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Mechanical Engineering Principles

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Mechanical Engineering PrinciplesJOHN BIRDCARL ROSSBSc, CEng, CMath, FIMA, MIEE, FCollP, FIIEBSc, PhD, DSc, CEng, FRINA, MSNAMENewnesOXFORDAUCKLAND BOSTONJOHANNESBURGMELBOURNE NEW DELHI

NewnesAn imprint of Butterworth-HeinemannLinacre House, Jordan Hill, Oxford OX2 8DP225 Wildwood Avenue, Woburn, MA 01801-2041A division of Reed Educational and Professional Publishing LtdFirst published 2002 John Bird and Carl Ross 2002All rights reserved. No part of this publicationmay be reproduced in any material form (includingphotocopying or storing in any medium by electronicmeans and whether or not transiently or incidentallyto some other use of this publication) without thewritten permission of the copyright holder exceptin accordance with the provisions of the Copyright,Designs and Patents Act 1988 or under the terms of alicence issued by the Copyright Licensing Agency Ltd,90 Tottenham Court Road, London, England W1P 9HE.Applications for the copyright holder’s written permissionto reproduce any part of this publication should be addressedto the publishersBritish Library Cataloguing in Publication DataA catalogue record for this book is available from the British LibraryLibrary of Congress Cataloguing in Publication DataA catalogue record for this book is available from the Library of CongressISBN 0 7506 5228 4For information on all Newnes publications visit our website at www.newnespress.comTypeset by Laserwords Pvt. Ltd., Chennai, IndiaPrinted and bound in

ContentsPrefaceixPart 1 Statics and strength of materials11 The effects of forces on materials1.1 Introduction1.2 Tensile force1.3 Compressive force1.4 Shear force1.5 Stress1.6 Strain1.7 Elasticity, limit ofproportionality and elastic limit1.8 Hooke’s law1.9 Ductility, brittlenessand malleability1.10 Modulus of rigidity1.11 Thermal strain1.12 Compound bars112222367111212132 Tensile testing2.1 The tensile test2.2 Worked problemson tensile testing2.3 Further worked problemson tensile testing18183 cting at a pointScalar and vector quantitiesCentre of gravity and equilibriumForcesThe resultant oftwo coplanar forcesTriangle of forces methodThe parallelogramof forces methodResultant of coplanarforces by calculationResultant of more thantwo coplanar forcesCoplanar forces in equilibriumResolution of forcesSummary4 Forces in structures4.1 Introduction1921272829293032343740404.2 Worked problems on mechanismsand pin-jointed trusses4.3 Graphical method4.4 Method of joints (amathematical method)4.5 The method of sections(a mathematical method)Assignment 141424652555 Simply supported beams5.1 The moment of a force5.2 Equilibrium and theprinciple of moments5.3 Simply supported beamshaving point loads5.4 Simply supportedbeams with couples57576 Bending moment andshear force diagrams6.1 Introduction6.2 Bending moment (M)6.3 Shearing force (F)6.4 Worked problems onbending moment andshearing force diagrams6.5 Uniformly distributed loads697 First and second moment of areas7.1 Centroids7.2 The first moment of area7.3 Centroid of area betweena curve and the x-axis7.4 Centroid of area betweena curve and the y-axis7.5 Worked problems oncentroids of simple shapes7.6 Further worked problems oncentroids of simple shapes7.7 Second moments ofarea of regular sections7.8 Second moment of areafor ‘built-up’ sectionsAssignment 25861646969697078848484848586878896102

