Machine Design - Bhandari (3rd Edition) PDF

Alive PDF Merger: Order full version from www.alivemedia.net to remove this watermark! Design of Machine Elements Third Edition  About the Au...

Alive PDF Merger: Order full version from www.alivemedia.net to remove this watermark! Design of Machine Elements Third Edition  About the Author V B Bhandari retired as Professor and Head, Department of Mechanical Engineering at Vishwakarma Institute of Technology, Pune. He holds First-Class BE and ME degrees in Mechanical Engineering from Pune University, and his teaching experience spans over 38 years in Government Colleges of Engineering at Pune, Karad and Aurangabad. He was also a postgraduate teacher of Pune University, Shivaji University and Marathwada University. Besides being a National Scholar, he has received five prizes from Pune University during his academic career. Professor Bhandari was a member of ‘Board of Studies in Mechanical Engineering’ and a member of ‘Faculty of Engineering’ of Pune University. He is a Fellow of Institution of Engineers (India), a Fellow of Institution of Mechanical Engineers (India) and a Senior Member of Computer Society of India. He was a Fellow of Institution of Production Engineers (India) and a Member of American Society of Mechanical Engineers (USA). He has presented and published twenty technical papers in national and international conferences and journals, and is also the author of Introduction to Machine Design published by Tata McGraw Hill Education Private Limited. Contents Preface xvii Visual Walkthrough xxi 1. Introduction 1 1.1 Machine Design 1 1.2 Basic Procedure of Machine Design 2 1.3 Basic Requirements of Machine Elements 3 1.4 Design of Machine Elements 4 1.5 Traditional Design Methods 8 1.6 Design Synthesis 8 1.7 Use of Standards in Design 9 1.8 Selection of Preferred Sizes 11 1.9 Aesthetic Considerations in Design 14 1.10 Ergonomic Considerations in Design 15 1.11 Concurrent Engineering 17 Short Answer Questions 19 Problems for Practice 19 2. Engineering Materials 20 2.1 Stress–Strain Diagrams 20 2.2 Mechanical Properties of Engineering Materials 23 2.3 Cast Iron 26 2.4 BIS System of Designation of Steels 29 2.5 Plain-carbon Steels 30 2.6 Free-cutting Steels 32 2.7 Alloy Steels 32 2.8 Overseas Standards 34 2.9 Heat Treatment of Steels 36 2.10 Case Hardening of Steels 37 2.11 Cast Steel 38 vi Contents 2.12 Aluminium Alloys 39 2.13 Copper Alloys 41 2.14 Die-casting Alloys 43 2.15 Ceramics 44 2.16 Plastics 45 2.17 Fibre-reinforced Plastics 48 2.18 Natural and Synthetic Rubbers 49 2.19 Creep 50 2.20 Selection of Material 51 2.21 Weighted Point Method 51 Short Answer Questions 53 3. Manufacturing Considerations in Design 55 3.1 Selection of Manufacturing Method 55 3.2 Design Considerations of Castings 57 3.3 Design Considerations of Forgings 59 3.4 Design Considerations of Machined Parts 61 3.5 Hot and Cold Working of Metals 62 3.6 Design Considerations of Welded Assemblies 62 3.7 Design for Manufacture and Assembly (DFMA) 64 3.8 Tolerances 65 3.9 Types of Fits 66 3.10 BIS System of Fits and Tolerances 67 3.11 Selection of Fits 69 3.12 Tolerances and Manufacturing Methods 69 3.13 Selective Assembly 70 3.14 Tolerances For Bolt Spacing 72 3.15 Surface Roughness 73 Short Answer Questions 73 Problems for Practice 74 4. Design Against Static Load 76 4.1 Modes of Failure 76 4.2 Factor of Safety 77 4.3 Stress–strain Relationship 79 4.4 Shear Stress and Shear Strain 80 4.5 Stresses Due To Bending Moment 81 4.6 Stresses Due To Torsional Moment 82 4.7 Eccentric Axial Loading 83 4.8 Design of Simple Machine Parts 84 4.9 Cotter Joint 85 4.10 Design Procedure for Cotter Joint 90 4.11 Knuckle Joint 94 4.12 Design Procedure for Knuckle Joint 99 4.13 Principal Stresses 104 4.14 Theories of Elastic Failure 106  Contents vii 4.15 Maximum Principal Stress Theory 107 4.16 Maximum Shear Stress Theory 108 4.17 Distortion-Energy Theory 110 4.18 Selection and Use of Failure Theories 112 4.19 Levers 117 4.20 Design of Levers 118 4.21 Fracture Mechanics 128 4.22 Curved Beams 130 4.23 Thermal Stresses 135 4.24 Residual Stresses 136 Short Answer Questions 137 Problems for Practice 138 5. Design Against Fluctuating Load 141 5.1 Stress Concentration 141 5.2 Stress Concentration Factors 142 5.3 Reduction of Stress Concentration 145 5.4 Fluctuating Stresses 149 5.5 Fatigue Failure 151 5.6 Endurance Limit 152 5.7 Low-cycle and High-cycle Fatigue 153 5.8 Notch Sensitivity 154 5.9 Endurance Limit—Approximate Estimation 155 5.10 Reversed Stresses—Design for Finite and Infinite Life 159 5.11 Cumulative Damage in Fatigue 166 5.12 Soderberg and Goodman Lines 167 5.13 Modified Goodman Diagrams 168 5.14 Gerber Equation 174 5.15 Fatigue Design under Combined Stresses 177 5.16 Impact Stresses 180 Short Answer Questions 182 Problems for Practice 182 6. Power Screws 184 6.1 Power Screws 184 6.2 Forms of Threads 185 6.3 Multiple Threaded Screws 187 6.4 Terminology of Power Screw 187 6.5 Torque Requirement—Lifting Load 189 6.6 Torque Requirement—Lowering Load 189 6.7 Self-locking Screw 190 6.8 Efficiency of Square Threaded Screw 190 6.9 Efficiency of Self-locking Screw 192 6.10 Trapezoidal and Acme Threads 192 6.11 Collar Friction Torque 193 6.12 Overall Efficiency 194 viii Contents 6.13 Coefficient of Friction 194 6.14 Design of Screw and Nut 194 6.15 Design of Screw Jack 206 6.16 Differential and Compound Screws 214 6.17 Recirculating Ball Screw 215 Short-Answer Questions 216 Problems for Practice 217 7. Threaded Joints 219 7.1 Threaded Joints 219 7.2 Basic Types of Screw Fastening 220 7.3 Cap Screws 222 7.4 Setscrews 223 7.5 Bolt of Uniform Strength 224 7.6 Locking Devices 225 7.7 Terminology of Screw Threads 227 7.8 ISO Metric Screw Threads 228 7.9 Materials and Manufacture 230 7.10 Bolted Joint—Simple Analysis 231 7.11 Eccentrically Loaded Bolted Joints in Shear 233 7.12 Eccentric Load Perpendicular to Axis of Bolt 235 7.13 Eccentric Load on Circular Base 242 7.14 Torque Requirement for Bolt Tightening 248 7.15 Dimensions of Fasteners 249 7.16 Design of Turnbuckle 251 7.17 Elastic Analysis of Bolted Joints 254 7.18 Bolted Joint under Fluctuating Load 257 Short-Answer Questions 269 Problems for Practice 269 8. Welded and Riveted Joints 272 8.1 Welded Joints 272 8.2 Welding Processes 273 8.3 Stress Relieving of Welded Joints 274 8.4 Butt Joints 274 8.5 Fillet Joints 275 8.6 Strength of Butt Welds 276 8.7 Strength of Parallel Fillet Welds 277 8.8 Strength of Transverse Fillet Welds 278 8.9 Maximum Shear Stress in Parallel Fillet Weld 281 8.10 Maximum Shear Stress in Transverse Fillet Weld 282 8.11 Axially Loaded Unsymmetrical Welded Joints 284 8.12 Eccentric Load in the Plane of Welds 285 8.13 Welded Joint Subjected to Bending Moment 290 8.14 Welded Joint Subjected to Torsional Moment 294 8.15 Strength of Welded Joints 295  Contents ix 8.16 Welded Joints Subjected to Fluctuating Forces 296 8.17 Welding Symbols 297 8.18 Weld Inspection 298 8.19 Riveted Joints 298 8.20 Types of Rivet Heads 301 8.21 Types of Riveted Joints 303 8.22 Rivet Materials 305 8.23 Types of Failure 306 8.24 Strength Equations 306 8.25 Efficiency of Joint 307 8.26 Caulking and Fullering 307 8.27 Longitudinal Butt Joint for Boiler Shell 311 8.28 Circumferential Lap Joint for Boiler Shells 318 8.29 Eccentrically Loaded Riveted Joint 321 Short-Answer Questions 325 Problems for Practice 325 9. Shafts, Keys and Couplings 330 9.1 Transmission Shafts 330 9.2 Shaft Design on Strength Basis 331 9.3 Shaft Design on Torsional Rigidity Basis 333 9.4 ASME Code for Shaft Design 334 9.5 Design of Hollow Shaft on Strength Basis 342 9.6 Design of Hollow Shaft on Torsional Rigidity Basis 344 9.7 Flexible Shafts 346 9.8 Keys 346 9.9 Saddle Keys 347 9.10 Sunk Keys 348 9.11 Feather Key 349 9.12 Woodruff Key 350 9.13 Design of Square and Flat Keys 350 9.14 Design of Kennedy Key 352 9.15 Splines 354 9.16 Couplings 356 9.17 Muff Coupling 357 9.18 Design Procedure for Muff Coupling 357 9.19 Clamp Coupling 359 9.20 Design Procedure for Clamp Coupling 360 9.21 Rigid Flange Couplings 362 9.22 Design Procedure for Rigid Flange Coupling 364 9.23 Bushed-pin Flexible Coupling 368 9.24 Design Procedure for Flexible Coupling 371 9.25 Design for Lateral Rigidity 376 9.26 Castigliano’s Theorem 380 x Contents 9.27 Area Moment Method 382 9.28 Graphical Integration Method 383 9.29 Critical Speed of Shafts 385 Short-Answer Questions 388 Problems for Practice 389 10. Springs 393 10.1 Springs 393 10.2 Types of Springs 393 10.3 Terminology of Helical Springs 395 10.4 Styles of End 396 10.5 Stress and Deflection Equations 397 10.6 Series and Parallel Connections 399 10.7 Spring Materials 401 10.8 Design of Helical Springs 403 10.9 Spring Design—Trial-and-Error Method 405 10.10 Design against Fluctuating Load 405 10.11 Concentric Springs 425 10.12 Optimum Design of Helical Spring 430 10.13 Surge in Spring 432 10.14 Helical Torsion Springs 433 10.15 Spiral Springs 435 