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• TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page i 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.i CIVIL ENGINEERING FORMULAS TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page iii 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.iii CIVIL ENGINEERING FORMULAS Tyler G. Hicks, P.E. International Engineering Associates Member: American Society of Mechanical Engineers United States Naval Institute McGRAW-HILL New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page vi40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.vi vi CONTENTS Chapter 3. Column Formulas 99 T C General Considerations / 100 C Short Columns / 102 R Eccentric Loads on Columns / 102 D Column Base Plate Design / 111 U American Institute of Steel Construction Allowable-Stress Design Approach / 113 W Composite Columns / 115 Elastic Flexural Buckling of Columns / 118 U Allowable Design Loads for Aluminum Columns / 121 Ultimate-Strength Design of Concrete Columns / 124 W U W Chapter 4. Piles and Piling Formulas 131 U W Allowable Loads on Piles / 132 F Laterally Loaded Vertical Piles / 133 F Toe Capacity Load / 134 S Groups of Piles / 136 C Foundation-Stability Analysis / 139 S Axial-Load Capacity of Single Piles / 143 B Shaft Settlement / 144 L Shaft Resistance to Cohesionless Soil / 145 S C C Chapter 5. Concrete Formulas 147 W Reinforced Concrete / 148 Water/Cementitious Materials Ratio / 148 C Job Mix Concrete Volume / 149 Modulus of Elasticity of Concrete / 150 G Tensile Strength of Concrete / 151 S Reinforcing Steel / 151 B Continuous Beams and One-Way Slabs / 151 B Design Methods for Beams, Columns, and Other Members / 153 C Properties in the Hardened State / 167 C TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page viii40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.viii viii CONTENTS Compression at Angle to Grain / 220 C Recommendations of the Forest Products Laboratory / 221 F Compression on Oblique Plane / 223 S Adjustments Factors for Design Values / 224 V Fasteners for Wood / 233 Adjustment of Design Values for Connections with Fasteners / 236 C Roof Slope to Prevent Ponding / 238 Bending and Axial Tension / 239 L Bending and Axial Compression / 240 A L A Chapter 7. Surveying Formulas 243 L A Units of Measurement / 244 S Theory of Errors / 245 B Measurement of Distance with Tapes / 247 C Vertical Control / 253 B Stadia Surveying / 253 P Photogrammetry / 255 L C W D Chapter 8. Soil and Earthwork Formulas 257 F C Physical Properties of Soils / 258 N Index Parameters for Soils / 259 P Relationship of Weights and Volumes in Soils / 261 Internal Friction and Cohesion / 263 Vertical Pressures in Soils / 264 Lateral Pressures in Soils, Forces on Retaining Walls / 265 C Lateral Pressure of Cohesionless Soils / 266 F Lateral Pressure of Cohesive Soils / 267 Water Pressure / 268 S Lateral Pressure from Surcharge / 268 A Stability of Slopes / 269 L Bearing Capacity of Soils / 270 A Settlement under Foundations / 271 S Soil Compaction Tests / 272 H TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page ix 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.ix CONTENTS ix Compaction Equipment / 275 Formulas for Earthmoving / 276 Scraper Production / 278 Vibration Control in Blasting / 280 Chapter 9. Building and Structures Formulas 283 Load-and-Resistance Factor Design for Shear in Buildings / 284 Allowable-Stress Design for Building Columns / 285 Load-and-Resistance Factor Design for Building Columns / 287 Allowable-Stress Design for Building Beams / 2873 Load-and-Resistance Factor Design for Building Beams / 290 Allowable-Stress Design for Shear in Buildings / 295 Stresses in Thin Shells / 297 Bearing Plates / 298 Column Base Plates / 300 Bearing on Milled Surfaces / 301 Plate Girders in Buildings / 302 Load Distribution to Bents and Shear Walls / 304 Combined Axial Compression or Tension and Bending / 306 Webs under Concentrated Loads / 308 Design of Stiffeners under Loads / 3117 Fasteners for Buildings / 312 Composite Construction / 313 Number of Connectors Required for Building Construction / 316 Ponding Considerations in Buildings / 318 Chapter 10. Bridge and Suspension-Cable Formulas 321 Shear Strength Design for Bridges / 322 Allowable-Stress Design for Bridge Columns / 323 Load-and-Resistance Factor Design for Bridge Columns / 324 Allowable-Stress Design for Bridge Beams / 325 Stiffeners on Bridge Girders / 327 Hybrid Bridge Girders / 329 TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page x40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.x x CONTENTS Load-Factor Design for Bridge Beams / 330 F Bearing on Milled Surfaces / 332 O Bridge Fasteners / 333 W Composite Construction in Highway Bridges / 333 P Number of Connectors in Bridges / 337 T Allowable-Stress Design for Shear in Bridges / 339 F Maximum Width/Thickness Ratios for Compression C Elements for Highway Bridges / 341 O Suspension Cables / 341 M General Relations for Suspension Cables / 345 H Cable Systems / 353 N W F Chapter 11. Highway and Road Formulas 355 P E M Circular Curves / 356 Parabolic Curves / 359 C Highway Curves and Driver Safety / 361 G Highway Alignments / 362 W Structural Numbers for Flexible Pavements / 365 F Transition (Spiral) Curves / 370 E Designing Highway Culverts / 371 V American Iron and Steel Institute (AISI) Design H Procedure / 374 Chapter 12. Hydraulics and Waterworks Formulas 381 Capillary Action / 382 Viscosity / 386 Pressure on Submerged Curved Surfaces / 387 Fundamentals of Fluid Flow / 388 Similitude for Physical Models / 392 Fluid Flow in Pipes / 395 Pressure (Head) Changes Caused by Pipe Size Change / 403 Flow through Oriﬁces / 406 TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page xi 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xi CONTENTS xi Fluid Jets / 409 Oriﬁce Discharge into Diverging Conical Tubes / 410 Water Hammer / 412 Pipe Stresses Perpendicular to the Longitudinal Axis / 412 Temperature Expansion of Pipe / 414 Forces Due to Pipe Bends / 414 Culverts / 417 Open-Channel Flow / 420 Manning’s Equation for Open Channels / 424 Hydraulic Jump / 425 Nonuniform Flow in Open Channels / 429 Weirs / 436 Flow Over Weirs / 4385 Prediction of Sediment-Delivery Rate / 440 Evaporation and Transpiration / 442 Method for Determining Runoff for Minor Hydraulic Structures / 443 Computing Rainfall Intensity / 443 Groundwater / 446 Water Flow for Fireﬁghting / 446 Flow from Wells / 447 Economical Sizing of Distribution Piping / 448 Venturi Meter Flow Computation / 448 Hydroelectric Power Generation / 449 Index 4511 TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page xiii 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xiii PREFACE This handy book presents more than 2000 needed formulas 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 formulas are also useful to design drafters, structural engineers, bridge engineers, foundation builders, ﬁeld engineers, professional-engineer license examination candidates, concrete specialists, timber-structure builders, and students in a variety of civil engineering pursuits. The book presents formulas needed in 12 different spe- cialized branches of civil engineering—beams and girders, columns, piles and piling, concrete structures, timber engi- neering, surveying, soils and earthwork, building struc- tures, bridges, suspension cables, highways and roads, and hydraulics and open-channel ﬂow. Key formulas are pre- sented for each of these topics. Each formula is explained so the engineer, drafter, or designer knows how, where, and when to use the formula in professional work. Formula units are given in both the United States Customary System (USCS) and System International (SI). Hence, the text is usable throughout the world. To assist the civil engineer using this material in worldwide engineering practice, a com- prehensive tabulation of conversion factors is presented in Chapter 1. In assembling this collection of formulas, the author was guided by experts who recommended the areas of TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page xiv40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xiv xiv PREFACE greatest need for a handy book of practical and applied civil engineering formulas. Sources for the formulas presented here include the var- A ious regulatory and industry groups in the ﬁeld of civil engi- neering, authors of recognized books on important 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 engineering. These sources are cited in the Acknowledgments. When using any of the formulas in this book that may come from an industry or regulatory code, the user M is cautioned to consult the latest version of the code. t Formulas may be changed from one edition of a code to a the next. In a work of this magnitude it is difﬁcult to H include the latest formulas from the numerous constant- a ly changing codes. Hence, the formulas given here are those current at the time of publication of this book. w In a work this large it is possible that errors may occur. F Hence, the author will be grateful to any user of the book p who detects an error and calls it to the author’s attention. w Just write the author in care of the publisher. The error will a be corrected in the next printing. L In addition, if a user believes that one or more important N formulas have been left out, the author will be happy to e consider them for inclusion in the next edition of the book. Again, just write him in care of the publisher. p F Tyler G. Hicks, P.E. a s f e w p t TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page xvi40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xvi xvi ACKNOWLEDGMENTS The author also thanks Larry Hager, Senior Editor, Pro- fessional Group, The McGraw-Hill Companies, for his excellent guidance and patience during the long preparation of the manuscript for this book. Finally, the author thanks his wife, Mary Shanley Hicks, a publishing professional, who always most willingly offered help and advice when needed. Speciﬁc publications consulted during the preparation of this text include: American Association of State Highway and Transportation Ofﬁcials (AASHTO) “Standard Speciﬁ- cations for Highway Bridges”; American Concrete Institute (ACI) “Building Code Requirements for Reinforced Con- crete”; American Institute of Steel Construction (AISC) T “Manual of Steel Construction,” “Code of Standard Prac- c tice,” and “Load and Resistance Factor Design Speciﬁca- d tions for Structural Steel Buildings”; American Railway Engineering Association (AREA) “Manual for Railway s Engineering”; American Society of Civil Engineers p (ASCE) “Ground Water Management”; American Water Works Association (AWWA) “Water Quality and Treat- n ment.” In addition, the author consulted several hundred T civil engineering reference and textbooks dealing with the i topics in the current book. The author is grateful to the S writers of all the publications cited here for the insight they gave him to civil engineering formulas. A number of these o works are also cited in the text of this book. d c c l e d i d a TLFeBOOK
• 40816_FM_pi-xvii 10/22/01 12:37 PM Page xvii 40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xvii-sn HOW TO USEsn , THIS BOOKfy-e-) The formulas presented in this book are intended for use by- civil engineers in every aspect of their professional work—- design, evaluation, construction, repair, etc.y To ﬁnd a suitable formula for the situation you face,y start by consulting the index. Every effort has been made tos present a comprehensive listing of all formulas in the book.r Once you ﬁnd the formula you seek, read any accompa-- nying text giving background information about the formula.d Then when you understand the formula and its applications,e insert the numerical values for the variables in the formula.e Solve the formula and use the results for the task at hand.y Where a formula may come from a regulatory code,e or where a code exists for the particular work being done, be certain to check the latest edition of the appli- cable code to see that the given formula agrees with the code formula. If it does 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, and build. Using the latest edition of any govern- ing code is the only sensible way to produce a safe and dependable design that you will be proud to be associ- ated with. Further, you will sleep more peacefully! TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_01_p1-14 10/22/01 12:38 PM Page 1 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CHAPTER 1 CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_01_p1-14 10/22/01 12:38 PM Page 2 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 2 CHAPTER ONE Civil engineers throughout the world accept both the T United States Customary System (USCS) and the System International (SI) units of measure for both applied and theoretical calculations. However, the SI units are much more widely used than those of the USCS. Hence, both the USCS and the SI units are presented for essentially every formula in this book. Thus, the user of the book can apply s the formulas with conﬁdence anywhere in the world. c To permit even wider use of this text, this chapter con- p tains the conversion factors needed to switch from one sys- tem to the other. For engineers unfamiliar with either p system of units, the author suggests the following steps for f becoming acquainted with the unknown system: k 1. Prepare a list of measurements commonly used in your g daily work. 2. Insert, opposite each known unit, the unit from the other k 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 w 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 c each appropriate value in Table 1.2 from Table 1.3. a 4. Apply the conversion values wherever necessary for the u formulas in this book. c 5. Recognize—here and now—that the most difﬁcult F aspect of becoming familiar with a new system of meas- d urement is becoming comfortable with the names and magnitudes of the units. Numerical conversion is simple, once you have set up your own conversion table. TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 3 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 3e TABLE 1.1 Commonly Used USCS and SI Units†md Conversion factorh (multiply USCS unite by this factor toy USCS unit SI unit SI symbol obtain SI unit)y square foot square meter m2 0.0929 cubic foot cubic meter m3 0.2831- pound per- square inch kilopascal kPa 6.894r pound force newton Nu 4.448r foot pound torque newton meter Nиm 1.356 kip foot kilonewton meter kNиm 1.355r gallon per minute liter per second L/s 0.06309r kip per squareh inch megapascal MPa 6.89l † This table is abbreviated. For a typical engineering practice, an actual tables would be many times this length.e, Be careful, when using formulas containing a numericalt constant, to convert the constant to that for the system you are using. You can, however, use the formula for the USCSe units (when the formula is given in those units) and then convert the ﬁnal result to the SI equivalent using Table 1.3.t For the few formulas given in SI units, the reverse proce-- dure should be used.d , TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 4 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 4 CHAPTER ONE TABLE 1.2 Typical Conversion Table† T To convert from To Multiply by‡ square foot square meter 9.290304 E Ϫ 02 a foot per second meter per second a squared squared 3.048 E Ϫ 01 a cubic foot cubic meter 2.831685 E Ϫ 02 a pound per cubic kilogram per cubic inch meter 2.767990 E ϩ 04 a gallon per minute liter per second 6.309 E Ϫ 02 pound per square inch kilopascal 6.894757 b pound force newton 4.448222 b kip per square foot pascal 4.788026 E ϩ 04 b acre foot per day cubic meter per E Ϫ 02 B second 1.427641 acre square meter 4.046873 E ϩ 03 B cubic foot per cubic meter per second second 2.831685 E Ϫ 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 B 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 con- version factor is to be multiplied before use. Thus, for the square foot con- B version factor, 9.290304 ϫ 1/100 ϭ 0.09290304, the factor to be used to convert square feet to square meters. For a positive exponent, as in convert- ing acres to square meters, multiply by 4.046873 ϫ 1000 ϭ 4046.8. Where a conversion factor cannot be found, simply use the dimensional substitution. Thus, to convert pounds per cubic inch to kilograms per cubic meter, ﬁnd 1 lb ϭ 0.4535924 kg and 1 in3 ϭ 0.00001638706 m3. Then, B 1 lb/in3 ϭ 0.4535924 kg/0.00001638706 m3 ϭ 27,680.01, or 2.768 E ϩ 4. TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 5 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 5 TABLE 1.3 Factors for Conversion to SI Units of Measurement To convert from To Multiply by2 acre foot, acre ft cubic meter, m3 1.233489 E ϩ 03 acre square meter, m2 4.046873 E ϩ 031 angstrom, Å meter, m 1.000000* E Ϫ 102 atmosphere, atm pascal, Pa 1.013250* E ϩ 05 (standard)4 atmosphere, atm pascal, Pa 9.806650* E ϩ 04 (technical2 ϭ 1 kgf/cm2) bar pascal, Pa 1.000000* E ϩ 05 barrel (for petroleum, cubic meter, m2 1.589873 E Ϫ 014 42 gal) board foot, board ft cubic meter, m3 2.359737 E Ϫ 032 British thermal unit, joule, J 1.05587 E ϩ 03 Btu, (mean)3 British thermal unit, watt per meter 1.442279 E Ϫ 01 Btu (International kelvin, W/(mиK)2 Table)иin/(h)(ft2) (°F) (k, thermal conductivity) British thermal unit, watt, W 2.930711 E Ϫ 01 Btu (Internationale Table)/h British thermal unit, watt per square 5.678263 E ϩ 00 Btu (International meter kelvin, Table)/(h)(ft2)(°F) W/(m2иK) (C, thermal conductance) British thermal unit, joule per kilogram, 2.326000* E ϩ 03 Btu (International J/kg Table)/lb TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 6 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 6 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement T (Continued) ( To convert from To Multiply by British thermal unit, joule per kilogram 4.186800* E ϩ 03 ( Btu (International kelvin, J/(kgиK) Table)/(lb)(°F) (c, heat capacity) British thermal unit, joule per cubic 3.725895 E ϩ 04 d cubic foot, Btu meter, J/m3 f (International f Table)/ft3 f bushel (U.S.) cubic meter, m3 3.523907 E Ϫ 02 f calorie (mean) joule, J 4.19002 E ϩ 00 candela per square candela per square 1.550003 E ϩ 03 s inch, cd/in2 meter, cd/m2 s centimeter, cm, of pascal, Pa 1.33322 E ϩ 03 mercury (0°C) centimeter, cm, of pascal, Pa 9.80638 E ϩ 01 s water (4°C) chain meter, m 2.011684 E ϩ 01 c circular mil square meter, m2 5.067075 E Ϫ 10 day second, s 8.640000* E ϩ 04 c day (sidereal) second, s 8.616409 E ϩ 04 degree (angle) radian, rad 1.745329 E Ϫ 02 c degree Celsius kelvin, K TK ϭ tC ϩ 273.15 degree Fahrenheit degree Celsius, °C tC ϭ (tF Ϫ 32)/1.8 f degree Fahrenheit kelvin, K TK ϭ (tF ϩ 459.67)/1.8 degree Rankine kelvin, K TK ϭ TR /1.8 (°F)(h)(ft2)/Btu kelvin square 1.761102 E Ϫ 01 f (International meter per watt, Table) (R, thermal Kиm2/W f resistance) TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 7 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 7 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by 2 (°F)(h)(ft )/(Btu kelvin meter per 6.933471 E ϩ 00 (International watt, Kиm/W Table)иin) (thermal resistivity) dyne, dyn newton, N 1.000000† E Ϫ 05 fathom meter, m 1.828804 E ϩ 00 foot, ft meter, m 3.048000† E Ϫ 01 foot, ft (U.S. survey) meter, m 3.048006 E Ϫ 01 foot, ft, of water pascal, Pa 2.98898 E ϩ 03 (39.2°F) (pressure) square foot, ft2 square meter, m2 9.290304† E Ϫ 02 square foot per hour, square meter per 2.580640† E Ϫ 05 ft2/h (thermal second, m2/s diffusivity) square foot per square meter per 9.290304† E Ϫ 02 second, ft2/s second, m2/s cubic foot, ft3 (volume cubic meter, m3 2.831685 E Ϫ 02 or section modulus) cubic foot per minute, cubic meter per 4.719474 E Ϫ 04 ft3/min second, m3/s cubic foot per second, cubic meter per 2.831685 E Ϫ 02 ft3/s second, m3/s foot to the fourth meter to the fourth 8.630975 E Ϫ 038 power, ft4 (area power, m4 moment of inertia) foot per minute, meter per second, 5.080000† E Ϫ 03 ft/min m/s foot per second, meter per second, 3.048000† E Ϫ 01 ft/s m/s TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 8 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 8 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement T (Continued) ( To convert from To Multiply by foot per second meter per second 3.048000† E Ϫ 01 h squared, ft/s2 squared, m/s2 h footcandle, fc lux, lx 1.076391 E ϩ 01 footlambert, fL candela per square 3.426259 E ϩ 00 h meter, cd/m2 foot pound force, ftиlbf joule, J 1.355818 E ϩ 00 h foot pound force per watt, W 2.259697 E Ϫ 02 minute, ftиlbf/min h foot pound force per watt, W 1.355818 E ϩ 00 second, ftиlbf/s h foot poundal, ft joule, J 4.214011 E Ϫ 02 h poundal h free fall, standard g meter per second 9.806650† E ϩ 00 i squared, m/s2 i gallon, gal (Canadian cubic meter, m3 4.546090 E Ϫ 03 liquid) i gallon, gal (U.K. cubic meter, m3 4.546092 E Ϫ 03 liquid) i gallon, gal (U.S. dry) cubic meter, m3 4.404884 E Ϫ 03 gallon, gal (U.S. cubic meter, m3 3.785412 E Ϫ 03 liquid) s gallon, gal (U.S. cubic meter per 4.381264 E Ϫ 08 c liquid) per day second, m3/s gallon, gal (U.S. cubic meter per 6.309020 E Ϫ 05 liquid) per minute second, m3/s i grad degree (angular) 9.000000† E Ϫ 01 grad radian, rad 1.570796 E Ϫ 02 grain, gr kilogram, kg 6.479891† E Ϫ 05 i gram, g kilogram, kg 1.000000† E Ϫ 03 TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 9 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 9 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by1 hectare, ha square meter, m 2 1.000000† E ϩ 04 horsepower, hp watt, W 7.456999 E ϩ 021 (550 ftиlbf/s)0 horsepower, hp watt, W 9.80950 E ϩ 03 (boiler)0 horsepower, hp watt, W 7.460000† E ϩ 022 (electric) horsepower, hp watt, W 7.46043† E ϩ 020 (water) horsepower, hp (U.K.) watt, W 7.4570 E ϩ 022 hour, h second, s 3.600000† E ϩ 03 hour, h (sidereal) second, s 3.590170 E ϩ 030 inch, in meter, m 2.540000† E Ϫ 02 inch of mercury, in Hg pascal, Pa 3.38638 E ϩ 033 (32°F) (pressure) inch of mercury, in Hg pascal, Pa 3.37685 E ϩ 033 (60°F) (pressure) inch of water, in pascal, Pa 2.4884 E ϩ 023 H2O (60°F)3 (pressure) square inch, in2 square meter, m2 6.451600† E Ϫ 048 cubic inch, in3 cubic meter, m3 1.638706 E Ϫ 05 (volume or section5 modulus) inch to the fourth meter to the fourth 4.162314 E Ϫ 071 power, in4 (area power, m42 moment of inertia)5 inch per second, in/s meter per second, 2.540000† E Ϫ 023 m/s TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 10 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 10 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement T (Continued) ( To convert from To Multiply by kelvin, K degree Celsius, °C tC ϭ TK Ϫ 273.15 m kilogram force, kgf newton, N 9.806650† E ϩ 00 m kilogram force meter, newton meter, 9.806650† E ϩ 00 m kgиm Nиm m kilogram force second kilogram, kg 9.806650† E ϩ 00 m squared per meter, m kgfиs2/m (mass) kilogram force per pascal, Pa 9.806650† E ϩ 04 m square centimeter, s kgf/cm2 kilogram force per pascal, Pa 9.806650† E ϩ 00 s square meter, kgf/m2 m kilogram force per pascal, Pa 9.806650† E ϩ 06 square millimeter, m kgf/mm2 kilometer per hour, meter per second, 2.777778 E Ϫ 01 m km/h m/s m kilowatt hour, kWh joule, J 3.600000† E ϩ 06 kip (1000 lbf) newton, N 4.448222 E ϩ 03 m kipper square inch, pascal, Pa 6.894757 E ϩ 06 m kip/in2 ksi m knot, kn (international) meter per second, 5.144444 E Ϫ 01 o m/s lambert, L candela per square 3.183099 E ϩ 03 o meter, cd/m liter cubic meter, m3 1.000000† E Ϫ 03 o maxwell weber, Wb 1.000000† E Ϫ 08 o mho siemens, S 1.000000† E ϩ 00 o TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 11 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 11 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by microinch, ␮in meter, m 2.540000† E Ϫ 080 micron, ␮m meter, m 1.000000† E Ϫ 060 mil, mi meter, m 2.540000† E Ϫ 05 mile, mi (international) meter, m 1.609344† E ϩ 030 mile, mi (U.S. statute) meter, m 1.609347 E ϩ 03 mile, mi (international meter, m 1.852000† E ϩ 03 nautical)4 mile, mi (U.S. nautical) meter, m 1.852000† E ϩ 03 square mile, mi2 square meter, m2 2.589988 E ϩ 06 (international)0 square mile, mi2 square meter, m2 2.589998 E ϩ 06 (U.S. statute) mile per hour, mi/h meter per second, 4.470400† E Ϫ 016 (international) m/s mile per hour, mi/h kilometer per hour, 1.609344† E ϩ 00 (international) km/h1 millibar, mbar pascal, Pa 1.000000† E ϩ 02 millimeter of mercury, pascal, Pa 1.33322 E ϩ 026 mmHg (0°C)3 minute (angle) radian, rad 2.908882 E Ϫ 046 minute, min second, s 6.000000† E ϩ 01 minute (sidereal) second, s 5.983617 E ϩ 011 ounce, oz kilogram, kg 2.834952 E Ϫ 02 (avoirdupois)3 ounce, oz (troy or kilogram, kg 3.110348 E Ϫ 02 apothecary)3 ounce, oz (U.K. ﬂuid) cubic meter, m3 2.841307 E Ϫ 058 ounce, oz (U.S. ﬂuid) cubic meter, m3 2.957353 E Ϫ 050 ounce force, ozf newton, N 2.780139 E Ϫ 01 TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 12 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 12 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement T (Continued) ( To convert from To Multiply by ounce forceиinch, newton meter, 7.061552 E Ϫ 03 p ozfиin Nиm ounce per square foot, kilogram per square 3.051517 E Ϫ 01 p oz (avoirdupois)/ft2 meter, kg/m2 ounce per square yard, kilogram per square 3.390575 E Ϫ 02 p oz (avoirdupois)/yd2 meter, kg/m2 perm (0°C) kilogram per pascal 5.72135 E Ϫ 11 p second meter, kg/(Paиsиm) p perm (23°C) kilogram per pascal 5.74525 E Ϫ 11 second meter, p kg/(Paиsиm) perm inch, permиin kilogram per pascal 1.45322 E Ϫ 12 p (0°C) second meter, kg/(Paиsиm) p perm inch, permиin kilogram per pascal 1.45929 E Ϫ 12 (23°C) second meter, p kg/(Paиsиm) pint, pt (U.S. dry) cubic meter, m3 5.506105 E Ϫ 04 p pint, pt (U.S. liquid) cubic meter, m3 4.731765 E Ϫ 04 poise, p (absolute pascal second, 1.000000† E Ϫ 01 p viscosity) Paиs p pound, lb kilogram, kg 4.535924 E Ϫ 01 p (avoirdupois) pound, lb (troy or kilogram, kg 3.732417 E Ϫ 01 p apothecary) pound square inch, kilogram square 2.926397 E Ϫ 04 p lbиin2 (moment of meter, kgиm2 inertia) p TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 13 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS 13 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by pound per footи pascal second, 1.488164 E ϩ 00 second, lb/ftиs Paиs pound per square kilogram per square 4.882428 E ϩ 00 foot, lb/ft2 meter, kg/m22 pound per cubic kilogram per cubic 1.601846 E Ϫ 01 foot, lb/ft3 meter, kg/m3 pound per gallon, kilogram per cubic 9.977633 E ϩ 01 lb/gal (U.K. liquid) meter, kg/m3 pound per gallon, kilogram per cubic 1.198264 E ϩ 021 lb/gal (U.S. liquid) meter, kg/m3 pound per hour, lb/h kilogram per 1.259979 E Ϫ 04 second, kg/s2 pound per cubic inch, kilogram per cubic 2.767990 E ϩ 04 lb/in3 meter, kg/m3 pound per minute, kilogram per 7.559873 E Ϫ 032 lb/min second, kg/s pound per second, kilogram per 4.535924 E Ϫ 01 lb/s second, kg/s4 pound per cubic yard, kilogram per cubic 5.932764 E Ϫ 014 lb/yd3 meter, kg/m31 poundal newton, N 1.382550 E Ϫ 01 poundиforce, lbf newton, N 4.448222 E ϩ 001 pound force foot, newton meter, 1.355818 E ϩ 00 lbfиft Nиm1 pound force per foot, newton per meter, 1.459390 E ϩ 01 lbf/ft N/m4 pound force per pascal, Pa 4.788026 E ϩ 01 square foot, lbf/ft2 pound force per inch, newton per meter, 1.751268 E ϩ 02 lbf/in N/m TLFeBOOK
• 40816_01_p1-14 10/22/01 12:38 PM Page 14 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 14 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by pound force per square pascal, Pa 6.894757 E ϩ 03 inch, lbf/in2 (psi) quart, qt (U.S. dry) cubic meter, m3 1.101221 E Ϫ 03 quart, qt (U.S. liquid) cubic meter, m3 9.463529 E Ϫ 04 rod meter, m 5.029210 E ϩ 00 second (angle) radian, rad 4.848137 E Ϫ 06 second (sidereal) second, s 9.972696 E Ϫ 01 square (100 ft2) square meter, m2 9.290304† E ϩ 00 ton (assay) kilogram, kg 2.916667 E Ϫ 02 ton (long, 2240 lb) kilogram, kg 1.016047 E ϩ 03 ton (metric) kilogram, kg 1.000000† E ϩ 03 ton (refrigeration) watt, W 3.516800 E ϩ 03 ton (register) cubic meter, m3 2.831685 E ϩ 00 ton (short, 2000 lb) kilogram, kg 9.071847 E ϩ 02 ton (long per cubic kilogram per cubic 1.328939 E ϩ 03 yard, ton)/yd3 meter, kg/m3 ton (short per cubic kilogram per cubic 1.186553 E ϩ 03 yard, ton)/yd3 meter, kg/m3 ton force (2000 lbf) newton, N 8.896444 E ϩ 03 tonne, t kilogram, kg 1.000000† E ϩ 03 watt hour, Wh joule, J 3.600000† E ϩ 03 yard, yd meter, m 9.144000† E Ϫ 01 square yard, yd2 square meter, m2 8.361274 E Ϫ 01 cubic yard, yd3 cubic meter, m3 7.645549 E Ϫ 01 year (365 days) second, s 3.153600† E ϩ 07 year (sidereal) second, s 3.155815 E ϩ 07 † Exact value. From E380, “Standard for Metric Practice,” American Society for Testing and Materials. TLFeBOOK
• 40816_02_p15-51 10/22/01 12:39 PM Page 16 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 16 408 16 CHAPTER TWO In analyzing beams of various types, the geometric proper- ties of a variety of cross-sectional areas are used. Figure 2.1 gives equations for computing area A, moment of inertia I, section modulus or the ratio S ϭ I/c, where c ϭ distance from the neutral axis to the outermost ﬁber of the beam or other member. Units used are inches and millimeters and their powers. The formulas in Fig. 2.1 are valid for both USCS and SI units. Handy formulas for some dozen different types of beams are given in Fig. 2.2. In Fig. 2.2, both USCS and SI units can be used in any of the formulas that are applicable to both steel and wooden beams. Note that W ϭ load, lb (kN); L ϭ length, ft (m); R ϭ reaction, lb (kN); V ϭ shear, lb (kN); M ϭ bending moment, lb и ft (N и m); D ϭ deﬂec- tion, ft (m); a ϭ spacing, ft (m); b ϭ spacing, ft (m); E ϭ modulus of elasticity, lb/in2 (kPa); I ϭ moment of inertia, in4 (dm4); Ͻ ϭ less than; Ͼ ϭ greater than. Figure 2.3 gives the elastic-curve equations for a variety of prismatic beams. In these equations the load is given as P, lb (kN). Spacing is given as k, ft (m) and c, ft (m). CONTINUOUS BEAMS Continuous beams and frames are statically indeterminate. Bending moments in these beams are functions of the geometry, moments of inertia, loads, spans, and modulus of elasticity of individual members. Figure 2.4 shows how any span of a continuous beam can be treated as a single beam, with the moment diagram decomposed into basic com- ponents. Formulas for analysis are given in the diagram. Reactions of a continuous beam can be found by using the formulas in Fig. 2.5. Fixed-end moment formulas for beams of constant moment of inertia (prismatic beams) for TLFeBOOK
• 40816_02_p15-51 10/22/01 12:39 PM Page 17 16 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 17 TLFeBOOK-1 ,erdh f Ieb , -ϭ ,ys FIGURE 2.1 Geometric properties of sections. .efy ,- .err 17
• 40816_02_p15-51 10/22/01 12:39 PM Page 18 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 18 408 TLFeBOOK 18
• 40816_02_p15-51 10/22/01 12:39 PM Page 1918 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 19 TLFeBOOK Geometric properties of sections. FIGURE 2.1 (Continued) 19
• 40816_02_p15-51 10/22/01 12:39 PM Page 20 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 20 408 TLFeBOOK 20
• 40816_02_p15-51 10/22/01 12:39 PM Page 2120 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 21 TLFeBOOK Geometric properties of sections. FIGURE 2.1 (Continued) 21
• 40816_02_p15-51 10/22/01 12:39 PM Page 22 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 22 408 TLFeBOOK 22
• 40816_02_p15-51 10/22/01 12:39 PM Page 2322 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 23 TLFeBOOK Geometric properties of sections. FIGURE 2.1 (Continued) 23
• 40816_02_p15-51 10/22/01 12:39 PM Page 24 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 24 408 TLFeBOOK 24
• 40816_02_p15-51 10/22/01 12:39 PM Page 2524 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 25 TLFeBOOK Geometric properties of sections. FIGURE 2.1 (Continued) 25
• 40816_02_p15-51 10/22/01 12:39 PM Page 26 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 26 408 TLFeBOOK s C ﬁ i m i i b g ⌺ m b d F Geometric properties of sections. b e e b l l m o d s F FIGURE 2.1 (Continued) f l t 26
• 40816_02_p15-51 10/22/01 12:39 PM Page 28 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 28 408 TLFeBOOK 28
• 28 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM29 Third Pass Page 29 bcj 7/19/01 p. 29 FIGURE 2.2 Beam formulas. (From J. Callender, Time-Saver Standards for Architectural Design Data, 6th ed., McGraw-Hill, N.Y.) TLFeBOOK
• 40816_02_p15-51 10/22/01 12:39 PM Page 30 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 30 408 TLFeBOOK 30
• 40816_02_p15-51 10/22/01 12:39 PM Page 3130 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 31 TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 31
• 40816_02_p15-51 10/22/01 12:39 PM Page 32 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 32 408 TLFeBOOK 32
• 40816_02_p15-51 10/22/01 12:39 PM Page 3332 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 33 TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 33
• 40816_02_p15-51 10/22/01 12:39 PM Page 34 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 34 408 TLFeBOOK 34
• 40816_02_p15-51 10/22/01 12:39 PM Page 3534 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 35 TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 35
• 40816_02_p15-51 10/22/01 12:39 PM Page 36 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 36 408 TLFeBOOK 36
• 40816_02_p15-51 10/22/01 12:39 PM Page 3736 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 37 TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 37
• 40816_02_p15-51 10/22/01 12:39 PM Page 38 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 38 408 TLFeBOOK 38
• 40816_02_p15-51 10/22/01 12:39 PM Page 3938 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 39 TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 39
• 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/0140 12:39 PM Second Pass Page 40 pb 7/3/01 p. 40 FIGURE 2.3 Elastic-curve equations for prismatic beams. (a) Shears, moments, deﬂections for full uni- form load on a simply supported prismatic beam. (b) Shears and moments for uniform load over part of a simply supported prismatic beam. (c) Shears, moments, deﬂections for a concentrated load at any point of a simply supported prismatic beam. TLFeBOOK 408
• 40 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM41 Third Pass Page 41 bcj 7/19/01 p. 41 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (d) Shears, moments, deﬂec- tions for a concentrated load at midspan of a simply supported prismatic beam. (e) Shears, moments, deﬂec- tions for two equal concentrated loads on a simply supported prismatic beam. ( f ) Shears, moments, deﬂections for several equal loads equally spaced on a simply supported prismatic beam. TLFeBOOK
• 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/0142 12:39 PM Second Pass Page 42 pb 7/3/01 p. 42 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (g) Shears, moments, deﬂec- tions for a concentrated load on a beam overhang. (h) Shears, moments, deﬂections for a concentrated load on the end of a prismatic cantilever. (i) Shears, moments, deﬂections for a uniform load over the full length of a beam with overhang. TLFeBOOK 408
• 42 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM43 Third Pass Page 43 bcj 7/19/01 p. 43 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (j) Shears, moments, deﬂections for uniform load over the full length of a cantilever. (k) Shears, moments, deﬂections for uniform load on a beam overhang. (l) Shears, moments, deﬂections for triangular loading on a prismatic cantilever. TLFeBOOK
• 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/0144 12:39 PM Second Pass Page 44 pb 7/3/01 p. 44 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (m) Simple beam—load increas- ing uniformly to one end. TLFeBOOK 408
• 44 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM45 Third Pass Page 45 bcj 7/19/01 p. 45 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (n) Simple beam—load increas- ing uniformly to center. TLFeBOOK
• 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/0146 12:39 PM Second Pass Page 46 pb 7/3/01 p. 46 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (o) Simple beam—uniform load partially distributed at one end. TLFeBOOK 408
• 46 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM47 Third Pass Page 47 bcj 7/19/01 p. 47 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (p) Cantilever beam—concen- trated load at free end. TLFeBOOK
• 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/0148 12:39 PM Second Pass Page 48 pb 7/3/01 p. 48 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (q) Beam ﬁxed at both ends— concentrated load at center. TLFeBOOK 408
• 48 40816_02_p15-51 40816 HICKS Mcghp Ch02 10/22/01 12:39 PM49 Third Pass Page 49 bcj 7/19/01 p. 49 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. TLFeBOOK
• 40816_02_p15-51 10/22/01 12:39 PM Page 50 40816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 50 408 50 CHAPTER TWO F c c I o f FIGURE 2.5 Reactions of continuous beam (a) found by making the beam statically determinate. (b) Deﬂections computed with interior supports removed. (c), (d ), and (e) Deﬂections calculated for unit load over each removed support, to obtain equations for w each redundant. i t TLFeBOOK
• 40816_02_p15-51 10/22/01 12:39 PM Page 51 50 40816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 51 BEAM FORMULAS 51 FIGURE 2.6 Fixed-end moments for a prismatic beam. (a) For a concentrated load. (b) For a uniform load. (c) For two equal con- centrated loads. (d) For three equal concentrated loads. In a similar manner the ﬁxed-end moment for a beam with one end hinged and the supports at different levels can be found from dg MF ϭ K (2.3)h Ldr where K is the actual stiffness for the end of the beam that is ﬁxed; for beams of variable moment of inertia K equals the ﬁxed-end stiffness times (1 Ϫ C FC F). L R TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 52 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 52 408 FIGURE 2.7 Chart for ﬁxed-end moments due to any type of loading. F C M b u FIGURE 2.8 Characteristics of loadings. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 53p. 52 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 53 BEAM FORMULAS 53f FIGURE 2.8 (Continued) Characteristics of loadings. ULTIMATE STRENGTH OF CONTINUOUS BEAMS Methods for computing the ultimate strength of continuous beams and frames may be based on two theorems that ﬁx upper and lower limits for load-carrying capacity: 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. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 54 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 54 408 54 CHAPTER TWO p t m t t r f U t i FIGURE 2.9 Moments due to deﬂection of a ﬁxed-end beam. C 2. Lower-bound theorem. The load corresponding to an o 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 A cases. k TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 55p. 54 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 55 BEAM FORMULAS 55 For the continuous beam in Fig. 2.10, the ratio of the plastic moment for the end spans is k times that for the cen- ter span (k Ͼ 1). Figure 2.10b shows the moment diagram for the beam made determinate by ignoring the moments at B and C and the moment diagram for end moments MB and MC applied to the determinate beam. Then, by using Fig. 2.10c, equilib- rium is maintained when wL2 1 1 MP ϭ Ϫ M Ϫ M 4 2 B 2 C wL2 ϭ 4 Ϫ kM P wL2 ϭ (2.4) 4(1 ϩ k) The mechanism method can be used to analyze rigid frames of constant section with ﬁxed bases, as in Fig. 2.11. Using this method with the vertical load at midspan equal to 1.5 times the lateral load, the ultimate load for the frame is 4.8MP /L laterally and 7.2M P /L vertically at midspan. Maximum moment occurs in the interior spans AB and CD when L M xϭ Ϫ (2.5) 2 wLn or ifse L kM Ps M ϭ kM P when xϭ Ϫ (2.6) 2 wLne A plastic hinge forms at this point when the moment equals kMP. For equilibrium, TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 56 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 56 408 56 CHAPTER TWO 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 redun- dants removed and for the redundants. The two moment diagrams are combined in (c), producing peaks at which plastic hinges are assumed to form. TLFeBOOK