viMECHANICAL ENGINEERING PRINCIPLES8 Bending of beams8.1 IntroductionMEσ yIR8.3 Worked problems onthe bending of beams8.2 To prove that9 Torque9.1 Couple and torque9.2 Work done and powertransmitted by a constant torque9.3 Kinetic energy andmoment of inertia9.4 Power transmission and efficiency10 Twisting of shafts10.1 122Assignment 3126120Part 2 Dynamics12711 Linear11.111.211.311.411.5127127127129130132and angular motionThe radianLinear and angular velocityLinear and angular accelerationFurther equations of motionRelative velocity12 Linear momentum and impulse12.1 Linear momentum12.2 Impulse and impulsive forces13613613913 Force,13.113.213.313.4144144144147mass and accelerationIntroductionNewton’s laws of motionCentripetal accelerationRotation of a rigidbody about a fixed axis13.5 Moment of inertia (I )14 Work,14.114.214.314.414.517017017017217316 Motion in a circle16.1 Introduction16.2 Motion on a curved banked track16.3 Conical pendulum16.4 Motion in a vertical circle16.5 Centrifugal clutch18218218418518718917 6173174175177116τTGθ rJLWorked problems onthe twisting of shaftsTo prove that15 Friction15.1 Introduction to friction15.2 Coefficient of friction15.3 Applications of friction15.4 Friction on an inclined plane15.5 Motion up a plane withthe pulling force Pparallel to the plane15.6 Motion down a planewith the pulling forceP parallel to the plane15.7 Motion up a plane dueto a horizontal force P15.8 The efficiency of a screw jack149149energy and powerWorkEnergyPowerPotential and kinetic energyKinetic energy of rotation153153157159162165Assignment 4169harmonic motionIntroductionSimple harmonic motion (SHM)The spring-mass systemThe simple pendulumThe compound pendulumTorsional vibrations18 Simple machines18.1 Machines18.2 Force ratio, movementratio and efficiency18.3 Pulleys18.4 The screw-jack18.5 Gear trains18.6 LeversAssignment 5198198198200202203205209Part 3 Heat transfer and fluid mechanics21119 Heat19.119.219.319.419.5211211212212214energy and transferIntroductionThe measurement of temperatureSpecific heat capacityChange of stateLatent heats of fusionand vaporisation19.6 A simple refrigerator19.7 Conduction, convection and radiation19.8 Vacuum flask215217217218

CONTENTS19.9Use of insulationin conserving fuel20 Thermal expansion20.1 Introduction20.2 Practical applicationsof thermal expansion20.3 Expansion and contraction of water20.4 Coefficient of linear expansion20.5 Coefficient of superficial expansion20.6 Coefficient of cubic expansionAssignment 621822122122122222222422522921 Hydrostatics21.1 Pressure21.2 Fluid pressure21.3 Atmospheric pressure21.4 Archimedes’ principle21.5 Measurement of pressure21.6 Barometers21.7 Absolute and gauge pressure21.8 The manometer21.9 The Bourdon pressure gauge21.10 Vacuum gauges21.11 Hydrostatic pressureon submerged surfaces21.12 Hydrostatic thruston curved surfaces21.13 Buoyancy21.14 The stability of floating bodies23023023123223323523523723723823922 247248249249250250250flowIntroductionDifferential pressure flowmetersOrifice plateVenturi tubeFlow nozzlePitot-static tubeMechanical flowmetersDeflecting vane flowmeterTurbine type .16Float and tapered-tube meterElectromagnetic flowmeterHot-wire anemometerChoice of flowmeterEquation of continuityBernoulli’s EquationImpact of a jet ona stationary plate23 Ideal gas laws23.1 Introduction23.2 Boyle’s law23.3 Charles’ law23.4 The pressure law23.5 Dalton’s law of partial pressure23.6 Characteristic gas equation23.7 Worked problems on thecharacteristic gas equation23.8 Further worked problems onthe characteristic gas equation24 The measurement of temperature24.1 Introduction24.2 Liquid-in-glass thermometer24.3 Thermocouples24.4 Resistance thermometers24.5 Thermistors24.6 Pyrometers24.7 Temperature indicatingpaints and crayons24.8 Bimetallic thermometers24.9 Mercury-in-steel thermometer24.10 Gas thermometers24.11 Choice of measuring deviceAssignment 63267267267268270272272274274274275275277A list of formulae279Greek alphabet283Answers to multiple-choice questions284Index287

