Structural Steel Design LRFD Approach Second Edition
J. C . Smith North Carolina SfafeUniversity
JohnWiley & Sons, Inc. New York Chichester Brisbane Toronfo Singapore
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Cliff Robichaud Catherine Beckham Debra Reigert Ken Santor Harry Nolan Michael Jung Dorothy Sinclair
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Smith, J.C., 1933Structural steel design : LRFD approach / J.C. Smith.Znd ed p. cm. Includes bibliographical references and index. ISBN 0471106933 (cloth: alk. paper) 1. Building, Iron and steel. 2. Steel, Structural. 3. Load factor design. I. Title. TA684S584 1996 624.1’821dc20 9536503 CIP
109 8 7 6 5 4 3
Preface
This book has been written to serve as the undergraduatelevel textbook for the first two structural steel design courses in Civil Engineering. In this edition, eachchapter was modified to reflect the changes made in the 1993 AISC LRFD Specification for Structural Steel Buildings and the 1994 LRFD Manual of Steel Construction, Second Edition, which consists of Volume I: Structural Members, Specifications, & Codes Volume 11: Connections The chapter on the behavior and design of tension members is located before the chapter on connections for tension members, which is separated from the chapter on other types of connections. Bolted connections for tension members are discussed before welded connections.The long examples in the first edition have been replaced by shorter ones. Each professor has particular course constraints and preferences of what to present in each course. Chapters 1 to 6 probably contain most of the material that is taughtin the first steeldesigncourse.Chapters7 t o l l containmaterial tomeet theother needs of each professor. Appendix B gives the review material needed for a thorough understanding of principal axes involved in column and beam behavior. Appendix C provides some formulas for the warping and torsional constants of open sections. The LRFD Specification requires a factored load analysis and permits either an elastic analysis or a plastic analysis. In our capstone structural design course, the students are required to design a steelframed building and a reinforcedconcreteframed building. Since the ACI Code permits only,an elastic analysis due to factored loads, I use only the elastic analysis approach in the capstone structural design course. Consequently, Chapter 6 and Appendix A give students a brief but realistic introduction to elastic analysis and design of unbraced frames in the LRFD approach. Chapter 11 should be adequate for those who wish to discuss plastic analysis and design. Appendix D provides some handbook information pertaining to plastic analysis. vi
I use the textual material associated with Appendix A in the classroom whenever appropriate. The reviewers of this edition were: P. R. Chakrabarti, California State University  Fullerton; W. S. Easterling, VirginiaTech;S. C. Goel,Universityof Michigan;R. B. McPherson,New Mexico State University;and A. C. Singhal, Arizona State University. I am appreciative of their comments, suggestions for improvement, constructive criticisms, and identified errors. J. C. Smith
Gontents Chapter 1 Introduction 1 1.1 Structural Steel 1 1.1.1 Composition and Types 1 1.1.2 Manufacturing Process 4 1.1.3 Strength and Ductility 5 1.1.4 Properties and Behavorial Characteristics of Steel 5 1.1.5 Residual Stresses 8 1.1.6 Effect of Residual Stresses on Tension Member Strength 9 1.1.7 Effect of Residual Stresses on Column Strength 11 1.2 Structural Behavior, Analysis, and Design 13 1.3 Idealized Analytical Models 15 1.4 Boundary Conditions 18 1.5 Interior Joints 20 1.6 Loads and Environmental Effects 23 1.6.1 DeadLoads 24 1.6.2 Live Loads 24 Occupancy Loads for Buildings 24 Traffic Loads for Bridges 24 1.6.3 Roof Loads 24 SnowLoads 25 Rain or Ice Loads 25 Roof Live Loads 26 1.6.4 Wind Loads 26 1.6.5 Earthquake Loads 26 1.6.6 Impact Loads 27 1.6.7 Water Pressure and Earth Pressure Loads 27 1.6.8 Induced Loads 27 1.7 Construction Process 27 1.8 Load and Resistance Factor Design 28 1.9 Structural Safety 34 1.10 SigruficantDigits and Computational Precision 40
Problems 41 vii
viii
Contents
Chapter 2 Tension Members 43 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
Introduction 43 Strength of a Tension Member with BoltedEnd Connections 44 Effect of Staggered Bolt Holes on Net Area 52 Design of a Tension Member with BoltedEnd Connections 59 Strength of a Tension Member with WeldedEnd Connections 63 Design of a Tension Member with WeldedEnd Connections 67 SingleAngle Members 69 ThreadedRods 70 Stiffness Considerations 71
Problems 73
Chapter 3 Connections for Tension Members 83 3.1 3.2 3.3 3.4
Introduction 83 Connectors Subjected to Concentric Shear 83 Bolting 84 Types of Connections 86 3.4.1 SlipCritical Connections 86 3.4.2 BearingType Connections 87 3.5 Bolts in a BearingType Connection 87 3.6 Bearing at the Bolt Holes 90 3.7 Connecting Elements in a Bolted Connection 92 3.8 Welding 96 3.9 Fillet Welds 97 3.9.1 Strength of Fillet Welds 97 3.9.2 Design of Fillet Welds 101 3.10 Connecting Elements in a Welded Connection 107 Problems 109
Chapter 4 Columns 118 4.1 Introduction 318 4.2 Elastic Euler Buckling of Columns 119 4.3 Effect of Initial Crookedness on Column Buckling 122 4.4 Inelastic Buckling of Columns 125 4.5 Effective Length 127 4.6 Local Buckling of the Crosssectional Elements 135 4.7 FlexuralTorsional Buckling of Columns 145 4.8 Builtup Columns 148 4.9 SingleAngleColumns 155 4.10 Story Design Strength 155
Problems 162
Contents ix
Chapter5 Beams 168 5.1 Introduction 168 5.2 Deflections 169 5.3 Shear 170 5.4 Bending Behavior of Beams 171 5.5 Plastic Bending 176 5.6 Limiting WidthThickness Ratios for Compression Elements 184 5.7 Lateral Support 185 5.8 Holes in Beam Flanges 187 5.9 Design Bending Strength 188 5.10 When Local Buckling Governs $Mnx 202 5.11 Builtup Beam Sections 209 5.12 Biaxial Bending of Symmetric Sections 221 5.13 Bending of Unsymmetric Sections 223 5.14 Web and Flanges Subjected to Concentrated Loads 227 5.15 Bearing Stiffeners 236 Problems 241