• p. 56 e e s s - 40816_02_p52-98 40816 HICKS Mcghp Ch02 10/22/01 12:41 PM57 Second Pass Page 57 7/3/01 pb p. 57 FIGURE 2.11 Ultimate-load possibilities for a rigid frame of constant section with ﬁxed bases. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 58 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 58 408 58 CHAPTER TWO w x kM P ϭ x (L Ϫ x) Ϫ kM P 2 L ϭ w 2 ΂ L Ϫ kM ΃ ΂ L ϩ KM ΃ Ϫ ΂ 1 Ϫ kM ΃ kM 2 wL P 2 wL P 2 wL P 2 P leading to F k 2M 2 P wL2 Ϫ 3kM P ϩ ϭ0 wL 2 4 (2.7) When the value of MP previously computed is substituted, s o 7k 2 ϩ 4k ϭ 4 or k (k ϩ 4͞7) ϭ 4͞7 o s from which k ϭ 0.523. The ultimate load is s 4M P (1 ϩ k) M f wL ϭ ϭ 6.1 P (2.8) L L m i In any continuous beam, the bending moment at any section is equal to the bending moment at any other section, plus the shear at that section times its arm, plus the product of all the intervening external forces times their respective arms. Thus, in Fig. 2.12, Vx ϭ R 1 ϩ R 2 ϩ R 3 Ϫ P1 Ϫ P2 Ϫ P3 M x ϭ R 1 (l 1 ϩ l 2 ϩ x) ϩ R 2 (l 2 ϩ x) ϩ R 3x Ϫ P1 (l 2 ϩ c ϩ x) Ϫ P2 (b ϩ x) Ϫ P3a M x ϭ M 3 ϩ V3x Ϫ P3a Table 2.1 gives the value of the moment at the various F supports of a uniformly loaded continuous beam over equal a TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 59p. 58 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 59 BEAM FORMULAS 59 FIGURE 2.12 Continuous beam.) , spans, and it also gives the values of the shears on each side of the supports. Note that the shear is of the opposite sign on either side 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 a uniformly loaded continuous beam of) four equal spans. (See Table 2.1.) Table 2.1 also gives the maximum bending moment that occurs between supports, in addition to the position of this moment and the points ofy ,tes FIGURE 2.13 Relation between moment and shear diagrams forl a uniformly loaded continuous beam of four equal spans. TLFeBOOK
• 40816_02_p52-98 40816 HICKS Mcghp Ch02 TABLE 2.1 Uniformly Loaded Continuous Beams over Equal Spans (Uniform load per unit length ϭ w; length of each span ϭ l) Distance to 10/22/01 point of Distance to Shear on each side max moment, point of Notation of support. L ϭ left, measured to inﬂection, Number of R ϭ right. Reaction at Moment Max right from measured to of support any support is L ϩ R over each moment in from right from60 12:41 PM supports of span L R support each span support support Third Pass 2 1 or 2 0 1 ͞2 0 0.125 0.500 None 3 1 0 3 ͞8 0 0.0703 0.375 0.750 2 5 ͞8 5 ͞8 1 ͞8 0.0703 0.625 0.250 ͞10 Page 60 4 4 1 0 0 0.080 0.400 0.800 7/19/01 bcj p. 60 2 ͞10 6 5 ͞10 1 ͞10 0.025 0.500 0.276, 0.724 1 0 11 ͞28 0 0.0772 0.393 0.786 5 2 17 ͞28 15 ͞28 3 ͞28 0.0364 0.536 0.266, 0.806 3 13 ͞28 13 ͞28 2 ͞28 0.0364 0.464 0.194, 0.734 1 0 15 ͞38 0 0.0779 0.395 0.789 TLFeBOOK 408 6 2 23 ͞ 20 ͞ 4 ͞ 0 0332 0 526 0 268 0 783
• ͞38 p. 60 40816_02_p52-98 40816 HICKS Mcghp Ch02 6 2 23 ͞38 20 ͞38 4 ͞38 0.0332 0.526 0.268, 0.783 3 18 ͞38 19 ͞38 3 ͞38 0.0461 0.500 0.196, 0.804 1 0 41 ͞104 0 0.0777 0.394 0.788 2 63 ͞104 55 ͞104 11 ͞104 0.0340 0.533 0.268, 0.790 7 3 49 ͞104 51 ͞104 8 ͞104 0.0433 0.490 0.196, 0.785 10/22/01 4 53 ͞104 53 ͞104 9 ͞104 0.0433 0.510 0.215, 0.804 1 0 56 ͞142 0 0.0778 0.394 0.789 8 2 86 ͞142 75 ͞142 15 ͞142 0.0338 0.528 0.268, 0.788 3 67 ͞142 70 ͞142 11 ͞142 0.0440 0.493 0.196, 0.790 4 72 ͞142 71 ͞142 12 ͞142 0.0405 0.500 0.215, 0.785 12:41 PM wl wl wl 2 wl 2 l l61 Values apply to Second Pass The numerical values given are coefﬁcients of the expressions at the foot of each column. Page 61 7/3/01 pb p. 61 TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 62 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 62 408 62 CHAPTER TWO t r t r t w FIGURE 2.14 Values of the functions for a uniformly loaded B continuous beam resting on three equal spans with four supports. B inﬂection. Figure 2.14 shows the values of the functions for s a uniformly loaded continuous beam resting on three equal l spans with four supports. M t Maxwell’s Theorem w a When a number of loads rest upon a beam, the deﬂection at i any point is equal to the sum of the deﬂections at this point due to each of the loads taken separately. Maxwell’s theo- t rem states that if unit loads rest upon a beam at two points, s A and B, the deﬂection at A due to the unit load at B equals t the deﬂection at B due to the unit load at A. e i Castigliano’s Theorem m 2 This theorem states that the deﬂection of the point of s application of an external force acting on a beam is equal a TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 63p. 62 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 63 BEAM FORMULAS 63 to the partial derivative of the work of deformation with respect to this force. Thus, if P is the force, f is the deﬂec- tion, and U is the work of deformation, which equals the resilience: dU ϭf dP According to the principle of least work, the deforma- tion of any structure takes place in such a manner that the work of deformation is a minimum.d BEAMS OF UNIFORM STRENGTH . Beams of uniform strength so vary in section that the unitr stress S remains constant, and I/c varies as M. For rectangu-l lar beams of breadth b and depth d, I/c ϭ I/c ϭ bd 2/6 and M ϭ Sbd2/6. For a cantilever beam of rectangular cross sec- tion, under a load P, Px ϭ Sbd 2/6. If b is constant, d 2 varies with x, and the proﬁle of the shape of the beam is a parabola, as in Fig. 2.15. If d is constant, b varies as x, and the beamt is triangular in plan (Fig. 2.16).t Shear at the end of a beam necessitates modiﬁcation of- the forms determined earlier. The area required to resist, shear is P/Sv in a cantilever and R/Sv in a simple beam. Dot-s ted 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 extra cost in fabricating, however, make many of the forms impractical, except for cast iron. Figure 2.17 shows some of the simple sections of uniformf strength. In none of these, however, is shear taken intol account. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 64 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 64 408 64 CHAPTER TWO FIGURE 2.15 Parabolic beam of uniform strength. SAFE LOADS FOR BEAMS OF VARIOUS TYPES Table 2.2 gives 32 formulas for computing the approximate safe loads on steel beams of various cross sections for an T allowable stress of 16,000 lb/in2 (110.3 MPa). Use these W formulas for quick estimation of the safe load for any steel p beam you are using in a design. l TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 65p. 64 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 65 BEAM FORMULAS 65 FIGURE 2.16 Triangular beam of uniform strength.e Table 2.3 gives coefﬁcients for correcting values inn Table 2.2 for various methods of support and loading.e When combined with Table 2.2, the two sets of formulasl provide useful time-saving means of making quick safe- load computations in both the ofﬁce and the ﬁeld. TLFeBOOK
• g y ͞2 p. 68 40816_02_p52-98 40816 HICKS Mcghp Ch02 Beam ﬁxed at one end, cantilever Load uniformly distributed over span 1 ͞4 2.40 Load concentrated at end 1 ͞8 3.20 Load increasing uniformly to ﬁxed end 3 ͞8 1.92 Beam continuous over two supports equidistant from ends 10/22/01 Load uniformly distributed over span 1. If distance a Ͼ 0.2071l l2/4a2 2. If distance a Ͻ 0.2071l l/(l – 4a) 3. If distance a ϭ 0.2071l 5.83 Two equal loads concentrated at ends l/4a 12:41 PM l ϭ length of beam; c ϭ distance from support to nearest concentrated load; a ϭ distance from support to end69 † Second Pass of beam. Page 69 7/3/01 pb p. 69 TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 70 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 70 408 TLFeBOOK 70
• 40816_02_p52-98 10/22/01 12:41 PM Page 71p. 70 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 71 TLFeBOOK FIGURE 2.17 Beams of uniform strength (in bending). 71
• 40816_02_p52-98 10/22/01 12:41 PM Page 72 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 72 408 TLFeBOOK 72
• 40816_02_p52-98 10/22/01 12:41 PM Page 73p. 72 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 73 TLFeBOOK Beams of uniform strength (in bending). FIGURE 2.17 (Continued) 73
• 40816_02_p52-98 10/22/01 12:41 PM Page 74 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 74 408 TLFeBOOK 74
• 40816_02_p52-98 10/22/01 12:41 PM Page 75p. 74 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 75 TLFeBOOK Beams of uniform strength (in bending). FIGURE 2.17 (Continued) 75
• 40816_02_p52-98 10/22/01 12:41 PM Page 76 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 76 408 TLFeBOOK 76
• 40816_02_p52-98 10/22/01 12:41 PM Page 77p. 76 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 77 TLFeBOOK Beams of uniform strength (in bending). FIGURE 2.17 (Continued) 77
• 40816_02_p52-98 10/22/01 12:41 PM Page 78 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 78 408 TLFeBOOK R R p w w b s m Beams of uniform strength (in bending). FIGURE 2.17 (Continued) a b 78
• 40816_02_p52-98 10/22/01 12:41 PM Page 79p. 78 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 79 BEAM FORMULAS 79 ROLLING AND MOVING LOADS Rolling and moving loads are loads that may change their position on a beam or beams. 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 diagram shows that the maximum moment occurs under a wheel. Thus, with x Ͻ a/2: ΂ R1 ϭ P 1 Ϫ 2x l ϩ a l ΃ M2 ϭ Pl 2 1Ϫ΂a l ϩ 2x a l l 4x 2 Ϫ 2 l ΃ ΂ R2 ϭ P 1 ϩ 2x l Ϫ a l ΃ Pl ΂1 Ϫ al Ϫ 2a ΃ 2 2x 3a 4x 2 M1 ϭ ϩ Ϫ 2 2 l 2 l l l M2 max when x ϭ 1͞4a M1 max when x ϭ 3͞4a Pl ΂1 Ϫ 2l ΃ ϭ 2l ΂l Ϫ 2 ΃ a P a 2 2 M max ϭ 2 Figure 2.19 shows the condition when two equal loads are equally distant on opposite sides of the center of the beam. The moment is then equal under the two loads. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 80 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 80 408 t r b t s b t p FIGURE 2.18 Two equal concentrated moving loads. FIGURE 2.19 Two equal moving loads equally distant on oppo- site sides of the center. F 80 TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 81p. 80 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 81 BEAM FORMULAS 81 If two moving loads are of unequal weight, the condi- tion for maximum moment is the maximum moment occur- ring under the heavier wheel when the center of the beam bisects the distance between the resultant of the loads and the heavier wheel. Figure 2.20 shows this position and the shear and moment diagrams. When several wheel loads constituting a system are on a beam or beams, the several wheels must be examined in turn to determine which causes the greatest moment. The position for the greatest moment that can occur under a given - FIGURE 2.20 Two moving loads of unequal weight. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 82 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 82 408 82 CHAPTER TWO wheel is, as stated earlier, when the center of the span bisects the distance between the wheel in question and the resultant of all loads then on the span. The position for max- imum shear at the support is when one wheel is passing off the span. CURVED BEAMS The application of the flexure formula for a straight beam to the case of a curved beam results in error. When all “fibers” of a member have the same center 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 at a point y units from the centroidal axis is Sϭ M AR ΄1 ϩ Z (Ryϩ y) ΅ Z M is the bending moment, positive when it increases curva- f ture; y is positive when measured toward the convex side; A is the cross-sectional area; R is the radius of the centroidal i axis; Z is a cross-section property deﬁned by Zϭ Ϫ 1 A ͵ y Rϩy dA TABLE 2 4 A l i l E Analytical expressions for Z of certain sections are given in Table 2.4. Z can also be found by graphical integration methods (see any advanced strength book). The neutral surface shifts toward the center of curvature, or inside ﬁber, an amount equal to e ϭ ZR/(Z ϩ 1). The Winkler–Bach TLFeBOOK
• m A p. 82 h n n h n e s - f - l l l l , , 40816_02_p52-98 40816 HICKS Mcghp Ch02 TABLE 2.4 Analytical Expressions for Z Section Expression R ΂ln R Ϫ C ΃ RϩC 10/22/01 Z ϭ Ϫ1 ϩ h ΂ R ΃ ΄ R Ϫ √΂ R ΃ Ϫ 1 ΅ 2 Z ϭ Ϫ1 ϩ 2 12:41 PM r r r83 Second Pass Z ϭ Ϫ1 ϩ R A ΄ t ln (R ϩ C1) ϩ (b Ϫ t) ln (R Ϫ C0) Ϫ b ln (R Ϫ C2) ΅ and A ϭ tC1 Ϫ (b Ϫ t) C3 ϩ bC2 Page 83 7/3/01 pb p. 83 Z ϭ Ϫ1 ϩ R A ΄ ΂ b ln R ϩ C2 R Ϫ C2 ΃ ϩ (t Ϫ b) ln ΂ R ϩ C ΃ ΅ RϪC 1 1 A ϭ 2 [(t Ϫ b)C1 ϩ bC2] TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 84 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 84 408 84 CHAPTER TWO B FIGURE 2.21 Curved beam. theory, though practically satisfactory, disregards radial stresses as well as lateral deformations and assumes pure B bending. The maximum stress occurring on the inside ﬁber is S ϭ Mhi /AeRi, whereas that on the outside ﬁber is S ϭ Mh0 /AeR0. The deﬂection in curved beams can be computed by means of the moment-area theory. The resultant deﬂection is then equal to ⌬0 ϭ √⌬2 ϩ ⌬2 x y in the direction deﬁned by tan ␪ ϭ ⌬y / ⌬x. Deﬂections can also be found conveniently by use of Castigliano’s theorem. It states that in an elastic system the displacement in the direction of a force (or couple) and due to that force (or cou- ple) is the partial derivative 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 F is applied in the radial direction at free-end B. Then, the deﬂection of B is F TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 85p. 84 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 85 BEAM FORMULAS 85 By moment area, y ϭ R sin ␪ x ϭ R(1 Ϫ cos ␪) ds ϭ Rd␪ M ϭ FR sin ␪ ␲FR 3 FR 3 B⌬x ϭ B⌬y ϭϪ 4EI 2EI FR3 ␲2 and ⌬B ϭ 2EI √ 1ϩ 4 at ␪x ϭ tan Ϫ1 Ϫ ΂ FR 3 2EI ϫ 4EI ␲FR 3 ΃ 2 ϭ tan Ϫ1 ␲ ϭ 32.5Њ le By Castigliano, rϭ ␲FR3 FR3 B⌬x ϭ B⌬y ϭϪ 4EI 2EIy 2 yn .e-tnt FIGURE 2.22 Quadrant with ﬁxed end. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 86 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 86 4 86 CHAPTER TWO Eccentrically Curved Beams T ( These beams (Fig. 2.23) are bounded by arcs having differ- ent centers of curvature. In addition, it is possible for either 1 radius to be the larger one. The one in which the section depth shortens as the central section is approached may be called the arch beam. When the central section is the largest, the beam is of the crescent type. Crescent I denotes the beam of larger outside radius and crescent II of larger inside radius. The stress at the central section of such beams may be found from S ϭ KMC/I. In the case of rectangular cross section, the equation 2 becomes S ϭ 6KM/bh2, where M is the bending moment, b is the width of the beam section, and h its height. The stress factors, K for the inner boundary, established from photo- elastic data, are given in Table 2.5. The outside radius 3 i c c FIGURE 2.23 Eccentrically curved beams. a TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 87p. 86 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 87 BEAM FORMULAS 87 TABLE 2.5 Stress Factors for Inner Boundary at Central Section (see Fig. 2.23)-r 1. For the arch-type beamsne h Ro ϩ Ri (a) K ϭ 0.834 ϩ 1.504 if Ͻ5e Ro ϩ Ri h h R ϩ Rid (b) K ϭ 0.899 ϩ 1.181 if 5Ͻ o Ͻ 10 Ro ϩ Ri h l (c) In the case of larger section ratios use the equivalent beamI. solutionn 2. For the crescent I-type beamsb s h Ro ϩ Ri (a) K ϭ 0.570 ϩ 1.536 if Ͻ2 - Ro ϩ Ri h s h Ro ϩ Ri (b) K ϭ 0.959 ϩ 0.769 if 2Ͻ Ͻ 20 Ro ϩ Ri h ΂R ϩR ΃ h Ro ϩ Ri 0.0298 (c) K ϭ 1.092 if Ͼ 20 o i h 3. For the crescent II-type beams h Ro ϩ Ri (a) K ϭ 0.897 ϩ 1.098 if Ͻ8 Ro ϩ Ri h ΂R ϩR ΃ h Ro ϩ Ri 0.0378 (b) K ϭ 1.119 if 8Ͻ Ͻ 20 o i h ΂ h ΃ Ro ϩ Ri 0.0270 (c) K ϭ 1.081 if Ͼ 20 Ro ϩ Ri h is denoted by Ro and the inside by Ri. The geometry of crescent beams is such that the stress can be larger in off- center sections. The stress at the central section determined above must then be multiplied by the position factor k, TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 88 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 88 408 88 CHAPTER TWO given in Table 2.6. As in the concentric beam, the neutral surface shifts slightly toward the inner boundary. (See Vidosic, “Curved Beams with Eccentric Boundaries,” Transactions of the ASME, 79, pp. 1317–1321.) w s ELASTIC LATERAL BUCKLING OF BEAMS m f When lateral buckling of a beam occurs, the beam under- goes a combination of twist and out-of-plane bending (Fig. 2.24). For a simply supported beam of rectangular T cross section subjected to uniform bending, buckling occurs at the critical bending moment, given by ( ␲ A Mcr ϭ √EIyGJ L where L ϭ unbraced length of the member E ϭ modulus of elasticity Iy ϭ moment of inertial about minor axis G ϭ shear modulus of elasticity J ϭ torsional constant The critical moment is proportional to both the lateral bending stiffness EIy /L and 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 tor- sional stiffness. Buckling of a simply supported beam of open † cross section subjected to uniform bending occurs at the b critical bending moment, given by b TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 89p. 88 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 89 BEAM FORMULAS 89le” Mcr ϭ ␲ L √ ΂ EIy GJ ϩ ECw ␲2 L2 where Cw is the warping constant, a function of cross- sectional shape and dimensions (Fig. 2.25). In the preceding equations, the distribution of bending moment is assumed to be uniform. For the case of a nonuni- form bending-moment gradient, buckling often occurs at-gr TABLE 2.6 Crescent-Beam Position Stress Factorss (see Fig. 2.23)† k Angle ␪, degree Inner Outer 10 1 ϩ 0.055 H/h 1 ϩ 0.03 H/h 20 1 ϩ 0.164 H/h 1 ϩ 0.10 H/h 30 1 ϩ 0.365 H / h 1 ϩ 0.25 H/h 40 1 ϩ 0.567 H / h 1 ϩ 0.467 H/h (0.5171 Ϫ 1.382 H/h)1/2 50 1.521 Ϫ 1 ϩ 0.733 H/h 1.382 (0.2416 Ϫ 0.6506 H/h)1/2 60 1.756 Ϫ 1 ϩ 1.123 H/h 0.6506 (0.4817 Ϫ 1.298 H/h)1/2 70 2.070 Ϫ 1 ϩ 1.70 H/hl 0.6492e (0.2939 Ϫ 0.7084 H/h)1/2 80 2.531 Ϫ 1 ϩ 2.383 H/h 0.3542r- 90 1 ϩ 3.933 H/hn † Note: All formulas are valid for 0 Ͻ H/h Յ 0.325. Formulas for the innere boundary, except for 40 degrees, may be used to H/h Յ 0.36. H ϭ distance between centers. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 90 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 90 408 90 CHAPTER TWO F A I Y FIGURE 2.24 (a) Simple beam subjected to equal end moments. (b) Elastic lateral buckling of the beam. a a larger critical moment. Approximation of this critical Ј bending moment, Mcr may be obtained by multiplying Mcr given by the previous equations by an ampliﬁcation factor M cr ϭ Cb Mcr Ј 12.5Mmax where Cb ϭ 2.5Mmax ϩ 3MA ϩ 4MB ϩ 3MC TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 91p. 90 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 91 BEAM FORMULAS 91 FIGURE 2.25 Torsion-bending constants for torsional buckling. A ϭ cross-sectional area; Ix ϭ moment of inertia about x–x axis; Iy ϭ moment of inertia about y–y axis. (After McGraw-Hill, New York). Bleich, F., Buckling Strength of Metal Structures.s. and Mmax ϭ absolute value of maximum moment in the unbraced beam segmentl MA ϭ absolute value of moment at quarter point ofr the unbraced beam segmentr MB ϭ absolute value of moment at centerline of the unbraced beam segment MC ϭ absolute value of moment at three-quarter point of the unbraced beam segment TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 92 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 92 408 92 CHAPTER TWO Cb equals 1.0 for unbraced cantilevers and for members where the moment within a signiﬁcant portion of the unbraced segment is greater than, or equal to, the larger of the segment end moments. w s a o COMBINED AXIAL AND BENDING LOADS a For short beams, subjected to both transverse and axial loads, the stresses are given by the principle of superposition if the deﬂection due to bending may be neglected without serious error. That is, the total stress is given with sufﬁcient w accuracy at any section by the sum of the axial stress and ( the bending stresses. The maximum stress, lb/in2 (MPa), equals U P Mc fϭ ϩ A I W where P ϭ axial load, lb (N ) c m A ϭ cross-sectional area, in2 (mm2) t a M ϭ maximum bending moment, in lb (Nm) p p c ϭ distance from neutral axis to outermost ﬁber at the section where maximum moment occurs, in (mm) I ϭ moment of inertia about neutral axis at that section, in4 (mm4) w When the deﬂection due to bending is large and the axial load produces bending stresses that cannot be neglect- ed, the maximum stress is given by TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 93p. 92 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 93 BEAM FORMULAS 93s P ce fϭ ϩ (M ϩ Pd) A If where d is the deﬂection of the beam. For axial compres- sion, the moment Pd should be given the same sign as M; and for tension, the opposite sign, but the minimum value of M ϩ Pd is zero. The deﬂection d for axial compression and bending can be closely approximated byl do dϭn 1 Ϫ (P/Pc)tt where do ϭ deﬂection for the transverse loading alone, ind (mm); and Pc ϭ critical buckling load ␲2EI / L2, lb (N). , UNSYMMETRICAL BENDING When a beam is subjected to loads that do not lie in a plane containing a principal axis of each cross section, unsym- metrical bending occurs. Assuming that the bending axis of the beam lies in the plane of the loads, to preclude torsion, and that the loads are perpendicular to the bending axis, to preclude axial components, the stress, lb/in2 (MPa), at any point in a cross section ist, Mx y M x fϭ ϩ y Ix Iyt where Mx ϭ bending moment about principal axis XX, in lb (Nm)e- My ϭ bending moment about principal axis YY, in lb (Nm) TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 94 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 94 408 94 CHAPTER TWO x ϭ distance from point where stress is to be computed to YY axis, in (mm) y ϭ distance from point to XX axis, in (mm) Ix ϭ moment of inertia of cross section about XX, F in (mm4) c Iy ϭ moment of inertia about YY, in (mm4) a s If the plane of the loads makes an angle ␪ with a principal f plane, the neutral surface forms an angle ␣ with the other n principal plane such that s D Ix tan ␣ ϭ tan ␪ Iy t s m t ECCENTRIC LOADING If an eccentric longitudinal load is applied to a bar in the P plane of symmetry, it produces a bending moment Pe, where e is the distance, in (mm), of the load P from the o centroidal axis. The total unit stress is the sum of this a moment and the stress due to P applied as an axial load: ( fϭ P A Ϯ Pec I ϭ P A ΂1 Ϯ ec ΃ r 2 where A ϭ cross-sectional area, in2 (mm2) w c ϭ distance from neutral axis to outermost ﬁber, in (mm) TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 95p. 94 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 95 BEAM FORMULAS 95 I ϭ moment of inertia of cross section about neutral axis, in4 (mm4) r ϭ radius of gyration ϭ √I/A, in (mm) Figure 2.1 gives values of the radius of gyration for several cross sections. If there is to be no tension on the cross section under a compressive load, e should not exceed r 2/c. For a rectangular section with width b, and depth d, the eccentricity, there-l fore, should be less than b/6 and d/6 (i.e., the load shouldr not be applied outside the middle third). For a circular cross section with diameter D, the eccentricity should not exceed D/8. When the eccentric longitudinal load produces a deﬂec- tion too large to be neglected in computing the bending stress, account must be taken of the additional bending moment Pd, where d is the deﬂection, in (mm). This deﬂec- tion may be closely approximated by 4eP/Pc dϭ ␲(1 Ϫ P/Pc)e Pc is the critical buckling load ␲2EI/L2, lb (N)., If the load P, does not lie in a plane containing an axise of symmetry, it produces bending about the two principals axes through the centroid of the section. The stresses, lb/in2 (MPa), are given by P Pexcx Peycy fϭ ϩ ϩ A Iy Ix where A ϭ cross-sectional area, in2 (mm2)n ex ϭ eccentricity with respect to principal axis YY, in (mm) TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 96 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 96 408 96 CHAPTER TWO ey ϭ eccentricity with respect to principal axis XX, in (mm) cx ϭ distance from YY to outermost ﬁber, in (mm) cy ϭ distance from XX to outermost ﬁber, in (mm) Ix ϭ moment of inertia about XX, in4 (mm4) Iy ϭ moment of inertia about YY, in4 (mm4) The principal axes are the two perpendicular axes through the centroid for which the moments of inertia are a maxi- mum or a minimum and for which the products of inertia are zero. NATURAL CIRCULAR FREQUENCIES AND NATURAL PERIODS OF VIBRATION OF PRISMATIC BEAMS Figure 2.26 shows the characteristic shape and gives con- stants for determination of natural circular frequency ␻ and natural period T, for the ﬁrst four modes of cantilever, simply supported, ﬁxed-end, and ﬁxed-hinged beams. To obtain ␻, select the appropriate constant from Fig. 2.26 and multiply it by √EI/wL4. To get T, divide the appropriate constant by √EI/wL4. In these equations, ␻ ϭ natural frequency, rad/s W ϭ beam weight, lb per linear ft (kg per linear m) L ϭ beam length, ft (m) TLFeBOOK
• p. 96 d o d h e a - - ) ) , , 40816_02_p52-98 40816 HICKS Mcghp Ch02 10/22/01 12:41 PM97 Second Pass Page 97 7/3/01 pb p. 97 FIGURE 2.26 Coefﬁcients for computing natural circular frequencies and natural periods of vibration of prismatic beams. TLFeBOOK
• 40816_02_p52-98 10/22/01 12:41 PM Page 98 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 98 98 CHAPTER TWO E ϭ modulus of elasticity, lb/in2 (MPa) I ϭ moment of inertia of beam cross section, in4 (mm4) T ϭ natural period, s To determine the characteristic shapes and natural peri- ods for beams with variable cross section and mass, use the Rayleigh method. Convert the beam into a lumped-mass system by dividing the span into elements and assuming the mass of each element to be concentrated at its center. Also, compute all quantities, such as deﬂection and bending moment, at the center of each element. Start with an assumed charac- teristic shape. ] TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 100 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 100 40816 100 CHAPTER THREE GENERAL CONSIDERATIONS Columns are structural members subjected to direct com- pression. All columns can be grouped into the following three classes: 1. Compression blocks are so short (with a slenderness ratio — that is, unsupported length divided by the least radius of gyration of the member — below 30) that bend- TABLE 3 1 Strength of Round-Ended Columns According to Euler’s Formula* ing is not potentially occurring. 2. Columns so slender that bending under load is a 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 is reached. Long columns are analyzed using Euler’s column formula, namely, n␲ 2EI n␲ 2EA Pcr ϭ ϭ l 2 (l/r)2 In this formula, the coefﬁcient n accounts for end condi- tions. When the column is pivoted at both ends, n ϭ 1; when one end is ﬁxed and the other end is rounded, n ϭ 2; when both ends are ﬁxed, n ϭ 4; and when one end is ﬁxed and the other is free, n ϭ 0.25. The slenderness ratio sepa- rating long columns from short columns depends on the modulus of elasticity and the yield strength of the column material. When Euler’s formula results in (Pcr /A) > Sy, strength instead of buckling causes failure, and the column ceases to be long. In quick estimating numbers, this critical slenderness ratio falls between 120 and 150. Table 3.1 gives additional column data based on Euler’s formula. TLFeBOOK
• n n d d n g e e s s - - - - l ; ; l t , . p. 100 40816_03_p99-130 40816 HICKS Mcghp Chap_03 TABLE 3.1 Strength of Round-Ended Columns According to Euler’s Formula* Low- Medium- Wrought carbon carbon Material† Cast iron iron steel steel 2 Ultimate compressive strength, lb/in 107,000 53,400 62,600 89,000 10/22/01 Allowable compressive stress, lb/in2 (maximum) 7,100 15,400 17,000 20,000 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 12:42 PM101 17,500,000 56,000,000 60,300,000 61,700,000 SECOND PASS 7/5/01 pb p. 101 Limit of ratio, l/r Ͼ 50.0 60.6 59.4 55.6 ΂ ΃ Rectangle r ϭ b√1͞12 , l/b Ͼ 14.4 17.5 17.2 16.0 Circle ΂r ϭ 1͞4 d΃, l/d Ͼ 12.5 15.2 14.9 13.9 Page 101 Circular ring of small thickness ΂r ϭ d √ ͞ ΃, l/d Ͼ 1 8 17.6 21.4 21.1 19.7 * (P ϭ allowable load, lb; l ϭ length of column, in; b ϭ smallest dimension of a rectangular section, in; d ϭ diameter of a circular section, in; r ϭ least radius of gyration of section.) † To convert to SI units, use: 1b/in2 ϫ 6.894 ϭ kPa; in4 ϫ (25.4)4 ϭ mm4. TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 102 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 102 40816 102 CHAPTER THREE FIGURE 3.1 L/r plot for columns. SHORT COLUMNS Stress in short columns can be considered to be partly due to compression and partly due to bending. Empirical, rational expressions for column stress are, in general, based on the assumption that the permissible stress must be reduced below that which could be permitted were it due to compression only. The manner in which this reduction is made determines the type of equation and the slenderness ratio beyond which the equation does not apply. Figure 3.1 shows the curves for this situation. Typical column formulas are given in Table 3.2. ECCENTRIC LOADS ON COLUMNS When 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 stress results. The maximum unit stress SM is the algebraic sum of these two unit stresses. TLFeBOOK
• w o n h n o e e s r l , . p. 102 40816_03_p99-130 40816 HICKS Mcghp Chap_03 TABLE 3.2 Typical Short-Column Formulas Formula Material Code Slenderness ratio 10/22/01 l ΂ ΃ l 2 Sw ϭ 17,000 Ϫ 0.485 Carbon steels AISC Ͻ 120 r r Sw ϭ 16,000 Ϫ 70΂ rl ΃ Carbon steels Chicago l r Ͻ 120 S ϭ 15,000 Ϫ 50 ΂ ΃ l l103 12:42 PM Carbon steels AREA Ͻ 150 SECOND PASS 7/5/01 pb p. 103 w r r S ϭ 19,000 Ϫ 100 ΂ ΃ l l w Carbon steels Am. Br. Co. 60 Ͻ Ͻ 120 r r c ΂r΃ 15.9 l 2 l † S ϭ 135,000 Ϫ Ͻ 65 Page 103 cr Alloy-steel tubing ANC √cr S ϭ 9,000 Ϫ 40 ΂ ΃ l l w Cast iron NYC Ͻ 70 r r TLFeBOOK
• 40816_03_p99-130 40816 HICKS Mcghp Chap_03 TABLE 3.2 Typical Short-Column Formulas (Continued) Formula Material Code Slenderness ratio † Scr ϭ 34,500 Ϫ 245 ΂ ΃ l 1 Ͻ 94 10/22/01 2017ST aluminum ANC √c r √cr ΂ rl ΃ 2 0.5 1 † Scr ϭ 5,000 Ϫ Spruce ANC Ͻ 72 c √cr104 ΄ Sy ΂ rl ΃ ΅ l 2n␲2E 2 SECOND PASS 7/5/01 pb p. 104 √ 12:42 PM † Scr ϭ Sy 1 Ϫ Steels Johnson Ͻ 4n␲ 2E r Sy Sy l ‡ Scr ϭ Steels Secant Ͻ critical r ec 1 ϩ 2 sec r ΂ l r √ ΃ P 4AE Page 104 † Scr ϭ theoretical maximum; c ϭ end ﬁxity coefﬁcient; c ϭ 2, both ends pivoted; c ϭ 2.86, one pivoted, other ﬁxed; c ϭ 4, both ends ﬁxed; c ϭ 1 one ﬁxed, one free. ‡ is initial eccentricity at which load is applied to center of column cross section. TLFeBOOK 40816 t y a t a d F
• 40816_03_p99-130 10/22/01 12:42 PM Page 105p. 104 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 105 COLUMN FORMULAS 105 FIGURE 3.2 Load plot for columns. In Fig. 3.2, a load, P, acts in a line of symmetry at the distance e from cg; r ϭ radius 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 ϭ Sc Ϯ Sb ϭ (P/A) Ϯ Pey/I ϭ Sc(1 Ϯ ey/r 2) y is positive for points on the same side of cg as P, and nega- tive on the opposite side. For a rectangular cross section of TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 106 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 106 40816 106 CHAPTER THREE width b, the maximum stress, SM ϭ Sc (1 ϩ 6e/b). When P is outside the middle third of width b and is a compressive m load, tensile stresses occur. s For a circular cross section of diameter d, SM ϭ Sc(1 ϩ t 8e/d). The stress due to the weight of the solid modiﬁes t these relations. 3 Note that in these formulas e is measured from the grav- 0 ity axis and gives tension when e is greater than one-sixth f the width (measured in the same direction as e), for rectan- p gular sections, and when greater than one-eighth the diam- t eter, for solid circular sections. ( a FIGURE 3.3 Load plot for columns. F TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 107 p. 106 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 107 COLUMN FORMULAS 107P If, as in certain classes of masonry construction, thee material cannot withstand tensile stress and, thus, no ten- sion can occur, the center of moments (Fig. 3.3) is taken atϩ the center of stress. For a rectangular section, P acts at dis- s tance k from the nearest edge. Length under compression ϭ 3k, and SM ϭ ͞3P/hk. For a circular section, SM ϭ [0.372 ϩ 2- 0.056(k/r)]P/k √rk , where r ϭ radius and k ϭ distance of Ph from circumference. For a circular ring, S ϭ average com-- pressive stress on cross section produced by P; e ϭ eccen-- tricity of P; z ϭ length of diameter under compression (Fig. 3.4). Values of z/r and of the ratio of Smax to average S are given in Tables 3.3 and 3.4. FIGURE 3.4 Circular column load plot. TLFeBOOK
• 40816_03_p99-130 40816 HICKS Mcghp Chap_03 TABLE 3.3 Values of the Ratio z/r (See Fig. 3.5) r1 e r e r 0.0 0.5 0.6 0.7 0.8 0.9 1.0 r 10/22/01 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 0.40108 0.45 1.37 1.61 1.71 1.81 1.90 0.45 SECOND PASS 7/5/01 pb p. 108 12:42 PM 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 Page 108 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 TLFeBOOK 40816
• p. 108 40816_03_p99-130 40816 HICKS Mcghp Chap_03 TABLE 3.4 Values of the Ratio Smax/Savg (In determining S average, use load P divided by total area of cross section) 10/22/01 r1 e r e r 0.0 0.5 0.6 0.7 0.8 0.9 1.0 r 0.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 0.00109 12:42 PM 0.05 1.20 1.16 1.15 1.13 1.12 1.11 1.10 0.05 SECOND PASS 7/5/01 pb p. 109 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 Page 109 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 TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 110 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 110 40816 110 CHAPTER THREE The kern is the area around the center of gravity of a cross section within which any load applied produces stress of only one sign throughout the entire cross section. Outside the kern, a load produces stresses of different sign. Figure 3.5 shows kerns (shaded) for various sections. c For a circular ring, the radius of the kern r ϭ 0 D[1ϩ(d/D)2]/8. w For a hollow square (H and h ϭ lengths of outer and inner sides), the kern is a square similar to Fig. 3.5a, where C B a t T u t w t FIGURE 3.5 Column characteristics. 1 TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 111 p. 110 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 111 COLUMN FORMULAS 111 ΄1 ϩ ΂ H ΃ ΅ ϭ 0.1179H ΄1 ϭ ΂ H ΃ ΅f H 1 h 2 h 2s rmin ϭs 6 √2fs For a hollow octagon, Ra and Ri ϭ radii of circles cir- cumscribing the outer and inner sides; thickness of wall ϭϭ 0.9239(Ra – Ri); and the kern is an octagon similar to Fig. 3.5c, where 0.2256R becomes 0.2256Ra[1 ϩ (Ri /Ra)2].de COLUMN BASE PLATE DESIGN Base plates are usually used to distribute column loads over a large enough area of supporting concrete construction that the design bearing strength of the concrete is not exceeded. The factored load, Pu, is considered to be uniformly distrib- uted under a base plate. The nominal bearing strength fp kip/in2 or ksi (MPa) of the concrete is given by A1 A2 fp ϭ 0.85f c Ј √ A1 and √ A1 Յ2 where f Ј ϭ speciﬁed compressive strength of concrete, c ksi (MPa) A1 ϭ area of the base plate, in2 (mm2) A2 ϭ area of the supporting concrete that is geo- metrically similar to and concentric with the loaded area, in2 (mm2) Ј In most cases, the bearing strength, fp is 0.85f c , when the concrete support is slightly larger than the base plate or Ј 1.7f c , when the support is a spread footing, pile cap, or mat TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 112 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 112 40816 112 CHAPTER THREE foundation. Therefore, the required area of a base plate for a factored load Pu is Pu A1 ϭ ␾c0.85 fcЈ where ␾c is the strength reduction factor ϭ 0.6. For a wide- ﬂange column, A1 should not be less than bf d, where bf is the ﬂange width, in (mm), and d is the depth of column, in (mm). The length N, in (mm), of a rectangular base plate for a wide-ﬂange column may be taken in the direction of d as N ϭ √A1 ϩ ⌬ Ͼ d or ⌬ ϭ 0.5(0.95d Ϫ 0.80bf) A C The width B, in (mm), parallel to the ﬂanges, then, is D A1 Bϭ N T a The thickness of the base plate tp, in (mm), is the largest m of the values given by the equations that follow: 2Pu tp ϭ m √ 0.9Fy BN 2Pu w tp ϭ n √ 0.9Fy BN p 2Pu i t p ϭ ␭nЈ √ 0.9Fy BN c t where m ϭ projection of base plate beyond the ﬂange s and parallel to the web, in (mm) t i ϭ (N Ϫ 0.95d)/2 c TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 113 p. 112 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 113 COLUMN FORMULAS 113r n ϭ projection of base plate beyond the edges of the ﬂange and perpendicular to the web, in (mm) ϭ (B Ϫ 0.80bf)/2 nЈ ϭ √(dbf)/4- ␭ ϭ (2√X)/[1 ϩ √(1 Ϫ X)] Յ 1.0s, X ϭ [(4 dbf)/(d ϩ bf)2][Pu/(␾ ϫ 0.85fЈ 〈1)] ca AMERICAN INSTITUTE OF STEEL CONSTRUCTION ALLOWABLE-STRESS DESIGN APPROACH The lowest columns of a structure usually are supported on a concrete foundation. The area, in square inches (squaret millimeters), required is found from: P Aϭ FP where P is the load, kip (N) and Fp is the allowable bearing pressure on support, ksi (MPa). The allowable pressure depends on strength of concrete in the foundation and relative sizes of base plate and con- crete support area. If the base plate occupies the full area of the support, Fp ϭ 0.35f c , where f c is the 28-day compres- Ј Ј sive strength of the concrete. If the base plate covers less than the full area, FP ϭ 0.35fc √A2/A1 Յ 0.70f c , where A1 Ј Ј is the base-plate area (B ϫ N), and A2 is the full area of the concrete support. TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 114 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 114 40816 114 CHAPTER THREE Eccentricity of loading or presence of bending moment T at the column base increases the pressure on some parts s of the base plate and decreases it on other parts. To com- b pute these effects, the base plate may be assumed com- b pletely rigid so that the pressure variation on the concrete t is linear. t Plate thickness may be determined by treating projections m and n of the base plate beyond the column as cantilevers. w c p s t C T f d t FIGURE 3.6 Column welded to a base plate. c TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 115 p. 114 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 115 COLUMN FORMULAS 115t The cantilever dimensions m and n are usually deﬁned ass shown in Fig. 3.6. (If the base plate is small, the area of the- base plate inside the column profile should be treated as a- beam.) Yield-line analysis shows that an equivalent can-e tilever dimension nЈ can be defined as nЈ ϭ 1͞4√dbf , and the required base plate thickness tp can be calculated froms. fp tp ϭ 2l √ Fy where l ϭ max (m, n, nЈ), in (mm) fp ϭ P/(BN) Յ Fp, ksi (MPa) Fy ϭ yield strength of base plate, ksi (MPa) P ϭ column axial load, kip (N) For columns subjected only to direct load, the welds of column to base plate, as shown in Fig. 3.6, are required principally for withstanding erection stresses. For columns subjected to uplift, the welds must be proportioned to resist the forces. COMPOSITE COLUMNS The AISC load-and-resistance factor design (LRFD) speci- fication for structural steel buildings contains provisions for design of concrete-encased compression members. It sets the following requirements for qualiﬁcation as a composite column: The cross-sectional area of the steel core—shapes, TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 116 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 116 40816 116 CHAPTER THREE pipe, or tubing—should be at least 4 percent of the total w composite area. The concrete should be reinforced with longitudinal load-carrying bars, continuous at framed levels, and lateral ties and other longitudinal bars to restrain the concrete; all should have at least 1 1͞2 in (38.1 mm) of clear concrete cover. The cross-sectional area of transverse 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 Ј the composite section. Strength of the concrete f c should be between 3 and 8 ksi (20.7 and 55.2 MPa) for normal- weight concrete and at least 4 ksi (27.6 MPa) for light- weight concrete. Speciﬁed minimum 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 should be at least b√Fy /3E or D√Fy /8E, where b is the width of the face of a rectangular section, D is the out- side 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 loaded composite column as ␾Pn, where ␾ ϭ 0.85 and Pn is determined from F a ␾Pn ϭ 0.85As Fcr 0 For ␭c Յ 1.5 s o b 2 Fcr ϭ 0.658␭c Fmy e For ␭c Ͼ 1.5 d t d 0.877 s Fcr ϭ Fmy ␭2 c A TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 117 p. 116 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 117 COLUMN FORMULAS 117l where ␭c ϭ (KL/rm␲)√Fmy /E mh , KL ϭ effective length of column in (mm)er As ϭ gross area of steel core in2 (mm2)d Fmy ϭ Fy ϩ c1Fyr(Ar /As) ϩ c2 fcЈ(Ac /As)2s Em ϭ E ϩ c3Ec(Ac /As)fd rm ϭ radius of gyration of steel core, in Յ 0.3 of the- overall thickness of the composite cross section- in the plane of buckling for steel shapesl4 Ac ϭ cross-sectional area of concrete in2 (mm2)h Ar ϭ area of longitudinal reinforcement in2 (mm2)s- Ec ϭ elastic modulus of concrete ksi (MPa)- Fyr ϭ speciﬁed minimum yield stress of longitudi-h nal reinforcement, ksi (MPa)ϭ For concrete-ﬁlled pipe and tubing, c1 ϭ 1.0, c2 ϭ 0.85, and c3 ϭ 0.4. For concrete-encased shapes, c1 ϭ 0.7, c2 ϭ 0.6, and c3 ϭ 0.2. When the steel core consists of two or more steel shapes, they should be tied together with lacing, tie plates, or batten plates to prevent buckling of individual shapes Ј before the concrete attains 0.75 f c . The portion of the required strength of axially loaded encased composite columns resisted by concrete should be developed by direct bearing at connections or shear connec- tors can be used to transfer into the concrete the load applied directly to the steel column. For direct bearing, the design strength of the concrete is 1.7␾c fcЈ Ab, where ␾c ϭ 0.65 and Ab ϭ loaded area, in2 (mm2). Certain restrictions apply. TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 118 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 118 40816 118 CHAPTER THREE ELASTIC FLEXURAL BUCKLING OF COLUMNS Elastic buckling is a state of lateral instability that occurs while the material is stressed below the yield point. It is of special importance in structures with slender members. Euler’s formula for pin-ended columns (Fig. 3.7) gives valid results for the critical buckling load, kip (N). This formula is, with L/r as the slenderness ratio of the column, ␲2EA Pϭ (L/r)2 where E ϭ modulus of elasticity of the column material, psi (Mpa) A ϭ column cross-sectional area, in2 (mm2) r ϭ radius of gyration of the column, in (mm) Figure 3.8 shows some ideal end conditions for slender columns and corresponding critical buckling loads. Elastic critical buckling loads may be obtained for all cases by sub- stituting an effective length KL for the length L of the pinned column, giving ␲2EA Pϭ (KL/r)2 F In some cases of columns with open sections, such as a l cruciform section, the controlling buckling mode may be one of twisting instead of lateral deformation. If the warp- w ing rigidity of the section is negligible, torsional buckling in a pin-ended column occurs at an axial load of GJA Pϭ Ip TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 119 p. 118 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 119 COLUMN FORMULAS 119sfssh,rc-e FIGURE 3.7 (a) Buckling of a pin-ended column under axiala load. (b) Internal forces hold the column in equilibrium.e- where G ϭ shear modulus of elasticityg J ϭ torsional constant A ϭ cross-sectional area Ip ϭ polar moment of inertia ϭ Ix ϩ Iy TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 120 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 120 40816 I r w s A A FIGURE 3.8 Buckling formulas for columns. E d s e E J u 120 TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 121 p. 120 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 121 COLUMN FORMULAS 121 If the section possesses a signiﬁcant amount of warping rigidity, the axial buckling load is increased to A ΂GJ ϩ ␲ L ΃ EC 2 w Pϭ Ip 2 where Cw is the warping constant, a function of cross-sectional shape and dimensions. ALLOWABLE DESIGN LOADS FOR ALUMINUM COLUMNSFIGURE 3.8 Buckling formulas for columns. Euler’s equation is used for long aluminum columns, and depending on the material, either Johnson’s parabolic or straight-line equation is used for short columns. These equations for aluminum follow: Euler’s equation: c␲2E Fe ϭ (L/␳)2 Johnson’s generalized equation: ΄ ΂ √ ΃΅ (L/␳) n Fc ϭ Fce 1 Ϫ K cE ␲ Fce The value of n, which determines whether the short col- umn formula is the straight-line or parabolic type, is selected TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 122 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 122 40816 122 CHAPTER THREE from Table 3.5. The transition from the long to the short column range is given by ΂L΃ ␳ cr ϭ␲ √ kcE Fce where Fe ϭ allowable column compressive stress Fce ϭ column yield stress and is given as a function of Fcy (compressive yield stress) L ϭ length of column ␳ ϭ radius of gyration of column E ϭ modulus of elasticity—noted on nomograms c ϭ column-end ﬁxity from Fig. 3.9 n, K, k ϭ constants from Table 3.5 FIGURE 3.9 Values of c, column-end ﬁxity, for determining the critical L/␳ ratio of different loading conditions. TLFeBOOK