Preface(i) National Certificate/Diploma courses inMechanical Engineering(ii) Mechanical Engineering Principles (Advanced GNVQ Unit 8)(iii) Further Mechanical Engineering Principles(Advanced GNVQ Unit 12)(iv) Any introductory/access/foundation courseinvolving Mechanical Engineering Principlesat University, and Colleges of Further andHigher education.Mathematics and Science, each topic considered inthe text is presented in a way that assumes thatthe reader has little previous knowledge of thattopic.Mechanical Engineering Principles contains over280 worked problems, followed by over 470 further problems (all with answers). The furtherproblems are contained within some 130 Exercises;each Exercise follows on directly from the relevant section of work, every few pages. In addition,the text contains 260 multiple-choice questions (allwith answers), and 260 short answer questions,the answers for which can be determined from thepreceding material in that particular chapter. Whereat all possible, the problems mirror practical situations found in mechanical engineering. 330 linediagrams enhance the understanding of the theory.At regular intervals throughout the text are some7 Assignments to check understanding. For example, Assignment 1 covers material contained inChapters 1 to 4, Assignment 2 covers the materialin Chapters 5 to 7, and so on. No answers are givenfor the questions in the assignments, but a lecturer’sguide has been produced giving full solutions andsuggested marking scheme. The guide is offered freeto those staff that adopt the text for their course.At the end of the text, a list of relevant formulaeis included for easy reference.‘Learning by Example’ is at the heart ofMechanical Engineering Principles.Although pre-requisites for the modules coveredin this book include GCSE/GNVQ intermediate inJohn Bird and Carl RossUniversity of PortsmouthMechanical Engineering Principles aims tobroaden the reader’s knowledge of the basicprinciples that are fundamental to mechanicalengineering design and the operation of mechanicalsystems.Modern engineering systems and products stillrely upon static and dynamic principles to makethem work. Even systems that appear to be entirelyelectronic have a physical presence governed by theprinciples of statics.For clarity, the text is divided into three sections,these being:Part 1 Statics and strength of materialsPart 2 DynamicsPart 3 Heat transfer and fluid mechanicsMechanical Engineering Principles covers thefollowing syllabuses:

Statics and strength ofmaterialsPart 11The effects of forces on materials perform calculations using Hooke’s lawAt the end of this chapter you should beable to: plot a load/extension graph from givendata define force and state its unit define ductility, brittleness and malleability, with examples of each recognise a tensile force and state relevantpractical examples recognise a compressive force and staterelevant practical examples define rigidity or shear modulus understand thermal stresses and strains calculates stresses in compound bars recognise a shear force and state relevantpractical examples define stress and state its unitF calculate stress σ from σ A define strainx calculate strain ε from ε L define elasticity, plasticity, limit of proportionality and elastic limit state Hooke’s law define Young’s modulus of elasticity Eand stiffness appreciate typical values for Eσ calculate E from E ε1.1 IntroductionA force exerted on a body can cause a change ineither the shape or the motion of the body. The unitof force is the newton, N.No solid body is perfectly rigid and when forcesare applied to it, changes in dimensions occur. Suchchanges are not always perceptible to the humaneye since they are so small. For example, the spanof a bridge will sag under the weight of a vehicleand a spanner will bend slightly when tighteninga nut. It is important for engineers and designers toappreciate the effects of forces on materials, togetherwith their mechanical properties.The three main types of mechanical force that canact on a body are: (i) tensile, (ii) compressive, and(iii) shear