10.16 Multi-Leaf Spring 437 10.17 Nipping of Leaf Springs 439 10.18 Belleville Spring 441 10.19 Shot Peening 443 Short-Answer Questions 443 Problems for Practice 444 11. Friction Clutches 448 11.1 Clutches 448 11.2 Torque Transmitting Capacity 450 11.3 Multi-disk Clutches 456 11.4 Friction Materials 459 11.5 Cone Clutches 461 11.6 Centrifugal Clutches 465 11.7 Energy Equation 467 11.8 Thermal Considerations 469 Short-Answer Questions 470 Problems for Practice 471 12. Brakes 472 12.1 Brakes 472 12.2 Energy Equations 472 12.3 Block Brake with Short Shoe 475 12.4 Block Brake with Long Shoe 480  Contents xi 12.5 Pivoted Block Brake with Long Shoe 482 12.6 Internal Expanding Brake 485 12.7 Band Brakes 490 12.8 Disk Brakes 493 12.9 Thermal Considerations 496 Short-Answer Questions 496 Problems for Practice 497 13. Belt Drives 499 13.1 Belt Drives 499 13.2 Belt Constructions 501 13.3 Geometrical Relationships 503 13.4 Analysis of Belt Tensions 504 13.5 Condition for Maximum Power 507 13.6 Condition for Maximum Power (Alternative Approach) 507 13.7 Characteristics of Belt Drives 509 13.8 Selection of Flat-belts from Manufacturer’s Catalogue 514 13.9 Pulleys for Flat Belts 517 13.10 Arms of Cast-iron Pulley 520 13.11 V-belts 522 13.12 Selection of V-belts 534 13.13 V-grooved Pulley 535 13.14 Belt-Tensioning Methods 540 13.15 Ribbed V-belts 540 Short-Answer Questions 542 Problems for Practice 542 14. Chain Drives 544 14.1 Chain Drives 544 14.2 Roller Chains 546 14.3 Geometric Relationships 548 14.4 Polygonal Effect 549 14.5 Power Rating of Roller Chains 549 14.6 Sprocket Wheels 551 14.7 Design of Chain Drive 553 14.8 Chain Lubrication 555 14.9 Silent Chain 562 Short-Answer Questions 562 Problems for Practice 563 15. Rolling Contact Bearings 564 15.1 Bearings 564 15.2 Types of Rolling-contact Bearings 565 15.3 Principle of Self-aligning Bearing 568 15.4 Selection of Bearing-type 569 15.5 Static Load Carrying Capacity 569 xii Contents 15.6 Stribeck’s Equation 569 15.7 Dynamic Load Carrying Capacity 571 15.8 Equivalent Bearing Load 571 15.9 Load-Life Relationship 572 15.10 Selection of Bearing Life 572 15.11 Load Factor 573 15.12 Selection of Bearing from Manufacturer’s Catalogue 573 15.13 Selection of Taper Roller Bearings 580 15.14 Design for Cyclic Loads and Speeds 588 15.15 Bearing with Probability of Survival other than 90 Per Cent 592 15.16 Needle Bearings 595 15.17 Bearing Failure—Causes and Remedies 596 15.18 Lubrication of Rolling Contact Bearings 596 15.19 Mounting of Bearing 597 Short-Answer Questions 598 Problems for Practice 599 16. Sliding Contact Bearings 601 16.1 Basic Modes of Lubrication 601 16.2 Viscosity 604 16.3 Measurement of Viscosity 605 16.4 Viscosity Index 605 16.5 Petroff’s Equation 606 16.6 McKee’s Investigation 607 16.7 Viscous Flow through Rectangular Slot 608 16.8 Hydrostatic Step Bearing 609 16.9 Energy Losses in Hydrostatic Bearing 611 16.10 Reynold’s Equation 619 16.11 Raimondi and Boyd Method 622 16.12 Temperature Rise 624 16.13 Bearing Design—Selection of Parameters 625 16.14 Bearing Constructions 634 16.15 Bearing Materials 635 16.16 Sintered Metal Bearings 637 16.17 Lubricating Oils 637 16.18 Additives for Mineral Oils 639 16.19 Selection of Lubricants 640 16.20 Greases 641 16.21 Bearing Failure—Causes and Remedies 641 16.22 Comparison of Rolling and Sliding Contact Bearings 642 Short-Answer Questions 643 Problems for Practice 644 17. Spur Gears 646 17.1 Mechanical Drives 646 17.2 Gear Drives 647 xii Contents 15.6 Stribeck’s Equation 569 15.7 Dynamic Load Carrying Capacity 571 15.8 Equivalent Bearing Load 571 15.9 Load-Life Relationship 572 15.10 Selection of Bearing Life 572 15.11 Load Factor 573 15.12 Selection of Bearing from Manufacturer’s Catalogue 573 15.13 Selection of Taper Roller Bearings 580 15.14 Design for Cyclic Loads and Speeds 588 15.15 Bearing with Probability of Survival other than 90 Per Cent 592 15.16 Needle Bearings 595 15.17 Bearing Failure—Causes and Remedies 596 15.18 Lubrication of Rolling Contact Bearings 596 15.19 Mounting of Bearing 597 Short-Answer Questions 598 Problems for Practice 599 16. Sliding Contact Bearings 601 16.1 Basic Modes of Lubrication 601 16.2 Viscosity 604 16.3 Measurement of Viscosity 605 16.4 Viscosity Index 605 16.5 Petroff’s Equation 606 16.6 McKee’s Investigation 607 16.7 Viscous Flow through Rectangular Slot 608 16.8 Hydrostatic Step Bearing 609 16.9 Energy Losses in Hydrostatic Bearing 611 16.10 Reynold’s Equation 619 16.11 Raimondi and Boyd Method 622 16.12 Temperature Rise 624 16.13 Bearing Design—Selection of Parameters 625 16.14 Bearing Constructions 634 16.15 Bearing Materials 635 16.16 Sintered Metal Bearings 637 16.17 Lubricating Oils 637 16.18 Additives for Mineral Oils 639 16.19 Selection of Lubricants 640 16.20 Greases 641 16.21 Bearing Failure—Causes and Remedies 641 16.22 Comparison of Rolling and Sliding Contact Bearings 642 Short-Answer Questions 643 Problems for Practice 644 17. Spur Gears 646 17.1 Mechanical Drives 646 17.2 Gear Drives 647  Contents xiii 17.3 Classification of Gears 647 17.4 Selection of Type of Gears 648 17.5 Law of Gearing 649 17.6 Terminology of Spur Gears 650 17.7 Standard Systems of Gear Tooth 653 17.8 Gear Trains 656 17.9 Interference and Undercutting 657 17.10 Backlash 658 17.11 Force Analysis 658 17.12 Gear Tooth Failures 665 17.13 Selection of Material 666 17.14 Gear Blank Design 667 17.15 Number of Teeth 670 17.16 Face Width 671 17.17 Beam Strength of Gear Tooth 672 17.18 Permissible Bending Stress 673 17.19 Effective Load on Gear Tooth 674 17.20 Estimation of Module Based on Beam Strength 677 17.21 Wear Strength of Gear Tooth 678 17.22 Estimation of Module Based on Wear Strength 680 17.23 Internal Gears 688 17.24 Gear Lubrication 690 Short-Answer Questions 690 Problems for Practice 690 18. Helical Gears 694 18.1 Helical Gears 694 18.2 Terminology of Helical Gears 694 18.3 Virtual Number of Teeth 695 18.4 Tooth Proportions 696 18.5 Force Analysis 697 18.6 Beam Strength of Helical Gears 702 18.7 Effective Load on Gear Tooth 702 18.8 Wear Strength of Helical Gears 703 18.9 Herringbone Gears 706 18.10 Crossed Helical Gears 708 Short-Answer Questions 709 Problems for Practice 710 19. Bevel Gears 711 19.1 Bevel Gears 711 19.2 Terminology of Bevel Gears 713 19.3 Force Analysis 715 19.4 Beam Strength of Bevel Gears 720 19.5 Wear Strength of Bevel Gears 722 19.6 Effective Load on Gear Tooth 722 xiv Contents 19.7 Spiral Bevel Gears 727 Short-Answer Questions 728 Problems for Practice 728 20. Worm Gears 730 20.1 Worm Gears 730 20.2 Terminology of Worm Gears 731 20.3 Proportions of Worm Gears 733 20.4 Force Analysis 735 20.5 Friction in Worm Gears 737 20.6 Selection of Materials 741 20.7 Strength Rating of Worm Gears 742 20.8 Wear Rating of Worm Gears 744 20.9 Thermal Considerations 745 Short-Answer Questions 747 Problems for Practice 747 21. Flywheel 749 21.1 Flywheel 749 21.2 Flywheel and Governor 750 21.3 Flywheel Materials 750 21.4 Torque Analysis 751 21.5 Coefficient of Fluctuation of Energy 752 21.6 Solid Disk Flywheel 753 21.7 Rimmed Flywheel 755 21.8 Stresses in Rimmed Flywheel 756 Short-Answer Questions 767 Problems for Practice 767 22. Cylinders and Pressure Vessels 768 22.1 Thin Cylinders 768 22.2 Thin Spherical Vessels 769 22.3 Thick Cylinders—Principal Stresses 770 22.4 Lame’s Equation 771 22.5 Clavarino’s and Birnie’s Equations 772 22.6 Cylinders with External Pressure 774 22.7 Autofrettage 775 22.8 Compound Cylinder 775 22.9 Gaskets 779 22.10 Gasketed Joint 780 22.11 Unfired Pressure Vessels 783 22.12 Thickness of Cylindrical and Spherical Shells 785 22.13 End Closures 785 22.14 Openings in Pressure Vessel 791 Short-Answer Questions 794 Problems for Practice 794  Contents xv 23. Miscellaneous Machine Elements 796 23.1 Oil Seals 796 23.2 Wire Ropes 797 23.3 Stresses in Wire Ropes 800 23.4 Rope Sheaves and Drums 804 23.5 Buckling of Columns 806 Short-Answer Questions 812 Problems for Practice 812 24. Statistical Considerations in Design 814 24.1 Frequency Distribution 814 24.2 Characteristics of Frequency Curves 816 24.3 Measures of Central Tendency and Dispersion 817 24.4 Probability 819 24.5 Probability Distribution 819 24.6 Normal Curve 821 24.7 Population Combinations 823 24.8 Design and Natural Tolerances 825 24.9 Reliability 829 24.10 Probabilistic Approach to Design 830 Short-Answer Questions 840 Problems for Practice 841 25. Design of IC Engine Components 843 25.1 Internal Combustion Engine 843 25.2 Cylinder and Cylinder Liner 844 25.3 Bore and Length of Cylinder 845 25.4 Thickness of Cylinder Wall 845 25.5 Stresses in Cylinder Wall 846 25.6 Cylinder Head 847 25.7 Design of Studs for Cylinder Head 847 25.8 Piston 853 25.9 Piston Materials 854 25.10 Thickness of Piston Head 854 25.11 Piston Ribs and Cup 855 25.12 Piston Rings 856 25.13 Piston Barrel 857 25.14 Piston Skirt 858 25.15 Piston Pin 858 25.16 Connecting Rod 867 25.17 Buckling of Connecting Rod 868 25.18 