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• I have a question relating to the slenderness of the beam. The bottom chord of Truss has some panels and has moment and axial force in plane of truss, so where is the actual length that to determine this slenderness, whether the panel length or the span of truss. In the out of plane do not any purlin or bracing or sheating in Bottom chord. Thanks!
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Civil engineering formulas_2009Document Transcript

• CIVILENGINEERING FORMULAS
• ABOUT THE AUTHORTyler G. Hicks, P.E., is a consulting engineer and a successful engi-neering book author. He has worked in plant design and operationin a variety of industries, taught at several engineering schools, andlectured both in the United States and abroad. Mr. Hicks holds abachelor’s degree in Mechanical Engineering from Cooper UnionSchool of Engineering in New York. He is the author of more than100 books in engineering and related fields.
• CIVIL ENGINEERING FORMULAS Tyler G. Hicks, P.E. International Engineering Associates Member: American Society of Mechanical Engineers United States Naval Institute Second EditionNew York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto
• CONTENTS Preface xi Acknowledgments xiii How to Use This Book xvChapter 1. Conversion Factors for CivilEngineering Practice 1Chapter 2. Beam Formulas 11Continuous Beams / 11Ultimate Strength of Continuous Beams / 46Beams of Uniform Strength / 52Safe Loads for Beams of Various Types / 53Rolling and Moving Loads / 53Curved Beams / 65Elastic Lateral Buckling of Beams / 69Combined Axial and Bending Loads / 72Unsymmetrical Bending / 73Eccentric Loading / 73Natural Circular Frequencies and Natural Periods of Vibration of Prismatic Beams / 74Torsion in Structural Members / 76Strain Energy in Structural Members / 76Fixed-End Moments in Beams / 79Chapter 3. Column Formulas 81General Considerations / 81Short Columns / 81Eccentric Loads on Columns / 83Columns of Special Materials / 88Column Base Plate Design / 90American Institute of Steel Construction Allowable-Stress Design Approach / 91Composite Columns / 92Elastic Flexural Buckling of Columns / 94Allowable Design Loads for Aluminum Columns / 96Ultimate Strength Design Concrete Columns / 97Design of Axially Loaded Steel Columns / 102 v
• vi CONTENTSChapter 4. Piles and Piling Formulas 105Allowable Loads on Piles / 105Laterally Loaded Vertical Piles / 105Toe Capacity Load / 107Groups of Piles / 107Foundation-Stability Analysis / 109Axial-Load Capacity of Single Piles / 112Shaft Settlement / 112Shaft Resistance in Cohesionless Soils / 113Chapter 5. Concrete formulas 115Reinforced Concrete / 115Water/Cementitious Materials Ratio / 115Job Mix Concrete Volume / 116Modulus of Elasticity of Concrete / 116Tensile Strength of Concrete / 117Reinforcing Steel / 117Continuous Beams and One-Way Slabs / 117Design Methods for Beams, Columns, and Other Members / 118Properties in the Hardened State / 127Tension Development Lengths / 128Compression Development Lengths / 128Crack Control of Flexural Members / 128Required Strength / 129Deflection Computations and Criteria for Concrete Beams / 130Ultimate-Strength Design of Rectangular Beams with Tension Reinforcement Only / 130Working-Stress Design of Rectangular Beams with Tension Reinforcement Only / 133Ultimate-Strength Design of Rectangular Beams with Compression Bars / 135Working-Stress Design of Rectangular Beams with Compression Bars / 136Ultimate-Strength Design of I- and T-beams / 138Working-Stress Design of I- and T-beams / 138Ultimate-Strength Design for Torsion / 140Working-Stress Design for Torsion / 141Flat-Slab Construction / 142Flat-Plate Construction / 142Shear in Slabs / 145Column Moments / 146Spirals / 147Braced and Unbraced Frames / 147Shear Walls / 148Concrete Gravity Retaining Walls / 150Cantilever Retaining Walls / 153Wall Footings / 155
• CONTENTS viiChapter 6. Timber Engineering Formulas 157Grading of Lumber / 157Size of Lumber / 157Bearing / 159Beams / 159Columns / 160Combined Bending and Axial Load / 161Compression at Angle to Grain / 161Recommendations of the Forest Products Laboratory / 162Compression on Oblique Plane / 163Adjustment Factors for Design Values / 164Fasteners for Wood / 169Adjustment of Design Values for Connections with Fasteners / 171Roof Slope to Prevent Ponding / 172Bending and Axial Tension / 173Bending and Axial Compression / 173Solid Rectangular or Square Columns with Flat Ends / 174Chapter 7. Surveying Formulas 177Units of Measurement / 177Theory of Errors / 178Measurement of Distance with Tapes / 179Vertical Control / 182Stadia Surveying / 183Photogrammetry / 184Chapter 8. Soil and Earthwork Formulas 185Physical Properties of Soils / 185Index Parameters for Soils / 186Relationship of Weights and Volumes in Soils / 186Internal Friction and Cohesion / 188Vertical Pressures in Soils / 188Lateral Pressures in Soils, Forces on Retaining Walls / 189Lateral Pressure of Cohesionless Soils / 190Lateral Pressure of Cohesive Soils / 191Water Pressure / 191Lateral Pressure from Surcharge / 191Stability of Slopes / 192Bearing Capacity of Soils / 192Settlement under Foundations / 193Soil Compaction Tests / 193Compaction Equipment / 195Formulas for Earthmoving / 196Scraper Production / 197Vibration Control in Blasting / 198
• viii CONTENTSChapter 9. Building and Structures Formulas 207Load-and-Resistance Factor Design for Shear in Buildings / 207Allowable-Stress Design for Building Columns / 208Load-and-Resistance Factor Design for Building Columns / 209Allowable-Stress Design for Building Beams / 209Load-and-Resistance Factor Design for Building Beams / 211Allowable-Stress Design for Shear in Buildings / 214Stresses in Thin Shells / 215Bearing Plates / 216Column Base Plates / 217Bearing on Milled Surfaces / 218Plate Girders in Buildings / 219Load Distribution to Bents and Shear Walls / 220Combined Axial Compression or Tension and Bending / 221Webs under Concentrated Loads / 222Design of Stiffeners under Loads / 224Fasteners in Buildings / 225Composite Construction / 225Number of Connectors Required for Building Construction / 226Ponding Considerations in Buildings / 228Lightweight Steel Construction / 228Choosing the Most Economic Structural Steel / 239Steel Carbon Content and Weldability / 240Statically Indeterminate Forces and Moments in Building Structures / 241Roof Live Loads / 244Chapter 10. Bridge and Suspension-Cable Formulas 249Shear Strength Design for Bridges / 249Allowable-Stress Design for Bridge Columns / 250Load-and-Resistance Factor Design for Bridge Columns / 250Additional Bridge Column Formulas / 251Allowable-Stress Design for Bridge Beams / 254Stiffeners on Bridge Girders / 255Hybrid Bridge Girders / 256Load-Factor Design for Bridge Beams / 256Bearing on Milled Surfaces / 258Bridge Fasteners / 258Composite Construction in Highway Bridges / 259Number of Connectors in Bridges / 261Allowable-Stress Design for Shear in Bridges / 262Maximum Width/Thickness Ratios for CompressionElements for Highway Bridges / 263Suspension Cables / 263General Relations for Suspension Cables / 267Cable Systems / 272Rainwater Accumulation and Drainage on Bridges / 273
• CONTENTS ixChapter 11. Highway and Road Formulas 275Circular Curves / 275Parabolic Curves / 277Highway Curves and Driver Safety / 278Highway Alignments / 279Structural Numbers for Flexible Pavements / 281Transition (Spiral) Curves / 284Designing Highway Culverts / 285American Iron and Steel Institute (AISI) Design Procedure / 286Chapter 12. Hydraulics and Waterworks Formulas 291Capillary Action / 291Viscosity / 291Pressure on Submerged Curved Surfaces / 295Fundamentals of Fluid Flow / 296Similitude for Physical Models / 298Fluid Flow in Pipes / 300Pressure (Head) Changes Caused by Pipe Size Change / 306Flow through Orifices / 308Fluid Jets / 310Orifice Discharge into Diverging Conical Tubes / 311Water Hammer / 312Pipe Stresses Perpendicular to the Longitudinal Axis / 312Temperature Expansion of Pipe / 313Forces due to Pipe Bends / 313Culverts / 315Open-Channel Flow / 318Manning’s Equation for Open Channels / 320Hydraulic Jump / 321Nonuniform Flow in Open Channels / 323Weirs / 329Flow over Weirs / 330Prediction of Sediment-Delivery Rate / 332Evaporation and Transpiration / 332Method for Determining Runoff for Minor Hydraulic Structures / 333Computing Rainfall Intensity / 333Groundwater / 334Water Flow for Fire Fighting / 335Flow from Wells / 335Economical Sizing of Distribution Piping / 336Venturi Meter Flow Computation / 336Hydroelectric Power Generation / 337Pumps and Pumping Systems / 338Hydraulic Turbines / 344Dams / 348
• x CONTENTSChapter 13. Stormwater, Sewage, SanitaryWastewater, and Environmental Protection 361Determining Storm Water Flow / 361Flow Velocity in Straight Sewers / 361Design of a Complete-Mix Activated Sludge Reactor / 364Design of a Circular Settling Tank / 368Sizing a Polymer Dilution/Feed System / 369Design of a Solid-Bowl Centrifuge for Sludge Dewatering / 369Design of a Trickling Filter Using the NRC Equations / 371Design of a Rapid-Mix Basin and Flocculation Basin / 373Design of an Aerobic Digester / 374Design of a Plastic Media Trickling Filter / 375Design of an Anaerobic Digestor / 377Design of a Chlorination System for Wastewater Disinfection / 379Sanitary Sewer System Design / 380Design of an Aerated Grit Chamber / 383 Index 385
• PREFACEThe second edition of this handy book presents some 2,500 formulas and calcu-lation guides for civil engineers to help them in the design office, in the field,and on a variety of construction jobs, anywhere in the world. These formulasand guides are also useful to design drafters, structural engineers, bridge engi-neers, foundation builders, field engineers, professional-engineer license exam-ination candidates, concrete specialists, timber-structure builders, and studentsin a variety of civil engineering pursuits. The book presents formulas needed in 13 different specialized branches ofcivil engineering—beams and girders, columns, piles and piling, concretestructures, timber engineering, surveying, soils and earthwork, building struc-tures, bridges, suspension cables, highways and roads, hydraulics and openchannel flow, stormwater, sewage, sanitary wastewater, and environmentalprotection. Some 500 formulas and guides have been added to this second edi-tion of the book. Key formulas are presented for each of the major topics listed above.Each formula is explained so the engineer, drafter, or designer knows how,where, and when to use the formula in professional work. Formula units aregiven in both the United States Customary System (USCS) and SystemInternational (SI). Hence, the content of this book is usable throughout theworld. To assist the civil engineer using these formulas in worldwide engineer-ing practice, a comprehensive tabulation of conversion factors is presentedin Chap. 1. New content is this second edition spans the world of civil engineering.Specific new topics include columns for supporting commercial wind turbinesused in onshore and offshore renewable energy projects, design of axiallyloaded steel columns, strain energy in structural members, shaft twist formulas,new retaining wall formulas and data, solid-wood rectangular column design,blasting operations for earth and rock removal or relocation, hydraulic turbinesfor power generation, dams of several types (arch, buttress, earth), comparisonsof key hydraulic formulas (Darcy, Manning, Hazen-Williams), and a completenew chapter on stormwater, sewage, sanitary wastewater, and environmentalprotection. In assembling this collection of formulas, the author was guided by expertswho recommended the areas of greatest need for a handy book of practical andapplied civil engineering formulas. Sources for the formulas presented here include the various regulatory andindustry groups in the field of civil engineering, authors of recognized books onimportant topics in the field, drafters, researchers in the field of civil engineer-ing, and a number of design engineers who work daily in the field of civil engi-neering. These sources are cited in the Acknowledgments. xi
• xii PREFACE When using any of the formulas in this book that may come from an indus-try or regulatory code, the user is cautioned to consult the latest version of thecode. Formulas may be changed from one edition of code to the next. In a workof this magnitude it is difficult to include the latest formulas from the numerousconstantly changing codes. Hence, the formulas given here are those current atthe time of publication of this book. In a work this large it is possible that errors may occur. Hence, the authorwill be grateful to any user of the book who detects an error and calls it to theauthor’s attention. Just write the author in care of the publisher. The error willbe corrected in the next printing. In addition, if a user believes that one or more important formulas have beenleft out, the author will be happy to consider them for inclusion in the next editionof the book. Again, just write to him in care of the publisher. Tyler G. Hicks, P.E.
• ACKNOWLEDGMENTSMany engineers, professional societies, industry associations, and governmen-tal agencies helped the author find and assemble the thousands of formulas pre-sented in this book. Hence, the author wishes to acknowledge this help andassistance. The author’s principal helper, advisor, and contributor was late FrederickS. Merritt, P.E., Consulting Engineer. For many years Fred and the author wereeditors on companion magazines at The McGraw-Hill Companies. Fred was aneditor on Engineering-News Record, whereas the author was an editor onPower magazine. Both lived on Long Island and traveled on the same railroadto and from New York City, spending many hours together discussing engi-neering, publishing, and book authorship. When the author was approached by the publisher to prepare this book, heturned to Fred Merritt for advice and help. Fred delivered, preparing many ofthe formulas in this book and giving the author access to many more in Fred’sextensive files and published materials. The author is most grateful to Fred forhis extensive help, advice, and guidance. Other engineers and experts to whom the author is indebted for formulasincluded in this book are Roger L. Brockenbrough, Calvin Victor Davis, F. E.Fahey, Gary B. Hemphill, P.E., Metcalf & Eddy, Inc., George Tchobanoglous,Demetrious E. Tonias, P.E., and Kevin D. Wills, P.E. Further, the author thanks many engineering societies, industry associations,and governmental agencies whose work is referred to in this publication. Theseorganizations provide the framework for safe design of numerous structures ofmany different types. The author also thanks Larry Hager, Senior Editor, Professional Group, TheMcGraw-Hill Companies, for his excellent guidance and patience during thelong preparation of the manuscript for this book. Finally, the author thanks hiswife, Mary Shanley Hicks, a publishing professional, who always most willinglyoffered help and advice when needed. Specific publications consulted during the preparation of this text includeAmerican Association of State Highway and Transportation Officials(AASHTO) “Standard Specifications for Highway Bridges”; American Con-crete Institute (ACI) “Building Code Requirements for Reinforced Concrete”;American Institute of Steel Construction (AISC) “Manual of Steel Construc-tion,” “Code of Standard Practice,” and “Load and Resistance Factor DesignSpecifications for Structural Steel Buildings”; American Railway Engineer-ing Association (AREA) “Manual for Railway Engineering”; American Societyof Civil Engineers (ASCE) “Ground Water Management”; and AmericanWater Works Association (AWWA) “Water Quality and Treatment.” In addition, xiii
• xiv ACKNOWLEDGMENTSthe author consulted several hundred civil engineering reference and textbooksdealing with the topics in the current book. The author is grateful to the writersof all the publications cited here for the insight they gave him to civil engi-neering formulas. A number of these works are also cited in the text of thisbook.
• HOW TO USE THIS BOOKThe formulas presented in this book are intended for use by civil engineersin every aspect of their professional work—design, evaluation, construction,repair, etc. To find a suitable formula for the situation you face, start by consulting theindex. Every effort has been made to present a comprehensive listing of all for-mulas in the book. Once you find the formula you seek, read any accompanying text giving back-ground information about the formula. Then when you understand the formulaand its applications, insert the numerical values for the variables in the formula.Solve the formula and use the results for the task at hand. Where a formula may come from a regulatory code, or where a code existsfor the particular work being done, be certain to check the latest edition of theapplicable code to see that the given formula agrees with the code formula. If itdoes not agree, be certain to use the latest code formula available. Remember,as a design engineer you are responsible for the structures you plan, design, andbuild. Using the latest edition of any governing code is the only sensible way toproduce a safe and dependable design that you will be proud to be associatedwith. Further, you will sleep more peacefully! xv
• CIVILENGINEERING FORMULAS
• CHAPTER 1 CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICECivil engineers throughout the world accept both the United States CustomarySystem (USCS) and the System International (SI) units of measure for bothapplied and theoretical calculations. However, the SI units are much morewidely used than those of the USCS. Hence, both the USCS and the SI units arepresented for essentially every formula in this book. Thus, the user of the bookcan apply the formulas with confidence anywhere in the world. To permit even wider use of this text, this chapter contains the conversionfactors needed to switch from one system to the other. For engineers unfamiliarwith either system of units, the author suggests the following steps for becom-ing acquainted with the unknown system:1. Prepare a list of measurements commonly used in your daily work.2. Insert, opposite each known unit, the unit from the other system. Table 1.1 shows such a list of USCS units with corresponding SI units and symbols prepared by a civil engineer who normally uses the USCS. The SI units shown in Table 1.1 were obtained from Table 1.3 by the engineer.3. Find, from a table of conversion factors, such as Table 1.3, the value used to convert from USCS to SI units. Insert each appropriate value in Table 1.2 from Table 1.3.4. Apply the conversion values wherever necessary for the formulas in this book.5. Recognize—here and now—that the most difficult aspect of becoming familiar with a new system of measurement is becoming comfortable with the names and magnitudes of the units. Numerical conversion is simple, once you have set up your own conversion table. Be careful, when using formulas containing a numerical constant, to convertthe constant to that for the system you are using. You can, however, use the for-mula for the USCS units (when the formula is given in those units) and thenconvert the final result to the SI equivalent using Table 1.3. For the few formu-las given in SI units, the reverse procedure should be used. 1
• 2 CHAPTER ONE TABLE 1.1 Commonly Used USCS and SI Units* Conversion factor (multiply USCS unit by this factor to USCS unit SI unit SI symbol obtain SI unit) Square foot Square meter m2 0.0929 Cubic foot Cubic meter m3 0.2831 Pound per Kilopascal kPa 6.894 square inch Pound force Newton N 4.448 Foot pound Newton meter Nm 1.356 torque Kip foot Kilonewton meter kN m 1.355 Gallon per Liter per second L/s 0.06309 minute Kip per square Megapascal MPa 6.89 inch *This table is abbreviated. For a typical engineering practice, an actual table would be many times this length.TABLE 1.2 Typical Conversion Table* To convert from To Multiply by†Square foot Square meter 9.290304 E 02Foot per second Meter per second squared squared 3.048 E 01Cubic foot Cubic meter 2.831685 E 02Pound per cubic inch Kilogram per cubic meter 2.767990 E 04Gallon per minute Liter per second 6.309 E 02Pound per square inch Kilopascal 6.894757Pound force Newton 4.448222Kip per square foot Pascal 4.788026 E 04Acre foot per day Cubic meter per second 1.427641 E 02Acre Square meter 4.046873 E 03Cubic foot per second Cubic meter per second 2.831685 E 02 *This table contains only selected values. See the U.S. Department of the Interior Metric Manual,or National Bureau of Standards, The International System of Units (SI), both available from the U.S.Government Printing Office (GPO), for far more comprehensive listings of conversion factors. † The E indicates an exponent, as in scientific notation, followed by a positive or negative number,representing the power of 10 by which the given conversion factor is to be multiplied before use. Thus,for the square foot conversion factor, 9.290304 1/100 0.09290304, the factor to be used to convertsquare feet to square meters. For a positive exponent, as in converting acres to square meters, multiplyby 4.046873 1000 4046.8. Where a conversion factor cannot be found, simply use the dimensional substitution. Thus, to con-vert pounds per cubic inch to kilograms per cubic meter, find 1 lb 0.4535924 kg and 1 in30.00001638706 m3. Then, 1 lb/in3 0.4535924 kg/0.00001638706 m3 27,680.01, or 2.768 E 4.
• CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE 3TABLE 1.3 Factors for Conversion to SI Units of Measurement To convert from To Multiply byAcre foot, acre ft Cubic meter, m3 1.233489 E 03Acre Square meter, m2 4.046873 E 03Angstrom, Å Meter, m 1.000000* E 10Atmosphere, atm Pascal, Pa 1.013250* E 05 (standard)Atmosphere, atm Pascal, Pa 9.806650* E 04 (technical 1 kgf/cm2)Bar Pascal, Pa 1.000000* E 05Barrel (for petroleum, Cubic meter, m2 1.589873 E 01 42 gal)Board foot, board ft Cubic meter, m3 2.359737 E 03British thermal unit, Joule, J 1.05587 E 03 Btu, (mean)British thermal unit, Watt per meter 1.442279 E 01 Btu (International kelvin, W/(m K) Table) in/(h)(ft2) (°F) (k, thermal conductivity)British thermal unit, Watt, W 2.930711 E 01 Btu (International Table)/hBritish thermal unit, Watt per square 5.678263 E 00 Btu (International meter kelvin, Table)/(h)(ft2)(°F) W/(m2 K) (C, thermal conductance)British thermal unit, Joule per kilogram, 2.326000* E 03 Btu (International J/kg Table)/lbBritish thermal unit, Joule per kilogram 4.186800* E 03 Btu (International kelvin, J/(kg K) Table)/(lb)(°F) (c, heat capacity)British thermal unit, Joule per cubic 3.725895 E 04 cubic foot, Btu meter, J/m3 (International Table)/ft3Bushel (U.S.) Cubic meter, m3 3.523907 E 02Calorie (mean) Joule, J 4.19002 E 00Candela per square Candela per square 1.550003 E 03 inch, cd/in2 meter, cd/m2Centimeter, cm, of Pascal, Pa 1.33322 E 03 mercury (0°C) (Continued)
• 4 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by Centimeter, cm, of Pascal, Pa 9.80638 E 01 water (4°C) Chain Meter, m 2.011684 E 01 Circular mil Square meter, m2 5.067075 E 10 Day Second, s 8.640000* E 04 Day (sidereal) Second, s 8.616409 E 04 Degree (angle) Radian, rad 1.745329 E 02 Degree Celsius Kelvin, K TK tC 273.15 Degree Fahrenheit Degree Celsius, °C tC (tF 32)/1.8 Degree Fahrenheit Kelvin, K TK (tF 459.67)/1.8 Degree Rankine Kelvin, K TK TR /1.8 (°F)(h)(ft2)/Btu Kelvin square 1.761102 E 01 (International meter per watt, Table) (R, thermal K m2/W resistance) (°F)(h)(ft2)/(Btu Kelvin meter per 6.933471 E 00 (International watt, K m/W Table) in) (thermal resistivity) Dyne, dyn Newton, N 1.000000† E 05 Fathom Meter, m 1.828804 E 00 Foot, ft Meter, m 3.048000† E 01 Foot, ft (U.S. survey) Meter, m 3.048006 E 01 Foot, ft, of water Pascal, Pa 2.98898 E 03 (39.2°F) (pressure) Square foot, ft2 Square meter, m2 9.290304† E 02 Square foot per hour, Square meter per 2.580640† E 05 ft2/h (thermal second, m2/s diffusivity) Square foot per Square meter per 9.290304† E 02 second, ft2/s second, m2/s Cubic foot, ft3 (volume Cubic meter, m3 2.831685 E 02 or section modulus) Cubic foot per minute, Cubic meter per 4.719474 E 04 ft3/min second, m3/s Cubic foot per second, Cubic meter per 2.831685 E 02 ft3/s second, m3/s Foot to the fourth Meter to the fourth 8.630975 E 03 power, ft4 (area power, m4 moment of inertia) Foot per minute, Meter per second, 5.080000† E 03 ft/min m/s Foot per second, Meter per second, 3.048000† E 01 ft/s m/s
• CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE 5TABLE 1.3 Factors for Conversion to SI Units of Measurement(Continued) To convert from To Multiply byFoot per second Meter per second 3.048000† E 01 squared, ft/s2 squared, m/s2Footcandle, fc Lux, lx 1.076391 E 01Footlambert, fL Candela per square 3.426259 E 00 meter, cd/m2Foot pound force, ft lbf Joule, J 1.355818 E 00Foot pound force per Watt, W 2.259697 E 02 minute, ft lbf/minFoot pound force per Watt, W 1.355818 E 00 second, ft lbf/sFoot poundal, ft Joule, J 4.214011 E 02 poundalFree fall, standard g Meter per second 9.806650† E 00 squared, m/s2Gallon, gal (Canadian Cubic meter, m3 4.546090 E 03 liquid)Gallon, gal (U.K. Cubic meter, m3 4.546092 E 03 liquid)Gallon, gal (U.S. dry) Cubic meter, m3 4.404884 E 03Gallon, gal (U.S. Cubic meter, m3 3.785412 E 03 liquid)Gallon, gal (U.S. Cubic meter per 4.381264 E 08 liquid) per day second, m3/sGallon, gal (U.S. Cubic meter per 6.309020 E 05 liquid) per minute second, m3/sGrad Degree (angular) 9.000000† E 01Grad Radian, rad 1.570796 E 02Grain, gr Kilogram, kg 6.479891† E 05Gram, g Kilogram, kg 1.000000† E 03Hectare, ha Square meter, m2 1.000000† E 04Horsepower, hp Watt, W 7.456999 E 02 (550 ft lbf/s)Horsepower, hp Watt, W 9.80950 E 03 (boiler)Horsepower, hp Watt, W 7.460000† E 02 (electric)horsepower, hp Watt, W 7.46043† E 02 (water)Horsepower, hp (U.K.) Watt, W 7.4570 E 02Hour, h Second, s 3.600000† E 03Hour, h (sidereal) Second, s 3.590170 E 03Inch, in Meter, m 2.540000† E 02 (Continued)
• 6 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by Inch of mercury, in Hg Pascal, Pa 3.38638 E 03 (32°F) (pressure) Inch of mercury, in Hg Pascal, Pa 3.37685 E 03 (60°F) (pressure) Inch of water, in Pascal, Pa 2.4884 E 02 H2O (60°F) (pressure) Square inch, in2 Square meter, m2 6.451600† E 04 Cubic inch, in3 Cubic meter, m3 1.638706 E 05 (volume or section modulus) Inch to the fourth Meter to the fourth 4.162314 E 07 power, in4 (area power, m4 moment of inertia) Inch per second, in/s Meter per second, 2.540000† E 02 m/s Kelvin, K Degree Celsius, °C tC TK 273.15 Kilogram force, kgf Newton, N 9.806650† E 00 Kilogram force meter, Newton meter, 9.806650† E 00 kg m Nm Kilogram force second Kilogram, kg 9.806650† E 00 squared per meter, kgf s2/m (mass) Kilogram force per Pascal, Pa 9.806650† E 04 square centimeter, kgf/cm2 Kilogram force per Pascal, Pa 9.806650† E 00 square meter, kgf/m2 Kilogram force per Pascal, Pa 9.806650† E 06 square millimeter, kgf/mm2 Kilometer per hour, Meter per second, 2.777778 E 01 km/h m/s Kilowatt hour, kWh Joule, J 3.600000† E 06 Kip (1000 lbf) Newton, N 4.448222 E 03 Kipper square inch, Pascal, Pa 6.894757 E 06 kip/in2 ksi Knot, kn (international) Meter per second, m/s 5.144444 E 01 Lambert, L Candela per square 3.183099 E 03 meter, cd/m2 Liter Cubic meter, m3 1.000000† E 03 Maxwell Weber, Wb 1.000000† E 08 Mho Siemens, S 1.000000† E 00
• CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE 7TABLE 1.3 Factors for Conversion to SI Units of Measurement(Continued) To convert from To Multiply byMicroinch, in Meter, m 2.540000† E 08Micron, m Meter, m 1.000000† E 06Mil, mi Meter, m 2.540000† E 05Mile, mi (international) Meter, m 1.609344† E 03Mile, mi (U.S. statute) Meter, m 1.609347 E 03Mile, mi (international Meter, m 1.852000† E 03 nautical)Mile, mi (U.S. nautical) Meter, m 1.852000† E 03Square mile, mi2 Square meter, m2 2.589988 E 06 (international)Square mile, mi2 Square meter, m2 2.589998 E 06 (U.S. statute)Mile per hour, mi/h Meter per second, 4.470400† E 01 (international) m/sMile per hour, mi/h Kilometer per hour, 1.609344† E 00 (international) km/hMillibar, mbar Pascal, Pa 1.000000† E 02Millimeter of mercury, Pascal, Pa 1.33322 E 02 mmHg (0°C)Minute, min (angle) Radian, rad 2.908882 E 04Minute, min Second, s 6.000000† E 01Minute, min (sidereal) Second, s 5.983617 E 01Ounce, oz Kilogram, kg 2.834952 E 02 (avoirdupois)Ounce, oz (troy or Kilogram, kg 3.110348 E 02 apothecary)Ounce, oz (U.K. fluid) Cubic meter, m3 2.841307 E 05Ounce, oz (U.S. fluid) Cubic meter, m3 2.957353 E 05Ounce force, ozf Newton, N 2.780139 E 01Ounce force inch, Newton meter, 7.061552 E 03 ozf in NmOunce per square foot, Kilogram per square 3.051517 E 01 oz (avoirdupois)/ft2 meter, kg/m2Ounce per square yard, Kilogram per square 3.390575 E 02 oz (avoirdupois)/yd2 meter, kg/m2Perm (0°C) Kilogram per pascal 5.72135 E 11 second meter, kg/(Pa s m)Perm (23°C) Kilogram per pascal 5.74525 E 11 second meter, kg/(Pa s m)Perm inch, perm in Kilogram per pascal 1.45322 E 12 (0°C) second meter, kg/(Pa s m) (Continued)
• 8 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by Perm inch, perm in Kilogram per pascal 1.45929 E 12 (23°C) second meter, kg/(Pa s m) Pint, pt (U.S. dry) Cubic meter, m3 5.506105 E 04 Pint, pt (U.S. liquid) Cubic meter, m3 4.731765 E 04 Poise, P (absolute Pascal second, 1.000000† E 01 viscosity) Pa s Pound, lb Kilogram, kg 4.535924 E 01 (avoirdupois) Pound, lb (troy or Kilogram, kg 3.732417 E 01 apothecary) Pound square inch, Kilogram square 2.926397 E 04 lb in2 (moment of meter, kg m2 inertia) Pound per foot Pascal second, 1.488164 E 00 second, lb/ft s Pa s Pound per square Kilogram per square 4.882428 E 00 foot, lb/ft2 meter, kg/m2 Pound per cubic Kilogram per cubic 1.601846 E 01 foot, lb/ft3 meter, kg/m3 Pound per gallon, Kilogram per cubic 9.977633 E 01 lb/gal (U.K. liquid) meter, kg/m3 Pound per gallon, Kilogram per cubic 1.198264 E 02 lb/gal (U.S. liquid) meter, kg/m3 Pound per hour, lb/h Kilogram per 1.259979 E 04 second, kg/s Pound per cubic inch, Kilogram per cubic 2.767990 E 04 lb/in3 meter, kg/m3 Pound per minute, Kilogram per 7.559873 E 03 lb/min second, kg/s Pound per second, Kilogram per 4.535924 E 01 lb/s second, kg/s Pound per cubic yard, Kilogram per cubic 5.932764 E 01 lb/yd3 meter, kg/m3 Poundal Newton, N 1.382550 E 01 Pound force, lbf Newton, N 4.448222 E 00 Pound force foot, Newton meter, 1.355818 E 00 lbf ft Nm Pound force per foot, Newton per meter, 1.459390 E 01 lbf/ft N/m Pound force per Pascal, Pa 4.788026 E 01 square foot, lbf/ft2 Pound force per inch, Newton per meter, 1.751268 E 02 lbf/in N/m Pound force per square Pascal, Pa 6.894757 E 03 inch, lbf/in2 (psi)
• CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE 9TABLE 1.3 Factors for Conversion to SI Units of Measurement(Continued) To convert from To Multiply byQuart, qt (U.S. dry) Cubic meter, m3 1.101221 E 03Quart, qt (U.S. liquid) Cubic meter, m3 9.463529 E 04Rod Meter, m 5.029210 E 00Second (angle) Radian, rad 4.848137 E 06Second (sidereal) Second, s 9.972696 E 01Square (100 ft2) Square meter, m2 9.290304† E 00Ton (assay) Kilogram, kg 2.916667 E 02Ton (long, 2240 lb) Kilogram, kg 1.016047 E 03Ton (metric) Kilogram, kg 1.000000† E 03Ton (refrigeration) Watt, W 3.516800 E 03Ton (register) Cubic meter, m3 2.831685 E 00Ton (short, 2000 lb) Kilogram, kg 9.071847 E 02Ton (long per cubic Kilogram per cubic 1.328939 E 03 yard, ton)/yd3 meter, kg/m3Ton (short per cubic Kilogram per cubic 1.186553 E 03 yard, ton)/yd3 meter, kg/m3Ton force (2000 lbf) Newton, N 8.896444 E 03Tonne, t Kilogram, kg 1.000000† E 03Watt hour, Wh Joule, J 3.600000† E 03Yard, yd Meter, m 9.144000† E 01Square yard, yd2 Square meter, m2 8.361274 E 01Cubic yard, yd3 Cubic meter, m3 7.645549 E 01Year (365 days) Second, s 3.153600† E 07Year (sidereal) Second, s 3.155815 E 07 *Commonly used in engineering practice. † Exact value. From E380, “Standard for Metric Practice,” American Society for Testing andMaterials.
• A = bd bd3 l1 = c3 12 3 c1 = d/2 bd3 l2 = c2 = d 3 c1 b3d3 bd 2d c3 = bd l3 = A= c1 = 6 (b2 + d2) 2 3 d 1 1 b2 + d2 c1 d bd3 bd3 bd2 b2d2 l1 = l2 = S1 = S3 = 1 1 36 12 6 6 b2 + d2 2 2 2 b 2 bd2 d b d r3 = bd S1 = r1 = 3 r1 = 24 18 12 6 (b2 + d2) Rectangle Triangle12 2 c2 2 2 3 2 A = bd c= d A= bd 3 5 3 c c1 d 8 b3d d 1 1 3 5 1 1 l1 = bd3 l2 = c1 = d c2 = b 175 30 5 8 2 3 3 2 8 19 3 16 b l1 = bd3 l2 = b l3 = bd bd 105 175 480 Parabola Half Parabola
• Section Moment of inertia Section modulus Radius of gyration Equilateral polygon I = I A = Area R = Rad circumscribed I = A (6R2 − a2) c r 24 6R2 − a2 ≈ R circle I 24 2 = r = Rad inscribed circle = A (12r2 + a2) 180° n = No. of sides 48 R cos n 12r2 + a2 a = Length of side 2 = AR (approx) 48 Axis as in preceding section 4 = AR (approx) of octagon 4 b 6b2 + 6bb1 + b12 3 I= h13 36 (2b + b1) 6b2 + 6bb1 + b12 2 c h I = h h 12b2 + 12bb1 + 2b12 1 3b + 2b1 c 12 (3b + 2b1) 6 (2b + b1) c= h b1 b1 3 2b + b1 2 2 FIGURE 2.1 Geometric properties of sections.
• Section Moment of inertia Section modulus Radius of gyration B b b b B B B BH3 + bh3 2 2 2 2 I = b 12 BH3 + bh3 BH3 + bh3 H I 12 (BH + bh) c = H H h h h 6H c b b b b 2 2 BH3 – bh3 I = 1214 BH3 – bh3 H H H h h h 3 3 = BH – bh 12 (BH – bh) I c c 6H B B B
• Section Moment of inertia and section modulus Radius of gyration b b 2 1 d1 I= (Bc13 – B1h3 + bc23 – b1h13) 3 c2 h1 a H 1 aH2 + B1d2 + b1d1 (2H – d1) I h c1 c1 = d 2 aH + B1d + b1d1 (Bd + bd1) + a (h + h1) B1 B1 2 2 B a a a a 1 2 2 I= (Bc13 – bh3 + ac23) 215 1 2 2 c1 = ali + bd h c1 c2 c2 H 2 ali + bd H H h c2 = II – c1 c1 d d d d b b I b b r= 2 2 [Bd + a (II – d)] B B B FIGURE 2.1 (Continued)
• Section Moment of inertia Section modulus Radius of gyration πd4 πr4 A 2 I πd3 πr3 A r d I= = = r c = 32 = 4 = 4 r = 64 4 4 2 4 r = 0.05d4 (approx) = 0.1d3 (approx) d π (D4 – d4) I π D4 – d4 I= c = 32 64 D r π R4 – r4 d π (R4 – r4) = = R4 + r2 D2 + d2 R 4 4 R =16 = A (R2 + r2) = 0.8dm2s (approx) 2 2 D 4 = 0.05 (D4 – d4) (approx) s 1 when is very small dm = (D + d) dm 2 1 s= (D – d) 2
• I 3 π _ 8 c2 = 0.1908r c2 I = r4 9π2 – 64 r = 0.264r 8 9π I 3 c1 = 0.2587r c1 6π = 0.1098r 4 r c1 = 0.4244r I = 0.1098 (R4 – r 4) 0.283R2r2 (R – r) 4 R2 + Rr + r 2 c2 – c1 = 2I c1 R+r 3π R+r π (R2 – r 2) t r r1 = 0.3tr13 (approx) c2 = R – c1 = 0.31r1 (approx) R t when is very small r117 a 3 I = πa2b = 0.7854a2b I = πa b = 0.7854a3b a 4 c 4 2 b I = π (a3b – a13b1) t 4 I a I = π a (a + 3b)t a1 t (πab – a1b1) π a2 (a + 3b)t c 4 = a a + 3b 4 (approx) = 2 a+b (approx) (approx) b1 b FIGURE 2.1 (Continued)
• Section Moment of inertia and section modulus Radius of gyration I = l 3π d4 + b (h3 – d3) + b3 (h – d) h I 12 16 d2 b π + 2b (h – d) I = l 3π d4 + b (h3 + d3) + b3 (h – d) 4 c 6h 16 (approx) d b B B 2 2 I= t πB3 πBh2 2 + B2h + + h3 4 16 2 3 I B 418 I = 2I πB H h=H– B 2 +h t h 2 4 c H+t B 4 t B I = 64 (b1h13 – b2h23), where h1 h2 105 H 1 1 b2 t h1 = (H + t) b1 = (B + 2.6t) 2 4 r= 3I b1 1 1 t (2B + 5.2H) h2 = (H – t) b2 = (B – 2.6t) B 2 4 Corrugated sheet iron, I 2I = parabolically curved c H+t
• Approximate values of least radius of gyration r D D D Deck D I-beam Channel D beam Carnegie Z-bar Phoenix column column r= 0.3636D 0.295D D/4.58 D/3.54 D/619 D Angle Angle D D D T-beam Equal legs Unequal legs D Cross B D r= D/4.74 D/5 BD/2.6 (B + D) D/4.74 FIGURE 2.1 (Continued)
• Shaft angle-of-twist section formulas θ = twist, radians T = torque, in . lb (N . m) L = length of shaft, in (mm) N = modulus of rigidity, psi (MPa) D, do, di, dm, ds, s, a, b = shaft sectional dimensions, in (mm) Shaft section Angle-of-twist θ = Shaft torque formulas and location of maximum shear stress in the shaft 32TL Location of Torque formulas D Shaft section max shear T= π D4N D Outer fiber πD3f 16 32TL π20 di do (d 4 – di4) π (d04 – d14)N Outer fiber 16 o di do f do ds 16 (dm2 + ds2)TL ds πdmdm2f Ends of minor axis 16 πdm3 ds3N dm dm 7.11TL 0.208S3f S Middle of sides s s4N a A2B2f 3.33 (a2 + b2)TL B Midpoint of major sides 3A + 1.8B b A a3b3N FIGURE 2.1 (Continued)
• CASE 2. Beam Supported Both Ends—Concentrated Load at Any Point At x: when x = a (a + 2b) + 3 and a > b R = Wb At point of load: L Wab D (max) = Wab (a + 2b) 3a (a + 2b) + 27 EIL M (max) = R1 = Wa L L At x: when x < a V (max) = R when a < b and R1 when a > b At x: when x < a Wbx D= 2L (L – x) – b2 – (L – x)2 Wb Wbx 6 EIL At x: V= M= L L At x: when x > a Wa (L – x) D= 2Lb – b2 – (L – x)2 6 EIL21 LW a b D R R1 x M V FIGURE 2.2 Beam formulas. (From J. Callender, Time-Saver Standards for Architectural Design Data, 6th ed., McGraw-Hill, N.Y.)