Chapter 6 Members Subject to Bending and Axial Force 247 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
Introduction 247 MemberSecondOrder (P8) Effects 248 SystemSecondOrder (PA) Effects 253 Elastic Factored Load Analyses 255 Members Subject to Bending and Axial Tension 258 BeamColumns 261 Braced Frame Examples 263 Unbraced Frame Examples 269 Preliminary Design 276
Problems 280
Chapter 7 Bracing Requirements 287 7.1 Introduction 287 7.2 Stability of a Braced Frame 287 7.2.1 Required Stiffness and Strength of Cross Braces 289 7.2.2 Required Stiffness and Strength of K Braces 298 7.3 WeakAxis Stability of a Column 299 7.3.1 Bracing Stiffness and Strength Requirements When h = L/2 301 7.3.2 Bracing Stiffness and Strength Requirements When h = L/3 302 7.3.3 Bracing Stiffness and Strength Requirements When h = L/4 303 7.3.4 BracingStiffnessand Strength RequirementsWhen h = L/n for Large n 304 7.3.5 When Point of Inflection Does Not Occur at a Braced Point 305 7.3.6 Example Problem 306 7.4 Lateral Stability of a Beam Compression Flange 309
x
Contents
Chapter 8 Connections 312 8.1 Introduction 312 8.2 Connectors Subjected to Eccentric Shear 312 8.2.1 A Bolt Group Subjected to Eccentric Shear 313 Ulitmate Strength Method 313 Elastic Method 316 8.2.2 A Weld Group Subjected to Eccentric Shear 321 Ulitmate Strength Method 321 Elastic Method 323 8.3 Bolts Subjected to Tension and Prying Action 327 8.4 Bolts Subjected to Tension and Shear 329 8.5 Connectors Subjected to Eccentric Tension and Shear 330 8.5.1 Weld Groups 331 8.5.2 Bolt Groups 333 Elastic Method 334 Ulitmate Strength Method 334 8.6 Truss Member Connections and Splices 336 8.7 Column Base Plates 337 Case 1: (e = MJP,) IH/6 338 Case 2: (e = MJP,) > H/6 339 8.8 ColumnSplices 342 8.9 Simple Shear Connections for Beams 343 8.9.1 Beam Web Connections 343 8.9.2 Unstiffened Beam Seats 344 8.9.3 Stiffened Beam Seats 350 8.9.4 Shear EndPlate Connections 353 8.9.5 Bracket Plates 353 8.10 Moment Connections for Beams 353 8.10.1 BeamtoBeam Connections and Splices 353 8.10.2 BeamtoColumn Connections 354 8.11 Knee or Corner Connections 356 Problems 360
Chapter 9 Plate Girders 366 9.1 Introduction 366 9.2 Conventional Design Method 370 9.2.1 Design Strength Definitions 371 9.2.2 Intermediate Stiffener Requirements 375 9.2.3 Design Examples 375 9.3 Tension Field Design Method 387 9.3.1 Design Strength Definitions 388 9.3.2 Intermediate Stiffener Requirements 389 9.3.3 Design Examples 391 Problems 397
Contents xi
Chapter 10 Composite Members 399 10.1 Introduction 399 10.2 Composite Columns 399 10.2.1 Limitations 402 10.2.2 Column Design Strength 403 10.3 Composite Beams with Shear Connectors 408 10.3.1 Composite Construction 408 10.3.2 Effective Concrete Flange Width 409 10.3.3 Shear Design Strength 410 10.3.4 Shear Connectors 410 10.3.5 Flexural Design Strength 418 Positive Moment Region 418 Negative Moment Region 419 10.4 ConcreteEncased Beams 429 10.5 Deflections of Composite Beams 431 10.6 Composite BeamColumns 435 10.7 Design Examples 438 Problems 438
Chapter 11 Plastic Analysis and Design 440 11.1 Introduction 440 11.2 Plastic Hinge 440 11.3 Plastic Collapse Mechanism 445 11.4 Equilibrium Method of Analysis 448 11.5 Virtual Work Method of Analysis 471 11.6 Jointsize 490
Problems 493
Appendix A ComputerOutput for an Elastic,Factored Load Analysis of a Plane Frame 499 Appendix B CrossSectional Properties and Flexure 508 Notation 508 Centroidalhes 509 Moments and Product of Inertia 509 Transfer Axes Formulas 509 Summation Formulas 510 Principal Axes 513 Using Mohr’s Circle to Find the Principal Axes 513 B.8 Radius of Gyration 515 B.9 Properties of a Steel L Section 516 B.10 Flexure Formula 518 B.11 Biaxial Bending 519 B.1 B.2 B.3 B.4 8.5 B.6 B.7
Problems 521
xii Contents
Appendix C Torsional Properties 524 Appendix D Plastic Analysis Formulas 526 References 531 Index 533
Introduction
1.1 STRUCTURAL STEEL Steel is extensively used for the frameworks of bridges, buildings, buses, cars, conveyors,cranes, pipelines, ships, storage tanks, towers, trucks, and other structures. 1.1.1 Composition and Types
Yield strength is the term used to denote the yield point (see Figure 1.1)of the common structural steels or the stress at a certain offset strain for steels not having a welldefined yield point. Prior to about 1960, steel used in building frameworks was ASTM (American Society for Testing and Materials) designation A7 with a yield strength of 33 ksi. Today, there are a variety of ASTM designations available with yield strengths ranging from 24 to 100 ksi. Steel is composed almost entirely of iron, but contains small amounts of other chemical elements to produce desired physical properties such as strength, hardness, diictility, toughness, and corrosion resistance. Carbon is the most important of the other elements. Increasing the carbon content produces an increase in strength and hardness, but decreases the ductility and toughness. Manganese, silicon, copper, chromium, columbium, molybdenum, nickel, phosphorus, vanadium, zirconium, and aluminum are some of the other elements that may be added to structural steel. Hotrolled structural steels may be classified as carbon sfeels, highstrength lowalloy steels, and dloy steels. Carboii steels contain the following maximum percentages of elements other than iron: 1.7%carbon, 1.65%manganese, 0.60% silicon, and 0.60% copper. Carbon and manganese are added to increase the strength of the pure iron. Carbon steels are divided into four categories: (1) low carbon (less than 0.1570);(2) mild carbon (0.150.29%); ( 3 ) medium carbon (0.300.5970);and, (4) high carbon (0.6&1.70%). Structural carbon steels are of the mildcarbon category and have a distinct yield point [see curve (a) of Figures 1.1and 1.21.The most common structural steel is A36, which has 1
2 lntroduction
Alloy steel [quenched and tempered (A514 and A709)]
80

(c> Highstrength,lowalloy steel ( A M , A441, A572)
Carbon steel (A36)
(For a coupon test specimen)
1 1 ‘sh
I
I
0.05
Strain ( i n h . )
I
I
I
I
0.10
I
I
I
I
0.20
0.15
I
I
0.25
b
0.30
FIGURE 1.1 Typical stressstrain curves for steel.
