rotation of Rigid Bodies—Velocity and acceleration—General plane motion. Total : 60 Periods 1 Basics and Statics of Particles

1.1. INTRODUCTION

Engineering mechanics is that branch of science which deals with the behaviour of a body when the body is at rest or in motion. The engineering mechanics may be divided into statics and dynamics. The branch of science, which deals with the study of a body when the body is at rest, is known as statics while the branch of science which deals with the study of a body when the body is in motion, is known as dynamics. Dynamics is further divided into kinematics and kinetics. The study of a body in motion, when the forces which cause the motion are not considered, is called kinematics and if the forces are also considered for the body in motion, that branch of science is called kinetics. The classification of Engineering Mechanics are shown in Fig. 1.1 below.

ENGINEERING MECHANICS

1. Statics 2. Dynamics (Body is at rest) (Body is in motion)

(i) Kinematics (ii) Kinetics

Fig. 1.1 Note. Statics deals with equilibrium of bodies at rest, whereas dynamics deals with the motion of bodies and the forces that cause them. The following terms are generally used in Mechanics : 1. Vector quantity, 2. Scalar quantity, 3. Particle, 4. Law of parallelogram of forces, 5. Triangle law and 6. Lame’s theorem. 1.1.1. Vector Quantity. A quantity which is completely specified by magnitude and direction, is known as a vector quantity. Some examples of vector quantities are : velocity, acceleration, force and momentum. A vector quantity is represented by means of a straight line with an arrow as shown in Fig. 1.2. The length of the straight line (i.e., AB) represents the magnitude and arrow represents the direction of the vector. → AB The symbol AB also represents this vector, which means it Fig. 1.2. Vector quantity is acting from A to B. 1 2 ENGINEERING MECHANICS

1.1.2. Scalar Quantity. A quantity, which is completely specified by magnitude only, is known as a scalar quantity. Some examples of scalar quantity are : mass, length, time and temperature. 1.1.3. A Particle. A particle is a body of infinitely small volume (or a particle is a body of negligible dimensions) and the mass of the particle is considered to be concentrated at a point. Hence a particle is assumed to a point and the mass of the particle is concentrated at this point. 1.1.4. Law of Parallelogram of Forces. The law of parallelogram of forces is used to determine the resultant* of two forces acting at a point in a plane. It states, “If two forces, acting at a point be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.” Let two forces P and Q act at a point O as shown in Fig. 1.3. The force P is represented in magnitude and direction by OA whereas the force Q is presented in magnitude and direction by OB. Let the angle between the two forces be ‘α’. The resultant of these two forces will be obtained in magnitude and direction by the diagonal (passing through O) of the parallelogram of which OA and OB are two adjacent sides. Hence draw the parallelogram with OA and OB as adjacent sides as shown in Fig. 1.4. The resultant R is represented by OC in magnitude and direction.

B B C

Q Q R

α α θ α

O P A O P A D Fig. 1.3 Fig. 1.4 Magnitude of Resultant (R) From C draw CD perpendicular to OA produced. Let α = Angle between two forces P and Q = ∠AOB Now ∠DAC = ∠AOB (Corresponding angles) = α In parallelogram OACB, AC is parallel and equal to OB. ∴ AC = Q. In triangle ACD, AD = AC cos α = Q cos α and CD = AC sin α = Q sin α. In triangle OCD, OC2 = OD2 + DC2.

*The resultant of a system of forces may be defined as a single force which has the same effect as system of forces acting on the body. BASICS AND STATICS OF PARTICLES 3

But OC = R, OD = OA + AD = P + Q cos α and DC = Q sin α. ∴ R2 = (P + Q cos α)2 + (Q sin α)2 = P2 + Q2 cos2 α + 2PQ cos α + Q2 sin2 α = P2 + Q2 (cos2 α + sin2 α) + 2PQ cos α = P2 + Q2 + 2PQ cos α (∵ cos2 α + sin2 α = 1)

∴ R = PQ22++2 PQcos α ...(1.1) Equation (1.1) gives the magnitude of resultant force R. Direction of Resultant Let θ = Angle made by resultant with OA. Then from triangle OCD, CD Q sin α tan θ = = OD PQ+ cos α F Q sin α I ∴θ = tan–1 ...(1.2) HG PQ+ cos αKJ Equation (1.2) gives the direction of resultant (R). C The direction of resultant can also be obtained by using sine rule [In triangle OAC, OA = P, AC = Q, OC = R, angle OAC = (180 – α), angle ACO = 180 – [θ + (–)αθ R 180 – α] = (α – θ)] Q θααθ− − sin= sin (180 )= sin ( ) (180–α ) α ACOCOA θ sinθααθ sin (180 − ) sin (− ) = = O P A QR P Fig. 1.4(a) Two cases are important. 1st Case. If the two forces P and Q act at right angles, then α = 90° From equation (1.1), we get the magnitude of resultant as