• CASE 3. Beam Supported Both Ends—Two Unequal Concentrated Loads, Unequally Distributed At point of load W: 1 W (L – a) + W b R= 1 M= a W (L – a) + W b 1 L L 1 Wa + W (L – b) At point of load W1: R1 = 1 L b Wa + W (L – b) M1 = 1 V (max) = Maximum reaction L At x: when x > a and < (L – b) At x: when x > a or < (L – b) V=R–W a bx M = W (L – x) + W1 L L L W W1 a b22 R R1 M M1 x V
• CASE 4. Beam Supported Both Ends—Three Unequal Concentrated Loads, Unequally Distributed Wb + W1b1 + W2b2 At x: when x = a R= L M = Ra Wa + W1a1 + W2a2 At x: when x = a1 R1 = L M1 = Ra1 – W (a1 – a) V (max) = Maximum reaction At x: when x = a2 M2 = Ra2 – W (a2 – a) – W1 (a2 – a1) At x: when x > a and < a1 M (max) = M when W = R or > R V=R–W W1 + W = R or > R At x: when x > a1 and < a2 M (max) = M1 when W1 + W2 = R1 or > R1 V = R – W – W1 M (max) = M2 when W2 = R1 or > R1 L23 a2 b2 a1 b1 W W W1 2 a b R R1 x M1 M M2 V FIGURE 2.2 (Continued)
• CASE 5. Beam Fixed Both Ends—Continuous Load, Uniformly Distributed At center: WL At center: M (max) = W 24 1 WL3 R = R1 = V (max) = D (max) = 2 At supports: 384 EI WL At x: M1 (max) = At x: W Wx 12 Wx2 V= – At x: D= (L2 – 2Lx + x2) 2 L 24 EIL W L2 M= – + Lx – x2 2L 6 L24 W D R1 R M M1 x V V
• CASE 6. Beam Fixed Both Ends—Concentrated Load at Any Point At support R: b2 (3a + b) ab2 R=W L3 M1 max neg. mom. = – W 2 At x: when x = 2 aL and a > b when b > a L 3a + b 2 At support R1: R1 = W a (3b + a) a2b 2 W a3b2 L3 M2 max neg. mom. = – W 2 D (max) = 3 EI (3a + b)2 when a > b L V (max) = R when a < b = R1 when a > b At point of load: when x < a ab2 W b2x2 M (max) = Ra + M1 = Ra – W 2 At x: when x < a L D= (3aL – 3ax – bx) V=R ab2 6 EIL3 At x: M = Rx – W 2 L L25 W a b R D R1 x M M1 M2 V FIGURE 2.2 (Continued)
• CASE 7. Beam Fixed at One End (Cantilever)—Continuous Load, Uniformly Distributed R1 = V (max) = W At fixed end: At free end: WL WL3 M (max) = D (max) = 2 8EI At x: At x: At x: V= Wx Wx2 W M= D= (x4 – 4L3x + 3L4) L 2L 24EIL L W26 D R1 Loading x M Bending V Shear
• CASE 8. Beam Fixed at One End (Cantilever)—Concentrated Load at Any Point At free end: R1 = V (max) = W At fixed end: WL3 3a a 3 D (max) = 2– + 6EI L L At x: when x > a M (max) = Wb At point of load: V=W W At x: when x > a D= (L – a)3 3EI At x: when x < a M = W (x = a) At x: when x > a V=0 W –3aL2 + 2L3 + x3 D= 6EI –3ax2 – 3L2x + 6aLx W L27 a b D R1 x M V FIGURE 2.2 (Continued)
• CASE 9. Beam Fixed at One End, Supported at Other—Concentrated Load at Any Point 3b2L – b3 At point of load: 2 3 At x: when x = a = 0.414L R=W 2L3 M (max) = Wa 3b L – b D (max) = 0.0098 WL3 2L 3 At fixed end: EI 3b2L – b3 – W (L – a) At x: when x < a R1 = W 3aL2 – a3 M1 (max) = WL 2L3 1 3RL2x – Rx3 – 2L3 D= 3 W (L – a)2 x At x: when x < a At x: when x < a 6EI 2 3 V=R M = Wx 3b L – b At x: when x > a 2L3 At x: when x > a At x: when x > a 2L – b3 1 R1(2L3 – 3L2x + x3) – V=R–W M = Wx 3b – W (x – a) D= 6EI 3 Wa (L – x)2 2L328 L a W b D R1 R M x M1 V
• CASE 10. Beam Fixed at One End, Supported at Other—Continuous Load, Uniformly Distributed 3 3 At x: when x = L R= 8 At x: when x = 0.4215L 8 W 9 WL3 M (max) = WL D (max) = 0.0054 128 EI 5 R1 = V (max) = At fixed end: 8W 1 M1 (max) = WL At x: At x: 8 D= Wx – 3Lx2 + 2x3 + L3 3 Wx At x: 48EIL V= W– Wx 3 1 8 L M= L– x L 8 2 L29 W D R R1 M x M1 V FIGURE 2.2 (Continued)
• CASE 11. Beam Overhanging Both Supported Unsymmetrically Placed—Continuous Load, Uniformly Distributed R = W = load per unit of length At x1: when x1 = w – a W a+L+b R M (max) = R –a R = w [(a + L)2 + b2] 2w 2L At R: 1 M1 = wa2 R1 = w [(b + L)2 + a2] 2 At R1: 1 2L M1 = wb2 V (max) = wa or R – wa 2 1 At x: when x < a M = w (a – x)2 At x: when x < a V = w (a – x) 2 At x1: when x1 < L V = R – w (a + x1) At x1: when x1 < L M = 1 w (a + x1)2 – Rx1 2 At x2: when x2 < b V = w (b – x2) 1 At x2: when x2 < b M = w (b – x2)2 230 W a L b R R1 x x1 x2 M M1 M1
• CASE 12. Beam Overhanging Both Supports, Symmetrically Placed—Two Equal Concentrated Loads at Ends At x1: when x1 < L At free ends: W Wa Wa2 (3L + 2a) R = R1 = V (max) = M (max) = D= 2 2 12EI At x: when x < a At x: when x < a At center: W W WaL2 V= M = (a – x) D= 2 2 16EI W W D 2 231 R R1 D a L a M x x1 V V FIGURE 2.2 (Continued)
• L P a b c kL (l – k)L k< 1 w 2 R1 = (l – k)P R2 = Pk wb L L wb R2 = (2c + b) R= (2c + b) w 2L 2L Load Load R = wL R = wL R1 2 2 Load R1 R1 – w (x – a) R2 1 Shear – x wL x R2 R 2 (a < x < a + b) x′L xL R1 x″L 1 R a+ w (x′ < k) x< (x″ < (1 – k)) 2 1 L Shear R1 x′ L k(1 – k)PL R2 x″ L 232 Shear 2 1 Moment x(l – x)wL2 2 PL3 2 3 2 k 1–k 3 dmax = PL 2 k (1 – k)2 wL 3EI 3EI 3 2 w (x – a)2 8 (x′ < a) (x″ < c) 2w R1(a + 1 PL2 R wL3 Moment c L x′ x″ PL 2 k(1 – k2) 6EI 2 24EI R1a R2c k(1 – k)(2 – k) 6EI PL3 kx″(1 – k – x″2) 2 R1x – 5wL4 PL3 6EI (1 – k)x′(2k – k2 – x′2) 384EI R1x′ R2x″ 6EI 2 L 1–k wL4 Elastic curve x(l – 2x2 + x3) 3 24EI Moment Elastic curve (a) (b) (c) FIGURE 2.3 Elastic-curve equations for prismatic beams: (a) Shears, moments, and deﬂections for full uniform load on a simply sup- ported prismatic beam. (b) Shears and moments for uniform load over part of a simply supported prismatic beam. (c) Shears, moments, and deﬂections for a concentrated load at any point of a simply supported prismatic beam.
• P L W = np L P kL (1 – 2k)L kL P 2 P P P 2P aL aL aL aL aL aL P 1 L L a= 2 2 R=P R=P n+1 R = 1 nP R = 1 nP L 2 2 L 1 1 Load Load R= P R= P 2 2 L P Load R P P P P P R P/2 Shear Shear P P/2 L L xL 2 2 (x < k) maL Shear PL m(n – m + 1) PxL 2 n+1 PkL xL33 1 x< PL (n + 1)(For n an Moment PL n(n + 2) 1 2 xPL PL Moment 8 8 n+1 2 4 odd number) L L′′ c L (For n an 2 2 even number) Moment PL3 x(3k – 3k2 – x2) PL3 6EI x(3 – 4x2) c L 48EI PL2 n(n + 2) PL2 24EI n + 1 k(1 – k) PL3 n(n + 2) (5n2 + 10n + 6) 2EI PL3 x′ 24EI k(3 – 4k2) 384EI (n + 1)3 PL2 PL3 (k < x′ < (1 – k)) (For n an even number) 16EI PL 3 PL3 5n2 + 10n + 1 (For n an odd number) 48EI k(3x′ – 3x′2 – k2) 6EI 384EI n+l Elastic curve Elastic curve Elastic curve (d) (e) (f) FIGURE 2.3 Elastic-curve equations for prismatic beams: (d) Shears, moments, and deﬂections for a concentrated load at midspan of a simply supported prismatic beam. (e) Shears, moments, and deﬂections for two equal concentrated loads on a simply supported prismatic beam. ( f ) Shears, moments, and deﬂec- tions for several equal loads equally spaced on a simply supported prismatic beam. (Continued)
• L L′ w w P w 2 w R1 = (L – L′2) R2 = (L + L′)2 2L 2L Load L′ L + L′ R1 = P R2 = P P L L L x′L′ L L′ xL R1 = wLx wL′x′ Load PL R1 R=P wL′ Load 2 w 2 R2 P L 1 – L′2 R2 (L – L′2) 2 L 2L R1 Shear w Shear (L + L′)2 (L – L′)2 wx (L2 – L′2 – xL2) 8L2 xL x′L′ R P 2 wL′2 wL′2 2 x′34 Shear 2 2 xPL′ Px′L′ PL′ xL Moment L′2 L 1– L PxL L2 PL Moment 3 Moment wL′ (4L′2 – L3 + 3L′3) PL′L2 24EI PL′2 PL3 wL2x 2 dmax = PL′L 2 (L + L′) (2 – 3x + x3) L (1 – 2x2 + x2) –2L′2(1 – x2) 9 3EI x(1 – x2) 3EI 3EI 24EI 6EI 2 PL′ (1 – x′) 2(L + L′) – L′x′ (1 + x′) PL3 6EI 3EI Elastic curve Elastic curve Elastic curve (g) (h) (i) FIGURE 2.3 Elastic-curve equations for prismatic beams: (g) Shears, moments, and deﬂections for a concentrated load on a beam overhang. (h) Shears, moments, and deﬂections for a concentrated load on the end of a prismatic cantilever. (i) Shears, moments, and deﬂections for a uniform load over the full length of a beam with overhang. (Continued)
• w L L′ L wL w W= w 3 2 2 xL wL wx 2 R = wL 2 WL L R1 = wL′ R2 = wL′ (2L + L′) 3 2L 2L R=W Load L Load Load wLx xL x′L′ wL′ R2 Wx2 wL Shear R1 wL′x′ Shear W Shear xL wL2 2 1 wL′2 x′2 wL′2x WL x335 x 2 2 2 3 1 wL2 1 WL 2 Moment Moment wL′2 3 Moment 2 xL L wL′2L2 3 x(1 – x2) WL3 wL4 12EI wL′3 (4 – 5x + x3) (3 – 4x + x4) (4L + 3L′) 60EI 24EI 24EI wL′2L2 Elastic curve wL4 dmax = 18 3EI WL3 8EI Elastic curve Elastic curve 15EI (j) (k) (l) FIGURE 2.3 Elastic-curve equations for prismatic beams: (j) Shears, moments, and deﬂections for uniform load over the full length of a cantilever. (k) Shears, moments, and deﬂections for uniform load on a beam overhang. (l) Shears, moments, and deﬂections for triangular loading on a prismatic cantilever. (Continued)
• R1 = V1 W l = x W 3 R2 = V2max 2W = R1 R2 3 W Wx2 Vx = – 2 0.5774l 3 l Mmax at x = W = 0.5774l 2Wl = 0.1283Wl V1 = 3 9 3 Shear V2 Wx 2 2 Mx = (l – x ) 3l2 ∆max at x = l 8 = 0.5193l Wl2 Mmax 1– = 0.01304 15 EI Wx Moment ∆x = (3x4 – 10l2x2 + 7l4) 180EIl236 (m) FIGURE 2.3 Elastic-curve equations for prismatic beams. (m) Simple beam—load increasing uniformly to one end. (Continued)
• BEAM FORMULAS 37 l x W W R=V = R R 2 l l Vx when x < l W = 2 (l2 – 4x2) 2 2 2 2l Mmax (at center) Wl V = 6 Shear Mx when x < l 1 – 2x2 V = Wx 2 2 3l2 Wl 3 ∆max (at center) = 60EIMmax Wx ∆x = (5l2 – 4x2)2 480EIl2 Moment (n)FIGURE 2.3 Elastic-curve equations for prismatic beams: (n) Simple beam—loadincreasing uniformly to center. (Continued) wa l R1 = V1max = (2l – a) 2l a wa 2 wa R2 = V2 = 2l V (when x < a) = R1 – wx R1 R2 2 R x Mmax at x = 1 = R1 w 2w 2V1 Mx (when x < a) = R1x – wx 2 R1 Mx (when x > a) = R2 (l – x) Shear wx W V2 ∆x (when x < a) = [a2(2l – a)2 24EIl – 2ax2 (2l – a) + lx3]Mmax wa2(l – x) ∆x (when x > a) = 24EIl Moment (4xl – 2x2 – a2) (o)FIGURE 2.3 Elastic-curve equations for prismatic beams: (o) Simple beam—uniformload partially distributed at one end. (Continued)
• 38 CHAPTER TWO l P R R=V =P Mmax (at fixed end) = Pl x Mx = Px Pl3 ∆max (at free end) = V 3EI P Shear ∆x = (2l3 – 3l2x + x3) 6EI Mmax Moment (p)FIGURE 2.3 Elastic-curve equations for prismatic beams: (p) Cantilever beam—con-centrated load at free end. (Continued) l x P P R=V = R R 2 Pl l l Mmax (at center and ends) = 8 2 2 P V Mx when x < l = (4x – l) 2 8 3 Shear Pl V ∆max (at center) = 192EI l Px2 4 Mmax ∆x = (3l – 4x) 48EIMmax Mmax Moment (q)FIGURE 2.3 Elastic-curve equations for prismatic beams: (q) Beam ﬁxed at bothends—concentrated load at center. (Continued)
• P A L R B C (a) P kL ML R MR VL L VR (b)39 ML MR ML MR L R = + + L R L R L R k(l – k)PL kL k(l – k)PL kL (c) FIGURE 2.4 Any span of a continuous beam (a) can be treated as a simple beam, as shown in (b) and (c). In (c), the moment diagram is decomposed into basic components.
• 40 CHAPTER TWO W L1 L2 L3 L4 R0 R1 R2 R3 R4 (a) d3 d1 d2 (b) 1 y31 y11 y21 (c) 1 y12 y32 y22 (d) 1 y13 y23 y33 (e) d1 = y11R1 + y12R2 + y13R3 d2 = y21R1 + y22R2 + y23R3 d3 = y31R1 + y32R2 + y33R3 FIGURE 2.5 Reactions of continuous beam (a) found by making the beam statically determinate. (b) Deﬂections computed with inte- rior supports removed. (c), (d ), and (e) Deﬂections calculated for unit load over each removed support, to obtain equations for each redundant.
• BEAM FORMULAS 41 W = wL W w kL L L–k(l – k)2 WL k2(l – k)WL WL WL – 12 12 1 k(l – k)WL WL 8 L 2 (a) (b) 1 1 1 W W W W W L 3 L3 L3 L kL 2 2 kL 4 4 4 4 L L 1 1 5– k(l – k)WL k(l – k)WL c L WL 2 2 48 5 – WL 48 1 1 1 1 kWL WL WL WL 2 8 6 8 kL kL (c) (d)FIGURE 2.6 Fixed-end moments for a prismatic beam. (a) For a concentrated load.(b) For a uniform load. (c) For two equal concentrated loads. (d) For three equalconcentrated loads.
• 42 CHAPTER TWO W MF aL m= WL 0 0.05 0.10 0.15 0.20 m R R for or m m 0 L) = for for G2 .05 05 f L) 05 05 m (–0 = 0. (0. = –0. S3 S3 G 2 G =0 2 =0 G2 = 0.10 =0 .05 S3 G2 =0 .10 G G 2 2 G = = 2 0. 0. =0 20 15 .25 2 =0 G 0 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0.9 1.0 0.5 1.0 0.9 0.8 0.7 0.6 0.4 0.3 0.2 0.1 0 Use upper line for MRF a Use lower line for ML F FIGURE 2.7 Chart for ﬁxed-end moments due to any type of loading.
• BEAM FORMULAS 43 W = (n + 1)P nP P yL W=P – xL – X=0 x= n y – 1+n G2 = 0 n y2 S3 = 0 G2 = (1 + n)2 n(n – 1) y3 S3 = (1 + n)3 Case 1 Case 2 W = wyL yL wyL – – W= xL xL 2w w yL x= 1 y – 2 x= 1 y – 2 = 1 y2 3 G 12 G 2 = 1 y2 S3 = 0 18 S3 = – 1 y3 135 Case 3 Case 4FIGURE 2.8 Characteristics of loadings.
• 44 CHAPTER TWO wyL 1 W= – W= wyL 2 xL 3 w w = ky2L2 1 yL – 1 x = yL yL 2 2 yL x= 1 y – x= 1 y – 4 2 G2 = 3 y2 G2 = 1 y2 80 24 S3 = 0 S3 = – 1 y3 160 Case 5 Case 6 y W = nP W = LÚ f(x′)dx′ 0 aL – – xL –xL xL P P P P P P P Ldx yL yL yL yL yL yL W = f(x′) x′L x= n–1 y – yL 2 2 2 y G2 = n – 1 y2 = n + 1 . a – LÚ wx′dx′ 0 12 n – 1 12 x= W S3 = 0 y L wx2dx 2 = Ú0 G W y LÚ wx3dx S3 = 0 W Case 7 Case 8 FIGURE 2.8 (Continued)
• BEAM FORMULAS 45S3 b 3 Pn/W. These values are given in Fig. 2.8 for some common types nof loading. Formulas for moments due to deﬂection of a ﬁxed-end beam are given inFig. 2.9. To use the modiﬁed moment distribution method for a ﬁxed-end beamsuch as that in Fig. 2.9, we must ﬁrst know the ﬁxed-end moments for a beamwith supports at different levels. In Fig. 2.9, the right end of a beam with span Lis at a height d above the left end. To ﬁnd the ﬁxed-end moments, we ﬁrstdeﬂect the beam with both ends hinged; and then ﬁx the right end, leaving theleft end hinged, as in Fig. 2.9(b). By noting that a line connecting the two sup-ports makes an angle approximately equal to d/L (its tangent) with the originalposition of the beam, we apply a moment at the hinged end to produce an endrotation there equal to d/L. By the deﬁnition of stiffness, this moment equalsthat shown at the left end of Fig. 2.9(b). The carryover to the right end is shownas the top formula on the right-hand side of Fig. 2.9(b). By using the law of R F MR d L MLF L (a) d d CRFKLF R L L d =CLFKRF d d L L KLF L (b) d d L KRF R L d CLFKRF L d L d = CRFKLF L (c) R KRF 1 + CLF d L L KLF 1 + CRF d L (d) FIGURE 2.9 Moments due to deﬂection of a ﬁxed-end beam.
• 46 CHAPTER TWOreciprocal deﬂections, we obtain the end moments of the deﬂected beam inFig. 2.9 as d MF L K F (1 L C F) R (2.1) L d MF R K F (1 R C F) L (2.2) LIn a similar manner the ﬁxed-end moment for a beam with one end hinged andthe supports at different levels can be found from d MF K (2.3) Lwhere K is the actual stiffness for the end of the beam that is ﬁxed; for beams ofvariable moment of inertia K equals the ﬁxed-end stiffness times (1 C FC F). L RULTIMATE STRENGTH OF CONTINUOUS BEAMSMethods for computing the ultimate strength of continuous beams and framesmay be based on two theorems that ﬁx upper and lower limits for load-carryingcapacity:1. Upper-bound theorem. A load computed on the basis of an assumed link mechanism is always greater than, or at best equal to, the ultimate load.2. Lower-bound theorem. The load corresponding to an equilibrium condition with arbitrarily assumed values for the redundants is smaller than, or at best equal to, the ultimate loading—provided that everywhere moments do not exceed MP. The equilibrium method, based on the lower bound theorem, is usually easier for simple cases. For the continuous beam in Fig. 2.10, the ratio of the plastic moment for theend spans is k times that for the center span (k 1). Figure 2.10(b) shows the moment diagram for the beam made determinateby ignoring the moments at B and C and the moment diagram for end momentsMB and MC applied to the determinate beam. Then, by using Fig. 2.10(c), equi-librium is maintained when wL2 1 1 MP M M 4 2 B 2 C wL2 4 kM P wL2 (2.4) 4(1 k)
• BEAM FORMULAS 47 2w w w A D B C L L L (a) Moment diagram for redundants MB MC wL2 wL2 8 8 wL2 4 Moment diagram for determinate beam (b) x kMP kMP x kMP kMP MP (c) FIGURE 2.10 Continuous beam shown in (a) carries twice as much uniform load in the center span as in the side span. In (b) are shown the moment diagrams for this loading condition with redundants removed and for the redundants. The two moment diagrams are combined in (c), producing peaks at which plastic hinges are assumed to form. The mechanism method can be used to analyze rigid frames of constant sec-tion with ﬁxed bases, as in Fig. 2.11. Using this method with the vertical load atmidspan equal to 1.5 times the lateral load, the ultimate load for the frame is4.8MP /L laterally and 7.2M P /L vertically at midspan. Maximum moment occurs in the interior spans AB and CD when L M x (2.5) 2 wL
• 1.5P L L P 2 2 B E C B C B E C B C L E E 2 A D A D A D A D L (a) (b) Beam (c) Frame (d) Combination mechanism mechanism mechanism θL L 1.5P θL θL 2 2 2 1.5P θ P θC48 B 2 B CP B C θ θ θ θ θ E L E 2 θL 2θ L 2θ L θ 2 2 2 θ θ θ A D A D L (e) (f) (g) FIGURE 2.11 Ultimate-load possibilities for a rigid frame of constant section with ﬁxed bases.
• BEAM FORMULAS 49or if L kM P M kM P when x (2.6) 2 wLA plastic hinge forms at this point when the moment equals kMP. For equilibrium, w x kM P x (L x) kM P 2 L w L kM P L KM P 1 kM P kM P 2 2 wL 2 wL 2 wL2leading to k 2M 2 wL2 P 3kM P 0 (2.7) wL2 4When the value of MP previously computed is substituted, 7k 2 4k 4 or k (k 4 7) 4 7from which k 0.523. The ultimate load is 4M P (1 k) MP wL 6.1 (2.8) L L In any continuous beam, the bending moment at any section is equal to thebending moment at any other section, plus the shear at that section times itsarm, plus the product of all the intervening external forces times their respectivearms. Thus, in Fig. 2.12 Vx R1 R2 R3 P1 P2 P3 (2.9) Mx R 1 (l 1 l2 x) R 2 (l 2 x) R 3x P1 (l 2 c x) P2 (b x) P3a (2.10) Mx M3 V3x P3a (2.11) Table 2.1 gives the value of the moment at the various supports of a uniformlyloaded continuous beam over equal spans, and it also gives the values of the shears P1 P2 P3 c b a R1 R2 R3 x R4 R5 l1 l2 l3 l4FIGURE 2.12 Continuous beam.
• TABLE 2.1 Uniformly Loaded Continuous Beams over Equal Spans (Uniform load per unit length w; length of each span l) Shear on each side Distance to point Notation of support. L left, of max moment, Distance to Number of R right. reaction at Moment Max. measured to point of inﬂection, of support any support is L R over each moment in right from measured to right supports of span L R support each span support from support 1 2 1 or 2 0 2 0 0.125 0.500 None 3 3 1 0 8 0 0.0703 0.375 0.750 5 5 1 2 8 8 8 0.0703 0.625 0.250 4 4 1 0 10 0 0.080 0.400 0.800 6 5 1 2 10 10 10 0.025 0.500 0.276, 0.724 11 1 0 28 0 0.0772 0.393 0.78650 17 15 3 5 2 28 28 28 0.0364 0.536 0.266, 0.806 13 13 2 3 28 28 28 0.0364 0.464 0.194, 0.734 15 1 0 38 0 0.0779 0.395 0.789 23 20 4 6 2 38 38 38 0.0332 0.526 0.268, 0.783 18 19 3 3 38 38 38 0.0461 0.500 0.196, 0.804 41 1 0 104 0 0.0777 0.394 0.788 63 55 11 2 104 104 104 0.0340 0.533 0.268, 0.790 49 51 8 7 3 104 104 104 0.0433 0.490 0.196, 0.785 53 53 9 4 104 104 104 0.0433 0.510 0.215, 0.804 56 1 0 142 0 0.0778 0.394 0.789 86 75 15 8 2 142 142 142 0.0338 0.528 0.268, 0.788 67 70 11 3 142 142 142 0.0440 0.493 0.196, 0.790 72 71 12 4 142 142 142 0.0405 0.500 0.215, 0.785 Values apply to wl wl wl 2 wl 2 l l The numerical values given are coefﬁcients of the expressions at the foot of each column.
• BEAM FORMULAS 51 Shear Moment FIGURE 2.13 Relation between moment and shear diagrams for a uni- formly loaded continuous beam of four equal spans.on each side of the supports. Note that the shear is of the opposite sign on eitherside of the supports and that the sum of the two shears is equal to the reaction. Figure 2.13 shows the relation between the moment and shear diagrams for auniformly loaded continuous beam of four equal spans. (See Table 2.1.) Table 2.1also gives the maximum bending moment that occurs between supports, in addi-tion to the position of this moment and the points of inﬂection. Figure 2.14 showsthe values of the functions for a uniformly loaded continuous beam resting onthree equal spans with four supports.Maxwell’s TheoremWhen a number of loads rest upon a beam, the deﬂection at any point is equalto the sum of the deﬂections at this point due to each of the loads taken separately. R1 = 0.4 wl R2 = 1.1 wl R3 = 1.1 wl R4 = 0.4 wl l l l M1 = 0.08 wl2 M2 = 0.025 wl2 Point of intersection Point of intersection 0.4 l 0.1 wl2 0.4 l 0.8 l 0.5 l 0.276 l 0.276 l 0.2 l Moment0.4 wl 0.5 wl 0.6 wl 0.6 wl 0.5 wl 0.4 wl ShearFIGURE 2.14 Values of the functions for a uniformly loaded continuous beam restingon three equal spans with four supports.
• 52 CHAPTER TWOMaxwell’s theorem states that if unit loads rest upon a beam at two points, Aand B, the deﬂection at A due to the unit load at B equals the deﬂection at B dueto the unit load at A.Castigliano’s TheoremThis theorem states that the deﬂection of the point of application of an externalforce acting on a beam is equal to the partial derivative of the work of defor-mation with respect to this force. Thus, if P is the force, f is the deﬂection, andU is the work of deformation, which equals the resilience: dU f (2.12) dP According to the principle of least work, the deformation of any structuretakes place in such a manner that the work of deformation is a minimum.BEAMS OF UNIFORM STRENGTHBeams of uniform strength so vary in section that the unit stress S remains con-stant, and I/c varies as M. For rectangular beams of breadth b and depth d, I/cbd 2/6 and M Sbd2/6. For a cantilever beam of rectangular cross section, undera load P, Px Sbd 2/6. If b is constant, d 2 varies with x, and the proﬁle of theshape of the beam is a parabola, as in Fig. 2.15. If d is constant, b varies as x, andthe beam is triangular in plan (Fig. 2.16). Plan b x Plan P P d x Elevation ElevationFIGURE 2.15 Parabolic beam of FIGURE 2.16 Triangular beam ofuniform strength. uniform strength.
• BEAM FORMULAS 53 Shear at the end of a beam necessitates modiﬁcation of the forms deter-mined earlier. The area required to resist shear is P/Sv in a cantilever and R/Sv ina simple beam. Dashed extensions in Figs. 2.15 and 2.16 show the changes nec-essary to enable these cantilevers to resist shear. The waste in material and extracost in fabricating, however, make many of the forms impractical, except forcast iron. Figure 2.17 shows some of the simple sections of uniform strength. Innone of these, however, is shear taken into account.SAFE LOADS FOR BEAMS OF VARIOUS TYPESTable 2.2 gives 32 formulas for computing the approximate safe loads onsteel beams of various cross sections for an allowable stress of 16,000 lb/in2(110.3 MPa). Use these formulas for quick estimation of the safe load for anysteel beam you are using in a design. Table 2.3 gives coefﬁcients for correcting values in Table 2.2 for variousmethods of support and loading. When combined with Table 2.2, the two sets offormulas provide useful time-saving means of making quick safe-load compu-tations in both the ofﬁce and the ﬁeld.ROLLING AND MOVING LOADSRolling and moving loads are loads that may change their position on a beam orbeams. Figure 2.18 shows a beam with two equal concentrated moving loads,such as two wheels on a crane girder, or the wheels of a truck on a bridge.Because the maximum moment occurs where the shear is zero, the shear diagramshows that the maximum moment occurs under a wheel. Thus, with x a/2: 2x a R1 P 1 (2.13) l l Pl a 2x a 4x 2 M2 1 (2.14) 2 l l l l2 2x a R2 P 1 (2.15) l l Pl a 2a2 2x 3a 4x 2 M1 1 (2.16) 2 l l2 l l l2 1 M2 max when x 4a 3 M1 max when x 4a 2 2 Pl a P a M max 1 l (2.17) 2 2l 2l 2
• 1. Fixed at One End, Load P Concentrated at Other End Elevation Beam Cross section Formulas and plan B x b l A 6P Elevation: y2 = x h 2 1 bSs y y 1, top, straight h line; bottom, h 6Pl P Rectangle: parabola. 2, h= bSs B l b width (b) con complete pa A x -stant, depth -rabola (y) variable Deflection at A: 2 2h y h h Plan: f = 8P l 3 y rectangle bE h54 P
• B A y 6P y= x Elevation: h2Ss h h Rectangle: rectangle 6Pl b= 2 l b width (y) var h Ss x Plan: P -iable, depth Deflection at A: (h) constant triangle b y f = 6P l 3 bE h B 2 h b 3 A Rectangle: Elevation: y3 = 6P x kSs width (z) var cubic parabola h h y y -iable, depth z = ky55 l P z (y) variable Plan: 3 6Pl x z = k (const) cubic parabola h= y bSs 2b b = kh b z 3 FIGURE 2.17 Beams of uniform strength (in bending).
• Elevation Beam Cross section Formulas and plan 1. Fixed at One End, Load P Concentrated at Other End (Continued) 2 d B 3 A l d Elevation: 32P x cubic parabola y3 = x Circle: πSs diam (y) d y variable Plan: 3 32Pl d= cubic parabola πSs y P B x 3P56 b Elevation: l A y=x Rectangle: triangle blS width (b) con 3Pl y y h h= -stant, depth h Plan: bSs (y) variable rectangle P l 3 f=6 bE h
• 2 B A y Elevation: y = 3Px2 lSsh rectangle b = 3Pl h Rectangle: width (z) var- Plan: Ssh2 l b x iable, depth two parabolic curves with Deflection at A: (h) constant b y f = 3P vertices at free l 3 end bE h B b A Elevation: 3Px2 Rectangle: y3 = semicubic kSsl h h h 3 y width (z) var- y parabola iable, depth z = ky l57 x z (y) variable, Plan: 3 3Pl z =k semicubic h= kSs b b 3 z y parabola b = kh FIGURE 2.17 (Continued)
• 2. Fixed at One End, Load P Uniformly Distributed Over l Elevation Beam Cross section and plan Formulas B A Elevation l d 16P 2 x semicubic y3 = x Circle: parabola πlSs diam (y) y d Plan: variable 3 16Pl semicubic d= πSs y parabola d 3 3. Supported at Both Ends, Load P Concentrated at Point C l P Elevation: 2 3P58 b two parabolas, y= x Ssb C Rectangle: vertices at width (b) con- points of support 3Pl h h y h= y stant, depth 2bSs x (y) variable Plan: l 3 f= P l rectangle 2Eb h
• l 2 P y Elevation: rectangle y = 3P2 x h C Ssh Rectangle: b width (y) vari- Plan: able, depth b = 3Pl2 two triangles, 2Ssh (h) constant vertices at 3 points of support f = 3Pl 3 y b 8Ebh x C Elevation: two parabolas, 6P(l – p) y2 =59 x A B Rectangle: vertices at blSs width (b) con- points of support 6Pp h h y12 = y1 x y blSs 1 h 2 h 2 y stant, depth x x (y or y1) Plan: 6P(l – p)p l–p b variable h= P rectangle blSs P FIGURE 2.17 (Continued)
• Elevation Beam Cross section and plan Formulas 3. Supported at Both Ends, Load P Moving Across Span b Elevation: x2 + y 2 = 1 Rectangle: ellipse l 2 3Pl Major axis = l h h width (b) con- 2 y y 2bSs h 2 A B Minor axis = 2h O x stant, depth l l (y) variable Plan: 3Pl h= 2 2 rectangle 2bSs 4. Supported at Both Ends, Load P Uniformly Distributed Over l60 x2 + y2 = 1 l 2 3Pl 2 4bSs b Rectangle: Elevation: 3Pl h= width (b) con- ellipse 4bSs h h y y stant, depth h 2 A B O x C (y) variable Plan: Deflection at O: l l rectangle 3 2 2 P f = 1 Pl 64 EI = 3 P l 3 16 bE h
• Elevation Beam Cross section and plan Formulas 4. Supported at Both Ends, Load P Uniformly Distributed Over l y Elevation: h rectangle x2 b Rectangle: y = 3P2 x – width (y) Ss h l variable, depth Plan: x (h) constant two parabolas b = 3Pl2 4Ssh with vertices at b y61 center of span l 2 FIGURE 2.17 (Continued)
• 64 CHAPTER TWO l2 1 P P R1 a R2 l x 2 Shear M1 M2 Moment FIGURE 2.18 Two equal concentrated moving loads. Figure 2.19 shows the condition when two equal loads are equally distant onopposite sides of the center of the beam. The moment is then equal under thetwo loads. If two moving loads are of unequal weight, the condition for maximummoment is the maximum moment occurring under the heavier wheel when thecenter of the beam bisects the distance between the resultant of the loads andthe heavier wheel. Figure 2.20 shows this position and the shear and momentdiagrams. l l 2 2 P P a a R1 = P 2 2 R2 = P Shear Moment FIGURE 2.19 Two equal moving loads equally distant on opposite sides of the center.
• BEAM FORMULAS 65 l l 2 a 2 6 2P P R1 a R2 Shear Moment FIGURE 2.20 Two moving loads of unequal weight. When several wheel loads constituting a system are on a beam or beams, theseveral wheels must be examined in turn to determine which causes the greatestmoment. The position for the greatest moment that can occur under a givenwheel is, as stated earlier, when the center of the span bisects the distancebetween the wheel in question and the resultant of all loads then on the span.The position for maximum shear at the support is when one wheel is passing offthe span.CURVED BEAMSThe application of the flexure formula for a straight beam to the case of acurved beam results in error. When all “fibers” of a member have the samecenter of curvature, the concentric or common type of curved beam exists(Fig. 2.21). Such a beam is defined by the Winkler-Bach theory. The stress ata point y units from the centroidal axis is M y S 1 (2.18) AR Z (R y)M is the bending moment, positive when it increases curvature; y is posi-tive when measured toward the convex side; A is the cross-sectional area;
• 66 CHAPTER TWO F F ho e +y a hi Neutral surface R Ri Ro F F FIGURE 2.21 Curved beam.R is the radius of the centroidal axis; Z is a cross-section property definedby 1 y Z dA (2.19) A R y Analytical expressions for Z of certain sections are given in Table 2.4. Z canalso be found by graphical integration methods (see any advanced strengthbook). The neutral surface shifts toward the center of curvature, or inside ﬁber,an amount equal to e ZR/(Z 1). The Winkler-Bach theory, though practi-cally satisfactory, disregards radial stresses as well as lateral deformations andassumes pure bending. The maximum stress occurring on the inside fiber isS Mhi /AeRi, whereas that on the outside ﬁber is S Mho/AeRo. The deﬂection in curved beams can be computed by means of the moment-area theory. 2 2 The resultant deﬂection is then equal to 0 x y in the directiondeﬁned by tan y / x. Deﬂections can also be found conveniently by useof Castigliano’s theorem. It states that in an elastic system the displacement inthe direction of a force (or couple) and due to that force (or couple) is the partialderivative of the strain energy with respect to the force (or couple). A quadrant of radius R is ﬁxed at one end as shown in Fig. 2.22. The force Fis applied in the radial direction at free-end B. Then, the deﬂection of B isBy moment area, y R sin x R(1 cos ) (2.20) ds Rd M FR sin (2.21) FR 3 FR 3 B x B y (2.22) 4EI 2EI
• TABLE 2.4 Analytical Expressions for Z Section Expression Section R R C h Z 1 ln h R C C R r B r 2 R R R Z 1 2 1 r r r R t67 b R C1 Z 1 t ln (R C1) (b t) ln (R C0) b ln (R C2) C3 C2 A R and A tC1 (b t) C3 bC2 t b C1 R R C2 R C1 C2 Z 1 b ln (t b) ln R A R C2 R C1 A 2 [(t b)C1 bC2]
• 68 CHAPTER TWO 2EI B FR3 2 and B 1 (2.23) 4 1 FR 3 4EI at x tan (2.24) 2EI FR 3 1 2 tan 32.5By Castigliano, FR3 FR3 B x B y (2.25) 4EI 2EIEccentrically Curved BeamsThese beams (Fig. 2.23) are bounded by arcs having different centers of cur-vature. In addition, it is possible for either radius to be the larger one. Theone in which the section depth shortens as the central section is approachedmay be called the arch beam. When the central section is the largest, thebeam is of the crescent type. Crescent I denotes the beam of larger outside radius and crescent II oflarger inside radius. The stress at the central section of such beams may befound from S KMC/I. In the case of rectangular cross section, the equationbecomes S 6KM/bh2, where M is the bending moment, b is the width of thebeam section, and h its height. The stress factors, K for the inner boundary,established from photoelastic data, are given in Table 2.5. The outside radiusis denoted by Ro and the inside by Ri. The geometry of crescent beams is suchthat the stress can be larger in off-center sections. The stress at the central Ro Ri A Ri Ro y R Ro Ro θ Ri Ri F B XFIGURE 2.22 Quadrant with fixed FIGURE 2.23 Eccentrically curved beams.end.
• BEAM FORMULAS 69 TABLE 2.5 Stress Factors for Inner Boundary at Central Section (see Fig. 2.23) 1. For the arch-type beams h Ro Ri (a) K 0.834 1.504 if 5 Ro Ri h h Ro Ri (b) K 0.899 1.181 if 5 10 Ro Ri h (c) In the case of larger section ratios use the equivalent beam solution 2. For the crescent I-type beams h Ro Ri (a) K 0.570 1.536 if 2 Ro Ri h h Ro Ri (b) K 0.959 0.769 if 2 20 Ro Ri h 0.0298 h Ro Ri (c) K 1.092 if 20 Ro Ri h 3. For the crescent II-type beams h Ro Ri (a) K 0.897 1.098 if 8 Ro Ri h 0.0378 h Ro Ri (b) K 1.119 if 8 20 Ro Ri h 0.0270 h Ro Ri (c) K 1.081 if 20 Ro Ri hsection determined above must then be multiplied by the position factor k,given in Table 2.6. As in the concentric beam, the neutral surface shiftsslightly toward the inner boundary. (See Vidosic, “Curved Beams with Eccen-tric Boundaries,” Transactions of the ASME, 79, pp. 1317–1321.)ELASTIC LATERAL BUCKLING OF BEAMSWhen lateral buckling of a beam occurs, the beam undergoes a combination oftwist and out-of-plane bending (Fig. 2.24). For a simply supported beam ofrectangular cross section subjected to uniform bending, buckling occurs at thecritical bending moment, given by Mcr L 2EIyGJ (2.26)
• 70 CHAPTER TWO TABLE 2.6 Crescent-Beam Position Stress Factors (see Fig. 2.23)* k Angle , degree Inner Outer 10 1 0.055 H/h 1 0.03 H/h 20 1 0.164 H/h 1 0.10 H/h 30 1 0.365 H/h 1 0.25 H/h 40 1 0.567 H/h 1 0.467 H/h (0.5171 1.382 H/h)1/2 50 1.521 1 0.733 H/h 1.382 (0.2416 0.6506 H/h)1/2 60 1.756 1 1.123 H/h 0.6506 (0.4817 1.298 H/h)1/2 70 2.070 1 1.70 H/h 0.6492 (0.2939 0.7084 H/h)1/2 80 2.531 1 2.383 H/h 0.3542 90 1 3.933 H/h *Note: All formulas are valid for 0 H/h 0.325. Formulas for the inner boundary, except for 40 degrees, may be used to H/h 0.36. H distance between centers. Compressive stress M M Tensile stress L (a) Mcr β Mcr (b) FIGURE 2.24 (a) Simple beam subjected to equal end moments. (b) Elastic lateral buckling of the beam.
• BEAM FORMULAS 71where L unbraced length of the member E modulus of elasticity Iy moment of inertial about minor axis G shear modulus of elasticity J torsional constantThe critical moment is proportional to both the lateral bending stiffness EIy /Land the torsional stiffness of the member GJ/L. For the case of an open section, such as a wide-ﬂange or I-beam section,warping rigidity can provide additional torsional stiffness. Buckling of a simplysupported beam of open cross section subjected to uniform bending occurs at thecritical bending moment, given by L B y 2 Mcr EI GJ ECw 2 (2.27) Lwhere Cw is the warping constant, a function of cross-sectional shape anddimensions (Fig. 2.25). In the preceding equations, the distribution of bending moment is assumed tobe uniform. For the case of a nonuniform bending-moment gradient, bucklingoften occurs at a larger critical moment. Approximation of this critical bending b Area A t b2 t t h w bt 3 3 Cw = A Cw = t (b13 + b23) Cw = b3t3 + h3w3 144 36 144 36(a) Equal-leg angle (b) Un-equal-leg angle (c) T Section y y – h x x x h x x y y 2 2 h2Iy Cw = h Iy + x2 A I – h A – Cw = 4 4Ix 4 (d) Channel (e) Symmetrical IFIGURE 2.25 Torsion-bending constants for torsional buckling. A cross-sectionalarea; Ix moment of inertia about x–x axis; Iy moment of inertia about y–y axis. (AfterMcGraw-Hill, New York). Bleich, F., Buckling Strength of Metal Structures.
• 72 CHAPTER TWOmoment, Mcr may be obtained by multiplying Mcr given by the previous equa-tions by an ampliﬁcation factor M cr Cb Mcr (2.28) 12.5Mmaxwhere Cb (2.29) 2.5Mmax 3MA 4MB 3MCand Mmax absolute value of maximum moment in the unbraced beam segment MA absolute value of moment at quarter point of the unbraced beam segment MB absolute value of moment at centerline of the unbraced beam segment MC absolute value of moment at three-quarter point of the unbraced beam segment Cb equals 1.0 for unbraced cantilevers and for members where the momentwithin a signiﬁcant portion of the unbraced segment is greater than, or equal to,the larger of the segment end moments.COMBINED AXIAL AND BENDING LOADSFor short beams, subjected to both transverse and axial loads, the stresses aregiven by the principle of superposition if the deﬂection due to bending may beneglected without serious error. That is, the total stress is given with sufﬁcientaccuracy at any section by the sum of the axial stress and the bending stresses.The maximum stress, lb/in2 (MPa), equals P Mc f (2.30) A Iwhere P axial load, lb (N ) A cross-sectional area, in2 (mm2) M maximum bending moment, in lb (Nm) c distance from neutral axis to outermost ﬁber at the section where max- imum moment occurs, in (mm) I moment of inertia about neutral axis at that section, in4 (mm4) When the deﬂection due to bending is large and the axial load producesbending stresses that cannot be neglected, the maximum stress is given by P c f (M Pd) (2.31) A Iwhere d is the deﬂection of the beam. For axial compression, the moment Pdshould be given the same sign as M; and for tension, the opposite sign, but the
• Type of Support Fundamental Mode Second Mode Third Mode Fourth Mode Cantilever 0.5L 0.132L 0.356L 0.094L L 0.774L ω wL4/EI = 20.0 125 350 0.644L 684 T EI/wL4 = 0.315 0.0503 0.0180 0.0092 0.5L L L L L Simple L 4 3 3 4 L 2 ω wL4/EI= 56.0 224 502 897 T EI/wL4 = 0.112 0.0281 0.0125 0.0070 L 0.359L 0.359L 0.278L Fixed L 275 L 2 ω wL4/EI = 127 350 684 1,133 0.278L T EI/wL4 = 0.0496 0.0180 0.0092 0.0056 Fixed-hinged 0.56L 0.384L 0.308L 0.294L 0.235L L 0.529L ω wL4/EI = 87.2 283 591 1,111 T EI/wL4 = 0.0722 0.0222 0.0106 0.0062 FIGURE 2.26 Coefﬁcients for computing natural circular frequencies and natural periods of vibration of prismatic beams.