a maximum carboncontent of 0.25 to 0.29%, depending on the thickness,and a yield strength of 36 ksi. The carbon steels of Table 1.1 are A36, A53, A500, A501, A529, A570, and A709 (grade 36); their yield strengths range from 25 to 100 h i . Highstrength lowalloysteels [see curve (b) of Figures 1.1 and 1.21 have a distinct yield point ranging from 40to 70 ksi. Alloy elements such as chromium, columbium,
(c)
5 = 100 ksi taken at 0.002 offset (b)
1
I I
0.005
0.010
0.0 15
0.020
Strain (inhn.) I
I
w
0.025
‘sh
Plastic region
Strain hardening(extends to 4 )
FIGURE 1.2 Enlargement of Figure 1.1 in vicinity of yield point.
b
1.1 Structural Steel
Table 1.1 Steels Used for Buildings and Bridges AsTh4 FY FYThickness Steel Type Designation (ksi) (hi) (in) Carbon
A36
HighStrength lowalloy
Corrosion resistant Highstrength, lowalloy
A572 Grade 42 Grade 50 Grade 60 Grade 65 A242
Over 8 To 8 To 0.5 To 1.5
40
60
48
42 46
50
63 67 70
42
60
50 60 65
65 80
42
63
46
100
67 70 63 67 70 100130 110130
1.54 0.751.5 To 0.75 To 6 To 2 To 1.25 To 1.25 1.54 0.751.5 To 0.75 58 45 To 4 2.56 To 2.5
General; buildings General; buildings Metal building systems Metal building systems Welded construction Welded construction Welded construction Welded construction Buildings; bridges Buildings; bridges Buildings; bridges Buildings; bridges Bridges Bridges Bridges Weathering steel weathering steel Weathering steelbridges Plates for welding Plates for welding
70
110190
To 4
Plates for welding
42 50
50 A588
42 46
50
Quenched A514 & tempered lowalloy Quenched A852 and tempered alloy
Common Usage
5880 5880 6085 70100
32
36 A529 Grade 42 Grade 50 A441
3
90
75
copper, manganese, molybdenum, nickel, phosphorus, vanadium, and zirconium are added to improve some of the mechanicalproperties of steel by producing a fine instead of a coarse microstructure obtained during cooling of the steel. The highstrength lowalloy steels of Table 1.1are A242,A441, A572, A588, A606, A607, A618, and A709 (grades 50 and 50W). Alloy steels [seecurve(c)ofFiguresl.l and1.2) donothaveadistinctyield point. Their yield strength is defined as the stress at an offset strain of 0.002 with yield strengths ranging from 80 to 110h i . These steels generally have a maximum carbon content of about 0.20%to limit the hardness that may occur during heat treating and welding. Heat treating consists of quenching (rapid cooling with water or oil from 1650°F to about 300°F) and tempering (reheating to 1150°F and cooling to room temperature).Temperingsomewhatreducesthe strengthand hardnessof the quenched material, but sigruficantly improves the ductility and toughness. The quenched and tempered alloy steels of Table 1.1are A514 and A709 (grades 100 and 100W).
Bolts and threaded fasteners are classified as: 1. A307 (lowcarbon) bolts, usually referred to as common or machine or unfinished bolts, do not have a distinct yield point (minimumyield strength of 60 ksi is taken at a strain of 0.002). Consequently,the Load and Resistance
4
Introduction Factor Design (LRFD)Specification [2]' doesnot permit these bolts to be used in a slipcritical connection [see LRFD J1.ll (p. 672),J3.l(p.679), and Table J3.2(p. 681)].However, they may be used in a bearingtype connection. 2. A325 (mediumcarbon; quenched and tempered with not more than 0.30% carbon) bolts have a 0.2% offset minimum yield strength of 92 ksi (0.51 in.diameter bolts) and 81 ksi (1.1251.5 in.diameter bolts) and an ultimate strength of 105 to 120 ksi. 3. A449 bolts have tensile strengths and yield strengths similar to A325 bolts, have longer thread lengths, and are available up to 3 in. in diameter. A449 bolts and threaded rods are permitted only where greater than 1.5411. diameter is needed. 4. A490 bolts are quenched and tempered, have alloy elements in amounts similar to A514 steels, have up to 0.53% carbon, and a 0.2% offset minimum yield strength of 115 ksi (2.54411. diameter) and 130 ksi (less than 2.5in. diameter). Weld electrodes are classified as E60XX, E70XX, E80XX, E90XX, ElOOXX, and EllOXX where E denotes electrodes, the digits denote the tensile strength in ksi, and XX represents characters indicating the usage of the electrode.