Cross-section for Connecting Rod 869 25.19 Big and Small End Bearings 871 xvi Contents 25.20 Big End Cap and Bolts 873 25.21 Whipping Stress 875 25.22 Crankshaft 880 25.23 Design of Centre Crankshaft 881 25.24 Centre Crankshaft at Top-Dead Centre Position 881 25.25 Centre Crankshaft at Angle of Maximum Torque 883 25.26 Side Crankshaft at Top-Dead Centre Position 892 25.27 Side Crankshaft at Angle of Maximum Torque 895 25.28 Valve-Gear Mechanism 903 25.29 Design of Valves 904 25.30 Design of Rocker Arm 906 25.31 Design of Valve Spring 910 25.32 Design of Push Rod 911 Short-Answer Questions 922 Problems for Practice 923 References 927 Index 930  Contents xvii Preface It was really a pleasure to receive an overwhelming response to the textbook Design of Machine Elements since it was published first in 1994. In fact, whenever I visit an engineering college in any part of the country, students and staff members of the Mechanical Engineering Department know me as the ‘Machine Design author’ and the book has become my identity. Machine design occupies a prominent position in the curriculum of Mechanical Engineering. It consists of applications of scientific principles, technical information and innovative ideas for the development of a new or improved machine. The task of a machine designer has never been easy, since he has to consider a number of factors, which are not always compatible with the present-day technology. In the context of today’s techni- cal and social climate, the designer’s task has become increasingly difficult. Today’s designer is required to account for many factors and considerations that are almost impossible for one individual to be thoroughly conversant with. At the same time, he cannot afford to play a role of something like that of a music director. He must have a special competence of his own and a reasonable knowledge of other ‘instruments.’ New to this Edition After the publication of the second edition in 2007, it was observed that there was a need to incorporate a broader coverage of topics in the textbook to suit the content of ‘Machine Design’ syllabi of various uni- versities in our country. One complete chapter on ‘Design of Engine Components’ (Chapter 25) and half a chapter on ‘Design of Riveted Joints’ (Chapter 8) are added to fulfill this requirement. Design of Engine Components includes cylinders, pistons, connecting rods, crankshafts and valve-gear mechanism. Design of Riveted Joints includes strength equations, eccentrically loaded joints and riveted joints in boiler shells. Another important feature of the current edition is changing the style of solutions to numerical examples. A ‘step-by-step’ approach is incorporated in all solved examples of the book. This will further simplify and clarify the understanding of the examples. Target Audience This book is intended to serve as a textbook for all the courses in Machine Design. It covers the syllabi of all universities, technical boards and professional examining bodies such as Institute of Engineers in the country. It is also useful for the preparation of competitive examinations like UPSC and GATE. This textbook is particularly written for the students of the Indian subcontinent, who find it difficult to adopt the textbooks written by foreign authors. Salient Features The main features of the book are the following: (i) SI system of units used throughout the book (ii) Indian standards used throughout the book for materials, tolerances, screw threads, springs, gears, wire ropes and pressure vessels xviii Preface (iii) The basic procedure for selection of a machine component from the manufacturer’s catalogue discussed with a particular reference to Indian products (iv) Step by step approach of problem solving Organization The book is divided into 25 chapters. Chapter 1 is an introductory chapter on machine design and discusses the various procedures, requirements, design methods and ergonomic considerations for design. Chapter 2 is on engineering materials and describes the different kinds of irons, steels and alloys used in engineering design. Chapter 3 explains in detail the manufacturing considerations in design. Chapters 4 and 5 discuss the various procedures for design against static load and fluctuating load correspondingly. Chapter 6 describes power screws in detail while chapters 7 and 8 specify the features and varieties of threaded joints, and welded and riveted joints in that order. Similarly, chapters 9 to 22 are each devoted to a particular design element, that is, shafts, keys and couplings; springs; friction clutches; brakes; belt drives; chain drives; rolling contact bearings; sliding contact bearings; spur gears; helical gears; bevel gears; worm gears; flywheel; cylinders and pressure vessels respectively. Chapter 23 describes miscellaneous machine elements like oil seals, wire ropes, rope sheaves and drums. Chapter 24 details the various statistical considerations in design. Finally, Chapter 25 explains the design of IC engine components. Web Resources The readers should note that there is a website of this textbook which can be accessed at http://www.mhhe.com/bhandari/dme3e that contains the following. For Instructors: (i) Solution Manual (ii) Power Point Lecture Slides For Students: (i) Interactive 643 Objective Type Questions (ii) 803 Short Answer Questions (iii) Glossary (iv) Bibliography The above additional information will be useful for students in preparing for competitive examinations. Acknowledgements For this textbook, information has been collected from various sources such as textbooks, handbooks, cata- logues and journals. I would like to express my gratitude to the authors, publishers and firms who have per- mitted me to use their valuable material in this textbook. The following firms and individuals need special mention: 1. A A Raimondi of Westinghouse Electric Corporation, USA for the data on ‘Dimensionless Performance Parameters of Hydrodynamic Bearings’ 2. George Sines, University of California, USA for the ‘Notch Sensitivity Charts’ in Fatigue Design 3. B K Sollars, President, Diamond Chain Company, USA, for his valuable suggestions and design data related to the selection of roller chains 4. McGraw-Hill Education, USA, for their permission to include the table of ‘Reliability Factors’ from their publication Mechanical Engineering Design by J E Shigley and ‘Surface Finish Factors’ in Fatigue Design from Engineering Considerations of Stress, Strain and Strength by R C Juvinall 5. Associated Bearing Company Limited, Mumbai, for their permission to include different tables for the selection of SKF bearings  Preface xix 6. The Dunlop Rubber Co. (India) Ltd., Kolkata, for their permission to include data for the ‘Selection of Dunlop belts’ 7. The Bureau of Indian Standards, New Delhi, for its permission to include extracts from standards- IS-4218, IS-7008, IS-919, IS-1570, IS-2644, IS- 733, IS-2403, IS-3681, IS-2266, IS-3973, IS-5129, IS-4454, IS-4694, IS-210, IS-1030, IS-617, IS-813, IS-25, IS-2825, IS-2365, IS-2494 and IS-7443 I acknowledge with a deep sense of gratitude, the encouragement and inspiration received from my stu- dents, readers and teachers. I would also like to thank the following reviewers of this edition whose names are given below. A Bhattacharya Institute of Technology, Banaras Hindu University Banaras, Uttar Pradesh A D Bhatt Motilal Nehru National Institute of Technology, Allahabad Allahabad, Uttar Pradesh Pratesh Jayaswal Madhav Institute of Technology and Science Gwalior, Madhya Pradesh Shivabrata Mojumdar Dr B C Roy Engineering College Durgapur, West Bengal Shashidhar K Kudari D Y Patil College of Engineering and Technology Kolhapur, Maharashtra P M Bapat Manohar Lal Patel Institute of Engineering and Technology Nagpur, Maharashtra A D Diwate Sinhagad Institute of Technology Pune, Maharashtra R Sesharajan Bharat Heary Electricals Limited Vela Murali College of Engineering, Guindy Guindy, Tamil Nadu K S Seetharama Adichunchanagiri Institute of technology Chikmagalur, Karnataka K Mallikarjuna Rao Jawaharlal Nehru Technological University College of Engineering Kakinada, Andhra Pradesh A C S Kumar Jawaharlal Nehru Technological University College of Engineering Hyderabad, Andhra Pradesh A special thanks to the Editorial and Production teams of Tata McGraw-Hill headed by Vibha Mahajan and her enthusiastic team members—Shalini Jha, Suman Sen, Devshree Lohchab, Sohini Mukherjee and P L Pandita. xx Preface Feedback Suggestions and comments for improvement of the book will be appreciated. They can be sent either to the publisher or to me at [email protected]. V B BHANDARI Publisher’s Note Do you have a feature request? A suggestion? We are always open to new ideas (the best ideas come from you!). You may send your comments to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.  