• 76 CHAPTER TWOmultiply it by EI/wL4. To get T, divide the appropriate constant by EI/wL4. In these equations, natural frequency, rad/s W beam weight, lb per linear ft (kg per linear m) L beam length, ft (m) E modulus of elasticity, lb/in2 (MPa) I moment of inertia of beam cross section, in4 (mm4) T natural period, s To determine the characteristic shapes and natural periods for beams withvariable cross section and mass, use the Rayleigh method. Convert the beaminto a lumped-mass system by dividing the span into elements and assumingthe mass of each element to be concentrated at its center. Also, compute all quanti-ties, such as deﬂection and bending moment, at the center of each element. Startwith an assumed characteristic shape.TORSION IN STRUCTURAL MEMBERSTorsion in structural members occurs when forces or moments twist the beamor column. For circular members, Hooke’s law gives the shear stress at any giv-en radius, r. Table 2.7 shows the polar moment of inertia, J, and the maximumshear for ﬁve different structural sections.STRAIN ENERGY IN STRUCTURAL MEMBERS*Strain energy is generated in structural members when they are acted on byforces, moments, or deformations. Formulas for strain energy, U, for shear,torsion and bending in beams, columns, and other structural members are:Strain Energy in Shear.For a member subjected to pure shear, strain energy is given by V 2L U (2.38) 2AG AG 2 U (2.39) 2Lwhere V shear load shear deformation L length over which the deformation takes place A shear area G shear modulus of elasticity *Brockenbrough and Merritt—Structural Steel Designer’s Handbook, McGraw-Hill.
• BEAM FORMULAS 77TABLE 2.7 Polar Moment of Intertia and Maximum Torsional Shear Polar moment of Maximum shear* inertia J vmax 1 4 2T πr 2 πr3 2r at periphery R = 2T 3 208a a 0.141a4 at midpoint of each side T(3a – 1.8b) b 1 – 0.21 b 1 – b4 a2b2 ab3 3 a 12a4 at midpoint a of longer sides 20T a3 a 0.0214a4 at midpoint of each side 1 2TR π(R4 – r4) π(R4 – r4) 2r 2R 2 at outer periphery *T twisting moment, or torque.Strain Energy in TorsionFor a member subjected to torsion T 2L U (2.40) 2JG JG 2 U (2.41) 2L
• 78 CHAPTER TWO P P 2 2 Pab Pa b PL PL 2 2 8 8 L a b L L/2 L/2 P P P P Pa(L – a) 2PL a L–2a a L L/3 L/3 L/3 9 P P P P P P P 5PL 2PL L/4 L/4 L/4 L/4 16 L/5 L/5 L/5 L/5 L/5 5 wk/t wk/t 2 2 2 wL wL wL 20 30 12 wk/t wk/t 2 2 2 3 2 2 2 wa (6 – 8a + 3a )L wa (4 – 3a)L 11wL 5wL 12 aL 12 192 L/2 192 wk/t wk/t 2 2 2 3 2 2 2wa (10 – 15a + 6a )L wa (5 – 4a)L wL 3wL 30 aL 20 30 L/2 160 k/t wk/t w 2 2 2 3 2 2 2wa (10 – 10a + 3a )L wa (5 – 3a)L 23wL 7wL 60 aL 60 960 L/2 960 k/t Parabola w wk/t 2 2 wL 5wL 15 L/2 L/2 96 wk/t wk/t Parabola 2 2 wL wL 60 32 L/2 L/2 wk/t wk/t wk/t wk/t 2 2 2 wa (3 – 2a)L 5wL 6 192 aL aL L/4 L/2 L/4 M M M(1 – a) (3a – 1) Ma(1 – 3a) – M M 4 4 aL L/2 L/2FIGURE 2.27 Fixed-end moments in beams. (Brockenbrongh and Merritt-StructuralDesigner’s Handbook, McGraw-Hill.)
• BEAM FORMULAS 79where T torque angle of twist L length over which the deformation takes place J polar moment of inertia G shear modulus of elasticityStrain Energy in BendingFor a member subjected to pure bending (constant moment) M 2L U (2.42) 2EI EI 2 U (2.43) 2Lwhere M bending moment angle through which one end of beam rotates with respect to the other end L length over which the deformation takes place I moment of inertia E modulus of elasticity For beams carrying transverse loads, the total strain energy is the sum of theenergy for bending and that for shear.FIXED-END MOMENTS IN BEAMSFigure 2.27 gives formulas for ﬁxed-end moments in a variety of beams. Theformat of this illustration gives easy access to these useful formulas.
• CHAPTER 3 COLUMN FORMULASGENERAL CONSIDERATIONSColumns are structural members subjected to direct compression. All columnscan be grouped into the following three classes:1. Compression blocks are so short (with a slenderness ratio — that is, unsup- ported length divided by the least radius of gyration of the member—below 30) that bending is not potentially occurring.2. Columns so slender that bending under load is given are termed long columns and are deﬁned by Euler’s theory.3. Intermediate-length columns, often used in structural practice, are called short columns. Long and short columns usually fail by buckling when their critical load isreached. Long columns are analyzed using Euler’s column formula, namely, n 2EI n 2EA (3.1) Pcr l2 (l/r)2In this formula, the coefﬁcient n accounts for end conditions. When the columnis pivoted at both ends, n 1; when one end is ﬁxed and the other end isrounded, n 2; when both ends are ﬁxed, n 4; and when one end is ﬁxedand the other is free, n 0.25. The slenderness ratio separating long columnsfrom short columns depends on the modulus of elasticity and the yield strengthof the column material. When Euler’s formula results in (Pcr /A) > Sy, strengthinstead of buckling causes failure, and the column ceases to be long. In quickestimating numbers, this critical slenderness ratio falls between 120 and 150.Table 3.1 gives additional column data based on Euler’s formula.SHORT COLUMNSStress in short columns can be considered to be partly due to compression andpartly due to bending. Empirical, rational expressions for column stress are, in 81
• TABLE 3.1 Strength of Round-Ended Columns According to Euler’s Formula* Low- Medium- Wrought carbon carbon Material† Cast iron iron steel steel Ultimate compressive strength, lb/in2 107,000 53,400 62,600 89,000 Allowable compressive stress, lb/in2 7,100 15,400 17,000 20,000 (maximum) Modulus of elasticity 14,200,000 28,400,000 30,600,000 31,300,000 Factor of safety 8 5 5 5 Smallest I allowable at worst section, in4 Pl 2 Pl 2 Pl 2 Pl 2 17,500,000 56,000,000 60,300,000 61,700,00082 Limit of ratio, l/r 50.0 60.6 59.4 55.6 Rectangle r bw1 12 , l/b 14.4 17.5 17.2 16.0 Circle r 1 4 d , l/d 12.5 15.2 14.9 13.9 Circular ring of small thickness 17.6 21.4 21.1 19.7 r d w1 8 , l/d *P allowable load, lb; l length of column, in; b smallest dimension of a rectangular section, in; d diameter of a circular sec- tion, in; r least radius of gyration of section. † To convert to SI units, use: 1b/in2 6.894 kPa; in4 (25.4)4 mm4.
• COLUMN FORMULAS 83 S Euler column Critical L/r Compression blocks Parabolic type Straight line type L/r Short Long FIGURE 3.1 L /r plot for columns.general, based on the assumption that the permissible stress must be reducedbelow that which could be permitted were it due to compression only. The mannerin which this reduction is made determines the type of equation and the slender-ness ratio beyond which the equation does not apply. Figure 3.1 shows the curvesfor this situation. Typical column formulas are given in Table 3.2.ECCENTRIC LOADS ON COLUMNSWhen short blocks are loaded eccentrically in compression or in tension, that is,not through the center of gravity (cg), a combination of axial and bending stressresults. The maximum unit stress SM is the algebraic sum of these two unit stresses. In Fig. 3.2, a load, P, acts in a line of symmetry at the distance e from cg; rradius of gyration. The unit stresses are (1) Sc, due to P, as if it acted through cg,and (2) Sb, due to the bending moment of P acting with a leverage of e about cg.Thus, unit stress, S, at any point y is S Sc Sb (3.2) (P/A) Pey/I Sc(1 ey/r 2)y is positive for points on the same side of cg as P, and negative on the oppositeside. For a rectangular cross section of width b, the maximum stress, SMSc (1 6e/b). When P is outside the middle third of width b and is a compres-sive load, tensile stresses occur. For a circular cross section of diameter d, SM Sc(1 8e/d). The stress due tothe weight of the solid modiﬁes these relations. Note that in these formulas e is measured from the gravity axis and givestension when e is greater than one-sixth the width (measured in the same directionas e), for rectangular sections, and when greater than one-eighth the diameter,for solid circular sections.
• 84 CHAPTER THREETABLE 3.2 Typical Short-Column Formulas Formula Material Code Slenderness ratio 2 l lSw 17,000 0.485 Carbon steels AISC 120 r r l lSw 16,000 70 Carbon steels Chicago 120 r r l lSw 15,000 50 Carbon steels AREA 150 r r l lSw 19,000 100 Carbon steels Am. Br. Co. 60 120 r r 2 15.9 l l*Scr 135,000 Alloy-steel ANC 65 c r tubing wcr l lSw 9,000 40 Cast iron NYC 70 r r 245 l 1*Scr 34,500 2017ST ANC 94 wc r aluminum wcr 2 0.5 l 1*Scr 5,000 Spruce ANC 72 c r wcr B Sy 2 Sy l l 2n 2E*Scr Sy 1 Steels Johnson 4n 2E r r† Sy l Scr Steels Secant critical r r B 4AE ec l P 1 sec r2 *Scr theoretical maximum; c end ﬁxity coefﬁcient; c 2, both ends pivoted; c 2.86, onepivoted, other ﬁxed; c 4, both ends ﬁxed; c 1 one ﬁxed, one free. † Is initial eccentricity at which load is applied to center of column cross section. If, as in certain classes of masonry construction, the material cannot withstandtensile stress and, thus, no tension can occur, the center of moments (Fig. 3.3) istaken at the center of stress. For a rectangular section, P acts at distance k from thenearest edge. Length under compression 3k, and SM 2/3 P/hk. For a circularsection, SM [0.372 0.056 (k/r)]P/k rk, where r radius and k distanceof P from circumference. For a circular ring, S average compressive stress oncross section produced by P; e eccentricity of P; z length of diameter undercompression (Fig. 3.4). Values of z/r and of the ratio of Smax to average S are givenin Tables 3.3 and 3.4.
• COLUMN FORMULAS 85 P k e P e b Pey P P P I A Pey A A I Pey Pey I I FIGURE 3.2 Load plot for columns. P k e r r1 S A Smax e b Z Z Smax O c. of g. 3k SMFIGURE 3.3 Load plot for columns. FIGURE 3.4 Circular column load plot.
• TABLE 3.3 Values of the Ratio z/r r1 r e e r 0.0 0.5 0.6 0.7 0.8 0.9 1.0 r 0.25 2.00 0.25 0.30 1.82 0.30 0.35 1.66 1.89 1.98 0.35 0.40 1.51 1.75 1.84 1.93 0.40 0.45 1.37 1.61 1.71 1.81 1.90 0.4586 0.50 1.23 1.46 1.56 1.66 1.78 1.89 2.00 0.50 0.55 1.10 1.29 1.39 1.50 1.62 1.74 1.87 0.55 0.60 0.97 1.12 1.21 1.32 1.45 1.58 1.71 0.60 0.65 0.84 0.94 1.02 1.13 1.25 1.40 1.54 0.65 0.70 0.72 0.75 0.82 0.93 1.05 1.20 1.35 0.70 0.75 0.59 0.60 0.64 0.72 0.85 0.99 1.15 0.75 0.80 0.47 0.47 0.48 0.52 0.61 0.77 0.94 0.80 0.85 0.35 0.35 0.35 0.36 0.42 0.55 0.72 0.85 0.90 0.24 0.24 0.24 0.24 0.24 0.32 0.49 0.90 0.95 0.12 0.12 0.12 0.12 0.12 0.12 0.25 0.95 (See Fig. 3.5)
• TABLE 3.4 Values of the Ratio Smax/Savg r1 r e e 0.0 0.5 0.6 0.7 0.8 0.9 1.0 r r 0.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00 0.05 1.20 1.16 1.15 1.13 1.12 1.11 1.10 0.0587 0.10 1.40 1.32 1.29 1.27 1.24 1.22 1.20 0.10 0.15 1.60 1.48 1.44 1.40 1.37 1.33 1.30 0.15 0.20 1.80 1.64 1.59 1.54 1.49 1.44 1.40 0.20 0.25 2.00 1.80 1.73 1.67 1.61 1.55 1.50 0.25 0.30 2.23 1.96 1.88 1.81 1.73 1.66 1.60 0.30 0.35 2.48 2.12 2.04 1.94 1.85 1.77 1.70 0.35 0.40 2.76 2.29 2.20 2.07 1.98 1.88 1.80 0.40 0.45 3.11 2.51 2.39 2.23 2.10 1.99 1.90 0.45 (In determining S average, use load P divided by total area of cross section)
• 88 CHAPTER THREE h h 3 6 h h 3 m 2r h . h in 6 h 3 h b b b .226R 3 3 3 R R b (a) (b) (c) 1h 2h 3 1 3 D 4h b b 2r 4 d 4 d r1 d r2 h (d) (e) (f) FIGURE 3.5 Column characteristics. The kern is the area around the center of gravity of a cross section withinwhich any load applied produces stress of only one sign throughout the entirecross section. Outside the kern, a load produces stresses of different sign.Figure 3.5 shows kerns (shaded) for various sections. For a circular ring, the radius of the kern r D[1 (d/D)2]/8. For a hollow square (H and h lengths of outer and inner sides), the kern isa square similar to Fig. 3.5(a), where 2 2 H 1 h h rmin 1 0.1179H 1 (3.3) 6 2 H H For a hollow octagon, Ra and Ri are the radii of circles circumscribing the outerand inner sides respectively; thickness of wall 0.9239(Ra – Ri); and the kern is anoctagon similar to Fig. 3.5(c), where 0.2256R becomes 0.2256Ra[1 (Ri /Ra)2].COLUMNS OF SPECIAL MATERIALS*Here are formulas for columns made of special materials. The nomenclature forthese formulas is: *Roark—Formulas for Stress and Strain, McGraw-Hill.
• COLUMN FORMULAS 89 Nomenclature for formulas Eqs. (3.4) through (3.12): Q allowable load, lb P ultimate load, lb A section area of column, sq in L length of column, in r least radium of gyration of column section, in Su ultimate strength, psi Sy yield point or yield strength of material, psi E modulus of elasticity of material, psi m factor of safety (L/r) critical slenderness ratio For columns of cast iron that are hollow, round, with ﬂat ends, used inbuildings: Q L 12,000 60 (3.4) A r Q L Max 10,000; max 100 (3.5) A r Q L 9000 40 (3.6) A r L Max 70 (3.7) r Min diameter 6 in; min thickness 1/2 in P L 34,000 88 (3.8) A r For structural aluminum columns with the following speciﬁcations: 6061 T6 6062 T6 sy 35,000 su 38,000 E 10,000,000used in nonwelded building structures in structural shapes or in fabricated formwith partial constraint: L Q 10 19,000 (3.9) r A L Q L 10 67 20,400 135 (3.10) r A r
• 90 CHAPTER THREE L Q 51,000,000 67 (3.11) r A (L/r)2 P Q 1.95 (3.12) A ACOLUMN BASE PLATE DESIGNBase plates are usually used to distribute column loads over a large enough areaof supporting concrete construction that the design bearing strength of the con-crete is not exceeded. The factored load, Pu, is considered to be uniformly dis-tributed under a base plate. The nominal bearing strength, fp, kip/in2 or ksi (MPa) of the concrete isgiven by B A1 B A1 A1 A2 fp 0.85f c and 2 (3.13)where fc speciﬁed compressive strength of concrete, ksi (MPa) A1 area of the base plate, in2 (mm2) A2 area of the supporting concrete that is geometrically similar to and concentric with the loaded area, in2 (mm2) In most cases, the bearing strength, fp is 0.85f c , when the concrete support isslightly larger than the base plate or 1.7f c , when the support is a spread footing,pile cap, or mat foundation. Therefore, the required area of a base plate for afactored load Pu is Pu A1 (3.14) c0.85 fcwhere c is the strength reduction factor 0.6. For a wide-ﬂange column, A1should not be less than bf d, where bf is the ﬂange width, in (mm), and d is thedepth of column, in (mm). The length N, in (mm), of a rectangular base plate for a wide-ﬂange columnmay be taken in the direction of d as N 2A1 d or 0.5(0.95d 0.80bf) (3.15)The width B, in (mm), parallel to the ﬂanges, then, is A1 B (3.16) N
• COLUMN FORMULAS 91 The thickness of the base plate tp, in (mm), is the largest of the values givenby the equations that follow B 0.9Fy BN 2Pu tp m (3.17) B 0.9Fy BN 2Pu tp n (3.18) B 0.9Fy BN 2Pu tp n (3.19)where m projection of base plate beyond the ﬂange and parallel to the web, in (mm) (N 0.95d)/2 n projection of base plate beyond the edges of the ﬂange and perpendi- cular to the web, in (mm) (B 0.80bf)/2 (dbf)/4 (2 2X)/[1 2(1 n X)] 1.0 X (4 dbf)/(d bf)2][Pu/( 0.85f c 1)AMERICAN INSTITUTE OF STEEL CONSTRUCTIONALLOWABLE-STRESS DESIGN APPROACHThe lowest columns of a structure usually are supported on a concrete founda-tion. The area, in square inches (square millimeters), required is found from P A (3.20) FPwhere P is the load, kip (N) and Fp is the allowable bearing pressure on sup-port, ksi (MPa). The allowable pressure depends on strength of concrete in the foundationand relative sizes of base plate and concrete support area. If the base plate occu-pies the full area of the support, Fp 0.35f c , where f c is the 28-day compres-sive strength of the concrete. If the base plate covers less than the full area,FP 0.35fc A2/A1 0.70f c, where A1 is the base-plate area (B N), and A2is the full area of the concrete support. Eccentricity of loading or presence of bending moment at the column baseincreases the pressure on some parts of the base plate and decreases it onother parts. To compute these effects, the base plate may be assumed com-pletely rigid so that the pressure variation on the concrete is linear. Plate thickness may be determined by treating projections m and n of the baseplate beyond the column as cantilevers. The cantilever dimensions m and n are
• 92 CHAPTER THREE (76.2 mm) 1/4 n 3" 0.80bf bf B 3" n 1/4 (76.2 mm) 2 For columns N 1/4 2 10" or larger d m 0.95d m FIGURE 3.6 Column welded to a base plate.usually deﬁned as shown in Fig. 3.6. (If the base plate is small, the area of thebase plate inside the column profile should be treated as a beam.) Yield-line 42dbf , and the required base plate thickness tp can be calculated fromanalysis shows that an equivalent cantilever dimension n can be defined as 1n B Fy fp tp 2l (3.21)where l max (m, n, n ), in (mm) fp P/(BN) Fp, ksi (MPa) Fy yield strength of base plate, ksi (MPa) P column axial load, kip (N) For columns subjected only to direct load, the welds of column to baseplate, as shown in Fig. 3.6, are required principally for withstanding erectionstresses. For columns subjected to uplift, the welds must be proportioned toresist the forces.COMPOSITE COLUMNSThe AISC load-and-resistance factor design (LRFD) speciﬁcation for structuralsteel buildings contains provisions for design of concrete-encased compressionmembers. It sets the following requirements for qualiﬁcation as a composite col-umn: The cross-sectional area of the steel core—shapes, pipe, or tubing—should
• COLUMN FORMULAS 93be at least 4 percent of the total composite area. The concrete should be rein-forced with longitudinal load-carrying bars, continuous at framed levels, andlateral ties and other longitudinal bars to restrain the concrete; all should haveat least 1 1 2 in (38.1 mm) of clear concrete cover. The cross-sectional area oftransverse and longitudinal reinforcement should be at least 0.007 in2(4.5 mm2) per in (mm) of bar spacing. Spacing of ties should not exceed two-thirds of the smallest dimension of the composite section. Strength of the con-crete f c should be between 3 and 8 ksi (20.7 and 55.2 MPa) for normal-weightconcrete and at least 4 ksi (27.6 MPa) for lightweight concrete. Speciﬁed mini-mum yield stress Fy of steel core and reinforcement should not exceed 60 ksi(414 MPa). Wall thickness of steel pipe or tubing ﬁlled with concrete shouldbe at least bwFy /3E or DwFy /8E , where b is the width of the face of a rectangularsection, D is the outside diameter of a circular section, and E is the elastic mod-ulus of the steel. The AISC LRFD speciﬁcation gives the design strength of an axially loadedcomposite column as Pn, where 0.85 and Pn is determined from Pn 0.85As Fcr (3.22)For c 1.5 Fcr 0.658l2 Fmy c (3.23)For c 1.5 0.877 Fcr 2 Fmy (3.24) cwhere c (KL/rm )wFmy /Em KL effective length of column, in (mm) As gross area of steel core, in2 (mm2) Fmy Fy c1Fyr(Ar /As) c2 fc (Ac /As) Em E c3Ec(Ac /As) rm radius of gyration of steel core, in 0.3 of the overall thickness of the composite cross section in the plane of buckling for steel shapes Ac cross-sectional area of concrete, in2 (mm2) Ar area of longitudinal reinforcement, in2 (mm2) Ec elastic modulus of concrete, ksi (MPa) Fyr speciﬁed minimum yield stress of longitudinal reinforcement, ksi (MPa)For concrete-ﬁlled pipe and tubing, c1 1.0, c2 0.85, and c3 0.4. Forconcrete-encased shapes, c1 0.7, c2 0.6, and c3 0.2. When the steel core consists of two or more steel shapes, they should be tiedtogether with lacing, tie plates, or batten plates to prevent buckling of individ-ual shapes before the concrete attains 0.75 f c .
• 94 CHAPTER THREE The portion of the required strength of axially loaded encased compositecolumns resisted by concrete should be developed by direct bearing at connec-tions or shear connectors can be used to transfer into the concrete the loadapplied directly to the steel column. For direct bearing, the design strength of theconcrete is 1.7 c fc Ab, where c 0.65 and Ab loaded area, in2 (mm2). Cer-tain restrictions apply.ELASTIC FLEXURAL BUCKLING OF COLUMNSElastic buckling is a state of lateral instability that occurs while the material isstressed below the yield point. It is of special importance in structures with slen-der members. Euler’s formula for pin-ended columns (Fig. 3.7) gives validresults for the critical buckling load, kip (N). This formula is, with L/r as theslenderness ratio of the column, 2 EA (3.25) P (L/r)2 P P y y(x) P M(X) L x P M(X) P P (a) (b) FIGURE 3.7 (a) Buckling of a pin-ended column under axial load. (b) Internal forces hold the column in equilibrium.
• COLUMN FORMULAS 95where E modulus of elasticity of the column material, psi (Mpa) A column cross-sectional area, in2 (mm2) r radius of gyration of the column, in (mm) Figure 3.8 shows some ideal end conditions for slender columns and corre-sponding critical buckling loads. Elastic critical buckling loads may be obtainedfor all cases by substituting an effective length KL for the length L of thepinned column, giving 2 EA P (3.26) (KL/r)2 In some cases of columns with open sections, such as a cruciform section,the controlling buckling mode may be one of twisting instead of lateraldeformation. If the warping rigidity of the section is negligible, torsional buck-ling in a pin-ended column occurs at an axial load of GJA P (3.27) Ip Type of column Effective length Critical buckling load π2EI L L L2 L/4 L/2 L/4 L 4π2EI 2 L2 0.3L 0.7L 2 ~0.7L ~ 2π EI L2 π2EI 2L 4L2 L FIGURE 3.8 Buckling formulas for columns.
• 96 CHAPTER THREEwhere G shear modulus of elasticity, psi (MPa) J torsional constant A cross-sectional area, in2 (mm2) Ip polar moment of inertia Ix Iy , in4 (mm4)If the section possesses a signiﬁcant amount of warping rigidity, the axial buck-ling load is increased to 2 A ECw P GJ (3.28) Ip L2where Cw is the warping constant, a function of cross-sectional shape anddimensions.ALLOWABLE DESIGN LOADS FOR ALUMINUM COLUMNSEuler’s equation is used for long aluminum columns, and depending on thematerial, either Johnson’s parabolic or straight-line equation is used for shortcolumns. These equations for aluminum followEuler’s equation c 2E Fe (3.29) (L/ )2Johnson’s generalized equation (L/ ) n (3.30) Fc Fce 1 K B Fce cE The value of n, which determines whether the short column formula is thestraight-line or parabolic type, is selected from Table 3.5. The transition from thelong to the short column range is given by B Fce L kcE cr (3.31)where Fe allowable column compressive stress, psi (MPa) Fce column yield stress and is given as a function of Fcy (compres- sive yield stress), psi (MPa) L length of column, ft (m) radius of gyration of column, in (mm) E modulus of elasticity—noted on nomograms, psi (MPa) c column-end ﬁxity from Fig. 3.9 n, K, k constants from Table 3.5
• COLUMN FORMULAS 97 P P P P P Restraining bulk- head partially fixed P c = 1.25 to 1.50 P P c = 1.00 c = 2.00 c = 4.00 FIGURE 3.9 Values of c, column-end ﬁxity, for determining the critical L/ ratio of dif- ferent loading conditions.ULTIMATE STRENGTH DESIGN CONCRETE COLUMNSAt ultimate strength Pu, kip (N), columns should be capable of sustaining loads asgiven by the American Concrete Institute required strength equations in Chap. 5,“Concrete Formulas” at actual eccentricities. Pu , may not exceed Pn , where isthe capacity reduction factor and Pn, kip (N), is the column ultimate strength. IfP0, kip (N), is the column ultimate strength with zero eccentricity of load, then P0 0.85 f c (Ag Ast) fy Ast (3.32)where fy yield strength of reinforcing steel, ksi (MPa) fc 28-day compressive strength of concrete, ksi (MPa) Ag gross area of column, in2 (mm2) Ast area of steel reinforcement, in2 (mm2)For members with spiral reinforcement then, for axial loads only, Pu 0.85 P0 (3.33)For members with tie reinforcement, for axial loads only, Pu 0.80 P0 (3.34) Eccentricities are measured from the plastic centroid. This is the centroid ofthe resistance to load computed for the assumptions that the concrete is stresseduniformly to 0.85 f c and the steel is stressed uniformly to fy.
• TABLE 3.5 Material Constants for Common Aluminum Alloys Average Values Fcy Fce Type Johnson Material psi MPa psi MPa K k n equation98 14S–T4 34,000 234.4 39,800 274.4 0.385 3.00 1.0 Straight line 24S–T3 and T4 40,000 275.8 48,000 330.9 0.385 3.00 1.0 Straight line 61S–T6 35,000 241.3 41,100 283.4 0.385 3.00 1.0 Straight line 14S–T6 57,000 393.0 61,300 422.7 0.250 2.00 2.0 Squared parabolic 75S–T6 69,000 475.8 74,200 511.6 0.250 2.00 2.0 Squared parabolic Ref: ANC-5.
• COLUMN FORMULAS 99 The axial-load capacity Pu kip (N), of short, rectangular members subject toaxial load and bending may be determined from Pu (0.85 f c ba A s fy As fs) (3.34) a Pue 0.85 fc ba d As fy (d d ) (3.35) 2where e eccentricity, in (mm), of axial load at end of member with respect to centroid of tensile reinforcement, calculated by conventional methods of frame analysis b width of compression face, in (mm) a depth of equivalent rectangular compressive-stress distribution, in (mm) As area of compressive reinforcement, in2 (mm2) As area of tension reinforcement, in2 (mm2) d distance from extreme compression surface to centroid of tensile reinforcement, in (mm) d distance from extreme compression surface to centroid of compres- sion reinforcement, in (mm) fs tensile stress in steel, ksi (MPa) The two preceding equations assume that a does not exceed the columndepth, that reinforcement is in one or two faces parallel to axis of bending, andthat reinforcement in any face is located at about the same distance from theaxis of bending. Whether the compression steel actually yields at ultimatestrength, as assumed in these and the following equations, can be veriﬁed bystrain compatibility calculations. That is, when the concrete crushes, the strainin the compression steel, 0.003 (c – d )/c, must be larger than the strain whenthe steel starts to yield, fy /Es. In this case, c is the distance, in (mm), from theextreme compression surface to the neutral axis and Es is the modulus of elastic-ity of the steel, ksi (MPa). The load, Pb for balanced conditions can be computed from the preceding Puequation with fs fy and a ab (3.36) 1cb 87,000 1d 87,000 fyThe balanced moment, in. kip (k . Nm), can be obtained from Mb Pbeb (3.37) ab 0.85 fc bab d d 2 As fy (d d d ) As fy d
• 100 CHAPTER THREEwhere eb is the eccentricity, in (mm), of the axial load with respect to the plasticcentroid and d is the distance, in (mm), from plastic centroid to centroid of ten-sion reinforcement. When Pu is less than Pb or the eccentricity, e, is greater than eb, tension gov-erns. In that case, for unequal tension and compression reinforcement, the ulti-mate strength is e Pu 0.85 fc bd m m 1 d B 2 e e d 1 2 ( m m) m 1 (3.38) d d dwhere m fy /0.85f c m m 1 As / bd As / bdSpecial Cases of ReinforcementFor symmetrical reinforcement in two faces, the preceding Pu equation becomes e Pu 0.85 fc bd 1 d B 2 e d e 1 2 m 1 (3.39) d d dColumn Strength When Compression GovernsFor no compression reinforcement, the Pu equation becomes e Pu 0.85 f cbd m 1 d B 2 e e m 1 2 (3.40) d d
• COLUMN FORMULAS 101 When Pu is greater than Pb, or e is less than eb, compression governs. In that case,the ultimate strength is approximately Mu Pu Po (Po Pb) (3.41) Mb Po Pu (3.42) 1 (Po /Pb 1)(e/eb)where Mu is the moment capacity under combined axial load and bending, inkip (kNm) and Po is the axial-load capacity, kip (N), of member when concen-trically loaded, as given. For symmetrical reinforcement in single layers, the ultimate strength whencompression governs in a column with depth, h, may be computed from As fy bhf c Pu (3.43) e/d d 0.5 3he/d 2 1.18Circular ColumnsUltimate strength of short, circular members with bars in a circle may be deter-mined from the following equations:When tension controls, B D 2 0.85e mDs 0.85e Pu 0.85 f c D2 0.38 1 0.38 2.5D D (3.44)where D overall diameter of section, in (mm) Ds diameter of circle through reinforcement, in (mm) t Ast /AgWhen compression governs, Ast fv Ag f c Pu (3.45) 3e/Ds 1 9.6De /(0.8D 0.67Ds)2 1.18The eccentricity for the balanced condition is given approximately by eb (0.24 0.39 t m)D (3.46)Short ColumnsUltimate strength of short, square members with depth, h, and with bars in acircle may be computed from the following equations:
• 102 CHAPTER THREEWhen tension controls, B h 2 e Ds e Pu 0.85bhf c 0.5 0.67 tm 0.5 (3.47) h hWhen compression governs, Ast fy Ag f c Pu (3.48) 3e/Ds 1 12he/(h 0.67Ds)2 1.18Slender ColumnsWhen the slenderness of a column has to be taken into account, the eccentricityshould be determined from e Mc /Pu, where Mc is the magniﬁed moment.DESIGN OF AXIALLY LOADED STEEL COLUMNS*Design of columns that are subjected to compression applied through the cen-troidal axis (axial compression) is based on the assumption of uniform stressover the gross area. This concept is applicable to both load and resistance factordesign (LRFD) and allowable stress design (ASD). Design of an axially loaded compression member or column for both LRFDand ASD utilizes the concept of effective column length KL. The buckling coef-ﬁcient K is the ratio of the effective column length to the unbraced length L.Values of K depend on the support conditions of the column to be designed.The AISC speciﬁcations for LRFD and ASD indicate the K should be taken asunity for columns in braced frames unless analysis indicates that a smallervalue is justiﬁed. Analysis is required for determination of K for unbracedframes, but K should not be less than unity. Design values for K recommendedby the Structural Stability Research Council for use with six idealized condi-tions of rotation and translation at column supports are illustrated in Fig. 9.1. The axially compression strength of a column depends on its stiffness meas-ured by the slenderness ratio KL/r, where r is the radius of gyration about theplane of buckling. For serviceability considerations, AISC recommends thatKL/r not exceed 200. LRFD strength for a compression member wf;Pn (kips) is given by Pn 0.85Ag Fcr (3.49)where LRFD resistance factor, less than unity Pn LFRD design strength (kips) of member (also called “maximum load” for columns, kips): *Brockenbrough and Merritt—Structural Steel Designer’s Handbook, McGraw-Hill.
• COLUMN FORMULAS 103with 0.85. For c 1.5 c2 Fcr 0.658 Fy (3.50)for c 1.5 0.877 Fcr Fy (3.51) l2 cwhere c (KL/r ) Fy /E Fy minimum speciﬁed yield stress of steel, ksi Ag gross area of member, in2 E elastic modulus of the steel 29,000 ksi For ASD, the allowable compression stress depends on whether bucklingwill be elastic or inelastic, as indicated by the slenderness ratio 22 E Fy 2 Cc (3.52)When KL/r < Cc, the allowable compression stress Fa (kips) on the gross sec-tion should be computed from 1 (KL r)2 2C2 c Fa 5 F (3.53) 3 3(KL r) 8Cc (KL r)3 8C3 y cWhen KL/r > Cc, the allowable compression stress is 12 2E Fa (3.54) 23(KL r)2Table of allowable loads for columns are contained in the AISC “Manual ofSteel Construction” for ASD and for LRFD.Columns Supporting Wind TurbinesWith increasing emphasis on renewable energy throughout the world, wind tur-bines are ﬁnding wider use. Today’s wind turbines are growing in generatingcapacity, with 5-mW the norm per unit, and 20-mW a near-time goal of turbinedesigners. As the electrical capacity of a wind turbine increases, so too does the directload of the nacelle on the supporting column and the wind loads on propellerblades. Both loads must be considered when designing the support column andthe foundation for the column.
• 104 CHAPTER THREE In the United States, land-based wind turbines (also called onshore turbines)have been the most popular type because there is sufﬁcient land area for, singleor multiple, wind turbine installations. In Europe, land scarcity led to offshorewind farms where the wind strength and dependability are also a direct beneﬁt. Designing columns for wind turbines involves two steps: (1) the columnfoundation for either land-based or sea-based wind turbines, and (2) the columnitself and the loads it must carry. Most land-based commercial wind turbines in the United States are sup-ported on a tubular steel column manufactured speciﬁcally for the site and theexpected wind velocities at the site. A concrete foundation for the column isgenerally used, depending on the soil conditions. In northern Europe, precastconcrete piles are popular for onshore wind-turbine bases, with the columnbeing tubular steel. Overturning moments are produced by the wind load on theturbine blades. Groundwater levels can be a consideration when designing thecolumn foundation. Dynamic loads also occur during wind gusts on the pro-peller blades. For sea-based commercial wind turbines, six types of support structures areavailable: (1) monopile driven into the sea bed; (2) gravity base which can be asteel or concrete caisson with suitable ballasting to resist the overturning momentcaused by the wind; (3) tripod with piles driven into the seabed; (4) suctionbucket in which an inverted type caisson is sunk into the sea bed using suction;(5) tension legs in which the vertical wind turbine column is supported by anunderwater ﬂoat anchored to the bottom by vertical anchors; and (6) ﬂoatingsupport—a concept still being worked on in various parts of the design world. Designers of sea-based wind turbine farms and individual units must beaware of the dangers to, and from, local shipping routes posed by the turbinestructure. Large European wind turbines have a total height of 650 ft (198 m),and diameter of 413 ft (126 m).
• PILES AND PILING FORMULAS 107of the pile head on y and M can be evaluated by substituting the value of Mtfrom the preceding equation into the earlier y and M equations. Note that, forthe fixed-head case, PtT 3 A By yf Ay (4.7) EI BTOE CAPACITY LOADFor piles installed in cohesive soils, the ultimate tip load may be computed from Qbu Ab q Ab Nc cu (4.8)where Ab end-bearing area of pile, ft2 (m2) q bearing capacity of soil, tons/ft2 (MPa) Nt bearing-capacity factor cu undrained shear strength of soil within zone 1 pile diameter above and 2 diameters below pile tip, psi (MPa)Although theoretical conditions suggest that Nc may vary between about 8 and12, Nc is usually taken as 9. For cohesionless soils, the toe resistance stress, q, is conventionallyexpressed by Eq. (4.1) in terms of a bearing-capacity factor Nq and the effectiveoverburden pressure at the pile tip vo q Nq vo ql (4.9) Some research indicates that, for piles in sands, q, like fs, reaches a quasi-constant value, ql , after penetrations of the bearing stratum in the range of 10 to20 pile diameters. Approximately ql 0.5Nq tan (4.10)where is the friction angle of the bearing soils below the critical depth. Val-ues of Nq applicable to piles are given in Fig. 4.1. Empirical correlations of soiltest data with q and ql have also been applied to predict successfully end-bearingcapacity of piles in sand.GROUPS OF PILESA pile group may consist of a cluster of piles or several piles in a row. Thegroup behavior is dictated by the group geometry and the direction and locationof the load, as well as by subsurface conditions. Ultimate-load considerations are usually expressed in terms of a groupefficiency factor, which is used to reduce the capacity of each pile in the
• 108 CHAPTER FOUR 1000 500 Bearing capacity factor, Nq 100 50 10 25° 30° 35° 40° 45° 50° Angle of internal friction, φ FIGURE 4.1 Bearing-capacity factor for granular soils related to angle of internal friction.group. The efficiency factor Eg is defined as the ratio of the ultimate groupcapacity to the sum of the ultimate capacity of each pile in the group. Eg is conventionally evaluated as the sum of the ultimate peripheral frictionresistance and end-bearing capacities of a block of soil with breadth B, widthW, and length L, approximately that of the pile group. For a given pile, spacingS and number of piles n, 2(BL WL) fs BWg Eg (4.11) nQuwhere fs is the average peripheral friction stress of block and Qu is the single-pilecapacity. The limited number of pile-group tests and model tests available suggestthat for cohesive soils, Eg 1 if S is more than 2.5 pile diameters D and for cohe-sionless soils, Eg 1 for the smallest practical spacing. A possible exceptionmight be for very short, heavily tapered piles driven in very loose sands. In practice, the minimum pile spacing for conventional piles is in the rangeof 2.5 to 3.0D. A larger spacing is typically applied for expanded-base piles. A very approximate method of pile-group analysis calculates the upper limitof group drag load, Qgd from Qgd AF FHF PHcu (4.12)where Hf , f , and AF represent the thickness, unit weight, and area of fill con-tained within the group. P, H, and cu are the circumference of the group, the
• PILES AND PILING FORMULAS 109thickness of the consolidating soil layers penetrated by the piles, and theirundrained shear strength, respectively. Such forces as Qgd could only beapproached for the case of piles driven to rock through heavily surcharged,highly compressible subsoils. Design of rock sockets is conventionally based on Qd ds Ls fR ds2qa (4.13) 4where Qd allowable design load on rock socket, psi (MPa) ds socket diameter, ft (m) Ls socket length, ft (m) fR allowable concrete-rock bond stress, psi (MPa) qa allowable bearing pressure on rock, tons/ft2 (MPa)Load-distribution measurements show, however, that much less of the loadgoes to the base than is indicated by Eq. (4.6). This behavior is demonstratedby the data in Table 4.1, where Ls /ds is the ratio of the shaft length to shaftdiameter and Er /Ep is the ratio of rock modulus to shaft modulus. The finite-element solution summarized in Table 4.1 probably reflects a realistic trend ifthe average socket-wall shearing resistance does not exceed the ultimate fRvalue; that is, slip along the socket side-wall does not occur. A simplified design approach, taking into account approximately the compati-bility of the socket and base resistance, is applied as follows:1. Proportion the rock socket for design load Qd with Eq. (4.6) on the assump- tion that the end-bearing stress is less than qa [say qa /4, which is equivalent to assuming that the base load Qb ( /4) d2qa /4]. s2. Calculate Qb RQd, where R is the base-load ratio interpreted from Table 4.1.3. If RQd does not equal the assumed Qb , repeat the procedure with a new qa value until an approximate convergence is achieved and q qa.The final design should be checked against the established settlement toleranceof the drilled shaft. Following the recommendations of Rosenberg and Journeaux, a more realis-tic solution by the previous method is obtained if fRu is substituted for fR. Ideally,fRu should be determined from load tests. If this parameter is selected from datathat are not site specific, a safety factor of at least 1.5 should be applied to fRu inrecognition of the uncertainties associated with the UC strength correlations.*FOUNDATION-STABILITY ANALYSISThe maximum load that can be sustained by shallow foundation elements atincipient failure (bearing capacity) is a function of the cohesion and frictionangle of bearing soils as well as the width B and shape of the foundation. *Rosenberg, P. and Journeaux, N. L., “Friction and End-Bearing Tests on Bedrock for High-Capacity Socket Design,” Canadian Geotechnical Journal, 13(3).