1.1.2 Manufacturing Process
At the steel mill, the manufacturing process begins at the blast furnace where iron ore, limestone, and coke are dumped in at the top and molten pig iron comes out at the bottom. Then the pig iron is converted into steel in basic oxygen furnaces. Oxygen is essential to oxidize the excess of carbon and other elements and must be highly controlled to avoid gas pockets in the steel ingots since gas pockets will become defects in the final rolled steel product. Silicon and aluminum are deoxidizers used to control the dissolved oxygen content. Steels are classified by the degree of deoxidation: (1) killed steel (highest); (2)semikilled steel (intermediate); and (3) rinrmed steel (lowest). Potential mechanical properties of steel are dictated by the chemical content, the rolling process, finishing temperature, cooling rate, and any subsequent heat treatment. In the rolling process, material is squeezed between two rollers revolving at the same speed in opposite directions. Thus, rolling produces the steel shape, reduces it in cross section, elongates it, and increases its strength. Ordinarily, ingots are poured from the basic oxygen furnaces, reheated in a soaking pit, rolled into slabs, billets, orblooms in the bloom mill, and then rolled intoshapes, bars, and plates in the breakdown mill and finishing mill. If the continuous casting process is used, the ingot stage is bypassed. A chemical analysis, also known as the heat or ladleanalysis, is made on samples of the molten metal and is reported on the mill test certificate for the heat or lot (50300 tons) of steel taken from each steelmaking unit. One to 8 hours are required to produce a heat of steel depending on the type of furnace being used. 'We assume that each reader has a copy of the LRFD Manual[2]. Throughout this text, each applicable specification and design aid in the LRFD Manual is cited. Also, to enable the reader to quickly locate these items, the corresponding page numbers are given
1.1 Structitral Steel
5
Mechanical properties (nludulus of elasticity, yield strength, tensile strength, and elongation to determine thedegreeufductility) of steel are determined from tensile tests of specimens taken from the final rolled product. These mechanical properties listed on the mill test certificate normally exceed the specified properties by a significant amount and merely certify that the test certificate meets prescribed steelmaking specifications. Each piece of steel made from the heat of steel covered by the mill test certificate does not have precisely the properties listed on the mill test certificate. Therefore, structural designers do not use the mill test certificate properties for design purposes. The minimum specified properties listed in the design specifications are used by the structural designer. 1.1.3
Strength and Ductility Strength and ductility are important characteristics of structural steel in the structural design process. Suppose identical members (same length and same crosssectional area) are made of wood, reinforced concrete, and steel. The steel member has the greatest strength and stiffness, which permit designers to use fewer columns in long clear spans of relatively small members to produce steel structures with minimum dead weight. Ductility, the ability of a material to undergo large deformations without fracture, permits a steel member to yield when overloaded and redistribute some of its load to other adjoining members in the structure. Without adequate ductility, (1) there is a greater possibility of a fatigue failure due to repeated loading and unloading of a member; and (2) a brittle fracture can occur. Strength and ductility are determined from data taken during a standard, tensile, loadelongation test. (We contend that more appropriately for a member subjected to bending, the area under the momentxurvature curve is a better measure of ductility due to bending.) A stressstrain curve such as Figure 1.1can be drawn using the loadelongation test data. On the stressstrain curve, after the peak or irltiniate stretzgfh F,,, is reached, a descending branch of the curve occurs for two reasons: 1. Stress is defined as the applied load divided by the original, unloaded, crosssectional area. However, the actual crosssectional area reduces rapidly after the ultimate strength is reached. 2. The load is hydraulically applied in the lab. If the load were applied by pouring beads of lead into a bucket, for example, no decrease in load would occur from the time the ultimate strength was obtained until the specimen fractured and a horizontal, straight line would occur on the usual stressstrain curve from the ultimate strength point to the fracture point.
1.1.4 Properties and Behavorial Characteristics of Steel
For purposes of most structural design calculations, the following values are used for steel:
1. Weight = 490 Ib/ft3. 2. Coefficient of thermal expansion, CTE = 0.0000065 strain/"F). 3. Poisson's ratio u = 0.3.
6
Introduction The stressstrain curves shown in Figure 1.1 are for roomtemperature conditions. As shown in Figure 1.3, after steel reaches a temperature of about 200"F, the yield strength, tensile strength, and modulus of elasticityare sigruficantlyinfluenced by the temperature of the steel. Also, at high temperatures, steel creeps (deformations increasewith respect to time under a constant load). Temperatures in the range shown in Figure 1.3can occur in members of a building in case of a fire, in the vicinity of welds, and in members over an open flame in a foundry, for example. Temperature and prior straining into the strainhardening region have an adverse effect on ductility.Fractures at temperatures sigruficantlybelow room temperatureare brittle instead of ductile. Toughness (ability to absorb a large amount of energy prior to fracture) is related to ductility.Toughness usually is measured in the lab by a Charpy Vnotch impact test in which a standard notched specimen chilled or heated to a specified temperatureis struckby a swingingpendulum. Toughness, as implied by the type of test for toughness, is important for structures subjected to impact loads (earthquakes,vertical motion of trucks on bridges, and onelevatorcables if an elevator suddenly stops). Killed steels and heattreated steels have the most toughness. As Elevated temperature property Room temperature property
A
1.24 1.21.o
0.8
0.6
0.4
0.2
I
I
1
400
I
I
800
I
I 1200
I
I )
1600
Temperature( O F ) Note: For temperatures below 32' F, the properties shown increase; however, ductility and toughness decrease. FIGURE 1.3 W section.