• n e s t p. 122 40816_03_p99-130 40816 HICKS Mcghp Chap_03 10/22/01 TABLE 3.5 Material Constants for Common Aluminum Alloys Average Values Fcy Fce Type Johnson Material psi MPa psi MPa K k n equation123 12:42 PM SECOND PASS 7/5/01 pb p. 123 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 Page 123 Ref: ANC-5. TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 124 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 124 40816 124 CHAPTER THREE ULTIMATE STRENGTH DESIGN CONCRETE COLUMNS At ultimate strength Pu, kip (N), columns should be capable of sustaining loads as given by the American Concrete Institute required strength equations in Chap. 5, “Concrete w Formulas” at actual eccentricities. Pu , may not exceed ␾Pn , where ␾ is the capacity reduction factor and Pn , kip (N), is the column ultimate strength. If Po, kip (N), is the column ultimate strength with zero eccentricity of load, then Po ϭ 0.85 f c (Ag Ϫ Ast) ϩ fy Ast Ј where fy ϭ yield strength of reinforcing steel, ksi (MPa) f c ϭ 28-day compressive strength of concrete, Ј ksi (MPa) Ag ϭ gross area of column, in2 (mm2) Ast ϭ area of steel reinforcement, in2 (mm2) For members with spiral reinforcement then, for axial loads only, Pu Յ 0.85␾Po For members with tie reinforcement, for axial loads only, Pu Յ 0.80␾Po e t Eccentricities are measured from the plastic centroid. m This is the centroid of the resistance to load computed for t the assumptions that the concrete is stressed uniformly to y Ј 0.85 f c and the steel is stressed uniformly to fy. l The axial-load capacity Pu kip (N), of short, rectangular c members subject to axial load and bending may be deter- i mined from t TLFeBOOK
• 40816_03_p99-130 10/22/01 12:42 PM Page 125 p. 124 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 125 COLUMN FORMULAS 125 Pu ϭ ␾(0.85 f c ba ϩ AЈ fy Ϫ As fs) Ј se ΄ ΂ PueЈ ϭ ␾ 0.85 f c ba d Ϫ Ј a 2 ΃ ϩ AЈ f (d Ϫ dЈ )΅ s yee where eЈ ϭ eccentricity, in (mm), of axial load at end of , member with respect to centroid of tensiles reinforcement, calculated by conventional meth-n ods of frame analysis b ϭ width of compression face, in (mm) a ϭ depth of equivalent rectangular compressive- stress distribution, in (mm)) AЈ ϭ area of compressive reinforcement, in2 (mm2) s , As ϭ area of tension reinforcement, in2 (mm2) d ϭ distance from extreme compression surface to centroid of tensile reinforcement, in (mm) dЈ ϭ distance from extreme compression surface tos centroid of compression reinforcement, in (mm) fs ϭ tensile stress in steel, ksi (MPa) The two preceding equations assume that a does not exceed the column depth, that reinforcement is in one or two faces parallel to axis of bending, and that reinforce- . ment in any face is located at about the same distance fromr the axis of bending. Whether the compression steel actuallyo yields at ultimate strength, as assumed in these and the fol- lowing equations, can be veriﬁed by strain compatibilityr calculations. That is, when the concrete crushes, the strain- in the compression steel, 0.003 (c – dЈ)/c, must be larger than the strain when the steel starts to yield, fy /Es. In this TLFeBOOK
• 40816_03_p99-130 10/22/01 12:43 PM Page 126 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 126 40816 126 CHAPTER THREE case, c is the distance, in (mm), from the extreme compres- w sion surface to the neutral axis and Es is the modulus of elasticity of the steel, ksi (MPa). The load Pb for balanced conditions can be computed from the preceding Pu equation with fs ϭ fy and a ϭ ab ϭ ␤1cb 87,000 ␤1d S ϭ 87,000 ϩ fy F The balanced moment, in. и kip (k и Nm), can be obtained P from Mb ϭ Pbeb P ΄ ΂ ϭ ␾ 0.85 fcЈ ba b d Ϫ dЈЈ Ϫ ab 2 ΃ ϩ AЈ fy (d Ϫ dЈ Ϫ dЈЈ) ϩ As fy dЈЈ s ΅ where eb is the eccentricity, in (mm), of the axial load with respect to the plastic centroid and dЈЈ is the distance, in (mm), C from plastic centroid to centroid of tension reinforcement. When Pu is less than Pb or the eccentricity, e, is greater F than eb, tension governs. In that case, for unequal tension and compression reinforcement, the ultimate strength is P Ά Pu ϭ 0.85 fcЈbd␾ ␳ЈmЈ Ϫ ␳m ϩ 1 Ϫ ΂ eЈ d ΃ √΂ eЈ ΃ ϩ 2 ΄(␳m Ϫ ␳ЈmЈ) eЈ ϩ ␳ЈmЈ ΂1 Ϫ ddЈ ΃΅· 2 ϩ 1Ϫ d d TLFeBOOK
• 40816_03_p99-130 10/22/01 12:43 PM Page 127 p. 126 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 127 COLUMN FORMULAS 127- where Ј m ϭ fyЈ /0.85f cf mЈ ϭ m Ϫ 1d ␳ ϭ As / bd ␳Ј ϭ AЈ / bd s Special Cases of Reinforcement For symmetrical reinforcement in two faces, the precedingd Pu equation becomes Ά Pu ϭ 0.85 fcЈbd␾ Ϫ␳ ϩ 1 Ϫ eЈ d √΂ eЈ ΃ ϩ 2␳ ΄mЈ΂1 Ϫ dЈ ΃ ϩ eЈ ΅· 2 ϩ 1Ϫ d d dh , Column Strength When Compression Governsr For no compression reinforcement, the Pu equation becomesn ΄ Pu ϭ 0.85 f cbd␾ Ϫ␳m ϩ 1 Ϫ Ј eЈ d· ϩ √΂1 Ϫ d ΃ ϩ 2 d eЈ 2 eЈ␳m ΅ TLFeBOOK
• 40816_03_p99-130 10/22/01 12:43 PM Page 128 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 128 40816 128 CHAPTER THREE When Pu is greater than Pb, or e is less than eb, compression w governs. In that case, the ultimate strength is approximately Mu Pu ϭ Po Ϫ (Po Ϫ Pb) Mb Po Pu ϭ W 1 ϩ (Po /Pb Ϫ 1)(e/eb) where Mu is the moment capacity under combined axial load and bending, in kip (k Nm) and Po is the axial-load capacity, kip (N), of member when concentrically loaded, as given. For symmetrical reinforcement in single layers, the ulti- mate strength when compression governs in a column with T depth, h, may be computed from m Pu ϭ ␾ ΂ e/d ϪAЈ f ϩ 0.5 ϩ 3he/dbhfϩЈ 1.18 ΃ dЈ s y 2 c S Circular Columns U Ultimate strength of short, circular members with bars in a a circle may be determined from the following equations: i When tension controls, W Pu ϭ 0.85 f c D 2␾ Ј ΄ √΂ D Ϫ 0.38΃ ϩ 2.5D 0.85e ␳ mD 2 1 s Ϫ ΂ 0.85e Ϫ 0.38΃ D ΅ TLFeBOOK
• 40816_03_p99-130 10/22/01 12:43 PM Page 129 p. 128 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 129 COLUMN FORMULAS 129n where D ϭ overall diameter of section, in (mm) Ds ϭ diameter of circle through reinforcement, in (mm) ␳t ϭ Ast /Ag When compression governs,ld Pu ϭ ␾ ΄ 3e/D fϩ 1 A st v s ,- ϩ Ј Ag f c 9.6De /(0.8D ϩ 0.67Ds)2 ϩ 1.18 ΅h The eccentricity for the balanced condition is given approxi- mately by eb ϭ (0.24 Ϫ 0.39 ␳t m)D Short Columns Ultimate strength of short, square members with depth, h,a and with bars in a circle may be computed from the follow- ing equations: When tension controls, Pu ϭ 0.85bhf c␾ Ј ΄√΂ h Ϫ 0.5΃ ϩ 0.67 h e 2 D s ␳t m΅ Ϫ ΂ h Ϫ 0.5΃ e ΅ TLFeBOOK
• 40816_03_p99-130 10/22/01 12:43 PM Page 130 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 130 130 CHAPTER THREE When compression governs, Pu ϭ ␾ ΄ 3e/D fϩ 1 A st y s ϩ Ј Ag f c 12he/(h ϩ 0.67Ds)2 ϩ 1.18 ΅ Slender Columns When the slenderness of a column has to be taken into account, the eccentricity should be determined from e ϭ Mc /Pu, where Mc is the magniﬁed moment. TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 131 40816 HICKS Mcghp Chap_04 SECOND PASS 5/5/01 pb p 131 CHAPTER 4 PILES AND PILING FORMULAS TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_04_p131-146 10/22/01 12:44 PM Page 134 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 134 408 134 CHAPTER FOUR TABLE 4.1 Percentage of Base Load Transmitted to Rock Socket w Er /Ep Ls /ds 0.25 1.0 4.0 0.5 54† 48 44 1.0 31 23 18 1.5 17† 12 8† 2.0 13† 8 4 † Estimated by interpretation of ﬁnite-element solution; for Poisson’s ratio ϭ 0.26. The negative moment imposed at the pile head by pile- A cap or another structural restraint can be evaluated as a b function of the head slope (rotation) from c A PT ␪ EI c ϪM t ϭ ␪ t Ϫ s t B␪ B␪T where ␪s rad represents the counterclockwise (ϩ) rotation of the pile head and A␪ and B␪ are coefﬁcients (see Table 4.1). The inﬂuence of the degrees of ﬁxity of the pile head on y and M can be evaluated by substituting the value of S ϪMt from the preceding equation into the earlier y and M r equations. Note that, for the ﬁxed-head case, b A yf ϭ PtT 3 EI ΂ A B Ay Ϫ ␪ y B␪ ΃ w TOE CAPACITY LOAD c F For piles installed in cohesive soils, the ultimate tip load q may be computed from i TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 135 34 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 135 PILES AND PILING FORMULAS 135 Q bu ϭ Ab q ϭAb Nc cu (4.1) where Ab ϭ end-bearing area of pile q ϭ bearing capacity of soil Nt ϭ bearing-capacity factor cu ϭ undrained shear strength of soil within zone 1 pile diameter above and 2 diameters below pile tip- Although theoretical conditions suggest that Nc may varya between about 8 and 12, Nc is usually taken as 9. For cohesionless soils, the toe resistance stress, q, is conventionally expressed by Eq. (4.1) in terms of a bearing- capacity factor Nq and the effective overburden pressure at the pile tip ␴Ј : von q ϭ Nq␴Ј Յ ql (4.2) voed f Some research indicates that, for piles in sands, q, like fs,M reaches a quasi-constant value, ql , after penetrations of the bearing stratum in the range of 10 to 20 pile diameters. Approximately: ql ϭ 0.5Nq tan ␾ (4.3) where ␾ is the friction angle of the bearing soils below the critical depth. Values of Nq applicable to piles are given in Fig. 4.1. Empirical correlations of soil test data with q andd ql have also been applied to predict successfully end-bear- ing capacity of piles in sand. TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 136 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 136 408 136 CHAPTER FOUR t g p a a i w a p c D c h FIGURE 4.1 Bearing-capacity factor for granular soils related to p angle of internal friction. c c GROUPS OF PILES A pile group may consist of a cluster of piles or several piles w in a row. The group behavior is dictated by the group geom- a etry and the direction and location of the load, as well as by t subsurface conditions. d Ultimate-load considerations are usually expressed in u terms of a group efficiency factor, which is used to reduce c the capacity of each pile in the group. The efﬁciency factor r Eg is defined as the ratio of the ultimate group capacity s TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 13736 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 137 PILES AND PILING FORMULAS 137 to the sum of the ultimate capacity of each pile in the group. Eg is conventionally evaluated as the sum of the ultimate peripheral friction resistance and end-bearing capacities of a block of soil with breadth B, width W, and length L, approximately that of the pile group. For a given pile, spac- ing S and number of piles n, 2(BL ϩ WL) fs ϩ BWg Eg ϭ (4.4) nQ u where fs is the average peripheral friction stress of block and Qu is the single-pile capacity. The limited number of pile-group tests and model tests available suggest that for cohesive soils Eg Ͼ 1 if S is more than 2.5 pile diameters D and for cohesionless soils Eg Ͼ 1 for the smallest practi- cal spacing. A possible exception might be for very short, heavily tapered piles driven in very loose sands. In practice, the minimum pile spacing for conventionalo piles is in the range of 2.5 to 3.0D. A larger spacing is typi- cally applied for expanded-base piles. A very approximate method of pile-group analysis cal- culates the upper limit of group drag load, Qgd from Q gd ϭ AF␥FHF ϩ PHcu (4.5)s where Hf , ␥f , and AF represent the thickness, unit weight,- and area of ﬁll contained within the group. P, H, and cu arey the circumference of the group, the thickness of the consoli- dating soil layers penetrated by the piles, and theirn undrained shear strength, respectively. Such forces as Qgde could only be approached for the case of piles driven tor rock through heavily surcharged, highly compressible sub-y soils. TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 138 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 138 408 138 CHAPTER FOUR Design of rock sockets is conventionally based on T s ␲ 2 Q d ϭ ␲d s L s fR ϩ d q (4.6) 4 s a J where Qd ϭ allowable design load on rock socket i d ds ϭ socket diameter f 1 Ls ϭ socket length t fR ϭ allowable concrete-rock bond stress b T qa ϭ allowable bearing pressure on rock d Load-distribution measurements show, however, that much less of the load goes to the base than is indicated by Eq. (4.6). This behavior is demonstrated by the data in Table F 4.1, where Ls /ds is the ratio of the shaft length to shaft diameter and Er /Ep is the ratio of rock modulus to shaft T modulus. The ﬁnite-element solution summarized in Table d 4.1 probably reﬂects a realistic trend if the average socket- f wall shearing resistance does not exceed the ultimate fR a value; that is, slip along the socket side-wall does not occur. b A simpliﬁed design approach, taking into account approxi- v mately the compatibility of the socket and base resistance, is applied as follows: 1. Proportion the rock socket for design load Qd with Eq. w (4.6) on the assumption that the end-bearing stress is less than qa [say qa /4, which is equivalent to assuming that the base load Q b ϭ (␲/4) d 2qa /4]. s 2. Calculate Qb ϭ RQd, where R is the base-load ratio interpreted from Table 4.1. 3. If RQd does not equal the assumed Qb , repeat the proce- dure with a new qa value until an approximate conver- gence is achieved and q Յ qa. TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 13938 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 139 PILES AND PILING FORMULAS 139 The ﬁnal design should be checked against the established settlement tolerance of the drilled shaft.) Following the recommendations of Rosenberg and Journeaux, a more realistic 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 data that are not site speciﬁc, a safety factor of at least 1.5 should be applied to fRu in recognition of the uncertain- ties associated with the UC strength correlations (Rosen- berg, P. and Journeaux, N. L., “Friction and End-Bearing Tests on Bedrock for High-Capacity Socket Design,” Cana- dian Geotechnical Journal, 13(3)).h .e FOUNDATION-STABILITY ANALYSIStt The maximum load that can be sustained by shallow foun-e dation elements at incipient failure (bearing capacity) is a- function of the cohesion and friction angle of bearing soilsR as well as the width B and shape of the foundation. The net. bearing capacity per unit area, qu, of a long footing is con-- ventionally expressed as, qu ϭ ␣f cu Nc ϩ ␴Ј Nq ϩ ␤f ␥BN␥ vo (4.7) . where ␣f ϭ 1.0 for strip footings and 1.3 for circular ands square footingsg cu ϭ undrained shear strength of soilo ␴Ј ϭ effective vertical shear stress in soil at level vo of bottom of footing-- ␤f ϭ 0.5 for strip footings, 0.4 for square foot- ings, and 0.6 for circular footings TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 140 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 140 408 140 CHAPTER FOUR ␥ ϭ unit weight of soil T S B ϭ width of footing for square and rectangular footings and radius of footing for circular footings Nc, Nq, N␥ ϭ bearing-capacity factors, functions of angle R of internal friction ␾ C For undrained (rapid) loading of cohesive soils, ␾ ϭ 0 and Eq. (4.7) reduces to † qu ϭ N c cu Ј (4.8) t ‡ where NЈ ϭ ␣f Nc. For drained (slow) loading of cohesive c soils, ␾ and cu are deﬁned in terms of effective friction angle ␾Ј and effective stress cЈ. u Modiﬁcations of Eq. (4.7) are also available to predict the s bearing capacity of layered soil and for eccentric loading. c Rarely, however, does qu control foundation design when m the safety factor is within the range of 2.5 to 3. (Should f creep or local yield be induced, excessive settlements may a occur. This consideration is particularly important when r selecting a safety factor for foundations on soft to ﬁrm a clays with medium to high plasticity.) f Equation (4.7) is based on an inﬁnitely long strip foot- t ing and should be corrected for other shapes. Correction factors by which the bearing-capacity factors should be multiplied are given in Table 4.2, in which L ϭ footing length. w The derivation of Eq. (4.7) presumes the soils to be homo- geneous throughout the stressed zone, which is seldom the case. Consequently, adjustments may be required for depar- tures from homogeneity. In sands, if there is a moderate vari- ation in strength, it is safe to use Eq. (4.7), but with bearing- F capacity factors representing a weighted average strength. 0 TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 14140 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 141 PILES AND PILING FORMULAS 141 TABLE 4.2 Shape Corrections for Bearing-Capacity Factors of Shallow Foundations†rr Correction factor Shape of foundation Nc Nq Ny Rectangle‡ 1ϩ΂ ΃΂ ΃B L Nq Nc 1ϩ ΂ ΃ tan ␾ B L 1 Ϫ 0.4 ΂B΃ L Circle and 1ϩ΂ ΃0 N q square 1 ϩ tan ␾ 0.60 N c † After De Beer, E. E., as modiﬁed by Vesic, A. S. See Fang, H. Y., Founda-) tion Engineering Handbook, 2d ed., Van Nostrand Reinhold, New York. ‡ No correction factor is needed for long-strip foundations.en Eccentric loading can have a signiﬁcant impact one selection of the bearing value for foundation design. The conventional approach is to proportion the foundation ton maintain the resultant force within its middle third. Thed footing is assumed to be rigid and the bearing pressure isy assumed to vary linearly as shown by Fig. (4.2b.) If then resultant lies outside the middle third of the footing, it ism assumed that there is bearing over only a portion of the footing, as shown in Fig. (4.2d.) For the conventional case,- the maximum and minimum bearing pressures areneg qm ϭ P BL ΂1 Ϯ 6e ΃ B (4.9) where B ϭ width of rectangular footing-e L ϭ length of rectangular footing- e ϭ eccentricity of loading-- For the other case (Fig. 4.3c), the soil pressure ranges from 0 to a maximum of TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 142 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 142 408 142 CHAPTER FOUR 2P F qm ϭ (4.10) t 3L(B/2 Ϫ e) e For square or rectangular footings subject to overturning about two principal axes and for unsymmetrical footings, m the loading eccentricities e1 and e2 are determined about the o two principal axes. For the case where the full bearing area ( of the footings is engaged, qm is given in terms of the dis- tances from the principal axes, c1 and c2, the radius of gyra- tion of the footing area about the principal axes r1 and r2, and the area of the footing A as A qm ϭ P A ΂1 ϩ erc 1 1 2 1 ϩ e2c2 r2 2 ΃ (4.11) P t e w p e a E t e t a o 1 FIGURE 4.2 Footings subjected to overturning. e TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 14342 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 143 PILES AND PILING FORMULAS 143 For the case where only a portion of the footing is bearing,) the maximum pressure may be approximated by trial and error.g For all cases of sustained eccentric loading, the maxi-s, mum (edge) pressures should not exceed the shear strength e of the soil and also the factor of safety should be at least 1.5 a (preferably 2.0) against overturning. - -2, AXIAL-LOAD CAPACITY OF SINGLE PILES) Pile capacity Qu may be taken as the sum of the shaft and toe resistances, Qsu and Qbu, respectively. The allowable load Qa may then be determined from either Eq. (4.12) or (4.13): Q su ϩ Q bu Qa ϭ (4.12) F Q su Q Qa ϭ ϩ bu (4.13) F1 F2 where F, F1, and F2 are safety factors. Typically, F for permanent structures is between 2 and 3, but may be larg- er, depending on the perceived reliability of the analysis and construction as well as the consequences of failure. Equation (4.13) recognizes that the deformations required to fully mobilize Qsu and Qbu are not compatible. For example, Qsu may be developed at displacements less than 0.25 in (6.35 mm), whereas Qbu may be realized at a toe displacement equivalent to 5 percent to 10 percent of the pile diameter. Consequently, F1 may be taken as 1.5 and F2 as 3.0, if the equivalent single safety factor equals F or larger. (If Q su/Q buϽ1.0, F is less than the TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 144 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 144 408 144 CHAPTER FOUR 2.0 usually considered as a major safety factor for perma- nent structures.) s t c SHAFT SETTLEMENT w Drilled-shaft settlements can be estimated by empirical cor- s relations or by load-deformation compatibility analyses. Other c methods used to estimate settlement of drilled shafts, singly w or in groups, are identical to those used for piles. These w include elastic, semiempirical elastic, and load-transfer solu- tions for single shafts drilled in cohesive or cohesionless soils. Resistance to tensile and lateral loads by straight-shaft drilled shafts should be evaluated as described for pile S foundations. For relatively rigid shafts with characteristic C length T greater than 3, there is evidence that bells increase the lateral resistance. The added ultimate resistance to uplift T of a belled shaft Qut can be approximately evaluated for f cohesive soils models for bearing capacity [Eq. (4.14)] and p friction cylinder [Eq. (4.15)] as a function of the shaft diameter D and bell diameter Db (Meyerhof, G. G. and Adams, J. I., “The Ultimate Uplift Capacity of Founda- tions,” Canadian Geotechnical Journal, 5(4):1968.) For the bearing-capacity solution, e a ␲ s Q ul ϭ (D 2 Ϫ D 2)Nc ␻cu ϩ Wp b (4.14) v 4 The shear-strength reduction factor ␻ in Eq. (4.14) consid- t ers disturbance effects and ranges from 1͞2 (slurry construc- w tion) to 3͞4 (dry construction). The cu represents the l undrained shear strength of the soil just above the bell sur- r face, and Nc is a bearing-capacity factor. p TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 14544 40816 HICKS Mcghp Ch04 Third Pass 7/19/01 bcj p 145 PILES AND PILING FORMULAS 145- The failure surface of the friction cylinder model is con- servatively assumed to be vertical, starting from the base of the bell. Qut can then be determined for both cohesive and cohesionless soils from Q ul ϭ ␲b L fut ϩ Ws ϩ Wp (4.15) where fut is the average ultimate skin-friction stress in ten-- sion developed on the failure plane; that is, fut ϭ 0.8cu forr clays or K ␴ vo tan ␾ for sands. Ws and Wp represent the Јy weight of soil contained within the failure plane and the shafte weight, respectively.-ste SHAFT RESISTANCE INc COHESIONLESS SOILSet The shaft resistance stress fs is a function of the soil-shaftr friction angle ␦, degree, and an empirical lateral earth-d pressure coefﬁcient K:td fs ϭ K ␴ vo tan ␦ Յ fl Ј (4.16)- At displacement-pile penetrations of 10 to 20 pile diam- eters (loose to dense sand), the average skin friction reaches a limiting value fl. Primarily depending on the relative den- sity 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 to 1.0 and ␦ is taken to be about ␾ Ϫ 5,- where ␾ is the angle of internal friction, degree. For pilese less than 50 ft (15.2 m) long, K is more likely to be in the- range of 1.0 to 2.0, but can be greater than 3.0 for tapered piles. TLFeBOOK
• 40816_04_p131-146 10/22/01 12:44 PM Page 146 40816 HICKS Mcghp Ch04 SECOND PASS 5/5/01 pb p 146 146 CHAPTER FOUR Empirical procedures have also been used to evaluate fs from in situ tests, such as cone penetration, standard pen- etration, and relative density tests. Equation (4.17), based on standard penetration tests, as proposed by Meyerhof, is generally conservative and has the advantage of simplicity: N fs ϭ (4.17) 50 where N ϭ average standard penetration resistance within the embedded length of pile and fs is given in tons/ft2. (Meyerhof, G. G., “Bearing Capacity and Settlement of Pile Foundations,” ASCE Journal of Geotechnical Engineering Division, 102(GT3):1976.) TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 148 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 148 148 CHAPTER FIVE REINFORCED CONCRETE a 0 When working with reinforced concrete and when design- r ing reinforced concrete structures, the American Concrete 2 Institute (ACI) Building Code Requirements for Reinforced c Concrete, latest edition, is widely used. Future references to v this document are denoted as the ACI Code. Likewise, pub- lications of the Portland Cement Association (PCA) ﬁnd w extensive use in design and construction of reinforced con- w crete structures. a Formulas in this chapter cover the general principles of v reinforced concrete and its use in various structural applica- l tions. Where code requirements have to be met, the reader ( must refer to the current edition of the ACI Code previously d mentioned. Likewise, the PCA publications should also be referred to for the latest requirements and recommendations. t i t e c WATER/CEMENTITIOUS MATERIALS RATIO The water/cementitious (w/c) ratio is used in both tensile and compressive strength analyses of Portland concrete J cement. This ratio is found from A w w m ϭ m m c wc v a where wm ϭ weight of mixing water in batch, lb (kg); and wc ϭ weight of cementitious materials in batch, lb (kg). The ACI Code lists the typical relationship between the w/c ratio by weight and the compressive strength of concrete. Ratios for non-air-entrained concrete vary between 0.41 for TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 149 148 40816 HICKS Mcghp Chap_05 Pgs 149 7/10/2001 CONCRETE FORMULAS 149 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 from 0.40 to 0.74 for 5000 lb/in2 (34 MPa) ande 2000 lb/in2 (14 MPa) compressive strength, respectively. Bed certain to refer to the ACI Code for the appropriate w/co value when preparing designs or concrete analyses.- Further, the ACI Code also lists maximum w/c ratiosd when strength data are not available. Absolute w/c ratios by- weight vary from 0.67 to 0.38 for non-air-entrained concrete and from 0.54 to 0.35 for air-entrained concrete. Thesef values are for a speciﬁed 28-day compressive strength fcЈ in- lb/in2 or MPa, of 2500 lb/in2 (17 MPa) to 5000 lb/in2r (34 MPa). Again, refer to the ACI Code before making anyy design or construction decisions.e Maximum w/c ratios for a variety of construction condi- . tions are also listed in the ACI Code. Construction conditions include concrete protected from exposure to freezing and thawing; concrete intended to be watertight; and concrete exposed to deicing salts, brackish water, seawater, etc. Appli- cation formulas for w/c ratios are given later in this chapter.ee JOB MIX CONCRETE VOLUME A trial batch of concrete can be tested to determine how much concrete is to be delivered by the job mix. To deter- mine the volume obtained for the job, add the absolute volume Va of the four components—cements, gravel, sand, and water.d Find the Va for each component frome. WL Va ϭr (SG)Wu TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 150 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 150 150 CHAPTER FIVE where Va ϭ absolute volume, ft3 (m3) T WL ϭ weight of material, lb (kg) T SG ϭ speciﬁc gravity of the material d s wu ϭ 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. R MODULUS OF ELASTICITY OF CONCRETE A ﬁ The modulus of elasticity of concrete Ec—adopted in modi- t ﬁed form by the ACI Code—is given by 1 2 Ec ϭ 33w1.5 √f cЈ c lb/in2 in USCS units 3 ϭ 0.043w1.5 √fcЈ c MPa in SI units 4 5 With normal-weight, normal-density concrete these two relations can be simpliﬁed to C Ec ϭ 57,000 √fcЈ lb/in2 in USCS units A ϭ 4700 √fcЈ MPa in SI units T s where Ec ϭ modulus of elasticity of concrete, lb/in2 (MPa); w and fcЈ ϭ speciﬁed 28-day compressive strength of con- a crete, lb/in2 (MPa). A TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 151150 40816 HICKS Mcghp Chap_05 Pgs 151 7/10/2001 CONCRETE FORMULAS 151 TENSILE STRENGTH OF CONCRETE The tensile strength of concrete is used in combined-stress design. In normal-weight, normal-density concrete the ten- sile strength can be found froms fr ϭ 7.5 √fcЈ lb/in2 in USCS units, fr ϭ 0.7 √fcЈ MPa in SI units REINFORCING STEEL American Society for Testing and Materials (ASTM) speci- ﬁcations cover renforcing steel. The most important proper-- ties of reinforcing steel are 1. 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 wireo CONTINUOUS BEAMS AND ONE-WAY SLABS The ACI Code gives approximate formulas for ﬁnding shear and bending moments in continuous beams and one-; way slabs. A summary list of these formulas follows. They- are equally applicable to USCS and SI units. Refer to the ACI Code for speciﬁc applications of these formulas. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 152 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 152 152 CHAPTER FIVE For Positive Moment E End spans R If discontinuous end is unrestrained wl 2 ͞11 a n If discontinuous end is integral with the support wl 2 ͞14 n Interior spans wl 2 ͞16 n D C For Negative Moment Negative moment at exterior face of ﬁrst interior A support r a Two spans wl 2 ͞9 n s More than two spans wl 2 ͞10 n s Negative moment at other faces of interior supports wl 2 ͞11 n C 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 B sum of column stiffness to beam stiffness exceeds 8 at each end of the span wl 2 ͞12 n C Negative moment at interior faces of exterior t supports, for members built integrally with ( their supports w Where the support is a spandrel beam or girder wl 2 ͞24 n R Where the support is a column wl 2 ͞16 n t e e Shear Forces c Shear in end members at ﬁrst interior support 1.15 wl n ͞2 Shear at all other supports wl n ͞2 TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 153152 40816 HICKS Mcghp Chap_05 Pgs 153 7/10/2001 CONCRETE FORMULAS 153 End Reactions Reactions to a supporting beam, column, or wall are obtained1 as the sum of shear forces acting on both sides of the support.46 DESIGN METHODS FOR BEAMS, COLUMNS, AND OTHER MEMBERS A number of different design methods have been used for reinforced concrete construction. The three most common are working-stress design, ultimate-strength design, and9 strength design method. Each method has its backers and0 supporters. For actual designs the latest edition of the ACI1 Code should be consulted. Beams2 Concrete beams may be considered to be of three principal types: (1) rectangular beams with tensile reinforcing only, (2) T beams with tensile reinforcing only, and (3) beams with tensile and compressive reinforcing.4 Rectangular Beams with Tensile Reinforcing Only. This6 type of beam includes slabs, for which the beam width b equals 12 in (305 mm) when the moment and shear are expressed per foot (m) of width. The stresses in the con- crete and steel are, using working-stress design formulas,2 2M M M fc ϭ fs ϭ ϭ2 kjbd 2 As jd pjbd 2 TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 154 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 154 154 CHAPTER FIVE where b ϭ width of beam [equals 12 in (304.8 mm) for slab], in (mm) d ϭ effective depth of beam, measured from com- pressive face of beam to centroid of tensile reinforcing (Fig. 5.1), in (mm) M ϭ bending moment, lb . in (k . Nm) fc ϭ compressive stress in extreme ﬁber of concrete, lb/in2 (MPa) fs ϭ stress in reinforcement, lb/in2 (MPa) As ϭ cross-sectional area of tensile reinforcing, in2 (mm2) j ϭ ratio of distance between centroid of compres- sion and centroid of tension to depth d k ϭ ratio of depth of compression area to depth d p ϭ ratio of cross-sectional area of tensile reinforcing to area of the beam (ϭ As /bd) For approximate design purposes, j may be assumed to be 7 ͞8 and k, 1͞3. For average structures, the guides in Table 5.1 to the depth d of a reinforced concrete beam may be used. For a balanced design, one in which both the concrete and the steel are stressed to the maximum allowable stress, the following formulas may be used: M 1 bd 2 ϭ Kϭ f kj ϭ pfs j K 2 e Values of K, k, j, and p for commonly used stresses are given in Table 5.2. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 155154 40816 HICKS Mcghp Chap_05 Pgs 155 7/10/2001 TLFeBOOKr-e,, FIGURE 5.1 Rectangular concrete beam with tensile reinforcing only.-dgeoe,e 155
• 40816_05a_p147-195 10/22/01 12:45 PM Page 156 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 156 156 CHAPTER FIVE TABLE 5.1 Guides to Depth d of B Reinforced Concrete Beam† w s Member d a Roof and floor slabs l/25 m Light beams l/15 s Heavy beams and girders l/12–l/10 b † l is the span of the beam or slab in inches (millimeters). The width of a beam should be at least l/32. T Beams with Tensile Reinforcing Only. When a concrete w slab is constructed monolithically with the supporting con- crete beams, a portion of the slab acts as the upper 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 16 times 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 ﬂange thickness not less than one-half the width of the web and a ﬂange width not more than four times the width of the web. For preliminary designs, the preceding formulas 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 tensile reinforcing is usually critical. C TABLE 5.2 Coefficients K, k, j, p for Rectangular Sections† p fsЈ n fs K k j p e a 2000 15 900 175 0.458 0.847 0.0129 p 2500 12 1125 218 0.458 0.847 0.0161 I 3000 10 1350 262 0.458 0.847 0.0193 s 3750 8 1700 331 0.460 0.847 0.0244 i † fs ϭ 16,000 lb/in2 (110 MPa). s TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 157 156 40816 HICKS Mcghp Chap_05 Pgs 157 7/10/2001 CONCRETE FORMULAS 157 Beams with Tensile and Compressive Reinforcing. Beams with compressive reinforcing are generally used when the size of the beam is limited. The allowable beam dimensions are used in the formulas given earlier to determine the moment that could be carried by a beam without compres- sive reinforcement. The reinforcing requirements may then be approximately determined from 8M M Ϫ MЈ As ϭ Asc ϭ 7fs d nfc de where As ϭ total cross-sectional area of tensile reinforcing,- in2 (mm2)f) Asc ϭ cross-sectional area of compressive reinforcing,b in2 (mm2) ,s M ϭ total bending moment, lbиin (KиNm)e MЈ ϭ bending moment that would be carried by beamb of balanced design and same dimensions withe tensile reinforcing only, lbиin (KиNm)ne n ϭ ratio of modulus of elasticity of steel to that ofe concrete Checking Stresses in Beams. Beams designed using the preceding approximate formulas should be checked to ensure that the actual stresses do not exceed the allowable, and that the reinforcing is not excessive. This can be accom-9 plished by determining the moment of inertia of the beam.1 In this determination, the concrete below the neutral axis3 should not be considered as stressed, whereas the reinforc-4 ing steel should be transformed into an equivalent concrete section. For tensile reinforcing, this transformation is made TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 158 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 158 158 CHAPTER FIVE by multiplying the area As by n, the ratio of the modulus of a elasticity of steel to that of concrete. For compressive rein- forcing, the area Asc is multiplied by 2(n – 1). This factor includes allowances for the concrete in compression replaced by the compressive reinforcing and for the plastic ﬂow of concrete. The neutral axis is then located by solving w ͞2 bc 2 ϩ 2(n Ϫ 1)Asccsc ϭ nAscs 1 c for the unknowns cc, csc, and cs (Fig. 5.2). The moment of inertia of the transformed beam section is I ϭ 1͞3bc 3 ϩ 2(n Ϫ 1)Ascc 2 ϩ nAs c 2 c sc s a S s c w FIGURE 5.2 Transformed section of concrete beam. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 159158 40816 HICKS Mcghp Chap_05 Pgs 159 7/10/2001 CONCRETE FORMULAS 159f and the stresses are-r Mcc 2nMcsc nMcs fc ϭ fsc ϭ fs ϭd I I If where fc, fsc, fs ϭ actual unit stresses in extreme ﬁber of con- crete, in compressive reinforcing steel, and in tensile reinforcing steel, respectively, lb/in2 (MPa)f cc, csc, cs ϭ distances from neutral axis to face of con- crete, to compressive reinforcing steel, and to tensile reinforcing steel, respectively, in (mm) I ϭ moment of inertia of transformed beam section, in4 (mm4) b ϭ 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 measure of diagonal tension, in a reinforced con- crete beam is V vϭ bd where v ϭ shearing unit stress, lb/in2 (MPa) V ϭ total shear, lb (N) b ϭ width of beam (for T beam use width of stem), in (mm) d ϭ effective depth of beam TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 160 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 160 160 CHAPTER FIVE If the value of the shearing stress as computed earlier exceeds the allowable shearing unit stress as speciﬁed by the ACI Code, web reinforcement should be provided. Such reinforcement usually consists of stirrups. The cross- sectional area required for a stirrup placed perpendicular to the longitudinal reinforcement is (V Ϫ VЈ)s Av ϭ fi d where Av ϭ cross-sectional area of web reinforcement in distance s (measured parallel to longitudinal reinforcement), in2 (mm2) fv ϭ allowable unit stress in web reinforcement, lb/in2 (MPa) V ϭ total shear, lb (N) VЈ ϭ shear that concrete alone could carry (ϭ vc bd), lb (N) s ϭ spacing of stirrups in direction parallel to that of longitudinal reinforcing, in (mm) d ϭ effective depth, in (mm) Stirrups should be so spaced that every 45° line extend- ing from the middepth of the beam to the longitudinal tension bars is crossed by at least one stirrup. If the total Ј shearing unit stress is in excess of 3 √f c lb/in2 (MPa), every such line should be crossed by at least two stirrups. The shear Ј stress at any section should not exceed 5 √f c lb/in2 (MPa). Bond and Anchorage for Reinforcing Bars. In beams in which the tensile reinforcing is parallel to the compression face, the bond stress on the bars is TLFeBOOK