• 110 CHAPTER FOURThe net bearing capacity per unit area, qu, of a long footing is convention-ally expressed as qu f cu Nc voNq f BN (4.14)where f 1.0 for strip footings and 1.3 for circular and square footings cu undrained shear strength of soil vo effective vertical shear stress in soil at level of bottom of footing f 0.5 for strip footings, 0.4 for square footings, and 0.6 for circular footings unit weight of soil B width of footing for square and rectangular footings and radius of footing for circular footings Nc, Nq, N bearing-capacity factors, functions of angle of internal friction For undrained (rapid) loading of cohesive soils, 0 and Eq. (4.7) reduces to qu N c cu (4.15)where Nc f Nc. For drained (slow) loading of cohesive soils, and cu aredefined in terms of effective friction angle and effective stress cu. Modifications of Eq. (4.7) are also available to predict the bearing capacity oflayered soil and for eccentric loading. Rarely, however, does qu control foundation design when the safety factor iswithin the range of 2.5 to 3. (Should creep or local yield be induced, excessivesettlements may occur. This consideration is particularly important when select-ing a safety factor for foundations on soft to firm clays with medium to highplasticity.) Equation (4.7) is based on an infinitely long strip footing and should becorrected for other shapes. Correction factors by which the bearing-capacityfactors should be multiplied are given in Table 4.2, in which L footinglength. The derivation of Eq. (4.7) presumes the soils to be homogeneous throughoutthe stressed zone, which is seldom the case. Consequently, adjustments may berequired for departures from homogeneity. In sands, if there is a moderate varia-tion in strength, it is safe to use Eq. (4.7), but with bearing-capacity factors repre-senting a weighted average strength. Eccentric loading can have a significant impact on selection of the bearingvalue for foundation design. The conventional approach is to proportion thefoundation to maintain the resultant force within its middle third. The footing isassumed to be rigid and the bearing pressure is assumed to vary linearly asshown by Fig. 4.2(b). If the resultant lies outside the middle third of the foot-ing, it is assumed that there is bearing over only a portion of the footing, asshown in Fig. 4.2(d). For the conventional case, the maximum and minimumbearing pressures are P 6e qm 1 (4.16) BL B
• PILES AND PILING FORMULAS 111 TABLE 4.2 Shape Corrections for Bearing-Capacity Factors of Shallow Foundations* Correction factor Shape of foundation Nc Nq Ny B Nq B B Rectangle† 1 1 tan 1 0.4 L Nc L L Circle and Nq 1 tan 0.60 1 square Nc *After De Beer, E. E., as modified by Vesic, A. S. See Fang, H. Y., Foundation Engineering Handbook, 2d ed., Van Nostrand Reinhold, New York. † No correction factor is needed for long-strip foundations.where B width of rectangular footing L length of rectangular footing e eccentricity of loadingFor the other case Fig. 4.2(c), the soil pressure ranges from 0 to a maximum of 2P (4.17) qm 3L(B/2 e) b b P M P M (a) P (c) e P e 3( b – e) 2 p2 P1 p (b) (d) FIGURE 4.2 Footings subjected to overturning.
• PILES AND PILING FORMULAS 113These include elastic, semiempirical elastic, and load-transfer solutions forsingle shafts drilled in cohesive or cohesionless soils. Resistance to tensile and lateral loads by straight-shaft drilled shafts shouldbe evaluated as described for pile foundations. For relatively rigid shafts withcharacteristic length T greater than 3, there is evidence that bells increase thelateral resistance. The added ultimate resistance to uplift of a belled shaft Qutcan be approximately evaluated for cohesive soils models for bearing capacity[Eq. (4.14)] and friction cylinder [Eq. (4.15)] as a function of the shaft diameterD and bell diameter Db.* For the bearing-capacity solution, Qul (D2 b D2)Nc cu Wp (4.21) 4The shear-strength reduction factor in Eq. (4.14) considers disturbanceeffects and ranges from 1 2 (slurry construction) to 3 4 (dry construction). The curepresents the undrained shear strength of the soil just above the bell surface,and Nc is a bearing-capacity factor. The failure surface of the friction cylinder model is conservatively assumedto be vertical, starting from the base of the bell. Qut can then be determined forboth cohesive and cohesionless soils from Qul b L fut Ws Wp (4.22)where fut is the average ultimate skin-friction stress in tension developed on thefailure plane; that is, fut 0.8cu for clays or K vo tan for sands. Ws and Wprepresent the weight of soil contained within the failure plane and the shaftweight, respectively.SHAFT RESISTANCE IN COHESIONLESS SOILSThe shaft resistance stress fs is a function of the soil-shaft friction angle ,degree, and an empirical lateral earth-pressure coefficient K: fs K vo tan fl (4.23) At displacement-pile penetrations of 10 to 20 pile diameters (loose to densesand), the average skin friction reaches a limiting value fl. Primarily dependingon the relative density and texture of the soil, fl has been approximated conser-vatively by using Eq. (4.16) to calculate fs. For relatively long piles in sand, K is typically taken in the range of 0.7 to1.0 and is taken to be about 5, where is the angle of internal friction, *Meyerhof, G. G. and Adams, J. I., “The Ultimate Uplift Capacity of Foundations,” CanadianGeotechnical Journal, 5(4):1968.
• 114 CHAPTER FOURdegree. For piles less than 50 ft (15.2 m) long, K is more likely to be in therange of 1.0 to 2.0, but can be greater than 3.0 for tapered piles. Empirical procedures have also been used to evaluate fs from in situ tests,such as cone penetration, standard penetration, and relative density tests. Equa-tion (4.17), based on standard penetration tests, as proposed by Meyerhof, isgenerally conservative and has the advantage of simplicity: N fs (4.24) 50where N average standard penetration resistance within the embedded lengthof pile and fs is given in tons/ft2.* *Meyerhof, G. G., “Bearing Capacity and Settlement of Pile Foundations,” ASCE Journal ofGeotechnical Engineering Division, 102(GT3):1976.
• CHAPTER 5 CONCRETE FORMULASREINFORCED CONCRETEWhen working with reinforced concrete and when designing reinforced con-crete structures, the American Concrete Institute (ACI) Building Code Require-ments for Reinforced Concrete, latest edition, is widely used. Future referencesto this document are denoted as the ACI Code. Likewise, publications of thePortland Cement Association (PCA) find extensive use in design and construc-tion of reinforced concrete structures. Formulas in this chapter cover the general principles of reinforced concreteand its use in various structural applications. Where code requirements have tobe met, the reader must refer to the current edition of the ACI Code previouslymentioned. Likewise, the PCA publications should also be referred to for the lat-est requirements and recommendations.WATER/CEMENTITIOUS MATERIALS RATIOThe water/cementitious (w/c) ratio is used in both tensile and compressivestrength analyses of Portland concrete cement. This ratio is found from w wm (5.1) c wcwhere wm weight of mixing water in batch, lb (kg); and wc weight ofcementitious materials in batch, lb (kg). The ACI Code lists the typical relationship between the w/c ratio by weightand the compressive strength of concrete. Ratios for non-air-entrained concretevary between 0.41 for a 28-day compressive strength of 6000 lb/in2 (41 MPa)and 0.82 for 2000 lb/in2 (14 MPa). Air-entrained concrete w/c ratios vary from0.40 to 0.74 for 5000 lb/in2 (34 MPa) and 2000 lb/in2 (14 MPa) compressivestrength, respectively. Be certain to refer to the ACI Code for the appropriatew/c value when preparing designs or concrete analyses. Further, the ACI Code also lists maximum w/c ratios when strength dataare not available. Absolute w/c ratios by weight vary from 0.67 to 0.38 for 115
• 116 CHAPTER FIVEnon-air-entrained concrete and from 0.54 to 0.35 for air-entrained concrete.These values are for a specified 28-day compressive strength fc in lb/in2 orMPa, of 2500 lb/in2 (17 MPa) to 5000 lb/in2 (34 MPa). Again, refer to the ACICode before making any design or construction decisions. Maximum w/c ratios for a variety of construction conditions are also listed inthe ACI Code. Construction conditions include concrete protected from exposureto freezing and thawing; concrete intended to be watertight; and concrete exposedto deicing salts, brackish water, seawater, etc. Application formulas for w/c ratiosare given later in this chapter.JOB MIX CONCRETE VOLUMEA trial batch of concrete can be tested to determine how much concrete is to bedelivered by the job mix. To determine the volume obtained for the job, add theabsolute volume Va of the four components—cements, gravel, sand, and water. Find the Va for each component from WL Va (5.2) (SG)Wuwhere Va absolute volume, ft3 (m3) WL weight of material, lb (kg) SG specific gravity of the material wu density of water at atmospheric conditions (62.4 lb/ft3; 1000 kg/m3)Then, job yield equals the sum of Va for cement, gravel, sand, and water.MODULUS OF ELASTICITY OF CONCRETEThe modulus of elasticity of concrete Ec—adopted in modified form by the ACICode—is given by 33w1.5 2f c lb/in2 in USCS units 0.043w1.5 2fc MPa in SI units Ec c (5.3) cWith normal-weight, normal-density concrete these two relations can be simpli-fied to 57,000 2fc lb/in2 in USCS units 4700 2fc MPa in SI units Ec (5.4)where Ec modulus of elasticity of concrete, lb/in2 (MPa); and fc specified28-day compressive strength of concrete, lb/in2 (MPa).
• CONCRETE FORMULAS 117TENSILE STRENGTH OF CONCRETEThe tensile strength of concrete is used in combined-stress design. In normal-weight, normal-density concrete the tensile strength can be found from 7.5 2fc 0.7 2fc fr lb/in2 in USCS units (5.5) fr MPa in SI unitsREINFORCING STEELAmerican Society for Testing and Materials (ASTM) specifications cover ren-forcing steel. The most important properties of reinforcing steel are1. Modulus of elasticity Es, lb/in2 (MPa)2. Tensile strength, lb/in2 (MPa)3. Yield point stress fy, lb/in2 (MPa)4. Steel grade designation (yield strength)5. Size or diameter of the bar or wireCONTINUOUS BEAMS AND ONE-WAY SLABSThe ACI Code gives approximate formulas for finding shear and bendingmoments in continuous beams and one-way slabs. A summary list of these for-mulas follows. They are equally applicable to USCS and SI units. Refer to theACI Code for specific applications of these formulas.For Positive Moment End spans If discontinuous end is unrestrained 2 wln 11 If discontinuous end is integral with the support 2 wln 14 Interior spans 2 wln 16For Negative Moment Negative moment at exterior face of first interior support Two spans 2 wln 9 More than two spans 2 wln 10 Negative moment at other faces of interior supports 2 wln 11
• 118 CHAPTER FIVE Negative moment at face of all supports for (a) slabs with spans not exceeding 10 ft (3 m) and (b) beams and girders where the ratio of sum of column stiffness to beam stiffness exceeds 8 at each end of the span 2 wln 12 Negative moment at interior faces of exterior supports, for members built integrally with their supports Where the support is a spandrel beam or girder 2 wln 24 Where the support is a column 2 wln 16Shear Forces Shear in end members at first interior support 1.15 wln 2 Shear at all other supports wln 2End ReactionsReactions to a supporting beam, column, or wall are obtained as the sum of shearforces acting on both sides of the support.DESIGN METHODS FOR BEAMS,COLUMNS, AND OTHER MEMBERSA number of different design methods have been used for reinforced concrete con-struction. The three most common are working-stress design, ultimate-strengthdesign, and strength design method. Each method has its backers and supporters.For actual designs the latest edition of the ACI Code should be consulted.BeamsConcrete beams may be considered to be of three principal types: (1) rectangu-lar beams with tensile reinforcing only, (2) T-beams with tensile reinforcingonly, and (3) beams with tensile and compressive reinforcing.Rectangular Beams with Tensile Reinforcing Only This type of beam includesslabs, for which the beam width b equals 12 in (305 mm) when the moment andshear are expressed per foot (m) of width. The stresses in the concrete and steelare, using working-stress design formulas, 2M M M fc fs (5.6) kjbd 2 As jd pjbd 2
• CONCRETE FORMULAS 119 Cross section of beam Stress diagram fc 1 C= f 2 c kbd kd d jd As T = As fs = fs pbd fs /n b M = 1 fc kjbd 2 = fs pjbd 2 2 FIGURE 5.1 Rectangular concrete beam with tensile reinforcing only.where b width of beam [equals 12 in (304.8 mm) for slab], in (mm) d effective depth of beam, measured from compressive face of beam to centroid of tensile reinforcing (Fig. 5.1), in (mm) M bending moment, lb . in (k . Nm) fc compressive stress in extreme fiber of concrete, lb/in2 (MPa) fs stress in reinforcement, lb/in2 (MPa) As cross-sectional area of tensile reinforcing, in2 (mm2) j ratio of distance between centroid of compression and centroid of tension to depth d k ratio of depth of compression area to depth d p ratio of cross-sectional area of tensile reinforcing to area of the beam ( As /bd) For approximate design purposes, j may be assumed to be 7 8 and k, 1 3. Foraverage structures, the guides in Table 5.1 to the depth d of a reinforced concretebeam may be used. TABLE 5.1 Guides to Depth d of Reinforced Concrete Beam* Member d Roof and floor slabs l/25 Light beams l/15 Heavy beams and girders l/12–l/10 *l is the span of the beam or slab in inches (millimeters). The width of a beam should be at least l/32.
• 120 CHAPTER FIVE TABLE 5.2 Coefficients K, k, j, and p for Rectangular Sections* fs n fs K k j p 2000 15 900 175 0.458 0.847 0.0129 2500 12 1125 218 0.458 0.847 0.0161 3000 10 1350 262 0.458 0.847 0.0193 3750 8 1700 331 0.460 0.847 0.0244 *fs 16,000 lb/in2 (110 MPa). For a balanced design, one in which both the concrete and the steel arestressed to the maximum allowable stress, the following formulas may be used: M 1 bd 2 K f kj pfs j (5.7) K 2 eValues of K, k, j, and p for commonly used stresses are given in Table 5.2.T-Beams with Tensile Reinforcing Only When a concrete slab is constructedmonolithically with the supporting concrete beams, a portion of the slab acts as theupper flange of the beam. The effective flange width should not exceed (1) one-fourth the span of the beam, (2) the width of the web portion of the beam plus 16times the thickness of the slab, or (3) the center-to-center distance between beams.T-beams where the upper flange is not a portion of a slab should have a flangethickness not less than one-half the width of the web and a flange width not morethan four times the width of the web. For preliminary designs, the precedingformulas given for rectangular beams with tensile reinforcing only can be used,because the neutral axis is usually in, or near, the flange. The area of tensilereinforcing is usually critical.Beams with Tensile and Compressive Reinforcing Beams with compressivereinforcing are generally used when the size of the beam is limited. Theallowable beam dimensions are used in the formulas given earlier to determinethe moment that could be carried by a beam without compressive reinforcement.The reinforcing requirements may then be approximately determined from 8M M M As Asc (5.8) 7fs d nfc dwhere As total cross-sectional area of tensile reinforcing, in2 (mm2) Asc cross-sectional area of compressive reinforcing, in2 (mm2) M total bending moment, lb in (K Nm) M bending moment that would be carried by beam of balanced design and same dimensions with tensile reinforcing only, lb in (K Nm) n ratio of modulus of elasticity of steel to that of concrete
• CONCRETE FORMULAS 121Checking Stresses in Beams Beams designed using the preceding approximateformulas should be checked to ensure that the actual stresses do not exceed theallowable, and that the reinforcing is not excessive. This can be accomplished bydetermining the moment of inertia of the beam. In this determination, theconcrete below the neutral axis should not be considered as stressed, whereas thereinforcing steel should be transformed into an equivalent concrete section. Fortensile reinforcing, this transformation is made by multiplying the area As by n,the ratio of the modulus of elasticity of steel to that of concrete. For compressivereinforcing, the area Asc is multiplied by 2(n – 1). This factor includes allowancesfor the concrete in compression replaced by the compressive reinforcing and forthe plastic flow of concrete. The neutral axis is then located by solving 1 2 2 bcc 2(n 1)Asccsc nAscs (5.9)for the unknowns cc, csc, and cs (Fig. 5.2). The moment of inertia of the trans-formed beam section is 1 3 I 3bcc 2(n 1)Ascc2 sc nAs c2 s (5.10)and the stresses are Mcc 2nMcsc nMcs fc fsc fs (5.11) I I I 2(n – 1) Asc Concrete in Cc Csc compression Neutral axis d Cs nAs b FIGURE 5.2 Transformed section of concrete beam.
• 122 CHAPTER FIVEwhere fc, fsc, fs actual unit stresses in extreme fiber of concrete, in compressive reinforcing steel, and in tensile reinforcing steel, respectively, lb/in2 (MPa) cc, csc, cs distances from neutral axis to face of concrete, to compressive reinforcing steel, and to tensile reinforcing steel, respectively, in (mm) I moment of inertia of transformed beam section, in4 (mm4) b beam width, in (mm)and As, Asc, M, and n are as defined earlier in this chapter.Shear and Diagonal Tension in Beams The shearing unit stress, as a measureof diagonal tension, in a reinforced concrete beam is V v (5.12) bdwhere v shearing unit stress, lb/in2 (MPa) V total shear, lb (N) b width of beam (for T-beam use width of stem), in (mm) d effective depth of beam If the value of the shearing stress as computed earlier exceeds the allowableshearing unit stress as specified by the ACI Code, web reinforcement should beprovided. Such reinforcement usually consists of stirrups. The cross-sectional arearequired for a stirrup placed perpendicular to the longitudinal reinforcement is (V V )s Av (5.13) fi dwhere Av cross-sectional area of web reinforcement in distance s (measured parallel to longitudinal reinforcement), in2 (mm2) fv allowable unit stress in web reinforcement, lb/in2 (MPa) V total shear, lb (N) V shear that concrete alone could carry ( vc bd), lb (N) s spacing of stirrups in direction parallel to that of longitudinal rein- forcing, in (mm) d effective depth, in (mm) Stirrups should be so spaced that every 45° line extending from the mid-depth of the beam to the longitudinal tension bars is crossed by at least one stir-rup. If the total shearing unit stress is in excess of 3 f c lb/in2 (MPa), everysuch line should be crossed by at least two stirrups. The shear stress at any sectionshould not exceed 5 f c lb/in2 (MPa).Bond and Anchorage for Reinforcing Bars In beams in which the tensilereinforcing is parallel to the compression face, the bond stress on the bars is V u (5.14) jd 0
• CONCRETE FORMULAS 123where u bond stress on surface of bar, lb/in2 (MPa) V total shear, lb (N) d effective depth of beam, in (mm) 0 sum of perimeters of tensile reinforcing bars, in (mm)For preliminary design, the ratio j may be assumed to be 7/8. Bond stressesmay not exceed the values shown in Table 5.3.ColumnsThe principal columns in a structure should have a minimum diameter of 10 in(255 mm) or, for rectangular columns, a minimum thickness of 8 in(203 mm) anda minimum gross cross-sectional area of 96 in2 (61,935 mm2). Short columns with closely spaced spiral reinforcing enclosing a circularconcrete core reinforced with vertical bars have a maximum allowable load of P Ag(0.25fc fs pg) (5.15)where P total allowable axial load, lb (N) Ag gross cross-sectional area of column, in2 (mm2) fc compressive strength of concrete, lb/in2 (MPa) fs allowable stress in vertical concrete reinforcing, lb/in2 (MPa), equal to 40 percent of the minimum yield strength, but not to exceed 30,000 lb/ in2 (207 MPa) pg ratio of cross-sectional area of vertical reinforcing steel to gross area of column AgThe ratio pg should not be less than 0.01 or more than 0.08. The minimum num-ber of bars to be used is six, and the minimum size is No. 5. The spiral reinforc-ing to be used in a spirally reinforced column is Ag fc ps 0.45 1 (5.16) Ac fywhere ps ratio of spiral volume to concrete-core volume (out-to-out spiral) Ac cross-sectional area of column core (out-to-out spiral), in2 (mm2) fy yield strength of spiral reinforcement, lb/in2 (MPa), but not to exceed 60,000 lb/in2 (413 MPa)The center-to-center spacing of the spirals should not exceed one-sixth of thecore diameter. The clear spacing between spirals should not exceed one-sixththe core diameter, or 3 in (76 mm), and should not be less than 1.375 in (35 mm),or 1.5 times the maximum size of coarse aggregate used.Short Columns with Ties The maximum allowable load on short columnsreinforced with longitudinal bars and separate lateral ties is 85 percent of that given
• TABLE 5.3 Allowable Bond Stresses* Horizontal bars with more than 12 in (30.5 mm) of concrete cast below the bar† Other bars† 3.4 2fc 4.8 2fc Tension bars with sizes and or 350, whichever is less or 500, whichever is less deformations conforming to D D 2.1 2fc 3 2fc ASTM A305124 Tension bars with sizes and deformations conforming to 6.5 2fc or 400, whichever is less 6.5 2fc or 400, whichever is less ASTM A408 Deformed compression bars Plain bars 1.7 2fc or 160, whichever is less 2.4 2fc or 160, whichever is less 2 * lb/in ( 0.006895 MPa). † fc compressive strength of concrete, lb/in2 (MPa); D nominal diameter of bar, in (mm).
• 126 CHAPTER FIVE t diameter of column or overall depth of column, in (mm) d distance from extreme compression fiber to centroid of tension rein- forcement, in (mm) fv yield point of reinforcement, lb/in2 (MPa) Design of columns controlled by compression is based on the followingequation, except that the allowable load N may not exceed the allowable load P,given earlier, permitted when the column supports axial load only: fa fbx fby 1.0 (5.21) Fa Fb Fbwhere fa axial load divided by gross concrete area, lb/in2 (MPa) fbx, fby bending moment about x and y axes, divided by section modulus of corresponding transformed uncracked section, lb/in2 (MPa) Fb allowable bending stress permitted for bending alone, lb/in2 (MPa) Fa 0.34(1 pgm) fc The allowable bending load on columns controlled by tension varies linearlywith the axial load from M0 when the section is in pure bending to Mb when theaxial load is Nb. For spiral columns, M0 0.12Ast fyDs (5.22) For tied columns, M0 0.40As fy(d d) (5.23)where Ast total area of longitudinal reinforcement, in2 (mm2) fv yield strength of reinforcement, lb/in2 (MPa) Ds diameter of circle through centers of longitudinal reinforcement, in (mm) As area of tension reinforcement, in2 (mm2) d distance from extreme compression fiber to centroid of tension rein- forcement, in (mm)Nb and Mb are the axial load and moment at the balanced condition (i.e., whenthe eccentricity e equals eb as determined). At this condition, Nb and Mb shouldbe determined from Mb Nbeb (5.24)When bending is about two axes, Mx My 1 (5.25) M0x M0y
• CONCRETE FORMULAS 127where Mz and My are bending moments about the x and y axes, and M0x and M0yare the values of M0 for bending about these axes.PROPERTIES IN THE HARDENED STATEStrength is a property of concrete that nearly always is of concern. Usually, it isdetermined by the ultimate strength of a specimen in compression, but sometimesflexural or tensile capacity is the criterion. Because concrete usually gainsstrength over a long period of time, the compressive strength at 28 days is com-monly used as a measure of this property. The 28-day compressive strength of concrete can be estimated from the7-day strength by a formula proposed by W. A. Slater: S28 S7 30 2S7 (5.26)where S28 28-day compressive strength, lb/in2 (MPa); and S7 7-day strength,lb/in2 (MPa). Concrete may increase significantly in strength after 28 days, particularly whencement is mixed with fly ash. Therefore, specification of strengths at 56 or 90 daysis appropriate in design. Concrete strength is influenced chiefly by the water/cement ratio; the higher thisratio is, the lower the strength. The relationship is approximately linear whenexpressed in terms of the variable C/W, the ratio of cement to water by weight. Fora workable mix, without the use of water reducing admixtures, C S28 2700 760 (5.27) W Tensile strength of concrete is much lower than compressive strength and,regardless of the types of test, usually has poor correlation with fc . As deter-mined in flexural tests, the tensile strength (modulus of rupture—not the truestrength) is about 7 fc for the higher strength concretes and 10 fc for thelower strength concretes. Modulus of elasticity Ec , generally used in design for concrete, is a secantmodulus. In ACI 318, “Building Code Requirements for Reinforced Concrete,”it is determined by Ec w1.533 2fc (5.28)where w weight of concrete, lb/ft3 (kg/m3); and fc specified compressivestrength at 28 days, lb/in2 (MPa). For normal-weight concrete, with w 145 lb/ft3 (kg/m3), Ec 57,000 2fc (5.29)The modulus increases with age, as does the strength.
• 128 CHAPTER FIVETENSION DEVELOPMENT LENGTHSFor bars and deformed wire in tension, basic development length is defined bythe equations that follow. For No. 11 and smaller bars, 0.04Ab fy ld (5.30) fcwhere Ab area of bar, in2 (mm2) fy yield strength of bar steel, lb/in2 (MPa) fc 28-day compressive strength of concrete, lb/in2 (MPa)However, ld should not be less than 12 in (304.8 mm), except in computation oflap splices or web anchorage. For No. 14 bars, fy ld 0.085 (5.31) fc For No. 18 bars, fy ld 0.125 (5.32) fcand for deformed wire, fy 20,000 Aw fy ld 0.03db 0.02 (5.33) fc Sw fcwhere Aw is the area, in2 (mm2); and Sw is the spacing, in (mm), of the wire to bedeveloped. Except in computation of lap splices or development of web rein-forcement, ld should not be less than 12 in (304.8 mm).COMPRESSION DEVELOPMENT LENGTHSFor bars in compression, the basic development length ld is defined as 0.02 fy db ld 0.0003db fy (5.34) fcbut ld not be less than 8 in (20.3 cm) or 0.0003fy db.CRACK CONTROL OF FLEXURAL MEMBERSBecause of the risk of large cracks opening up when reinforcement is subjected tohigh stresses, the ACI Code recommends that designs be based on a steel yieldstrength fy no larger than 80 ksi (551.6 MPa). When design is based on a yieldstrength fy greater than 40 ksi (275.8 MPa), the cross sections of maximum
• 130 CHAPTER FIVEDEFLECTION COMPUTATIONS AND CRITERIA FORCONCRETE BEAMSThe assumptions of working-stress theory may also be used for computingdeflections under service loads; that is, elastic-theory deflection formulas may beused for reinforced-concrete beams. In these formulas, the effective moment ofinertia Ic is given by 3 3 Mcr Mcr Ic Ig 1 Icr Ig (5.40) Ma Mawhere Ig moment of inertia of the gross concrete section, in4 (mm4) Mcr cracking moment, lb.in (K.Nm) Ma moment for which deflection is being computed, lb.in (K.Nm) Icr cracked concrete (transformed) section, in4 (mm4) If yt is taken as the distance from the centroidal axis of the gross section, neglect-ing the reinforcement, to the extreme surface in tension, the cracking moment maybe computed from fr Ig Mcr (5.41) ytwith the modulus of rupture of the concrete fr 7.5 2fc . The deflections thus calculated are those assumed to occur immediately onapplication of load. Additional long-time deflections can be estimated by multiply-ing the immediate deflection by 2 when there is no compression reinforcement orby 2 1.2As /As 0.6, where As is the area of compression reinforcement and As isthe area of tension reinforcement.ULTIMATE-STRENGTH DESIGN OF RECTANGULAR BEAMSWITH TENSION REINFORCEMENT ONLYGenerally, the area As of tension reinforcement in a reinforced-concrete beam isrepresented by the ratio As /bd, where b is the beam width and d is the dis-tance from extreme compression surface to the centroid of tension reinforce-ment. At ultimate strength, the steel at a critical section of the beam is at itsyield strength fy if the concrete does not fail in compression first. Total tensionin the steel then will be As f y fy bd. It is opposed, by an equal compressiveforce: 0.85 fc ba 0.85 fc b 1c (5.42)where fc 28-day strength of the concrete, ksi (MPa) a depth of the equivalent rectangular stress distribution c distance from the extreme compression surface to the neutral axis 1 a constant
• CONCRETE FORMULAS 131Equating the compression and tension at the critical section yields pfy c d (5.43) 0.85 1 fc The criterion for compression failure is that the maximum strain in the con-crete equals 0.003 in/in (0.076 mm/mm). In that case, 0.003 c d (5.44) fs /Es 0.003where fs steel stress, ksi (MPa) Es modulus of elasticity of steel 29,000 ksi (199.9 GPa)Balanced ReinforcingUnder balanced conditions, the concrete reaches its maximum strain of 0.003when the steel reaches its yield strength fy. This determines the steel ratio forbalanced conditions: 0.85 1 fc 87,000 b (5.45) fy 87,000 fyMoment CapacityFor such underreinforced beams, the bending-moment capacity of ultimatestrength is Mu 0.90[bd 2 fc (1 0.59 )] (5.46) a 0.90 As fy d (5.47) 2where fy /fc and a As fy /0.85fc .Shear ReinforcementThe ultimate shear capacity Vn of a section of a beam equals the sum of thenominal shear strength of the concrete Vc and the nominal shear strength pro-vided by the reinforcement Vs; that is, Vn Vc Vs. The factored shear force Vuon a section should not exceed Vn (Vc Vs) (5.48)where capacity reduction factor (0.85 for shear and torsion). Except forbrackets and other short cantilevers, the section for maximum shear may betaken at a distance equal to d from the face of the support.
• 132 CHAPTER FIVE The shear strength Vc carried by the concrete alone should not exceed2 fc bw d, where bw is the width of the beam web and d, the depth of the centroidof reinforcement. (As an alternative, the maximum value for Vc may be taken as Vud Vc 1.9 fc 2500 w b d 3.5 fc bwd (5.49) Mu wwhere w As/bwd and Vu and Mu are the shear and bending moment, respec-tively, at the section considered, but Mu should not be less than Vud.) When Vu is larger than Vc, the excess shear has to be resisted by web rein-forcement. The area of steel required in vertical stirrups, in2 (mm2), per stirrup, with aspacing s, in (mm), is Vs S Av (5.50) fy dwhere fy is the yield strength of the shear reinforcement and Av is the area ofthe stirrups cut by a horizontal plane. Vs should not exceed 8 fc bw d in sec-tions with web reinforcement, and fy should not exceed 60 ksi (413.7 MPa).Where shear reinforcement is required and is placed perpendicular to the axisof the member, it should not be spaced farther apart than 0.5d, or more than24 in (609.6 mm) c to c. When Vs exceeds 4 fc bw d , however, the maximumspacing should be limited to 0.25d. Alternatively, for practical design, to indicate the stirrup spacing s for thedesign shear Vu, stirrup area Av, and geometry of the member bw and d, Av fyd s (5.51) Vu 2 fc bw d The area required when a single bar or a single group of parallel bars are allbent up at the same distance from the support at angle with the longitudinalaxis of the member is Vs Av (5.52) fy sinin which Vs should not exceed 3 fc bw d . Av is the area cut by a plane normal tothe axis of the bars. The area required when a series of such bars are bent up atdifferent distances from the support or when inclined stirrups are used is Vs s Av (5.53) (sin cos )fy d A minimum area of shear reinforcement is required in all members, exceptslabs, footings, and joists or where Vu is less than 0.5Vc.
• CONCRETE FORMULAS 133Development of Tensile ReinforcementAt least one-third of the positive-moment reinforcement in simple beams and one-fourth of the positive-moment reinforcement in continuous beams should extendalong the same face of the member into the support, in both cases, at least 6 in(152.4 mm) into the support. At simple supports and at points of inflection, thediameter of the reinforcement should be limited to a diameter such that the develop-ment length ld satisfies Mn ld la (5.54) Vuwhere Mn computed flexural strength with all reinforcing steel at section stressed to fy Vu applied shear at section la additional embedment length beyond inflection point or center of supportAt an inflection point, la is limited to a maximum of d, the depth of the centroidof the reinforcement, or 12 times the reinforcement diameter.Hooks on BarsThe basic development length for a hooked bar with fy 60 ksi (413.7 MPa) isdefined as 1200db lhb (5.55) fcwhere db is the bar diameter, in (mm), and fc is the 28-day compressivestrength of the concrete, lb/in2 (MPa).WORKING-STRESS DESIGN OF RECTANGULAR BEAMSWITH TENSION REINFORCEMENT ONLYFrom the assumption that stress varies across a beam section with the distancefrom the neutral axis, it follows that n fc k (5.56) fs 1 kwhere n modular ratio Es /Ec Es modulus of elasticity of steel reinforcement, ksi (MPa) Ec modulus of elasticity of concrete, ksi (MPa) fc compressive stress in extreme surface of concrete, ksi (MPa) fs stress in steel, ksi (MPa)
• 134 CHAPTER FIVE kd distance from extreme compression surface to neutral axis, in (mm) d distance from extreme compression to centroid of reinforcement, in (mm) When the steel ratio As /bd, where As area of tension reinforcement,in2 (mm2); and b beam width, in (mm), is known, k can be computed from k 22n (n )2 n (5.57)Wherever positive-moment steel is required, should be at least 200/fy, wherefy is the steel yield stress. The distance jd between the centroid of compressionand the centroid of tension, in (mm), can be obtained from k j 1 (5.58) 3Allowable Bending MomentThe moment resistance of the concrete, in kip (k Nm) is 1 2 Mc 2 fckjbd Kcbd 2 (5.59) 1where Kc 2 fc kj. The moment resistance of the steel is Ms fs As jd fs jbd 2 Ksbd 2 (5.60)where Ks fs j.Allowable ShearThe nominal unit shear stress, v acting on a section with shear V is V v (5.61) bdAllowable shear stresses are 55 percent of those for ultimate-strength design.Otherwise, designs for shear by the working-stress and ultimate-strengthmethods are the same. Except for brackets and other short cantilevers, thesection for maximum shear may be taken at a distance d from the face of thesupport. In working-stress design, the shear stress vc carried by the concretealone should not exceed 1.1 fc . (As an alternative, the maximum for vcmay be taken as fc 1300 Vd/M , with a maximum of 1.9 fc ; fc is the28-day compressive strength of the concrete, lb/in2 (MPa); and M is the bend-ing moment at the section but should not be less than Vd.) At cross sections where the torsional stress vt exceeds 0.825 fc , vc should 1.1 2f cnot exceed 21 vc (5.62) (vt /1.2v)2
• CONCRETE FORMULAS 135The excess shear v vc should not exceed 4.4 fc in sections with web rein-forcement. Stirrups and bent bars should be capable of resisting the excessshear V V vc bd. The area required in the legs of a vertical stirrup, in2 (mm2), is Vs Av (5.63) fv dwhere s spacing of stirrups, in (mm); and fv allowable stress in stirrupsteel, (lb/in2) (MPa). For a single bent bar or a single group of parallel bars all bent at an anglewith the longitudinal axis at the same distance from the support, the requiredarea is V Av (5.64) fv sinFor inclined stirrups and groups of bars bent up at different distances from thesupport, the required area is Vs Av (5.65) fv d(sin cos )Stirrups in excess of those normally required are provided each way from thecutoff for a distance equal to 75 percent of the effective depth of the member.Area and spacing of the excess stirrups should be such that bws Av 60 (5.66) fywhere Av stirrup cross-sectional area, in2 (mm2) bw web width, in (mm) s stirrup spacing, in (mm) fy yield strength of stirrup steel, (lb/in2) (MPa)Stirrup spacing s should not exceed d/8 b, where b is the ratio of the area ofbars cut off to the total area of tension bars at the section and d is the effec-tive depth of the member.ULTIMATE-STRENGTH DESIGN OF RECTANGULARBEAMS WITH COMPRESSION BARSThe bending-moment capacity of a rectangular beam with both tension andcompression steel is a Mu 0.90 (As As) fy d As fy (d d) (5.67) 2
• 136 CHAPTER FIVEwhere a depth of equivalent rectangular compressive stress distribution (As As)fy /fc b b width of beam, in (mm) d distance from extreme compression surface to centroid of tensile steel, in (mm) d distance from extreme compression surface to centroid of compres- sive steel, in (mm) As area of tensile steel, in2 (mm2) As area of compressive steel, in2 (mm2) fy yield strength of steel, ksi (MPa) fc 28-day strength of concrete, ksi (MPa)This is valid only when the compressive steel reaches fy and occurs when fc d 87,000 ( ) 0.85 1 (5.68) fy d 87,000 fywhere As /bd As /bd 1 a constantWORKING-STRESS DESIGN OF RECTANGULAR BEAMSWITH COMPRESSION BARSThe following formulas, based on the linear variation of stress and strain withdistance from the neutral axis, may be used in design: 1 k (5.69) 1 fs /nfcwhere fs stress in tensile steel, ksi (MPa) fc stress in extreme compression surface, ksi (MPa) n modular ratio, Es / Ec kd d fs 2f (5.70) d kd swhere fs stress in compressive steel, ksi (MPa) d distance from extreme compression surface to centroid of tensile steel, in (mm) d distance from extreme compression surface to centroid of compres- sive steel, in (mm) The factor 2 is incorporated into the preceding equation in accordance withACI 318, “Building Code Requirements for Reinforced Concrete,” to account
• CONCRETE FORMULAS 137for the effects of creep and nonlinearity of the stress–strain diagram for con-crete. However, fs should not exceed the allowable tensile stress for the steel. Because total compressive force equals total tensile force on a section, C Cc Cs T (5.71)where C total compression on beam cross section, kip (N) Cc total compression on concrete, kip (N) at section Cs force acting on compressive steel, kip (N) T force acting on tensile steel, kip (N) fs k (5.72) fc 2[ (kd d )/(d kd )]where As /bd and As /bd . For reviewing a design, the following formulas may be used: B d k 2n n2( )2 n( ) (5.73) d (k3d / 3) 4n d [k (d /d )] z jd d z (5.74) k2 4n [k (d /d )]where jd is the distance between the centroid of compression and the centroidof the tensile steel. The moment resistance of the tensile steel is M Ms Tjd As fs jd fs (5.75) As jdwhere M is the bending moment at the section of beam under consideration.The moment resistance in compression is 1 d Mc f jbd 2 k 2n 1 (5.76) 2 c kd 2M fc (5.77) jbd 2{k 2n [1 (d /kd )]} Computer software is available for the preceding calculations. Many design-ers, however, prefer the following approximate formulas: 1 kd M1 f bkd d (5.78) 2 c 3 Ms M M1 2 fs As(d d ) (5.79)
• 138 CHAPTER FIVEwhere M bending moment Ms moment-resisting capacity of compressive steel M1 moment-resisting capacity of concreteULTIMATE-STRENGTH DESIGN OF I- AND T-BEAMSWhen the neutral axis lies in the flange, the member may be designed as a rec-tangular beam, with effective width b and depth d. For that condition, theflange thickness t will be greater than the distance c from the extreme compres-sion surface to the neutral axis, 1.18 d c (5.80) 1where 1 constant As fy /bd fc As area of tensile steel, in2 (mm2) fy yield strength of steel, ksi (MPa) fc 28-day strength of concrete, ksi (MPa)When the neutral axis lies in the web, the ultimate moment should not exceed a t Mu 0.90 (As Asf ) fy d Asf fy d (5.81) 2 2where Asf area of tensile steel required to develop compressive strength of overhanging flange, in2 (mm2) 0.85(b bw)tfc / fy bw width of beam web or stem, in (mm) a depth of equivalent rectangular compressive stress distribution, in (mm) (As Asf)fy / 0.85 fc bwThe quantity w f should not exceed 0.75 b, where b is the steel ratio forbalanced conditions w As /bwd and f Asf /bw d.WORKING-STRESS DESIGN OF I- AND T-BEAMSFor T-beams, effective width of compression flange is determined by the samerules as for ultimate-strength design. Also, for working-stress design, two casesmay occur: the neutral axis may lie in the flange or in the web. (For negativemoment, a T-beam should be designed as a rectangular beam with width bequal to that of the stem.) If the neutral axis lies in the flange, a T-or I-beam may be designed as a rec-tangular beam with effective width b. If the neutral axis lies in the web or stem,
• CONCRETE FORMULAS 139an I- or T-beam may be designed by the following formulas, which ignore thecompression in the stem, as is customary: I k (5.82) 1 fs /nfcwhere kd distance from extreme compression surface to neutral axis, in (mm) d distance from extreme compression surface to centroid of tensile steel, in (mm) fs stress in tensile steel, ksi (MPa) fc stress in concrete at extreme compression surface, ksi (MPa) n modular ratio Es / EcBecause the total compressive force C equals the total tension T, 1 bt C f (2kd t) T As fs (5.83) 2 c kd 2ndAs bt 2 kd (5.84) 2nAs 2btwhere As area of tensile steel, in2 (mm2); and t flange thickness, in (mm). The distance between the centroid of the area in compression and the cen-troid of the tensile steel is t (3kd 2t) jd d z z (5.85) 3(2kd t)The moment resistance of the steel is Ms Tjd As fs jd (5.86)The moment resistance of the concrete is fc btjd Mc Cjd (2kd t) (5.87) 2kdIn design, Ms and Mc can be approximated by t Ms As fs d (5.88) 2 1 t Mc f bt d (5.89) 2 c 2derived by substituting d t/2 for jd and fc /2 for fc(1 t/2kd), the averagecompressive stress on the section.