1.1 Structural Steel
7
Stress
G' Increase in l$ (strain aging)
*I 14
\
1
E
I I I I I I I I I Reloading from E) I I I I I I , I
Ductility after strain hardening
ductility of virgin steel
D
Strail
I
A, B, C, D, F is the virgin steel curve D, E, D, F is unloading and immediate reloading curve E, F , G' is longtimedelaybeforereloading curve
FIGURE 1.4 Tensile test of a W section member. shown in Figure 1.4, ductility is sigruficantly reduced after a structure has been overloaded into the strainhardening region. Overloaded was chosen by the author as the descriptor since a building framework does not experience strains in the strainhardeningregion under normal serviceconditionloads except for severeearthquakes, for example. However, corners (bends of 90"or more at room temperature) in coldformed steel sections are strained into the strainhardening range. Corrosion resistance increasesas the temperature increases up to about 1000°F.In the welding process, a temperature of about 6500°F occurs at the electric arc tip of a welding electrode.Thus, high temperatures due to welding OCCUT and subsequently dissipate in a member in the vicinity of welds. Highstrength lowalloy steels have several times more resistance to rusting than carbon steels. Weathering steels form a crust of rust that protects the structure from further exposure to oxidation. Weldability (relative ease of producing a satisfactory, crackfree, structurally sound joint) is an important factor in structural steel design since most connections in the fabrication shop are made by welding using automated, highspeed welding procedures wherever possible. The temperature of the electric arc increases as the speed of welding increases, and more of the structural steel mixes
8
lnfroducfion with the weld. Steels with a carbon content 50.30% are well suited to highspeed welding. Steels with a carbon content > 0.35% require special care during welding. Members and their connections in a highway or railway bridge truss, for example, may be repeatedly loaded and unloaded millions of times during the life of the bridge. Some of the diagonal truss members may be in tension and later on in compression as a truck traverses the bridge. Even if the yield point of the steel in a member or its connections is never exceeded during the repeated loading and unloading occurrences, a fracture can occur and is called afatiguefracture. Anything that reduces the ductility of the steel in a member or its connections increases the chances of a brittle, fatigue fracture. Thus,fatigue strength may dictate the definition of nominal strength of members and connections that are repeatedly loaded and unloaded a very large number of times during the life of the structure. Indeed, the life of a repeatedly loaded and unloaded structure may be primarily dependent on the fatigue strength of its members and connections.
1.1.5 Residual Stresses
Residual stresses exist in a member due to: 1. the uneuen cooling to room temperature of a hotrolled steel product, 2. cold bending (process used in straightening a crooked member and in making coldformed steel sections), and 3. welding two or more sectionsor plates together to form a builtup section (e.g., four plates interconnected to form a box section). Figure 1.5shows a cross section of a steel rolled shape designated as a W section, which is the most common shape used in structural steel design as a beam (bending member), a colunzn (axiallyloaded compression member), and a beamcolunrn (bcnding plus axial compression member).
t FIGUR.E 1.5 Local buckling and column buckling.
2 . 2 Structural Steel
9
Consider a hotrolled W section after it leaves the rollers for the last time. Consider any cross section along the length of the W section product. The flange tips and the middle of the web cool under roomtemperature conditions at a faster rate than the junction regions of the flanges and the web. Steel shrinks as it cools. The flange tips and the middle of the web shrink freely when they cool since the other regions of the cross section have yet to cool. When the junction regions of the flanges and the web shrink, they are not completely free to shrink since they are interconnected to the flange tips and the middle of the web regions, which have already cooled. Thus, the lasttocool regions of the cross section contain residual tensile stresses, whereas the firsttocool regions of the cross section contain residual compressive stresses. These residual stresses, caused by shrinkage of the lasttocool portions of the cross section and their being interconnected to regions that are already cool, have a symmetrical pattern with respect to the principal axes of the cross section of the W section. Therefore, the residual stresses are selfequilibrating and do not cause any bending about either principal axis of a cross section at any point along the length direction of the member. Residual stresses in a W section are in the range of 10 to 20 ksi, regardless of the yield strength of the steel. 1.1.6 Effect of Residual Stresses on Tension Member Strength
Consider a laboratory tension test of a particular W section member. Some W sections have the residual stress pattern shown in Figure 1.6(a),which illustrates that: 1. The maximum residual compressive stress f,,occurs at the flange tips and at midheight of the web. 2. the maximum residual tensile stressf,, occurs at the junction of the flanges and web.
The residual stresses vary through the thickness of the flanges and web. Crosssectional geometry (flange thickness and width, web thickness and depth) influences the cooling rate and residual stress pattern. Some W sections are configured to be efficient as axial compression members, and other W sections are configured to be efficient as bending members. Depending on the crosssectional geometry, some W sections have only residual tensile stresses in the web, with the maximum value occurring at the junction of the flanges and web. Furthermore, the magnitude of the residual stresses is smaller for quenched and tempered members. Thus, the residual stress pattern as well as average values off,, and f,, through the thickness are dependent on several variables. Residual stress magnitudes on the order of 10 to 15 ksi or more occur if the member is not quenched and tempered. As shown in Figure 1.6(c),in a tensile test of a W section member:
v,,
+ T / A J = F,, ; that is, at locations where the residual tensile stress and applied tensile stress add up to the yield stress. 2. All fibers in the cross section yield before the firsttoyield fibers begin to strain harden. 1. Fibers in the cross section begin to yield when
10 Introduction
frc
L /r frc
J
e
frc
rc
(a) Residual stresses
(b) Tensile test stress
Stress
t
For a member1
/
r Foracoupon
member
For a coupon (specimen cut from web or a flange of W section)
b
(c) Stressstrain curve
Strain
FIGURE 1.6 Effect of temperature on properties of steel.
