ENGINEERING MECHANICS

Size: px

Start display at page:

Download "ENGINEERING MECHANICS"

Agnes Grant
6 years ago
Views:

1 ENGINEERING MECHANICS

3 ENGINEERING MECHANICS (In SI Units) For BE/B.Tech. Ist YEAR Strictly as per the latest syllabus prescribed by Mahamaya Technical University, Noida By Dr. R.K. BANSAL B.Sc. Engg. (Mech.), M.Tech., Hons. (IIT, 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 Delhi LAXMI PUBLICATIONS (P) LTD BANGALORE CHENNAI COCHIN GUWAHATI HYDERABAD JALANDHAR KOLKATA LUCKNOW MUMBAI RANCHI NEW DELHI BOSTON, USA

4 All rights reserved with the Publisher and Authors. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of authors and the publisher. Published by LAXMI PUBLICATIONS (P) LTD 113, Golden House, Daryaganj, New Delhi Phone : Fax : info@laxmipublications.com Compiled by : Smt. Nirmal Bansal Price : ` Only. First Edition : 2013 OFFICES Bangalore Jalandhar Chennai Kolkata Cochin , Lucknow Guwahati , Mumbai , Hyderabad Ranchi EEM ENGG MECHANICS (MTU)-BAN Typeset at : Excellent Graphics, Delhi. C 2179/010/09 Printed at : Ajit Printers

5 Contents Chapters Pages UNIT-I 1. Two Dimensional Concurrent Force Systems Basic Concepts Units (System of Units) Force Systems Laws of Motion Moment and Couple Vectors Vector Representation of Forces and Moments Vector Operations The Principle of Transmissibility of Forces Resultant of a Force System Equilibrium and Equations of Equilibrium Equilibrium Conditions Free Body Diagram Determination of Reaction Resultant of Two Dimensional Concurrent Forces Applications of Concurrent Forces Highlights Exercise UNIT-II 2. Two Dimensional Non-Concurrent Force Systems Basic Concepts Varignon s Theorem (or Principle of Moments) Transfer of a Force to Parallel Position Distributed Force System Highlights 2A.. 79 Exercise 2A Types of Supports and Their Reactions Converting Force into Couples and Vice-Versa Highlights 2B Exercise 2B ( v )

6 ( vi ) Chapters Pages 3. Friction Introduction Laws of Coulomb Friction Equilibrium of Bodies Involving Dry Friction Equilibrium of a Body Lying on a Rough Inclined Plane Ladder Friction Belt Friction Highlights Exercise Truss (or Frames) Introduction Types of Frames Perfect Frame Imperfect Frame Assumptions in Truss Analysis Analysis of a Frame Method of Joints Method of Sections Highlights Exercise UNIT-III 5. Centroid and Moment of Inertia Centre of Gravity Centroid Centroid or Centre of Gravity of Simple Plane Figures Centroid of Composite Bodies Problems of Finding Centroid or Centre of Gravity of Areas by Integration Method Centroid of Volume Moment of Inertia of Plane Area Theorem of the Perpendicular Axis Theorem of Parallel Axis Determination of Area Moment of Inertia Mass Moment of Inertia Mass Moment of Inertia of a Rectangular Plate Mass Moment of Inertia of Circular Ring Mass Moment of Inertia of a Disc or Circular Plate Mass Moment of Inertia of a Circular Cylinder Mass Moment of Inertia of a Sphere

7 Chapters ( vii ) Pages Mass Moment of Inertia of a Cone about its Axis of Symmetry Pappus Theorems Polar Moment of Inertia Highlights Exercise UNIT-IV 6. Kinematics of Rigid Bodies Introduction Plane Rectilinear Motion Rectilinear Motion Under Constant Acceleration Rectilinear Motion Under Variable Acceleration Highlights 6A Exercise 6A Plane Curvilinear Motion of Rigid Body Curvilinear Motion Under Constant Angular Acceleration Curvilinear Motion Under Variable Angular Acceleration Types of Motion Velocity of Link Having Translation and Rotation Motion by Instantaneous Centre Velocity and Acceleration of a Body Under Translation and Rotational Motion or Kinematics of Plane Motion Relative Motion Highlights 6B Exercise 6B UNIT-V 7. Kinetics of Rigid Bodies Introduction Force, Mass and Acceleration Newton s Laws of Motion Kinetics of Rigid Bodies Momentum and Angular Momentum (or Moment of Momentum) Laws for Rotary Motion Work and Energy Work-Energy Principle Law of Conservation of Energy Impulse and Momentum D Alembert s Principle and Dynamic Equilibrium Highlights Exercise

8 ( viii ) 8. Virtual Work Virtual Displacement and Virtual Work Principle of Virtual Work Stability of Equilibrium The Principle of Virtual Work for Problems on Framed Structure Application of Virtual Work on Ladders The Principle of Virtual Work for Problems on Lifting Machines Highlights Exercise Objective Type Questions Additional Questions