• 140 CHAPTER FIVEULTIMATE-STRENGTH DESIGN FOR TORSIONWhen the ultimate torsion Tu is less than the value calculated from the Tu equa-tion that follows, the area Av of shear reinforcement should be at least bw s Av 50 (5.90) fy However, when the ultimate torsion exceeds Tu calculated from the Tu equa-tion that follows, and where web reinforcement is required, either nominally orby calculation, the minimum area of closed stirrups required is 50bw s Av 2At (5.91) fywhere At is the area of one leg of a closed stirrup resisting torsion within adistance s. Torsion effects should be considered whenever the ultimate torsion exceeds Tu 0.5 fc x2y (5.92)where capacity reduction factor 0.85 Tu ultimate design torsional moment x2y sum for component rectangles of section of product of square of shorter side and longer side of each rectangle (where T section applies, overhanging flange width used in design should not exceed three times flange thickness)The torsion Tc carried by the concrete alone should not exceed 0.8 2f c 21 x2y Tc (5.93) (0.4Vu /Ct Tu )2where Ct bwd / x2y. Spacing of closed stirrups for torsion should be computed from At fy t x1 y1 s (5.94) (Tu Tc)where At area of one leg of closed stirrup t 0.66 0.33y1/x1 but not more than 1.50 fy yield strength of torsion reinforcement x1 shorter dimension c to c of legs of closed stirrup y1 longer dimension c to c of legs of closed stirrup
• CONCRETE FORMULAS 141The spacing of closed stirrups, however, should not exceed (x1 y1)/4 or 12 in(304.8 mm). Torsion reinforcement should be provided over at least a distanceof d b beyond the point where it is theoretically required, where b is thebeam width. At least one longitudinal bar should be placed in each corner of the stirrups.Size of longitudinal bars should be at least No. 3, and their spacing around theperimeters of the stirrups should not exceed 12 in (304.8 mm). Longitudinal barslarger than No. 3 are required if indicated by the larger of the values of Al com-puted from the following two equations: x1 y1 Al 2At (5.95) s 400xs Tu x1 y1 Al 2At (5.96) fy (Tu Vu /3Ct) sIn the second of the preceding two equations, 50bws /fy may be substituted for2At. The maximum allowable torsion is Tu 5Tc.WORKING-STRESS DESIGN FOR TORSIONTorsion effects should be considered whenever the torsion T due to serviceloads exceeds T 0.55(0.5 fc x2y) (5.97)where x2y sum for the component rectangles of the section of the productof the square of the shorter side and the longer side of each rectangle. Theallowable torsion stress on the concrete is 55 percent of that computed from thepreceding Tc equation. Spacing of closed stirrups for torsion should be computedfrom 3At t x1 y1 fv s (5.98) (vt vtc) x2ywhere At area of one leg of closed stirrup 0.33y1 t 0.66 , but not more than 1.50 x1 tc allowable torsion stress on concrete x1 shorter dimension c to c of legs of closed stirrup y1 longer dimension c to c of legs of closed stirrup
• 142 CHAPTER FIVEFLAT-SLAB CONSTRUCTIONSlabs supported directly on columns, without beams or girders, are classified asflat slabs. Generally, the columns flare out at the top in capitals (Fig. 5.3).However, only the portion of the inverted truncated cone thus formed that liesinside a 90° vertex angle is considered effective in resisting stress. Sometimes,the capital for an exterior column is a bracket on the inner face. The slab may be solid, hollow, or waffle. A waffle slab usually is the mosteconomical type for long spans, although formwork may be more expensivethan for a solid slab. A waffle slab omits much of the concrete that would be intension and thus is not considered effective in resisting stresses. To controldeflection, the ACI Code establishes minimum thicknesses for slabs, as indicatedby the following equation: ln(0.8 fy /200,000) ln(0.8 fy /200,000) h (5.99) 36 5 [ m 0.12(1 1/ )] 36 9where h slab thickness, in (mm) ln length of clear span in long direction, in (mm) fy yield strength of reinforcement, ksi (MPa) ratio of clear span in long direction to clear span in the short direction m average value of for all beams on the edges of a panel ratio of flexural stiffness Ecb Ib of beam section to flexural stiffness Ecs Is of width of slab bounded laterally by centerline of adjacent panel, if any, on each side of beam Ecb modulus of elasticity of beam concrete Ecs modulus of elasticity of slab concrete Ib moment of inertia about centroidal axis of gross section of beam, including that portion of slab on each side of beam that extends a distance equal to the projection of the beam above or below the slab, whichever is greater, but not more than four times slab thick- ness, in4 (mm4) Is moment of inertia about centroidal axis of gross section of slab h3/12 times slab width specified in definition of , in4 (mm4)Slab thickness h, however, need not be larger than (ln /36) (0.8 fy /200,000).FLAT-PLATE CONSTRUCTIONFlat slabs with constant thickness between supports are called flat plates. Gen-erally, capitals are omitted from the columns. Exact analysis or design of flat slabs or flat plates is very complex. It is com-mon practice to use approximate methods. The ACI Code presents two suchmethods: direct design and equivalent frame.
• c Columns L c Columns L B C A B– A A A c Columns 4 2 4 4 L A 4 Slab Drop panel Half column Middle Column Capital strip strip strip A 2 A Column (a)143 c Columns A 4 L Drop panel (b) FIGURE 5.3 Concrete flat slab: (a) Vertical section through drop panel and column at a support. (b) Plan view indicates division of slab into column and middle strips.
• 144 CHAPTER FIVE In both methods, a flat slab is considered to consist of strips parallel to columnlines in two perpendicular directions. In each direction, a column strip spansbetween columns and has a width of one-fourth the shorter of the two perpendic-ular spans on each side of the column centerline. The portion of a slab betweenparallel column strips in each panel is called the middle strip (see Fig. 5.3).Direct Design MethodThis may be used when all the following conditions exist: The slab has three or more bays in each direction. Ratio of length to width of panel is 2 or less. Loads are uniformly distributed over the panel. Ratio of live to dead load is 3 or less. Columns form an approximately rectangular grid (10 percent maximum offset). Successive spans in each direction do not differ by more than one-third of the longer span. When a panel is supported by beams on all sides, the relative stiffness of the beams satisfies 2 1 l2 0.2 5 (5.100) 2 l1where 1 in direction of l1 2 in direction of l2 relative beam stiffness defined in the preceding equation l1 span in the direction in which moments are being determined, c to c of supports l2 span perpendicular to l1, c to c of supports The basic equation used in direct design is the total static design moment ina strip bounded laterally by the centerline of the panel on each side of the center-line of the supports: wl2l2 Mo n (5.101) 8where w uniform design load per unit of slab area and ln clear span indirection moments are being determined. The strip, with width l2, should be designed for bending moments for whichthe sum in each span of the absolute values of the positive and average negativemoments equals or exceeds Mo.
• CONCRETE FORMULAS 1451. The sum of the flexural stiffnesses of the columns above and below the slab Kc should be such that Kc c min (5.102) (Ks Kb)where Kc flexural stiffness of column EccIc Ecc modulus of elasticity of column concrete Ic moment of inertia about centroidal axis of gross section of column Ks Ecs Is Kb Ecb Ib min minimum value of c as given in engineering handbooks2. If the columns do not satisfy condition 1, the design positive moments in the panels should be multiplied by the coefficient: 2 a c s 1 1 (5.103) 4 a minSHEAR IN SLABSSlabs should also be investigated for shear, both beam type and punching shear. Forbeam-type shear, the slab is considered as a thin, wide rectangular beam. The crit-ical section for diagonal tension should be taken at a distance from the face ofthe column or capital equal to the effective depth d of the slab. The criticalsection extends across the full width b of the slab. Across this section, thenominal shear stress vu on the unreinforced concrete should not exceed the ulti-mate capacity 2 f c or the allowable working stress 1.1 fc , where fc is the28-day compressive strength of the concrete, lb/in2 (MPa). Punching shear may occur along several sections extending completelyaround the support, for example, around the face of the column, or column cap-ital, or around the drop panel. These critical sections occur at a distance d/2from the faces of the supports, where d is the effective depth of the slab or droppanel. Design for punching shear should be based on Vn (Vc VS) (5.104)where capacity reduction factor (0.85 for shear and torsion), with shearstrength Vn taken not larger than the concrete strength Vc calculated from 2f c bo d 4 2f c bo d 4 Vc 2 (5.105) c
• 146 CHAPTER FIVEwhere bo perimeter of critical section and c ratio of long side to short sideof critical section. However, if shear reinforcement is provided, the allowable shear may beincreased a maximum of 50 percent if shear reinforcement consisting of bars isused and increased a maximum of 75 percent if shearheads consisting of twopairs of steel shapes are used. Shear reinforcement for slabs generally consists of bent bars and is designed inaccordance with the provisions for beams with the shear strength of the concrete atcritical sections taken as 2 f c bo d at ultimate strength and Vn 6 fc bo d .Extreme care should be taken to ensure that shear reinforcement is accuratelyplaced and properly anchored, especially in thin slabs.COLUMN MOMENTSAnother important consideration in design of two-way slab systems is the transferof moments to columns. This is generally a critical condition at edge columns,where the unbalanced slab moment is very high due to the one-sided panel. The unbalanced slab moment is considered to be transferred to the columnpartly by flexure across a critical section, which is d/2 from the periphery of thecolumn, and partly by eccentric shear forces acting about the centroid of thecritical section. That portion of unbalanced slab moment Mu transferred by the eccentricityof the shear is given by Mu: 1 v 1 (5.106) 3 B b2 2 b1 1where b1 width, in (mm), of critical section in the span direction for whichmoments are being computed; and b2 width, in (mm), of critical section inthe span direction perpendicular to b1. As the width of the critical section resisting moment increases (rectangularcolumn), that portion of the unbalanced moment transferred by flexure alsoincreases. The maximum factored shear, which is determined by combining thevertical load and that portion of shear due to the unbalanced moment beingtransferred, should not exceed Vc, with Vc given by preceding the Vc equation.The shear due to moment transfer can be determined at the critical section bytreating this section as an analogous tube with thickness d subjected to a bend-ing moment Mu. The shear stress at the crack, at the face of the column or bracket support, islimited to 0.2 fc or a maximum of 800 Ac, where Ac is the area of the concretesection resisting shear transfer. The area of shear-friction reinforcement Avf required in addition to reinforcementprovided to take the direct tension due to temperature changes or shrinkage should be
• CONCRETE FORMULAS 147computed from Vu Avf (5.107) fywhere Vu is the design shear, kip (kN), at the section; fy is the reinforcementyield strength, but not more than 60 ksi (413.7 MPa); and , the coefficient offriction, is 1.4 for monolithic concrete, 1.0 for concrete placed against hardenedconcrete, and 0.7 for concrete placed against structural rolled-steel members.The shear-friction reinforcement should be well distributed across the face of thecrack and properly anchored at each side.SPIRALSThis type of transverse reinforcement should be at least 3/8 in (9.5 mm) indiameter. A spiral may be anchored at each of its ends by 11/2 extra turns ofthe spiral. Splices may be made by welding or by a lap of 48 bar diameters,but at least 12 in (304.8mm). Spacing (pitch) of spirals should not exceed 3 in(76.2 mm), or be less than 1 in (25.4 mm). Clear spacing should be at least 11/3times the maximum size of coarse aggregate. The ratio of the volume of spiral steel/volume of concrete core (out to out ofspiral) should be at least Ag fc s 0.45 1 (5.108) Ac fywhere Ag gross area of column Ac core area of column measured to outside of spiral fy spiral steel yield strength fc 28-day compressive strength of concreteBRACED AND UNBRACED FRAMESAs a guide in judging whether a frame is braced or unbraced, note that the com-mentary on ACI 318–83 indicates that a frame may be considered braced if thebracing elements, such as shear walls, shear trusses, or other means resistinglateral movement in a story, have a total stiffness at least six times the sum ofthe stiffnesses of all the columns resisting lateral movement in that story. The slenderness effect may be neglected under the two following conditions: For columns braced against sidesway, when klu M1 34 12 (5.109) r M2
• CONCRETE FORMULAS 149where Vu total design shear force capacity reduction factor 0.85 d 0.8lw h overall thickness of wall lw horizontal length of wall The shear Vc carried by the concrete depends on whether Nu, the design axi-al load, lb (N), normal to the wall horizontal cross section and occurring simul-taneously with Vu at the section, is a compression or tension force. When Nu isa compression force, Vc may be taken as 2 fc hd , where fc is the 28-daystrength of concrete, lb/in2 (MPa). When Nu is a tension force, Vc should be tak-en as the smaller of the values calculated from 2fc hd Nu d Vc 3.3 (5.113) 4lw lw(1.25 2fc hd 0.6 2fc 0.2Nu /lwh) Vc (5.114) Mu /Vu lw /2This equation does not apply, however, when Mu/Vu lw /2 is negative. When the factored shear Vu is less than 0.5 Vc, reinforcement should beprovided as required by the empirical method for bearing walls. When Vu exceeds 0.5 Vc, horizontal reinforcement should be provided with VsAv fy d/s2, where s2 spacing of horizontal reinforcement and Av reinforcementarea. Also, the ratio h of horizontal shear reinforcement to the gross concrete area ofthe vertical section of the wall should be at least 0.0025. Spacing of horizontal shearbars should not exceed lw /5, 3h, or 18 in (457.2 mm). In addition, the ratio of verticalshear reinforcement area to gross concrete area of the horizontal section of wall doesnot need to be greater than that required for horizontal reinforcement but should notbe less than hw n 0.0025 0.5 2.5 (5.115) lw ( h 0.0025) 0.0025where hw total height of wall. Spacing of vertical shear reinforcement should notexceed lw /3, 3h, or 18 in (457.2 mm). In no case should the shear strength Vn be taken greater than 10 f c hd atany section. Bearing stress on the concrete from anchorages of posttensioned members withadequate reinforcement in the end region should not exceed fb calculated from B Ab Ab fb 0.8 fc 0.2 1.25 fci (5.116) 2fc B A Ab fb 0.6 fc (5.117) b
• 150 CHAPTER FIVEwhere Ab bearing area of anchor plate, and Ab maximum area of portionof anchorage surface geometrically similar to and concentric with area ofanchor plate. A more refined analysis may be applied in the design of the end-anchorageregions of prestressed members to develop the ultimate strength of the tendons.should be taken as 0.90 for the concrete.CONCRETE GRAVITY RETAINING WALLSForces acting on gravity walls include the weight of the wall, weight of the earthon the sloping back and heel, lateral earth pressure, and resultant soil pressure onthe base. It is advisable to include a force at the top of the wall to account forfrost action, perhaps 700 lb/linear ft (1042 kg/m). A wall, consequently, may failby overturning or sliding, overstressing of the concrete or settlement due tocrushing of the soil. Design usually starts with selection of a trial shape and dimensions, and thisconfiguration is checked for stability. For convenience, when the wall is of con-stant height, a 1-ft (0.305-m) long section may be analyzed. Moments are takenabout the toe. The sum of the righting moments should be at least 1.5 times thesum of the overturning moments. To prevent sliding, Rv 1.5Ph (5.118)where coefficient of sliding friction Rv total downward force on soil, lb (N) Ph horizontal component of earth thrust, lb (N) Next, the location of the vertical resultant Rv should be found at various sec-tions of the wall by taking moments about the toe and dividing the sum by Rv.The resultant should act within the middle third of each section if there is to beno tension in the wall. Finally, the pressure exerted by the base on the soil should be computed toensure that the allowable pressure is not exceeded. When the resultant is with-in the middle third, the pressures, lb/ft2 (Pa), under the ends of the base aregiven by Rv Mc Rv 6e p 1 (5.119) A I A Lwhere A area of base, ft2 (m2) L width of base, ft (m) e distance, parallel to L, from centroid of base to Rv, ft (m)Figure 5.4(b) shows the pressure distribution under a 1-ft (0.305-m) strip ofwall for e L/2 a, where a is the distance of Rv from the toe. When Rv isexactly L/3 from the toe, the pressure at the heel becomes zero. When Rv
• CONCRETE FORMULAS 151falls outside the middle third, the pressure vanishes under a zone around theheel, and pressure at the toe is much larger than for the other cases. The variables in the five formulas in Fig. 5.4 areP1 and P2 the pressure, lb/ft2 (MPa), at the locations shown L and a dimensions, ft (m), at the locations shown Rv the total downward force on the soil behind the retaining wall, lb (N) R resultant, lb(N) Earth backfill Stem Stem Front Back K Toe Heel Heel D C B A Earth in front Footing or base T- shape wall L-shape wall Stem Stem Counterfort Toe Toe Heel Reversed Counterfort wall L-shape wall Buttress Buttress Stem Stem Toe Toe Heel Buttressed wall (a) FIGURE 5.4 (a) Six types of retaining walls. (b) Soil-pressure variation in retaining walls. (Merritt-Building Construction Handbook, McGraw-Hill.)
• 152 CHAPTER FIVE Rv R Rv P1 = (4L – 6a) L/3 L2 Rv a P2 = (6a – 2L) 2 L P2 P1 L When a = , 2 Rv P1 = P2 = L L (a) Resultant in middle third Rv R 2Rv L/3 P1 = L a P2 P2 = O P1 (b) Resultant at edge middle third Rv R 2Rv L/3 P= 3a a 3a P (c) Resultant outside middle third (b) FIGURE 5.4 (Continued)
• CONCRETE FORMULAS 153 In usual design work on retaining walls the sum of the righting momentsand the sum of the overturing moments about the toe are found. It is assumedby designers that if the retaining wall is overturned, it will overturn about thetoe of the retaining wall. Designers then apply a safety factor thus: Retaining wall righting moment 1.5 (overturning moment)The 1.5 safety factor is a common value amongst designers.CANTILEVER RETAINING WALLSThis type of wall resists the lateral thrust of earth pressure through cantileveraction of a vertical stem and horizontal base (Fig. 5.5). Cantilever wallsgenerally are economical for heights from 10 to 20 ft (3 to 6 m). For lowerwalls, gravity walls may be less costly; for taller walls, counterforts (Fig. 5.6)may be less expensive. 41,000 8" 0 #4 @3–0" #8 @18" Resisting moment #4 @12" 13–0" #8@18" #8 Dowels @9" Stem stop alternate bars 2–2" and 5–3" above top of base Bending moment 4–3" 18"2–0" Toe heel #4 @12" 1–6" 0 133,000 270,000 1–3" #7 @9" 9" 3–9" 1–6" 4–9" 10 (a) Typical wall section (b) Moment diagramFIGURE 5.5 Cantilever retaining wall. (a) Vertical section shows main reinforcing steelplaced vertically in the stem. (b) Moment diagram.
• 154 CHAPTER FIVE Counterfort A A Toe Heel (a) Section B-B B B (b) Plan A-A FIGURE 5.6 Counterfort retaining wall. (a) Vertical section. (b) Horizontal section. Shear unit stress on a horizontal section of a counterfort may be computedfrom vc V1/bd, where b is the thickness of the counterfort and d is the hori-zontal distance from face of wall to main steel, M V1 V (tan tan ) (5.120) dwhere V shear on section M bending moment at section angle earth face of counterfort makes with vertical angle wall face makes with verticalFor a vertical wall face, 0 and V1 V (M/d)tan . The critical sectionfor shear may be taken conservatively at a distance up from the base equal to dsin cos , where d is the depth of counterfort along the top of the base.
• CONCRETE FORMULAS 155 Wall FIGURE 5.7 Concrete wall footing.WALL FOOTINGSThe spread footing under a wall (Fig. 5.7) distributes the wall load horizontallyto preclude excessive settlement. The footing acts as a cantilever on opposite sides of the wall under down-ward wall loads and upward soil pressure. For footings supporting concretewalls, the critical section for bending moment is at the face of the wall; forfootings under masonry walls, halfway between the middle and edge of thewall. Hence, for a 1-ft (0.305-m) long strip of symmetrical concrete-wall footing,symmetrically loaded, the maximum moment, ft lb (N m), is p M (L a)2 (5.121) 8where p uniform pressure on soil, lb/ft2 (Pa) L width of footing, ft (m) a wall thickness, ft (m)If the footing is sufficiently deep that the tensile bending stress at the bottom,6M/t2, where M is the factored moment and t is the footing depth, in (mm), doesnot exceed 5 f c , where fc is the 28-day concrete strength, lb/in2 (MPa) and 0.90, the footing does not need to be reinforced. If the tensile stress islarger, the footing should be designed as a 12-in (305-mm) wide rectangular,reinforced beam. Bars should be placed across the width of the footing, 3 in(76.2 mm) from the bottom. Bar development length is measured from thepoint at which the critical section for moment occurs. Wall footings also maybe designed by ultimate-strength theory.
• b h X X h/2 TABLE 6.1 Properties of Sections for Standard Lumber Sizes [Dressed (S4S) sizes, moment of inertia, and section modulus are given with respect to xx axis, with dimensions b and h, as shown on sketch] Moment of Section158 Standard Nominal dressed size Area of inertia modulus Board feet size S4S section bh 2 bh 2 per linear I S b h b h A bh 2 12 6 foot of piece 2 4 15 8 35 8 5.89 6.45 3.56 2 3 2 6 1 5 8 51 2 8.93 22.53 8.19 1 2 8 15 8 71 2 12.19 57.13 15.23 11 3 Source: National Lumber Manufacturers Association.
• TIMBER ENGINEERING FORMULAS 159BEARINGThe allowable unit stresses given for compression perpendicular to the grainapply to bearings of any length at the ends of beams and to all bearings 6 in(152.4 mm) or more in length at other locations. When calculating the requiredbearing area at the ends of beams, no allowance should be made for the fact that,as the beam bends, the pressure upon the inner edge of the bearing is greater thanat the end of the beam. For bearings of less than 6 in (152.4 mm) in length andnot nearer than 3 in (76.2 mm) to the end of the member, the allowable stress forcompression perpendicular to the grain should be modified by multiplying by thefactor (l 3 8)/l, where l is the length of the bearing in inches (mm) measuredalong the grain of the wood.BEAMSThe extreme fiber stress in bending for a rectangular timber beam is f 6M/bh2 (6.1) M/SA beam of circular cross section is assumed to have the same strength in bend-ing as a square beam having the same cross-sectional area. The horizontal shearing stress in a rectangular timber beam is H 3V /2bh (6.2)For a rectangular timber beam with a notch in the lower face at the end, thehorizontal shearing stress is H (3V /2bd1) (h /d1) (6.3)A gradual change in cross section, instead of a square notch, decreases theshearing stress nearly to that computed for the actual depth above the notch. Nomenclature for the preceding equations follows: f maximum fiber stress, lb/in2 (MPa) M bending moment, lb in (Nm) h depth of beam, in (mm) b width of beam, in (mm) S section modulus ( bh2/6 for rectangular section), in3 (mm3) H horizontal shearing stress, lb/in2 (MPa) V total shear, lb (N) d1 depth of beam above notch, in (mm) l span of beam, in (mm) P concentrated load, lb (N)
• 162 CHAPTER SIXwhere c allowable unit stress at angle to grain, lb/in2 (MPa) c allowable unit stress parallel to grain, lb/in2 (MPa) c allowable unit stress perpendicular to grain, lb/in2 (MPa) angle between direction of load and direction of grainRECOMMENDATIONS OF THE FOREST PRODUCTSLABORATORYThe Wood Handbook gives advice on the design of solid wood columns. (WoodHandbook, USDA Forest Products Laboratory, Madison, Wisc., 1999.) Columns are divided into three categories, short, intermediate, and long. LetK denote a parameter defined by the equation 1/2 E K 0.64 (6.11) fcThe range of the slenderness ratio and the allowable stress assigned to each cat-egory are next. Short column, L 11 f fc (6.12) d Intermediate column, 4 L 1 L/d 11 K f fc 1 (6.13) d 3 K Long column, L 0.274E K f (6.14) d (L /d )2The maximum L /d ratio is set at 50. The National Design Specification covers the design of solid columns. Theallowable stress in a rectangular section is as follows: 0.30E f but f fc (6.15) (L /d)2 The notational system for the preceding equations is P allowable load A sectional area L unbraced length
• TIMBER ENGINEERING FORMULAS 163 d smaller dimension of rectangular section E modulus of elasticity fc allowable compressive stress parallel to grain in short column of given species f allowable compressive stress parallel to grain in given columnCOMPRESSION ON OBLIQUE PLANEConsider that a timber member sustains a compressive force with an action linethat makes an oblique angle with the grain. Let P allowable compressive stress parallel to grain Q allowable compressive stress normal to grain N allowable compressive stress inclined to grain angle between direction of stress N and direction of grain By Hankinson’s equation, PQ N (6.16) P sin2 Q cos2 In Fig. 6.1, member M1 must be notched at the joint to avoid removing anexcessive area from member M2. If the member is cut in such a manner that ACand BC make an angle of /2 with vertical and horizontal planes, respectively,the allowable bearing pressures at these faces are identical for the two mem-bers. Let A sectional area of member M1 f1 pressure at AC f2 pressure at BCIt may be readily shown that sin ( /2) cos ( /2) AC b BC b (6.17) sin sin F sin F sin tan ( /2) f1 f2 (6.18) A tan ( /2) AThis type of joint is often used in wood trusses.
• 164 CHAPTER SIX F b φ/2 M1 φ A B φ/2 f1 C M2 f2 FIGURE 6.1 Timber joint.ADJUSTMENT FACTORS FOR DESIGN VALUESDesign values obtained by the methods described earlier should be multiplied byadjustment factors based on conditions of use, geometry, and stability. Theadjustments are cumulative, unless specifically indicated in the following. The adjusted design value Fb for extreme-fiber bending is given by Fb FbCDCMCtCLCFCVCfuCrCcCf (6.19)where Fb design value for extreme-fiber bending CD load-duration factor CM wet-service factor Ct temperature factor CL beam stability factor CF size factor—applicable only to visually graded, sawn lumber and round timber flexural members Cv volume factor—applicable only to glued-laminated beams Cfu flat-use factor—applicable only to dimension-lumber beams 2 to 4 in (50.8 to 101.6 mm) thick and glued-laminated beams Cr repetitive-member factor—applicable only to dimension-lumber beams 2 to 4 in (50.8 to 101.6 mm) thick Cc curvature factor—applicable only to curved portions of glued- laminated beams Cf form factorFor glued-laminated beams, use either CL or Cv (whichever is smaller), not both. The adjusted design value for tension Ft is given by Ft FtCDCMCtCF (6.20)where Ft is the design value for tension.
• TIMBER ENGINEERING FORMULAS 165 For shear, the adjusted design value FV is computed from FV FVCDCMCtCH (6.21)where FV is the design value for shear and CH is the shear stress factor 1—pemitted for FV parallel to the grain for sawn lumber members. For compression perpendicular to the grain, the adjusted design value Fc isobtained from Fc Fc CMCtCb (6.22)where Fc is the design value for compression perpendicular to the grain and Cbis the bearing area factor. For compression parallel to the grain, the adjusted design value Fc is givenby Fc FcCDCMCtCFCp (6.23)where Fc is the design value for compression parallel to grain and Cp is the col-umn stability factor. For end grain in bearing parallel to the grain, the adjusted design value, Fg iscomputed from Fg FgCDCt (6.24)where Fg is the design value for end grain in bearing parallel to the grain. The adjusted design value for modulus of elasticity, E is obtained from E ECMCT C (6.25)where E design value for modulus of elasticity CT buckling stiffness factor—applicable only to sawn-lumber truss compression chords 2 4 in (50.8 101.6 mm) or smaller, when subject to combined bending and axial compression and plywood sheathing 3 8 in (9.5 mm) or more thick is nailed to the narrow face C other appropriate adjustment factorsSize and Volume FactorsFor visually graded dimension lumber, design values Fb, Ft , and Fc for allspecies and species combinations, except southern pine, should be multipliedby the appropriate size factor Cf , given in reference data to account for theeffects of member size. This factor and the factors used to develop size-specific
• TIMBER ENGINEERING FORMULAS 167modification for duration of load. If these values are exceeded, mechanical rein-forcement sufficient to resist all radial tensile stresses is required. When M is in the direction tending to increase curvature (decrease theradius), the stress is compressive across the grain. For this condition, the designvalue is limited to that for compression perpendicular to grain for all species. For the curved portion of members, the design value for wood in bendingshould be modified by multiplication by the following curvature factor: 2 t Cc 1 2000 (6.29) Rwhere t is the thickness of lamination, in (mm); and R is the radius of curvatureof lamination, in (mm). Note that t/R should not exceed 1/100 for hardwoodsand southern pine or 1/125 for softwoods other than southern pine. The curva-ture factor should not be applied to stress in the straight portion of an assembly,regardless of curvature elsewhere.Bearing Area FactorDesign values for compression perpendicular to the grain Fc apply to bear-ing surfaces of any length at the ends of a member and to all bearings 6 in(152.4 mm) or more long at other locations. For bearings less than 6 in(152.4 mm) long and at least 3 in (76.2 mm) from the end of a member, Fcmay be multiplied by the bearing area factor: Lb 0.375 (6.30) Cb Lbwhere Lb is the bearing length, in (mm) measured parallel to grain. Equation(6.30) yields the values of Cb for elements with small areas, such as plates andwashers, listed in reference data. For round bearing areas, such as washers, Lbshould be taken as the diameter.Column Stability and Buckling Stiffness FactorsDesign values for compression parallel to the grain Ft should be multiplied bythe column stability factor Cp given by Eq. (6.31): 1 (FcE /Fc* ) B CP 1 (FcE /Fc*) 2 (FcE /Fc* ) (6.31) 2c 2c cwhere F * c design value for compression parallel to the grain multiplied by all applicable adjustment factors except Cp FcE K cE E /(L e /d)2 E modulus of elasticity multiplied by adjustment factors
• 168 CHAPTER SIX KcE 0.3 for visually graded lumber and machine-evaluated lumber 0.418 for products with a coefficient of variation less than 0.11 c 0.80 for solid-sawn lumber 0.85 for round timber piles 0.90 for glued-laminated timberFor a compression member braced in all directions throughout its length to pre-vent lateral displacement, Cp 1.0. The buckling stiffness of a truss compression chord of sawn lumber subjectedto combined flexure and axial compression under dry service conditions maybe increased if the chord is 2 4 in (50.8 101.6 mm) or smaller and has thenarrow face braced by nailing to plywood sheathing at least 3 8 in (9.5 mm)thick in accordance with good nailing practice. The increased stiffness may beaccounted for by multiplying the design value of the modulus of elasticity E bythe buckling stiffness factor CT in column stability calculations. When theeffective column length Le, in (mm), is 96 in (2.38 m) or less, CT may be com-puted from KM Le CT 1 (6.32) KT Ewhere KM 2300 for wood seasoned to a moisture content of 19 percent or less at time of sheathing attachment 1200 for unseasoned or partly seasoned wood at time of sheathing attachment KT 0.59 for visually graded lumber and machine-evaluated lumber 0.82 for products with a coefficient of variation of 0.11 or lessWhen Le is more than 96 in (2.38 m), CT should be calculated from Eq. (6.32)with Le 96 in (2.38 m). For additional information on wood trusses with metal-plate connections, see design standards of the Truss Plate Institute, Madison,Wisconsin. The slenderness ratio RB for beams is defined by B b2 Le d RB (6.33)The slenderness ratio should not exceed 50. The effective length Le for Eq. (6.33) is given in terms of unsupported lengthof beam in reference data. Unsupported length is the distance between supportsor the length of a cantilever when the beam is laterally braced at the supports toprevent rotation and adequate bracing is not installed elsewhere in the span.When both rotational and lateral displacements are also prevented at intermedi-ate points, the unsupported length may be taken as the distance between pointsof lateral support. If the compression edge is supported throughout the length ofthe beam and adequate bracing is installed at the supports, the unsupportedlength is zero.
• TIMBER ENGINEERING FORMULAS 169 The beam stability factor CL may be calculated from * 1 (FbE /Fb ) B * 2 * CL 1 (FbE /Fb ) FbE /Fb (6.34) 1.9 1.9 0.95 *where F b design value for bending multiplied by all applicable adjustment factors, except Cfu, CV, and CL 2 FbE K bE E /R B K bE 0.438 for visually graded lumber and machine-evaluated lumber 0.609 for products with a coefficient of variation of 0.11 or less E design modulus of elasticity multiplied by applicable adjustment factorsFASTENERS FOR WOODNails and SpikesThe allowable withdrawal load per inch (25.4 mm) of penetration of a commonnail or spike driven into side grain (perpendicular to fibers) of seasoned wood,or unseasoned wood that remains wet, is p 1,380G5/2D (6.35)where p allowable load per inch (mm) of penetration into member receiving point, lb (N) D diameter of nail or spike, in (mm) G specific gravity of wood, oven dry The total allowable lateral load for a nail or spike driven into side grain ofseasoned wood is p CD 3/2 (6.36)where p allowable load per nail or spike, lb (N) D diameter of nail or spike, in (mm) C coefficient dependent on group number of wood (see Table 6.1)Values of C for the four groups into which stress-grade lumber is classified are Group I: C 2,040 Group II: C 1,650 Group III: C 1,350 Group IV: C 1,080
• 170 CHAPTER SIXThe loads apply where the nail or spike penetrates into the member, receiving itspoint at least 10 diameters for Group I species, 11 diameters for Group IIspecies, 13 diameters for Group III species, and 14 diameters for Group IVspecies. Allowable loads for lesser penetrations are directly proportional to thepenetration, but the penetration must be at least one-third that specified.Wood ScrewsThe allowable withdrawal load per inch (mm) of penetration of the threadedportion of a wood screw into side grain of seasoned wood that remains dry is p 2,850G2D (6.37)where p allowable load per inch (mm) of penetration of threaded portion into member receiving point, lb (N) D diameter of wood screw, in (mm) G specific gravity of wood, oven dry (see Table 6.1)Wood screws should not be loaded in withdrawal from end grain. The total allowable lateral load for wood screws driven into the side grain ofseasoned wood which remains dry is p CD2 (6.38)where p allowable load per wood screw, lb (N) D diameter of wood screw, in (mm) C coefficient dependent on group number of wood (Table 6.2)Values of C for the four groups into which stress-grade lumber is classified are Group I: C 4,800 Group II: C 3,960 Group III: C 3,240 Group IV: C 2,520 TABLE 6.2 Specific Gravity and Group Number for Common Species of Lumber Group Specific Species number gravity, G G2 G5/2 Douglas fir II 0.51 0.260 0.186 Pine, southern II 0.59 0.348 0.267 Hemlock, western III 0.44 0.194 0.128 Hemlock, eastern IV 0.43 0.185 0.121 Pine, Norway III 0.47 0.221 0.151 Redwood III 0.42 0.176 0.114 Spruce IV 0.41 0.168 0.108
• 172 CHAPTER SIX For wood screws, W WCDCMCt (6.44) Z ZCD CM Ct Cd Ceg (6.45)where Ceg is the end-grain factor. For lag screws, W WCD CM Ct Ceg (6.46) Z ZCD CM Ct CgC Cd Ceg (6.47) For metal plate connectors, Z ZCD CM Ct (6.48) For drift bolts and drift pins, W WCD CM Ct Ceg (6.49) Z ZCD CM Ct Cg C Cd Ceg (6.50) For spike grids, Z ZCD CM Ct C (6.51)ROOF SLOPE TO PREVENT PONDINGRoof beams should have a continuous upward slope equivalent to 1 4 in/ft(20.8 mm/m) between a drain and the high point of a roof, in addition tominimum recommended camber to avoid ponding. When flat roofs haveinsuffic-ient slope for drainage (less than 1 4 in/ft) (20.8 mm/m) the stiffnessof supporting members should be such that a 5-lb/ft2 (239.4 N/mm2) loadcauses no more than 1 2 -in (12.7 mm) deflection. Because of ponding, snow loads or water trapped by gravel stops, parapetwalls, or ice dams magnify stresses and deflections from existing roof loads* by 1 Cp (6.52) 1 W L3/ 4EIwhere Cp factor for multiplying stresses and deflections under existing loads to determine stresses and deflections under existing loads plus ponding W weight of 1 in (25.4 mm) of water on roof area supported by beam, lb (N) L span of beam, in (mm) E modulus of elasticity of beam material, lb/in2 (MPa) I moment of inertia of beam, in4 (mm4) *Kuenzi and Bohannan, “Increases in Deflection and Stresses Caused by Ponding of Water onRoofs,” Forest Products Laboratory, Madison, Wisconsin.
• 174 CHAPTER SIXFor either uniaxial or biaxial bending, fc should not exceed K cE E FcE1 (6.56) (L e1/d 1)2where E is the modulus of elasticity multiplied by adjustment factors. Also, forbiaxial bending, fc should not exceed K cE E FcE2 (6.57) (L e2 /d 2)2and fb1 should not be more than K bE E FbE (6.58) R2Bwhere d1 is the width of the wide face and d2 is the width of the narrow face.Slenderness ratio RB for beams is given earlier in this section. KbE is definedearlier in this section. The effective column lengths Le1 for buckling in the d1 direc-tion and Le2 for buckling in the d2 direction, E , FcE1, and FcE2 should be determinedas shown earlier. As for the case of combined bending and axial tension, Fc , F b1 , and Fb2should be adjusted for duration of load by applying CD. Nomenclature for Formulas given in Eqs. (6.59) through (6.57): Q allowable load, lb P ultimate load, lb A section area of column, sq in L length of column, in r least radius of gyration of column section, in Su ultimate strength, psi Sy yield point or yield strength of material, psi E modulus of elasticity of material, psi m factor of safety (L/r) critical slenderness ratioSOLID RECTANGULAR OR SQUARECOLUMNS WITH FLAT ENDS*For select structural-grade lumber in general structural use under continuously dryconditions, the following formulas can be used for the allowable unit load, Q/A: BS 4 Q 1 L L E S 1 up to K 0.64 (6.59) A 3 Kd d *Roark—“Formulas for Stress and Strain,” McGraw-Hill.
• TIMBER ENGINEERING FORMULAS 175 Q 0.274E L 2 for K (6.60) A L d d Q L S up to 11 (6.61) A dThe ultimate unit load is: P Q 4 (6.62) A A For lumber of various types, the following formulas can be used, whereS allowable compressive stress; and d least dimension of the lumbercross section. Hemlock 4 Q L S 700 700 1 0.00000097 ,K 24.2 A d E 1,100,000 (6.63) Longleaf yellow pine 4 Q L S 1450 1450 1 0.00000162 ,K 21.23 A d E 1,600,000 (6.64) Southern cypress 4 Q L S 1100 1100 1 0.00000168 ,K 21.10 A d E 1,200,000 (6.65) Douglas fir 4 Q L S 1250 1200 1 0.00000112 ,K 23.35 A d E 1,600,000 (6.66) For general structural use under continuously dry conditions for laminatedwood with cover plates or boxed around a solid core, square or rectangular withends flat: Ratio of Q/A to Q/A for solid column of same dimensions depends on L/dand is as follows: L 6 10 14 18 22 26 d Ratio 0.82 0.77 0.71 0.65 0.74 0.82 (6.67)
• 176 CHAPTER SIX For solid circular columns with ends flat in general structural use under con-tinuously dry conditions: Q/A is same as for square column of equal area. For tapered round column,d and A are taken as for section distant 1/3 L from smaller end. Q/A at smallend must not exceed S. Note: Where numeric constants appear in formulas, compute in USCS val-ues and then convert to SI using the conversion factors in Chap. 1.
• CHAPTER 7 SURVEYING FORMULASUNITS OF MEASUREMENTUnits of measurement used in past and present surveys are For construction work: feet, inches, fractions of inches (m, mm) For most surveys: feet, tenths, hundredths, thousandths (m, mm) For National Geodetic Survey (NGS) control surveys: meters, 0.1, 0.01, 0.001 mThe most-used equivalents are 1 meter 39.37 in (exactly) 3.2808 ft 1 rod 1 pole 1 perch 161 2 ft (5.029 m) 1 engineer’s chain 100 ft 100 links (30.48 m) 1 Gunter’s chain 66 ft (20.11 m) 100 Gunter’s links (lk) 4 rods 1 80 mi (0.020 km) 1 acre 100,000 sq (Gunter’s) links 43,560 ft2 160 rods2 10 sq 2 (Gunter’s) chains 4046.87 m 0.4047 ha 1 rood 3 4 acre (1011.5 m2) 40 rods2 (also local unit 51 2 to 8 yd) (5.029 to 7.315 m) 1 ha 10,000 m2 107,639.10 ft2 2.471 acres 1 arpent about 0.85 acre, or length of side of 1 square arpent (varies) (about 3439.1 m2) 1 statute mi 5280 ft 1609.35 m 1 mi2 640 acres (258.94 ha) 1 nautical mi (U.S.) 6080.27 ft 1853.248 m 1 fathom 6 ft (1.829 m) 1 cubit 18 in (0.457 m) 1 vara 33 in (0.838 m) (Calif.), 331 3 in (0.851 m) (Texas), varies 1 degree 1 360 circle 60 min 3600 s 0.01745 rad sin 1 0.01745241 177
• 178 CHAPTER SEVEN 1 rad 57 17 44.8 or about 57.30 1 grad (grade) 1 400 circle 1 100 quadrant 100 centesimal min 104 cen- tesimals (French) 1 mil 1 6400 circle 0.05625 1 military pace (milpace) 21 2 ft (0.762 m)THEORY OF ERRORSWhen a number of surveying measurements of the same quantity have beenmade, they must be analyzed on the basis of probability and the theory oferrors. After all systematic (cumulative) errors and mistakes have been elimi-nated, random (compensating) errors are investigated to determine the mostprobable value (mean) and other critical values. Formulas determined fromstatistical theory and the normal, or Gaussian, bell-shaped probability distrib-ution curve, for the most common of these values follow. Standard deviation of a series of observations is Bn 1 d2 s (7.1)where d residual (difference from mean) of single observation and n num-ber of observations. The probable error of a single observation is PE s 0.6745 s (7.2)(The probability that an error within this range will occur is 0.50.) The probability that an error will lie between two values is given by theratio of the area of the probability curve included between the values to the totalarea. Inasmuch as the area under the entire probability curve is unity, there is a100 percent probability that all measurements will lie within the range of thecurve. The area of the curve between s is 0.683; that is, there is a 68.3 percentprobability of an error between s in a single measurement. This error rangeis also called the one-sigma or 68.3 percent confidence level. The area of thecurve between 2 s is 0.955. Thus, there is a 95.5 percent probability of anerror between 2 s and 2 s that represents the 95.5 percent error (two-sigma or 95.5 percent confidence level). Similarly, 3 s is referred to as the99.7 percent error (three-sigma or 99.7 percent confidence level). For practicalpurposes, a maximum tolerable level often is assumed to be the 99.9 percenterror. Table 7.1 indicates the probability of occurrence of larger errors in a sin-gle measurement. The probable error of the combined effects of accidental errors from differ-ent causes is Esum 2E 2 1 E2 2 E2 3 (7.3)
• SURVEYING FORMULAS 179 TABLE 7.1 Probability of Error in a Single Measurement Probability Confidence of larger Error level, % error Probable (0.6745 s) 50 1 in 2 Standard deviation ( s) 68.3 1 in 3 90% (1.6449 s) 90 1 in 10 2 s or 95.5% 95.5 1 in 20 3 s or 97.7% 99.7 1 in 370 Maximum (3.29 s) 99.9 1 in 1000where E1, E2, E3 . . . are probable errors of the separate measurements. Error of the mean is Esum Es n Es Em (7.4) n n nwhere Es specified error of a single measurement. Probable error of the mean is B n(n PEs d2 PEm 0.6745 (7.5) n 1)MEASUREMENT OF DISTANCE WITH TAPESReasonable precisions for different methods of measuring distances are Pacing (ordinary terrain): 1 50 to 1 100 Taping (ordinary steel tape): 1 1000 to 1 10,000 (Results can be improved by use of tension apparatus, transit alignment, leveling.) Baseline (invar tape): 1 50,000 to 1 1,000,000 Stadia: 1 300 to 1 500 (with special procedures) Subtense bar: 1 1000 to 1 7000 (for short distances, with a 1-s theodolite, averag- ing angles taken at both ends) Electronic distance measurement (EDM) devices have been in use since themiddle of the twentieth century and have now largely replaced steel tape mea-surements on large projects. The continued development, and the resulting dropin prices, are making their use widespread. A knowledge of steel-taping errorsand corrections remains important, however, because use of earlier survey datarequires a knowledge of how the measurements were made, common sources forerrors, and corrections that were typically required.