Thus, the second phenomenon is the same condition that occurs in a coupon test specimen torchcut from a flange or the web of a W section, properly machined, and laboratory tested to determine the stressstrain curve. Cutting the coupon specimen from the member’s flange or web completely removes the residual stresses from the coupon specimen. From a comparison of the tension tests on the coupon and the member in Figure 1.6 (c), the yield strength is
1.1 Structural Steel
11
identical. Therefore, the tension member strength is not affected by the presence of residual stresses. The fatigue strength of a tension member is determined by alternately loading (stretching)and unloading the member repeatedly until fracture occurs at the crosssectional locationwhere the tensile stress (frt + TIA,) is maximum during the loading cycle. Fatigue fractures can occur when the maximum tensile stress (f,, + T / A , ) is much less than Fv . Therefore, the fatigue strength of a tension member is affected by the presence of residual stresses. 1.1.7 Effect of Residual Stresses on Column Strength
In Figure 1.5,we see that a W section is Ishaped and composed of five elements (one vertical element and four horizontal elements).Each pair of horizontal elements is called af2ange and the vertical element is called the web. Suppose Figure 1.5 is the cross section of a column (an axially loaded compression member), of length L. Let A, denote the crosssectional area, and let P denote the axial compression force applied at each end of the column [see Figure 1.7(a)].If a W section is used as a column, the cross section is composed of five compression elements, each of which is subjected to a uniform compressive stress of P/A,. Each compression element in a cross section is classified as being either stiffened or unstiffeened (projecting). A stiffened compression element is attached on both ends to other crosssectional elements. An unstiffened compression element is not attached to anything on one end and is attached to another crosssectional element on the other end. When a W section is used as a column, the web is a stiffened compression element and each flangeis composed of two unstiffened compression elements. Each of the five compression elements of the cross section in Figure 1.5 essentially is a rectangle. The longer side of the rectangle is the width and the other side is the thickness. Each compression element has a property known as the widththickness ratio or b/t in mathematical terms. For each of the four unstiffened elements in Figure 1.5, b = 0.5bfand t = tf,where bfis the overall or total width of each top and bottom flange and t is the flange thickness. For the stiffened element (the web) in Figure 1.5, b = h, andt = t,, where t, is the thickness of the web and, for a W section, h, is the clear height of the web. If b/t does not exceed the limiting value stipulated in LRFD 85 (p. 636)for each compression element in a W section, local buckling does not occur before column buckling occurs. If b/t of a compression element exceeds the stipulated Kiting value, local buckling ofthe compression element occursas shown in Figure 1.7before the column buckles and affects the column buckling strength. Column buckling strength is affected by the presence of residual stresses. If a W section column buckles inelastically, the firsttocool regions of the cross section yield in compression when Vrc+ P/A,)= F,. However, the lasttocool regions of the cross section contain residual tension stresses and the applied compressive stress (P/A,). Consequently, some portions of these lasttocool regions of the cross section are still elastic when inelastic column buckling occurs; that is, (frt+ P/A,)< F,, where the negative sign indicates a tension stress and the compression stresses are positive.
12
Introduction
I
4J
(a) Column or member buckling
Number of half sine waves is a function of a/b and b/t of flange. (b) Flange local buckling of a W section column
(c) Section 1 1
(d) Section 22
Number of half sine waves is a function of d b and b/t of web. (e) Web local buckling of a W section column FIGURE 1.7 Stressstrain curves.
1.2 Structural Behavior, Analysis, And Design
13
1.2 STRUCTURAL BEHAVIOR, ANALYSIS, AND DESIGN A structure is an assembly of members interconnected by joints. A member spans
between two joints. The points at which two or more members of a structure are connected are called joints. Each support for the structure is a boundary joint that is prevented from moving in certain directions as defined by the structural designer. Structural behavior is the response of a structure to applied loads and environmental effects (wind, earthquakes, temperature changes, snow, ice, rain). Sfructural analysis is the determination of the reactions, member forces, and deformations of the structure due to applied loads and environmental effects.
Structural design involves: 1. Arranging the general layout of the structure to satisfy the owner’s functional requirements (for nonindustrial buildings, an architect usually does this part) 2. Conducting preliminary cost studiesof alternative structural framing schemes and/or materials of construction 3. Performing preliminary analyses and designs for one or more of the possible alternatives studied in item 2 4. Choosing the alternative to be used in the final design 5. Performing the final design, which involves the following:
(a) Choosing the analytical model to use in the analyses (b) Determining the loads (c) Performing the analyses using assumed member sizes that were obtained in the preliminary design phase (d) Using the analysis results to determine if the trial member sizes satisfy the design code requirements (e) Resizing the members, if necessary, and repeating items (c) and (d) if necessary 6. Checking the steel fabricator’s shop drawings to ensure that the fabricated pieces will fit together properly and behave properly after they are assembled 7. Inspecting the structure as construction progresses to ensure that the erected structure conforms to the structural design drawings and specifications Structural analysis is performed for structural design purposes. In the design process, members must be chosen such that design specifications for deflection, shear, bending moment, and axial force are not violated. Design specifications are written in such a manner that separate analyses are needed for dead loads (permanent loads), live loads (position and/or magnitude vary with time), snow loads, and effects due to wind and earthquakes. Influence lines may be needed for positioning live loads to cause their maximum effect. In addition, the structural designer may need to consider the effects due to fabrication and construction tolerances being exceeded, temperature changes, and differential settlement of supports. Numerical values of E and I must be known to perform continuousbeam analyses due to differential settlement of supports, but only relative values of E I are needed to perform analyses due to loads.