Contents xxi Visual Walkthrough Introduction Each chapter begins with an Introduction of the Machine Element designed in the chapter and its functions. This helps the reader in gaining an overview of the machine element. Theoretical Considerations Basic equations for design are derived from first principle, with a step-by- step approach. xxii Visual Walkthrough Properties of Materials Exhaustive tables are provided from Indian Standards for Mechanical Properties of Engineering Materials. Indian Standards Indian Standards are used for Machine Elements like screw threads, belts, springs, gears, wire ropes and pressure vessels.  Visual Walkthrough xxiii Selection Procedure When a machine component is to be selected from manufacturer’s catalogue, the selection processes are discussed with a particular reference to Indian products. Free-Body Diagram of Forces Whenever required, free-body diagrams are constructed to help the reader understand the forces acting on individual components. Fatigue Diagrams Fatigue diagrams are constructed for design of machine components subjected to fluctuating loads. xxiv Visual Walkthrough Isometric Views When it is difficult to understand the forces in three dimensions, isometric views are given for clear understanding. Statistical Considerations in Design A separate chapter on Statistical Considerations in Design is included and examples are solved on the basis of reliability. Numerical Examples Numerical Examples solved by step by step approach are provided in sufficient number in each chapter to help the reader understand the design procedures.  Visual Walkthrough xxv Short-Answer Questions At the end of each chapter, Short- Answer Questions are provided for the students for preparation of oral and theory examinations. Problems for Practice At the end of each chapter, a set of examples with answers is given as exercise to students. It is also helpful to teachers in setting classwork and homework assignments. References The list of textbooks, journals and company catalogues is provided at the end of respective pages for quick reference. Introduction Chapter 1 If the point of contact between the product and people becomes a point of friction, then the industrial designer has failed. On the other hand, if people are made safer, more efficient, more comfortable—or just plain happier—by contact with the product, then the designer has succeeded. Henry Dreyfuss1 1.1 MACHINE DESIGN belt and gear drives, bearings, oil seals and gaskets, springs, shafts, keys, couplings, Machine design is defined as the use of scientific and so on. A machine is a combination of principles, technical information and imagination these basic elements. The designer knows in the description of a machine or a mechanical the relative advantages and disadvantages of system to perform specific functions with maximum these basic elements and their suitability in economy and efficiency. This definition of machine different applications. design contains the following important features: (iii) The designer uses his skill and imagination (i) A designer uses principles of basic and to produce a configuration, which is a engineering sciences such as physics, combination of these basic elements. mathematics, statics and dynamics, However, this combination is unique thermodynamics and heat transfer, vibrations and different in different situations. The and fluid mechanics. Some of the examples intellectual part of constructing a proper of these principles are configuration is creative in nature. (a) Newton’s laws of motion, (iv) The final outcome of the design process (b) D’ Alembert’s principle, consists of the description of the machine. (c) Boyle’s and Charles’ laws of gases, The description is in the form of drawings of (d) Carnot cycle, and assembly and individual components. (e) Bernoulli’s principle. (v) A design is created to satisfy a recognised (ii) The designer has technical information of need of customer. The need may be to the basic elements of a machine. These perform a specific function with maximum elements include fastening devices, chain, economy and efficiency. 1 Henry Dreyfuss–The Profile of Industrial Designer—Machine Design, July 22, 1967. 2 Design of Machine Elements Machine design is the creation of plans for include the output capacity of the machine, and its a machine to perform the desired functions. service life, cost and reliability. In some cases, the The machine may be entirely new in concept, overall dimensions and weight of the product are performing a new type of work, or it may more specified. For example, while designing a scooter, economically perform the work that can be done the list of specifications will be as follows: by an existing machine. It may be an improvement (i) Fuel consumption = 40 km/l or enlargement of an existing machine for better (ii) Maximum speed = 85 km/hr economy and capability. (iii) Carrying capacity = two persons with 10 kg luggage 1.2 BASIC PROCEDURE OF MACHINE (iv) Overall dimensions DESIGN Width = 700 mm Length = 1750 mm The basic procedure of machine design consists of Height = 1000 mm a step-by-step approach from given specifications (v) Weight = 95 kg about the functional requirements of a product to (vi) Cost = Rs 40000 to Rs 45000 the complete description in the form of drawings In consumer products, external appearance, of the final product. A logical sequence of steps, noiseless performance and simplicity in operation usually common to all design projects, is illustrated of controls are important requirements. Depending in Fig. 1.1. These steps are interrelated and upon the type of product, various requirements are interdependent, each reflecting and affecting all given weightages and a priority list of specifications is prepared. Step 2: Selection of Mechanism After careful study of the requirements, the designer prepares rough sketches of different possible mechanisms for the product. For example, while designing a blanking or piercing press, the following mechanisms are possible: (i) a mechanism involving the crank and connecting rod, converting the rotary motion of the electric motor into the reciprocating motion of the punch; (ii) a mechanism involving nut and screw, which is a simple and cheap configuration but having poor efficiency; and (iii) a mechanism consisting of a hydraulic cylinder, piston and valves which is a costly configuration but highly efficient. The alternative mechanisms are compared Fig. 1.1 The Design Process with each other and also with the mechanism of the products that are available in the market. other steps. The following steps are involved in the An approximate estimation of the cost of each process of machine design. alternative configuration is made and compared Step 1: Product Specifications with the cost of existing products. This will reveal the competitiveness of the product. While selecting The first step consists of preparing a complete list of the final configuration, the designer should the requirements of the product. The requirements  Introduction 3 consider whether the raw materials and standard ultimate tensile strength, endurance limit or parts required for making the product are available permissible deflection. in the market. He should also consider whether (iv) Determine the geometric dimensions of the the manufacturing processes required to fabricate component using a suitable factor of safety the non-standard components are available in the and modify the dimensions from assembly factory. Depending upon the cost-competitiveness, and manufacturing considerations. availability of raw materials and manufacturing This stage involves detailed stress and deflection facility, the best possible mechanism is selected for analysis. The subjects ‘Machine Design’ or the product. ‘Elements of Machine Design’ cover mainly the design of machine elements or individual Step 3: Layout of Configuration components of the machine. Section 1.4 on Design The next step in a design procedure is to prepare of Machine Elements, elaborates the details of this a block diagram showing the general layout of the important step in design procedure. selected configuration. For example, the layout of an Electrically-operated Overhead Travelling (EOT) Step 5: Preparation of Drawings crane will consist of the following components: The last stage in a design process is to prepare (i) electric motor for power supply; drawings of the assembly and the individual (ii) flexible coupling to connect the motor shaft components. On these drawings, the material of to the clutch shaft; the component, its dimensions, tolerances, surface (iii) clutch to connect or disconnect the electric finish grades and machining symbols are specified. motor at the will of the operator; The designer prepares two separate lists of (iv) gear box to reduce the speed from 1440 rpm components—standard components to be purchased to about 15 rpm; directly from the market and special components (v) rope drum to convert the rotary motion