• 180 CHAPTER SEVEN For ordinary taping, a tape accurate to 0.01 ft (0.00305 m) should be used. Thetension of the tape should be about 15 lb (66.7 N). The temperature should bedetermined within 10°F (5.56°C); and the slope of the ground, within 2 percent;and the proper corrections, applied. The correction to be applied for tempera-ture when using a steel tape is Ct 0.0000065s(T T0) (7.6)The correction to be made to measurements on a slope is Ch s (1 cos ) exact (7.7) 2or 0.00015s approximate (7.8) 2or h /2s approximate (7.9)where Ct temperature correction to measured length, ft (m) Ch correction to be subtracted from slope distance, ft (m) s measured length, ft (m) T temperature at which measurements are made, F ( C) T0 temperature at which tape is standardized, F ( C) h difference in elevation at ends of measured length, ft (m) slope angle, degree In more accurate taping, using a tape standardized when fully supported through-out, corrections should also be made for tension and for support conditions. The cor-rection for tension is (Pm Ps)s Cp (7.10) SEThe correction for sag when not fully supported is w 2L3 Cs 2 (7.11) 24P mwhere Cp tension correction to measured length, ft (m) Cs sag correction to measured length for each section of unsupported tape, ft (m) Pm actual tension, lb (N) Ps tension at which tape is standardized, lb (N) (usually 10 lb) (44.4 N) S cross-sectional area of tape, in2 (mm2) E modulus of elasticity of tape, lb/in2 (MPa) [29 million lb/in2 (MPa) for steel] (199,955 MPa) w weight of tape, lb/ft (kg/m) L unsupported length, ft (m)Slope CorrectionsIn slope measurements, the horizontal distance H L cos x, where Lslope distance and x vertical angle, measured from the horizontal—a simple
• SURVEYING FORMULAS 181hand calculator operation. For slopes of 10 percent or less, the correction to beapplied to L for a difference d in elevation between tape ends, or for a horizon-tal offset d between tape ends, may be computed from d2 Cs (7.12) 2LFor a slope greater than 10 percent, Cs may be determined from d2 d4 Cs (7.13) 2L 8L3Temperature CorrectionsFor incorrect tape length: (actual tape length nominal tape length)L Ct (7.14) nominal tape lengthFor nonstandard tension: (applied pull standard tension)L Ct (7.15) AEwhere A cross-sectional area of tape, in2 (mm2); and E modulus of elas-ticity 29,000,00 lb/in2 for steel (199,955 MPa).For sag correction between points of support, ft (m): w2 L3 s C (7.16) 24P2where w weight of tape per foot, lb (N) Ls unsupported length of tape, ft (m) P pull on tape, lb (N)Orthometric CorrectionThis is a correction applied to preliminary elevations due to flattening of theearth in the polar direction. Its value is a function of the latitude and elevationof the level circuit. Curvature of the earth causes a horizontal line to depart from a level sur-face. The departure Cf , ft; or Cm, (m), may be computed from Cf 0.667M 2 0.0239F 2 (7.17) 2 Cm 0.0785K (7.18)
• 182 CHAPTER SEVENwhere M, F, and K are distances in miles, thousands of feet, and kilometers,respectively, from the point of tangency to the earth. Refraction causes light rays that pass through the earth’s atmosphere tobend toward the earth’s surface. For horizontal sights, the average angular dis-placement (like the sun’s diameter) is about 32 min. The displacement Rf, ft, orRm, m, is given approximately by Rf 0.093M 2 0.0033F 2 (7.19) 2 Rm 0.011K (7.20)To obtain the combined effect of refraction and curvature of the earth, subtractRf from Cf or Rm from Cm. Borrow-pit or cross-section leveling produces elevations at the corners ofsquares or rectangles with sides that are dependent on the area to be covered,type of terrain, and accuracy desired. For example, sides may be 10, 20, 40,50, or 100 ft (3.048, 6.09, 12.19, 15.24, or 30.48 m). Contours can be locatedreadily, but topographic features, not so well. Quantities of material to beexcavated or filled are computed, in yd3 (m3), by selecting a grade elevation orfinal ground elevation, computing elevation differences for the corners, andsubstituting in nxA Q (7.21) 108where n number of times a particular corner enters as part of a division block x difference in ground and grade elevation for each corner, ft (m) A area of each block, ft2 (m2)VERTICAL CONTROLThe NGS provides vertical control for all types of surveys. NGS furnishesdescriptions and elevations of bench marks on request. As given in “Standardsand Specifications for Geodetic Control Networks,” Federal Geodetic ControlCommittee, the relative accuracy C, mm, required between directly connectedbench marks for the three orders of leveling is First order: C 0.5 2K for Class I and 0.7 2K for Class II (7.22) Second order: C 1.0 2K for Class I and 1.3 2K for Class II (7.23) Third order: C 2.0 2K (7.24)where K is the distance between bench marks, km.
• SURVEYING FORMULAS 183STADIA SURVEYINGIn stadia surveying, a transit having horizontal stadia crosshairs above andbelow the central horizontal crosshair is used. The difference in the rod read-ings at the stadia crosshairs is termed the rod intercept. The intercept maybe converted to the horizontal and vertical distances between the instrumentand the rod by the following formulas: H Ki (cos a)2 (f c) cos a (7.25) 1 V Ki (sin 2a) (f c) sin a (7.26) 2where H horizontal distance between center of transit and rod, ft (m) V vertical distance between center of transit and point on rod inter- sected by middle horizontal crosshair, ft (m) K stadia factor (usually 100) i rod intercept, ft (m) a vertical inclination of line of sight, measured from the horizontal, degree f c instrument constant, ft (m) (usually taken as 1 ft) (0.3048 m)In the use of these formulas, distances are usually calculated to the foot (meter)and differences in elevation to tenths of a foot (meter). Figure 7.1 shows stadia relationships for a horizontal sight with theolder type of external-focusing telescope. Relationships are comparable forthe internal-focusing type. For horizontal sights, the stadia distance, ft, (m) (from instrument spindle torod), is f D R C (7.27) i f1 f2 C c f d B a´ a m i R b b´ F D A FIGURE 7.1 Distance D is measured with an external-focusing telescope by determining interval R intercepted on a rod AB by two horizontal sighting wires a and b.
• 184 CHAPTER SEVENwhere R intercept on rod between two sighting wires, ft (m) f focal length of telescope, ft (m) (constant for specific instrument) i distance between stadia wires, ft (m) C f c (7.28) c distance from center of spindle to center of objective lens, ft (m)C is called the stadia constant, although c and C vary slightly. The value of f/i, the stadia factor, is set by the manufacturer to be about 100,but it is not necessarily 100.00. The value should be checked before use onimportant work, or when the wires or reticle are damaged and replaced.PHOTOGRAMMETRYPhotogrammetry is the art and science of obtaining reliable measurements byphotography (metric photogrammetry) and qualitative evaluation of image data(photo interpretation). It includes use of terrestrial, close-range, aerial, vertical,oblique, strip, and space photographs along with their interpretation. Scale formulas are as follows: Photo scale photo distance (7.29) Map scale map distance ab f Photo scale (7.30) AB H h1where f focal length of lens, in (m) H flying height of airplane above datum (usually mean sea level), ft (m) h1 elevation of point, line, or area with respect to datum, ft (m)
• CHAPTER 8 SOIL AND EARTHWORK FORMULASPHYSICAL PROPERTIES OF SOILSBasic soil properties and parameters can be subdivided into physical, index, andengineering categories. Physical soil properties include density, particle size anddistribution, specific gravity, and water content. The water content w of a soil sample represents the weight of free watercontained in the sample expressed as a percentage of its dry weight. The degree of saturation S of the sample is the ratio, expressed as percent-age, of the volume of free water contained in a sample to its total volume ofvoids Vv. Porosity n, which is a measure of the relative amount of voids, is the ratio ofvoid volume to the total volume V of soil: Vv n (8.1) VThe ratio of Vv to the volume occupied by the soil particles Vs defines the voidratio e. Given e, the degree of saturation may be computed from wGs S (8.2) ewhere Gs represents the specific gravity of the soil particles. For most inorganicsoils, Gs is usually in the range of 2.67 0.05. The dry unit weight d of a soil specimen with any degree of saturation maybe calculated from w Gs S d (8.3) 1 wGswhere w is the unit weight of water and is usually taken as 62.4 lb/ft3 (1001 kg/m3)for freshwater and 64.0 lb/ft3 (1026.7 kg/m3) for seawater. 185
• 186 CHAPTER EIGHTINDEX PARAMETERS FOR SOILSIndex parameters of cohesive soils include liquid limit, plastic limit, shrinkagelimit, and activity. Such parameters are useful for classifying cohesive soils andproviding correlations with engineering soil properties. The liquid limit of cohesive soils represents a near-liquid state, that is, anundrained shear strength about 0.01 lb/ft2 (0.0488 kg/m2). The water content atwhich the soil ceases to exhibit plastic behavior is termed the plastic limit. Theshrinkage limit represents the water content at which no further volume changeoccurs with a reduction in water content. The most useful classification and cor-relation parameters are the plasticity index Ip, the liquidity index Il, the shrinkageindex Is, and the activity Ac. These parameters are defined in Table 8.1. Relative density Dr of cohesionless soils may be expressed in terms of voidratio e or unit dry weight d: emax e0 Dr (8.4) emax emin 1/ min 1/ d Dr (8.5) 1/ min 1/ maxDr provides cohesionless soil property and parameter correlations, includingfriction angle, permeability, compressibility, small-strain shear modulus, cyclicshear strength, and so on.RELATIONSHIP OF WEIGHTS ANDVOLUMES IN SOILSThe unit weight of soil varies, depending on the amount of water contained inthe soil. Three unit weights are in general use: the saturated unit weight sat, thedry unit weight dry, and the buoyant unit weight b:TABLE 8.1 Soil Indices Index Definition* CorrelationPlasticity Ip Wl Wp Strength, compressibility, compactibility, and so forth Wn WpLiquidity Il Compressibility and stress rate IpShrinkage Is Wp Ws Shrinkage potential IpActivity Ac Swell potential, and so forth *Wl liquid limit; Wp plastic limit; Wn moisture content, %; Ws shrinkage limit; percentof soil finer than 0.002 mm (clay size).
• SOIL AND EARTHWORK FORMULAS 187 (G e) 0 (1 w)G 0 sat S 100% (8.6) 1 e 1 e G 0 dry S 0% (8.7) (1 e) (G 1) 0 b S 100% (8.8) 1 eUnit weights are generally expressed in pound per cubic foot or gram per cubiccentimeter. Representative values of unit weights for a soil with a specific grav-ity of 2.73 and a void ratio of 0.80 are sat 122 lb/ft3 1.96 g/cm3 (8.9) dry 95 lb/ft3 1.52 g/cm3 (8.10) b 60 lb/ft3 0.96 g/cm3 (8.11)The symbols used in the three preceding equations and in Fig. 8.1 are G specific gravity of soil solids (specific gravity of quartz is 2.67; for majority of soils specific gravity ranges between 2.65 and 2.85; organic soils would have lower specific gravities) 0 unit weight of water, 62.4 lb/ft3 (1.0 g/cm3) e voids ratio, volume of voids in mass of soil divided by volume of solids in same mass; also equal to n /(1 n), where n is porosity—volume of voids in mass of soil divided by total volume of same mass S degree of saturation, volume of water in mass of soil divided by volume of voids in same mass w water content, weight of water in mass of soil divided by weight of solids in same mass; also equal to Se /G Gas V(voids) W(total) (for unit V) = γf 1+e W(water) = 1 Gγ0 = Se γ0 V(water) = e 1+e 1+e = Se Water = 1 + w Gγ0 = G + Se γ0V=1 1+e 1+e W(solids) V(solids) 1+e 1+e Solids = 1 Total volume (solids + water + gas) = 1FIGURE 8.1 Relationship of weights and volumes in soil.
• 188 CHAPTER EIGHTINTERNAL FRICTION AND COHESIONThe angle of internal friction for a soil is expressed by tan (8.12)where angle of internal friction tan coefficient of internal friction normal force on given plane in cohesionless soil mass shearing force on same plane when sliding on plane is impendingFor medium and coarse sands, the angle of internal friction is about 30° to 35°.The angle of internal friction for clays ranges from practically 0° to 20°. The cohesion of a soil is the shearing strength that the soil possesses byvirtue of its intrinsic pressure. The value of the ultimate cohesive resis-tance of a soil is usually designated by c. Average values for c are given inTable 8.2.VERTICAL PRESSURES IN SOILSThe vertical stress in a soil caused by a vertical, concentrated surface load maybe determined with a fair degree of accuracy by the use of elastic theory. Twoequations are in common use, the Boussinesq and the Westergaard. The Boussi-nesq equation applies to an elastic, isotropic, and homogeneous mass thatextends infinitely in all directions from a level surface. The vertical stress at apoint in the mass is 2 5/2 3P r z 1 (8.13) 2 z2 z TABLE 8.2 Cohesive Resistance of Various Soil Types Cohesion c General soil type lb/ft2 (kPa) Almost-liquid clay 100 (4.8) Very soft clay 200 (9.6) Soft clay 400 (19.1) Medium clay 1000 (47.8) Damp, muddy sand 400 (19.1)
• SOIL AND EARTHWORK FORMULAS 189 The Westergaard equation applies to an elastic material laterally reinforcedwith horizontal sheets of negligible thickness and infinite rigidity, which preventthe mass from undergoing lateral strain. The vertical stress at a point in themass, assuming a Poisson’s ratio of zero, is 2 3/2 P r z 1 2 (8.14) z2 zwhere z vertical stress at a point, lb/ft2 (kPa) P total concentrated surface load, lb (N) z depth of point at which z acts, measured vertically downward from surface, ft (m) r horizontal distance from projection of surface load P to point at which z acts, ft (m) For values of r/z between 0 and 1, the Westergaard equation gives stressesappreciably lower than those given by the Boussinesq equation. For values ofr /z greater than 2.2, both equations give stresses less than P/100z2.LATERAL PRESSURES IN SOILS,FORCES ON RETAINING WALLSThe Rankine theory of lateral earth pressures, used for estimating approxi-mate values for lateral pressures on retaining walls, assumes that the pres-sure on the back of a vertical wall is the same as the pressure that wouldexist on a vertical plane in an infinite soil mass. Friction between the walland the soil is neglected. The pressure on a wall consists of (1) the lateralpressure of the soil held by the wall, (2) the pressure of the water (if any)behind the wall, and (3) the lateral pressure from any surcharge on the soilbehind the wall. Symbols used in this section are as follows: unit weight of soil, lb/ft3 (kg/m3) (saturated unit weight, dry unit weight, or buoyant unit weight, depending on conditions) P total thrust of soil, lb/linear ft (kg/linear m) of wall H total height of wall, ft (m) angle of internal friction of soil, degree i angle of inclination of ground surface behind wall with horizontal; also angle of inclination of line of action of total thrust P and pres- sures on wall with horizontal KA coefficient of active pressure KP coefficient of passive pressure c cohesion, lb/ft2 (kPa)
• 190 CHAPTER EIGHTLATERAL PRESSURE OFCOHESIONLESS SOILSFor walls that retain cohesionless soils and are free to move an appreciableamount, the total thrust from the soil is 2(cos i)2 2(cos i)2 1 cos i (cos )2 P H 2 cos i (8.15) 2 cos i (cos )2When the surface behind the wall is level, the thrust is 1 P 2 H 2 KA (8.16) 2where KA tan 45 (8.17) 2The thrust is applied at a point H/3 above the bottom of the wall, and the pressuredistribution is triangular, with the maximum pressure of 2P/H occurring at thebottom of the wall. For walls that retain cohesionless soils and are free to move only a slightamount, the total thrust is 1.12P, where P is as given earlier. The thrust isapplied at the midpoint of the wall, and the pressure distribution is trapezoidal,with the maximum pressure of 1.4P /H extending over the middle six-tenth ofthe height of the wall. For walls that retain cohesionless soils and are completely restrained (veryrare), the total thrust from the soil is 2(cos i) 2(cos i)2 2 1 cos i (cos )2 P H 2 cos i (8.18) 2 cos i (cos )2When the surface behind the wall is level, the thrust is 1 P 2 H 2KP (8.19) 2where KP tan 45 (8.20) 2The thrust is applied at a point H /3 above the bottom of the wall, and the pres-sure distribution is triangular, with the maximum pressure of 2P / H occurringat the bottom of the wall.
• SOIL AND EARTHWORK FORMULAS 191LATERAL PRESSURE OF COHESIVE SOILSFor walls that retain cohesive soils and are free to move a considerable amountover a long period of time, the total thrust from the soil (assuming a level sur-face) is 1 P 2 H 2 KA 2cH KA (8.21)or, because highly cohesive soils generally have small angles of internalfriction, 1 P 2 H2 2cH (8.22)The thrust is applied at a point somewhat below H /3 from the bottom of thewall, and the pressure distribution is approximately triangular. For walls that retain cohesive soils and are free to move only a smallamount or not at all, the total thrust from the soil is 1 P 2 H 2K P (8.23)because the cohesion would be lost through plastic flow.WATER PRESSUREThe total thrust from water retained behind a wall is 1 2 P 2 0H (8.24)where H height of water above bottom of wall, ft (m); and 0 unit weight ofwater, lb/ft3 62.4 lb/ft3 (1001 kg/m3) for freshwater and 64 lb/ft3 (1026.7 kg/m3)for saltwater. The thrust is applied at a point H /3 above the bottom of the wall, and thepressure distribution is triangular, with the maximum pressure of 2P / H occur-ring at the bottom of the wall. Regardless of the slope of the surface behind thewall, the thrust from water is always horizontal.LATERAL PRESSURE FROM SURCHARGEThe effect of a surcharge on a wall retaining a cohesionless soil or an unsatu-rated cohesive soil can be accounted for by applying a uniform horizontal loadof magnitude KA p over the entire height of the wall, where p is the surcharge inpound per square foot (kilopascal). For saturated cohesive soils, the full valueof the surcharge p should be considered as acting over the entire height of thewall as a uniform horizontal load. KA is defined earlier.
• 192 CHAPTER EIGHTSTABILITY OF SLOPESCohesionless SoilsA slope in a cohesionless soil without seepage of water is stable if i (8.25) With seepage of water parallel to the slope, and assuming the soil to be satu-rated, an infinite slope in a cohesionless soil is stable if b tan i tan (8.26) satwhere i slope of ground surface angle of internal friction of soil b, sat unit weights, lb/ft3 (kg/m3)Cohesive SoilsA slope in a cohesive soil is stable if C H (8.27) Nwhere H height of slope, ft (m) C cohesion, lb/ft2 (kg/m2) unit weight, lb/ft3 (kg/m3) N stability number, dimensionless For failure on the slope itself, without seepage water, N (cos i)2 (tan i tan ) (8.28) Similarly, with seepage of water, b N (cos i)2 tan i tan (8.29) sat When the slope is submerged, is the angle of internal friction of the soiland is equal to b. When the surrounding water is removed from a submergedslope in a short time (sudden drawdown), is the weighted angle of internalfriction, equal to ( b / sat) , and is equal to sat.BEARING CAPACITY OF SOILSThe approximate ultimate bearing capacity under a long footing at the surfaceof a soil is given by Prandtl’s equation as
• SOIL AND EARTHWORK FORMULAS 193 2Kp (Kp e c 1 qu dry b tan 1) (8.30) tan 2where qu ultimate bearing capacity of soil, lb/ft2 (kg/m2) c cohesion, lb/ft2 (kg/m2) angle of internal friction, degree dry unit weight of dry soil, lb/ft3 (kg/m3) b width of footing, ft (m) d depth of footing below surface, ft (m) Kp coefficient of passive pressure 2 tan 45 2 e 2.718 . . . For footings below the surface, the ultimate bearing capacity of the soil maybe modified by the factor 1 Cd/b. The coefficient C is about 2 for cohesion-less soils and about 0.3 for cohesive soils. The increase in bearing capacity withdepth for cohesive soils is often neglected.SETTLEMENT UNDER FOUNDATIONSThe approximate relationship between loads on foundations and settlement is q 2d C2 C1 1 (8.31) P b bwhere q load intensity, lb/ft2 (kg/m2) P settlement, in (mm) d depth of foundation below ground surface, ft (m) b width of foundation, ft (m) C1 coefficient dependent on internal friction C2 coefficient dependent on cohesionThe coefficients C1 and C2 are usually determined by bearing-plate loading tests.SOIL COMPACTION TESTSThe sand-cone method is used to determine in the field the density of com-pacted soils in earth embankments, road fill, and structure backfill, as well asthe density of natural soil deposits, aggregates, soil mixtures, or other similarmaterials. It is not suitable, however, for soils that are saturated, soft, or friable(crumble easily).
• 194 CHAPTER EIGHT Characteristics of the soil are computed from Volume of soil, ft 3 (m3) weight of sand filling hole, lb (kg) (8.32) density of sand, lb/ft 3 (kg/m 3) 100(weight of moist soil weight of dry soil) % Moisture (8.33) weight of dry soil weight of soil, lb (kg) Field density, lb/ft 3 (kg /m3) (8.34) volume of soil, ft 3 (m3) field density Dry density (8.35) 1 % moisture/100 100(dry density) % Compaction (8.36) max dry densityMaximum density is found by plotting a density–moisture curve.Load-Bearing TestOne of the earliest methods for evaluating the in situ deformability ofcoarse-grained soils is the small-scale load-bearing test. Data developedfrom these tests have been used to provide a scaling factor to express the set-tlement of a full-size footing from the settlement 1 of a 1-ft2 (0.0929-m2)plate. This factor / 1 is given as a function of the width B of the full-sizebearing plate as 2 2B 1 (8.37) 1 BFrom an elastic half-space solution, Es can be expressed from results of a plateload test in terms of the ratio of bearing pressure to plate settlement kv as 2 k v (1 ) /4 Es (8.38) 4B/(1 B)2where represents Poisson’s ratio, usually considered to range between0.30 and 0.40. The Es equation assumes that 1 is derived from a rigid, 1-ft(0.3048-m)-diameter circular plate and that B is the equivalent diameter ofthe bearing area of a full-scale footing. Empirical formulations, such as the / 1 equation, may be significantly in error because of the limited footing-size range used and the large scatter of the database. Furthermore, considera-tion is not given to variations in the characteristics and stress history of thebearing soils.
• SOIL AND EARTHWORK FORMULAS 195California Bearing RatioThe California bearing ratio (CBR) is often used as a measure of the quality ofstrength of a soil that underlies a pavement, for determining the thickness of thepavement, its base, and other layers. F (8.39) CBR F0where F force per unit area required to penetrate a soil mass with a 3-in2(1935.6-mm2) circular piston (about 2 in (50.8 mm) in diameter) at the rate of0.05 in/min (1.27 mm/min); and F0 force per unit area required for corre-sponding penetration of a standard material. Typically, the ratio is determined at 0.10-in (2.54-mm) penetration, althoughother penetrations sometimes are used. An excellent base course has a CBR of100 percent. A compacted soil may have a CBR of 50 percent, whereas aweaker soil may have a CBR of 10.Soil PermeabilityThe coefficient of permeability k is a measure of the rate of flow of waterthrough saturated soil under a given hydraulic gradient i, cm/cm, and is definedin accordance with Darcy’s law as V kiA (8.40)where V rate of flow, cm3/s; and A cross-sectional area of soil conveyingflow, cm2. Coefficient k is dependent on the grain-size distribution, void ratio, and soilfabric and typically may vary from as much as 10 cm/s for gravel to less than10 –7 for clays. For typical soil deposits, k for horizontal flow is greater than kfor vertical flow, often by an order of magnitude.COMPACTION EQUIPMENTA wide variety of equipment is used to obtain compaction in the field. Sheeps-foot rollers generally are used on soils that contain high percentages of clay.Vibrating rollers are used on more granular soils. To determine maximum depth of lift, make a test fill. In the process, themost suitable equipment and pressure to be applied, lb/in2 (kPa), for groundcontact also can be determined. Equipment selected should be able to producedesired compaction with four to eight passes. Desirable speed of rolling also canbe determined. Average speeds, mi/h (km/h), under normal conditions are givenin Table 8.3.
• 196 CHAPTER EIGHT TABLE 8.3 Average Speeds of Rollers Type mi/h (km/h) Grid rollers 12 (19.3) Sheepsfoot rollers 3 (4.8) Tamping rollers 10 (16.1) Pneumatic rollers 8 (12.8) Compaction production can be computed from 16WSLFE yd3/h (m3/h) (8.41) Pwhere W width of roller, ft (m) S roller speed, mi/h (km/h) L lift thickness, in (mm) F ratio of pay yd3 (m3) to loose yd3 (m3) E efficiency factor (allows for time losses, such as those due to turns): 0.90, excellent; 0.80, average; 0.75, poor P number of passesFORMULAS FOR EARTHMOVINGExternal forces offer rolling resistance to the motion of wheeled vehicles, suchas tractors and scrapers. The engine has to supply power to overcome this resis-tance; the greater the resistance is, the more power needed to move a load.Rolling resistance depends on the weight on the wheels and the tire penetrationinto the ground: R Rf W R p pW (8.42)where R rolling resistance, lb (N) Rf rolling-resistance factor, lb/ton (N/tonne) W weight on wheels, ton (tonne) Rp tire-penetration factor, lb/ton in (N/tonne mm) penetration p tire penetration, in (mm)Rf usually is taken as 40 lb/ton (or 2 percent lb/lb) (173 N/t) and Rp as 30 lb/ton in (1.5% lb/lb in) (3288 N/t mm). Hence, Eq. (8.42) can be written as R (2% 1.5%p) W RW (8.43)where W weight on wheels, lb (N); and R 2% 1.5%p.
• 198 CHAPTER EIGHT The load, or amount of material a machine carries, can be determined byweighing or estimating the volume. Payload estimating involves determinationof the bank cubic yards (cubic meters) being carried, whereas the excavatedmaterial expands when loaded into the machine. For determination of bankcubic yards (cubic meters) from loose volume, the amount of swell or the loadfactor must be known. Weighing is the most accurate method of determining the actual load. This isnormally done by weighing one wheel or axle at a time with portable scales,adding the wheel or axle weights, and subtracting the weight of the vehicle whenempty. To reduce error, the machine should be relatively level. Enough loadsshould be weighed to provide a good average: weight of load, lb (kg) Bank yd3 (8.50) density of material, lb/bank yd3 (kg/m3)Equipment RequiredTo determine the number of scrapers needed on a job, required production mustfirst be computed: quantity, bank yd3 (m3) Production required, yd3 /h (m3/ h) (8.51) working time, h production required, yd3/h (m3/h) No. of scrapers needed (8.52) production per unit, yd3/h (m3/h) scraper cycle time, min No. of scrapers a pusher can load (8.53) pusher cycle time, min Because speeds and distances may vary on haul and return, haul and returntimes are estimated separately. haul distance, ft return distance, ft Variable time, min (8.54) 88 speed, mi/ h 88 speed, mi/ h haul distance, m return distance, m 16.7 speed, km/ h 16.7 speed, km/ hHaul speed may be obtained from the equipment specification sheet when thedrawbar pull required is known.VIBRATION CONTROL IN BLASTINGExplosive users should take steps to minimize vibration and noise from blastingand protect themselves against damage claims.
• SOIL AND EARTHWORK FORMULAS 199 Vibrations caused by blasting are propagated with a velocity V, ft/s (m/s);frequency f, Hz; and wavelength L, ft (m), related by V L (8.55) fVelocity v, in/s (mm/s), of the particles disturbed by the vibrations depends onthe amplitude of the vibrations A, in (mm): v 2 fA (8.56)If the velocity v1 at a distance D1 from the explosion is known, the velocity v2 ata distance D2 from the explosion may be estimated from 1.5 D1 v2 v1 (8.57) D2The acceleration a, in/s2 (mm/s2), of the particles is given by 2 2 a 4 f A (8.58)For a charge exploded on the ground surface, the overpressure P, lb/in2 (kPa),may be computed from W1/3 1.407 P 226.62 (8.59) Dwhere W maximum weight of explosives, lb (kg) per delay; and D distance,ft (m), from explosion to exposure. The sound pressure level, decibels, may be computed from 0.084 P dB 28 (8.60) 6.95 10For vibration control, blasting should be controlled with the scaled-distanceformula: D v H (8.61) Wwhere constant (varies for each site) and H constant (varies for each site). Distance to exposure, ft (m), divided by the square root of maximum pounds(kg) per delay is known as scaled distance. Most courts have accepted the fact that a particle velocity not exceeding 2 in/s(50.8 mm/s) does not damage any part of any structure. This implies that, for thisvelocity, vibration damage is unlikely at scaled distances larger than 8.
• 200 CHAPTER EIGHTBlasting OperationsBlasting operations are used in many civil engineering projects—for basementsin new buildings, bridge footings, canal excavation, dam construction, and so on.Here are a number of key formulas used in blasting operations of many types.*Borehole DiameterGenerally there are three criteria in determining the borehole diameter to be used:the availability of equipment, the depth of the cut, and the distance to the neareststructure. To reduce the amount of drilling, the blaster will usually use the largesthole that the depth of the cut and the proximity of structures will permit. Themaximum borehole diameter that can be effectively used depends on the depth ofthe hole. Conversely, the minimum depth to which a hole can be drilled is depen-dent on the diameter and can generally be represented by the formula: L 2Dh (8.62)where L is the minimum length of the borehold in feet and Dh is the diameter ofthe borehole in inches. (For metric, multiply the hole diameter by 25.4 to obtainthe minimum hole depth in millimeters.)Burden and Spacing DeterminationThe burden is the distance from the blast hole to the nearest perpendicular freeface. The true burden can vary depending upon the delay system used for theblast; therefore the delay design should be determined before the drill pattern islaid out.Anderson Formula One of the first of the modern blasting formulas wasdeveloped by O. Andersen. He developed his formula on the premise that burdenis a function of the hole diameter and the length of the hole. The Andersenformula is B dL (8.63)where B burden, ft d diameter of borehole, in L length of borehole, ft We have learned since this formula was developed that there are more fac-tors involved in burden determination than just hole diameter and length. How-ever, Andersen was quite correct in his assumption that length is a factor in *The formulas presented here are the work of Gary B. Hemphill, P.E., who at the time of their pre-sentation was Construction Manager, Hecla Mining Company, and presented in Blasting Operations,McGraw-Hill.
• SOIL AND EARTHWORK FORMULAS 201determining the burden. The length of the burden relative to the depth of the cuthas a significant effect on fragmentation.Langefors’ Formula Langefors suggested that the burden determination wasbased on more factors, including diameter of hole, weight strength of explosives,degree of packing, a rock constant, and the degree of fracture. Langefors’ formula for burden determination is B cf(E V) Ps V (db 33) (8.64)where V burden, m db diameter of drill bit, mm P the degree of packing 1.0 to 1.6 kg/dm3 s weight strength of explosive (1.3 for gelatin) c rock constant, generally 0.45 f 1 degree of fraction, for straight hole 1 E spacing E/V ratio of spacing to burdenKonya Formula Currently the best formula for burden determination is onedeveloped by C. J. Konya. This formula uses the diameter of the explosives inrelation to the specific gravity of the explosive and of the rock. The Konya formula is B SGr 3 SGe B 3.15 De (8.65)where B burden, ft De diameter of the explosive, in SGe specific gravity of the explosive SGr specific gravity of the rockThe optimum length-to-burden ratio (L/B) is 3.Spacing DeterminationThe spacing is calculated in relation to the burden length; that is, it is necessaryto complete the burden calculations before determining the spacing. Spacing isthe distance between blast holes fired, on the same delay or greater delay, in thesame row. For a single-row instantaneous blast, the spacing is usually 1.8 times the bur-den; that is, for a burden of 5 ft (1.5 m) the spacing would be 9 ft (2.7 m). Formultiple simultaneous (same-delay) blasting where the ratio of length of bore-hole to burden (L/B) is less than 4, the spacing can be determined by the formula S BL (8.66)
• 202 CHAPTER EIGHTwhere B burden, ft L length of borehole, ft If the length-to-burden ratio is greater than 4, then the spacing is twice theburden. Therefore, if the L/B is 5 the spacing is determined by S = 2B. Forexample, if the burden were equal to 7 ft (2.1 m), the spacing would be S 2(7) ft 14 ftagain showing the relationship of the length of borehole to the burden and spac-ing dimensions.StemmingThere must be some substance put into the top of every borehole to preventthe explosive gases from escaping prematurely. This substance is called“stemming.” Generally the amount of stemming required will range from 0.7B to 1B.Therefore, in the case of burdens of 8 ft (2.4 m) the stemming will be some-where between 5.6 ft (1.7 m) and 8 ft (2.4 m). That is, the ratio of stemming toburden can vary through this range depending on existing conditions. The con-dition most likely to affect the amount of stemming is the structural integrity ofthe material in the area of the borehole collar. If the material is very competent,as is a homogeneous granite, the stemming will approach 0.7B. However, if thematerial is fractured rock with many fissures and dirt seams, the required ratioof stemming to burden will be approximately 1. In material other than rock,such as overburden, the ratio of stemming to burden will be even greater. Over-burden is generally treated in a 2:1 ratio over rock; that is, 2 ft of overburden isapproximately equal to 1 ft of rock for stemming purposes. Therefore, if thereis 4 ft (1.2 m) of overburden on a shot that requires a burden of 7 ft (2.1 m), thestemming is calculated as follows. Using the ratio of 0.7B, the stemming would ordinarily be 5 ft (1.5 m).However, with 4 ft (1.2 m) of overburden, for stemming purposes equal to 2 ft(0.6 m) of rock, the actual amount of stemming is 7 ft (2.1 m). Another way ofexpressing it is to compute the required stemming and then add one-half theoverburden depth. Thus OB Stemming 0.7B (8.67) 2where B burden OB overburden To compute stemming for a side hill cut, use a right triangle and 2 to 1 ratio,as shown in Fig. 8.2.
• SOIL AND EARTHWORK FORMULAS 203 Stemming = 10 ft Nearest free Top of explosive column face 5 2 to 1 10 5 FIGURE 8.2 To compute the stemming for a side hill cut, use a right triangle, as shown here. (Hemphill-Blasting operations McGraw-Hill.)Air ConcussionAir concussion, or air blast, is a pressure wave traveling through the air; it isgenerally not a problem in construction blasting. The type of damage created byair concussion is broken windows. However, it must be realized that a properlyset window glass can generally tolerate pressures of up to 2 lb/in2 (13.8 kN/m2),whereas a wind of 100 mi/h (161 km/h) produces a pressure of only 0.35 lb/in2(2.4 kN/m2). Air concussion is caused by the movement of a pressure wave generallycaused by one or more of three things: a direct surface energy release, a releaseof inadequately confined gases, and a shock wave from a large free face (seeFig. 8.3). Direct surface energy release is caused by detonation of an explosionon the surface. Air concussion is also referred to as “overpressure,” that is, the air pressureover and above normal air pressure. The difference between noise and concus-sion is the frequency: air blast frequencies below 20 hertz (Hz), or cycles persecond, are concussions, because they are inaudible to human ears: whereasany air blast above 20 Hz is noise, because it is audible. Overpressure can be measured in two ways: in pounds of pressure persquare inch (lb/in2) or in decibels (dB). Decibels are a measurement or expres-sion of the relative difference of power of sound waves. Pounds per square inchand decibels can be made relative by the following equation: p overpressure (dB) 20 log (8.68) p0where overpressure (dB) overpressure in decibels log the common logarithm p overpressure in lb/in2 p0 3 10–9 lb/in2
• 204 CHAPTER EIGHT Inadequate stemming of mudseam Surface detonation Mudseam Long free face FIGURE 8.3 Three causes of air blast. (Hemphill-Blasting operations, McGraw-Hill.)Vibration Control in Blasting*Vibrations caused by blasting are propagated with a velocity V, ft/s; frequencyf, Hz; and wavelength L, ft, related by V L (8.69) fVelocity v, in/s, of the particles disturbed by the vibrations depends on theamplitude of the vibrations A, in v 2 fA (8.70)If the velocity v1 at a distance D1 from the explosion is known, the velocity v2 ata distance D2 from the explosion may be estimated from 1.5 D1 v2 v1 (8.71) D2 *Merritt—Standard Handbook of Civil Engineering, McGraw-Hill.
• SOIL AND EARTHWORK FORMULAS 205The acceleration a, in/s2, of the particles is given by a 4 2f 2A (8.72)For a charge exploded on the ground surface, the overpressure P, psi, may becomputed from W1 3 1.407 P 226.62 (8.73) Dwhere W maximum weight of explosives, lb per delay D distance, ft, from explosion to exposureThe sound pressure level, decibels, may be computed from 0.084 P dB 28 (8.74) 6.95 10For vibration control, blasting should be controlled with the scaled-distanceformula: β D v H (8.75) Wwhere constant (varies for each site) H constant (varies for each site)Distance to exposure, ft, divided by the square root of maximum pounds perdelay is known as scaled distance.
• CHAPTER 9 BUILDING AND STRUCTURES FORMULASLOAD-AND-RESISTANCE FACTORDESIGN FOR SHEAR IN BUILDINGSBased on the American Institute of Steel Construction (AISC) specifications forload-and-resistance factor design (LRFD) for buildings, the shear capacity Vu,kip (kN 4.448 kip), of flexural members may be computed from the fol-lowing: h Vu 0.54Fyw Aw when (9.1) tw 0.54 Fyw Aw h Vu when 1.25 (9.2) h/tw tw 23,760kAw h Vu when 1.25 (9.3) (h/t w)2 twwhere Fyw specified minimum yield stress of web, ksi (MPa 6.894 ksi) 187 2k /Fyw Aw web area, in2 (mm2) dtw k 5 if a/h exceeds 3.0 or 67,600/(h/tw)2, or if stiffeners are not required 5 5/(a /h)2, otherwise Stiffeners are required when the shear exceeds Vu. In unstiffened girders,h/tw may not exceed 260. In girders with stiffeners, maximum h/tw permitted is2,000/ Fyf for a/h 1.5 or 14,000/ Fyf (Fyf 16.5) for a/h > 1.5, where Fyfis the specified minimum yield stress, ksi, of the flange. For shear capacity withtension-field action, see the AISC specification for LRFD. 207
• 208 CHAPTER NINEALLOWABLE-STRESS DESIGN FORBUILDING COLUMNSThe AISC specification for allowable-stress design (ASD) for buildings pro-vides two formulas for computing allowable compressive stress Fa, ksi(MPa), for main members. The formula to use depends on the relationshipof the largest effective slenderness ratio Kl/r of the cross section of anyunbraced segment to a factor Cc defined by the following equation andTable 9.1: B Fy 2 2E 756.6 Cc (9.4) Fywhere E modulus of elasticity of steel 29,000 ksi (128.99 GPa) Fy yield stress of steel, ksi (MPa)When Kl/r is less than Cc , (Kl/r)2 1 Fy 2C2 Fa c (9.5) F.S. 5 3(Kl/r) (Kl/r)3where F.S. safety factor . 3 8Cc 8C3cWhen Kl /r exceeds Cc, 12 2E 150,000 Fa (9.6) 23(Kl/r)2 (Kl/r)2 The effective-length factor K, equal to the ratio of effective-column lengthto actual unbraced length, may be greater or less than 1.0. Theoretical K valuesfor six idealized conditions, in which joint rotation and translation are eitherfully realized or nonexistent, are tabulated in Fig. 9.1. TABLE 9.1 Values of Cc Fy Cc 36 126.1 50 107.0
• BUILDING AND STRUCTURES FORMULAS 209 (a) (b) (c) (d) (e) (f ) Buckled shape of column is shown by dashed line Theoretical K value 0.5 0.7 1.0 1.0 2.0 2.0 Recommended design value when 0.65 0.80 1.2 1.0 2.10 2.0 ideal conditions are approximated Rotation fixed and translation fixed Rotation free and translation fixed End condition Rotation fixed and translation free Rotation free and translation freeFIGURE 9.1 Values of effective-length factor K for columns.LOAD-AND-RESISTANCE FACTORDESIGN FOR BUILDING COLUMNSPlastic analysis of prismatic compression members in buildings is permitted if Fy (l/r) does not exceed 800 and Fu 65 ksi (448 MPa). For axially loadedmembers with b/t r, the maximum load Pu, ksi (MPa 6.894 ksi), may becomputed from Pu 0.85AgFcr (9.7)where Ag gross cross-sectional area of the member Fcr 0.658 Fy for 2.25 0.877 Fy / for 2.25 (Kl/r)2(Fy /286,220)The AISC specification for LRFD presents formulas for designing memberswith slender elements.ALLOWABLE-STRESS DESIGN FOR BUILDING BEAMSThe maximum fiber stress in bending for laterally supported beams and girdersis Fb 0.66Fy if they are compact, except for hybrid girders and members withyield points exceeding 65 ksi (448.1 MPa). Fb 0.60Fy for noncompact
• 210 CHAPTER NINE TABLE 9.2 Allowable Bending Stresses in Braced Beams for Buildings Yield strength, Compact, Noncompact, ksi (MPa) 0.66Fy (MPa) 0.60Fy (MPa) 36 (248.2) 24 (165.5) 22 (151.7) 50 (344.7) 33 (227.5) 30 (206.8)sections. Fy is the minimum specified yield strength of the steel, ksi (MPa).Table 9.2 lists values of Fb for two grades of steel. The allowable extreme-fiber stress of 0.60Fy applies to laterally supported,unsymmetrical members, except channels, and to noncompact box sections.Compression on outer surfaces of channels bent about their major axis shouldnot exceed 0.60Fy or the value given by Eq. (9.12). The allowable stress of 0.66Fy for compact members should be reduced to0.60Fy when the compression flange is unsupported for a length, in (mm),exceeding the smaller of 76.0bf lmax (9.8) Fy 20,000 lmax (9.9) Fyd/Afwhere bf width of compression flange, in (mm) d beam depth, in (mm) Af area of compression flange, in2 (mm2)The allowable stress should be reduced even more when l/rT exceeds certainlimits, where l is the unbraced length, in (mm), of the compression flange, andrT is the radius of gyration, in (mm), of a portion of the beam consisting of thecompression flange and one-third of the part of the web in compression. For 102,000Cb /Fy l/rT 510,00Cb /Fy , use 2 Fy(l/rT)2 Fb Fy (9.10) 3 1,530,000CbFor l/rT 510,000Cb /Fy , use 170,000Cb Fb (9.11) (l/rT )2where Cb modifier for moment gradient (Eq. 9.13).