14 Introduction
Structural engineers deal with the analysis and design of buildings, bridges, conveyor support structures, cranes, dams, offshore oil platforms, pipelines, stadiu m , transmission towers, storage tanks, tunnels, pavement slabs for airports and highways, and structural components of airplanes, spacecraft, automobiles, buses, and ships. The same basic principles of analysis are applicable to each of these structures. Architectural, heating, air conditioning, and other requirements by the owner impose constraints on the structural designer’s choice of the structural system for a building. The owner wants a durable, serviceable, and lowmaintenancestructure, and possibly a structure that can be easily remodeled. The structural designer’s choice of the structural framingscheme and the structural material are influenced by these factors. Sometimes, a special architectural effect dictates the choice of the material and framing scheme. The engineer in charge of the structural design must 1. Decide how the structure is to behave when it is subjected to applied loads and environmental effects. 2. Ensure that the structure is designed to behave that way. Otherwise, a designed structure must be studied to determine how it responds to applied loads and environmental effects. These studies may involve making and testing a smallscalemodel of the actual structure to determine the structural behavior (this approach is warranted for a uniquely designed structureno one has ever designed one like it before). Fullscale tests to collapse are not economicallyfeasiblefor oneofakindstructures. For massproduced structures such as airplanes, automobiles, and multipleunit (repetitive)construction, the optimum design is needed, and fullscale tests are routinely made to gather valuable data that are used in defining the analyticalmodel employed in computerized solutions.
Analyficul models (some analysts prefer to call them mathematical models) are studied to determine which analytical model best predicts the desired behavior of the structure due to applied loads and environmental effects. Determination of the applied loads and the effects due to the environment is a function of the structural behavior, any available experimental data, and the designer’s judgment based on experience. A properly designed structure must have adequate strength, stiffness, stability, and durabizity. The applicable structural design code is used to determine if a structural component has adequate strength to resist the forces required of it, based on the results obtained from structural analyses. Adequate stifiess is required, for example, to prevent excessive deflections and undesirable structural vibrations. There are two types of possible instability: 1. A structure may not be adequately configured either externally or internally
to resist a completely general set of applied loads. 2. A structure may buckle due to excessive compressive axial forces in one or
more members.
1.3 ldealized Analytical Models
15
Overall internal structural stability of determinate frames may be achieved by designing either trusstype bracing schemes or shear walls to resist the applied lateral loads. In the trusstype bracing schemes, members that are required to resist axial compression forces must be adequately designed to prevent buckling; otherwise, the integrity of the bracing scheme is destroyed. Indeterminate structural frames do not need shearwalls or trusstype bracing schemes to provide the lateral stability resistance required to resist the applied lateral loads. However, indeterminate frames canbecome unstable due to sidesway buckling of the structure. In the course work that an aspiring structural engineer takes, the traditional approach has been to teach at least one course in structural analysis and to require that course as a prerequisite for the first course in structural member behavior and design. This traditional approach of separately teaching analysis and design is the proper one in our opinion,but in this approach, the student is not exposed to the true role of a structural engineer unless the student takes a structural design course that deals with the design of an entire structure. In the design of an entire structure, it becomes obvious that structural behavior, analysis, and design are interrelated. A bothersome thing to the student in the first design of an entire structure using plane frame analyses is the determination of the loads and how they are transferred from floor slab to beams, from beams to girders, from girders to columns, and from columns to supports. Transferralof the loads is dependent on the analytical models that are deemed to best represent the behavior of the structure. Consequently, in the first structural design courses, the analytical model and the applied loads are given information, and the focus is on structural behavior and learning how to obtain member sizes that satisfy the design specifications. 1.3 IDEALIZED ANALYTICAL MODELS
Structuralanalyses are conducted on an analyticalmodel that is an idealization of the actual structure. Engineering judgment must be used in defining the idealized structure such that it represents the actual structural behavior as accurately as is practically possible. Certain assumptions have to be made for practical reasons: Idealized material properties are used, estimations of the effects of boundary conditions must be considered,and complex structural details that have little effect on the overall structural behavior can be ignored (or studied later as a localized effect after the overall structural analysis is obtained). All structures are threedimensional, but in many cases it is possible to analyze the structure as being twodimensional in two mutually perpendicular directions. This text deals only with truss and frame structures. If a structure must be treated as being threedimensional, in this text it is classified as being either a space truss or space frame. If all members of a structure lie in the same plane, the structure is a twodimensional or planar structure. Examples of planar structures shown in Figure 1.8are a plane truss, a beam, plane frames, and a plane grid. In Figure 1.8 each member is represented by only one straight line between two joints. Each joint is assumed to be a point that has no size. Members have dimensions of depth and width, but a single line is chosen for graphical convenience to represent the member spanning between two joints. Thus, the idealized structure is a line diagram configuration. The length of each line
16
Introduction
(a) Plane truss lying in XY plane
I t
+ +
i

(b) Beam lying in X Y plane
Y
4
( c ) One story plane frames lying in X Y plane FIGURE 1.8 Examples of planar structures.
defines the span length of a member, and usually each line is the trace along the member’s length of the intersecting point of the centroidal axes of the member’s cross section. A plane truss [see Figure 1.8(a)]is a structural system of members lying in one plane that are assumed to be pinconnected at their ends. Truss members are designed to resist only axial forces and truss joints are designed to simulate a no moment resistance capacity.
A pZaneframe [see Figures 1.8 (bd)] is a structural system of members lying in one plane. Each member end is connected to a joint capable of receiving member end moments and capable of transferring member end moments between two or more memberends at a common point.
1.3 Idealized Amlyticnl Models
17
(d) Multistory, multibay plane frame lying in X Y plane
JY (e) Plane grid lying in X Y planeall
loads in Z direction
FIGURE 1.8 (continued)
A plane grid [see Figure 1.8(e)] is a structural system of members lying in one plane that are connected at their ends to joints capable of receiving and transferring memberend moments and torques between two or more member ends at a common point. Note that all members of a plane grid lie in the same plane, but all loads are applied perpendicular to that plane. For all other planar structures in Figure 1.8, all applied loads and all members of the structure lie in the same plane.