of the to be machined in the factory. In many cases, a shaft to the linear motion of the wire rope; prototype model is prepared for the product and (vi) wire rope and pulley with the crane hook to thoroughly tested before finalising the assembly attach the load; and drawings. (vii) brake to stop the motion. It is seen that the process of machine design In this step, the designer specifies the joining involves systematic approach from known methods, such as riveting, bolting or welding to specifications to unknown solutions. Quite connect the individual components. Rough sketches often, problems arise on the shop floor during of shapes of the individual parts are prepared. the production stage and design may require modifications. In such circumstances, the designer Step 4: Design of Individual Components has to consult the manufacturing engineer and find The design of individual components or machine out the suitable modification. elements is an important step in a design process. It consists of the following stages: 1.3 BASIC REQUIREMENTS OF (i) Determine the forces acting on the MACHINE ELEMENTS component. (ii) Select proper material for the component A machine consists of machine elements. Each part depending upon the functional requirements of a machine, which has motion with respect to some such as strength, rigidity, hardness and wear other part, is called a machine element. It is important resistance. to note that each machine element may consist of (iii) Determine the likely mode of failure for the several parts, which are manufactured separately. component and depending upon it, select the For example, a rolling contact bearing is a machine criterion of failure, such as yield strength, element and it consists of an inner race, outer race, 4 Design of Machine Elements cage and rolling elements like balls. Machine elements (v) Manufacturability: Manufacturability is the can be classified into two groups—general-purpose ease of fabrication and assembly. The shape and and special-purpose machine elements. General- material of the machine part should be selected in purpose machine elements include shafts, couplings, such a way that it can be produced with minimum clutches, bearings, springs, gears and machine frames labour cost. Special-purpose machine elements include pistons, (vi) Safety: The shape and dimensions of the valves or spindles. Special-purpose machine elements machine parts should ensure safety to the operator are used only in certain types of applications. On the of the machine. The designer should assume the contrary, general-purpose machine elements are used worst possible conditions and apply ‘fail-safe’ or in a large number of machines. ‘redundancy’ principles in such cases. The broad objective of designing a machine element is to ensure that it preserves its operating (vii) Conformance to Standards: A machine part capacity during the stipulated service life with should conform to the national or international minimum manufacturing and operating costs. standard covering its profile, dimensions, grade and In order to achieve this objective, the machine material. element should satisfy the following basic (viii) Reliability: Reliability is the probability that requirements: a machine part will perform its intended functions under desired operating conditions over a specified (i) Strength: A machine part should not fail under period of time. A machine part should be reliable, the effect of the forces that act on it. It should have that is, it should perform its function satisfactorily sufficient strength to avoid failure either due to over its lifetime. fracture or due to general yielding. (ix) Maintainability: A machine part should be (ii) Rigidity: A machine component should be rigid, maintainable. Maintainability is the ease with that is, it should not deflect or bend too much due which a machine part can be serviced or repaired. to forces or moments that act on it. A transmission shaft in many times designed on the basis of lateral (x) Minimum: Life-cycle Cost: Life-cycle cost of and torsional rigidities. In these cases, maximum the machine part is the total cost to be paid by the permissible deflection and permissible angle of purchaser for purchasing the part and operating and twist are the criteria for design. maintaining it over its life span. It will be observed that the above mentioned (iii) Wear Resistance: Wear is the main reason for requirements serve as the basis for design projects putting the machine part out of order. It reduces in many cases. useful life of the component. Wear also leads to the loss of accuracy of machine tools. There are 1.4 DESIGN OF MACHINE ELEMENTS different types of wear such as abrasive wear, corrosive wear and pitting. Surface hardening Design of machine elements is the most important can increase the wear resistance of the machine step in the complete procedure of machine design. components, such as gears and cams. In order to ensure the basic requirements of machine elements, calculations are carried out to (iv) Minimum Dimensions and Weight: A machine find out the dimensions of the machine elements. part should be sufficiently strong, rigid and wear- These calculations form an integral part of the resistant and at the same time, with minimum design of machine elements. The basic procedure possible dimensions and weight. This will result in of the design of machine elements is illustrated in minimum material cost. Fig. 1.2. It consists of the following steps:  Introduction 5 (i) The external force due to energy, power or torque transmitted by the machine part, often called ‘useful’ load (ii) Static force due to deadweight of the machine part (iii) Force due to frictional resistance (iv) Inertia force due to change in linear or angular velocity (v) Centrifugal force due to change in direction of velocity (vi) Force due to thermal gradient or variation in temperature (vii) Force set up during manufacturing the part resulting in residual stresses (viii) Force due to particular shape of the part such as stress concentration due to abrupt change in cross-section For every machine element, all forces in this list may not be applicable. They vary depending on the application. There is one more important Fig. 1.2 Basic Procedure of Design of Machine consideration. The force acting on the machine Element part is either assumed to be concentrated at some point in the machine part or distributed over a Step 1: Specification of Function particular area. Experience is essential to make The design of machine elements begins with the such assumptions in the analysis of forces. specification of the functions of the element. The functions of some machine elements are as follows: Step 3: Selection of Material (i) Bearing To support the rotating shaft and Four basic factors, which are considered in selecting confine its motion the material, are availability, cost, mechanical (ii) Key To transmit the torque between the properties and manufacturing considerations. shaft and the adjoining machine part like For example, flywheel, housing of gearbox gear, pulley or sprocket or engine block have complex shapes. These (iii) Spring in Clock To store and release the components are made of cast iron because the energy casting process produces complicated shapes without (iv) Spring in Spring Balance To measure the involving machining operations. Transmission shafts force are made of plain carbon steels, because they are (v) Screw Fastening To hold two or more available in the form of rods, besides their higher machine parts together strength. The automobile body and hood are made (vi) Power Screw To produce uniform and of low carbon steels because their cold formability is slow motion and to transmit the force essential to press the parts. Free cutting steels have excellent machinability due to addition of sulphur. Step 2: Determination of Forces They are ideally suitable for bolts and studs because In many cases, a free-body diagram of forces of the ease with which the thread profiles can be is constructed to determine the forces acting on machined. The crankshaft and connecting rod are different parts of the machine. The external and subjected to fluctuating forces and nickel–chromium internal forces that act on a machine element are as steel is used for these components due to its higher follows: fatigue strength. 