R = PQ22++2290 PQcosα = PQ 22 ++ PQ cos °

= PQ22+ (∵ cos 90° = 0) ...(1.2A) From equation (1.2), the direction of resultant is obtained as F Q sin α I θ = tan–1 HG PQ+ cos αKJ ° F Q sin 90 I −1 Q = tan–1 = tan (∵ sin 90° = 1 and cos 90° = 0) HG PQ+°cos 90 KJ P 2nd Case. The two forces P and Q are equal and are acting at an angle α between them. Then the magnitude and direction of resultant is given as

R = PQ22++22 PQcosαα = PP 22 ++×× PP cos (∵ P = Q)

= 22PP22+=+cosαα 21 P 2 ( cos ) 4 ENGINEERING MECHANICS

α α 22 F 2 I = 22P × cos G∵ 12+=cosα cos J 2 H 2K αα = 4PP22cos= 2 cos ...(1.3) 2 2 Q sin α − P sin α and θ = tan–1 F I = tan 1 (∵ P = Q) HG PQ+ cos αKJ PP+ cos α P sin α − sin α = tan–1 = tan 1 P(cos)11+ α + cos α αα 2 sin cos F ααI = tan–1 22 G∵ sinα = 2 sin cos J α H 22K 2 cos2 2 α sin αα = tan–1 2 = −1 F I = ...(1.4) α tanG tan J cos H 22K 2 It is not necessary that one of two forces, should be along x-axis. The forces P and Q may be in any direc- Q tion as shown in Fig. 1.5. If the angle between the two forces is ‘α’, then their resultant will be given by equa- R tion (1.1). The direction of the resultant would be obtained from equation (1.2). But angle θ will be the angle made by resultant with the direction of P.

1.1.5. Law of Triangle of Forces. It states that, α θ P “if three forces acting at a point be represented in magnitude and direction by the three sides of a triangle, O taken in order, they will be in equilibrium.” Fig. 1.5 1.1.6. Lame’s Theorem. It states that, “If there forces acting at a point are in equilibrium, each force will be proportional to the sine of the angle between the other two forces.” Suppose the three forces P, Q and R are act- Q ing at a point O and they are in equilibrium as shown in Fig. 1.6. Let α = Angle between force P and Q. β = Angle between force Q and R. β α γ = Angle between force R and P. Then according to Lame’s theorem, O P P is proportional sine of angle between Q and γ R α sin β. P R ∴ = constant sin β Fig. 1.6 Q R Similarly = constant and = constant sin γ sin α P Q R or == ...(1.5) sinβγα sin sin BASICS AND STATICS OF PARTICLES 5

Proof of Lame’s Theorem. The three forces β acting on a point, are in equilibrium and hence they can be represented by the three sides of the triangle taken in the same order. Now draw the force triangle as shown in Fig. 1.6 (a). β R Now applying sine rule, we get (180 – ) Q P Q R = = α − βγα− − (180 – ) sin (180 ) sin ( 180 ) sin ( 180 ) (180 –γ ) α This can also be written O P P ==Q R γ sinβγα sin sin Fig. 1.6(a) This is same equation as equation (1.5). Note. All the three forces should be acting either towards the point or away from the point.

1.2. UNITS AND DIMENSIONS

The following units of different systems are mostly used : 1. C.G.S. (i.e., Centimetre-Gram Second) system of units. 2. M.K.S. (i.e., Metre-Kilogram-Second) system of units. 3. S.I. (i.e., International) system of units. 1.2.1. C.G.S. System of Units. In this system, length is expressed in centimetre, mass in gram and time in second. The unit of force in this system is dyne, which is defined as the force acting on a mass of one gram and producing an acceleration of one centimetre per second square. 1.2.2. M.K.S. System of Units. In this system, length is expressed in metre, mass in kilogram and time in second. The unit of force in this system is expressed as kilogram force and is represented as kgf. 1.2.3. S.I. System of Units. S.I. is abbreviation for ‘The System International Units’. It is also called the International System of Units. In this system length is expressed in metre mass in kilogram and time in second. The unit of force in this system is Newton and is represented N. Newton is the force acting on a mass of one kilogram and producing an acceleration of one metre per second square. The relation between newton (N) and dyne is obtained as One metre One Newton = One kilogram mass × s2 100 cm = 1000 gm × (∵ one kg = 1000 gm) s2 gm× cm = 1000 × 100 × s2 gm× cm 5 R∵ = U = 10 dyne S dyne 2 V T s W When the magnitude of forces is very large, then the unit of force like kilo-newton and mega-newton is used. Kilo-newton is represented by kN. One kilo-newton = 103 newton or 1 kN = 103 N and One mega newton = 106 Newton Engineering Mechanics(Anna Univ) By R. K. Bansal, Sanjay Bansal

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