2MECHANICAL ENGINEERING PRINCIPLESRivet1.2 Tensile forceForceTension is a force that tends to stretch a material,as shown in Figure 1.1. For example,(i) the rope or cable of a crane carrying a load isin tension(ii)rubber bands, when stretched, are in tensionForceFigure 1.3(ii)a guillotine cutting sheet metal, or gardenshears, each provide a shear force(iii)a horizontal beam is subject to shear force(iv)transmission joints on cars are subject to shearforcesForceA shear force can cause a material to bend, slide ortwist.Figure 1.1(iii)Forcewhen a nut is tightened, a bolt is under tensionProblem 1. Figure 1.4(a) represents a craneand Figure 1.4(b) a transmission joint. Statethe types of forces acting, labelled A to F .A tensile force, i.e. one producing tension, increasesthe length of the material on which it acts.BC1.3 Compressive forceACompression is a force that tends to squeezeor crush a material, as shown in Figure 1.2. Forexample,DEForceFLoad(b)(a)Figure 1.4ForceForceFigure 1.2(i) a pillar supporting a bridge is in compression(ii)the sole of a shoe is in compression(iii)the jib of a crane is in compressionA compressive force, i.e. one producing compression, will decrease the length of the material onwhich it acts.1.4 Shear forceShear is a force that tends to slide one face of thematerial over an adjacent face. For example,(i) a rivet holding two plates together is inshear if a tensile force is applied between theplates — as shown in Figure 1.3(a) For the crane, A, a supporting member, isin compression, B, a horizontal beam, is inshear, and C, a rope, is in tension.(b)For the transmission joint, parts D and F arein tension, and E, the rivet or bolt, is inshear.1.5 StressForces acting on a material cause a change in dimensions and the material is said to be in a state ofstress. Stress is the ratio of the applied force F tocross-sectional area A of the material. The symbolused for tensile and compressive stress is σ (Greekletter sigma). The unit of stress is the Pascal, Pa,where 1 Pa 1 N/m2 . Henceσ FPaA

THE EFFECTS OF FORCES ON MATERIALSwhere F is the force in Newton’s and A is thecross-sectional area in square metres. For tensileand compressive forces, the cross-sectional area isthat which is at right angles to the direction of theforce. For a shear force the shear stress is equalto F /A, where the cross-sectional area A is thatwhich is parallel to the direction of the force. Thesymbol used for shear stress is the Greek lettertau, τ .Problem 2. A rectangular bar having across-sectional area of 75 mm2 has a tensileforce of 15 kN applied to it. Determine thestress in the bar.Now try the following exerciseExercise 1A rectangular bar having a cross-sectionalarea of 80 mm2 has a tensile force of20 kN applied to it. Determine the stressin the bar.[250 MPa]2.A circular cable has a tensile force of1 kN applied to it and the force producesa stress of 7.8 MPa in the cable. Calculatethe diameter of the cable.[12.78 mm]3.A square-sectioned support of side 12 mmis loaded with a compressive force of10 kN. Determine the compressive stressin the support.[69.44 MPa]4.A bolt having a diameter of 5 mm isloaded so that the shear stress in it is120 MPa. Determine the value of theshear force on the bolt.[2.356 kN]5.A split pin requires a force of 400 N toshear it. The maximum shear stress beforeshear occurs is 120 MPa. Determine theminimum diameter of the pin.[2.06 mm]6.A tube of outside diameter 60 mm andinside diameter 40 mm is subjected to aload of 60 kN. Determine the stress in thetube.[38.2 MPa]F15 103 NStress in bar, σ A75 10 6 m2Problem 3. A circular wire has a tensileforce of 60.0 N applied to it and this forceproduces a stress of 3.06 MPa in the wire.Determine the diameter of the wire.Force F 60.0 N andstress σ 3.06 MPa 3.06 106 PaSinceσ FAthen area,A 60.0 NF σ3.06 106 Pa1.6 Strain 19.61 10 6 m2 19.61 mm22Cross-sectional area A πd4 ;hence 19.61 πd 2, from which,44 19.61from which, d d2 πi.e. diameter of wire 5.0 mmFurther problems on stress1.Cross-sectional area A 75 mm2 75 10 6 m2and force F 15 kN 15 103 N 0.2 109 Pa 200 MPa3The fractional change in a dimension of a materialproduced by a force is called the strain. For a tensileor compressive force, strain is the ratio of the changeof length to the original length. The symbol used forstrain is ε (Greek epsilon). For a material of lengthL metres which changes in length by an amount xmetres when subjected to stress,ε 4 19.61π xLStrain is dimension-less and is often expressed as apercentage, i.e.percentage strain x 100L