9 Preface The course contents of this edition of the book entitled, Engineering Mechanics are planned in such a way that the book covers the complete content of first year course of all disciplines of Mahamaya Technical University, Noida strictly as per the latest syllabus prescribed by Mahamaya Technical University, Noida. The book contains following five units : Unit I : Two Dimensional Concurrent Force Systems, Basic Concepts, Resultant of a Force System, Free Body Diagrams, Equilibrium and Equation of Equilibrium. Unit II : Two Dimensional Non-Concurrent Force Systems, Basic Concepts, Varignon s Theorem, Types of Supports and their Reactions, Friction and Structure (Truss) Unit III : Centroid and Moment of Inertia. Unit IV : Kinematics of Rigid Body. Unit V : Kinetics of Rigid Body. 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 problems with answers are given for the students to solve them. Mrs Nirmal Bansal deserves special credit as she not only provided an ideal atmosphere at home for book writing but also gave inspiration and valuable suggestions. Though every care has been taken in checking the manuscripts 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. Dr. R.K. BANSAL ( ix )

10 SYLLABUS (EME-102 / 202 : ENGINEERING MECHANICS) UNIT I Two Dimensional Concurrent 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 UNIT II Friction : Introduction, Laws of coulomb friction, Equilibrium of bodies involving dryfriction, Belt friction, Application. 3 Trusses : Introduction, Simple Truss and Solution of Simple Truss, Method of Joints and Method of Sections. 3 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 of Inertia, Mass Moment of Inertia of Circular Ring, Disc, Cylinder, Sphere and Cone about their Axis of Symmetry. 6 UNIT IV Kinematics of Rigid Body : Introduction, Plane Motion of Rigid Body, Velocity and Acceleration under Translation and Rotational Motion, Relative Velocity. 4 UNIT V Kinetics of Rigid Body : Introduction, Force, Mass and Acceleration, Work and Energy, Impulse and Momentum, D Alembert s Principles and Dynamic Equilibrium. 4 ( x )

11 UNIT I 1. Two Dimensional Concurrent Force Systems

13 1 Two Dimensional Concurrent Force Systems 1.1. BASIC CONCEPTS 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 (Body is at rest) 2. Dynamics (Body is in motion) ( i) Kinematics (Forces which cause motion are not considered) Fig. 1.1 ( ii) Kinetics (Forces are considered) Note. Statics deals with equilibrium of bodies at rest, whereas dynamics deals with the motion of bodies and the forces that cause them 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 The length of the straight line (i.e., AB) A B represents the magnitude and arrow represents the direction of the vector. The symbol AB Fig Vector Quantity. also represents this vector, which means it is acting from A to B. 3

14 4 ENGINEERING MECHANICS 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 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 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 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 The resultant R is represented by OC in magnitude and direction. B B C Q Q R O α P A α θ α 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, OC 2 = OD 2 + DC 2. *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.

15 TWO DIMENSIONAL CONCURRENT FORCE SYSTEMS 5 But OC = R, OD = OA + AD = P + Q cos α and DC = Q sin α. R 2 = (P + Q cos α) 2 + (Q sin α) 2 = P 2 + Q 2 cos 2 α + 2PQ cos α + Q 2 sin 2 α = P 2 + Q 2 (cos 2 α + sin 2 α) + 2PQ cos α = P 2 + Q 2 + 2PQ cos α ( cos 2 α + sin 2 α = 1) 2 2 R = P + Q + 2PQcos 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, tan θ = CD Q sin α = OD P + Qcos α θ = tan 1 Equation (1.2) gives the direction of resultant (R). 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 α] = (α θ)] F HG I Q sin α...(1.2) P + Qcos αkj C sin q sin (180 - a) sin ( a - q) = = AC OC OA O sin q sin (180 - a) sin ( a - q) = = Q R P 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 = P + Q + 2PQcos α = P + Q + 2PQcos 90 = 2 2 P + Q ( cos 90 = 0)...(1.2 A) From equation (1.2), the direction of resultant is obtained as θ = tan 1 = tan 1 F HG F HG Q sin α P + Qcos α Q sin 90 P + Qcos 90 I KJ I = KJ tan 1 Q P ( sin 90 = 1 and cos 90 = 0) 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 = P + Q + 2PQcos α = P + P + 2P P cos α ( P = Q) = 2P + 2P cos α = 2P ( 1+ cos α) R (180 ) P A Fig. 1.4 (a) ( ) Q

16 Engineering Mechanics (MTU) By Dr. R.K. Bansal 40% OFF Publisher : Laxmi Publications ISBN : Author : Dr. R.K. Bansal Type the URL : 10 Get this ebook

A TEXTBOOK APPLIED MECHANICS

A TEXTBOOK APPLIED MECHANICS A TEXTBOOK OF APPLIED MECHANICS A TEXTBOOK OF APPLIED MECHANICS (Including Laboratory Practicals) S.I. UNITS By R.K. RAJPUT M.E. (Heat Power Engg.) Hons. Gold Medallist ; Grad. (Mech. Engg. & Elect. Engg.)