6 Design of Machine Elements Step 4: Failure Criterion shape which will match with the adjoining belt. The Before finding out the dimensions of the component, profile of the teeth of sprocket wheel should match it is necessary to know the type of failure that the the roller, bushing, inner and outer link plates of the component may fail when put into service. The roller chain. Depending on the operating conditions machine component is said to have ‘failed’ when it and shape of the adjoining element, the shape of is unable to perform its functions satisfactorily. The the machine element is decided and a rough sketch three basic types of failure are as follows: is prepared. (i) failure by elastic deflection; The geometric dimensions of the component (ii) failure by general yielding; and are determined on the basis of failure criterion. In (iii) failure by fracture. simple cases, the dimensions are determined on the In applications like transmission shaft, which basis of allowable stress or deflection. For example, is used to support gears, the maximum force a tension rod, illustrated in Fig. 1.3, is subjected to acting on the shaft is limited by the permissible a force of 5 kN. The rod is made of plain carbon deflection. When this deflection exceeds a particular value (usually, 0.001 to 0.003 times of span length between two bearings), the meshing between teeth of gears is affected and the shaft cannot perform its function properly. In this case, the shaft is said to have ‘failed’ due to elastic deflection. Components made of ductile materials like steel lose their engineering usefulness due to large amount of plastic deformation. This type of failure is called failure by yielding. Components made of brittle materials like cast iron fail because of sudden fracture without any plastic deformation. There are two basic modes of gear-tooth failure— breakage of tooth due to static and dynamic load and surface pitting. The surface of the gear tooth is covered with small ‘pits’ resulting in rapid wear. Pitting is a surface fatigue failure. The components of ball bearings such as rolling elements, inner and outer races fail due to fatigue cracks after certain number of revolutions. Sliding contact bearings fail due to corrosion and abrasive wear by foreign particles. Step 5: Determination of Dimensions The shape of the machine element depends on two factors, viz., the operating conditions and the shape of the adjoining machine element. For example, involute profile is used for gear teeth because it satisfies the fundamental law of gearing. A V-belt has a trapezoidal cross-section because it results Fig. 1.3 Tension Rod in wedge action and increases the force of friction steel and the permissible tensile stress is 80 N/mm2. between the surfaces of the belt and the pulley. On The diameter of the rod is determined on the basis the other hand, the pulley of a V-belt should have a of allowable stress using the following expression:  Introduction 7 by the factor of safety. Therefore, yield force (5 ¥ 103 ) strength is the criterion of design. In case stress = or 80 = area Ê pd2 ˆ of a transmission shaft, lateral deflection or Á 4 ˜ rigidity is the criterion of design. Therefore, Ë ¯ Therefore, modulus of elasticity is an important d = 8.92 or 10 mm property for finding out the dimensions of As a second example, consider a transmission the shaft. shaft, shown in Fig. 1.4, which is used to support Determination of geometric dimensions is an a gear. The shaft is made of steel and the modulus important step while designing machine elements. Various criteria such as yield strength, ultimate tensile strength, torsional or lateral deflection and permissible bearing pressure are used to find out these dimensions. 5 kN Step 6: Design Modifications The geometric dimensions of the machine element are modified from assembly and manufacturing considerations. For example, the transmission shaft illustrated in Fig. 1.4 is provided with steps and shoulders for proper mounting of gear and bearings. d Revised calculations are carried out for operating capacity, margin of safety at critical cross-sections and resultant stresses taking into consideration the effect of stress concentration. When these values differ from desired values, the dimensions of the component are modified. The process is continued 100 100 till the desired values of operating capacity, factor Fig. 1.4 Transmission Shaft of safety and stresses at critical cross-sections are obtained. of elasticity is 207 000 N/mm2. For proper meshing between gear teeth, the permissible deflection at the Step 7: Working Drawing gear is limited to 0.05 mm. The deflection of the The last step in the design of machine elements shaft at the centre is given by, is to prepare a working drawing of the machine Pl 3 (5 ¥ 103 )(200)3 element showing dimensions, tolerances, surface d = or 0.05 = finish grades, geometric tolerances and special 48 EI Ê pd4 ˆ 48(207 000) Á ˜ production requirements like heat treatment. Ë 64 ¯ The working drawing must be clear, concise and Therefore, complete. It must have enough views and cross- d = 35.79 or 40 mm sections to show all details. The main view of the The following observations are made from the machine element should show it in a position, it above two examples: is required to occupy in service. Every dimension (i) Failure mode for the tension rod is general must be given. There should not be scope for yielding while elastic deflection is the failure guesswork and a necessity for scaling the drawing. criterion for the transmission shaft. All dimensions that are important for proper (ii) The permissible tensile stress for tension rod assembly and interchangeability must be provided is obtained by dividing the yield strength with tolerances. 8 Design of Machine Elements 1.5 TRADITIONAL DESIGN METHODS (iii) When the product is to be developed by trial and error, the process is carried out on There are two traditional methods of design— a drawing board instead of shop floor. The design by craft evolution and design by drawing. drawings of the product are modified and Bullock cart, rowing boat, plow and musical developed prior to manufacture. instruments are some of the products, which are In this method, much of the intellectual activity produced by the craft-evolution process. The salient is taken away from the shop floor and assigned to features of this age-old technique are as follows: design engineers. (i) The craftsmen do not prepare dimensioned drawings of their products. They cannot 1.6 DESIGN SYNTHESIS offer adequate justification for the designs they make. Design synthesis is defined as the process of (ii) These products are developed by trial and creating or selecting configurations, materials, error over many centuries. Any modification shapes and dimensions for a product. It is a in the product is costly, because the decision making process with the main objective craftsman has to experiment with the of optimisation. There is a basic difference product itself. Moreover, only one change between design analysis and design synthesis. In at a time can be attempted and complete design analysis, the designer assumes a particular reorganization of the product is difficult. mechanism, a particular material and mode of (iii) The essential information of the product failure for the component. With the help of this such as materials, dimensions of parts, information, he determines the dimensions of manufacturing methods and assembly the product. However, design synthesis does techniques is transmitted from place to place not permit such assumptions. Here, the designer and time to time by two ways. First, the selects the optimum configuration from a number product, which basically remains unchanged, of alternative solutions. He decides the material is the main source of information. The exact for the component from a number of alternative memory of the sequence of operations materials. He determines the optimum shape required to make the product is second and dimensions of the component on the basis of source of information. There is no symbolic mathematical analysis. medium to record the design information of In design synthesis, the designer has to fix the the product. objective. The objective can be minimum cost, With all these weaknesses, the craft-evolution minimum weight or volume, maximum reliability process has successfully developed some of the or maximum life. The second step is mathematical complex structures. The craft-evolution method has formulation of these objectives and requirements. become obsolete due to two reasons. This method The final step is mathematical analysis for cannot adapt to sudden changes in requirement. Secondly, the product cannot be manufactured on a optimisation and interpretation of the results. In mass scale. order to illustrate the process of design synthesis, The essential features of design by drawing let us consider a problem of designing cylindrical method are as follows: cans. The requirements are as follows: (i) The dimensions of the product are specified (i) The cylindrical can is completely enclosed in advance of its manufacture. and the cost of its material should be (ii) The complete manufacturing of the product minimum. can be subdivided into separate pieces, which (ii) The cans are to be stored on a shelf and the can be made by different people. This division dimensions of the shelf are such that the of work is not possible with craft-evolution. radius of the can should not exceed Rmax.  