4MECHANICAL ENGINEERING PRINCIPLESdetermine the minimum breadth of the barwhen loaded with a force of 3 kN.(b) If the bar in (a) is 2 m long anddecreases in length by 0.25 mm when theforce is applied, determine the strain and thepercentage strain.xAppliedforceγLReactionforce(a)Figure 1.5For a shear force, strain is denoted by the symbol γ (Greek letter gamma) and, with reference toFigure 1.5, is given by: 150 10 6 m2 150 mm2xγ LProblem 4. A bar 1.60 m long contractsaxially by 0.1 mm when a compressive loadis applied to it. Determine the strain and thepercentage strain.0.1 mmcontraction original length1.60 103 mm0.1 0.00006251600Strain ε Percentage strain 0.0000625 100 0.00625%Problem 5. A wire of length 2.50 m has apercentage strain of 0.012% when loadedwith a tensile force. Determine the extensionof the wire.force Farea AF3000 Nthen, area, A σ20 106 PaSince stress, σ Cross-sectional area width breadth, hence150area 15 mmwidth10contractionε original lengthbreadth (b)Strain,0.25 0.0001252000Percentage strain 0.000125 100 0.0125%Problem 7. A pipe has an outside diameterof 25 mm, an inside diameter of 15 mm andlength 0.40 m and it supports a compressiveload of 40 kN. The pipe shortens by 0.5 mmwhen the load is applied. Determine (a) thecompressive stress, (b) the compressivestrain in the pipe when supporting this load.Original length of wire 2.50 m 2500 mm0.012 0.00012100extension xStrain ε original length Land strain hence, extension x εL (0.00012)(2500) 0.30 mmProblem 6. (a) A rectangular metal bar hasa width of 10 mm and can support amaximum compressive stress of 20 MPa;Compressive force F 40 kN 40000 N,and cross-sectional area A π 2(D d 2 ),4where D outside diameter 25 mm andd inside diameter 15 mm. Henceπ(252 152 ) mm24π (252 152 ) 10 6 m24A 3.142 10 4 m2

THE EFFECTS OF FORCES ON MATERIALS(a) Compressive stress, 40000 NF σ A3.142 10 4 m20.001963 m2 80.03 MPa,which is the compressive stress in the punch.Contraction of pipe when loaded,x 0.5 mm 0.0005 m, and original lengthL 0.40 m. Hence, compressive strain,ε 157.1 103 N 8.003 107 Pa 12.73 107 Pa 127.3 MPa(b)5Problem 9. A rectangular block of plasticmaterial 500 mm long by 20 mm wide by300 mm high has its lower face glued to abench and a force of 200 N is applied to theupper face and in line with it. The upper facemoves 15 mm relative to the lower face.Determine (a) the shear stress, and (b) theshear strain in the upper face, assuming thedeformation is uniform.0.0005x L0.4 0.00125 (or 0.125%)Problem 8. A circular hole of diameter50 mm is to be punched out of a 2 mm thickmetal plate. The shear stress needed to causefracture is 500 MPa. Determine (a) theminimum force to be applied to the punch,and (b) the compressive stress in the punchat this value.(a)forcearea parallel to the forceShear stress, τ Area of any face parallel to the force 500 mm 20 mm(a) The area of metal to be sheared, A perimeterof hole thickness of plate. (0.5 0.02) m2 0.01 m2Hence, shear stress,Perimeter of hole πd π(50 10 3 )τ 0.1571 m.200 N0.01 m2 20000 Pa or 20 kPa 3Hence, shear area, A 0.1571 2 10 3.142 10 4 m2(b)x(see side viewLin Figure 1.6)15 0.05 (or 5%) 300Shear strain, γ force,areashear force shear stress areaSince shear stress (500 106 3.142 10 4 ) N 157.1 kN,which is the minimum force to be applied tothe punch.(b)π(0.050)2πd 2 Area of punch 44 0.001963 m2forceCompressive

vi MECHANICAL ENGINEERING PRINCIPLES 8 Bending of beams 103 8.1 Introduction 103 8.2 To prove that σ y M I E R 103 8.3 Worked problems on the bending of beams 105 9 Torque 109 9.1 Couple and torque 109 9.2 Work done and power transmitted by a constant torque 110 9.3 Kinetic energy

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