• n n y n o y r - - r l l t l . , , . 160 40816_05a_p147-195 40816 HICKS Mcghp Chap_05 Pgs 161 7/10/2001 TABLE 5.3 Allowable Bond Stresses† Horizontal bars with more than 12 in (30.5 mm) of concrete 10/22/01 cast below the bar‡ Other bars‡ 3.4√fcЈ 4.8√fcЈ Tension bars with sizes and or 350, whichever is less or 500, whichever is less D D deformations conforming to161 ASTM A305 12:45 PM Tension bars with sizes and 2.1√fcЈ 3√fcЈ deformations conforming to ASTM A408 Deformed compression bars 6.5√fcЈ or 400, whichever is less 6.5√fcЈ or 400, whichever is less Page 161 Plain bars 1.7√fcЈ or 160, whichever is less 2.4√fcЈ or 160, whichever is less † lb/in (ϫ 0.006895 ϭ MPa). 2 ‡ fcЈ ϭ compressive strength of concrete, lb/in2 (MPa); D ϭ nominal diameter of bar, in (mm). TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 162 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 162 162 CHAPTER FIVE V uϭ jd⌺0 where u ϭ bond stress on surface of bar, lb/in2 (MPa) V ϭ total shear, lb (N) d ϭ effective depth of beam, in (mm) T ⌺0 ϭ sum of perimeters of tensile reinforcing bars, T in (mm) m a For preliminary design, the ratio j may be assumed to be 7/8. Bond stresses may not exceed the values shown in Table 5.3. w Columns The principal columns in a structure should have a mini- mum diameter of 10 in (255 mm) or, for rectangular columns, a minimum thickness of 8 in (203 mm) and a minimum gross cross-sectional area of 96 in2 (61,935 mm2). Short columns with closely spaced spiral reinforcing enclosing a circular concrete core reinforced with vertical T bars have a maximum allowable load of e b e P ϭ Ag(0.25fcЈ ϩ fs pg) ( g where P ϭ total allowable axial load, lb (N) Ag ϭ gross cross-sectional area of column, in2 (mm2) S o f c ϭ compressive strength of concrete, lb/in2 (MPa) Ј r TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 163162 40816 HICKS Mcghp Chap_05 Pgs 163 7/10/2001 CONCRETE FORMULAS 163 fs ϭ 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 ϭ ratio of cross-sectional area of vertical rein- forcing steel to gross area of column Ag The ratio pg should not be less than 0.01 or more than 0.08., The minimum number of bars to be used is six, and the minimum size is No. 5. The spiral reinforcing to be used in a spirally reinforced column ise Ag Ј fcn ps ϭ 0.45 ΂ Ϫ 1΃ Ac fy where ps ϭ ratio of spiral volume to concrete-core volume (out-to-out spiral) Ac ϭ cross-sectional area of column core (out-to-- out spiral), in2 (mm2), fy ϭ yield strength of spiral reinforcement, lb/in2s (MPa), but not to exceed 60,000 lb/in2 (413 MPa)gl The center-to-center spacing of the spirals should not exceed one-sixth of the core diameter. The clear spacing between spirals should not exceed one-sixth the core diam- eter, 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 aggre- gate used.) Short Columns with Ties. The maximum allowable load on short columns reinforced with longitudinal bars and sepa-) rate lateral ties is 85 percent of that given earlier for spirally TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 164 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 164 164 CHAPTER FIVE reinforced columns. The ratio pg for a tied column should not l be less than 0.01 or more than 0.08. Longitudinal reinforcing a should consist of at least four bars; minimum size is No. 5. C Long Columns. Allowable column loads where compres- a sion governs design must be adjusted for column length as a follows: t t 1. If the ends of the column are ﬁxed so that a point of contraﬂexure occurs between the ends, the applied axial load and moments should be divided by R from (R can- not exceed 1.0) 0.006h R ϭ 1.32 Ϫ r 2. If the relative lateral displacement of the ends of the w columns is prevented and the member is bent in a single curvature, applied axial loads and moments should be divided by R from (R cannot exceed 1.0) 0.008h R ϭ 1.07 Ϫ r where h ϭ unsupported length of column, in (mm) r ϭ radius of gyration of gross concrete area, in (mm) ϭ 0.30 times depth for rectangular column ϭ 0.25 times diameter for circular column R ϭ long-column load reduction factor Applied axial load and moment when tension governs design should be similarly adjusted, except that R varies TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 165 164 40816 HICKS Mcghp Chap_05 Pgs 165 7/10/2001 CONCRETE FORMULAS 165t linearly with the axial load from the values given at the bal-g anced condition. . Combined Bending and Compression. The strength of- a symmetrical column is controlled by compression if the equiv-s alent axial load N has an eccentricity e in each principal direc- tion no greater than given by the two following equations and by tension if e exceeds these values in either principal direction.fl For spiral columns,- eb ϭ 0.43 pg mDs ϩ 0.14t For tied columns, eb ϭ (0.67pg m ϩ 0.17)de where e ϭ eccentricity, in (mm)ee eb ϭ maximum permissible eccentricity, in (mm) N ϭ eccentric load normal to cross section of column pg ϭ ratio of area of vertical reinforcement to gross concrete area m ϭ fy /0.85 fcЈ) Ds ϭ diameter of circle through centers of longitu- dinal reinforcement, in (mm) t ϭ diameter of column or overall depth of col- umn, in (mm) d ϭ distance from extreme compression ﬁber to centroid of tension reinforcement, in (mm)ss fy ϭ yield point of reinforcement, lb/in2 (MPa) TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 166 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 166 166 CHAPTER FIVE Design of columns controlled by compression is based on the following equation, except that the allowable load N may not exceed the allowable load P, given earlier, permit- ted when the column supports axial load only: fa fbx fby ϩ ϩ Յ 1.0 c Fa Fb Fb m where fa ϭ axial load divided by gross concrete area, f lb/in2 (MPa) fbx, fby ϭ bending moment about x and y axes, divided by section modulus of corresponding transformed W uncracked section, lb/in2 (MPa) Fb ϭ allowable bending stress permitted for bend- ing alone, lb/in2 (MPa) w Fa ϭ 0.34(1 ϩ pgm) fcЈ y a The allowable bending load on columns controlled by tension varies linearly with the axial load from M0 when the section is in pure bending to Mb when the axial load is Nb. P For spiral columns, M0 ϭ 0.12Ast fyDs S c For tied columns, a M0 ϭ 0.40As fy(d Ϫ dЈ) c s where Ast ϭ total area of longitudinal reinforcement, in2 (mm2) a fy ϭ yield strength of reinforcement, lb/in2 (MPa) e b Ds ϭ diameter of circle through centers of longitu- dinal reinforcement, in (mm) TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 167 166 40816 HICKS Mcghp Chap_05 Pgs 167 7/10/2001 CONCRETE FORMULAS 167d As ϭ area of tension reinforcement, in2 (mm2)N - d ϭ distance from extreme compression ﬁber to centroid of tension reinforcement, in (mm) Nb and Mb are the axial load and moment at the balanced condition (i.e., when the eccentricity e equals eb as deter- mined). At this condition, Nb and Mb should be determined , from Mb ϭ Nbebyd When bending is about two axes, Mx My ϩ Յ1 - M0x M0y where Mz and My are bending moments about the x and y axes, and M0x and M0y are the values of M0 for bending about these axes.ye . PROPERTIES IN THE HARDENED STATE Strength is a property of concrete that nearly always is of concern. Usually, it is determined by the ultimate strength of a specimen in compression, but sometimes ﬂexural or tensile capacity is the criterion. Because concrete usually gains strength over a long period of time, the compressive strength ) at 28 days is commonly used as a measure of this property. The 28-day compressive strength of concrete can be ) estimated from the 7-day strength by a formula proposed by W. A. Slater: - S28 ϭ S7 ϩ 30√S7 TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 168 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 168 168 CHAPTER FIVE where S28 ϭ 28-day compressive strength, lb/in2 (MPa), and T S7 ϭ 7-day strength, lb/in2 (MPa). Concrete may increase signiﬁcantly in strength after 28 days, F particularly when cement is mixed with ﬂy ash. Therefore, spec- l iﬁcation of strengths at 56 or 90 days is appropriate in design. a Concrete strength is inﬂuenced chieﬂy by the water/cement ratio; the higher this ratio is, the lower the strength. The rela- tionship is approximately linear when expressed in terms of the variable C/W, the ratio of cement to water by weight. For a workable mix, without the use of water reducing admixtures, w C S28 ϭ 2700 Ϫ 760 W Tensile strength of concrete is much lower than com- pressive strength and, regardless of the types of test, usually H has poor correlation with fcЈ. As determined in ﬂexural tests, i the tensile strength (modulus of rupture—not the true strength) is about 7√fcЈ for the higher strength concretes and 10√fcЈ for the lower strength concretes. Modulus of elasticity Ec , generally used in design for concrete, is a secant modulus. In ACI 318, “Building Code Requirements for Reinforced Concrete,” it is determined by Ec ϭ w1.533 √fcЈ where w ϭ weight of concrete, lb/ft3 (kg/m3); and fcЈ ϭ speci- fied compressive strength at 28 days, lb/in2 (MPa). For a normal-weight concrete, with w ϭ 145 lb/ft3 (kg/m3), Ec ϭ 57,000 √fcЈ w The modulus increases with age, as does the strength. ( TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 169 168 40816 HICKS Mcghp Chap_05 Pgs 169 7/10/2001 CONCRETE FORMULAS 169d TENSION DEVELOPMENT LENGTHS, For bars and deformed wire in tension, basic development- length is deﬁned by the equations that follow. For No. 11. and smaller bars,t- 0.04Ab fyf ld ϭr √fcЈ, where Ab ϭ area of bar, in2 (mm2) fy ϭ yield strength of bar steel, lb/in2 (MPa) fcЈ ϭ 28-day compressive strength of concrete,- lb/in2 (MPa)y However, ld should not be less than 12 in (304.8 mm), except , in computation of lap splices or web anchorage.) For No. 14 bars, fyr ld ϭ 0.085e √fcЈy For No. 18 bars, fy ld ϭ 0.125 √fcЈ-r and for deformed wire, fy Ϫ 20,000 A fy ld ϭ 0.03db Ն 0.02 w √fcЈ Sw √fcЈ where Aw is the area, in2 (mm2); and sw is the spacing, in (mm), of the wire to be developed. Except in computation of TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 170 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 170 170 CHAPTER FIVE lap splices or development of web reinforcement, ld should not be less than 12 in (304.8 mm). COMPRESSION DEVELOPMENT LENGTHS For bars in compression, the basic development length ld is deﬁned as T 0.02 fy db e ld ϭ Ն 0.0003db fy √fcЈ e 0 but ld not be less than 8 in (20.3 cm) or 0.0003fy db. e f m m CRACK CONTROL OF y FLEXURAL MEMBERS Because of the risk of large cracks opening up when rein- R forcement is subjected to high stresses, the ACI Code recom- mends that designs be based on a steel yield strength fy no larger than 80 ksi (551.6 MPa). When design is based on F a yield strength fy greater than 40 ksi (275.8 MPa), the cross s sections of maximum positive and negative moment should m be proportioned for crack control so that speciﬁc limits are r satisﬁed by 3 z ϭ fs √dc A w where fs ϭ calculated stress, ksi (MPa), in reinforcement v at service loads l TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 171170 40816 HICKS Mcghp Chap_05 Pgs 171 7/10/2001 CONCRETE FORMULAS 171d dc ϭ thickness of concrete cover, in (mm), meas- ured from extreme tension surface to center of bar closest to that surface A ϭ effective tension area of concrete, in2 (mm2) per bar. This area should be taken as that sur- rounding main tension reinforcement, having the same centroid as that reinforcement, multi- plied by the ratio of the area of the largest bars used to the total area of tension reinforcement These limits are z Յ 175 kip/in (30.6 kN/mm) for interior exposures and z Յ 145 kip/in (25.3 kN/mm) for exterior exposures. These correspond to limiting crack widths of 0.016 to 0.013 in (0.406 to 0.33 mm), respectively, at the extreme tension edge under service loads. In the equation for z, fs should be computed by dividing the bending moment by the product of the steel area and the internal moment arm, but fs may be taken as 60 percent of the steel yield strength without computation.- REQUIRED STRENGTH-on For combinations of loads, the ACI Code requires that as structure and its members should have the following ulti-d mate strengths (capacities to resist design loads and theire related internal moments and forces): With wind and earthquake loads not applied, U ϭ 1.4D ϩ 1.7L where D ϭ effect of basic load consisting of dead load plust volume change (shrinkage, temperature) and L ϭ effect of live load plus impact. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 172 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 172 172 CHAPTER FIVE When wind loads are applied, the largest of the preceed- ing equation and the two following equations determine the required strength: U ϭ 0.75(1.4D ϩ 1.7L ϩ 1.7W ) U ϭ 0.9D ϩ 1.3W g where W ϭ effect of wind load. f If the structure can be subjected to earthquake forces E, substitute 1.1E for W in the preceding equation. Where the effects of differential settlement, creep, shrink- age, or temperature change may be critical to the structure, they should be included with the dead load D, and the w strength should be at least equal to i U ϭ 0.75(1.4D ϩ 1.7L) Ն 1.4(D ϩ T ) d d where T ϭ cumulative effects of temperature, creep, o shrinkage, and differential settlement. s DEFLECTION COMPUTATIONS AND U CRITERIA FOR CONCRETE BEAMS O T The assumptions of working-stress theory may also be used for computing deﬂections under service loads; that is, elastic- theory deﬂection formulas may be used for reinforced-concrete G beams. In these formulas, the effective moment of inertia Ic f is given by w c ΂ M ΃ I ϩ ΄1 Ϫ ΂ M ΃ ΅ I 3 3 m cr cr Ie ϭ g cr Յ Ig t Ma M a f where Ig ϭ moment of inertia of the gross concrete section b TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 173172 40816 HICKS Mcghp Chap_05 Pgs 173 7/10/2001 CONCRETE FORMULAS 173- Mcr ϭ cracking momente Ma ϭ moment for which deﬂection is being computed Icr ϭ cracked concrete (transformed) section If yt is taken as the distance from the centroidal axis of the gross section, neglecting the reinforcement, to the extreme sur- face in tension, the cracking moment may be computed from, fr Ig Mcr ϭ- yt,e with the modulus of rupture of the concrete fr ϭ 7.5√fcЈ . The deﬂections thus calculated are those assumed to occur immediately on application of load. Additional long-time deﬂections can be estimated by multiplying the immediate deﬂection by 2 when there is no compression reinforcement, or by 2 Ϫ 1.2AЈ/As Ն 0.6, where AЈ is the area of compres- s s sion reinforcement and As is the area of tension reinforcement. ULTIMATE-STRENGTH DESIGN OF RECTANGULAR BEAMS WITH TENSION REINFORCEMENT ONLYd-e Generally, the area As of tension reinforcement in a rein- forced-concrete beam is represented by the ratio ␳ ϭ As /bd,c where b is the beam width and d is the distance 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 its yield strength fy if the concrete does not fail in compression first. Total tension in the steel then willn be As f y ϭ ␳fy bd. It is opposed, by an equal compressive force: TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 174 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 174 174 CHAPTER FIVE 0.85 fcЈba ϭ 0.85 fcЈb␤1c where fcЈ ϭ 28-day strength of the concrete, ksi (MPa) a ϭ depth of the equivalent rectangular stress distribution M c ϭ distance from the extreme compression sur- F face to the neutral axis i ␤1 ϭ a constant Equating the compression and tension at the critical section yields pfy cϭ d w 0.85␤1 fcЈ The criterion for compression failure is that the maximum strain in the concrete equals 0.003 in/in (0.076 mm/mm). In S that case, 0.003 T cϭ d t fs /Es ϩ 0.003 a m where fs ϭ steel stress, ksi (MPa) a Es ϭ modulus of elasticity of steel ϭ 29,000 ksi (199.9 GPa) w t s Balanced Reinforcing e T Under balanced conditions, the concrete reaches its maxi- e mum strain of 0.003 when the steel reaches its yield strength a fy. This determines the steel ratio for balanced conditions: a TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 175174 40816 HICKS Mcghp Chap_05 Pgs 175 7/10/2001 CONCRETE FORMULAS 175 0.85␤1 fcЈ 87,000 ␳b ϭ fy 87,000 ϩ fys Moment Capacity- For such underreinforced beams, the bending-moment capac- ity of ultimate strength is M u ϭ 0.90[bd 2 fcЈ␻(1 Ϫ 0.59␻)]n ΄ ΂ ϭ 0.90 As fy d Ϫ a 2 ΃΅ where ␻ ϭ ␳fy /fcЈ and a ϭ As fy /0.85fcЈ.mn Shear Reinforcement The ultimate shear capacity Vn of a section of a beam equals the sum of the nominal shear strength of the concrete Vc and the nominal shear strength provided by the reinforce- ment Vs; that is, Vn ϭ Vc ϩ Vs. The factored shear force Vu on a section should not exceed ␾Vn ϭ ␾(Vc ϩ Vs) where ␾ ϭ capacity reduction factor (0.85 for shear and torsion). Except for brackets and other short cantilevers, the section for maximum shear may be taken at a distance equal to d from the face of the support. The shear Vc carried by the concrete alone should not- exceed 2√fcЈ bw d, where bw is the width of the beam webh and d, the depth of the centroid of reinforcement. (As an alternative, the maximum for Vc may be taken as TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 176 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 176 176 CHAPTER FIVE ΂ Vc ϭ 1.9 √fcЈϩ2500␳w Vud Mu ΃b d w i Յ 3.5 √fcЈbw d b where ␳w ϭ As/bwd and Vu and Mu are the shear and bend- w ing moment, respectively, at the section considered, but Mu f should not be less than Vud.) When Vu is larger than ␾Vc, the excess shear has to be resisted by web reinforcement. The area of steel required in vertical stirrups, in2 (mm2), per stirrup, with a spacing s, in (mm), is a VS V Av ϭ s fy d where fy ϭ yield strength of the shear reinforcement. Av is D the area of the stirrups cut by a horizontal plane. Vs should not exceed 8√fcЈ bw d in sections with web reinforcement, A and fy should not exceed 60 ksi (413.7 MPa). Where shear s reinforcement is required and is placed perpendicular to the f axis of the member, it should not be spaced farther apart f than 0.5d, or more than 24 in (609.6 mm) c to c. When Vs ( exceeds 4√fcЈ bw d, however, the maximum spacing should o be limited to 0.25d. t Alternatively, for practical design, to indicate the stirrup spacing s for the design shear Vu, stirrup area Av, and geom- etry of the member bw and d, w Av␾fyd sϭ Vu Ϫ 2␾√fcЈ bw d The area required when a single bar or a single group of parallel bars are all bent up at the same distance from the sup- port at angle ␣ with the longitudinal axis of the member is TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 177 176 40816 HICKS Mcghp Chap_05 Pgs 177 7/10/2001 CONCRETE FORMULAS 177 Vs Av ϭ fy sin ␣ in which Vs should not exceed 3√fcЈ bw d. Av is the area cut by a plane normal to the axis of the bars. The area required - when a series of such bars are bent up at different distances from the support or when inclined stirrups are used is u Vs s e Av ϭ (sin ␣ ϩ cos ␣)fy d , A minimum area of shear reinforcement is required in all members, except slabs, footings, and joists or where Vu is less than 0.5Vc. s Development of Tensile Reinforcement d , At least one-third of the positive-moment reinforcement in r simple beams and one-fourth of the positive-moment rein- e forcement in continuous beams should extend along the same t face of the member into the support, in both cases, at least 6 inVs (152.4 mm) into the support. At simple supports and at points d of inﬂection, the diameter of the reinforcement should be limi- ted to a diameter such that the development length ld satisﬁes p Mn - ld ϭ ϩ la Vu where Mn ϭ computed ﬂexural strength with all reinforc- ing steel at section stressed to fy Vu ϭ applied shear at section f - la ϭ additional embedment length beyond inﬂec- s tion point or center of support TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 178 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 178 178 CHAPTER FIVE At an inﬂection point, la is limited to a maximum of d, the depth of the centroid of the reinforcement, or 12 times the reinforcement diameter. Hooks on Bars The basic development length for a hooked bar with fy ϭ 60 ksi (413.7 MPa) is deﬁned as 1200db lhb ϭ s √fcЈ ( where db is the bar diameter, in (mm), and fcЈ is the 28-day compressive strength of the concrete, lb/in2 (MPa). W l b WORKING-STRESS DESIGN OF t RECTANGULAR BEAMS WITH TENSION REINFORCEMENT ONLY From the assumption that stress varies across a beam sec- tion with the distance from the neutral axis, it follows that A n fc k ϭ T fs 1Ϫk where n ϭ modular ratio Es /Ec w Es ϭ modulus of elasticity of steel reinforcement, ksi (MPa) Ec ϭ modulus of elasticity of concrete, ksi (MPa) w TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 179178 40816 HICKS Mcghp Chap_05 Pgs 179 7/10/2001 CONCRETE FORMULAS 179e fc ϭ compressive stress in extreme surface of con-e crete, ksi (MPa) fs ϭ stress in steel, ksi (MPa) kd ϭ distance from extreme compression surface to neutral axis, in (mm)0 d ϭ distance from extreme compression to cen- troid of reinforcement, in (mm) When the steel ratio ␳ ϭ As /bd, where As ϭ area of ten- sion reinforcement, in2 (mm2), and b ϭ beam width, in (mm), is known, k can be computed fromy k ϭ √2n␳ ϩ (n␳)2 Ϫ n␳ Wherever positive-moment steel is required, ␳ should be at least 200/fy, where fy is the steel yield stress. The distance jd between the centroid of compression and the centroid of tension, in (mm), can be obtained from k jϭ1Ϫ 3-t Allowable Bending Moment The moment resistance of the concrete, inиkip (k иNm) is M c ϭ 1͞2 fckjbd 2 ϭ K cbd 2 where Kc ϭ 1͞2 fc kj. The moment resistance of the steel is, M s ϭ fs As jd ϭ fs␳jbd 2 ϭ K sbd 2) where Ks ϭ fs␳j. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 180 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 180 180 CHAPTER FIVE Allowable Shear w s The nominal unit shear stress acting on a section with shear V is a d V vϭ bd Allowable shear stresses are 55 percent of those for ultimate-strength design. Otherwise, designs for shear by F the working-stress and ultimate-strength methods are the d same. Except for brackets and other short cantilevers, the section for maximum shear may be taken at a distance d from the face of the support. In working-stress design, the shear stress vc carried by the concrete alone should not exceed 1.1 √fcЈ . (As an alternative, the maximum for vc may be taken as √fcЈ ϩ 1300␳Vd/M , with a maximum of 1.9 √fcЈ ; fcЈ is S the 28-day compressive strength of the concrete, lb/in2 e (MPa), and M is the bending moment at the section but o should not be less than Vd.) t At cross sections where the torsional stress vt exceeds 0.825√fcЈ , vc should not exceed 1.1√f cЈ vc ϭ w √1 ϩ (vt /1.2v)2 The excess shear v Ϫ vc should not exceed 4.4√fcЈ in sec- tions with web reinforcement. Stirrups and bent bars should be capable of resisting the excess shear VЈ ϭ V Ϫ vc bd. The area required in the legs of a vertical stirrup, in2 (mm2), is S t VsЈ s Av ϭ fv d m TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 181180 40816 HICKS Mcghp Chap_05 Pgs 181 7/10/2001 CONCRETE FORMULAS 181 where s ϭ spacing of stirrups, in (mm); and fv ϭ allowable stress in stirrup steel, (lb/in2) (MPa).r For a single bent bar or a single group of parallel bars all bent at an angle ␣ with the longitudinal axis at the same distance from the support, the required area is VЈ Av ϭ fv sin ␣ry For inclined stirrups and groups of bars bent up at differente distances from the support, the required area ised VsЈe Av ϭd fv d(sin ␣ ϩ cos ␣)ns Stirrups in excess of those normally required are provided2 each way from the cutoff for a distance equal to 75 percentt of the effective depth of the member. Area and spacing of the excess stirrups should be such thats bws Av Ն 60 fy where Av ϭ stirrup cross-sectional area, in2 (mm2) bw ϭ web width, in (mm)-d s ϭ stirrup spacing, in (mm)2 fy ϭ 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 of bars cut off to the total area of ten- sion bars at the section and d is the effective depth of the member. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 182 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 182 182 CHAPTER FIVE ULTIMATE-STRENGTH DESIGN OF w RECTANGULAR BEAMS WITH COMPRESSION BARS The bending-moment capacity of a rectangular beam with both tension and compression steel is W ΄ M u ϭ 0.90 (As Ϫ AЈ) fy d Ϫ s ΂ a 2 ΃ ϩ AЈ f (d Ϫ dЈ)΅ s y R C where a ϭ depth of equivalent rectangular compressive stress distribution T s ϭ (As Ϫ AЈ)fy /fcЈb s u b ϭ width of beam, in (mm) d ϭ distance from extreme compression surface to centroid of tensile steel, in (mm) w dЈ ϭ distance from extreme compression surface to centroid of compressive steel, in (mm) As ϭ area of tensile steel, in2 (mm2) AЈ ϭ area of compressive steel, in2 (mm2) s fy ϭ yield strength of steel, ksi (MPa) fcЈ ϭ 28-day strength of concrete, ksi (MPa) w This is valid only when the compressive steel reaches fy and occurs when Ј f c dЈ 87,000 (␳ Ϫ ␳Ј) Ն 0.85␤1 fy d 87,000 Ϫ fy TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 183182 40816 HICKS Mcghp Chap_05 Pgs 183 7/10/2001 CONCRETE FORMULAS 183 where ␳ ϭ As /bd ␳Ј ϭ AЈ /bd s ␤1 ϭ a constanth WORKING-STRESS DESIGN OF RECTANGULAR BEAMS WITH COMPRESSION BARSe The following formulas, based on the linear variation of stress and strain with distance from the neutral axis, may be used in design: 1 kϭo 1 ϩ fs /nfc where fs ϭ stress in tensile steel, ksi (MPa)o fc ϭ stress in extreme compression surface, ksi (MPa) n ϭ modular ratio, Es /Ec kd Ϫ dЈ fsЈ ϭ 2f d Ϫ kd s where fsЈ ϭ stress in compressive steel, ksi (MPa)d d ϭ distance from extreme compression surface to centroid of tensile steel, in (mm) d Ј ϭ distance from extreme compression surface to centroid of compressive steel, in (mm) TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 184 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 184 184 CHAPTER FIVE The factor 2 is incorporated into the preceding equation w in accordance with ACI 318, “Building Code Requirements s for Reinforced Concrete,” to account for the effects of r creep and nonlinearity of the stress–strain diagram for con- crete. However, fsЈ should not exceed the allowable tensile stress for the steel. Because total compressive force equals total tensile force on a section, w u C ϭ Cc ϩ Cs ϭ T Ј s where C ϭ total compression on beam cross section, kip (N) Cc ϭ total compression on concrete, kip (N) at section CsЈ ϭ force acting on compressive steel, kip (N) T ϭ force acting on tensile steel, kip (N) l fs k m ϭ fc 2[␳ Ϫ ␳Ј(kd Ϫ dЈ)/(d Ϫ kd )] where ␳ ϭ As /bd and ␳Ј ϭ AЈ/bd. s For reviewing a design, the following formulas may be used: √΂ ΃ ϩ n (␳ ϩ ␳Ј) Ϫ n(␳ ϩ ␳Ј) dЈ w kϭ 2n ␳ ϩ ␳Ј 2 2 d (k 3d / 3) ϩ 4n␳ЈdЈ[k Ϫ (dЈ/d )] zϭ jd ϭ d Ϫ z k 2 ϩ 4n␳Ј[k Ϫ (dЈ/d )] TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 185184 40816 HICKS Mcghp Chap_05 Pgs 185 7/10/2001 CONCRETE FORMULAS 185n where jd is the distance between the centroid of compres-s sion and the centroid of the tensile steel. The momentf resistance of the tensile steel is-e M Ms ϭ Tjd ϭ As fs jd fs ϭ As jde where M is the bending moment at the section of beam under consideration. The moment resistance in compres- sion is, Mc ϭ 1 ΄ f jbd 2 k ϩ 2n␳Ј 1 Ϫ 2 c ΂dЈ kd ΃΅t 2M fc ϭ jbd {k ϩ 2n␳Ј[1 Ϫ dЈ/kd )]} 2 Computer software is available for the preceding calcu- lations. Many designers, however, prefer the following approxi- mate formulas: M1 ϭ 1 2 ΂ fc bkd d Ϫ kd 3 ΃e MЈ ϭ M Ϫ M 1 ϭ 2 fsЈAЈ(d Ϫ dЈ ) s s where M ϭ bending moment MsЈ ϭ moment-resisting capacity of compressive steel M1 ϭ moment-resisting capacity of concrete TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 186 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 186 186 CHAPTER FIVE ULTIMATE-STRENGTH DESIGN OF T I AND T BEAMS t ␳ When the neutral axis lies in the ﬂange, the member may be designed as a rectangular beam, with effective width b and depth d. For that condition, the ﬂange thickness t will be W greater than the distance c from the extreme compression I surface to the neutral axis, 1.18␻ d F cϭ d ␤1 d where ␤1 ϭ constant o ( ␻ ϭ As fy /bd fcЈ r As ϭ area of tensile steel, in2 (mm2) b fy ϭ yield strength of steel, ksi (MPa) t b fcЈ ϭ 28-day strength of concrete, ksi (MPa) c When the neutral axis lies in the web, the ultimate moment should not exceed ΄ ΂ Mu ϭ 0.90 (As Ϫ Asf ) fy d Ϫ a 2 ΃ ΂ ϩ Asf fy d Ϫ t 2 ΃΅ (8.51) w where Asf ϭ area of tensile steel required to develop com- pressive strength of overhanging ﬂange, in2 (mm2) ϭ 0.85(b Ϫ bw)tfcЈ/ fy bw ϭ width of beam web or stem, in (mm) a ϭ depth of equivalent rectangular compressive stress distribution, in (mm) ϭ (As Ϫ Asf)fy / 0.85 fcЈ bw TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 187186 40816 HICKS Mcghp Chap_05 Pgs 187 7/10/2001 CONCRETE FORMULAS 187 The quantity ␳w Ϫ ␳f should not exceed 0.75␳b, where ␳b is the steel ratio for balanced conditions ␳w ϭ As /bwd, and ␳f ϭ Asf /bw d.ede WORKING-STRESS DESIGN OFn I AND T BEAMS For T beams, effective width of compression ﬂange is determined by the same rules as for ultimate-strength design. Also, for working-stress design, two cases may occur: the neutral axis may lie in the ﬂange or in the web. (For negative moment, a T beam should be designed as a rectangular beam with width b equal to that of the stem.) If the neutral axis lies in the ﬂange, a T or I beam may be designed as a rectangular beam with effective width b. If the neutral axis lies in the web or stem, an I or T beam may be designed by the following formulas, which ignore the compression in the stem, as is customary:t I kϭ 1 ϩ fs /nfc where kd ϭ distance from extreme compression surface to neutral axis, in (mm)- d ϭ distance from extreme compression surface2 to centroid of tensile steel, in (mm) fs ϭ stress in tensile steel, ksi (MPa)e fc ϭ stress in concrete at extreme compression surface, ksi (MPa) n ϭ modular ratio ϭ Es /Ec TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 188 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 188 188 CHAPTER FIVE Because the total compressive force C equals the total d tension T, f 1 bt Cϭ f (2kd Ϫ t) ϭ T ϭ As fs 2 c kd U 2ndAs ϩ bt 2 F kd ϭ 2nAs ϩ 2bt W l where As ϭ area of tensile steel, in2 (mm2); and t ϭ ﬂange r thickness, in (mm). The distance between the centroid of the area in com- pression and the centroid of the tensile steel is t (3kd Ϫ 2t) jd ϭ d Ϫ z zϭ 3(2kd Ϫ t) l The moment resistance of the steel is r t Ms ϭ Tjd ϩ As fs jd The moment resistance of the concrete is fc btjd Mc ϭ Cjd ϭ (2kd Ϫ t) w 2kd t In design, Ms and Mc can be approximated by m ΂ Ms ϭ As fs d Ϫ t 2 ΃ w Mc ϭ 1 2 ΂ fc bt d Ϫ t 2 ΃ TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 189188 40816 HICKS Mcghp Chap_05 Pgs 189 7/10/2001 CONCRETE FORMULAS 189l derived by substituting d Ϫ t/2 for jd and fc /2 for fc(1 Ϫ t/2kd), the average compressive stress on the section. ULTIMATE-STRENGTH DESIGN FOR TORSION When the ultimate torsion Tu is less than the value calcu- lated from the Tu equation that follows, the area Av of sheare reinforcement should be at least- bw s Av ϭ 50 fy However, when the ultimate torsion exceeds Tu calcu- lated from the Tu equation that follows, and where web reinforcement is required, either nominally or by calcula- tion, the minimum area of closed stirrups required is 50bw s Av ϩ 2At ϭ fy where At is the area of one leg of a closed stirrup resisting torsion within a distance s. Torsion effects should be considered whenever the ulti- mate torsion exceeds Tu ϭ ␾ ΂0.5 √fcЈ ͚x2y΃ where ␾ ϭ capacity reduction factor ϭ 0.85 Tu ϭ ultimate design torsional moment TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 190 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 190 190 CHAPTER FIVE ⌺x2y ϭ sum for component rectangles of section of product of square of shorter side and longer n side of each rectangle (where T section applies, N overhanging ﬂange width used in design should s not exceed three times ﬂange thickness) t o The torsion Tc carried by the concrete alone should not exceed 0.8√f cЈ ͚x2y Tc ϭ √1 ϩ (0.4Vu /Ct Tu )2 where Ct ϭ bwd /⌺x2y. Spacing of closed stirrups for torsion should be com- puted from At ␾ fy ␣t x1 y1 sϭ (Tu Ϫ ␾Tc) I where At ϭ area of one leg of closed stirrup b ␣t ϭ 0.66 ϩ 0.33y1/x1 but not more than 1.50 fy ϭ yield strength of torsion reinforcement x1 ϭ shorter dimension c to c of legs of closed W stirrup T y1 ϭ longer dimension c to c of legs of closed T stirrup The spacing of closed stirrups, however, should not exceed (x1 ϩ y1)/4 or 12 in (304.8 mm). Torsion reinforcement w should be provided over at least a distance of d ϩ b beyond t the point where it is theoretically required, where b is the l beam width. o TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 191 190 40816 HICKS Mcghp Chap_05 Pgs 191 7/10/2001 CONCRETE FORMULAS 191f At least one longitudinal bar should be placed in each cor-r ner of the stirrups. Size of longitudinal bars should be at least , No. 3, and their spacing around the perimeters of the stirrupsd should not exceed 12 in (304.8 mm). Longitudinal bars larger than No. 3 are required if indicated by the larger of the values of Al computed from the following two equations:t x 1 ϩ y1 Al ϭ 2At s- Al ϭ ΄ 400xs ΂ (T ϩTV /3C ) ΃ fy u u u t Ϫ 2At ΅΂ x ϩ y ΃ 1 s 1 In the second of the preceding two equations 50bws /fy may be substituted for 2At. The maximum allowable torsion is Tu ϭ ␾5Tc.d WORKING-STRESS DESIGN FOR TORSION Torsion effects should be considered whenever the torsiond T due to service loads exceeds T ϭ 0.55(0.5 fcЈ ͚x2y)dt where ⌺x2y ϭ sum for the component rectangles of the sec-d tion of the product of the square of the shorter side and thee longer side of each rectangle. The allowable torsion stress on the concrete is 55 percent of that computed from the TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 192 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 192 192 CHAPTER FIVE preceding Tc equation. Spacing of closed stirrups for tor- sion should be computed from 3At ␣t x1 y1 fv sϭ (vt Ϫ vtc)͚x2y where At ϭ area of one leg of closed stirrup 0.33y1 ␣t ϭ 0.66 ϩ , but not more than 1.50 x1 ␯tc ϭ allowable torsion stress on concrete x1 ϭ shorter dimension c to c of legs of closed stirrup y1 ϭ longer dimension c to c of legs of closed stirrup FLAT-SLAB CONSTRUCTION Slabs supported directly on columns, without beams or girders, are classiﬁed as ﬂat 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 lies inside 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 most economical type for long spans, although formwork may be more expensive than for a solid slab. A waffle slab omits much of the concrete that would be in ten- sion and thus is not considered effective in resisting stresses. TLFeBOOK
• A h b g d d a e s s - r - . 192 40816_05a_p147-195 40816 HICKS Mcghp Chap_05 Pgs 193 7/10/2001 10/22/01193 12:45 PM Page 193 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. TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 194 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 p. 194 194 CHAPTER FIVE To control deﬂection, the ACI Code establishes minimum thicknesses for slabs, as indicated by the following equation: l n(0.8 ϩ fy /200,000) S hϭ 36 ϩ 5␤[␣m Ϫ 0.12(1 ϩ 1/␤)] ( l n(0.8 ϩ fy /200,000) Ն 36 ϩ 9␤ F where h ϭ slab thickness, in (mm) ln ϭ length of clear span in long direction, in (mm) F c fy ϭ yield strength of reinforcement, ksi (MPa) c ␤ ϭ ratio of clear span in long direction to clear c span in the short direction T e ␣m ϭ average value of ␣ for all beams on the edges of a panel s ␣ ϭ ratio of ﬂexural stiffness Ecb Ib of beam section t to ﬂexural stiffness Ecs Is of width of slab c bounded laterally by centerline of adjacent t panel, if any, on each side of beam l e Ecb ϭ modulus of elasticity of beam concrete Ecs ϭ modulus of elasticity of slab concrete D Ib ϭ moment of inertia about centroidal axis of gross section of beam, including that portion of T slab on each side of beam that extends a dis- tance equal to the projection of the beam above or below the slab, whichever is greater, but not more than four times slab thickness TLFeBOOK