• BUILDING AND STRUCTURES FORMULAS 211 When, however, the compression flange is solid and nearly rectangular incross section, and its area is not less than that of the tension flange, the allow-able stress may be taken as 12,000Cb Fb (9.12) ld/AfWhen Eq. (9.12) applies (except for channels), Fb should be taken as the largerof the values computed from Eqs. (9.12) and (9.10) or (9.11), but not more than0.60Fy. The moment-gradient factor Cb in Eqs. (9.8) to (9.12) may be computed from 2 M1 M1 Cb 1.75 1.05 0.3 2.3 (9.13) M2 M2where M1 smaller beam end moment and M2 larger beam end moment. The algebraic sign of M1 /M2 is positive for double-curvature bending andnegative for single-curvature bending. When the bending moment at any pointwithin an unbraced length is larger than that at both ends, the value of Cbshould be taken as unity. For braced frames, Cb should be taken as unity forcomputation of Fbx and Fby. Equations (9.11) and (9.12) can be simplified by introducing a new term: (l/rT)2Fy Q (9.14) 510,000CbNow, for 0.2 Q 1, (2 Q)Fy (9.15) Fb 3For Q 1: Fy Fb (9.16) 3Q As for the preceding equations, when Eq. (9.8) applies (except for chan-nels), Fb should be taken as the largest of the values given by Eqs. (9.8) and(9.15) or (9.16), but not more than 0.60Fy.LOAD-AND-RESISTANCE FACTOR DESIGNFOR BUILDING BEAMSFor a compact section bent about the major axis, the unbraced length Lb of thecompression flange, where plastic hinges may form at failure, may not exceedLpd , given by Eqs. (9.17) and (9.18) that follow. For beams bent about theminor axis and square and circular beams, Lb is not restricted for plastic analysis.
• 212 CHAPTER NINE For I-shaped beams, symmetrical about both the major and the minor axis orsymmetrical about the minor axis but with the compression flange larger thanthe tension flange, including hybrid girders, loaded in the plane of the web: 3600 2200(M1/Mp) Lpd ry (9.17) Fycwhere Fyc minimum yield stress of compression flange, ksi (MPa) M1 smaller of the moments, in kip (mm MPa) at the ends of the unbraced length of beam Mp plastic moment, in kip (mm MPa) ry radius of gyration, in (mm), about minor axisThe plastic moment Mp equals Fy Z for homogeneous sections, where Z plas-tic modulus, in3 (mm3); and for hybrid girders, it may be computed from thefully plastic distribution. M1/Mp is positive for beams with reverse curvature. For solid rectangular bars and symmetrical box beams: 5000 3000(M1/Mp) ry Lpd ry 3000 (9.18) Fy Fy The flexural design strength 0.90Mn is determined by the limit state of lateral-torsional buckling and should be calculated for the region of the last hinge toform and for regions not adjacent to a plastic hinge. The specification gives for-mulas for Mn that depend on the geometry of the section and the bracing pro-vided for the compression flange. For compact sections bent about the major axis, for example, Mn depends onthe following unbraced lengths: Lb the distance, in (mm), between points braced against lateral displace- ment of the compression flange or between points braced to prevent twist Lp limiting laterally unbraced length, in (mm), for full plastic-bending capacity 300ry / Fyf for I shapes and channels 3750(ry /Mp)/ JA for solid rectangular bars and box beams Fyf flange yield stress, ksi (MPa) J torsional constant, in4 (mm4) (see AISC “Manual of Steel Construction” on LRFD) A cross-sectional area, in2 (mm2) Lr limiting laterally unbraced length, in (mm), for inelastic lateral buckling For I-shaped beams symmetrical about the major or the minor axis, or sym-metrical about the minor axis with the compression flange larger than the ten-sion flange and channels loaded in the plane of the web: 21 ry x1 2 Lr 1 X2 FL (9.19) Fyw Fr
• BUILDING AND STRUCTURES FORMULAS 213where Fyw specified minimum yield stress of web, ksi (MPa) Fr compressive residual stress in flange 10 ksi (68.9 MPa) for rolled shapes, 16.5 ksi (113.6 MPa), for welded sections FL smaller of Fyf Fr or Fyw ( /Sx) 2EGJA/2 Fyf specified minimum yield stress of flange, ksi (MPa) X1 X2 (4Cw /Iy) (Sx /GJ)2 E elastic modulus of the steel G shear modulus of elasticity Sx section modulus about major axis, in3 (mm3) (with respect to the compression flange if that flange is larger than the tension flange) Cw warping constant, in6 (mm6) (see AISC manual on LRFD) Iy moment of inertia about minor axis, in4 (mm4)For the previously mentioned shapes, the limiting buckling moment Mr, ksi (MPa),may be computed from Mr FL Sx (9.20) For compact beams with Lb Lr, bent about the major axis, Lb Lp Mn Cb Mp (Mp Mr) Mp (9.21) Lr Lpwhere Cb 1.75 1.05(M1 / M2) 0.3(M1 / M2) 2.3, where M1 is the smallerand M2 the larger end moment in the unbraced segment of the beam; M1/M2 ispositive for reverse curvature and equals 1.0 for unbraced cantilevers andbeams with moments over much of the unbraced segment equal to or greaterthan the larger of the segment end moments. (See Galambos, T. V., Guide toStability Design Criteria for Metal Structures, 4th ed., John Wiley & Sons, NewYork, for use of larger values of Cb.) For solid rectangular bars bent about the major axis, ry Lr 57,000 JA (9.22) Mrand the limiting buckling moment is given by: Mr Fy Sx (9.23)For symmetrical box sections loaded in the plane of symmetry and bent aboutthe major axis, Mr should be determined from Eq. (9.20) and Lr from Eq. (9.22) For compact beams with Lb > Lr, bent about the major axis, Mn M cr Cb M r (9.24)where Mcr critical elastic moment, kip in (MPa mm).
• 214 CHAPTER NINE For shapes to which Eq. (9.24) applies, Lb B y 2 E Mcr Cb EI GJ Iy Cw (9.25) LbFor solid rectangular bars and symmetrical box sections, 57,000Cb JA Mcr (9.26) Lb / ryFor determination of the flexural strength of noncompact plate girders andother shapes not covered by the preceding requirements, see the AISC manualon LRFD.ALLOWABLE-STRESS DESIGN FOR SHEAR IN BUILDINGSThe AISC specification for ASD specifies the following allowable shearstresses Fv, ksi (ksi 6.894 MPa): h 380 (9.27) Fv 0.40Fy tw Fy Cv Fy h 380 Fv 0.40Fy (9.28) 289 tw Fy 236,000kv /Fy(h/tw)where Cv 45,000kv /Fy(h/tw)2 for Cv 0.8 2 for Cv 0.8 2 kv 4.00 5.34/(a /h) for a /h 1.0 5.34 4.00/(a /h)2 for a /h 1.0 a clear distance between transverse stiffenersThe allowable shear stress with tension-field action is Fy 1 Cv Fv Cv 0.40Fy (9.29) 289 1.15√1 (a/h)2where Cv 1,When the shear in the web exceeds Fv , stiffeners are required. Within the boundaries of a rigid connection of two or more members withwebs lying in a common plane, shear stresses in the webs generally are high.The commentary on the AISC specification for buildings states that such websshould be reinforced when the calculated shear stresses, such as those along
• BUILDING AND STRUCTURES FORMULAS 215 Upper-story shear, Vs ΣF A A d1 d2 M1 M2 dc Lower-story shear FIGURE 9.2 Rigid connection of steel members with webs in a common plane.plane AA in Fig. 9.2, exceed Fv, that is, when F is larger than dc tw Fv, where dcis the depth and tw is the web thickness of the member resisting F. The shearmay be calculated from M1 M2 F Vs (9.30) 0.95d1 0.95d2where Vs shear on the section M1 M1L M1G M1L moment due to the gravity load on the leeward side of the connection M1G moment due to the lateral load on the leeward side of the connection M2 M2L M2G M2L moment due to the lateral load on the windward side of the connection M2G moment due to the gravity load on the windward side of the connectionSTRESSES IN THIN SHELLSResults of membrane and bending theories are expressed in terms of unit forcesand unit moments, acting per unit of length over the thickness of the shell. Tocompute the unit stresses from these forces and moments, usual practice is to
• 216 CHAPTER NINEassume normal forces and shears to be uniformly distributed over the shellthickness and bending stresses to be linearly distributed. Then, normal stresses can be computed from equations of the form: Nx Mx fx z (9.31) t t 3/12where z distance from middle surface t shell thickness Mx unit bending moment about an axis parallel to direction of unit nor- mal force NxSimilarly, shearing stresses produced by central shears T and twisting momentsD may be calculated from equations of the form: T D vxy z (9.32) t t 3/12 Normal shearing stresses may be computed on the assumption of a parabolicstress distribution over the shell thickness: V t2 vxz 3 z2 (9.33) t /6 4where V unit shear force normal to middle surface.BEARING PLATESTo resist a beam reaction, the minimum bearing length N in the direction of thebeam span for a bearing plate is determined by equations for prevention oflocal web yielding and web crippling. A larger N is generally desirable but islimited by the available wall thickness. When the plate covers the full area of a concrete support, the area, in2 (mm2),required by the bearing plate is R A1 (9.34) 0.35 f cwhere R beam reaction, kip (kN), fc specified compressive strength of theconcrete, ksi (MPa). When the plate covers less than the full area of the con-crete support, then, as determined from Table 9.3 2 R A1 (9.35) 0.35 fc A2where A2 full cross-sectional area of concrete support, in2 (mm2).
• BUILDING AND STRUCTURES FORMULAS 217 TABLE 9.3 Allowable Bearing Stress, Fp, on Concrete and Masonry* Full area of concrete support 0.35f c B A2 Less than full area of concrete A1 0.35f c 0.70f c support Sandstone and limestone 0.40 Brick in cement mortar 0.25 *Units in MPa 6.895 ksi. With N established, usually rounded to full inches (millimeters), the mini-mum width of plate B, in (mm), may be calculated by dividing A1 by N andthen rounded off to full inches (millimeters), so that BN A1. Actual bearingpressure fp, ksi (MPa), under the plate then is R fp (9.36) BNThe plate thickness usually is determined with the assumption of cantileverbending of the plate: B Fb 1 3fp t B k (9.37) 2where t minimum plate thickness, in (mm) k distance, in (mm), from beam bottom to top of web fillet Fb allowable bending stress of plate, ksi (MPa)COLUMN BASE PLATESThe area A1, in2 (mm2), required for a base plate under a column supported byconcrete should be taken as the larger of the values calculated from the equa-tion cited earlier, with R taken as the total column load, kip (kN), or R A1 (9.38) 0.70 fcUnless the projections of the plate beyond the column are small, the plate maybe designed as a cantilever assumed to be fixed at the edges of a rectangle withsides equal to 0.80b and 0.95d, where b is the column flange width, in (mm), andd is the column depth, in (mm).
• 218 CHAPTER NINE To minimize material requirements, the plate projections should be nearlyequal. For this purpose, the plate length N, in (mm) (in the direction of d), maybe taken as N A1 0.5(0.95d 0.80b) (9.39)The width B, in (mm), of the plate then may be calculated by dividing A1 by N.Both B and N may be selected in full inches (millimeters) so that BN A1. Inthat case, the bearing pressure fp, ksi (MPa), may be determined from the preced-ing equation. Thickness of plate, determined by cantilever bending, is given by B Fy fp t 2p (9.40)where Fy minimum specified yield strength, ksi (MPa), of plate; and plarger of 0.5(N 0.95d) and 0.5(B 0.80b). When the plate projections are small, the area A2 should be taken as the maxi-mum area of the portion of the supporting surface that is geometrically similar toand concentric with the loaded area. Thus, for an H-shaped column, the columnload may be assumed distributed to the concrete over an H-shaped area with flangethickness L, in (mm), and web thickness 2L: 4 B 1 1 4R L (d b) (d b)2 (9.41) 4 Fpwhere Fp allowable bearing pressure, ksi (MPa), on support. (If L is an imaginarynumber, the loaded portion of the supporting surface may be assumed rectangularas discussed earlier.) Thickness of the base plate should be taken as the larger of thevalues calculated from the preceding equation and B Fb 3fp t L (9.42)BEARING ON MILLED SURFACESIn building construction, allowable bearing stress for milled surfaces, includingbearing stiffeners, and pins in reamed, drilled, or bored holes, is Fp 0.90Fy,where Fy is the yield strength of the steel, ksi (MPa). For expansion rollers and rockers, the allowable bearing stress, kip/linear in(kN/mm), is Fy 13 Fp 0.66d (9.43) 20where d is the diameter, in (mm), of the roller or rocker. When parts in contacthave different yield strengths, Fy is the smaller value.
• BUILDING AND STRUCTURES FORMULAS 219PLATE GIRDERS IN BUILDINGSFor greatest resistance to bending, as much of a plate girder cross section aspracticable should be concentrated in the flanges, at the greatest distance fromthe neutral axis. This might require, however, a web so thin that the girderwould fail by web buckling before it reached its bending capacity. To precludethis, the AISC specification limits h/t. For an unstiffened web, this ratio should not exceed h 14,000 t Fy (Fy 16.5) (9.44)where Fy yield strength of compression flange, ksi (MPa). Larger values of h/t may be used, however, if the web is stiffened at appro-priate intervals. For this purpose, vertical angles may be fastened to the web or vertical plateswelded to it. These transverse stiffeners are not required, though, when h/t is lessthan the value computed from the preceding equation or Table 9.4. With transverse stiffeners spaced not more than 1.5 times the girder depthapart, the web clear-depth/thickness ratio may be as large as h 2000 (9.45) t Fy If, however, the web depth/thickness ratio h/t exceeds 760 / Fb, where Fb,ksi (MPa), is the allowable bending stress in the compression flange that wouldordinarily apply, this stress should be reduced to Fb , given by the followingequations: Fb RPG Re Fb (9.46) A h 760 RPG 1 0.0005 w 1.0 (9.47) Af t Fb 3 12 (Aw /Af)(3 ) Re 1.0 (9.48) 12 2(Aw /Af) TABLE 9.4 Critical h/t for Plate Girders in Buildings 2Fy(Fy 2Fy 14,000 2,000 Fy, ksi (MPa) 16.5) 36 (248) 322 333 50 (345) 243 283
• 220 CHAPTER NINEwhere Aw web area, in2 (mm2) Af area of compression flange, in2 (mm2) 0.6Fyw /Fb 1.0 Fyw minimum specified yield stress, ksi, (MPa), of web steel In a hybrid girder, where the flange steel has a higher yield strength than theweb, the preceding equation protects against excessive yielding of the lowerstrength web in the vicinity of the higher strength flanges. For nonhybrid gird-ers, Re 1.0.LOAD DISTRIBUTION TO BENTS AND SHEAR WALLSProvision should be made for all structures to transmit lateral loads, such asthose from wind, earthquakes, and traction and braking of vehicles, to foun-dations and their supports that have high resistance to displacement. For thispurpose, various types of bracing may be used, including struts, tension ties,diaphragms, trusses, and shear walls.Deflections of Bents and Shear WallsHorizontal deflections in the planes of bents and shear walls can be computedon the assumption that they act as cantilevers. Deflections of braced bents canbe calculated by the dummy-unit-load method or a matrix method. Deflectionsof rigid frames can be computed by adding the drifts of the stories, as deter-mined by moment distribution or a matrix method. For a shear wall (Fig. 9.3), the deflection in its plane induced by a load in itsplane is the sum of the flexural deflection as a cantilever and the deflection dueto shear. Thus, for a wall with solid rectangular cross section, the deflection atthe top due to uniform load is 3 1.5wH H H (9.49) Et L Lwhere w uniform lateral load H height of the wall E modulus of elasticity of the wall material t wall thickness L length of wallFor a shear wall with a concentrated load P at the top, the deflection at the top is 3 4P H H c 0.75 (9.50) Et L LIf the wall is fixed against rotation at the top, however, the deflection is 3 P H H f 3 (9.51) Et L L
• BUILDING AND STRUCTURES FORMULAS 221 Shear walls Spandrel beam Columns Wind bent (a) (b) (c) (d) (e) (f) (g) FIGURE 9.3 Building frame resists lateral forces with (a) wind bents or (b) shear walls or a combination of the two. Bents may be braced in any of several ways, including (c) X bracing, (d ) K bracing, (e) inverted V bracing, ( f ) knee bracing, and (g) rigid connections. Units used in these equations are those commonly applied in United StatesCustomary System (USCS) and the System International (SI) measurements,that is, kip (kN), lb/in2 (MPa), ft (m), and in (mm). Where shear walls contain openings, such as those for doors, corridors, or win-dows, computations for deflection and rigidity are more complicated. Approxi-mate methods, however, may be used.COMBINED AXIAL COMPRESSIONOR TENSION AND BENDINGThe AISC specification for allowable stress design for buildings includes threeinteraction formulas for combined axial compression and bending. When the ratio of computed axial stress to allowable axial stress fu /Faexceeds 0.15, both of the following equations must be satisfied: fa Cmx fbx Cmy fby 1 (9.52) Fa (1 fa /Fex)Fbx (1 fa /Fey)Fby fa fbx fby 1 (9.53) 0.60Fy Fbx Fby
• BUILDING AND STRUCTURES FORMULAS 223 N k N + 5k Toe of fillet N + 2.5k k N FIGURE 9.4 For investigating web yielding, stresses are assumed to be distributed over lengths of web indicated at the bearings, where N is the length of bearing plates, and k is the distance from outer surface of beam to the toe of the fillet. To prevent web crippling, the AISC specification requires that bearing stiff-eners be provided on webs where concentrated loads occur when the compres-sive force exceeds R, kip (kN), computed from the following:For a concentrated load applied at a distance from the beam end of at least d/2,where d is the depth of beam, 2Fyw tf /tw 1.5 N tw R 67.5t2 1 w 3 (9.57) d tfwhere tf flange thickness, in (mm).For a concentrated load applied closer than d/2 from the beam end, 2Fyw tf /tw 1.5 N tw R 34t2 1 w 3 (9.58) d tfIf stiffeners are provided and extend at least one-half of the web, R need not becomputed. Another consideration is prevention of sidesway web buckling. The AISCspecification requires bearing stiffeners when the compressive force from a con-centrated load exceeds limits that depend on the relative slenderness of web andflange rwf and whether or not the loaded flange is restrained against rotation: dc /tw rwf (9.59) l /bfwhere l largest unbraced length, in (mm), along either top or bottom flange at point of application of load bf flange width, in (mm) dc web depth clear of fillets d 2k
• 224 CHAPTER NINEStiffeners are required if the concentrated load exceeds R, kip (kN), computedfrom 6800t3 w R (1 0.4r3 ) wf (9.60) hwhere h clear distance, in (mm), between flanges, and rwf is less than 2.3when the loaded flange is restrained against rotation. If the loaded flange is notrestrained and rwf is less than 1.7, 6800t3 w R 0.4r3 wf (9.61) hR need not be computed for larger values of rwf.DESIGN OF STIFFENERS UNDER LOADSAISC requires that fasteners or welds for end connections of beams, girders,and trusses be designed for the combined effect of forces resulting frommoment and shear induced by the rigidity of the connection. When flanges ormoment-connection plates for end connections of beams and girders are weldedto the flange of an I- or H-shape column, a pair of column-web stiffeners havinga combined cross-sectional area Ast not less than that calculated from the follow-ing equations must be provided whenever the calculated value of Ast is positive: Pbf Fyc twc(tb 5K) Ast (9.62) Fystwhere Fyc column yield stress, ksi (MPa) Fyst stiffener yield stress, ksi (MPa) K distance, in (mm), between outer face of column flange and web toe of its fillet, if column is rolled shape, or equivalent distance if col- umn is welded shape Pbf computed force, kip (kN), delivered by flange of moment- 5 connection plate multiplied by 3, when computed force is due to 4 live and dead load only, or by 3, when computed force is due to live and dead load in conjunction with wind or earthquake forces twc thickness of column web, in (mm) tb thickness of flange or moment-connection plate delivering concen- trated force, in (mm) Notwithstanding the preceding requirements, a stiffener or a pair of stiffen-ers must be provided opposite the beam-compression flange when the column-web depth clear of fillets dc is greater than 4100t3 wc Fyc dc (9.63) Pbf
• BUILDING AND STRUCTURES FORMULAS 225and a pair of stiffeners should be provided opposite the tension flange when thethickness of the column flange tf is less than B Fyc Pbf tf 0.4 (9.64)Stiffeners required by the preceding equations should comply with the follow-ing additional criteria:1. The width of each stiffener plus half the thickness of the column web should not be less than one-third the width of the flange or moment-connection plate delivering the concentrated force.2. The thickness of stiffeners should not be less than tb /2.3. The weld-joining stiffeners to the column web must be sized to carry the force in the stiffener caused by unbalanced moments on opposite sides of the column.FASTENERS IN BUILDINGSThe AISC specification for allowable stresses for buildings specifies allowableunit tension and shear stresses on the cross-sectional area on the unthreadedbody area of bolts and threaded parts. (Generally, rivets should not be used indirect tension.) When wind or seismic loads are combined with gravity loads,the allowable stresses may be increased by one-third. Most building construction is done with bearing-type connections. Allowablebearing stresses apply to both bearing-type and slip-critical connections. In build-ings, the allowable bearing stress Fp, ksi (MPa), on projected area of fasteners is Fp 1.2Fu (9.65)where Fu is the tensile strength of the connected part, ksi (MPa). Distance mea-sured in the line of force to the nearest edge of the connected part (end distance)should be at least 1.5d, where d is the fastener diameter. The center-to-centerspacing of fasteners should be at least 3d.COMPOSITE CONSTRUCTIONIn composite construction, steel beams and a concrete slab are connected sothat they act together to resist the load on the beam. The slab, in effect, servesas a cover plate. As a result, a lighter steel section may be used.Construction in BuildingsThere are two basic methods of composite construction.Method 1 The steel beam is entirely encased in the concrete. Composite actionin this case depends on the steel-concrete bond alone. Because the beam is
• 226 CHAPTER NINEcompletely braced laterally, the allowable stress in the flanges is 0.66Fy, whereFy is the yield strength, ksi (MPa), of the steel. Assuming the steel to carry thefull dead load and the composite section to carry the live load, the maximum unitstress, ksi (MPa), in the steel is MD ML fs 0.66Fy (9.66) Ss Strwhere MD dead-load moment, in kip (kN mm) ML live-load moment, in kip (kN mm) Ss section modulus of steel beam, in3 (mm3) Str section modulus of transformed composite section, in3 (mm3) An alternative, shortcut method is permitted by the AISC specification. Itassumes that the steel beam carries both live and dead loads and compensatesfor this by permitting a higher stress in the steel: MD ML fs 0.76Fy (9.67) SsMethod 2 The steel beam is connected to the concrete slab by shear connectors.Design is based on ultimate load and is independent of the use of temporaryshores to support the steel until the concrete hardens. The maximum stress in thebottom flange is MD ML fs 0.66Fy (9.68) Str To obtain the transformed composite section, treat the concrete above theneutral axis as an equivalent steel area by dividing the concrete area by n, theratio of modulus of elasticity of steel to that of the concrete. In determination ofthe transformed section, only a portion of the concrete slab over the beam maybe considered effective in resisting compressive flexural stresses (positive-moment regions). The width of slab on either side of the beam centerline thatmay be considered effective should not exceed any of the following:1. One-eighth of the beam span between centers of supports2. Half the distance to the centerline of the adjacent beam3. The distance from beam centerline to edge of slab (Fig. 9.5)NUMBER OF CONNECTORS REQUIREDFOR BUILDING CONSTRUCTIONThe total number of connectors to resist Vh is computed from Vh /q, where q isthe allowable shear for one connector, kip (kN). Values of q for connectors inbuildings are given in structural design guides. The required number of shear connectors may be spaced uniformly betweenthe sections of maximum and zero moment. Shear connectors should have at
• BUILDING AND STRUCTURES FORMULAS 227 C ≤ 1 Span 8 C C Concrete ≤1a 2 slab a a FIGURE 9.5 Limitations on effective width of concrete slab in a composite steel-concrete beam.least 1 in (25.4 mm) of concrete cover in all directions; and unless studs arelocated directly over the web, stud diameters may not exceed 2.5 times thebeam-flange thickness. With heavy concentrated loads, the uniform spacing of shear connectorsmay not be sufficient between a concentrated load and the nearest point ofzero moment. The number of shear connectors in this region should be at least N1[(M /Mmax) 1] N2 (9.69) 1where M moment at concentrated load, ft kip (kN m) Mmax maximum moment in span, ft kip (kN m) N1 number of shear connectors required between Mmax and zero moment Str /Ss or Seff /Ss, as applicable Seff effective section modulus for partial composite action, in3 (mm3)Shear on ConnectorsThe total horizontal shear to be resisted by the shear connectors in building con-struction is taken as the smaller of the values given by the following two equations: 0.85 fc Ac Vh (9.70) 2 As Fy Vh (9.71) 2where Vh total horizontal shear, kip (kN), between maximum positive moment and each end of steel beams (or between point of maximum posi- tive moment and point of contraflexure in continuous beam) fc specified compressive strength of concrete at 28 days, ksi (MPa) Ac actual area of effective concrete flange, in2 (mm2) As area of steel beam, in2 (mm2)
• 228 CHAPTER NINE In continuous composite construction, longitudinal reinforcing steel may beconsidered to act compositely with the steel beam in negative-moment regions.In this case, the total horizontal shear, kip (kN), between an interior support andeach adjacent point of contraflexure should be taken as Asr Fyr Vh (9.72) 2where Asr area of longitudinal reinforcement at support within effective area,in2 (mm2); and Fyr specified minimum yield stress of longitudinal reinforce-ment, ksi (MPa).PONDING CONSIDERATIONS IN BUILDINGSFlat roofs on which water may accumulate may require analysis to ensure thatthey are stable under ponding conditions. A flat roof may be considered stableand an analysis does not need to be made if both of the following two equationsare satisfied: Cp 0.9Cs 0.25 (9.73) Id 25S4/106 (9.74)where Cp 32Ls L4 /107Ip p Cs 32SL4 /107Is s Lp length, ft (m), of primary member or girder Ls length, ft (m), of secondary member or purlin S spacing, ft (m), of secondary members Ip moment of inertia of primary member, in4 (mm4) Is moment of inertia of secondary member, in4 (mm4) Id moment of inertia of steel deck supported on secondary members, in4/ft (mm4/m)For trusses and other open-web members, Is should be decreased 15 percent. Thetotal bending stress due to dead loads, gravity live loads, and ponding should notexceed 0.80Fy, where Fy is the minimum specified yield stress for the steel.LIGHTWEIGHT STEEL CONSTRUCTIONLightweight steel construction is popular today for a variety of building compo-nents (floors, roofs, walls. etc.) and for entire buildings.* Such constructiontakes many forms, and a number of different designs of structural members andframing systems have been developed. *From Merritt—Building Construction Handbook, McGraw-Hill. Formulas in this section arebased on the work of F. E. Fahy, who, at the time of their preparation, was Chief Engineer, ProductEngineering, Bethlehem Steel Company.
• BUILDING AND STRUCTURES FORMULAS 229 w w t t w t w t Stiffening lip Unstiffened Stiffened element element (a) (b) FIGURE 9.6 Typical lightweight steel compression elements. (Merritt–Building Construction Handbook, McGraw-Hill.) For structural design computations, flat compression elements of cold-formed structural members can be divided into two kinds—stiffened elementsand unstiffened elements. Stiffened compression elements are flat compressionelements; that is, plane compression flanges of flexural members and planewebs and flanges of compression members, of which both edges parallel to thedirection of stress are stiffened by a web, flange, stiffening lip, or the like (AISISpecification for the Design of Light Gage Steel Structural Members). A flatelement stiffened at only one edge parallel to the direction of stress is called anunstiffened element. If the sections shown in Fig. 9.6 are used as compressionmembers, the webs are considered stiffened compression elements. In order that a compression element may qualify as a stiffened compressionelement, its edge stiffeners should comply with the following: B 2 w Imin 1.83t4 144 (9.75) tbut not less than 9.2t4.where w/t flat-width ratio of stiffened element Imin minimum allowable moment of inertia of stiffener (of any shape) about its own centroidal axis parallel to stiffened element Where the stiffener consists of a simple lip bent at right angles to the stiff-ened element, the required overall depth d of such a lip may be determined withsatisfactory accuracy from the following formula: B t 2 6 w d 2.8t 144 (9.76)
• 230 CHAPTER NINEbut not less than 4.8t. A simple lip should not be used as an edge stiffener forany element having a flat-width ratio greater than 60. For safe-load determination, that is, in computing effective area and sectionmodulus: w 4,020 (9.77) t lim fwhere f computed unit stress in psi in the element based upon effective width. Equation (9.77) is based on a safety factor of about 1.65 against yield stressat the outer fiber of the section. For any other safety factor m multiply the right-hand side of Eq. (9.77) by 1.65/m. For deflection determinations, that is, in computing moment of inertia to beused in deflection calculations or other calculations involving stiffness: w 5,160 (9.78) t lim f For a flat-width ratio w/t greater than 10 but not over 25, the compressivestress should not exceed: 5 1 w fc f 8,640 (f 12,950) (9.79) 3 b 15 b twhere fb basic design stress, not to exceed 30,000 psi w width of element t thicknessIf fb 20,000 psi, this formula reduces to w fc 24,700 470 (9.80) tFigure 9.7 shows a number compression elements used in lightweight steel con-struction. For w/t greater than 25, not over 60, the compressive stress should notexceed fc as given by Eqs. (9.79) and (9.80). For angle struts, 8,090,000 fc (9.81) (w/t)2For other sections, w fc 20,000 282 (9.82) tStiffened Light-gage Elements Subject to Local Bucking Compute sectionproperties based on an effective width of each stiffened compression element(Fig. 9.8). Determine the effective width b by means of Eqs. (9.83) and(9.84).
• BUILDING AND STRUCTURES FORMULAS 231 FIGURE 9.7 Typical compression elements used in light- weight steel construction. (Merritt–Building Construction Handbook, McGraw-Hill.) For safe-load determinations, that is, in computing effective area and sectionmodulus b 8,040 2,010 1 (9.83) t f (w/t) fwhere f unit stress in psi in the element computed on basis of reduced section, psi (MPa) w width of the element, in (mm) t thickness, in (mm)
• 232 CHAPTER NINE w w w w 1b 1b 1b 1b 1b 1b 1b 1b 2 2 2 2 2 2 2 2 t t t t (a) (b) (c) w w 1b 1b 1b 1b 2 2 2 2 t (d) Beams top flange in compression w1 w1 w1 w1 1b 1b 1b 1b 1b 1b 1b 1b 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1b 1b 1b 2 2 2 2 2 2 w2w2 w2 1b 1b 1b 2 2 t 2 2 t t 2 2 t (e) (f) (g) Columns effective area for computing column factor QaFIGURE 9.8 Effective width of stiffened compression elements. (Merritt–Building Con-struction Handbook, McGraw-Hill.)
• BUILDING AND STRUCTURES FORMULAS 233 For deflection determinations, that is, in computing moment of inertia to beused in deflection calculations or in other calculations involving stiffness, b 10,320 2,580 1 (9.84) t f (w t) fwith f, w, and t the same as for Eq. (9.83).Web Stresses in Light-gage Members The shear on the gross web section inlight-gage cold-formed flexural members is usually limited to two-thirds of thebasic working stress in tension and bending, or 64,000,000 Shear, psi,v (9.85) (h/t)2whichever is smaller. In this expression h/t is the ratio of depth to web thick-ness, or of stiffener spacing to web thickness, as for plate-girder construction.Where the web consists of two sheets, as in the case of two channels fastenedback to back to form an I section, each sheet should be considered as a separateweb carrying its share of the shear. Maximum End Reaction or Concentrated Load on End of Cantilever for Single Unreinforced Web of Light-gage Steel (Grade C steel, corner radii = thickness of material) Pmax (1) BOverall depth Overall h h H depth B H Pmax (1) t = web thickness, mm h = clear distance between flanges, mm B = length of bearing, mm H = overall depth, mm B B h h Pmax(1) 100t2 980 42 0.22 0.11 (9.86) t t t tIn solving this formula, B should not be assigned any value greater than h.
• 234 CHAPTER NINE Maximum Interior Reaction or Concentrated Load on Single Unreinforced Web of Light-gage Steel (Grade C steel, corner radii = thickness of material) Pmax (2) Pmax (2) B X X BOverall depth h Overall h H depth X H Pmax (1) Pmax (2) t = web thickness, mm h = clear distance between flanges, mm B = length of bearing, mm H = overall depth, mm B B h h Pmax(2) 100t2 3,050 23 0.09 5 (9.87) t t t t NOTE: In solving this formula, B should not be assigned any value greaterthan h. This formula applies only where the distance x is greater than 1.5h. Other-wise, Eq. (9.86) governs. Maximum Reactions and Concentrated Loads Bearing on Restrained Web of Light-gage Steel (Grade C steel, corner radii = thickness of material) Pmax (3) B t h h B t Pmax (3) t = web thickness (each web sheet), mm h = clear distance between flanges, mm Bt B Pmax(3) 20,000t2 7.4 0.93 (9.88)
• BUILDING AND STRUCTURES FORMULAS 235 Pmax (4) Pmax (4) B X B t h h X t Pmax (4) Pmax (4) t = web thickness (each web sheet), mm h = clear distance between flanges, mm Bt B P max(4) 20,000t2 11.1 2.41 (9.89)Values of P max (4) apply only where the distance x is greater than 1.5h. Other-wise use P max (3). The effective length of bearing B for substitution in the above formulasshould not be taken as greater than h. The column-design formulas recommended by American Iron and SteelInstitute specifications for light-gage cold-formed sections consist of a familyof Johnson parabolas all tangent to a single Euler curve. It can be shown that aninfinite number of such parabolas can be drawn, all having the form P fy fy2 L 2 (9.90) A C D rand all tangent to a single Euler curve represented by P D/4C 2 (9.91) A (L/r)2In Eqs. (9.90) and (9.91), C and D are constants that depend on the safety factoreccentricity allowance, and end fixity, and fy is the yield point of the material.Observe that fy does not appear in Eq. (9.91). The point of tangency between theequations is always at a P/A value equal to half the initial value (fy /2C). If a form factor or bucking factor Q is introduced such that fcr Q (9.92) fywhere fcr represents the reduced strength of the section due to the presence of widethin elements that buckle locally at stresses below the yield point of the material,then it is a simple matter to transform Eq. (9.90) into a consistent set of columncurves applicable to any value of Q, all tangent to a single curve of Eq. (9.91): P fy fy2 2 L 2 Q Q (9.93) A C D r
• 236 CHAPTER NINE (When the effective-width treatment of stiffened elements is used, Q can bedefined as the ratio between effective area and total area of the cross section,and A in the quantity P/A can mean total area of section.) Since fcr can never be greater than fy, the value of Q as a form factor orbuckling factor can never exceed 1.0. For any section that does not contain anyfor unstiffened elements and 4,020/ 2f for stiffened elements), Qelement with a flat-width ratio exceeding that for full effectiveness (w/t 10 1.0 anddisappears from the equation.Unit Stress for Axially Loaded Cold-formed Members For cold-formed,axially loaded compression members of Grade C steel, the allowable unit stressFor L/r less than 132/ 2Q,P/A shall be 2 P L 17,000 0.485Q2 (9.94) A rFor L/r greater than 132/ 2Q, P 149,000,000 (9.95) A (L/r)2(AISI) Specification for the Design of Light Gage Steel Structural Members)where P total allowable load, lb A full, unreduced cross-sectional area of member, sq in (mm2) L unsupported length of member, in (mm) r radius of gyration of full, unreduced cross section, in (mm) Q a factor determined as follows: a. For members composed entirely of stiffened elements, Q is the ratio between the effective design area, as determined from the effective-design widths of such elements, and the full or gross area of the cross section. The effective design area used in deter- mining Q is to be based on the basic design stress allowed in ten- sion and bending—20,000 psi for Grade C steel. b. For members composed entirely of unstiffened elements, Q is the ratio between the allowable compression stress for the weakest element of the cross section (the element having the largest flat- width ratio) and the basic design stress. c. For members composed of both stiffened and unstiffened ele- ments, the factor Q is to be the product of a stress factor Q, com- puted as outlined in b above and an area factor Qa computed as outlined in a above. However, the stress on which Qa is to be based shall be that value of the unit stress fc used in computing Qs; and the effective area to be used in computing Qa shall include the full area of all unstiffened elements. It is recommended that L/r not exceed 200, except that during construction a value of 300 may be allowed.
• BUILDING AND STRUCTURES FORMULAS 239 For continuous attachment, use n equal to four times the attached lengthalong each faying surface.Bolting Light-gage Members Bolting is employed as a common means ofmaking field connections in light-gage steel construction. The AISI Specificationfor the Design of Light-gage Steel Structural Members requires that the distancebetween bolt centers, in line of stress, and the distance from bolt center to edgeof sheet, in line of stress, shall not be less than 11/2 times the bolt diameter nor P (9.103) fbtwhere P load on bolt, lb (N) t thickness of thinnest connected sheet, in (mm) fb basic design stress, psi (MPa)That specification also recommends a limit of 3.5fb for the bearing stress, and amaximum allowable tension stress on net section of d 0.1 3 fb (9.104) swhere d bolt diameter, in (mm) s spacing perpendicular to line of stress, in (mm)CHOOSING THE MOST ECONOMIC STRUCTURAL STEEL*Structural steel is available in different strengths and grades. So when choosingsteel members for a structure, the designer must compare their relative costbased on cross-sectional areas and prices. For two tension members of the samelength but of different steel strengths, their material-cost ratio C2/C1 is: C2 A2 p2 (9.105) C1 A1 p1where A1 and A2 are the cross-sectional areas and p1 and p2 are the materialprices per unit weight. If the members are designed to carry the same load at astress that is a fixed percentage of the yield point, the cross-sectional areas areinversely proportional to the yield stresses. Therefore, their relative materialcost can be expressed as C2 Fy1 p2 (9.106) C1 Fy2 p1where Fy1 and Fy2 are the yield stresses of the two steels. The ratio p2/p1 is therelative price factors. Values of this factor for several steels are given in engi-neering handbooks.* *Brockenbrough and Merritt, Structural Steel Designer’s Handbook, McGraw-Hill.
• 240 CHAPTER NINEBeams The optimal section modulus for an elastically designed I-shaped beamresults when the area of both flanges equals half the total cross-sectional area ofthe member. Assume now two members made of steels having different yieldpoints and designed to carry the same bending moment, each beam beinglaterally braced and proportioned for optimal section modulus. Their relativeweight W2/W1 and relative cost C2/C1 are influenced by the web depth-to-thickness ratio d/t. For example, if the two members have the same d/t values,such as a maximum value imposed by the manufacturing process for rolledbeams, the relationships are 2/3 W2 Fy1 (9.107) W1 Fy2 2/3 C2 p2 Fy1 (9.108) C1 p1 Fy2If each of the two members has the maximum d/t value that precludes elasticweb buckling, a condition of interest in designing fabricated plate girders, therelationships are 1/2 W2 Fy1 (9.109) W1 Fy2 1/2 C2 p2 Fy1 (9.110) C1 p1 Fy2Relative steel costs and weights are given in engineering handbooks.* In makingthese calculations, the designer must remember that members with the same bend-ing moment must be compared. Further, the relative costs of girders used for longspans, may be considerably different from those for conventional structural steel.Columns For columns, the relative material cost for two columns of differentsteels carrying the same load1 is: C2 Fc1 p2 Fc1/p1 (9.111) C1 Fc2 p1 Fc2/p2where Fc1 and Fc2 are the column buckling stresses for the two members. Rela-tive costs for various structural steels are given in Brockenbrough and Merritt,Structural Steel Designer’s Handbook, McGraw-Hill.STEEL CARBON CONTENT AND WELDABILITYThe chemical composition of steel and its welability are often expressed interms of the carbon equivalent of the structural steel, given by* *Brockenbrough and Merritt, Structural Steel Designer’s Handbook, McGraw-Hill.
• BUILDING AND STRUCTURES FORMULAS 241 Mn (Cr Mo V) (Ni Cu) Ceq C (9.112) 6 5 15where C carbon content, % Mn manganese content, % Cr chromium content, % Mo molybdenum, % V vanadium, % Ni nickel content, % Cu copper, %To prevent underbead cracking, the weld must be cooled at an acceptable rate.A higher carbon equivalent requires a longer cooling rate. Refer to ASTM 6 forfurther requirements.STATICALLY INDETERMINATE FORCES AND MOMENTSIN BUILDING STRUCTURES*Shear, moment, and deflection formulas for statically indeterminate forcesin rigid frames as present by Roark have the following notation: W load,lb (kg); w unit load, lb/linear in (kg/cm); M is positive when clockwise;V is positive when upward; y is positive when upward. Constrainingmoments, applied couples, loads, and reactions are positive when acting asshown. All forces are in pounds; all moments are in inch-pounds; all deflec-tions and dimensions are in inches. is in radians and tan . Imoment of inertia of the section of the beam with respect to the neutral axisof the beam, in4 (mm4). With pinned supports, concentrated load on the horizontal member, Fig. 9.9: 1 L1L2b 2L2L3b 3L2b2 (b3/L3)(L1 L2) H W (9.113) 2 L1L2L3 L1L3 L2L3 L1 (I3/I1) L3(I3/I2) 2 2 3 2 wb H(L2 L1) V1 (9.114) L3 With pinned supports, concentrated load on one vertical member, Fig. 9.10: 2L3 1 2a3 3ab2 2L2L3 1 2L1L3b L1L2L3 L2L3b I1 I1 I1 I3 I3 I3 I3H2 W 2L3 1 2 2L2L3 2L1L2L3 2L3 1 2L2L3 1 I2 I3 I3 I1 I3 (9.115) Wa H2(L1 L2) V (9.116) L3 *Roark—Formulas for Stress and Strain, McGraw-Hill.
• 242 CHAPTER NINE w a b b 3 3 w 2 H a 1 2 H2 1 H H1 V2 V V1 V FIGURE 9.9 Pinned supports, con- FIGURE 9.10 Pinned supports, con- centrated load on the horizontal centrated load on one vertical member. member. (Roark–Formulas for (Roark–Formulas for Stress and Strain, Stress and Stress, McGraw-Hill.) McGraw-Hill.) With pinned supports and a uniform load on the horizontal member, Fig. 9.11: L2(L1 3 L2) H W 8L2L3 1 4L1L2L3 3 8L1(I3/I1) 8L3(I3/I2) 2 8L2L3 2 4L1L2L3 (9.117) 1 2 WL3 H(L1 L2) V1 (9.118) L3 With pinned supports and a uniform load on the vertical member, Fig. 9.12: 1 5L3(I3/I1) 2L1L2L3 1 4L2L3 1 H2 W 3 (9.119) 8 L1 (I3/I1) L2(I3/I2) L2L3 3 1 L2L3 2 L1L2L3 1 2 WL1 H2(L1 L2) V (9.120) L3 W 3 3 2 W 2 H H2 1 1 H H1 V2 V V1 V FIGURE 9.11 Pinned supports, FIGURE 9.12 Pinned supports, uni- uniform load on horizontal member. form load on vertical member. (Roark–Formulas for Stress and (Roark–Formulas for Stress and Strain, Strain, McGraw-Hill.) McGraw-Hill.)