Introduction 9 The following notations are used in the analysis: r = radius of can h = height of can A = surface area of can V = volume of can Therefore, A = 2pr2 + 2prh (a) V = pr2h (b) Substituting Eq. (b) in Eq. (a), 2V A = 2pr2 + (c) r For minimum cost of material of the can, dA 2V =0 or 4p r - =0 dr r2 1/ 3 Ê V ˆ or r =Á. Ë 2p ˜¯ Let us call this radius as r1 giving the condition of minimum material. Therefore, 1/ 3 ÊV ˆ (d) r1 = Á Ë 2p ˜¯ In order to satisfy the second requirement, Fig. 1.5 Optimum Solution to Can Radius 0 < r < Rmax. (e) 1.7 USE OF STANDARDS IN DESIGN In Eqs (d) and (e), r1 and Rmax. are two independent variables and there will be two Standardization is defined as obligatory norms, to separate cases as shown in Fig. 1.5. which various characteristics of a product should Case (a) conform. The characteristics include materials, r1 > Rmax. dimensions and shape of the component, method of testing and method of marking, packing and storing The optimum radius will be, of the product. The following standards are used in r = Rmax. mechanical engineering design: (i) (i) Standards for Materials, their Chemical Case (b) Compositions, Mechanical Properties and Heat r1 < Rmax. Treatment For example, Indian standard IS 210 specifies seven grades of grey cast iron designated The optimum radius will be as FG 150, FG 200, FG 220, FG 260, FG 300, FG r = r1 (ii) 350 and FG 400. The number indicates ultimate It is seen from the above example, that design tensile strength in N/mm2. IS 1570 (Part 4) synthesis begins with the statement of requirements, specifies chemical composition of various grades of which are then converted into mathematical alloy steel. For example, alloy steel designated by expressions and finally, equations are solved for 55Cr3 has 0.5–0.6% carbon, 0.10–0.35% silicon, optimisation. 0.6–0.8% manganese and 0.6–0.8% chromium. 10 Design of Machine Elements (ii) Standards for Shapes and Dimensions of items to a reasonable level. On the other hand, Commonly used Machine Elements The machine a code is defined as a set of specifications for the elements include bolts, screws and nuts, rivets, analysis, design, manufacture, testing and erection belts and chains, ball and roller bearings, wire of the product. The purpose of a code is to achieve ropes, keys and splines, etc. For example, IS 2494 a specified level of safety. (Part 1) specifies dimensions and shape of the cross- There are three types of standards used in design section of endless V-belts for power transmission. office. They are as follows: The dimensions of the trapezoidal cross-section of (i) Company standards They are used in a particular the belt, viz. width, height and included angle are company or a group of sister concerns. specified in this standard. The dimensions of rotary shaft oil seal units are given in IS 5129 (Part 1). (ii) National standards These are the IS (Bureau These dimensions include inner and outer diameters of Indian Standards), DIN (German), AISI or SAE and width of oil seal units. (USA) or BS (UK) standards. (iii) Standards for Fits, Tolerances and Surface (iii) International standards These are prepared by Finish of Component For example, selection of the the International Standards Organization (ISO). type of fit for different applications is illustrated in IS Standardization offers the following advantages: 2709 on ‘Guide for selection of fits’. The tolerances or (a) The reduction in types and dimensions of upper and lower limits for various sizes of holes and identical components to a rational number shafts are specified in IS 919 on ‘Recommendations makes it possible to manufacture the standard for limits and fits for engineering’. IS 10719 explains component on a mass scale in a centralised method for indicating surface texture on technical process. For example, a specialised factory drawings. The method of showing geometrical like Associated Bearing Company (SKF) tolerances is explained in IS 8000 on ‘Geometrical manufactures ball and roller bearings, which tolerancing on technical drawings’. are required by all engineering industries. Manufacture of a standard component on (iv) Standards for Testing of Products These mass production basis reduces the cost. standards, sometimes called ‘codes’, give (b) Since the standard component is manufactured procedures to test the products such as pressure by a specialised factory, it relieves the vessel, boiler, crane and wire rope, where safety machine-building plant of the laborious work of the operator is an important consideration. For of manufacturing that part. Availability of example, IS 807 is a code of practice for design, standard components like bearings, seals, manufacture, erection and testing of cranes and knobs, wheels, roller chains, belts, hydraulic hoists. The method of testing of pressure vessels is cylinders and valves has considerably explained in IS 2825 on ‘Code for unfired pressure reduced the manufacturing facilities required vessels’. by the individual organisation. (v) Standards for Engineering Drawing of (c) Standard parts are easy to replace when Components For example, there is a special worn out due to interchangeability. This publication SP46 prepared by Bureau of Indian facilitates servicing and maintenance of Standards on ‘Engineering Drawing Practice for machines. Availability of standard spare Schools and Colleges’ which covers all standards parts is always assured. The work of related to engineering drawing. servicing and maintenance can be carried There are two words—standard and code— out even at an ordinary service station. which are often used in standards. A standard These factors reduce the maintenance cost is defined as a set of specifications for parts, of machines. materials or processes. The objective of a standard (d) The application of standard machine is to reduce the variety and limit the number of elements and especially the standard units  Introduction 11 (e.g. couplings, cocks, pumps, pressure French balloonist and engineer Charles reducing valves and electric motors) reduce Renard first introduced preferred numbers in the the time and effort needed to design a new 19th century. The system is based on the use of machine. It is no longer necessary to design, geometric progression to develop a set of numbers. manufacture and test these elements and There are five basic series2, denoted as R5, R10, units, and all that the designer has to do is R20, R40 and R80 series, which increase in steps to select them from the manufacturer’s of 58%, 26%, 12%, 6%, and 3%, respectively. Each catalogues. On the other hand, enormous series has its own series factor. The series factors amount of work would be required to design are given in Table 1.1. a machine if all the screws, bolts, nuts, bearings, etc., had to be designed anew each Table 1.1 Series factors time. Standardization results in substantial R5 Series 5 saving in the designer’s effort. 10 = 1.58 (e) The standards of specifications and testing R10 Series 10 10 = 1.26 procedures of machine elements improve R20 Series 20 their quality and reliability. Standard 10 = 1.12 components like SKF bearings, Dunlop belts R40 Series 40 or Diamond chains have a long-standing 10 = 1.06 reputation for their reliability in engineering R80 Series 80 10 = 1.03 industries. Use of standard components improves the quality and reliability of the The series is established by taking the first machine to be designed. number and multiplying it by a series factor to get In design, the aim is to use as many standard the second number. The second number is again components as possible for a given machine. The multiplied by a series factor to get the third number. selection of standard parts in no way restricts the This procedure is continued until the complete creative initiative of the designer and does not prevent series is built up. The resultant numbers are rounded him from finding better and more rational solutions. and shown in Table 1.2. As an example, consider a manufacturer of lifting tackles who wants to 1.8 SELECTION OF PREFERRED SIZES introduce nine different models of capacities ranging from about 15 to 100 kN. Referring to the In engineering design, many a times, the designer R10 series, the capacities of different models of the has to specify the size of the product. The ‘size’ lifting tackle will be 16, 20, 25, 31.5, 40, 50, 63, 80 of the product is a general term, which includes and 100 kN. different parameters like power transmitting capacity, load carrying capacity, speed, dimensions Table 1.2 Preferred numbers of the component such as height, length and R5 R10 R20 R40 width, and volume or weight of the product. 