• 40816_05a_p147-195 10/22/01 12:45 PM Page 195 194 40816 HICKS Mcghp Chap_05 Pgs 195 7/10/2001 CONCRETE FORMULAS 195m Is ϭ moment of inertia about centroidal axis of : gross section of slab ϭ h3/12 times slab width speciﬁed in deﬁnition of ␣ Slab thickness h, however, need not be larger than (ln /36) (0.8 ϩ fy /200,000). FLAT-PLATE CONSTRUCTION ) Flat slabs with constant thickness between supports are called ﬂat plates. Generally, capitals are omitted from the columns. Exact analysis or design of ﬂat slabs or ﬂat plates is very r complex. It is common practice to use approximate methods. The ACI Code presents two such methods: direct design and equivalent frame. s In both methods, a ﬂat slab is considered to consist of strips parallel to column lines in two perpendicular direc-n tions. In each direction, a column strip spans betweenb columns and has a width of one-fourth the shorter of thet two perpendicular spans on each side of the column center- line. The portion of a slab between parallel column strips in each panel is called the middle strip (see Fig. 5.3). Direct Design Methodff This may be used when all the following conditions exist:- The slab has three or more bays in each direction.et Ratio of length to width of panel is 2 or less. Loads are uniformly distributed over the panel. TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 196 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 196 196 CHAPTER FIVE Ratio of live to dead load is 3 or less. v Columns form an approximately rectangular grid o (10 percent maximum offset). 1 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 ␣1 ΂ ll ΃ Յ 5 2 2 w 0.2 Յ ␣2 1 where ␣1 ϭ ␣ in direction of l1 ␣2 ϭ ␣ in direction of l2 ␣ ϭ relative beam stiffness deﬁned in the preced- ing equation l1 ϭ span in the direction in which moments are being determined, c to c of supports l2 ϭ span perpendicular to l1, c to c of supports 2 The basic equation used in direct design is the total static design moment in a strip bounded laterally by the cen- terline of the panel on each side of the centerline of the supports: wl2l2 n Mo ϭ 8 where w ϭ uniform design load per unit of slab area and S ln ϭ clear span in direction moments are being determined. The strip, with width l2, should be designed for bending S moments for which the sum in each span of the absolute a TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 197s 196 40816 HICKS Mcghp Chap_05 Pgs 197 7/10/2001 CONCRETE FORMULAS 197 values of the positive and average negative moments equalsd or exceeds Mo. 1. The sum of the ﬂexural stiffnesses of the columns abovee and below the slab ⌺Kc should be such thate ͚Kc ␣c ϭ Ն ␣min ͚(Ks ϩ Kb) where Kc ϭ ﬂexural stiffness of column ϭ EccIc Ecc ϭ modulus of elasticity of column concrete Ic ϭ moment of inertia about centroidal axis of gross section of column Ks ϭ Ecs Is- Kb ϭ Ecb Ibe ␣min ϭ minimum value of ␣c as given in engineering handbooks 2. If the columns do not satisfy condition 1, the design positive moments in the panels should be multiplied byl the coefficient:-e ␦s ϭ 1 ϩ 2 Ϫ ␤a 4 ϩ ␤a ΂1 Ϫ ␣␣ ΃ c mind SHEAR IN SLABS .g Slabs should also be investigated for shear, both beam typee and punching shear. For beam-type shear, the slab is considered TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 198 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 198 198 CHAPTER FIVE as a thin, wide rectangular beam. The critical section for b diagonal tension should be taken at a distance from the t face of the column or capital equal to the effective depth d E of the slab. The critical section extends across the full m width b of the slab. Across this section, the nominal shear t stress vu on the unreinforced concrete should not exceed the ultimate capacity 2√f cЈ or the allowable working stress 1.1√fcЈ , where fcЈ is the 28-day compressive strength of the concrete, lb/in2 (MPa). C Punching shear may occur along several sections extending completely around the support, for example, A around the face of the column or column capital or around s the drop panel. These critical sections occur at a distance a d/2 from the faces of the supports, where d is the effective a depth of the slab or drop panel. Design for punching shear should be based on f t ␾Vn ϭ ␾(Vc ϩ VS) p t where ␾ ϭ capacity reduction factor (0.85 for shear and torsion), with shear strength Vn taken not larger than the b concrete strength Vc calculated from ΂ Vc ϭ 2 ϩ 4 ␤c ΃ √f Ј b d Յ 4 √f Ј b d c o c o where bo ϭ perimeter of critical section and ␤c ϭ ratio of long side to short side of critical section. However, if shear reinforcement is provided, the allow- w able shear may be increased a maximum of 50 percent if d shear reinforcement consisting of bars is used and increased b a maximum of 75 percent if shearheads consisting of two t pairs of steel shapes are used. Shear reinforcement for slabs generally consists of bent i bars and is designed in accordance with the provisions for a TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 199s 198 40816 HICKS Mcghp Chap_05 Pgs 199 7/10/2001 CONCRETE FORMULAS 199r beams with the shear strength of the concrete at critical sec-e tions taken as 2√f cЈ bo d at ultimate strength and Vn Յ 6√fcЈbo d.d Extreme care should be taken to ensure that shear reinforce-l ment is accurately placed and properly anchored, especially inr thin slabs.dse COLUMN MOMENTSs , Another important consideration in design of two-way slabd systems is the transfer of moments to columns. This is gener-e ally a critical condition at edge columns, where the unbal-e anced slab moment is very high due to the one-sided panel.r The unbalanced slab moment is considered to be trans- ferred to the column partly by ﬂexure across a critical sec- tion, which is d/2 from the periphery of the column, and partly by eccentric shear forces acting about the centroid ofd the critical section.e That portion of unbalanced slab moment Mu transferred by the eccentricity of the shear is given by ␥␯ Mu: 1 ␥v ϭ 1 Ϫf 1ϩ ΂ ΃√ b 2 3 b 1 2- where b1 ϭ width, in (mm), of critical section in the spanf direction for which moments are being computed; andd b2 ϭ width, in (mm), of critical section in the span direc-o tion perpendicular to b1. As the width of the critical section resisting momentt increases (rectangular column), that portion of the unbal-r anced moment transferred by ﬂexure also increases. The TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 200 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 200 200 CHAPTER FIVE maximum factored shear, which is determined by combin- b ing the vertical load and that portion of shear due to the a unbalanced moment being transferred, should not exceed n ␾Vc, with Vc given by preceding the Vc equation. The shear C due to moment transfer can be determined at the critical s section by treating this section as an analogous tube with thickness d subjected to a bending moment ␥␯ Mu. c The shear stress at the crack, at the face of the column or bracket support, is limited to 0.2 fcЈ or a maximum of 800 Ac, where Ac is the area of the concrete section resisting shear transfer. The area of shear-friction reinforcement Avf required in addi- tion to reinforcement provided to take the direct tension due to w temperature changes or shrinkage should be computed from Vu Avf ϭ ␾fy␮ where Vu is the design shear, kip (kN), at the section; fy is the reinforcement yield strength, but not more than 60 ksi (413.7 MPa); and ␮, the coefficient of friction, is 1.4 for monolithic concrete, 1.0 for concrete placed against hard- ened concrete, and 0.7 for concrete placed against structural rolled-steel members. The shear-friction reinforcement should B be well distributed across the face of the crack and properly anchored at each side. A u c e SPIRALS r a This type of transverse reinforcement should be at least c ͞8 in (9.5 mm) in diameter. A spiral may be anchored at 3 each of its ends by 11͞2 extra turns of the spiral. Splices may f TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 201s 200 40816 HICKS Mcghp Chap_05 Pgs 201 7/10/2001 CONCRETE FORMULAS 201- be made by welding or by a lap of 48 bar diameters, bute at least 12 in (304.8 mm). Spacing (pitch) of spirals shouldd not exceed 3 in (76.2 mm), or be less than 1 in (25.4 mm).r Clear spacing should be at least 11͞3 times the maximuml size of coarse aggregate.h The ratio of the volume of spiral steel/volume of con- crete core (out to out of spiral) should be at leastn0g ␳s ϭ 0.45 ΂A A g c Ϫ1΃ ffЈ c y-o where Ag ϭ gross area of column Ac ϭ core area of column measured to outside of spiral fy ϭ spiral steel yield strengths fcЈ ϭ 28-day compressive strength of concreteir-ld BRACED AND UNBRACED FRAMESy As a guide in judging whether a frame is braced or unbraced, note that the commentary on ACI 318Ϫ83 indi- cates that a frame may be considered braced if the bracing elements, such as shear walls, shear trusses, or other means resisting lateral movement in a story, have a total stiffness at least six times the sum of the stiffnesses of all thet columns resisting lateral movement in that story.t The slenderness effect may be neglected under the twoy following conditions: TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 203s 202 40816 HICKS Mcghp Chap_05 Pgs 203 7/10/2001 CONCRETE FORMULAS 203 where fcЈ ϭ 28-day compressive strength of concrete, ksi (MPa) Ag ϭ gross area of wall section, in2 (mm2) s ␾ ϭ strength reduction factor ϭ 0.70h - lc ϭ vertical distance between supports, in (mm)ϭn h ϭ overall thickness of wall, in (mm) k ϭ effective-length factor For a wall supporting a concentrated load, the length of wall effective for the support of that concentrated load should be taken as the smaller of the distance center to cen- ter between loads and the bearing width plus 4h.or SHEAR WALLSe- Walls subject to horizontal shear forces in the plane of the wall should, in addition to satisfying ﬂexural requirements,d be capable of resisting the shear. The nominal shear stresse can be computed fromr Vud vu ϭ- ␾hd f where Vu ϭ total design shear force ␾ ϭ capacity reduction factor ϭ 0.85 d ϭ 0.8lw TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 204 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 204 204 CHAPTER FIVE h ϭ overall thickness of wall lw ϭ horizontal length of wall The shear Vc carried by the concrete depends on whether Nu, the design axial load, lb (N), normal to the wall w horizontal cross section and occurring simultaneously with f Vu at the section, is a compression or tension force. When Nu is a compression force, Vc may be taken as 2√fcЈ hd, t where fcЈ is the 28-day strength of concrete, lb/in2 (MPa). When Nu is a tension force, Vc should be taken as the smaller t of the values calculated from r Nu d Vc ϭ 3.3 √fcЈ hd Ϫ 4lw ΄ Vc ϭ hd 0.6√fcЈ ϩ lw(1.25 √fcЈ Ϫ 0.2Nu /lwh) Mu /Vu Ϫ lw /2 ΅ w This equation does not apply, however, when Mu/Vu Ϫ lw /2 m is negative. s When the factored shear Vu is less than 0.5␾Vc, rein- forcement should be provided as required by the empirical t method for bearing walls. t When Vu exceeds 0.5␾Vc, horizontal reinforcement should f be provided with Vs ϭ Av fy d/s2, where s2 ϭ spacing of hori- zontal reinforcement, and Av ϭ reinforcement area. Also, the ratio ␳h of horizontal shear reinforcement, to the gross concrete area of the vertical section of the wall should be at least 0.0025. C Spacing of horizontal shear bars should not exceed lw /5, 3h, or 18 in (457.2 mm). In addition, the ratio of vertical shear rein- F forcement area to gross concrete area of the horizontal section w of wall does not need to be greater than that required for hori- p zontal reinforcement but should not be less than a TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 205s 204 40816 HICKS Mcghp Chap_05 Pgs 205 7/10/2001 CONCRETE FORMULAS 205 ␳n ϭ 0.0025 ϩ 0.5 2.5 Ϫ ΂ hw lw ΃ (␳h Ϫ 0.0025) Յ 0.0025nl where hw ϭ total height of wall. Spacing of vertical shear rein-h forcement should not exceed lw /3, 3h, or 18 in (457.2 mm).n In no case should the shear strength Vn be taken greater , than 10√f cЈ hd at any section. . Bearing stress on the concrete from anchorages of post-r tensioned members with adequate reinforcement in the end region should not exceed fb calculated from Ab fb ϭ 0.8 fcЈ √ Ab Ϫ 0.2 Յ 1.25 fciЈ Ab fb ϭ 0.6 √fcЈ √ AbЈ Յ fcЈ where Ab ϭ bearing area of anchor plate, and AbЈ ϭ maxi-2 mum area of portion of anchorage surface geometrically similar to and concentric with area of anchor plate.- A more reﬁned analysis may be applied in the design ofl the end-anchorage regions of prestressed members to develop the ultimate strength of the tendons. ␾ should be taken as 0.90d for the concrete.-ee . CONCRETE GRAVITY RETAINING WALLSr- Forces acting on gravity walls include the weight of the wall,n weight of the earth on the sloping back and heel, lateral earth- pressure, and resultant soil pressure on the base. It is advis- able to include a force at the top of the wall to account for TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 206 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 206 206 CHAPTER FIVE frost action, perhaps 700 lb/linear ft (1042 kg/m). A wall, F consequently, may fail by overturning or sliding, overstress- ( ing of the concrete or settlement due to crushing of the soil. t Design usually starts with selection of a trial shape and t dimensions, and this conﬁguration is checked for stability. For convenience, when the wall is of constant height, a 1-ft (0.305 m) long section may be analyzed. Moments are taken about the toe. The sum of the righting moments should be at least 1.5 times the sum of the overturning moments. To pre- vent sliding, ␮R v Ն 1.5Ph where ␮ ϭ coefficient of sliding friction Rv ϭ total downward force on soil, lb (N) Ph ϭ horizontal component of earth thrust, lb (N) Next, the location of the vertical resultant Rv should be found at various sections 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 be no tension in the wall. Finally, the pressure exerted by the base on the soil should be computed to ensure that the allowable pressure is not exceeded. When the resultant is within the middle third, the pressures, lb/ft2 (Pa), under the ends of the base are given by pϭ Rv A Ϯ Mc I R ϭ v A ΂1 Ϯ 6e ΃ L where A ϭ area of base, ft2 (m2) L ϭ width of base, ft (m) F e ϭ distance, parallel to L, from centroid of base w to Rv, ft (m) c TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 207s 206 40816 HICKS Mcghp Chap_05 Pgs 207 7/10/2001 CONCRETE FORMULAS 207 , Figure 5.4 shows the pressure distribution under a 1-ft- (0.305-m) strip of wall for e ϭ L/2 Ϫa, where a is the dis- . tance of Rv from the toe. When Rv is exactly L/3 from thed toe, the pressure at the heel becomes zero (Fig. 5.4c). When .tnt-)estsdtey FIGURE 5.4 Diagrams for pressure of the base of a concrete gravitye wall on the soil below. (a) Vertical section through the wall. (b) Signiﬁ- cant compression under the entire base. TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 208 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 208 208 CHAPTER FIVE FIGURE 5.4 (Continued) Diagrams for pressure of the base of a concrete wall on the soil below. (c) No compression along one edge of F the base. (d) Compression only under part of the base. No support from r the soil under the rest of the beam. b Rv falls outside the middle third, the pressure vanishes f under a zone around the heel, and pressure at the toe is much g larger than for the other cases (Fig. 5.4d). f m o CANTILEVER RETAINING WALLS o This type of wall resists the lateral thrust of earth pressure through cantilever action of a vertical stem and horizontal TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 209s 208 40816 HICKS Mcghp Chap_05 Pgs 209 7/10/2001 CONCRETE FORMULAS 209a f FIGURE 5.5 Cantilever retaining wall. (a) Vertical section shows mainm reinforcing steel placed vertically in the stem. (b) Moment diagram. base (Fig. 5.5). Cantilever walls generally are economicals for heights from 10 to 20 ft (3 to 6 m). For lower walls,h gravity walls may be less costly; for taller walls, counter- forts (Fig. 5.6) may be less expensive. Shear unit stress on a horizontal section of a counterfort may be computed from vc ϭ V1/bd, where b is the thickness of the counterfort and d is the horizontal distance from face of wall to main steel,e M V1 ϭ V Ϫ (tan ␪ ϩ tan ␾)l d TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 210 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 210 210 CHAPTER FIVE w F T a i W T w w F f u o m m w FIGURE 5.6 Counterfort retaining wall. (a) Vertical section. (b) Horizontal section. TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 211s 210 40816 HICKS Mcghp Chap_05 Pgs 211 7/10/2001 CONCRETE FORMULAS 211 where V ϭ shear on section M ϭ bending moment at section ␪ ϭ angle earth face of counterfort makes with vertical ␾ ϭ angle wall face makes with vertical For a vertical wall face, ␾ ϭ 0 and V1 ϭ V Ϫ (M/d)tan ␪. The critical section for shear may be taken conservatively at a distance up from the base equal to dЈ sin ␪ cos ␪, where dЈ is the depth of counterfort along the top of the base. WALL FOOTINGS The spread footing under a wall (Fig. 5.7) distributes the wall load horizontally to preclude excessive settlement. The footing acts as a cantilever on opposite sides of the wall under downward wall loads and upward soil pressure. For footings supporting concrete walls, the critical section for bending moment is at the face of the wall; for footings under masonry walls, halfway between the middle and edge of the wall. Hence, for a 1-ft (0.305-m) long strip of sym- metrical concrete-wall footing, symmetrically loaded, the maximum moment, ftиlb (Nиm), is p Mϭ (L Ϫ a)2 8 where p ϭ uniform pressure on soil, lb/ft2 (Pa) L ϭ width of footing, ft (m) a ϭ wall thickness, ft (m) TLFeBOOK
• 40816_05b_p196-212 10/22/01 12:46 PM Page 212 40816 HICKS Mcghp Ch05 Third Pass bcj 7/19/01 Pgs 212 212 CHAPTER FIVE FIGURE 5.7 Concrete wall footing. 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), does not exceed 5␾√f cЈ , where fcЈ is the 28-day concrete strength, lb/in2 (MPa) and ␾ ϭ 0.90, the footing does not need to be reinforced. If the tensile stress is larger, 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 the point at which the critical section for moment occurs. Wall footings also may be designed by ultimate- strength theory. TLFeBOOK
• d n h n g y e f - ) - r - t , . , , 40816_06_p213-242 40816 HICKS 10/22/01 Mcghp 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) Chap_06215 12:47 PM Standard Moment of Section Nominal dressed size Area of inertia modulus Board feet size S4S section bh 2 bh 2 per linear Pgs 214 – 242 Iϭ Sϭ b h b h A ϭ bh 2 12 6 foot of piece 2ϫ4 15͞8 ϫ 35͞8 ͞3 2 Page 215 5.89 6.45 3.56 2ϫ6 1 ͞8 ϫ 5 ͞2 5 1 8.93 22.53 8.19 1 2ϫ8 7/10/2001 15͞8 ϫ 71͞2 12.19 57.13 15.23 11͞3 Source: National Lumber Manufacturers Association. TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 216 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 216 CHAPTER SIX BEARING A The allowable unit stresses given for compression perpendic- n ular to the grain apply to bearings of any length at the ends f of beams and to all bearings 6 in (152.4 mm) or more in length at other locations. When calculating the required bearing 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 than at the end of the beam. For bearings of less than 6 in (152.4 mm) in length and not nearer than 3 in (76.2 mm) to the end of the member, the allowable stress for compression perpendicular to the grain should be modiﬁed by multiplying by the factor (l ϩ 3͞8)/l, where l is the length of the bearing in inches (mm) measured along the grain of the wood. BEAMS The extreme ﬁber stress in bending for a rectangular timber beam is f ϭ 6M/bh2 ϭ M/S A beam of circular cross section is assumed to have the same strength in bending as a square beam having the same cross-sectional area. The horizontal shearing stress in a rectangular timber beam is f H ϭ 3V/2bh (6.1) o For a rectangular timber beam with a notch in the lower s face at the end, the horizontal shearing stress is o TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 217 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 217 H ϭ (3V/2bd1) (h/d1) (6.2) A gradual change in cross section, instead of a square- notch, decreases the shearing stress nearly to that computeds for the actual depth above the notch.n Nomenclature for the preceding equations follows:de f ϭ maximum ﬁber stress, lb/in2 (MPa)n M ϭ bending moment, lbиin (Nm)fn h ϭ depth of beam, in (mm)er b ϭ width of beam, in (mm)rs S ϭ section modulus (ϭbh2/6 for rectangular section), in3 (mm3) H ϭ horizontal shearing stress, lb/in2 (MPa) V ϭ total shear, lb (N) d1 ϭ depth of beam above notch, in (mm)r l ϭ span of beam, in (mm) P ϭ concentrated load, lb (N) V1 ϭ modiﬁed total end shear, lb (N)e W ϭ total uniformly distributed load, lb (N)e x ϭ distance from reaction to concentrated load in (mm)r For simple beams, the span should be taken as the distance from face to face of supports plus one-half the required length) of bearing at each end; and for continuous beams, the spanr should be taken as the distance between the centers of bearing on supports. TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 218 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 218 CHAPTER SIX When determining V, neglect all loads within a distance from either support equal to the depth of the beam. In the stress grade of solid-sawn beams, allowances for checks, end splits, and shakes have been made in the assigned unit stresses. For such members, Eq. (6.1) does not indicate f the actual shear resistance because of the redistribution of shear stress that occurs in checked beams. For a solid-sawn beam that does not qualify using Eq. (6.1) and the H values given in published data for allowable unit stresses, the modiﬁed reaction V1 should be determined as shown next. b For concentrated loads, 10P(l Ϫ x) (x /h)2 V1 ϭ (6.3) 9l[2 ϩ (x /h)2] T c For uniform loading, V m V1 ϭ W 2 ΂1 Ϫ 2h ΃ l (6.4) The sum of the V 1 values from Eqs. (6.3) and (6.4) should be substituted for V in Eq. (6.1), and the resulting H values should be checked against those given in tables of allow- able unit stresses for end-grain bearing. Such values should be adjusted for duration of loading. COLUMNS The allowable unit stress on timber columns consisting of a single piece of lumber or a group of pieces glued together s to form a single member is i TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 219 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 219e P 3.619E ϭ (6.5) A (l/r)2rd For columns of square or rectangular cross section, thise formula becomesfn P 0.30E ϭ (6.6)s A (l/d)2d For columns of circular cross section, the formula becomes P 0.22E ϭ (6.7) A (l/d)2) The allowable unit stress, P/A, may not exceed the allowable compressive stress, c. The ratio, 1/d, must not exceed 50. Values of P/A are subject to the duration of loading adjust- ment given previously.) Nomenclature for Eqs. (6.5) to (6.7) follows: P ϭ total allowable load, lb (N)d A ϭ area of column cross section, in2 (mm2)s- c ϭ allowable unit stress in compression paralleld to grain, lb/in2 (MPa) d ϭ dimension of least side of column, in (mm) l ϭ unsupported length of column between points of lateral support, in (mm) E ϭ modulus of elasticity, lb/in2 (MPa) r ϭ least radius of gyration of column, in (mm)f For members loaded as columns, the allowable unitr stresses for bearing on end grain (parallel to grain) are given in data published by lumber associations. These allowable TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 220 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 220 CHAPTER SIX stresses apply provided there is adequate lateral support and w end cuts are accurately squared and parallel. When stresses exceed 75 percent of values given, bearing must be on a snug-ﬁtting metal plate. These stresses apply under conditions continuously dry, and must be reduced by 27 percent for glued-laminated lumber and lumber 4 in (102 mm) or less in thickness and by 9 percent for sawn lumber more than 4 in (102 mm) in thickness, for lumber exposed to weather. COMBINED BENDING AND AXIAL LOAD Members under combined bending and axial load should be R so proportioned that the quantity F Pa /P ϩ M a /M Ͻ 1 (6.8) T w where Pa ϭ total axial load on member, lb (N) L P ϭ total allowable axial load, lb (N) m Ma ϭ total bending moment on member, lb in (Nm) e M ϭ total allowable bending moment, lb in (Nm) T COMPRESSION AT ANGLE TO GRAIN a The allowable unit compressive stress when the load is at an angle to the grain is cЈ ϭ c (cЌ)/[c (sin ␪)2 ϩ (cЌ) (cos ␪)2] (6.9) TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 221 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 221d where cЈ ϭ allowable unit stress at angle to grain, lb/in2s (MPa)as c ϭ allowable unit stress parallel to grain, lb/in2r (MPa)sn cЌ ϭ allowable unit stress perpendicular to grain, lb/in2 (MPa) ␪ ϭ angle between direction of load and direction of graine RECOMMENDATIONS OF THE FOREST PRODUCTS LABORATORY) The Wood Handbook gives advice on the design of solid wood columns. (Wood Handbook, USDA Forest Products Laboratory, Madison, Wisc., 1999.) Columns are divided into three categories, short, inter- mediate, and long. Let K denote a parameter deﬁned by the) equation ΂E΃) 1/2 K ϭ 0.64 (6.10) fc The range of the slenderness ratio and the allowable stress assigned to each category are next.t Short column, L Յ 11 f ϭ fc (6.11)) d TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 222 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 222 CHAPTER SIX Intermediate column, C L ΄ ΂ L/d ΃ ΅ 4 1 C 11 Ͻ ՅK f ϭ fc 1 Ϫ (6.12) d 3 K f t Long column, L 0.274E ϾK fϭ (6.13) d (L /d )2 The maximum L /d ratio is set at 50. The National Design Speciﬁcation covers the design of solid columns. The allowable stress in a rectangular section is as follows: 0.30E fϭ but f Յ fc (6.14) (L /d)2 a m a The notational system for the preceding equations is t f P ϭ allowable load A ϭ sectional area L ϭ unbraced length d ϭ smaller dimension of rectangular section E ϭ modulus of elasticity I fc ϭ allowable compressive stress parallel to grain in short column of given species f ϭ allowable compressive stress parallel to grain in given column TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 223 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 223 COMPRESSION ON OBLIQUE PLANE Consider that a timber member sustains a compressive) force with an action line that makes an oblique angle with the grain. Let P ϭ allowable compressive stress parallel to grain Q ϭ allowable compressive stress normal to grain) N ϭ allowable compressive stress inclined to grain ␪ ϭ angle between direction of stress N and direction of grainf By Hankinson’s equation,n PQ Nϭ (6.15) P sin 2 ␪ ϩ Q cos2 ␪) In Fig. 6.1, member M1 must be notched at the joint to avoid removing an excessive area from member M2. If the member is cut in such a manner that AC and BC make an angle of ␾/2 with vertical and horizontal planes, respectively, the allowable bearing pressures at these faces are identical for the two members. Let A ϭ sectional area of member M1 f1 ϭ pressure at AC f2 ϭ pressure at BC It may be readily shown that sin (␾ /2) cos (␾ /2) AC ϭ b BC ϭ b (6.16) sin ␾ sin ␾ TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 224 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 224 CHAPTER SIX w FIGURE 6.1 Timber joint. F sin ␾ F sin ␾ tan (␾ /2) f1 ϭ f2 ϭ (6.17) A tan (␾/2) A This type of joint is often used in wood trusses. ADJUSTMENT FACTORS FOR DESIGN VALUES F Design values obtained by the methods described earlier should i be multiplied by adjustment factors based on conditions of use, geometry, and stability. The adjustments are cumula- tive, unless speciﬁcally indicated in the following: Ј The adjusted design value Fb for extreme-ﬁber bending is given by w TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 225 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 225 Fb ϭ FbCDCMCtCLCF CV CfuCrCcCf Ј (6.18) where Fb ϭ design value for extreme-ﬁber bending CD ϭ load-duration factor CM ϭ wet-service factor Ct ϭ temperature factor CL ϭ beam stability factor CF ϭ size factor—applicable only to visually graded, sawn lumber and round timber ﬂexural members C v ϭ volume factor—applicable only to glued- laminated beams Cfu ϭ 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 ϭ repetitive-member factor—applicable only to dimension-lumber beams 2 to 4 in (50.8 to 101.6 mm) thick Cc ϭ curvature factor—applicable only to curved portions of glued-laminated beams Cf ϭ form factor For glued-laminated beams, use either CL or Cv (whicheverd is smaller), not both.f The adjusted design value for tension FtЈ is given by- FtЈ ϭ FtCDCMCtCF (6.19)g where Ft is the design value for tension. TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 226 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 226 CHAPTER SIX For shear, the adjusted design value Fv is computed from w FV ϭ FVCDCMCtCH Ј (6.20) where Fv is the design value for shear and CH is the shear stress factor ≥1—permitted for Fv parallel to the grain for sawn lumber members. For compression perpendicular to the grain, the adjusted design value FcЈЌ is obtained from FcЈЌ ϭ FcЌCMCtCb (6.21) S where FcЌ is the design value for compression perpendicu- F lar to the grain and Cb is the bearing area factor. a For compression parallel to the grain, the adjusted design s value FcЈ is given by f o FcЈ ϭ FcCDCMCtCFCp (6.22) s a a where Fc is the design value for compression parallel to g grain and Cp is the column stability factor. r For end grain in bearing parallel to the grain, the adjusted Ј design value, Fg is computed from o e FgЈ ϭ FgCDCt (6.23) b where Fg is the design value for end grain in bearing paral- lel to the grain. D The adjusted design value for modulus of elasticity, EЈ is s obtained from EЈ ϭ ECMCT C и и и (6.24) TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 227 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 227m where E ϭ design value for modulus of elasticity CT ϭ buckling stiffness factor—applicable only to) sawn-lumber truss compression chords 2 ϫ 4 in (50.8 ϫ 101.6 mm) or smaller, when subjectr to combined bending and axial compressionr and plywood sheathing 3͞8 in (9.5 mm) or more thick is nailed to the narrow faced C и и и ϭ other appropriate adjustment factors) Size and Volume Factors- For visually graded dimension lumber, design values Fb, Ft , and Fc for all species and species combinations, exceptn southern pine, should be multiplied by the appropriate size factor Cf , given in reference data to account for the effects of member size. This factor and the factors used to develop) size-speciﬁc values for southern pine are based on the adjustment equation given in American Society for Testing and Materials (ASTM) D1990. This equation, based on in-o grade test data, accounts for differences in Fb, Ft, and Fc related to width and in Fb and Ft related to length (test span).d For visually graded timbers [5 ϫ 5 in (127 ϫ 127 mm) or larger], when the depth d of a stringer beam, post, or timber exceeds 12 in (304.8 mm), the design value for bending should) be adjusted by the size factor CF ϭ (12 /d)1/9 (6.25)- Design values for bending Fb for glued-laminated beamss should be adjusted for the effects of volume by multiplying by ΄΂ 21 ΃΂ 12 ΃΂ 5.125 ΃΅ 1/x CV ϭ K L (6.25)) L d b TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 228 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 228 CHAPTER SIX where L ϭ length of beam between inﬂection points, ft (m) ( d ϭ depth, in (mm), of beam F g b ϭ width, in (mm), of beam z ϭ width, in (mm), of widest piece in multiple-piece f layups with various widths; thus, b ≤ 10.75 in T (273 mm) f ﬁ x ϭ 20 for southern pine m s ϭ 10 for other species ( KL ϭ loading condition coefﬁcient F For glulam beams, the smaller of Cv and the beam stability p factor CL should be used, not both. w t Radial Stresses and Curvature Factor The radial stress induced by a bending moment in a mem- ber of constant cross section may be computed from w r s 3M 1 fr ϭ (6.26) 2Rbd f o where M ϭ bending moment, inиlb (Nиm) R ϭ radius of curvature at centerline of member, in (mm) B b ϭ width of cross section, in (mm) D F d ϭ depth of cross section, in (mm) a TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 229 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 229) When M is in the direction tending to decrease curvature (increase the radius), tensile stresses occur across the grain. For this condition, the allowable tensile stress across the grain is limited to one-third the allowable unit stress in hori- zontal shear for southern pine for all load conditions ande for Douglas ﬁr and larch for wind or earthquake loadings.n The limit is 15 lb/in2 (0.103 MPa) for Douglas ﬁr and larch for other types of loading. These values are subject to modi- ﬁcation for duration of load. If these values are exceeded, mechanical reinforcement sufﬁcient to resist all radial ten- sile stresses is required. When M is in the direction tending to increase curvature (decrease the radius), the stress is compressive across the grain. For this condition, the design value is limited to that for com-y pression perpendicular to grain for all species. For the curved portion of members, the design value for wood in bending should be modiﬁed by multiplication by the following curvature factor: ΂R΃ t 2 Cc ϭ 1 Ϫ 2000 (6.27)- where t is the thickness of lamination, in (mm), and R is the radius of curvature of lamination, in (mm). Note that t/R should not exceed 1͞100 for hardwoods and southern pine or) 1 ͞125 for softwoods other than southern pine. The curvature factor should not be applied to stress in the straight portion of an assembly, regardless of curvature elsewhere., Bearing Area Factor Design values for compression perpendicular to the grain FcЌ apply to bearing surfaces of any length at the ends of a member and to all bearings 6 in (152.4 mm) or more long TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 230 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 230 CHAPTER SIX 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, FcЌ may be multiplied by the bearing area factor: L b ϩ 0.375 Cb ϭ (6.28) Lb where Lb is the bearing length, in (mm) measured parallel to grain. Equation (6.28) yields the values of Cb for elements with small areas, such as plates and washers, listed in refer- ence data. For round bearing areas, such as washers, Lb should be taken as the diameter. F o s Column Stability and Buckling Stiffness Factors p Design values for compression parallel to the grain Ft t should be multiplied by the column stability factor Cp given t by Eq. (6.29): l p m 1 ϩ (FcE /F c*) b CP ϭ 2c c (6.29) 9 √΄ ΅Ϫ Ϫ 1ϩ (FcE /F c*) 2 (FcE /Fc* ) 2c c where F* c ϭ design value for compression parallel to the grain multiplied by all applicable adjustment factors except Cp w FcE ϭ K cE EЈ/(L e /d) 2 EЈ ϭ modulus of elasticity multiplied by adjust- ment factors TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 231 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 231 ) KcE ϭ 0.3 for visually graded lumber and machine- , evaluated lumber ϭ 0.418 for products with a coefﬁcient of varia- tion less than 0.11 ) c ϭ 0.80 for solid-sawn lumber l ϭ 0.85 for round timber piles s - ϭ 0.90 for glued-laminated timber b For a compression member braced in all directions through- out its length to prevent lateral displacement, Cp ϭ 1.0. The buckling stiffness of a truss compression chord of sawn lumber subjected to combined ﬂexure and axial com- pression under dry service conditions may be increased ifFt the chord is 2 ϫ 4 in (50.8 ϫ 101.6 mm) or smaller and hasn the narrow face braced by nailing to plywood sheathing at least 3͞8 in (9.5 mm) thick in accordance with good nailing practice. The increased stiffness may be accounted for by multiplying the design value of the modulus of elasticity E by the buckling stiffness factor CT in column stability cal- culations. When the effective column length Le, in (mm), is ) 96 in (2.38 m) or less, CT may be computed from KM Le CT ϭ 1 ϩ (6.30) KT Eet where KM ϭ 2300 for wood seasoned to a moisture con- tent of 19 percent or less at time of sheathing attachment ϭ 1200 for unseasoned or partly seasoned wood at time of sheathing attachment TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 232 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 232 CHAPTER SIX KT ϭ 0.59 for visually graded lumber and machine- w evaluated lumber ϭ 0.82 for products with a coefﬁcient of varia- tion of 0.11 or less When Le is more than 96 in (2.38 m), CT should be calculated from Eq. (6.30) with Le ϭ 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 Le d RB ϭ √ b2 (6.31) The slenderness ratio should not exceed 50. The effective length Le for Eq. (6.31) is given in terms of unsupported length of beam in reference data. Unsupported F length is the distance between supports or the length of a cantilever when the beam is laterally braced at the supports to prevent rotation and adequate bracing is not installed N elsewhere in the span. When both rotational and lateral dis- T placements are also prevented at intermediate points, the e unsupported length may be taken as the distance between ( points of lateral support. If the compression edge is supported w throughout the length of the beam and adequate bracing is installed at the supports, the unsupported length is zero. The beam stability factor CL may be calculated from 1 ϩ (FbE /F b ) * CL ϭ w 1.9 (6.32) √΄ ΅Ϫ Ϫ 1ϩ * (FbE /F b ) 2 FbE /F b * 1.9 0.95 TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 233 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 233- where F b ϭ design value for bending multiplied by all * applicable adjustment factors, except Cfu, CV, and CL- FbE ϭ K bE EЈ/R B 2d K bE ϭ 0.438 for visually graded lumber and machine-l evaluated lumber , , ϭ 0.609 for products with a coefﬁcient of vari- ation of 0.11 or less EЈ ϭ design modulus of elasticity multiplied by applicable adjustment factors)fd FASTENERS FOR WOODasd Nails and Spikes- The allowable withdrawal load per inch (25.4 mm) of pen-e etration of a common nail or spike driven into side grainn (perpendicular to ﬁbers) of seasoned wood, or unseasonedd wood that remains wet, iss p ϭ 1,380G5/2D (6.33) where p ϭ allowable load per inch (mm) of penetration into member receiving point, lb (N)) D ϭ diameter of nail or spike, in (mm) G ϭ speciﬁc gravity of wood, oven dry TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 234 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 234 CHAPTER SIX The total allowable lateral load for a nail or spike driven into side grain of seasoned wood is w p ϭ CD 3/2 (6.34) where p ϭ allowable load per nail or spike, lb (N) D ϭ diameter of nail or spike, in (mm) C ϭ coefﬁcient dependent on group number of wood (see Table 6.1) W g Values of C for the four groups into which stress-grade i lumber is classiﬁed are Group I: C ϭ 2,040 Group II: C ϭ 1,650 w Group III: C ϭ 1,350 Group IV: C ϭ 1,080 The loads apply where the nail or spike penetrates into the member, receiving its point at least 10 diameters for Group I species, 11 diameters for Group II species, 13 diameters for T Group III species, and 14 diameters for Group IV species. S Allowable loads for lesser penetrations are directly propor- tional to the penetration, but the penetration must be at least one-third that speciﬁed. D P H Wood Screws H P The allowable withdrawal load per inch (mm) of penetra- R tion of the threaded portion of a wood screw into side grain S of seasoned wood that remains dry is TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 235 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 235n p ϭ 2,850G2D (6.35) where p ϭ allowable load per inch (mm) of penetration of) threaded portion into member receiving point, lb (N) D ϭ diameter of wood screw, in (mm) G ϭ 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 drivene into the side grain of seasoned wood which remains dry is p ϭ CD2 (6.36) where p ϭ allowable load per wood screw, lb (N) D ϭ diameter of wood screw, in (mm) C ϭ coefﬁcient dependent on group number ofe wood (Table 6.2)Ir TABLE 6.2 Speciﬁc Gravity and Group Number for Common. Species of Lumber-t 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 0.114n Spruce IV 0.41 0.168 0.108 TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 237 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 237e QЈ ϭ adjusted value for loading normal to grain Q ϭ nominal value for loading normal to grain For bolts, ZЈ ϭ ZCDCMCtCgD⌬ where CD ϭ load-duration factor, not to exceed 1.6 ford connections CM ϭ wet-service factor, not applicable to toenails loaded in withdrawal Ct ϭ temperature factor Cg ϭ group-action factors C⌬ ϭ geometry factor- For split-ring and shear-plate connectors,l .o PЈ ϭ PCDCMCt CgC⌬Cd Cston QЈ ϭ QCDCMCtCgC⌬Cddg where Cd is the penetration-depth factor and Cst is the metal-side-plate factor. For nails and spikes, WЈ ϭ WCD CM Ct Ctn ZЈ ϭ ZCD CM Ct Cd Ceg Cdi Ctn where Cdi ϭ is the diaphragm factor and Ctn ϭ toenail factor. TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 238 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 238 CHAPTER SIX For wood screws, t a WЈ ϭ WCDCMCt ( ZЈ ϭ ZCD CM Ct Cd Ceg s t where Ceg is the end-grain factor. For lag screws, WЈ ϭ WCD CM Ct Ceg w ZЈ ϭ ZCD CM Ct CgC⌬Cd Ceg For metal plate connectors, ZЈ ϭ ZCD CM Ct For drift bolts and drift pins, WЈ ϭ WCD CM Ct Ceg ZЈ ϭ ZCD CM Ct Cg C⌬Cd Ceg For spike grids, ( C ZЈ ϭ ZCD CM Ct C⌬ L B ROOF SLOPE TO PREVENT PONDING M s Roof beams should have a continuous upward slope equiv- alent to 1͞4 in/ft (20.8 mm/m) between a drain and the high point of a roof, in addition to minimum recommended camber to avoid ponding. When flat roofs have insuffic- ient slope for drainage (less than 1͞4 in/ft) (20.8 mm/m) a TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 239 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 239 the stiffness of supporting members should be such that a 5-lb/ft2 (239.4 N/mm2) load causes no more than 1͞2 -in (12.7-mm) deflection. Because of ponding, snow loads or water trapped by gravel stops, parapet walls, or ice dams magnify stresses and deﬂec- tions from existing roof loads by 1 Cp ϭ 1 Ϫ WЈL3/␲ 4EI where Cp ϭ factor for multiplying stresses and deﬂections under existing loads to determine stresses and deﬂections under existing loads plus ponding W ϭ weight of 1 in (25.4 mm) of water on roof area supported by beam, lb (N) L ϭ span of beam, in (mm) E ϭ modulus of elasticity of beam material, lb/in2 (MPa) I ϭ moment of inertia of beam, in4 (mm4) (Kuenzi and Bohannan, “Increases in Deﬂection and Stresses Caused by Ponding of Water on Roofs,” Forest Products Laboratory, Madison, Wisconsin.) BENDING AND AXIAL TENSION Members subjected to combined bending and axial tension should be proportioned to satisfy the interaction equations-h ft fd ϩ b* Յ 1 FcЈ Fb-) and TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 240 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 240 CHAPTER SIX ( fb Ϫ ft ) w Յ1 F ** b where ft ϭ tensile stress due to axial tension acting alone fb ϭ bending stress due to bending moment alone FЈ ϭ design value for tension multiplied by applica- t ble adjustment factors F b ϭ design value for bending multiplied by appli- * cable adjustment factors except CL F ** ϭ design value for bending multiplied by applica- b ble adjustment factors except Cv The load duration factor CD associated with the load of shortest duration in a combination of loads with differing duration may be used to calculate FЈ and F b. All applicable t * load combinations should be evaluated to determine the critical load combination. F BENDING AND AXIAL COMPRESSION Members subjected to a combination of bending and axial w compression (beam columns) should be proportioned to m satisfy the interaction equation ΂ FЈ ΃ ϩ [1 Ϫ ( f f/F f 2 c b1 c c cE1)]FЈ1 b a fb2 ϩ Յ1 [1 Ϫ ( fc /FcE2) Ϫ ( fb1/FbE )2]FЈ2 b TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 241 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 TIMBER ENGINEERING FORMULAS 241 where fc ϭ compressive stress due to axial compression acting alone FЈ ϭ design value for compression parallel to grain ce multiplied by applicable adjustment factors,e including the column stability factor- fb1 ϭ bending stress for load applied to the narrow face of the member fb2 ϭ bending stress for load applied to the wide- face of the member FЈ ϭ design value for bending for load applied to b1- the narrow face of the member multiplied by applicable adjustment factors, including the column stability factorfg FЈ ϭ design value for bending for load applied to b2e the wide face of the member multiplied bye applicable adjustment factors, including the column stability factor For either uniaxial or biaxial bending, fc should not exceed K cE EЈ FcE1 ϭ (L e1/d 1)2l where EЈ is the modulus of elasticity multiplied by adjust-o ment factors. Also, for biaxial bending, fc should not exceed K cE EЈ FcE2 ϭ (L e2 /d 2)2 and fb1 should not be more than K bE EЈ FbE ϭ R2B TLFeBOOK