• BUILDING AND STRUCTURES FORMULAS 243 W a b M1 M2 b 3 M1 3 M2 W 2 2 H a 1 H2 1 H H1 V2 V V1 V FIGURE 9.13 Fixed supports, con- FIGURE 9.14 Fixed supports, concen- centrated load on horizontal member. trated load on one vertical member. (Roark–Formulas for Stress and (Roark–Formulas for Stress and Strain, Strain, McGraw-Hill.) McGraw-Hill.) With fixed supports and a concentrated load on the horizontal member,Fig. 9.13: 1 3 1 2 1 3 1 2 3 HL1 2 M1L1 3 HL2 2 M2L2 (9.121) I1 I1 I2 I2 1 2 1 1 1 2 HL1 M1L1 3 M1L3 6 W(bL3 b3/L3) 6 M2L3 (9.122) I1 I1 I3 I3 I31 2 1 1 12 HL2 M2L2 3 M2L3 6 M1L3 6 W[2bL3 (b3/L3) 3b2] (9.123) I2 I2 I3 I3 I3 With fixed supports and and a concentrated load on one vertical member,Fig. 9.14: 1 3 1 2 1 3 1 2 1 3 1 2 3 Wa 2 Wa b 3 H2L1 2 M1L1 3 H2L2 2 M2L2 (9.124) I1 I1 I1 I1 I2 I2 1 2 1 2 1 1 2 Wa 2 H2L1 M1L1 3 M1L3 6 M2L3 (9.125) I1 I1 I1 I3 I3 1 2 1 1 2 HL2 M2L2 3 M2L3 6 M1L3 (9.126) I2 I2 I3 I3 With fixed supports and a uniform load on the horizontal member, Fig. 9.15: 1 2 1 3 1 3 1 2 2 M1L1 3 HL1 3 HL2 2 M2L2 (9.127) I1 I1 I2 I2 1 2 1 1 1 2 2 HL1 M1L1 3 M1L3 6 M2L3 24 WL3 (9.128) I1 I1 I3 I3 I3 1 2 1 1 1 2 2 HL2 M2L2 3 M2L3 6 M1L3 24 WL3 (9.129) I2 I2 I3 I3 I3
• 244 CHAPTER NINE W M1 M2 M1 3 M2 3 W 2 H 1 H2 1 2 H H1 V2 V V1 V FIGURE 9.15 Fixed supports, FIGURE 9.16 Fixed supports, uni- uniform load on horizontal member. form load on one vertical member. (Roark–Formulas for Stress and (Roark–Formulas for Stress and Strain, Strain, McGraw-Hill.) McGraw-Hill.) With fixed supports and a uniform load on the vertical member, Fig. 9.16: 1 3 1 2 1 3 1 3 1 2 8 WL1 2 M1L1 3 H2L1 3 H2L2 2 M2L2 (9.130) I1 I1 I1 I2 I2 1 2 1 2 1 1 6 WL1 2 H2L1 M1L1 3 M1L3 6 M2L3 (9.131) I1 I1 I1 I3 I3 1 2 1 1 2 H2L2 M2L2 3 M2L3 6 M1L3 (9.132) I2 I2 I3 I3ROOF LIVE LOADS*Some building codes specify that design of flat, curved, or pitched roofs shouldtake into account the effects of occupancy and rain loads and be designed forminimum live loads. Other codes require that structural members in flat,pitched, or curved roofs be designed for a live load Lr (lb per ft2 of horizontalprojection) computed from Lr 20R1R2 12 (9.133)where R1 reduction factor for size of tributary area 1 for At 200 1.2 0.001At for 200 < At < 600 0.6 for At 600 Al tributary area, or area contributing load to the structural member, ft2 R2 reduction factor for slope of roof 1 for F 4 0.6 for F 12 F rate of rise for a pitched roof, in/ft rise-to-span ratio multiplied by 32 for an arch or dome *Brokenbrough and Merritt—Structural Steel Designer’s Handbook, McGraw-Hill.
• 246 CHAPTER NINEwhere Kz velocity exposure coefficient evaluated at height z Kzt topographic factor Kd wind directionality factor I importance factor V basic wind speed corresponding to a 3-s gust speed at 33 ft above the ground in exposure C Velocity pressures due to wind to be used in building design vary with typeof terrain, distance above ground level, importance of building, likelihood ofhurricanes, and basic wind speed recorded near the building site. The windpressures are assumed to act horizontally on the building area projected on avertical plane normal to the wind direction. ASCE 7 permits the use of either Method I or Method II to define the designwind loads. Method I is a simplified procedure and may be used for enclosed orpartially enclosed buildings. ASCE 7 Method II is a rigorous computation procedure that accounts for theexternal, and internal pressure variation as well as gust effects. The following isthe general equation for computing the design wind pressure, p: p qGCp qi(GCpt) (9.138)where q and qi velocity pressure as given by ASCE 7 G gust effect factor as given by ASCE 7 Cp external pressure coefficient as given by ASCE 7 GCpt internal pressure coefficient as given by ASCE 7 Codes and standards may present the gust factors and pressure coefficientsin different formats. Coefficients from different codes and standards should notbe mixed.Seismic LoadsThe engineering approach to seismic design differs from that for other load types.For live, wind or snow loads, the intent of a structural design is to preclude struc-tural damage. However, to achieve an economical seismic design, codes and stan-dards permit local yielding of a structure during a major earthquake. Local yield-ing absorbs energy but results in permanent deformations of structures. Thusseismic design incorporates not only application of anticipated seismic forces butalso use of structural details that ensure adequate ductility to absorb the seismicforces without compromising the stability of structures. Provisions for this areincluded in the AISC specifications for structural steel for buildings. The forces transmitted by an earthquake to a structure result from vibratoryexcitation of the ground. The vibration has both vertical and horizontal compo-nents. However, it is customary for building design to neglect the vertical com-ponent because most structures have reserve strength in the vertical directiondue to gravity-load design requirements. Seismic requirements in building codes and standards attempt to translatethe complicated dynamic phenomenon of earthquake force into a simplifiedequivalent static force to be applied to a structure for design purposes. For
• BUILDING AND STRUCTURES FORMULAS 247example, ASCE 7-95 stipulates that the total lateral force, or base shear, V(kips) acting in the direction of each of the principal axes of the main structuralsystem should be computed from V CsW (9.139)where Cs seismic response coefficient W total dead load and applicable portions of other loads The seismic coefficient, Cs, is determined by the following equation: Cs 1.2Cv /RT2/3 (9.140)where Cv seismic coefficient for acceleration dependent (short period) struc- tures R response modification factor T fundamental period, s Alternatively, Cs need not be greater than Cs 2.5Ca/R (9.141)where Ca seismic coefficient for velocity dependent (intermediate and longperiod) structures. A rigorous evaluation of the fundamental elastic period, T, requires consid-eration of the intensity of loading and the response of the structure to the load-ing. To expedite design computations, T may be determined by the following: Ta CT h3/4 n (9.142)where CT 0.035 for steel frames CT 0.030 for reinforced concrete frames CT 0.030 steel eccentrically braced frames CT 0.020 all other buildings hn height above the basic to the highest level of the building, ft For vertical distribution of seismic forces, the lateral force, V, should be dis-tributed over the height of the structure as concentrated loads at each floor levelor story. The lateral seismic force, Fx, at any floor level is determined by thefollowing equation: Fx CuxV (9.143)where the vertical distribution factor is given by wxhk x Cux n k (9.144) i 1wihiwhere wx and wi height from the base to level x or i k 1 for building having period of 0.5 s or less 2 for building having period of 2.5 s or more use linear interpolation for building periods between 0.5 and 2.5 s
• 248 CHAPTER NINE For horizontal shear distribution, the seismic design story shear in any story,Vx, is determined by the following: n Vx i 1Fi (9.145)where Fi the portion of the seismic base shear induced at level i. The seismicdesign story shear is to be distributed to the various elements of the force-resisting system in a story based on the relative lateral stiffness of the verticalresisting elements and the diaphragm. Provision also should be made in design of structural framing for horizontaltorsion, overturning effects, and the building drift.
• CHAPTER 10 BRIDGE AND SUSPENSION- CABLE FORMULASSHEAR STRENGTH DESIGN FOR BRIDGESBased on the American Association of State Highway and Transportation Offi-cials (AASHTO) specifications for load-factor design (LFD), the shear capacity,kip (kN), may be computed from Vu 0.58Fy htwC (10.1)for flexural members with unstiffened webs with h /tw 150 or for girders withstiffened webs but a/h exceeding 3 or 67,600(h / tw)2: h C 1.0 when tw h when 1.25 h/t w tw 45,000k h when 1.25 Fy (h/t w)2 twFor girders with transverse stiffeners and a/h less than 3 and 67,600(h/tw)2, theshear capacity is given by 1 C Vu 0.58Fy dtw C (10.2) 1.15 1 (a/h)2Stiffeners are required when the shear exceeds Vu. Refer to Chap. 9, “Building and Structures Formulas,” for symbols used inthe preceding equations. 249
• 250 CHAPTER TEN TABLE 10.1 Column Formulas for Bridge Design Yield strength, Allowable stress, ksi (MPa) ksi (MPa) Cc Kl/r Cc Kl/r Cc 2 36 (248) 126.1 16.98–0.00053(Kl/r) 50 (345) 107.0 23.58–0.00103(Kl/r)2 135,000 90 (620) 79.8 42.45–0.00333(Kl/r)2 (Kl/r)2 100 (689) 75.7 47.17–0.00412(Kl/r)2ALLOWABLE-STRESS DESIGN FOR BRIDGE COLUMNSIn the AASHTO bridge-design specifications, allowable stresses in concentri-cally loaded columns are determined from the following equations: When Kl /r is less than Cc, Fy (Kl/r)2 Fa 1 (10.3) 2.12 2C 2 c When Kl /r is equal to or greater than Cc, 2 E 135,000 Fa (10.4) 2.12(Kl/r)2 (Kl/r)2See Table 10.1.LOAD-AND-RESISTANCE FACTORDESIGN FOR BRIDGE COLUMNSCompression members designed by LFD should have a maximum strength, kip(kN), Pu 0.85As Fcr (10.5)where As gross effective area of column cross section, in2 (mm2). For KLc /r 22 2E/Fy , 2 Fy KLc Fcr Fy 1 (10.6) 4 2E r For KLc /r 22 2E/Fy , 2 E 286,220 Fcr (10.7) (KLc /r)2 (KLc /r)2
• BRIDGE AND SUSPENSION-CABLE FORMULAS 251where Fcr buckling stress, ksi (MPa) Fy yield strength of the steel, ksi (MPa) K effective-length factor in plane of buckling Lc length of member between supports, in (mm) r radius of gyration in plane of buckling, in (mm) E modulus of elasticity of the steel, ksi (MPa) The preceding equations can be simplified by introducing a Q factor: 2 KLc Fy Q (10.8) r 2 2EThen, the preceding equations can be rewritten as shown next: For Q 1.0, Q Fcr 1 Fy (10.9) 2 For Q 1.0, Fy Fcr (10.10) 2QADDITIONAL BRIDGE COLUMN FORMULAS*The nomenclature for the bridge column formulas is shown at the end of theirexplanation of their applications. For bridges using structural carbon steel withsy 33,000 psi for the tensile yield strength, or for structural shapes havingtheir ends riveted, the following formulas apply, using the nomenclature givenbelow for formulas from Eqs. (10.11) through (10.33) are known as the ‘secantformula’ when the ‘sec’ expression appears in their denominator: Q allowable load, 1b P ultimate load, 1b A section area of column, sq in L length of column, in r least radius of gyration of column section, in Su ultimate strength, psi Sy yield point or yield strength of material, psi *Roark—Formulas for Stress and Strain, McGraw-Hill.
• 252 CHAPTER TEN E modulus of elasticity of material, psi m factor of safety (L/r) critical slenderness ratio Q Sy /m (10.11) 2r B EA A 0.75L mQ 1 0.25 sec Suggested Sy 32,000; suggested m 1.7 P Sy (10.12) 2 B EA A 0.75L P 1 0.25 secor 2 P L L 25,600 0.425 up to 160 (10.13) A r r 2 Q 1 L L 15,000 up to 140 (10.14) A 4 r r Q 18,750 L for 140 (10.15) 2r B EA A 0.75L 1.76Q r 1 0.25 sec When the column ends are pinned, the following formulas apply: 2 Q 1 L L 15,000 up to 140 (10.16) A 3 r r Q 18,750 L for 140 (10.17) 2r B EA A 0.875L 1.76Q r 1 0.25 sec P Sy (10.18) 2r B EA A 0.85L P 1 0.25 sec 2 P L L 25,600 0.566 up to 140 (10.19) A r r For structural silicon steel with Su 80,000 psi and Sy 45,000 psi, usingstructural shapes, or being fabricated with the ends riveted, 2 Q L L 20,000 0.46 up to 130 (10.20) A r r
• BRIDGE AND SUSPENSION-CABLE FORMULAS 253 Q 25,000 L for 130 (10.21) 2r B EA A 0.75L 1.8Q r 1 0.25 sec Q L Q 27,000 80 max 23,000 (10.22) A r A P Q 1.8 (10.23) A A When the column ends are pinned, 2 Q L L 20,000 0.61 up to 130 (10.24) A r r Q 25,000 L for 130 (10.25) 2r B EA A 0.875L 1.8Q r 1 0.25 sec P Q 1.8 (10.26) A A For structural nickel steel with Sy 55,000 psi, using structural shapes, orbeing fabricated with column ends being riveted, 2 Q L L 24,000 0.66 up to 120 (10.27) A r r Q 30,000 L for 120 (10.28) 2r B EA A 0.75L 1.83Q r 1 0.25 sec P Q 1.83 (10.29) A A Q L 24,000 98 (10.30) A r When the column ends are pinned, 2 Q L L 24,000 0.86 up to 120 (10.31) A r r Q 30,000 L for 120 (10.32) 2r B EA A 0.875L 1.83Q r 1 0.25 sec P Q 1.83 (10.33) A A
• 254 CHAPTER TEN TABLE 10.2 Allowable Bending Stress in Braced Bridge Beams* Fy Fb 36 (248) 20 (138) 50 (345) 27 (186) *Units in ksi (MPa).ALLOWABLE-STRESS DESIGN FOR BRIDGE BEAMSAASHTO gives the allowable unit (tensile) stress in bending as Fb 0.55Fy. Thesame stress is permitted for compression when the compression flange is support-ed laterally for its full length by embedment in concrete or by other means. When the compression flange is partly supported or unsupported in a bridge,the allowable bending stress, ksi (MPa), is (Table 10.2): 5 107Cb Iyc Fb Sxc L B Iyc 2 0.772J d 9.87 0.55Fy (10.34) Lwhere L length, in (mm), of unsupported flange between connections of lat- eral supports, including knee braces Sxc section modulus, in3 (mm3), with respect to the compression flange Iyc moment of inertia, in4 (mm4) of the compression flange about the vertical axis in the plane of the web J 1/3(bc tc btt3 Dt 3 ) 3 c w bc width, in (mm), of compression flange bt width, in (mm), of tension flange tc thickness, in (mm), of compression flange tt thickness, in (mm), of tension flange tw thickness, in (mm), of web D depth, in (mm), of web d depth, in (mm), of flexural memberIn general, the moment-gradient factor Cb may be computed from the nextequation. It should be taken as unity, however, for unbraced cantilevers andmembers in which the moment within a significant portion of the unbracedlength is equal to, or greater than, the larger of the segment end moments. Ifcover plates are used, the allowable static stress at the point of cutoff should becomputed from the preceding equation. The moment-gradient factor may be computed from 2 M1 M1 Cb 1.75 1.05 0.3 2.3 (10.35) M2 M2
• BRIDGE AND SUSPENSION-CABLE FORMULAS 255where M1 smaller beam end moment; and M2 larger beam end moment.The algebraic sign of M1 /M2 is positive for double-curvature bending and nega-tive for single-curvature bending.STIFFENERS ON BRIDGE GIRDERSThe minimum moment of inertia, in4 (mm4), of a transverse stiffener should beat least I a0 t3J (10.36)where J 2.5h2/a0 2 0.5 2 h clear distance between flanges, in (mm) a0 actual stiffener spacing, in (mm) t web thickness, in (mm)For paired stiffeners, the moment of inertia should be taken about the centerlineof the web; for single stiffeners, about the face in contact with the web. The gross cross-sectional area of intermediate stiffeners should be at least V 2 A 0.15BDtw (1 C) 18tw (10.37) Vuwhere is the ratio of web-plate yield strength to stiffener-plate yield strength,B 1.0 for stiffener pairs, 1.8 for single angles, and 2.4 for single plates; and Cis defined in the earlier section, “Allowable-Stress Design for Bridge Columns.”Vu should be computed from the previous section equations in “Shear StrengthDesign for Bridges.” The width of an intermediate transverse stiffener, plate, or outstanding legof an angle should be at least 2 in (50.8 mm), plus 1 30 of the depth of the girderand preferably not less than 1 4 of the width of the flange. Minimum thickness is1 16 of the width.Longitudinal StiffenersThese should be placed with the center of gravity of the fasteners h/5 from thetoe, or inner face, of the compression flange. Moment of inertia, in4 (mm4),should be at least a2 0 I ht3 2.4 0.13 (10.38) h2where a0 actual distance between transverse stiffeners, in (mm); and t webthickness, in (mm).
• 256 CHAPTER TEN Thickness of stiffener, in (mm), should be at least b fb /71.2, where b is the stiff-ener width, in (mm); and fb is the flange compressive bending stress, ksi (MPa).The bending stress in the stiffener should not exceed that allowable for the material.HYBRID BRIDGE GIRDERSThese may have flanges with larger yield strength than the web and may becomposite or noncomposite with a concrete slab, or they may utilize anorthotropic-plate deck as the top flange. Computation of bending stresses and allowable stresses is generally thesame as that for girders with uniform yield strength. The bending stress in theweb, however, may exceed the allowable bending stress if the computed flangebending stress does not exceed the allowable stress multiplied by (1 )2 (3 ) R 1 (10.39) 6 (3 )where ratio of web yield strength to flange yield strength distance from outer edge of tension flange or bottom flange of orthotropic deck to neutral axis divided by depth of steel section ratio of web area to area of tension flange or bottom flange of orthotropic-plate bridgeLOAD-FACTOR DESIGN FOR BRIDGE BEAMSFor LFD of symmetrical beams, there are three general types of members toconsider: compact, braced noncompact, and unbraced sections. The maximumstrength of each (moment, in kip) (mm kN) depends on member dimensionsand unbraced length, as well as on applied shear and axial load (Table 10.3). The maximum strengths given by the formulas in Table 10.3 apply onlywhen the maximum axial stress does not exceed 0.15Fy A, where A is thearea of the member. Symbols used in Table 10.3 are defined as follows: Fy steel yield strength, ksi (MPa) Z plastic section modulus, in3 (mm3) S section modulus, in3 (mm3) b width of projection of flange, in (mm) d depth of section, in (mm) h unsupported distance between flanges, in (mm) M1 smaller moment, in kip (mm kN), at ends of unbraced length of member Mu Fy Z M1 /Mu is positive for single-curvature bending.
• TABLE 10.3 Design Criteria for Symmetrical Flexural Sections for Load-Factor Design of Bridges Maximum Flange minimum Web minimum Maximum Type of bending strength Mu, thickness tf, thickness tu, unbraced length lb, section in kip (mm kN) in (mm) in (mm) in (mm) Compact* Fy Z b 2Fy d 2Fy [3600 2200(M1/Mu)]ry 65.0 608 2Fy Fy b h 20,000 Af Braced Fy S noncompact† 69.6 150 Fy d257 Unbraced ___________________________________ See AASHTO specification ________________________________ *Straight-line interpolation between compact and braced noncompact moments may be used for intermediate criteria, except that tw d Fy /608 should be maintained as well as the following: For compact sections, when both b /tf and d/tw exceed 75% of the limits for these ratios, the following interaction equation applies: d b 1064 9.35 tw tf Fyf where Fyf is the yield strength of the flange, ksi (MPa); tw is the web thickness, in (mm); and tf flange thickness, in (mm).
• 258 CHAPTER TEN TABLE 10.4 Allowable Bearing Stresses on Pins* Bridges Pins subject Pins not subject Buildings to rotation to rotation Fy Fp 0.90Fy Fp 0.40Fy Fp 0.80Fy 36 (248) 33 (227) 14 (96) 29 (199) 50 (344) 45 (310) 20 (137) 40 (225) * Units in ksi (MPa).BEARING ON MILLED SURFACESFor highway design, AASHTO limits the allowable bearing stress on milledstiffeners and other steel parts in contact to Fp 0.80Fu. Allowable bearingstresses on pins are given in Table 10.4. The allowable bearing stress for expansion rollers and rockers used inbridges depends on the yield point in tension Fy of the steel in the roller or thebase, whichever is smaller. For diameters up to 25 in (635 mm) the allowablestress, kip/linear in (kN/mm), is Fy 13 p 0.6d (10.40) 20For diameters from 25 to 125 in (635 to 3175 mm), Fy 13 p 3 d (10.41) 20where d diameter of roller or rocker, in (mm).BRIDGE FASTENERSFor bridges, AASHTO specifies the working stresses for bolts. Bearing-typeconnections with high-strength bolts are limited to members in compressionand secondary members. The allowable bearing stress is Fp 1.35Fu (10.42)where Fp allowable bearing stress, ksi (MPa); and Fu tensile strength ofthe connected part, ksi (MPa) (or as limited by allowable bearing on the fas-teners). The allowable bearing stress on A307 bolts is 20 ksi (137.8 MPa) andon structural-steel rivets is 40 ksi (275.6 MPa).
• BRIDGE AND SUSPENSION-CABLE FORMULAS 259COMPOSITE CONSTRUCTION IN HIGHWAY BRIDGESShear connectors between a steel girder and a concrete slab in compositeconstruction in a highway bridge should be capable of resisting both hori-zontal and vertical movement between the concrete and steel. Maximumspacing for shear connectors generally is 24 in (609.6 mm), but wider spacingmay be used over interior supports, to avoid highly stressed portions of thetension flange (Fig. 10.1). Clear depth of concrete cover over shear connec-tors should be at least 2 in (50.8 mm), and they should extend at least 2 in(50.8 mm) above the bottom of the slab.Span/Depth RatiosIn bridges, for composite beams, preferably the ratio of span/steel beam depthshould not exceed 30 and the ratio of span/depth of steel beam plus slab shouldnot exceed 25.Effective Width of SlabsFor a composite interior girder, the effective width assumed for the concreteflange should not exceed any of the following:1. One-fourth the beam span between centers of supports2. Distance between centerlines of adjacent girders3. Twelve times the least thickness of the slabFor a girder with the slab on only one side, the effective width of slab shouldnot exceed any of the following:1. One-twelfth the beam span between centers of supports2. Half the distance to the centerline of the adjacent girder3. Six times the least thickness of the slab Half space Shear connector 3" 24" spacing 24" MinC BearingL Max Max 2"Min 2" Min Studs 1" Min clearance Full penetration butt weldFIGURE 10.1 Maximum pitch for stud shear connectors in composite beams: 1 in (25.4 mm),2 in (50.8 mm), 3 in (76.2 mm), and 24 in (609.6 mm).
• 260 CHAPTER TENBending StressesIn composite beams in bridges, stresses depend on whether or not the members areshored; they are determined as for beams in buildings (see “Composite Construc-tion” in Chap. 9,“Building and Structures Formulas”), except that the stresses in thesteel may not exceed 0.55Fy. (See the following equations.) For unshored members, MD ML fs 0.55Fy (10.43) Ss Strwhere fy yield strength, ksi (MPa). For shored members, MD ML fs 0.55Fy (10.44) Strwhere fs stress in steel, ksi (MPa) MD dead-load moment, in.kip (kN.mm) ML live-load moment, in.kip (kN mm) Ss section modulus of steel beam, in3 (mm3) Str section modulus of transformed composite section, in3 (mm3) Vr shear range (difference between minimum and maximum shears at the point) due to live load and impact, kip (kN) Q static moment of transformed compressive concrete area about neu- tral axis of transformed section, in3 (mm3) I moment of inertia of transformed section, in4 (mm4)Shear RangeShear connectors in bridges are designed for fatigue and then are checked forultimate strength. The horizontal-shear range for fatigue is computed from VrQ Sr (10.45) Iwhere Sr horizontal-shear range at the juncture of slab and beam at point underconsideration, kip/linear in (kN/linear mm). The transformed area is the actual concrete area divided by n (Table 10.5). The allowable range of horizontal shear Zr, kip (kN), for an individual con-nector is given by the next two equations, depending on the connector used. For channels, with a minimum of 3 16 in (4.76 mm) fillet welds along heel and toe, Zr Bw (10.46)
• BRIDGE AND SUSPENSION-CABLE FORMULAS 261 TABLE 10.5 Ratio of Moduli of Elasticity of Steel and Concrete for Bridges fc for Es n concrete Ec 2.0–2.3 11 2.4–2.8 10 2.9–3.5 9 3.6–4.5 8 4.6–5.9 7 6.0 and over 6where w channel length, in (mm), in transverse direction on girder flange; and B cyclic variable 4.0 for 100,000 cycles, 3.0 for 500,000 cycles, 2.4 for 2 million cycles, and 2.1 for over 2 million cycles. For welded studs (with height/diameter ratio H/d 4): Zr d2 (10.47)where d stud diameter, in (mm); and cyclic variable 13.0 for 100,000 cycles, 10.6 for 500,000 cycles, 7.85 for 2 million cycles, and 5.5 for over 2 million cycles. Required pitch of shear connectors is determined by dividing the allowablerange of horizontal shear of all connectors at one section Zr, kip (kN), by thehorizontal range of shear Sr, kip per linear in (kN per linear mm).NUMBER OF CONNECTORS IN BRIDGESThe ultimate strength of the shear connectors is checked by computation of thenumber of connectors required from P N (10.48) Suwhere N number of shear connectors between maximum positive moment and end supports Su ultimate shear connector strength, kip (kN) [see Eqs. (10.1) and (10.2) that follow and AASHTO data] reduction factor 0.85 P force in slab, kip (kN)
• 262 CHAPTER TEN At points of maximum positive moments, P is the smaller of P1 and P2,computed from P1 As Fy (10.49) P2 0.85 fc Ac (10.50)where Ac effective concrete area, in2 (mm2) fc 28-day compressive strength of concrete, ksi (MPa) As total area of steel section, in2 (mm2) Fy steel yield strength, ksi (MPa) The number of connectors required between points of maximum positivemoment and points of adjacent maximum negative moment should equal orexceed N2, given by P P3 N2 (10.51) Su At points of maximum negative moments, the force in the slab P3, is com-puted from P3 AsrFyr (10.52)where Asr area of longitudinal reinforcing within effective flange, in (mm2); 2and Fyr reinforcing steel yield strength, ksi (MPa).Ultimate Shear Strength of Connectors in BridgesFor channels, t Su 17.4 h w fc (10.53) 2where h average channel-flange thickness, in (mm) t channel-web thickness, in (mm) w channel length, in (mm) For welded studs (H/d 4 in (101.6 mm), Su 0.4d 2 2fc Ec (10.54)ALLOWABLE-STRESS DESIGN FOR SHEAR IN BRIDGESBased on the AASHTO specification for highway bridges, the allowable shearstress, ksi (MPa), may be computed from Fy Fy Fv C (10.55) 3 3
• BRIDGE AND SUSPENSION-CABLE FORMULAS 263for flexural members with unstiffened webs with h/tw 150 or for girders withstiffened webs with a/h exceeding 3 and 67,600(h/tw)2: h C 1.0 when tw h when 1.25 h/t w tw 45,000k h when 1.25 Fy (h/t w)2 tw 2 a h k 5 if exceeds 3 or 67,600 h tw or stiffeners are not required 5 5 otherwise (a/h)2 B Fy k 190For girders with transverse stiffeners and a/h less than 3 and 67,600(h/tw)2, theallowable shear stress is given by Fy 1 C Fv C (10.56) 3 1.15 1 (a/h)2Stiffeners are required when the shear exceeds Fv.MAXIMUM WIDTH/THICKNESS RATIOSFOR COMPRESSION ELEMENTS FOR HIGHWAY BRIDGESTable 10.6 gives a number of formulas for maximum width/thickness ratios forcompression elements for highway bridges. These formulas are valuable forhighway bridge design.SUSPENSION CABLESParabolic Cable Tension and LengthSteel cables are often used in suspension bridges to support the horizon-tal roadway load (Fig. 10.2). With a uniformly distributed load along the
• TABLE 10.6 Maximum Width/Thickness Ratios b /t a for Compression Elements for Highway Bridgesb Load-and-resistance-factor designc Description of element Compact Noncompactd 2Fy 2Fy 65 70 e Flange projection of rolled or fabricated I-shaped beams 2Fy 608 Webs in flexural compression 150264 Allowable-stress design f fa 0.44Fy Description of element fa 0.44Fy Fy 36 ksi (248 MPa) Fy 50 ksi (344.5 MPa) Plates supported in one side and outstanding legs of angles 51 In main members 12 11 wfa 12
• In bracing and other 51 12 11 secondary members wfa 16 Plates supported on two edges or 126 32 27 webs of box shapesg wfa 45 Solid cover plates supported on 158 40 34 two edges or solid websh wfa 50 Perforated cover plates supported 190 48 41 on two edges for box shapes wfa 55265 a b width of element or projection; t thickness. The point of support is the inner line of fasteners or fillet welds con- necting a plate to the main segment or the root of the flange of rolled shapes. In LRFD, for webs of compact sections, b d, the beam depth; and for noncompact sections, b D, the unsupported distance between flange components. b As required in AASHTO “Standard Specification for Highway Bridges.” The specifications also provide special limita- tions on plate-girder elements. c Fy specified minimum yield stress, ksi (MPa), of the steel. d Elements with width/thickness ratios that exceed the noncompact limits should be designed as slender elements. e When the maximum bending moment M is less than the bending strength Mu, b/t in the table may be multiplied by Mu /M . f fa computed axial compression stress, ksi (MPa). g For box shapes consisting of main plates, rolled sections, or component segments with cover plates. h For webs connecting main members or segments for H or box shapes.
• 266 CHAPTER TEN L A B d C Unit load = w T (a) A C H (b) FIGURE 10.2 Cable supporting load uniformly distributed along the horizontal.horizontal, the cable assumes the form of a parabolic arc. Then, the ten-sion at midspan is wL2 H (10.57) 8dwhere H midspan tension, kip (N) w load on a unit horizontal distance, klf (kN/m) L span, ft (m) d sag, ft (m) The tension at the supports of the cable is given by 2 0.5 wL T H2 (10.58) 2where T tension at supports, kip (N); and other symbols are as before. Length of the cable, S, when d/L is 1/20, or less, can be approximated from 8d 2 S L (10.59) 3Lwhere S cable length, ft (m).Catenary Cable Sag and Distance between SupportsA cable of uniform cross section carrying only its own weight assumes theform of a catenary. By using the same previous notation, the catenary parame-ter, c, is found from
• BRIDGE AND SUSPENSION-CABLE FORMULAS 267 T d c (10.60) wThen 2 0.5 S c (d c)2 (10.61) 2 Sag d c ft (m) (10.62)Span length then is L 2c, with the previous same symbols.GENERAL RELATIONS FOR SUSPENSION CABLESCatenaryFor any simple cable (Fig. 10.3) with a load of qo per unit length of cable, kip/ft(N/m), the catenary length s, ft (m), measured from the low point of the cableis, with symbols as given in Fig. 10.3, ft (m), 2 H q x 1 qo s sinh o x x3 (10.63) qo H 3! HTension at any point is T 2H 2 q2 s2 o H qoy (10.64) y L a b L L 2 2 TR VR TL h/2 R H h VL H L fR Cable s L y M yM P x C x (a) (b)FIGURE 10.3 Simple cables: (a) shape of cable with concentrated load; (b) shape ofcable with supports at different levels.
• 268 CHAPTER TENThe distance from the low point C to the left support is H qo a cosh 1 f 1 (10.65) qo H Lwhere fL vertical distance from C to L, ft (m). The distance from C to the rightsupport R is H qo b cosh 1 f 1 (10.66) qo H Rwhere fR vertical distance from C to R. Given the sags of a catenary fL and fR under a distributed vertical load qo, thehorizontal component of cable tension H may be computed from qo l 1 qo fL 1 qo fR cosh 1 cosh 1 (10.67) H H Hwhere l span, or horizontal distance between supports L and R a b. Thisequation usually is solved by trial. A first estimate of H for substitution in theright-hand side of the equation may be obtained by approximating the catenary bya parabola. Vertical components of the reactions at the supports can be com-puted from qo a qo b RL H sinh RR H sinh (10.68) H HParabolaUniform vertical live loads and uniform vertical dead loads other than cableweight generally may be treated as distributed uniformly over the horizontalprojection of the cable. Under such loadings, a cable takes the shape of aparabola. Take the origin of coordinates at the low point C (Fig. 10.3). If wo is theload per foot (per meter) horizontally, the parabolic equation for the cableslope is wo x 2 y (10.69) 2HThe distance from the low point C to the left support L is l Hh a (10.70) 2 wo l
• BRIDGE AND SUSPENSION-CABLE FORMULAS 269where l span, or horizontal distance between supports L and R a b; hvertical distance between supports. The distance from the low point C to the right support R is 1 Hh b (10.71) 2 wo lSupports at Different LevelsThe horizontal component of cable tension H may be computed from wo l 2 h wo l 2 H fR fL fR (10.72) h2 2 8fwhere fL vertical distance from C to L fR vertical distance from C to R f sag of cable measured vertically from chord LR midway between supports (at x Hh/wol)As indicated in Fig. 10.3(b), h f fL yM (10.73) 2where yM Hh2/2wo l 2. The minus sign should be used when low point C isbetween supports. If the vertex of the parabola is not between L and R, the plussign should be used. The vertical components of the reactions at the supports can be computedfrom wol Hh VL wo a (10.74) 2 l wol Hh Vr wo b (10.75) 2 lTension at any point is 2H 2 T wo x 2 2 (10.76)Length of parabolic arc RC is 2 B 2 b wob H wb LRC 1 sinh o (10.77) H 2wo H 2 1 wo (10.78) b b3 6 H
• 270 CHAPTER TENLength of parabolic arc LC is 2 B 2 a wo a H wa LLC 1 sinh o (10.79) H 2wo H 2 1 wo a a3 (10.80) 6 HSupports at Same LevelIn this case, fL fR f, h 0, and a b l/2. The horizontal component ofcable tension H may be computed from wo l 2 H (10.81) 8fThe vertical components of the reactions at the supports are wo l VL VR (10.82) 2Maximum tension occurs at the supports and equals 2 B wo l l2 TL TR 1 (10.83) 16 f 2Length of cable between supports is 2 B 2 1 wo l H w l L 1 sinh o (10.84) 2H wo 2H 8 f2 32 f 4 256 f 6 l 1 (10.85) 3 l2 5 l4 7 l6If additional uniformly distributed load is applied to a parabolic cable, thechange in sag is approximately 15 l L f (10.86) 16 f 5 24 f 2/l 2For a rise in temperature t, the change in sag is about 15 l 2ct 8 f2 f 1 (10.87) 16 f (5 24 f 2/l 2 ) 3 l2where c coefficient of thermal expansion.
• 272 CHAPTER TEN If elastic elongation and can be ignored, 1 ML dx 1 0 3 HL 1 ML dx (10.95) 2 fl 0 yD dx 0Thus, for a load uniformly distributed horizontally wL, 1 wL l 3 ML dx (10.96) 0 12and the increase in the horizontal component of tension due to live load is 3 wL l 3 wL l 2 wL l 2 8HD HL (10.97) 2 f l 12 8f 8 wDl 2 wL H (10.98) wD DCABLE SYSTEMSThe cable that is concave downward (Fig. 10.4) usually is considered the load-carrying cable. If prestress in that cable exceeds that in the other cable, the nat-ural frequencies of vibration of both cables always differ for any value of liveload. To avoid resonance, the difference between the frequencies of the cablesshould increase with increase in load. Thus, the two cables tend to assume dif-ferent shapes under specific dynamic loads. As a consequence, the resultingflow of energy from one cable to the other dampens the vibrations of bothcables. Natural frequency, cycles per second, of each cable may be estimated from l Bw n Tg n (10.99) (a) (b) (c) (d) FIGURE 10.4 Planar cable systems: (a) completely separated cables; (b) cables intersecting at midspan; (c) crossing cables; (d) cables meeting at supports.
• BRIDGE AND SUSPENSION-CABLE FORMULAS 273where n integer, 1 for fundamental mode of vibration, 2 for second mode, . . . l span of cable, ft (m) w load on cable, kip/ft (kN/m) g acceleration due to gravity 32.2 ft/s2 T cable tension, kip (N) The spreaders of a cable truss impose the condition that under a given loadthe change in sag of the cables must be equal. Nevertheless, the changes in ten-sion of the two cables may not be equal. If the ratio of sag to span f/l is small(less than about 0.1), for a parabolic cable, the change in tension is givenapproximately by 16 AEf H f (10.100) 3 l2where f change in sag A cross-sectional area of cable E modulus of elasticity of cable steelRAINWATER ACCUMULATIONAND DRAINAGE ON BRIDGESRainwater accumulation and drainage are important considerations in highwaybridge design. The runoff rate of rainwater from a bridge during a rainstorm isgiven by:* Q kCiA (10.101)where Q peak runoff rate (ft3/s) C runoff coefficient i average rainfall intensity (in/h) A drainage area (acres) k 1.00083 The runoff coefficient, C, ranges from 0.70 to 0.95 for pavements made ofasphalt, concrete, or brick. Specific values are available in Tonias—Bridge Engi-neering, McGraw-Hill. Most bridge designers base the rainfall intensity on a once-in-10-year stormlasting for 5 min. Historic rainfall intensities can be determined from local munic-ipal records available from the city or state in which a bridge will be located. To drain the water from a highway bridge the sheet flow of the rainwatermust be studied. Determine the deck width for handling the rainwater runoff tothe drain scuppers by using:* 1 W Shoulder Width (Traffic Lane) (10.102) 3where W width of deck used in analysis *Tonias—Bridge Engineering, McGraw-Hill.
• 274 CHAPTER TEN The drain should have a large enough area to handle the maximum runoffanticipated. Rainwater runoff from a bridge should be led a specially designedand built, or existing, sewerage system. Where a new sewerage system is beingdesigned, the bridge must be considered as an additional water load source thatmust be included in the system load calculations.
• CHAPTER 11 HIGHWAY AND ROAD FORMULASCIRCULAR CURVESCircular curves are the most common type of horizontal curve used to con-nect intersecting tangent (or straight) sections of highways or railroads. Inmost countries, two methods of defining circular curves are in use: the first,in general use in railroad work, defines the degree of curve as the centralangle subtended by a chord of 100 ft (30.48 m) in length; the second, usedin highway work, defines the degree of curve as the central angle subtendedby an arc of 100 ft (30.48 m) in length. The terms and symbols generally used in reference to circular curves arelisted next and shown in Figs. 11.1 and 11.2. PC point of curvature, beginning of curve PI point of intersection of tangents PT point of tangency, end of curve R radius of curve, ft (m) D degree of curve I deflection angle between tangents at PI, also central angle of curve T tangent distance, distance from PI to PC or PT, ft (m) L length of curve from PC to PT measured on 100-ft (30.48-m) chord for chord definition, on arc for arc definition, ft (m) C length of long chord from PC to PT, ft (m) E external distance, distance from PI to midpoint of curve, ft (m) M midordinate, distance from midpoint of curve to midpoint of long chord, ft (m) d central angle for portion of curve (d D) l length of curve (arc) determined by central angle d, ft (m) c length of curve (chord) determined by central angle d, ft (m) a tangent offset for chord of length c, ft (m) b chord offset for chord of length c, ft (m) 275
• 276 CHAPTER ELEVEN PI I E T T M PC PT I C 2 R R I I 2 FIGURE 11.1 Circular curve. a C b C d d FIGURE 11.2 Offsets to circular curve.Equations of Circular Curves 5,729.578 exact for arc definition, approxi- R (11.1) D mate for chord definition 50 exact for chord definition sin 1 2 D T R tan 1 2 I exact (11.2) 1 1 E R exsec 2I R (sec 2 I 1) exact (11.3) 1 1 M R vers 2I R (1 cos 2 I) exact (11.4)
• HIGHWAY AND ROAD FORMULAS 277 C 2R sin 1 2 I exact (11.5) 100I L exact (11.6) D L3 C3 L C approximate (11.7) 24R 2 24R 2 Dl d exact for arc definition (11.8) 100 Dc approximate for chord definition (11.9) 100 d c sin exact for chord definition (11.10) z 2R c2 a approximate (11.11) 2R c2 b approximate (11.12) RPARABOLIC CURVESParabolic curves are used to connect sections of highways or railroads of differ-ing gradient. The use of a parabolic curve provides a gradual change in directionalong the curve. The terms and symbols generally used in reference to paraboliccurves are listed next and are shown in Fig. 11.3. PVC point of vertical curvature, beginning of curve PVI point of vertical intersection of grades on either side of curve L L 2 2 PVI (elev. V) G1 X Elev. Ex G2 L PVC (elev. E0) PVT (elev. Et) FIGURE 11.3 Vertical parabolic curve (summit curve).
• 278 CHAPTER ELEVEN PVT point of vertical tangency, end of curve G1 grade at beginning of curve, ft/ft (m/m) G2 grade at end of curve, ft/ft (m/m) L length of curve, ft (m) R rate of change of grade, ft/ft2 (m/m2) V elevation of PVI, ft (m) E0 elevation of PVC, ft (m) Et elevation of PVT, ft (m) x distance of any point on the curve from the PVC, ft (m) Ex elevation of point x distant from PVC, ft (m) xs distance from PVC to lowest point on a sag curve or highest point on a summit curve, ft (m) Es elevation of lowest point on a sag curve or highest point on a summit curve, ft (m)Equations of Parabolic CurvesIn the parabolic-curve equations given next, algebraic quantities should alwaysbe used. Upward grades are positive and downward grades are negative. G2 G1 R (11.13) L 1 E0 V 2 LG1 (11.14) 1 2 Ex E0 G1x 2 Rx (11.15) G1 xs (11.16) R G2 1 Es E0 (11.17) 2R Note: If xs is negative or if xs L, the curve does not have a high point ora low point.HIGHWAY CURVES AND DRIVER SAFETYFor the safety and comfort of drivers, provision usually is made for gradualchange from a tangent to the start of a circular curve. As indicated in Fig. 11.4, typically the outer edge is raised first until the outerhalf of the cross section is levelled with the crown (point B). Then, the outer edgeis raised farther until the cross section is straight (point C). From there on, theentire cross section is rotated until the full superelevation is attained (point E).