1.00 1.00 1.00 1.00 These parameters are expressed numerically, e.g., 1.06 5 kW, 10 kN or 1000 rpm. Often, the product is 1.12 1.12 manufactured in different sizes or models; for 1.18 instance, a company may be manufacturing seven 1.25 1.25 1.25 different models of electric motors ranging from 1.32 0.5 to 50 kW to cater to the need of different customers. Preferred numbers are used to specify 1.40 1.40 1.50 the ‘sizes’ of the product in these cases. (Contd) 2 IS 1076–1985: Preferred Numbers (in three parts). 12 Design of Machine Elements Table 1.2 Contd numbers minimise unnecessary variation in sizes. They assist the designer in avoiding selection of R5 R10 R20 R40 sizes in an arbitrary manner. The complete range 1.60 1.60 1.60 1.60 is covered by minimum number of sizes, which is 1.70 advantageous to the producer and consumer. 1.80 1.80 There are two terms, namely, ‘basic series’ 1.90 and ‘derived series’, which are frequently used in 2.00 2.00 2.00 relation to preferred numbers. R5, R10, R20, R40 2.12 and R80 are called basic series. Any series that 2.24 2.24 is formed on the basis of these five basic series 2.36 is called derived series. In other words, derived 2.50 2.50 2.50 2.50 series are derived from basic series. There are 2.65 two methods of forming derived series, namely, 2.80 2.80 reducing the numbers of a particular basic series or 3.00 increasing the numbers. 3.15 3.15 3.15 In the first method, a derived series is obtained 3.35 by taking every second, third, fourth or pth term 3.55 3.55 of a given basic series. Such a derived series 3.75 is designated by the symbol of the basic series followed by the number 2, 3, 4 or p and separated 4.00 4.00 4.00 4.00 4.25 by ‘/’ sign. If the series is limited, the designation also includes the limits inside the bracket. If the 4.50 4.50 series is unlimited, at least one of the numbers of 4.75 that series is mentioned inside the bracket. Let us 5.00 5.00 5.00 consider the meaning of these designations. 5.30 (i) Series R 10/3 (1, … ,1000) indicates a derived 5.60 5.60 series comprising of every third term of the 6.00 R10 series and having the lower limit as 1 6.30 6.30 6.30 6.30 and higher limit as 1000. 6.70 (ii) Series R 20/4 (…, 8, …) indicates a derived 7.10 7.10 series comprising of every fourth term of 7.50 the R20 series, unlimited in both sides and 8.00 8.00 8.00 having the number 8 inside the series. 8.50 (iii) Series R 20/3 (200, …) indicates a derived 9.00 9.00 series comprising of every third term of the 9.50 R20 series and having the lower limit as 200 10.00 10.00 10.00 10.00 and without any higher limit. (iv) Series R 20/3 (…200) indicates a derived It is observed from Table 1.2 that small sizes series comprising of every third term of the differ from each other by small amounts, while R20 series and having the higher limit as large sizes by large amounts. In the initial stages, 200 and without any lower limit. the product is manufactured in a limited quantity In the second method, the derived series is and use is made of the R5 series. As the scale of obtained by increasing the numbers of a particular production is increased, a change over is made basic series. Let us consider an example of a from R5 to R10 series, introducing new sizes derived series of numbers ranging from 1 to of intermediate values of R10 series. Preferred 1000 based on the R5 series. From Table 1.2, the  Introduction 13 numbers belonging to the R5 series from 1 to 10 In above calculations, the rounded numbers are are as follows: shown in brackets. 1, 1.6, 2.5, 4, 6.3, 10 Example 1.2 Find out the numbers of R20/4(100, The next numbers are obtained by multiplying the above numbers by 10. They are as follows: …, 1000) derived series. 16, 25, 40, 63, 100 Solution The same procedure is repeated and the next Step I Calculation of series factor numbers are obtained by multiplying the above numbers by 10. The series factor for the R20 series is given by 20 160, 250, 400, 630, 1000 10 = 1.122 Therefore, the complete derived series on the Step II Calculation of ratio factor basis of R5 series is as follows: Since every fourth term of the R20 series is 1, 1.6, 2.5, 4, 6.3, 10, 16, 25, 40, 63, 100, 160, selected, the ratio factor (f) is given by, 250, 400, 630, 1000 f = (1.122) 4 = 1.5848 The advantage of derived series is that one can obtain geometric series for any range of numbers, Step III Calculation of numbers that is, with any value of the first and the last First number = 100 numbers. Also, one can have any intermediate Second number = 100(1.5848) = 158.48 = (160) numbers between these two limits. Third number = 100(1.5848)(1.5848) = 100(1.5848)2 = 251.16 = (250) Example 1.1 Find out the numbers of the R5 basic Fourth number = 100(1.5848)2(1.5848) series from 1 to 10. = 100(1.5848)3 = 398.04 = (400) Fifth number = 100(1.5848)3(1.5848) Solution = 100(1.5848)4 = 630.81= (630) Step I Calculation of series factor Sixth number = 100(1.5848)4(1.5848) The series factor for the R5 series is given by = 100(1.5848)5 = 999.71 = (1000) 5 10 = 1.5849 In the above calculations, the rounded numbers Step II Calculation of numbers are shown in brackets. The complete series is given The series R5 is established by taking the first by number and multiplying it by a series factor to get 100, 160, 250, 400, 630 and 1000 the second number. The second number is again Example 1.3 A manufacturer is interested multiplied by a series factor to get the third number. in starting a business with five different models This procedure is continued until the complete of tractors ranging from 7.5 to 75 kW capacities. series is built up. The numbers thus obtained are Specify power capacities of the models. There is rounded. an expansion plan to further increase the number First number = 1 of models from five to nine to fulfill the requirement Second number = 1 (1.5849) = 1.5849 = (1.6) of farmers. Specify the power capacities of the Third number = (1.5849)(1.5849) = (1.5849)2 additional models. = 2.51 = (2.5) Fourth number = (1.5849)2(1.5849) = (1.5849)3 Solution = 3.98 = (4) Part I Starting Plan Fifth number = (1.5849)3(1.5849) = (1.5849)4 Step I Calculation of ratio factor = (6.3) Let us denote the ratio factor as (f). The derived Sixth number = (1.5849)4(1.5849) = (1.5849)5 series is based on geometric progression. The = (10) power rating of five models will as follows, 7.5(f)0, 7.5(f)1, 7.5(f)2, 7.5(f)3 and 7.5(f)4 14 Design of Machine Elements The maximum power rating is 75 kW. Example 1.4 It is required to standardize eleven Therefore, shafts from 100 to 1000 mm diameter. Specify their 1/ 4 diameters. Ê 75 ˆ 7.5(f)4 = 75 or f = Á Ë 7.5 ˜¯ Solution = (10)1 / 4 = 4 10 = 1.7783 Step I Calculation of ratio factor The diameters of shafts will be as follows: Step II Power rating of models 100(f)0, 100(f)1, 100(f)2, 100(f)3, …, 100(f)10 Rating of first model = (7.5) kW The maximum diameter is 1000 mm. Therefore, Rating of second model = 7.5(1.7783) = 13.34 1 / 10 = (13) kW Ê 1000 ˆ 100(f)10 = 1000 or f =Á Rating of third model = 7.5(1.7783)2 = 23.72 Ë 100 ˜¯ = (24) kW = (10)1 / 10 = 10 10 Rating of fourth model = 7.5(1.7783)3 = 42.18 = (42) kW Therefore the diameters belong to the R10 Rating of fifth model = 7.5(1.7783)4 = 75.0 series. = (75) kW Step II Calculation of shaft diameters Since the minimum diameter is 100 mm, the values Part II Expansion Plan of the R10 series given in Table 1.2 are multiplied Step III Calculation of ratio factor by 100. The diameter series is written as follows: When the number of models is increased to nine, 100, 125, 160, 200, 250, 315, 400, 500, 630, 800 the power rating of nine models will be as follows: and 1000 mm 7.5(f)0, 7.5(f)1, 7.5(f)2, 7.5(f)3, 7.5(f)4, …, 7.5(f)8 1.9 AESTHETIC CONSIDERATIONS The maximum power rating is 75 kW. IN DESIGN Therefore, 1/ 8 Each product has a definite purpose. It has to Ê 75 ˆ perform specific functions to the satisfaction of 7.5(f)8 = 75 or f =Á Ë 7.5 ˜¯ customer. The contact between the product and = (10)1/8 = 1.3335 the people arises due to the sheer necessity of this Step IV Power rating of models functional requirement. The functional requirement The power rating of the nine models will be as of an automobile car is to carry four passengers at a speed of 60 km/hr. There are people in cities follows: who want to go to their office at a distance of 15 First model = 7.5 (1.3335)0 = (7.5) kW km in 15 minutes. So they purchase a car. The Second model = 7.5 (1.3335)1 = 10.00 = (10) kW specific function of a domestic refrigerator is to Third model = 7.5 (1.3335)2 = 13.34 = (13) kW preserve vegetables and fruits for a week. There is Fourth model = 7.5 (1.3335)3 = 17.78 = (18) kW a housewife in the city who cannot go to the market Fifth model = 7.5 (1.3335)4 = 23.72 = (24) kW daily and purchase fresh vegetables. Therefore, Sixth model = 7.5 (1.3335)5 = 31.62 = (32) kW she purchases the refrigerator. It is seen that such Seventh model = 7.5 (1.3335)6 = 42.17 = (42) kW functional requirements bring products and people Eighth model = 7.5 (1.3335)7 = 56.24 = (56) kW together. Ninth model = 7.5 (1.3335)8 = 74.99 = (75) kW However, whe