• 40816_06_p213-242 10/22/01 12:47 PM Page 242 40816 HICKS Mcghp Chap_06 Pgs 214 – 242 7/10/2001 242 CHAPTER SIX where 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 deﬁned earlier in this section. The effective column lengths Le1 for buckling in the d1 direction and Le2 for buckling in the d2 direction, EЈ, FcE1, and FcE2 should be determined as shown earlier. As for the case of combined bending and axial tension, FЈ, FЈ , and FЈ2 should be adjusted for duration of load by c b1 b applying CD. TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 244 40816 HICKS Mcghp Chap_07 pgs 244 7/12/2001 244 CHAPTER SEVEN UNITS OF MEASUREMENT Units 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 m The most-used equivalents are 1 meter ϭ 39.37 in (exactly) ϭ 3.2808 ft 1 rod ϭ 1 pole ϭ 1 perch ϭ 161͞2 ft (5.029 m) 1 engineer’s chain ϭ 100 ft ϭ 100 links (30.48 m) 1 Gunter’s chain ϭ 66 ft (20.11 m) ϭ 100 Gunter’s links (lk) ϭ 4 rods ϭ 1͞80 mi (0.020 km) 1 acre ϭ 100,000 sq (Gunter’s) links ϭ 43,560 ft2 ϭ 160 rods2 ϭ 10 sq (Gunter’s) chains ϭ 4046.87 m2 ϭ 0.4047 ha T 1 rood ϭ 3͞4 acre (1011.5 m2) ϭ 40 rods2 (also local unit ϭ 51͞2 to 8 yd) (5.029 to 7.315 m) W q 1 ha ϭ 10,000 m2 ϭ 107,639.10 ft2 ϭ 2.471 acres b s 1 arpent ϭ about 0.85 acre, or length of side of n 1 square arpent (varies) (about 3439.1 m2) d 1 statute mi ϭ 5280 ft ϭ 1609.35 m v t 1 mi2 ϭ 640 acres (258.94 ha) t TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 245 40816 HICKS Mcghp Chap_07 pgs 245 7/12/2001 SURVEYING FORMULAS 245 1 nautical mi (U.S.) ϭ 6080.27 ft ϭ 1853.248 m 1 fathom ϭ 6 ft (1.829 m)e 1 cubit ϭ 18 in (0.457 m)s 1 vara ϭ 33 in (0.838 m) (Calif.), 331͞3 in (0.851 m)s (Texas), varies 1 degree ϭ 1͞360 circle ϭ 60 min ϭ 3600 s ϭ: 0.01745 rad sin 1Њ ϭ 0.01745241 1 rad ϭ 57Њ 17Ј 44.8Љ or about 57.30Њ 1 grad (grade) ϭ 1͞400 circle ϭ 1͞100 quadrant ϭ 100 centesimal min ϭ 104 centesimals (French) 1 mil ϭ 1͞6400 circle ϭ 0.05625Њ 1 military pace (milpace) ϭ 21͞2 ft (0.762 m) THEORY OF ERRORS When a number of surveying measurements of the same quantity have been made, they must be analyzed on the basis of probability and the theory of errors. After all systematic (cumulative) errors and mistakes have been elimi- nated, random (compensating) errors are investigated to determine the most probable value (mean) and other critical values. Formulas determined from statistical theory and the normal, or Gaussian, bell-shaped probability distribu- tion curve, for the most common of these values follow. TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 246 40816 HICKS Mcghp Chap_07 pgs 246 7/12/2001 246 CHAPTER SEVEN Standard deviation of a series of observations is ⌺d 2 ␴s ϭ Ϯ √ nϪ1 where d ϭ residual (difference from mean) of single obser- vation and n ϭ number of observations. The probable error of a single observation is PE s ϭ Ϯ0.6745␴s (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 the ratio of the area of the probability curve w included between the values to the total area. Inasmuch as the m area under the entire probability curve is unity, there is a 100 percent probability that all measurements will lie within the range of the curve. The area of the curve between Ϯ␴s is 0.683; that is, there is a 68.3 percent probability of an error between Ϯ␴s in a single measurement. This error range is also called the w one-sigma or 68.3 percent conﬁdence level. The area of the curve between Ϯ2␴s is 0.955. Thus there is a 95.5 percent probability of an error between Ϯ2␴s and Ϯ2␴s that represents the 95.5 percent error (two-sigma or 95.5 percent con- ﬁdence level). Similarly, Ϯ3␴s is referred to as the 99.7 percent error (three-sigma or 99.7 percent conﬁdence level). For practical purposes, a maximum tolerable level often is assumed to be the 99.9 percent error. Table 7.1 indicates the probability of occurrence of larger errors in a single M measurement. The probable error of the combined effects of accidental R errors from different causes is d TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 247 40816 HICKS Mcghp Chap_07 pgs 247 7/12/2001 SURVEYING FORMULAS 247 TABLE 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) 99.9ϩ 1 in 1000ss Esum ϭ √E 2 ϩ E2 ϩ E2 ϩ и и и 1 2 3e where E1, E2, E3 . . . are probable errors of the separatee measurements.a Error of the mean isn Esum E √n E , Em ϭ ϭ s ϭ s n n √nse where Es ϭ speciﬁed error of a single measurement.e Probable error of the mean ists PEs ⌺d 2-7 PEm ϭ √n ϭ Ϯ0.6745 √ n(n Ϫ 1) .nse MEASUREMENT OF DISTANCE WITH TAPESl Reasonable precisions for different methods of measuring distances are TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 248 40816 HICKS Mcghp Chap_07 pgs 248 7/12/2001 248 CHAPTER SEVEN Pacing (ordinary terrain): 1͞50 to 1͞100 w Taping (ordinary steel tape): 1͞1000 to 1͞10,000 (Results can be improved by use of tension apparatus, transit alignment, leveling.) Baseline (invar tape): 1͞50,000 to 1͞1,000,000 Stadia: 1͞300 to 1͞500 (with special procedures) Subtense bar: 1͞1000 to 1͞7000 (for short distances, with a 1-s theodolite, averaging angles taken at both ends) Electronic distance measurement (EDM) devices have been in use since the middle of the twentieth century and have now largely replaced steel tape measurements on large projects. The continued development, and the resulting drop in prices, are making their use widespread. A knowledge of steel-taping errors and corrections remains important, how- ever, because use of earlier survey data requires a knowl- edge of how the measurements were made, common sources f for errors, and corrections that were typically required. t For ordinary taping, a tape accurate to 0.01 ft (0.00305 m) should be used. The tension of the tape should be about 15 lb (66.7 N). The temperature should be determined within 10°F (5.56°C); and the slope of the ground, within 2 per- T cent; and the proper corrections, applied. The correction to be applied for temperature when using a steel tape is Ct ϭ 0.0000065s(T Ϫ T0) w The correction to be made to measurements on a slope is Ch ϭ s (1 Ϫ cos ␪) exact or ϭ 0.00015s␪ 2 approximate or ϭ h 2 /2s approximate TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 249 40816 HICKS Mcghp Chap_07 pgs 249 7/12/2001 SURVEYING FORMULAS 249 where Ct ϭ temperature correction to measured length, ft (m)e Ch ϭ correction to be subtracted from slope dis-, tance, ft (m) s ϭ measured length, ft (m) T ϭ temperature at which measurements are made,s ЊF (ЊC)e T0 ϭ temperature at which tape is standardized, ЊF (ЊC)d h ϭ difference in elevation at ends of measurede length, ft (m)pf ␪ ϭ slope angle, degree-- In more accurate taping, using a tape standardized whens fully supported throughout, corrections should also be made for tension and for support conditions. The correction for tension is) (Pm Ϫ Ps)s5 Cp ϭn SE- The correction for sag when not fully supported iso w 2L3 Cs ϭ 2 24P m where Cp ϭ tension correction to measured length, ft (m) Cs ϭ sag correction to measured length for each section of unsupported tape, ft (m) Pm ϭ actual tension, lb (N) Ps ϭ tension at which tape is standardized, lb (N) (usually 10 lb) (44.4 N) TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 250 40816 HICKS Mcghp Chap_07 pgs 250 7/12/2001 250 CHAPTER SEVEN S ϭ cross-sectional area of tape, in2 (mm2) F E ϭ modulus of elasticity of tape, lb/in (MPa) 2 (29 million lb/in2 (MPa) for steel) (199,955 MPa) w ϭ weight of tape, lb/ft (kg/m) w L ϭ unsupported length, ft (m) E ( Slope Corrections In slope measurements, the horizontal distance H ϭ L cos x, where L ϭ slope distance and x ϭ vertical angle, measured from the horizontal—a simple hand calculator operation. For slopes of 10 percent or less, the correction to w be applied to L for a difference d in elevation between tape ends, or for a horizontal offset d between tape ends, may be computed from d2 Cs ϭ 2L For a slope greater than 10 percent, Cs may be determined from O d2 d4 T Cs ϭ ϩ t 2L 8L3 a Temperature Corrections f b For incorrect tape length: (actual tape length Ϫ nominal tape length)L Ct ϭ nominal tape length TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 251 40816 HICKS Mcghp Chap_07 pgs 251 7/12/2001 SURVEYING FORMULAS 251 For nonstandard tension:) (applied pull Ϫ standard tension)L) Ct ϭ AE where A ϭ cross-sectional area of tape, in2 (mm2); and E ϭ modulus of elasticity ϭ 29,000,00 lb/in2 for steel (199,955 MPa). For sag correction between points of support, ft (m):L w2 L3 s Cϭ Ϫ , 24P2ro where w ϭ weight of tape per foot, lb (N)ee Ls ϭ unsupported length of tape, ft (m) P ϭ pull on tape, lb (N)m Orthometric Correction This is a correction applied to preliminary elevations due to ﬂattening of the earth in the polar direction. Its value is a function of the latitude and elevation of the level circuit. Curvature of the earth causes a horizontal line to depart from a level surface. The departure Cf , ft, or Cm, (m), may be computed from Cf ϭ 0.667M 2 ϭ 0.0239F 2 Cm ϭ 0.0785K 2 TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 252 40816 HICKS Mcghp Chap_07 pgs 252 7/12/2001 252 CHAPTER SEVEN where M, F, and K are distances in miles, thousands of feet, V and kilometers, respectively, from the point of tangency to the earth. T Refraction causes light rays that pass through the earth’s N atmosphere to bend toward the earth’s surface. For horizon- o tal sights, the average angular displacement (like the sun’s G diameter) is about 32 min. The displacement Rf, ft, or Rm, C m, is given approximately by d R f ϭ 0.093M 2 ϭ 0.0033F 2 l R m ϭ 0.011K 2 To obtain the combined effect of refraction and curvature of the earth, subtract Rf from Cf or Rm from Cm. Borrow-pit or cross-section leveling produces elevations w at the corners of squares 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 located readily, but topographic features, not so well. S Quantities of material to be excavated or ﬁlled are com- puted, in yd3 (m3), by selecting a grade elevation or ﬁnal I ground elevation, computing elevation differences for the c corners, and substituting in i c nxA b Qϭ 108 b f where n ϭ number of times a particular corner enters as part of a division block x ϭ difference in ground and grade elevation for each corner, ft (m) A ϭ area of each block, ft2 (m2) TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 253 40816 HICKS Mcghp Chap_07 pgs 253 7/12/2001 SURVEYING FORMULAS 253 , VERTICAL CONTROLo The NGS provides vertical control for all types of surveys.s NGS furnishes descriptions and elevations of bench marks- on request. As given in “Standards and Speciﬁcations fors Geodetic Control Networks,” Federal Geodetic Control, Committee, the relative accuracy C, mm, required between directly connected bench marks for the three orders of leveling is First order: C ϭ 0.5√K for Class I and 0.7√K for Class II Second order: C ϭ 1.0√K for Class I and 1.3√K for Class IIf Third order: C ϭ 2.0√Ks where K is the distance between bench marks, km.ed,s. STADIA SURVEYING-l In stadia surveying, a transit having horizontal stadiae crosshairs above and below the central horizontal crosshair is used. The difference in the rod readings at the stadia crosshairs is termed the rod intercept. The intercept may be converted to the horizontal and vertical distances between the instrument and the rod by the following formulas:s H ϭ Ki(cos a)2 ϩ ( f ϩ c) cos ar 1 Vϭ Ki(sin 2a) ϩ ( f ϩ c) sin a 2 TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 254 40816 HICKS Mcghp Chap_07 pgs 254 7/12/2001 254 CHAPTER SEVEN where H ϭ horizontal distance between center of transit and rod, ft (m) V ϭ vertical distance between center of transit and point on rod intersected by middle horizontal crosshair, ft (m) K ϭ stadia factor (usually 100) i ϭ rod intercept, ft (m) a ϭ vertical inclination of line of sight, measured F from the horizontal, degree t h f ϩ c ϭ 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 the older type of external-focusing telescope. C Relationships are comparable for the internal-focusing type. For horizontal sights, the stadia distance, ft, (m) (from f instrument spindle to rod), is T o f DϭR ϩC i P where R ϭ intercept on rod between two sighting wires, ft (m) P f ϭ focal length of telescope, ft (m) (constant for m speciﬁc instrument) q i i ϭ distance between stadia wires, ft (m) s TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 255 40816 HICKS Mcghp Chap_07 pgs 255 7/12/2001 SURVEYING FORMULAS 255 tdld 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. s Cϭfϩcda c ϭ distance from center of spindle to center of objective lens, ft (m) l . C is called the stadia constant, although c and C vary slightly. . The value of f/i, the stadia factor, is set by the manu-m facturer to be about 100, but it is not necessarily 100.00. The value should be checked before use on important work, or when the wires or reticle are damaged and replaced. PHOTOGRAMMETRY , Photogrammetry is the art and science of obtaining reliable r measurements by photography (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. TLFeBOOK
• 40816_07_p243-256 10/22/01 12:47 PM Page 256 40816 HICKS Mcghp Chap_07 pgs 256 7/12/2001 256 CHAPTER SEVEN Scale formulas are as follows: Photo scale photo distance ϭ Map scale map distance ab f Photo scale ϭ ϭ AB H Ϫ h1 where f ϭ focal length of lens, in (m) H ϭ ﬂying height of airplane above datum (usually mean sea level), ft (m) h1 ϭ elevation of point, line, or area with respect to datum, ft (m) TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 258 40816 HICKS Mcghp Chap_8 Pgs 258 7/12/2001 258 CHAPTER EIGHT PHYSICAL PROPERTIES OF SOILS w a Basic soil properties and parameters can be subdivided into ( physical, index, and engineering categories. Physical soil prop- erties include density, particle size and distribution, speciﬁc gravity, and water content. I The water content w of a soil sample represents the weight of free water contained in the sample expressed as a percentage of its dry weight. I The degree of saturation S of the sample is the ratio, p expressed as percentage, of the volume of free water con- a tained in a sample to its total volume of voids Vv. r Porosity n, which is a measure of the relative amount of voids, is the ratio of void volume to the total volume V of s soil: ( t Vv s nϭ (8.1) f V c p The ratio of Vv to the volume occupied by the soil particles t Vs deﬁnes the void ratio e. Given e, the degree of saturation a may be computed from wGs i Sϭ (8.2) e where Gs represents the speciﬁc gravity of the soil particles. For most inorganic soils, Gs is usually in the range of 2.67 Ϯ 0.05. The dry unit weight ␥d of a soil specimen with any degree of saturation may be calculated from D ␥w Gs S l ␥d ϭ (8.3) 1 ϩ wGs s TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 259 40816 HICKS Mcghp Chap_8 Pgs 259 7/12/2001 SOIL AND EARTHWORK FORMULAS 259 where ␥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/ft3o (1026.7 kg/m3) for seawater.-c INDEX PARAMETERS FOR SOILSes Index parameters of cohesive soils include liquid limit,, plastic limit, shrinkage limit, and activity. Such parameters- are useful for classifying cohesive soils and providing cor- relations with engineering soil properties.f The liquid limit of cohesive soils represents a near-liquidf state, that is, an undrained shear strength about 0.01 lb/ft2 (0.0488 kg/m2). The water content at which the soil ceases to exhibit plastic behavior is termed the plastic limit. The shrinkage limit represents the water content at which no) further volume change occurs with a reduction in water content. The most useful classiﬁcation and correlation parameters are the plasticity index Ip, the liquidity index Il,s the shrinkage index Is, and the activity Ac. These parametersn are deﬁned in Table 8.1. Relative density Dr of cohesionless soils may be expressed in terms of void ratio e or unit dry weight ␥d:) emax Ϫ eo Dr ϭ (8.4). emax Ϫ eminf 1/␥min Ϫ 1/␥d Dr ϭ (8.5)y 1/␥min Ϫ 1/␥max Dr provides cohesionless soil property and parameter corre-) lations, including friction angle, permeability, compressibility, small-strain shear modulus, cyclic shear strength, and so on. TLFeBOOK
• 40816_08_p257-282 40816 HICKS Mcghp Chap_8 Pgs 260 7/12/2001 TABLE 8.1 Soil Indices Index Deﬁnition† Correlation 10/22/01 Plasticity Ip ϭ Wl Ϫ Wp Strength, compressibility, compactibility, and so forth Wn Ϫ Wp Liquidity Il ϭ Compressibility and stress rate Ip260 Shrinkage Is ϭ Wp Ϫ Ws Shrinkage potential 12:48 PM Ip Activity Ac ϭ Swell potential, and so forth ␮ † Wt ϭ liquid limit; Wp ϭ plastic limit; Wn ϭ moisture content, %; Ws ϭ shrinkage limit; ␮ ϭ percent of soil Page 260 ﬁner than 0.002 mm (clay size). TLFeBOOK a u f U ␥ e o T V R
• 40816_08_p257-282 10/22/01 12:48 PM Page 261 40816 HICKS Mcghp Chap_8 Pgs 261 7/12/2001 SOIL AND EARTHWORK FORMULAS 261 RELATIONSHIP OF WEIGHTS AND VOLUMES IN SOILS The unit weight of soil varies, depending on the amount of water contained in the soil. Three unit weights are in gen- eral use: the saturated unit weight ␥sat, the dry unit weight ␥dry, and the buoyant unit weight ␥b: (G ϩ e)␥0 (1 ϩ w)G␥0 ␥sat ϭ ϭ S ϭ 100% 1ϩe 1ϩe G␥0 ␥dry ϭ S ϭ 0% (1 ϩ e) (G Ϫ 1)␥0 ␥b ϭ S ϭ 100% 1ϩe Unit weights are generally expressed in pound per cubic foot or gram per cubic centimeter. Representative values of unit weights for a soil with a speciﬁc gravity of 2.73 and a void ratio of 0.80 are ␥sat ϭ 122 lb/ft3 ϭ 1.96 g/cm3 ␥dry ϭ 95 lb/ft3 ϭ 1.52 g/cm3 ␥b ϭ 60 lb/ft3 ϭ 0.96 g/cm3 TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 262 40816 HICKS Mcghp Chap_8 Pgs 262 7/12/2001 262 CHAPTER EIGHT The symbols used in the three preceding equations and in I Fig. 8.1 are G ϭ speciﬁc gravity of soil solids (speciﬁc gravity of T 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 ϭ unit weight of water, 62.4 lb/ft3 (1.0 g/cm3) w e ϭ voids ratio, volume of voids in mass of soil divided by volume of solids in same mass; also equal to n /(1 Ϫ n), where n is porosity—volume of voids in mass of soil divided by total volume of same mass S ϭ degree of saturation, volume of water in mass of soil divided by volume of voids in same mass w ϭ water content, weight of water in mass of soil F divided by weight of solids in same mass; also i equal to Se/ G r s o n FIGURE 8.1 Relationship of weights and volumes in soil. TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 263 40816 HICKS Mcghp Chap_8 Pgs 263 7/12/2001 SOIL AND EARTHWORK FORMULAS 263n INTERNAL FRICTION AND COHESION The angle of internal friction for a soil is expressed by ␶d tan ␾ ϭ ␴ where ␾ ϭ angle of internal friction tan ␾ ϭ coefﬁcient of internal friction ␴ ϭ normal force on given plane in cohesionless soil mass ␶ ϭ shearing force on same plane when sliding on plane is impending For 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 by virtue of its intrinsic pressure. The value of the ultimate cohesive resistance of a soil is usually desig- nated by c. Average values for c are given in Table 8.2. 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) TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 264 40816 HICKS Mcghp Chap_8 Pgs 264 7/12/2001 264 CHAPTER EIGHT VERTICAL PRESSURES IN SOILS b 2 The vertical stress in a soil caused by a vertical, concentrated surface load may be determined with a fair degree of accuracy by the use of elastic theory. Two equations are in common use, the Boussinesq and the Westergaard. L The Boussinesq equation applies to an elastic, isotropic, F homogeneous mass that extends inﬁnitely in all directions from a level surface. The vertical stress at a point in the mass is T e r ΄1 ϩ ΂ rz ΃ ΅ 2 5/2 3P a ␴z ϭ 2␲z 2 o t The Westergaard equation applies to an elastic material c laterally reinforced with horizontal sheets of negligible w thickness and inﬁnite rigidity, which prevent the mass from a undergoing lateral strain. The vertical stress at a point in the b mass, assuming a Poisson’s ratio of zero, is P ΄ ΂ rz ΃ ΅ 2 3/2 ␴z ϭ 1ϩ2 ␲z 2 where ␴z ϭ vertical stress at a point, lb/ft2 (kPa) P ϭ total concentrated surface load, lb (N) z ϭ depth of point at which ␴z acts, measured vertically downward from surface, ft (m) r ϭ 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 stresses appreciably lower than those given TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 265 40816 HICKS Mcghp Chap_8 Pgs 265 7/12/2001 SOIL AND EARTHWORK FORMULAS 265 by the Boussinesq equation. For values of r /z greater than 2.2, both equations give stresses less than P/100z2.dfn . LATERAL PRESSURES IN SOILS, , FORCES ON RETAINING WALLSse The Rankine theory of lateral earth pressures, used for estimating approximate values for lateral pressures on retaining walls, assumes that the pressure on the back of a vertical wall is the same as the pressure that would exist on a vertical plane in an inﬁnite soil mass. Friction between the wall and the soil is neglected. The pressure on a wall l consists of (1) the lateral pressure of the soil held by thee wall, (2) the pressure of the water (if any) behind the wall,m and (3) the lateral pressure from any surcharge on the soile behind the wall. Symbols used in this section are as follows: ␥ ϭ unit weight of soil, lb/ft3 (kg/m3) (saturated unit weight, dry unit weight, or buoyant unit weight, depending on conditions) P ϭ total thrust of soil, lb/linear ft (kg/linear m) of wall H ϭ total height of wall, ft (m) ␾ ϭ angle of internal friction of soil, degree i ϭ angle of inclination of ground surface behinde wall with horizontal; also angle of inclination of line of action of total thrust P and pressures ond wall with horizontaln KA ϭ coefﬁcient of active pressure TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 266 40816 HICKS Mcghp Chap_8 Pgs 266 7/12/2001 266 CHAPTER EIGHT KP ϭ coefﬁcient of passive pressure c ϭ cohesion, lb/ft2 (kPa) p LATERAL PRESSURE OF W COHESIONLESS SOILS For walls that retain cohesionless soils and are free to move an appreciable amount, the total thrust from the soil is w 1 cos i Ϫ √(cos i)2 Ϫ (cos ␾)2 Pϭ ␥H 2 cos i 2 cos i ϩ √(cos i)2 Ϫ (cos ␾)2 T o When the surface behind the wall is level, the thrust is t t P ϭ 1͞2 ␥ H 2 KA ΄ ΂ ␾ ΃΅ 2 where KA ϭ tan 45ЊϪ L 2 F The thrust is applied at a point H/3 above the bottom of the a wall, and the pressure distribution is triangular, with the maxi- t mum pressure of 2P/H occurring at the bottom of the wall. For walls that retain cohesionless soils and are free to move only a slight amount, the total thrust is 1.12P, where P is as given earlier. The thrust is applied at the midpoint of o the wall, and the pressure distribution is trapezoidal, with a the maximum pressure of 1.4P/H extending over the mid- dle six-tenth of the height of the wall. TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 267 40816 HICKS Mcghp Chap_8 Pgs 267 7/12/2001 SOIL AND EARTHWORK FORMULAS 267 For walls that retain cohesionless soils and are com- pletely restrained (very rare), the total thrust from the soil is 1 cos i ϩ √(cos i)2 Ϫ (cos ␾)2 Pϭ ␥H 2 cos i 2 cos i Ϫ √(cos i)2 Ϫ (cos ␾)2 When the surface behind the wall is level, the thrust is P ϭ 1͞2 ␥H 2 K Pe ΄ ΂ ␾ ΃΅ 2 where KP ϭ tan 45ЊϪ 2 The thrust is applied at a point H /3 above the bottom of the wall, and the pressure distribution is triangular, with the maximum pressure of 2P / H occurring at the bottom of the wall. LATERAL PRESSURE OF COHESIVE SOILS For walls that retain cohesive soils and are free to movee a considerable amount over a long period of time, the total- thrust from the soil (assuming a level surface) iso P ϭ 1͞2 ␥H 2 KA Ϫ 2cH √KAef or, because highly cohesive soils generally have smallh angles of internal friction,- P ϭ 1͞2␥H 2 Ϫ 2cH TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 268 40816 HICKS Mcghp Chap_8 Pgs 268 7/12/2001 268 CHAPTER EIGHT The thrust is applied at a point somewhat below H/3 from p the bottom of the wall, and the pressure distribution is approxi- s mately triangular. a For walls that retain cohesive soils and are free to move z only a small amount or not at all, the total thrust from the soil is P ϭ 1͞2 ␥H 2K P S because the cohesion would be lost through plastic ﬂow. C A WATER PRESSURE s The total thrust from water retained behind a wall is P ϭ 1͞2␥0 H 2 t s where H ϭ height of water above bottom of wall, ft (m); and ␥0 ϭ unit weight of water, 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 the pressure distribution is triangular, with the w maximum pressure of 2P/H occurring at the bottom of the wall. Regardless of the slope of the surface behind the wall, the thrust from water is always horizontal. LATERAL PRESSURE FROM SURCHARGE C The effect of a surcharge on a wall retaining a cohesionless A soil or an unsaturated cohesive soil can be accounted for by applying a uniform horizontal load of magnitude KA p over the entire height of the wall, where p is the surcharge in TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 269 40816 HICKS Mcghp Chap_8 Pgs 269 7/12/2001 SOIL AND EARTHWORK FORMULAS 269m pound per square foot (kilopascal). For saturated cohesive - soils, the full value of the surcharge p should be considered as acting over the entire height of the wall as a uniform hori-e zontal load. KA is deﬁned earlier.e STABILITY OF SLOPES Cohesionless Soils A slope in a cohesionless soil without seepage of water is stable if iϽ␾ With seepage of water parallel to the slope, and assuming the soil to be saturated, an inﬁnite slope in a cohesionless soil is stable ifd ΂ ␥␥ ΃ tan ␾r b tan i Ͻ satfe where i ϭ slope of ground surfacee, ␾ ϭ angle of internal friction of soil ␥b, ␥sat ϭ unit weights, lb/ft3 (kg/m3) Cohesive Soilss A slope in a cohesive soil is stable ify Cr HϽn ␥N TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 270 40816 HICKS Mcghp Chap_8 Pgs 270 7/12/2001 270 CHAPTER EIGHT where H ϭ height of slope, ft (m) w C ϭ cohesion, lb/ft (kg/m ) 2 2 ␥ ϭ unit weight, lb/ft3 (kg/m3) N ϭ stability number, dimensionless For failure on the slope itself, without seepage water, N ϭ (cos i)2 (tan i Ϫ tan ␾) Similarly, with seepage of water, N ϭ (cos i)2 tan i Ϫ΄ ΂ ␥␥ ΃ tan ␾΅ b sat When the slope is submerged, ␾ is the angle of internal c friction of the soil and ␥ is equal to ␥b. When the surround- T ing water is removed from a submerged slope in a short 0 time (sudden drawdown), ␾ is the weighted angle of internal d friction, equal to (␥b /␥sat)␾, and ␥ is equal to ␥sat. S BEARING CAPACITY OF SOILS T a The approximate ultimate bearing capacity under a long footing at the surface of a soil is given by Prandtl’s equa- tion as w qu ϭ ΂ c tan ␾ ΃ ϩ 1 ␥ b √Kp (Kp e␲ tan ␾ Ϫ 1) 2 dry TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 271 40816 HICKS Mcghp Chap_8 Pgs 271 7/12/2001 SOIL AND EARTHWORK FORMULAS 271 where qu ϭ ultimate bearing capacity of soil, lb/ft2 (kg/m2) c ϭ cohesion, lb/ft2 (kg/m2) ␾ ϭ angle of internal friction, degree ␥dry ϭ unit weight of dry soil, lb/ft3 (kg/m3) b ϭ width of footing, ft (m) d ϭ depth of footing below surface, ft (m) Kp ϭ coefﬁcient of passive pressure ΄ ΂ ␾ ΃΅ 2 ϭ tan 45 ϩ 2 e ϭ 2.718 и и и For footings below the surface, the ultimate bearingl capacity of the soil may be modiﬁed by the factor 1 ϩ Cd/b.- The coefﬁcient C is about 2 for cohesionless soils and aboutt 0.3 for cohesive soils. The increase in bearing capacity withl depth for cohesive soils is often neglected. SETTLEMENT UNDER FOUNDATIONS The approximate relationship between loads on foundations and settlement isg- q P ϭ C1 1 ϩ΂2d b ΃ϩ C b 2 where q ϭ load intensity, lb/ft2 (kg/m2) P ϭ settlement, in (mm) TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 272 40816 HICKS Mcghp Chap_8 Pgs 272 7/12/2001 272 CHAPTER EIGHT d ϭ depth of foundation below ground surface, ft (m) b ϭ width of foundation, ft (m) C1 ϭ coefﬁcient dependent on internal friction C2 ϭ coefﬁcient dependent on cohesion M The coefﬁcients C1 and C2 are usually determined by bearing- c plate loading tests. L SOIL COMPACTION TESTS O d The sand-cone method is used to determine in the ﬁeld the l density of compacted soils in earth embankments, road ﬁll, b and structure backﬁll, as well as the density of natural soil m deposits, aggregates, soil mixtures, or other similar materials. 1 It is not suitable, however, for soils that are saturated, soft, o or friable (crumble easily). Characteristics of the soil are computed from Volume of soil, ft 3 (m3) weight of sand filling hole, lb (kg) F ϭ f density of sand, lb/ft 3 (kg/m 3) i % Moisture 100(weight of moist soil Ϫ weight of dry soil) ϭ weight of dry soil w r weight of soil, lb (kg) ␳ Field density, lb/ft 3 (kg /m3) ϭ volume of soil, ft 3 (m3) p TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 273 40816 HICKS Mcghp Chap_8 Pgs 273 7/12/2001 SOIL AND EARTHWORK FORMULAS 273) field density Dry density ϭ 1 ϩ % moisture/100 100(dry density) % Compaction ϭ max dry density Maximum density is found by plotting a density–moisture- curve. Load-Bearing Test One of the earliest methods for evaluating the in situ deformability of coarse-grained soils is the small-scalee load-bearing test. Data developed from these tests have, been used to provide a scaling factor to express the settle-l ment ␳ 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-size bearing plate as ␳ ΂ 1 2B B ΃ 2 ϭ ␳1 ϩ From an elastic half-space solution, EsЈ can be expressed from results of a plate load test in terms of the ratio of bear- ing pressure to plate settlement kv as k v (1 Ϫ ␮2)␲/4 EsЈ ϭ 4B/(1 ϩ B)2 where ␮ represents Poisson’s ratio, usually considered to range between 0.30 and 0.40. The EsЈ equation assumes that ␳1 is derived from a rigid, 1-ft (0.3048-m)-diameter circular plate and that B is the equivalent diameter of the bearing TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 274 40816 HICKS Mcghp Chap_8 Pgs 274 7/12/2001 274 CHAPTER EIGHT area of a full-scale footing. Empirical formulations, such as g the ␳/␳1 equation, may be signiﬁcantly in error because of the l limited footing-size range used and the large scatter of the database. Furthermore, consideration is not given to varia- tions in the characteristics and stress history of the bearing w soils. o v California Bearing Ratio m t The California bearing ratio (CBR) is often used as a meas- f ure of the quality of strength of a soil that underlies a pavement, for determining the thickness of the pavement, its base, and other layers. C F CBR ϭ F0 A i where F ϭ force per unit area required to penetrate a soil t mass with a 3-in2 (1935.6-mm2) circular piston (about 2 in u (50.8 mm) in diameter) at the rate of 0.05 in/min (1.27 mm/min); and F0 ϭ force per unit area required for corre- t sponding penetration of a standard material. a Typically, the ratio is determined at 0.10-in (2.54-mm) m penetration, although other penetrations sometimes are used. c An excellent base course has a CBR of 100 percent. A com- r pacted soil may have a CBR of 50 percent, whereas a u weaker soil may have a CBR of 10. Soil Permeability w The coefﬁcient of permeability k is a measure of the rate of ﬂow of water through saturated soil under a given hydraulic TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 275 40816 HICKS Mcghp Chap_8 Pgs 275 7/12/2001 SOIL AND EARTHWORK FORMULAS 275s gradient i, cm/cm, and is deﬁned in accordance with Darcy’se law ase- V ϭ kiAg where V ϭ rate of ﬂow, cm3/s, and A ϭ cross-sectional area of soil conveying ﬂow, cm2. Coefﬁcient k is dependent on the grain-size distribution, void ratio, and soil fabric and typically may vary from as much as 10 cm/s for gravel to less than 10 –7 for clays. For typical soil deposits, k for horizontal ﬂow is greater than k- for vertical ﬂow, often by an order of magnitude. ,d COMPACTION EQUIPMENT A wide variety of equipment is used to obtain compaction in the ﬁeld. Sheepsfoot rollers generally are used on soilsl that contain high percentages of clay. Vibrating rollers aren used on more granular soils.7 To determine maximum depth of lift, make a test ﬁll. In- the process, the most suitable equipment and pressure to be applied, lb/in2 (kPa), for ground contact also can be deter-) mined. Equipment selected should be able to produce desired. compaction with four to eight passes. Desirable speed of- rolling also can be determined. Average speeds, mi/h (km/h),a under normal conditions are given in Table 8.3. Compaction production can be computed from 16WSLFE yd3/h (m3/h) ϭ P where W ϭ width of roller, ft (m)fc S ϭ roller speed, mi/h (km/h) TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 276 40816 HICKS Mcghp Chap_8 Pgs 276 7/12/2001 276 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) R a H L ϭ lift thickness, in (mm) F ϭ ratio of pay yd3 (m3) to loose yd3 (m3) E ϭ efﬁciency factor (allows for time losses, such w as those due to turns): 0.90, excellent; 0.80, average; 0.75, poor t w P ϭ number of passes w FORMULAS FOR EARTHMOVING External forces offer rolling resistance to the motion of wheeled vehicles, such as tractors and scrapers. The engine has to supply power to overcome this resistance; the greater the resistance is, the more power needed to move a load. T Rolling resistance depends on the weight on the wheels and r the tire penetration into the ground: R ϭ R f W ϩ R p pW (8.6) where R ϭ rolling resistance, lb (N) I p Rf ϭ rolling-resistance factor, lb/ton (N/tonne) 1 TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 277 40816 HICKS Mcghp Chap_8 Pgs 277 7/12/2001 SOIL AND EARTHWORK FORMULAS 277 W ϭ weight on wheels, ton (tonne) Rp ϭ tire-penetration factor, lb/tonиin (N/tonneиmm) penetration p ϭ 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/tonиin (1.5% lb/lbиin) (3288 N/tиmm). Hence, Eq. (8.6) can be written as R ϭ (2% ϩ 1.5%p) WЈ ϭ RЈWЈ (8.7)h where WЈ ϭ weight on wheels, lb (N); and RЈ ϭ 2% ϩ 1.5%p. , Additional power is required to overcome rolling resis- tance on a slope. Grade resistance also is proportional to weight: G ϭ R g sW (8.8) where G ϭ grade resistance, lb (N) Rg ϭ grade-resistance factor ϭ 20 lb/ton (86.3 N/t) ϭ 1% lb/lb (N/N)f s ϭ percent grade, positive for uphill motion,e negative for downhillr . Thus, the total road resistance is the algebraic sum of thed rolling and grade resistances, or the total pull, lb (N), required: T ϭ (RЈ ϩ R g s)WЈ ϭ (2% ϩ 1.5%p ϩ 1%s)WЈ (8.9)) In addition, an allowance may have to be made for loss of power with altitude. If so, allow 3 percent pull loss for each 1000 ft (305 m) above 2500 ft (762 m). TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 278 40816 HICKS Mcghp Chap_8 Pgs 278 7/12/2001 278 CHAPTER EIGHT Usable pull P depends on the weight W on the drivers: P PϭfW (8.10) where f ϭ coefﬁcient of traction. Earth Quantities Hauled b When soils are excavated, they increase in volume, or swell, l because of an increase in voids: y 100 m Vb ϭ VL L ϭ V (8.11) m 100 ϩ % swell L u where Vb ϭ original volume, yd3 (m3), or bank yards a VL ϭ loaded volume, yd3 (m3), or loose yards a w L ϭ load factor t When soils are compacted, they decrease in volume: b Vc ϭ Vb S (8.12) where Vc ϭ compacted volume, yd3 (m3); and S ϭ shrinkage factor. Bank yards moved by a hauling unit equals weight of load, lb (kg), divided by density of the material in place, lb (kg), per E bank yard (m3). T r SCRAPER PRODUCTION Production is measured in terms of tons or bank cubic yards (cubic meters) of material a machine excavates and dis- charges, under given job conditions, in 1 h. TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 279 40816 HICKS Mcghp Chap_8 Pgs 279 7/12/2001 SOIL AND EARTHWORK FORMULAS 279: Production, bank yd3/h (m3/h) ϭ load, yd3 (m3) ϫ trips per hour) working time, min/h Trips per hour ϭ cycle time, min The load, or amount of material a machine carries, can be determined by weighing or estimating the volume. Pay-, load estimating involves determination of the bank cubic yards (cubic meters) being carried, whereas the excavated material expands when loaded into the machine. For deter-) mination of bank cubic yards (cubic meters) from loose vol- ume, the amount of swell or the load factor must be known. Weighing is the most accurate method of determining the actual load. This is normally done by weighing one wheel or axle at a time with portable scales, adding the wheel or axle weights, and subtracting the weight empty. To reduce error, the machine should be relatively level. Enough loads should be weighed to provide a good average:) weight of load, lb (kg)e Bank yd3 ϭ density of material, lb/bank yd3 (kg/m3),r Equipment Required To determine the number of scrapers needed on a job, required production must ﬁrst be computed: Production required, yd3 /h (m3/ h)s- quantity, bank yd3 (m3) ϭ working time, h TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 280 40816 HICKS Mcghp Chap_8 Pgs 280 7/12/2001 280 CHAPTER EIGHT No. of scrapers needed production required, yd3/h (m3/h) ϭ production per unit, yd3/h (m3/h) V v No. of scrapers a pusher can load i scraper cycle time, min ϭ pusher cycle time, min I Because speeds and distances may vary on haul and k return, haul and return times are estimated separately. m Variable time, min haul distance, ft return distance, ft ϭ ϩ 88 ϫ speed, mi/ h 88 ϫ speed, mi/ h T haul distance, m return distance, m ϭ ϩ 16.7 ϫ speed, km/ h 16.7 ϫ speed, km/ h F Haul speed may be obtained from the equipment speciﬁ- s cation sheet when the drawbar pull required is known. VIBRATION CONTROL IN BLASTING w a Explosive users should take steps to minimize vibration and noise from blasting and protect themselves against damage claims. Vibrations caused by blasting are propagated with a velocity V, ft/s (m/s), frequency f, Hz, and wavelength L, F ft (m), related by s TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 281 40816 HICKS Mcghp Chap_8 Pgs 281 7/12/2001 SOIL AND EARTHWORK FORMULAS 281 V Lϭ f Velocity v, in/s (mm/s), of the particles disturbed by the vibrations depends on the amplitude of the vibrations A, in (mm): v ϭ 2␲fA If the velocity v1 at a distance D1 from the explosion isd known, the velocity v2 at a distance D2 from the explosion may be estimated from ΂D ΃ 1.5 1 v2 Ϸ v1 D 2 The acceleration a, in/s (mm/s ), of the particles is given by 2 2 a ϭ 4␲ 2 f 2A For a charge exploded on the ground surface, the overpres-- sure P, lb/in2 (kPa), may be computed from ΂ WD ΃ 1/3 1.407 P ϭ 226.62 where W ϭ maximum weight of explosives, lb (kg) per delay; and D ϭ distance, ft (m), from explosion to exposure. The sound pressure level, decibels, may be computed fromd ΂ 6.95 ϫ 10 ΃ P 0.084e dB ϭ Ϫ28a, For vibration control, blasting should be controlled with the scaled-distance formula: TLFeBOOK
• 40816_08_p257-282 10/22/01 12:48 PM Page 282 40816 HICKS Mcghp Chap_8 Pgs 282 7/12/2001 282 CHAPTER EIGHT ΂ √W ΃ Ϫ␤ D vϭH where ␤ ϭ constant (varies for each site), and H ϭ 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 this velocity, vibration damage is unlikely at scaled distances larger than 8. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 283 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 283 CHAPTER 9 BUILDING AND STRUCTURES FORMULAS TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_09_p283-320 10/22/01 12:49 PM Page 284 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 284 4081 284 CHAPTER NINE LOAD-AND-RESISTANCE FACTOR F DESIGN FOR SHEAR IN BUILDINGS s Based on the American Institute of Steel Construction (AISC) speciﬁcations for load-and-resistance factor design (LRFD) for buildings, the shear capacity Vu, kip (kN ϭ 4.448 ϫ kip), A of ﬂexural members may be computed from the following: B h T Vu ϭ 0.54Fyw Aw when Յ␣ b tw c 0.54␣Fyw Aw h f Vu ϭ when ␣Ͻ Յ 1.25␣ h/t w tw e u 23,760kAw h e Vu ϭ when Ͼ 1.25␣ (h/t w)2 tw where Fyw ϭ specified minimum yield stress of web, ksi (MPa ϭ 6.894 ϫ ksi) Aw ϭ web area, in2 (mm2) ϭ dtw w ␣ ϭ 187√k /Fyw k ϭ 5 if a/h exceeds 3.0 or 67,600/(h/tw)2, or if stiffeners are not required W ϭ 5 ϩ 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 is 2,000/√Fyf for a/h Յ 1.5 or 14,000/√Fyf (Fyf ϩ 16.5) for a/h > 1.5, where Fyf is the speciﬁed minimum yield stress, ksi, of the ﬂange. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 285p 284 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 285 BUILDING AND STRUCTURES FORMULAS 285 For shear capacity with tension-ﬁeld action, see the AISC speciﬁcation for LRFD.)), ALLOWABLE-STRESS DESIGN FOR: BUILDING COLUMNS The AISC speciﬁcation for allowable-stress design (ASD) for buildings provides two formulas for computing allowable compressive stress Fa, ksi (MPa), for main members. The formula to use depends on the relationship of the largest effective slenderness ratio Kl/r of the cross section of any unbraced segment to a factor Cc deﬁned by the following equation and Table 9.1: 2␲2E 756.6 , Cc ϭ √ Fy ϭ √Fy where E ϭ modulus of elasticity of steel ϭ 29,000 ksi (128.99 GPa)f Fy ϭ yield stress of steel, ksi (MPa) When Kl/r is less than Cc ,n TABLE 9.1 Values of Ccs Fy Ccre 36 126.1 50 107.0 . TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 286 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 286 4081 286 CHAPTER NINE ΄1 Ϫ (Kl/r) ΅ F 2 L 2C 2 c y F F ϭ a F.S. P 5 3(Kl/r) (Kl/r)3 b where F.S. ϭ safety factor ϭ ϩ Ϫ . 3 8Cc 8C3c F When Kl /r exceeds Cc, b m 12␲ 2E 150,000 Fa ϭ ϭ 23(Kl/r)2 (Kl/r)2 The effective-length factor K, equal to the ratio of effec- w tive-column length to actual unbraced length, may be greater or less than 1.0. Theoretical K values for six ideal- ized conditions, in which joint rotation and translation are either fully realized or nonexistent, are tabulated in Fig. 9.1. T d A B T p p p c o FIGURE 9.1 Values of effective-length factor K for columns. g TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 287p 286 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 287 BUILDING AND STRUCTURES FORMULAS 287 LOAD-AND-RESISTANCE FACTOR DESIGN FOR BUILDING COLUMNS Plastic analysis of prismatic compression members in buildings is permitted if √Fy (l/r) does not exceed 800 and Fu Յ 65 ksi (448 MPa). For axially loaded members with b/t Յ ␭r, the maximum load Pu, ksi (MPa ϭ 6.894 ϫ ksi), may be computed from Pu ϭ 0.85AgFcr- where Ag ϭ gross cross-sectional area of the membere- Fcr ϭ 0.658␭Fy for ␭ Յ 2.25e ϭ 0.877 Fy /␭ for ␭ Ͼ 2.25. ␭ ϭ (Kl/r)2(Fy /286,220) The AISC speciﬁcation for LRFD presents formulas for designing members with slender elements. ALLOWABLE-STRESS DESIGN FOR BUILDING BEAMS The maximum ﬁber stress in bending for laterally sup- ported beams and girders is Fb ϭ 0.66Fy if they are com- pact, except for hybrid girders and members with yield points exceeding 65 ksi (448.1 MPa). Fb ϭ 0.60Fy for non- compact 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. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 288 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 288 4081 288 CHAPTER NINE TABLE 9.2 Allowable Bending Stresses in Braced t Beams for Buildings i 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) F The allowable extreme-ﬁber stress of 0.60Fy applies to laterally supported, unsymmetrical members, except chan- nels, and to noncompact box sections. Compression on outer surfaces of channels bent about their major axis should not exceed 0.60Fy or the value given by Eq. (9.5). w The allowable stress of 0.66Fy for compact members should be reduced to 0.60Fy when the compression ﬂange is n unsupported for a length, in (mm), exceeding the smaller of t t 76.0bf lmax ϭ (9.1) √Fy W 20,000 t lmax ϭ (9.2) Fyd/Af a where bf ϭ width of compression ﬂange, in (mm) m d ϭ beam depth, in (mm) Af ϭ area of compression ﬂange, in2 (mm2) w The allowable stress should be reduced even more when b l/rT exceeds certain limits, where l is the unbraced length, in (mm), of the compression ﬂange, and rT is the radius of c gyration, in (mm), of a portion of the beam consisting of W TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 289p 288 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 289 BUILDING AND STRUCTURES FORMULAS 289 the compression ﬂange and one-third of the part of the web in compression. For √102,000Cb /Fy Յ l/rT Յ √510,00Cb /Fy, use ΄ 2 Ϫ 1,530,000C ΅ F F (l/r ) 2 y T Fb ϭ y (9.3) 3 b For l/rT Ͼ √510,000Cb /Fy, useo- 170,000Cbr Fb ϭ (9.4) (l/rT )2t where Cb ϭ modiﬁer for moment gradient (Eq. 9.6).s When, however, the compression ﬂange is solid ands nearly rectangular in cross section, and its area is not lessf than that of the tension ﬂange, the allowable stress may be taken as 12,000Cb) Fb ϭ (9.5) ld/Af When Eq. (9.5) applies (except for channels), Fb should be) taken as the larger of the values computed from Eqs. (9.5) and (9.3) or (9.4), but not more than 0.60Fy. The moment-gradient factor Cb in Eqs. (9.1) to (9.5) may be computed from M1 ΂ M ΃ Յ 2.3 2 1 Cb ϭ 1.75 ϩ 1.05 ϩ 0.3 (9.6) M2 M 2 where M1 ϭ smaller beam end moment, and M2 ϭ largern beam end moment. , The algebraic sign of M1 /M2 is positive for double-f curvature bending and negative for single-curvature bending.f When the bending moment at any point within an unbraced TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 290 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 290 4081 290 CHAPTER NINE length is larger than that at both ends, the value of Cb should be taken as unity. For braced frames, Cb should be a taken as unity for computation of Fbx and Fby. w Equations (9.4) and (9.5) can be simpliﬁed by introduc- i ing a new term: (l/rT)2Fy Qϭ (9.7) 510,000Cb w Now, for 0.2 Յ Q Յ 1, (2 Ϫ Q)Fy Fb ϭ (9.8) 3 For Q Ͼ 1: Fy Fb ϭ (9.9) 3Q T As for the preceding equations, when Eq. (9.1) applies t (except for channels), Fb should be taken as the largest of g the values given by Eqs. (9.1) and (9.8) or (9.9), but not t more than 0.60Fy. LOAD-AND-RESISTANCE FACTOR DESIGN FOR BUILDING BEAMS l For a compact section bent about the major axis, the l unbraced length Lb of the compression ﬂange, where plastic n hinges may form at failure, may not exceed Lpd , given by m Eqs. (9.10) and (9.11) that follow. For beams bent about the t minor axis and square and circular beams, Lb is not restricted for plastic analysis. e TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 291p 290 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 291 BUILDING AND STRUCTURES FORMULAS 291b For I-shaped beams, symmetrical about both the majore and the minor axis or symmetrical about the minor axis but with the compression ﬂange larger than the tension ﬂange,- including hybrid girders, loaded in the plane of the web: 3600 ϩ 2200(M1/Mp) Lpd ϭ ry (9.10)) Fyc where Fyc ϭ minimum yield stress of compression ﬂange, ksi (MPa)) M1 ϭ smaller of the moments, inиkip (mmиMPa) at the ends of the unbraced length of beam Mp ϭ plastic moment, inиkip (mmиMPa)) ry ϭ radius of gyration, in (mm), about minor axis The plastic moment Mp equals Fy Z for homogeneous sec-s tions, where Z ϭ plastic modulus, in3 (mm3); and for hybridf girders, it may be computed from the fully plastic distribu-t tion. M1/Mp is positive for beams with reverse curvature. For solid rectangular bars and symmetrical box beams: 5000 ϩ 3000(M1/Mp) ry Lpd ϭ ry Ն 3000 (9.11) Fy Fy The ﬂexural design strength 0.90Mn is determined by the limit state of lateral-torsional buckling and should be calcu-e lated for the region of the last hinge to form and for regionsc not adjacent to a plastic hinge. The speciﬁcation gives for-y mulas for Mn that depend on the geometry of the section ande the bracing provided for the compression ﬂange.d For compact sections bent about the major axis, for example, Mn depends on the following unbraced lengths: TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 292 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 292 4081 292 CHAPTER NINE Lb ϭ the distance, in (mm), between points braced against lateral displacement of the compression ﬂange or between points braced to prevent twist Lp ϭ limiting laterally unbraced length, in (mm), for full plastic-bending capacity ϭ 300ry / √Fyf for I shapes and channels ϭ 3750(ry /Mp)/√JA for solid rectangular bars and box beams Fyf ϭ ﬂange yield stress, ksi (MPa) J ϭ torsional constant, in4 (mm4) (see AISC “Manual of Steel Construction” on LRFD) A ϭ cross-sectional area, in2 (mm2) Lr ϭ limiting laterally unbraced length, in (mm), for inelastic lateral buckling F For I-shaped beams symmetrical about the major or the m minor axis, or symmetrical about the minor axis with the compression ﬂange larger than the tension ﬂange and chan- nels loaded in the plane of the web: ry x1 Lr ϭ √1 ϩ √1 ϩ X2 FL 2 (9.12) a Fyw Ϫ Fr where Fyw ϭ specified minimum yield stress of web, ksi (MPa) Fr ϭ compressive residual stress in ﬂange w ϭ 10 ksi (68.9 MPa) for rolled shapes, 16.5 w ksi (113.6 MPa), for welded sections t FL ϭ smaller of Fyf Ϫ Fr or Fyw r a Fyf ϭ specified minimum yield stress of flange, m ksi (MPa) m TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 293p 292 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 293 BUILDING AND STRUCTURES FORMULAS 293 X1 ϭ (␲/Sx)√EGJA/2 X2 ϭ (4Cw /Iy) (Sx /GJ)2 E ϭ elastic modulus of the steel G ϭ shear modulus of elasticity Sx ϭ section modulus about major axis, in3 (mm3) (with respect to the compression ﬂange if that ﬂange is larger than the tension ﬂange) Cw ϭ warping constant, in6 (mm6) (see AISC manual on LRFD) Iy ϭ moment of inertia about minor axis, in4 (mm4) For the previously mentioned shapes, the limiting bucklinge moment Mr, ksi (MPa), may be computed frome- M r ϭ FL Sx (9.13) For compact beams with Lb Յ Lr, bent about the major) axis:, ΄ Mn ϭ Cb Mp Ϫ (Mp Ϫ Mr) Lb Ϫ Lp Lr Ϫ Lp ΅ՅM p (9.14) where Cb ϭ 1.75 ϩ 1.05(M1 /M2) ϩ 0.3(M1 /M2) Յ 2.3,5 where M1 is the smaller and M2 the larger end moment in the unbraced segment of the beam; M1/M2 is positive for reverse curvature and equals 1.0 for unbraced cantilevers and beams with moments over much of the unbraced seg-, ment equal to or greater than the larger of the segment end moments. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 294 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 294 4081 294 CHAPTER NINE (See Galambos, T. V., Guide to Stability Design Criteria A for Metal Structures, 4th ed., John Wiley & Sons, New York, I for use of larger values of Cb.) For solid rectangular bars bent about the major axis: T a Lr ϭ 57,000 ΂ ΃ √JA ry Mr (9.15) and the limiting buckling moment is given by: M r ϭ Fy Sx (9.16) For symmetrical box sections loaded in the plane of sym- w metry and bent about the major axis, Mr should be deter- mined from Eq. (9.13) and Lr from Eq. (9.15) For compact beams with Lb > Lr, bent about the major axis: M n ϭ M cr Յ Cb M r (9.17) where Mcr ϭ critical elastic moment, kipиin (MPaиmm). T For shapes to which Eq. (9.17) applies: ␲ ΂ ␲E ΃ 2 Mcr ϭ Cb Lb √ EIy GJ ϩ Iy Cw Lb (9.18) For solid rectangular bars and symmetrical box sections: w 57,000Cb √JA W Mcr ϭ (9.19) Lb / ry m For determination of the ﬂexural strength of noncompact s plate girders and other shapes not covered by the preceding t requirements, see the AISC manual on LRFD. s TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 295p 294 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 295 BUILDING AND STRUCTURES FORMULAS 295a ALLOWABLE-STRESS DESIGN FOR SHEAR , IN BUILDINGS The AISC speciﬁcation for ASD speciﬁes the following allowable shear stresses Fv, ksi (ksi ϫ 6.894 ϭ MPa):) h 380 Fv ϭ 0.40Fy Յ tw √Fy Cv Fy h 380 Fv ϭ Յ 0.40Fy Ͼ) 289 tw √Fy- where Cv ϭ 45,000kv /Fy(h/tw)2 for Cv Ͻ 0.8- ϭ √36,000kv /Fy(h/tw) 2 for Cv Ͼ 0.8r kv ϭ 4.00 ϩ 5.34/(a/h) 2 for a/h Ͻ 1.0 ϭ 5.34 ϩ 4.00/(a/ h) 2 for a/h Ͼ 1.0) a ϭ clear distance between transverse stiffeners The allowable shear stress with tension-ﬁeld action is) Fv ϭ Fy 289 ΄C ϩ 1.15√1ϪϩC(a/h) ΅ Յ 0.40F v 1 v 2 y where Cv Յ 1) When the shear in the web exceeds Fv , stiffeners are required. Within the boundaries of a rigid connection of two or more members with webs lying in a common plane, sheart stresses in the webs generally are high. The commentary ong the AISC speciﬁcation for buildings states that such webs should be reinforced when the calculated shear stresses, TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 296 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 296 4081 296 CHAPTER NINE S R FIGURE 9.2 Rigid connection of steel members with webs in a t common plane. l s such as those along plane AA in Fig. 9.2, exceed Fv, that is, a when ⌺F is larger than dc tw Fv, where dc is the depth and tw u is the web thickness of the member resisting ⌺F. The shear e may be calculated from o M1 M2 ⌺F ϭ ϩ Ϫ Vs 0.95d 1 0.95d 2 where Vs ϭ shear on the section w M1 ϭ M1L ϩ M1G M1L ϭ moment due to the gravity load on the leeward side of the connection TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 297p 296 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 297 BUILDING AND STRUCTURES FORMULAS 297 M1G ϭ moment due to the lateral load on the leeward side of the connection M2 ϭ M2L Ϫ M2G M2L ϭ moment due to the lateral load on the wind- ward side of the connection M2G ϭ moment due to the gravity load on the wind- ward side of the connection STRESSES IN THIN SHELLS Results of membrane and bending theories are expressed ina terms of unit forces and unit moments, acting per unit of length over the thickness of the shell. To compute the unit stresses from these forces and moments, usual practice is to, assume normal forces and shears to be uniformly distrib-w uted over the shell thickness and bending stresses to be lin-r early distributed. Then, normal stresses can be computed from equations of the form: Nx M fx ϭ ϩ 3 x z (9.20) t t /12 where z ϭ distance from middle surface t ϭ shell thicknessd Mx ϭ unit bending moment about an axis parallel to direction of unit normal force Nx TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 298 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 298 4081 298 CHAPTER NINE Similarly, shearing stresses produced by central shears T T and twisting moments D may be calculated from equations M of the form: F T D vxy ϭ Ϯ 3 z (9.21) t t /12 L Normal shearing stresses may be computed on the assumption of a parabolic stress distribution over the shell S thickness: B vxz ϭ 3 V t /6 ΂ t2 4 Ϫ z2΃ (9.22) † where V ϭ unit shear force normal to middle surface. BEARING PLATES w i To resist a beam reaction, the minimum bearing length N in the direction of the beam span for a bearing plate is deter- m mined by equations for prevention of local web yielding c and web crippling. A larger N is generally desirable but is i limited by the available wall thickness. s When the plate covers the full area of a concrete sup- port, the area, in2 (mm2), required by the bearing plate is R A1 ϭ Ј 0.35 f c T where R ϭ beam reaction, kip (kN), f cЈ ϭ speciﬁed com- t pressive strength of the concrete, ksi (MPa). When the plate covers less than the full area of the concrete support, then, as determined from Table 9.3, TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 299p 298 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 299 BUILDING AND STRUCTURES FORMULAS 299T TABLE 9.3 Allowable Bearing Stress, Fp, on Concrete ands Masonry† Full area of concrete support 0.35f cЈ) A1e Less than full area of concrete support 0.35f cЈ √ A2 Յ 0.70f cЈl Sandstone and limestone 0.40 Brick in cement mortar 0.25) † Units in MPa ϭ 6.895 ϫ ksi. ΂ 0.35 fRЈ √A ΃ 2 A1 ϭ c 2 where A2 ϭ full cross-sectional area of concrete support, in2 (mm2).n With N established, usually rounded to full inches (milli-- meters), the minimum width of plate B, in (mm), may beg calculated by dividing A1 by N and then rounded off to fulls inches (millimeters), so that BN Ն A1. Actual bearing pres- sure fp, ksi (MPa), under the plate then is- R fp ϭ BN The plate thickness usually is determined with the assump-- tion of cantilever bending of the plate:e, tϭ ΂ 1 B Ϫ k΃ √ 3f 2 F b p TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 300 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 300 4081 300 CHAPTER NINE where t ϭ minimum plate thickness, in (mm) e b k ϭ distance, in (mm), from beam bottom to top of web ﬁllet Fb ϭ allowable bending stress of plate, ksi (MPa) w p COLUMN BASE PLATES t s l The area A1, in2 (mm2), required for a base plate under a col- m umn supported by concrete should be taken as the larger of a the values calculated from the equation cited earlier, with R taken as the total column load, kip (kN), or R A1 ϭ w 0.70 fcЈ ( Unless the projections of the plate beyond the column are p small, the plate may be designed as a cantilever assumed to l be ﬁxed at the edges of a rectangle with sides equal to o 0.80b and 0.95d, where b is the column ﬂange width, in (mm), and d is the column depth, in (mm). To minimize material requirements, the plate projections should be nearly equal. For this purpose, the plate length N, in (mm) (in the direction of d), may be taken as N ϭ √A1 ϩ 0.5(0.95d Ϫ 0.80b) B The width B, in (mm), of the plate then may be calculated I by dividing A1 by N. Both B and N may be selected in full s inches (millimeters) so that BN Ն A1. In that case, the bearing d pressure fp, ksi (MPa), may be determined from the preceding s TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 301p 300 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 301 BUILDING AND STRUCTURES FORMULAS 301 equation. Thickness of plate, determined by cantilever bending, is given byf fp t ϭ 2p √ Fy where Fy ϭ minimum speciﬁed yield strength, ksi (MPa), of plate; and p ϭ larger of 0.5(N Ϫ 0.95d) and 0.5(B Ϫ 0.80b). When the plate projections are small, the area A2 should be taken as the maximum area of the portion of the supporting surface that is geometrically similar to and concentric with the loaded area. Thus, for an H-shaped column, the column load- may be assumed distributed to the concrete over an H-shapedf area with ﬂange thickness L, in (mm), and web thickness 2L:R 1 1 4R Lϭ 4 (d ϩ b) Ϫ 4 √ (d ϩ b)2 Ϫ Fp where Fp ϭ allowable bearing pressure, ksi (MPa), on support. (If L is an imaginary number, the loaded portion of the sup-e porting surface may be assumed rectangular as discussed ear-o lier.) Thickness of the base plate should be taken as the largero of the values calculated from the preceding equation andn 3fps, tϭL √ Fb BEARING ON MILLED SURFACESd In building construction, allowable bearing stress for milledl surfaces, including bearing stiffeners, and pins in reamed,g drilled, or bored holes, is Fp ϭ 0.90Fy, where Fy is the yieldg strength of the steel, ksi (MPa). TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 302 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 302 4081 302 CHAPTER NINE For expansion rollers and rockers, the allowable bearing stress, kip/linear in (kN/mm), is Fy Ϫ 13 Fp ϭ 0.66d 20 where d is the diameter, in (mm), of the roller or rocker. When parts in contact have different yield strengths, Fy is the smaller value. t m PLATE GIRDERS IN BUILDINGS For greatest resistance to bending, as much of a plate girder cross section as practicable should be concentrated in the ﬂanges, at the greatest distance from the neutral axis. This might require, however, a web so thin that the girder would 7 fail by web buckling before it reached its bending capacity. s To preclude this, the AISC speciﬁcation limits h/t. a For an unstiffened web, this ratio should not exceed l h 14,000 ϭ t √Fy (Fy ϩ 16.5) where Fy ϭ yield strength of compression ﬂange, ksi (MPa). Larger values of h/t may be used, however, if the web is stiffened at appropriate intervals. For this purpose, vertical angles may be fastened to the web or vertical plates welded to it. These transverse stiffeners w are not required, though, when h/t is less than the value computed from the preceding equation or Table 9.4. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 303p 302 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 303 BUILDING AND STRUCTURES FORMULAS 303g TABLE 9.4 Critical h/t for Plate Girders in Buildings 14,000 2,000 Fy, ksi (MPa) √Fy(Fy ϩ 16.5) √Fy 36 (248) 322 333 50 (345) 243 283.s With transverse stiffeners spaced not more than 1.5 times the girder depth apart, the web clear-depth/thickness ratio may be as large as h 2000 ϭ t √Fyres If, however, the web depth/thickness ratio h/t exceedsd 760 / √Fb, where Fb, ksi (MPa), is the allowable bending . stress in the compression flange that would ordinarily Ј apply, this stress should be reduced to Fb , given by the fol- lowing equations: Fb ϭ RPG Re Fb Ј ΄ RPG ϭ 1 Ϫ 0.0005 Aw Af ΂ ht Ϫ 760 ΃΅ Յ 1.0 √F b. ΄ 12 ϩ12(Aϩ/A )(3␣ Ϫ ␣ ) ΅ Յ 1.0s w f 3 Re ϭ 2(A /A ) w fes where Aw ϭ web area, in (mm ) 2 2e Af ϭ area of compression ﬂange, in2 (mm2) TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 304 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 304 4081 304 CHAPTER NINE ␣ ϭ 0.6Fyw /Fb Յ 1.0 Fyw ϭ minimum speciﬁed yield stress, ksi, (MPa), of web steel In a hybrid girder, where the ﬂange steel has a higher yield strength than the web, the preceding equation protects against excessive yielding of the lower strength web in the vicinity of the higher strength ﬂanges. For nonhybrid girders, Re ϭ 1.0. LOAD DISTRIBUTION TO BENTS AND SHEAR WALLS F Provision should be made for all structures to transmit lateral o loads, such as those from wind, earthquakes, and traction a and braking of vehicles, to foundations and their supports V that have high resistance to displacement. For this purpose, various types of bracing may be used, including struts, d tension ties, diaphragms, trusses, and shear walls. T d Deﬂections of Bents and Shear Walls Horizontal deﬂections in the planes of bents and shear walls can be computed on the assumption that they act as w cantilevers. Deﬂections of braced bents can be calculated by the dummy-unit-load method or a matrix method. Deﬂections of rigid frames can be computed by adding the drifts of the stories, as determined by moment distribution or a matrix method. For a shear wall (Fig. 9.3), the deﬂection in its plane induced by a load in its plane is the sum of the ﬂexural TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 305p 304 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 305 BUILDING AND STRUCTURES FORMULAS 305fdtf . FIGURE 9.3 Building frame resists lateral forces with (a) wind bentsl or (g) shear walls or a combination of the two. Bents may be braced inn any of several ways, including (b) X bracing, (c) K bracing, (d) inverteds V bracing, (e) knee bracing, and ( f ) rigid connections. , , deﬂection as a cantilever and the deﬂection due to shear. Thus, for a wall with solid rectangular cross section, the deﬂection at the top due to uniform load is ΄΂ H ΃ ϩ H ΅ 3 1.5wH ␦ϭ Et L Lrs where w ϭ uniform lateral loadd . H ϭ height of the walle E ϭ modulus of elasticity of the wall materialn t ϭ wall thicknessel L ϭ length of wall TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 306 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 306 4081 306 CHAPTER NINE For a shear wall with a concentrated load P at the top, the deﬂection at the top is ΄΂ H ΃ ϩ 0.75 H ΅ 3 4P ␦c ϭ Et L L If the wall is ﬁxed against rotation at the top, however, the W deﬂection is i P ΄΂ H ΃ ϩ 3 H ΅ 3 ␦f ϭ Et L L Units used in these equations are those commonly applied in United States Customary System (USCS) and the I System International (SI) measurements, that is, kip (kN), a lb/in2 (MPa), ft (m), and in (mm). Where shear walls contain openings, such as those for doors, corridors, or windows, computations for deﬂection and rigidity are more complicated. Approximate methods, however, may be used. COMBINED AXIAL COMPRESSION OR TENSION AND BENDING The AISC speciﬁcation for allowable stress design for buildings includes three interaction formulas for combined axial compression and bending. When the ratio of computed axial stress to allowable axial stress fu /Fa exceeds 0.15, both of the following equa- tions must be satisﬁed: TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 307p 306 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 307 BUILDING AND STRUCTURES FORMULAS 307e fa Cmx fbx Cmy fby ϩ ϩ Յ1 Fa (1 Ϫ fa /FЈ )Fbx ex (1 Ϫ fa /FЈ )Fby ey fa f f ϩ bx ϩ by Յ 1 0.60Fy Fbx Fbye When fa /Fa Յ 0.15, the following equation may be used instead of the preceding two: fa f f ϩ bx ϩ by Յ 1 Fa Fbx Fbyye In the preceding equations, subscripts x and y indicate the , axis of bending about which the stress occurs, andr Fa ϭ axial stress that would be permitted if axial forcen alone existed, ksi (MPa) , Fb ϭ compressive bending stress that would be permit- ted if bending moment alone existed, ksi (MPa) Fe ϭ 149,000/(Klb /rb)2, ksi (MPa); as for Fa, Fb, and Ј 0.6Fy, FeЈ may be increased one-third for wind and seismic loads lb ϭ actual unbraced length in plane of bending, in (mm) rb ϭ radius of gyration about bending axis, in (mm)r K ϭ effective-length factor in plane of bendingd fa ϭ computed axial stress, ksi (MPa)e fb ϭ computed compressive bending stress at point- under consideration, ksi (MPa) Cm ϭ adjustment coefﬁcient TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 308 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 308 4081 308 CHAPTER NINE WEBS UNDER CONCENTRATED LOADS w Criteria for Buildings The AISC speciﬁcation for ASD for buildings places a limit on compressive stress in webs to prevent local web yield- ing. For a rolled beam, bearing stiffeners are required at a concentrated load if the stress fa, ksi (MPa), at the toe of the web ﬁllet exceeds Fa ϭ 0.66Fyw, where Fyw is the minimum speciﬁed yield stress of the web steel, ksi (MPa). In the cal- F culation of the stressed area, the load may be assumed dis- tributed over the distance indicated in Fig. 9.4. For a concentrated load applied at a distance larger than the depth of the beam from the end of the beam: R r fa ϭ c t w (N ϩ 5k) e F e w F b 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 I beam to the toe of the ﬁllet. w TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 309p 308 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 309 BUILDING AND STRUCTURES FORMULAS 309 where R ϭ concentrated load of reaction, kip (kN) tw ϭ web thickness, in (mm) N ϭ length of bearing, in (mm), (for end reaction, t not less than k) -a k ϭ distance, in (mm), from outer face of ﬂange toe web toe of ﬁlletm - For a concentrated load applied close to the beam end: - R fa ϭn t w (N ϩ 2.5k) To prevent web crippling, the AISC speciﬁcation requires that bearing stiffeners be provided on webs where concentrated loads occur when the compressive 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: ΄ ΂ N ΃΂ tt ΃ ΅ √F 1.5 w R ϭ 67.5t2 1 ϩ 3 w yw tf /tw d f where tf ϭ ﬂange thickness, in (mm) For a concentrated load applied closer than d/2 from the beam end: ΄ ΂ N ΃΂ tt ΃ ΅ √F 1.5 w R ϭ 34t2 1 ϩ 3 w yw tf / two d fsf If stiffeners are provided and extend at least one-half of the web, R need not be computed. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 310 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 310 4081 310 CHAPTER NINE Another consideration is prevention of sidesway web D buckling. The AISC speciﬁcation requires bearing stiffeners when the compressive force from a concentrated load A exceeds limits that depend on the relative slenderness of o web and ﬂange rwf and whether or not the loaded ﬂange is e restrained against rotation: b c dc /tw a rwf ϭ l /bf c a e where l ϭ largest unbraced length, in (mm), along either o top or bottom ﬂange at point of application of load bf ϭ ﬂange width, in (mm) dc ϭ web depth clear of ﬁllets ϭ d Ϫ 2k w Stiffeners are required if the concentrated load exceeds R, kip (kN), computed from 6800t 3 w Rϭ (1 ϩ 0.4r 3 ) wf h where h ϭ clear distance, in (mm), between ﬂanges, and rwf is less than 2.3 when the loaded ﬂange is restrained against rotation. If the loaded ﬂange is not restrained and rwf is less than 1.7, 6800t3 w R ϭ 0.4r3 wf h R need not be computed for larger values of rwf. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 311p 310 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 311 BUILDING AND STRUCTURES FORMULAS 311b DESIGN OF STIFFENERS UNDER LOADSsd AISC requires that fasteners or welds for end connectionsf of beams, girders, and trusses be designed for the combineds effect of forces resulting from moment and shear induced by the rigidity of the connection. When ﬂanges or moment- connection plates for end connections of beams and girders are welded to the ﬂange of an I- or H-shape column, a pair of column-web stiffeners having a combined cross-sectional area Ast not less than that calculated from the following equations must be provided whenever the calculated valuer of Ast is positive:f Pbf Ϫ Fyc twc(tb ϩ 5K) Ast ϭ Fyst where Fyc ϭ column yield stress, ksi (MPa), Fyst ϭ stiffener yield stress, ksi (MPa) K ϭ distance, in (mm), between outer face of col- umn ﬂange and web toe of its ﬁllet, if col- umn is rolled shape, or equivalent distance if column is welded shape Pbf ϭ computed force, kip (kN), delivered byf ﬂange of moment-connection plate multi-t plied by 5͞3, when computed force is due tos live and dead load only, or by 4͞3, when com- puted force is due to live and dead load in conjunction with wind or earthquake forces twc ϭ thickness of column web, in (mm) tb ϭ thickness of ﬂange or moment-connection plate delivering concentrated force, in (mm) TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 312 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 312 4081 312 CHAPTER NINE Notwithstanding the preceding requirements, a stiffener ( or a pair of stiffeners must be provided opposite the beam- W compression ﬂange when the column-web depth clear of l ﬁllets dc is greater than c 4100t3 √Fyc wc b dc ϭ Pbf a f and a pair of stiffeners should be provided opposite the ten- sion ﬂange when the thickness of the column ﬂange tf is less than w Pbf tf ϭ 0.4 √ Fyc ( e 1 Stiffeners required by the preceding equations should com- s ply with the following 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 C of the ﬂange or moment-connection plate delivering the concentrated force. I 2. The thickness of stiffeners should not be less than tb /2. a 3. The weld-joining stiffeners to the column web must be t sized to carry the force in the stiffener caused by unbal- r anced moments on opposite sides of the column. C FASTENERS IN BUILDINGS T The AISC speciﬁcation for allowable stresses for buildings M speciﬁes allowable unit tension and shear stresses on the c cross-sectional area on the unthreaded body area of bolts s and threaded parts. b TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 313p 312 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 313 BUILDING AND STRUCTURES FORMULAS 313r (Generally, rivets should not be used in direct tension.)- When wind or seismic load are combined with gravityf loads, the allowable stresses may be increased one-third. Most building construction is done with bearing-type connections. Allowable bearing stresses apply to both bearing-type and slip-critical connections. In buildings, the allowable bearing stress Fp, ksi (MPa), on projected area of fasteners is-s Fp ϭ 1.2Fu where Fu is the tensile strength of the connected part, ksi (MPa). Distance measured 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-center- spacing of fasteners should be at least 3d.eh COMPOSITE CONSTRUCTIONe In composite construction, steel beams and a concrete slab. are connected so that they act together to resist the load one the beam. The slab, in effect, serves as a cover plate. As a- result, a lighter steel section may be used. Construction In Buildings There are two basic methods of composite construction.s Method 1. The steel beam is entirely encased in thee concrete. Composite action in this case depends on thes steel-concrete bond alone. Because the beam is completely braced laterally, the allowable stress in the ﬂanges is TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 314 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 314 4081 314 CHAPTER NINE 0.66Fy, where Fy is the yield strength, ksi (MPa), of the b steel. Assuming the steel to carry the full dead load and the e composite section to carry the live load, the maximum unit t stress, ksi (MPa), in the steel is o c MD M T fs ϭ ϩ L Յ 0.66Fy Ss Str m f where MD ϭ dead-load moment, inиkip (kNиmm) ML ϭ live-load moment, inиkip (kNиmm) 1 Ss ϭ section modulus of steel beam, in3 (mm3) 2 Str ϭ section modulus of transformed composite 3 section, in3 (mm3) An alternative, shortcut method is permitted by the AISC speciﬁcation. It assumes that the steel beam carries both live and dead loads and compensates for this by per- mitting a higher stress in the steel: MD ϩ ML fs ϭ Յ 0.76Fy Ss Method 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 temporary shores to sup- port the steel until the concrete hardens. The maximum stress in the bottom ﬂange is MD ϩ ML fs ϭ Յ 0.66Fy Str To obtain the transformed composite section, treat the F concrete above the neutral axis as an equivalent steel area c TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 315p 314 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 315 BUILDING AND STRUCTURES FORMULAS 315e by dividing the concrete area by n, the ratio of modulus ofe elasticity of steel to that of the concrete. In determination oft the transformed section, only a portion of the concrete slab over the beam may be considered effective in resisting compressive ﬂexural stresses (positive-moment regions). The width of slab on either side of the beam centerline that may be considered effective should not exceed any of the following: 1. One-eighth of the beam span between centers of sup- ports 2. Half the distance to the centerline of the adjacent beame 3. The distance from beam centerline to edge of slab (Fig. 9.5)es-ed -me FIGURE 9.5 Limitations on effective width of concrete slab in aa composite steel-concrete beam. TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 316 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 316 4081 316 CHAPTER NINE NUMBER OF CONNECTORS REQUIRED S FOR BUILDING CONSTRUCTION T n The total number of connectors to resist Vh is computed t from Vh /q, where q is the allowable shear for one connector, kip (kN). Values of q for connectors in buildings are given in structural design guides. The required number of shear connectors may be spaced uniformly between the sections of maximum and zero moment. Shear connectors should have at least 1 in (25.4 mm) of concrete cover in all directions; and unless studs are located directly over the web, stud diameters may not exceed 2.5 times the beam-ﬂange thickness. w With heavy concentrated loads, the uniform spacing of shear connectors may not be sufficient between a concentrated load and the nearest point of zero moment. The number of shear connectors in this region should be at least N1[(M␤/Mmax) Ϫ 1] N2 ϭ ␤Ϫ1 where M ϭ moment at concentrated load, ftиkip (kNиm) Mmax ϭ maximum moment in span, ftиkip (kNиm) f N1 ϭ number of shear connectors required between s Mmax and zero moment t a ␤ ϭ Str /Ss or Seff /Ss, as applicable Seff ϭ effective section modulus for partial composite action, in3 (mm3) TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 317p 316 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 317 BUILDING AND STRUCTURES FORMULAS 317 Shear on Connectors The total horizontal shear to be resisted by the shear con- nectors in building construction is taken as the smaller ofd the values given by the following two equations: ,n 0.85 fcЈ Acd Vh ϭ 2o4 As Fys Vh ϭ 2t where Vh ϭ total horizontal shear, kip (kN), between maxi-g mum positive moment and each end of steela beams (or between point of maximum positive . moment and point of contraﬂexure in continu-e ous beam) fcЈ ϭ speciﬁed compressive strength of concrete at 28 days, ksi (MPa) Ac ϭ actual area of effective concrete ﬂange, in2 (mm2) As ϭ area of steel beam, in2 (mm2)) In continuous composite construction, longitudinal rein- forcing steel may be considered to act compositely with then steel beam in negative-moment regions. In this case, the total horizontal shear, kip (kN), between an interior support and each adjacent point of contraﬂexure should be taken ase Asr Fyr Vh ϭ 2 TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 318 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 318 4081 318 CHAPTER NINE where Asr ϭ area of longitudinal reinforcement at support F within effective area, in2 (mm2); and Fyr ϭ speciﬁed mini- d mum yield stress of longitudinal reinforcement, ksi (MPa). l 0 t PONDING CONSIDERATIONS IN BUILDINGS Flat roofs on which water may accumulate may require analysis to ensure that they are stable under ponding condi- tions. A ﬂat roof may be considered stable and an analysis does not need to be made if both of the following two equa- tions are satisﬁed: Cp ϩ 0.9Cs Յ 0.25 Id Ն 25S4/106 where Cp ϭ 32Ls L4 /107Ip p Cs ϭ 32SL4 /107Is s Lp ϭ length, ft (m), of primary member or girder Ls ϭ length, ft (m), of secondary member or purlin S ϭ spacing, ft (m), of secondary members Ip ϭ moment of inertia of primary member, in4 (mm4) Is ϭ moment of inertia of secondary member, in4 (mm4) Id ϭ moment of inertia of steel deck supported on secondary members, in4/ft (mm4/m) TLFeBOOK
• 40816_09_p283-320 10/22/01 12:49 PM Page 319p 318 40816 HICKS Mcghp Ch09 THIRD PASS bcj 07/12/01 p 319 BUILDING AND STRUCTURES FORMULAS 319t For trusses and other open-web members, Is should be- decreased 15 percent. The total bending stress due to dead. loads, gravity live loads, and ponding should not exceed 0.80Fy, where Fy is the minimum speciﬁed yield stress for the steel.e-s-rn44n TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 321 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p321 CHAPTER 10 BRIDGE AND SUSPENSION- CABLE FORMULAS TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use.
• 40816_10_p321-354 10/22/01 12:49 PM Page 322 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p322 408 322 CHAPTER TEN SHEAR STRENGTH DESIGN FOR BRIDGES A F Based on the American Association of State Highway and Transportation Ofﬁcials (AASHTO) speciﬁcations for load- I factor design (LFD), the shear capacity, kip (kN), may be s computed from f Vu ϭ 0.58Fy htwC for ﬂexural members with unstiffened webs with h/ tw Ͻ 150 or for girders with stiffened webs but a/h exceeding 3 or 67,600(h/ tw)2: h C ϭ 1.0 when Ͻ␤ tw ␤ h ϭ when ␤Յ Յ 1.25␤ h/t w tw 45,000k h S ϭ when Ͼ 1.25␤ Fy (h/t w)2 tw For girders with transverse stiffeners and a/h less than 3 and 67,600(h / tw)2, the shear capacity is given by T Vu ϭ 0.58Fy dtw C ϩ ΄ 1ϪC 1.15 √1 ϩ (a/h)2 ΅ S Stiffeners are required when the shear exceeds Vu. Chap. 9, “Building and Structures Formulas,” for sym- bols used in the preceding equations. TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 323p322 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p323 BRIDGE AND SUSPENSION-CABLE FORMULAS 323 ALLOWABLE-STRESS DESIGN FOR BRIDGE COLUMNSd- In the AASHTO bridge-design speciﬁcations, allowablee stresses in concentrically loaded columns are determined from the following equations: When Kl/r is less than Cc,0 Fy ΄1 Ϫ (Kl/r) ΅ 2r Fa ϭ 2 2.12 2C c When Kl / r is equal to or greater than Cc, ␲2E 135,000 Fa ϭ ϭ 2.12(Kl/r)2 (Kl/r)2 See Table 10.1.3 TABLE 10.1 Column Formulas for Bridge Design Yield Allowable Stress, ksi (MPa) Strength, ksi (MPa) Cc Kl/r Ͻ Cc Kl/r Ն Cc 36 (248) 126.1 16.98–0.00053(Kl/r)2 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)2 TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 324 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p324 408 324 CHAPTER TEN LOAD-AND-RESISTANCE FACTOR DESIGN FOR BRIDGE COLUMNS i Compression members designed by LFD should have a maximum strength, kip (kN), T Pu ϭ 0.85As Fcr n where As ϭ gross effective area of column cross section, in2 (mm2). For KLc /r Յ √2␲2E/Fy : ΄ Fy ΂ KL ΃ ΅ 2 c Fcr ϭ Fy 1 Ϫ 4␲2E r For KLc /r Ͼ √2␲2E/Fy : ␲2E 286,220 Fcr ϭ ϭ (KLc /r)2 (KLc /r)2 where Fcr ϭ buckling stress, ksi (MPa) A Fy ϭ yield strength of the steel, ksi (MPa) F K ϭ effective-length factor in plane of buckling A a Lc ϭ length of member between supports, in s (mm) f r ϭ radius of gyration in plane of buckling, in m (mm) u E ϭ modulus of elasticity of the steel, ksi (MPa) ( TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 325p324 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p325 BRIDGE AND SUSPENSION-CABLE FORMULAS 325 The preceding equations can be simpliﬁed by introduc- ing a Q factor: ΂ KL ΃ Fy 2a Qϭ c r 2␲2E Then, the preceding equations can be rewritten as shown next:2 For Q Յ 1.0: ΂ Fcr ϭ 1 Ϫ Q 2 ΃Fy For Q Ͼ 1.0: Fy Fcr ϭ 2Q ALLOWABLE-STRESS DESIGN FOR BRIDGE BEAMS AASHTO gives the allowable unit (tensile) stress in bending as Fb ϭ 0.55Fy. The same stress is permitted for compres- sion when the compression flange is supported 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): TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 326 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p326 408 326 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). I p h t ΂ 5 ϫS10 C ΃ ΂ IL ΃ 7 l b yc Fb ϭ e xc s p ΂ L ΃ Յ 0.55F d 2 0.772J ϫ √ Iyc ϩ 9.87 y where L ϭ length, in (mm), of unsupported ﬂange between connections of lateral supports, including knee braces w Sxc ϭ section modulus, in3 (mm3), with respect to b the compression ﬂange d b Iyc ϭ moment of inertia, in4 (mm4) of the compres- sion ﬂange about the vertical axis in the plane of the web J ϭ 1/3(bc t3 ϩ bt t3 ϩ Dt3 ) c c w S bc ϭ width, in (mm), of compression ﬂange T bt ϭ width, in (mm), of tension ﬂange s tc ϭ thickness, in (mm), of compression ﬂange TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 327p326 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p327 BRIDGE AND SUSPENSION-CABLE FORMULAS 327 tt ϭ thickness, in (mm), of tension ﬂange tw ϭ thickness, in (mm), of web D ϭ depth, in (mm), of web d ϭ depth, in (mm), of ﬂexural member In general, the moment-gradient factor Cb may be com- puted from the next equation. It should be taken as unity, however, for unbraced cantilevers and members in which the moment within a signiﬁcant portion of the unbraced length is equal to, or greater than, the larger of the segment end moments. If cover plates are used, the allowable static stress at the point of cutoff should be computed from the preceding equation. The moment-gradient factor may be computed from M1 ΂ M ΃ Յ 2.3 2 1n Cb ϭ 1.75 ϩ 1.05 ϩ 0.3e M2 M 2 where M1 ϭ smaller beam end moment, and M2 ϭ larger beam end moment. The algebraic sign of M1 /M2 is positive for double-curvature bending and negative for single-curvature bending.-e STIFFENERS ON BRIDGE GIRDERS The minimum moment of inertia, in4 (mm4), of a transverse stiffener should be at least I ϭ ao t3J TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 328 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p328 408 328 CHAPTER TEN where J ϭ 2.5h2/a2 Ϫ 2 Ն 0.5 o h ϭ clear distance between ﬂanges, in (mm) ao ϭ actual stiffener spacing, in (mm) w ( t ϭ web thickness, in (mm) w For paired stiffeners, the moment of inertia should be taken c about the centerline of the web; for single stiffeners, about t the face in contact with the web. The gross cross-sectional area of intermediate stiffeners should be at least H ΄ A ϭ 0.15BDtw (1 Ϫ C) V Vu Ϫ 18t2 ⌼ w ΅ T w where ⌼ is the ratio of web-plate yield strength to stiffener- c plate yield strength, B ϭ 1.0 for stiffener pairs, 1.8 for t single angles, and 2.4 for single plates; and C is deﬁned in the earlier section, “Allowable-Stress Design for Bridge i Columns.” Vu should be computed from the previous sec- s tion equations in “Shear Strength Design for Bridges.” e The width of an intermediate transverse stiffener, plate, b or outstanding leg of an angle should be at least 2 in (50.8 p mm), plus 1͞30 of the depth of the girder and preferably not less than 1͞4 of the width of the ﬂange. Minimum thickness is 1͞16 of the width. w Longitudinal Stiffeners These should be placed with the center of gravity of the fas- teners h/5 from the toe, or inner face, of the compression ﬂange. Moment of inertia, in4 (mm4), should be at least TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 329p328 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p329 BRIDGE AND SUSPENSION-CABLE FORMULAS 329 ΂ I ϭ ht3 2.4 a2 o h2 Ϫ 0.13 ΃ where ao ϭ actual distance between transverse stiffeners, in (mm); and t ϭ web thickness, in (mm). Thickness of stiffener, in (mm), should be at least b√fb /71.2, where b is the stiffener width, in (mm), and fb is the ﬂangen compressive bending stress, ksi (MPa). The bending stress int the stiffener should not exceed that allowable for the material.s HYBRID BRIDGE GIRDERS These may have ﬂanges with larger yield strength than the web and may be composite or noncomposite with a con-- crete slab, or they may utilize an orthotropic-plate deck asr the top ﬂange.n Computation of bending stresses and allowable stressese is generally the same as that for girders with uniform yield- strength. The bending stress in the web, however, may exceed the allowable bending stress if the computed ﬂange , bending stress does not exceed the allowable stress multi-8 plied byts ␤␺(1 Ϫ ␣)2 (3 Ϫ ␺ ϩ ␺␣) Rϭ1Ϫ 6 ϩ ␤␺(3 Ϫ ␺) where ␣ ϭ ratio of web yield strength to ﬂange yield strength- ␺ ϭ distance from outer edge of tension ﬂange orn bottom ﬂange of orthotropic deck to neutral axis divided by depth of steel section TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 333p332 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p333 BRIDGE AND SUSPENSION-CABLE FORMULAS 333 BRIDGE FASTENERSg For bridges, AASHTO speciﬁes the working stresses foro bolts. Bearing-type connections with high-strength bolts aren limited to members in compression and secondary mem- bers. The allowable bearing stress isd-s Fp ϭ 1.35Fue where Fp ϭ allowable bearing stress, ksi (MPa); and Fu ϭ tensile strength of the connected part, ksi (MPa) (or as limited by allowable bearing on the fasteners). The allow- able bearing stress on A307 bolts is 20 ksi (137.8 MPa) and on structural-steel rivets is 40 ksi (275.6 MPa). COMPOSITE CONSTRUCTION IN HIGHWAY BRIDGES Shear connectors between a steel girder and a concrete slab in composite construction in a highway bridge should be capable of resisting both horizontal and verti- cal movement between the concrete and steel. Maximumt spacing for shear connectors generally is 24 in (609.6 mm), but wider spacing may be used over interior supports, to avoid highly stressed portions of the tension flange (Fig. 10.1). Clear depth of concrete cover over shear connectors 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. TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 330 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p330 408 330 CHAPTER TEN ␤ ϭ ratio of web area to area of tension ﬂange or bottom ﬂange of orthotropic-plate bridge LOAD-FACTOR DESIGN FOR BRIDGE BEAMS For LFD of symmetrical beams, there are three general types of members to consider: compact, braced noncom- pact, and unbraced sections. The maximum strength of each (moment, inиkip) (mmиkN) depends on member dimensions and 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 only when the maximum axial stress does not exceed 0.15F y A, where A is the area of the member. Symbols used in Table 10.3 are defined as follows: Fy ϭ steel yield strength, ksi (MPa) Z ϭ plastic section modulus, in3 (mm3) S ϭ section modulus, in3 (mm3) bЈ ϭ width of projection of ﬂange, in (mm) d ϭ depth of section, in (mm) h ϭ unsupported distance between ﬂanges, in (mm) M1 ϭ smaller moment, inиkip (mmиkN), at ends of unbraced length of member Mu ϭ Fy Z M1 /Mu is positive for single-curvature bending. TLFeBOOK
• n h p330 e r s s s - l l 40816_10_p321-354 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p331 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 inиkip (mmиkN) in (mm) in (mm) in (mm) 10/22/01 bЈ√Fy d√Fy [3600 Ϫ 2200(M1/Mu)]ry Compact† Fy Z 65.0 608 Fy bЈ√Fy h 20,000 Af Braced Fy S331 69.6 150 Fy d 12:49 PM noncompact† Unbraced ________________________ See AASHTO speciﬁcation ________________________ † Straight-line interpolation between compact and braced noncompact moments may be used for intermediate criteria, except that tw Յ d√Fy /608 should be maintained as well as the following: For compact sections, when both bЈ/tf and d/tw exceed 75% of the limits for these ratios, the following interaction equation applies: Page 331 d bЈ 1064 ϩ 9.35 Յ tw tf √Fyf where Fyf is the yield strength of the ﬂange, ksi (MPa); tw is the web thickness, in (mm); and tf ϭ ﬂange thickness, in (mm). TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 332 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p332 408 332 CHAPTER TEN BEARING ON MILLED SURFACES B For highway design, AASHTO limits the allowable bearing F stress on milled stiffeners and other steel parts in contact to b Fp ϭ 0.80Fu. Allowable bearing stresses on pins are given l in Table 10.4. b The allowable bearing stress for expansion rollers and rockers used in bridges depends on the yield point in ten- sion Fy of the steel in the roller or the base, whichever is smaller. For diameters up to 25 in (635 mm) the allowable stress, kip/linear in (kN/mm), is w Fy Ϫ 13 F pϭ 0.6d l 20 a For diameters from 25 to 125 in (635 to 3175 mm), o Fy Ϫ 13 pϭ 3√d 20 C where d ϭ diameter of roller or rocker, in (mm). I S TABLE 10.4 Allowable Bearing Stresses on Pins† s s Bridges c Pins subject Pins not subject s Buildings to rotation to rotation b Fy Fp ϭ 0.90Fy Fp ϭ 0.40Fy Fp ϭ 0.80Fy t ( 36 (248) 33 (227) 14 (96) 29 (199) c 50 (344) 45 (310) 20 (137) 40 (225) s † Units in ksi (MPa). o TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 333p332 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p333 BRIDGE AND SUSPENSION-CABLE FORMULAS 333 BRIDGE FASTENERSg For bridges, AASHTO speciﬁes the working stresses foro bolts. Bearing-type connections with high-strength bolts aren limited to members in compression and secondary mem- bers. The allowable bearing stress isd-s Fp ϭ 1.35Fue where Fp ϭ allowable bearing stress, ksi (MPa); and Fu ϭ tensile strength of the connected part, ksi (MPa) (or as limited by allowable bearing on the fasteners). The allow- able bearing stress on A307 bolts is 20 ksi (137.8 MPa) and on structural-steel rivets is 40 ksi (275.6 MPa). COMPOSITE CONSTRUCTION IN HIGHWAY BRIDGES Shear connectors between a steel girder and a concrete slab in composite construction in a highway bridge should be capable of resisting both horizontal and verti- cal movement between the concrete and steel. Maximumt spacing for shear connectors generally is 24 in (609.6 mm), but wider spacing may be used over interior supports, to avoid highly stressed portions of the tension flange (Fig. 10.1). Clear depth of concrete cover over shear connectors 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. TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 334 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p334 408 334 CHAPTER TEN B I n i i s FIGURE 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). Span/Depth Ratios w In bridges, for composite beams, preferably the ratio of span/steel beam depth should not exceed 30 and the ratio of span/depth of steel beam plus slab should not exceed 25. Effective Width of Slabs w For a composite interior girder, the effective width assumed for the concrete ﬂange should not exceed any of the fol- lowing: 1. One-fourth the beam span between centers of supports 2. Distance between centerlines of adjacent girders 3. Twelve times the least thickness of the slab For a girder with the slab on only one side, the effective width of slab should not exceed any of the following: 1. One-twelfth the beam span between centers of supports 2. Half the distance to the centerline of the adjacent girder 3. Six times the least thickness of the slab TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 335p334 40816 HICKS Mcghp Ch. 10 REVISE PASS rg 07/27/01 p335 BRIDGE AND SUSPENSION-CABLE FORMULAS 335 Bending Stresses In composite beams in bridges, stresses depend on whether or not the members are shored; they are determined as for beams in buildings (see “Composite Construction” in Chap. 9,“Build- ing and Structures Formulas”), except that the stresses in the steel may not exceed 0.55Fy. (See the following equations.)e6 For unshored members: MD M fs ϭ ϩ L Յ 0.55Fy Ss Str where fy ϭ yield strength, ksi (MPa).f For shored members:f MD ϩ ML fs ϭ Յ 0.55Fy Str where fs ϭ stress in steel, ksi (MPa)d MD ϭ dead-load moment, in.kip (kN.mm)- ML ϭ live-load moment, in.kip (kNиmm) Ss ϭ section modulus of steel beam, in3 (mm3)s Str ϭ section modulus of transformed composite section, in3 (mm3)e Vr ϭ shear range (difference between minimum and maximum shears at the point) due to live load and impact, kip (kN)s Q ϭ static moment of transformed compressiver concrete area about neutral axis of transformed section, in3 (mm3) TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 336 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p336 408 336 CHAPTER TEN I ϭ moment of inertia of transformed section, in4 (mm4) Shear Range w Shear connectors in bridges are designed for fatigue and then are checked for ultimate strength. The horizontal-shear range for fatigue is computed from VrQ Sr ϭ I where Sr ϭ horizontal-shear range at the juncture of slab and beam at point under consideration, kip/linear in (kN/linear mm). w 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 connector is given by the next two equations, depending on the connector used. d n TABLE 10.5 Ratio of Moduli of Elasticity o of Steel and Concrete for Bridges fcЈ for Es nϭ concrete Ec N 2.0–2.3 11 2.4–2.8 10 T 2.9–3.5 9 c 3.6–4.5 8 4.6–5.9 7 6.0 and over 6 TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 337p336 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p337 BRIDGE AND SUSPENSION-CABLE FORMULAS 3374 For channels, with a minimum of 3͞16-in (4.76-mm) ﬁllet welds along heel and toe: Z r ϭ Bw where w ϭ channel length, in (mm), in transverse directiond on girder ﬂange; andr B ϭ cyclic variable ϭ 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 Ն 4):d Z r ϭ ␣d 2r where d ϭ stud diameter, in (mm); andd ␣ ϭ cyclic variable ϭ 13.0 for 100,000 cycles, 10.6r 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 allowable range of horizontal shear of all con- nectors at one section Zr, kip (kN), by the horizontal range of shear Sr, kip per linear in (kN per linear mm). NUMBER OF CONNECTORS IN BRIDGES The ultimate strength of the shear connectors is checked by computation of the number of connectors required from P Nϭ ␾Su TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 338 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p338 408 338 CHAPTER TEN where N ϭ number of shear connectors between maxi- mum positive moment and end supports t Su ϭ ultimate shear connector strength, kip (kN) [see Eqs. (10.1) and (10.2) that follow and AASHTO data] w t ␾ ϭ reduction factor ϭ 0.85 s P ϭ force in slab, kip (kN) At points of maximum positive moments, P is the U smaller of P1 and P2, computed from i F P1 ϭ As Fy P2 ϭ 0.85 fcЈAc w where Ac ϭ effective concrete area, in (mm ) 2 2 fcЈ ϭ 28-day compressive strength of concrete, ksi (MPa) As ϭ total area of steel section, in2 (mm2) Fy ϭ steel yield strength, ksi (MPa) The number of connectors required between points of maximum positive moment and points of adjacent maxi- mum negative moment should equal or exceed N2, given A by F P ϩ P3 B N2 ϭ ␾Su a TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 339p338 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p339 BRIDGE AND SUSPENSION-CABLE FORMULAS 339- At points of maximum negative moments, the force in the slab P3, is computed from) P3 ϭ AsrFyrd where Asr ϭ area of longitudinal reinforcing within effec- tive ﬂange, in2 (mm2); and Fyr ϭ reinforcing steel yield strength, ksi (MPa).e Ultimate Shear Strength of Connectors in Bridges For channels: ΂ Su ϭ 17.4 h ϩ t 2 ΃ w √f Ј c (10.1) where h ϭ average channel-ﬂange thickness, in (mm) t ϭ channel-web thickness, in (mm), w ϭ channel length, in (mm) For welded studs (H/d Ն 4 in (101.6 mm): Su ϭ 0.4d 2 √fcЈEc (10.2)f-n ALLOWABLE-STRESS DESIGN FOR SHEAR IN BRIDGES Based on the AASHTO speciﬁcation for highway bridges, the allowable shear stress, ksi (MPa), may be computed from TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 340 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p340 408 340 CHAPTER TEN Fy Fy M Fv ϭ CՅ 3 3 F H for ﬂexural members with unstiffened webs with h/tw Ͻ 150 or for girders with stiffened webs with a/h exceeding 3 and 67,600(h/tw)2: T w b h d C ϭ 1.0 when Յ␤ tw ␤ h ϭ when ␤Ͻ Յ 1.25␤ S h/t w tw 45,000k h ϭ when Ͼ 1.25␤ P Fy (h/t w)2 tw a ΂ th ΃ 2 S k ϭ 5 if exceeds 3 or 67,600 t h w d or stiffeners are not required f 5 ϭ5ϩ otherwise (a/h)2 w k ␤ ϭ 190 √ Fy For girders with transverse stiffeners and a/h less than 3 and 67,600(h/tw)2, the allowable shear stress is given by Fv ϭ Fy 3 ΄C ϩ 1.15 √1ϪϩC(a/h) ΅ 1 2 Stiffeners are required when the shear exceeds Fv. TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 341p340 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p341 BRIDGE AND SUSPENSION-CABLE FORMULAS 341 MAXIMUM WIDTH/THICKNESS RATIOS FOR COMPRESSION ELEMENTS FOR HIGHWAY BRIDGES0d Table 10.6 gives a number of formulas for maximum width/thickness ratios for compression elements for highway bridges. These formulas are valuable for highway bridge design. SUSPENSION CABLES Parabolic Cable Tension and Length Steel cables are often used in suspension bridges to support the horizontal roadway load (Fig. 10.2). With a uniformly distributed load along the horizontal, the cable assumes the form of a parabolic arc. Then, the tension at midspan is wL2 Hϭ 8d where H ϭ midspan tension, kip (N) w ϭ load on a unit horizontal distance, klf (kN/m) L ϭ span, ft (m)ny d ϭ sag, ft (m) The tension at the supports of the cable is given by ΄ ΂ wL ΃ ΅ 2 0.5 T ϭ H2 ϩ 2 TLFeBOOK
• 40816_10_p321-354 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p342 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 65 70 e 10/22/01 Flange projection of rolled or fabricated I-shaped beams √Fy √Fy 608 Webs in ﬂexural compression 150342 √Fy 12:49 PM Allowable-stress design f fa ϭ 0.44Fy Description of element fa Ͻ 0.44Fy Fy ϭ 36 ksi (248 MPa) Fy ϭ 50 ksi (344.5 MPa) Plates supported in one side and Page 342 outstanding legs of angles 51 In main members 12 11 √fa Յ 12 TLFeBOOK 408
• p342 40816_10_p321-354 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p343 51 In bracing and other 12 11 secondary members √fa Յ 16 126 Plates supported on two edges or 32 27 webs of box shapesg √fa Յ 45 10/22/01 158 Solid cover plates supported on 40 34 two edges or solid websh √fa Յ 50 190 Perforated cover plates supported 48 41 on two edges for box shapes √fa Յ 55343 12:49 PM a b ϭ width of element or projection; t ϭ thickness. The point of support is the inner line of fasteners or ﬁllet welds connect- ing a plate to the main segment or the root of the ﬂange of rolled shapes. In LRFD, for webs of compact sections, b ϭ d, the beam depth; and for noncompact sections, b ϭ D, the unsupported distance between ﬂange components. b As required in AASHTO “Standard Speciﬁcation for Highway Bridges.” The speciﬁcations also provide special limitations on plate-girder elements. c Fy ϭ speciﬁed minimum yield stress, ksi (MPa), of the steel. Page 343 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 ϭ 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. TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 344 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p344 408 344 CHAPTER TEN T S G FIGURE 10.2 Cable supporting load uniformly distributed along S the horizontal. C where T ϭ tension at supports, kip (N); and other symbols are as before. F Length of the cable, S, when d/L is 1/20, or less, can be l approximated from m a 8d 2 SϭLϩ 3L where S ϭ cable length, ft (m). T Catenary Cable Sag and Distance between Supports T A cable of uniform cross section carrying only its own weight assumes the form of a catenary. By using the same previous notation, the catenary parameter, c, is found from TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 345p344 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p345 BRIDGE AND SUSPENSION-CABLE FORMULAS 345 T dϩcϭ w Then ΄ ΂S΃ ΅ 2 0.5 c ϭ (d ϩ c)2 Ϫ 2 Sag ϭ d ϩ c ft (m) Span length then is L ϭ 2c, with the previous same symbols. GENERAL RELATIONS FORg SUSPENSION CABLES Catenarys For any simple cable (Fig. 10.3) with a load of qo per unite length of cable, kip/ft (N/m), the catenary length s, ft (m), measured from the low point of the cable is, with symbols as given in Fig. 10.3, ft (m), H q x ΂q ΃ x ϩ иии 2 1 sinh o ϭ x ϩ o sϭ 3 qo H 3! H Tension at any point is T ϭ √H 2 ϩ q2 s2 ϭ H ϩ qoy o The distance from the low point C to the left support isnem aϭ H qo coshϪ1 ΂q f H o L ϩ1 ΃ TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 346 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p346 408 346 CHAPTER TEN b s P FIGURE 10.3 Simple cables. (a) Shape of cable with concen- U trated load; (b) shape of cable with supports at different levels. l d t where fL ϭ vertical distance from C to L, ft (m). The dis- a tance from C to the right support R is 1 p bϭ H qo coshϪ1 ΂q f H o R ϩ1΃ where fR ϭ vertical distance from C to R. T Given the sags of a catenary fL and fR under a distributed vertical load qo, the horizontal component of cable tension H may be computed from qo l H coshϪ1 ΂ qHf o L ΃ ϩ 1 ϩ coshϪ1 ΂ qHfo R ϩ1 ΃ w L where l ϭ span, or horizontal distance between supports L R and R ϭ a ϩ b. This equation usually is solved by trial. A ﬁrst estimate of H for substitution in the right-hand side of the equation may be obtained by approximating the catenary TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 347p346 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p347 BRIDGE AND SUSPENSION-CABLE FORMULAS 347 by a parabola. Vertical components of the reactions at the supports can be computed from qo a qo b RL ϭ H sinh RR ϭ H sinh H H Parabola- Uniform vertical live loads and uniform vertical dead loads other than cable weight generally may be treated as distributed uniformly over the horizontal projection of the cable. Under such loadings, a cable takes the shape of- a parabola. Take the origin of coordinates at the low point C (Fig. 10.3). If wo is the load per foot (per meter) horizontally, the parabolic equation for the cable slope is wo x 2 yϭ 2H The distance from the low point C to the left support L isdn l Hh aϭ Ϫ 2 wo l where l ϭ span, or horizontal distance between supports L and R ϭ a ϩ b; h ϭ vertical distance between supports. The distance from the low point C to the right supportL R is .f 1 Hh bϭ ϩy 2 wo l TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 348 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p348 408 348 CHAPTER TEN Supports at Different Levels T The horizontal component of cable tension H may be com- puted from L wo l ΂f h ΃ w l 2 2 Hϭ R Ϫ Ϯ √fL fR ϭ o h2 2 8f where fL ϭ vertical distance from C to L fR ϭ vertical distance from C to R f ϭ sag of cable measured vertically from chord L LR midway between supports (at x ϭ Hh/wol) As indicated in Fig. 10.3b: h f ϭ fL ϩ Ϫ yM 2 where yM ϭ Hh2/2wo l 2. The minus sign should be used when low point C is between supports. If the vertex of the S parabola is not between L and R, the plus sign should be I used. z The vertical components of the reactions at the supports f can be computed from wol Hh VL ϭ wo a ϭ Ϫ 2 l T wol Hh Vr ϭ wo b ϭ ϩ 2 l TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 349p348 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p349 BRIDGE AND SUSPENSION-CABLE FORMULAS 349 Tension at any point is- T ϭ √H 2 ϩ wo x 2 2 Length of parabolic arc RC is √ ΂ ΃ b wob 2 H wb LRC ϭ 1ϩ ϩ sinh o 2 H 2wo H ΂w ΃ b ϩ иии 2 1 o ϭbϩ 3 6 H Length of parabolic arc LC is) a ΂ wHa ΃ ϩ 2w H wo a 2 LLC ϭ 2 √1ϩ o o sinh H ΂w ΃ a ϩ иии 2 1 o ϭaϩ 3 6 Hde Supports at Same Levele In this case, fL ϭ fR ϭ f, h ϭ 0, and a ϭ b ϭ l/2. The hori- zontal component of cable tension H may be computeds from wo l 2 Hϭ 8f The vertical components of the reactions at the supports are wo l VL ϭ VR ϭ 2 TLFeBOOK
• 40816_10_p321-354 10/22/01 12:49 PM Page 350 40816 HICKS Mcghp Ch. 10 REVISE PASS bzm 6/7/01 p350 408 350 CHAPTER TEN Maximum tension occurs at the supports and equals I e wo l l 2 TL ϭ TR ϭ 2 √ 1ϩ 16 f 2 f Length of cable between supports is ΂w l΃ ϩ w H wo l 2 1 Lϭ 2 √ 1ϩ 2H o o sinh 2H i ΂ 8 f 32 f 256 f ΃ 2 4 6 ϭl 1ϩ Ϫ ϩ ϩ иии 3 l2 5 l4 7 l6 If additional uniformly distributed load is applied to a para- bolic cable, the change in sag is approximately w d 15 l ⌬L p ⌬fϭ 16 f 5 Ϫ 24 f 2/l 2 t For a rise in temperature t, the change in sag is about ⌬fϭ 15 l 2ct 16 f (5 Ϫ 24 f /l ) 2 2 1ϩ 8 f2 3 l2 ΂ ΃ w where c ϭ coefﬁcient of thermal expansion. Elastic elongation of a parabolic cable is approximately Hl ΂1 ϩ 16 lf ΃ 2 ⌬L ϭ AE 3 2 where A ϭ cross-sectional area of cable S E ϭ modulus of elasticity of cable steel H ϭ horizontal component of tension in cable TLFeBOOK