18
Introduction
1.4 BOUNDARY CONDITIONS
For simplicity purposes in the followingdiscussion, the structure is assumed to be a plane frame. At one or more points on the structure, the structure must be connected either to a foundation or another structure. These points are called support joints (or boundary joints, or exterior joints). The manner in which the structure is connected to the foundation and the behavior of the foundation influence the number and type of restraints provided by the support joints. Since the support joints are on the boundary of a structure and specialconditionscan exist at the support joint locations, the term boundary conditions is used for brevity to embody the special conditions that exist at the support joints. The various idealized boundary condition symbols for the line diagram structure are shown in Figure 1.9and discussed in the following paragraphs.
A hinge [Figure 1.9(a)]represents a structural part that is pinconnected to a foundation that does not allow translational movements in two mutually perpendicular directions. The pin connection is assumed to be frictionless. Therefore, the attached structural part is completely free to rotate with respect to the foundation. Since many of the applied loads on the structure are caused by and act in the direction of gravity, one of the two mutually perpendicular support directions is chosen to be parallel to the gravity direction. In conducting a structural analysis, the analyst assumes that the correct direction of this support force component is either opposite to the direction of the forces caused by gravity or in the same direction as the forces caused by gravity. In Figure 1.9, the reaction components are shown as vectors whose arrow indicates our choice for the assumed direction of each vector. A rolIer [Figure 1.9@)] represents a foundation that permits the attached structural part to rotate freely with respect to the foundation and to translate freely in the direction parallel to the foundation surface,but does not permit any translational movement in any other direction. To avoid any ambiguity for a roller on an inclined surface [Figure 1.9(c)J,we prefer to use a different roller symbol than used on a horizontal surface. A link is defined as being a fictitious, weightless, nondeformable, pinnedended member that never has any loads applied to it except at the ends of the member. Some analysts prefer to use a link [Figure1.9(d)]instead of a roller to represent the boundary condition described at the beginning of this paragraph. Afixed support [Figure 1.9(e)]represents a bedrock type of foundation that does not deform in any manner whatsoever, and the structural part is attached to the foundation such that no relative movements can occur between the foundation and the attached structural part. AtransZationalspring[Figure 1.9(Q]isa linkthatcandeformonlyalongitslength. Th~ssymbol is used to represent either a joint in another structure or a foundation resting on a deformable soil. A rotational spring [Figure 1.9(g)] represents a support that provides some rotational restraint for the attached structural part, but does not provide any translational restraint. The support can be either a joint in another structure or
1.4Boundary Conditions
19
t
t
(b) Roller support
(a) Hinge support
(c) Inclined roller support (d) Link support (equivalent to Figure 1 . 8 ~ )
t
A
(e) Fixed support
t I
t
(g) Rotational spring
( f )Translational spring
t
1
(i) Prescribed support movement
(h) Rotational and translational springs FIGURE 1.9 Boundary condition symbols and reaction components.
a foundation resting on a deformablesoil. Generally, as shown in Figures 1.9(g) and (h),a rotational spring is used in conjunctionwith either a hinge, or a roller, or a roller plus a translational spring, or a translational spring, or two mutually perpendicular translational springs. The soil beneath each individual foundation is compressed by the weight of the structure. Soil conditions beneath all individual foundations are not identical. The
20
lntroducfion weights acting on the foundations are not identical and vary with respect to time. Therefore, nonuniform or differential settlement of the structure occurs at the support joints. Estimated differential settlements of the supports are made by the foundation engineer and treated as prescribed support movements by the structural engineer. Figure 1.9(i)shows a prescribed support movement.
1.5 INTERIOR JOINTS
For simplicity and generality purposes in the following discussion, the structure is assumed to be a plane frame. On a line diagram structure, an interior joint is a point at which two or more member length axes intersect. For example, in Figure 1.10, points 2,4,5,7,8, and 10 are interior joints, whereas points 1,3,6, and 9 are support joints (or exterior joints, or boundary joints). The manner in which the member ends are connected at an interior joint must be accounted for on the line diagram. The types of connections for a structure composed of steel members can be broadly categorized as being one of the following types: 1. A shear connection develops no appreciable moment. If the connection at joint 10 of Figure 1.10 is as shown in Figure 1.11,it is classified by designers as being a shear connection. Thus, an internal hinge is shown on the line
diagram at joint 10 of Figure 1.10 to indicate that no moment can be transferred between the ends of members 2 and 10 at joint 10. However, the internal hinge is capable of transferring translationaltype memberend forces (axial forces and shears) between the ends of members 2 and 10 at joint 10. Note that this type of connection can transfer a small amount of moment, but the moment is small and can be ignored in design. 2. A rigid connection fully transfers all memberend forces. If the connection at joint 7 of Figure 1.10is as shown in Figure 1.12, it is classified by designers as a joint that behaves like a rigid (nondeformable) body. Thus, if joint 7 of Figure 1.10 rotates 5" in the counterclockwise direction, the ends of members 1,2,8, and 9 at joint 7 also rotate 5" in the counterclockwise direction. 3. A semirigid connection is a partial memberend moment transferral connection. If the beamtocolumn connection at joint 4 of Figure 1.10 is as shown in
o 3 5
4
Hinge 1
2
At the joint 4 end of member 1, there is an internal hinge plus a rotational spring spanning across the hinge. FIGURE 1.10 Idealized interior joint conditions.
1.5Interior /oinks
T 0
kL Section AA FIGURE 1.11 Web connection (shear connection).
Lightly shaded areas are column web stiffeners (each side of web)
FIGURE 1.12 Rigid connection: fully welded plus web stiffeners
21
22 lntroduction
Section AA (a) Side elevation and sectional view
(b) Assumed behavior
(c) Deformation of connection (separated for clarity)
FIGURE 1.13 Behavior of semirigid connection.
Figure 1.13, it is classified by designers as being a semirigid connection. (Webster's dictionary definitionof semirigid is "rigid to some degree or in some parts.") The top and bottom flange angles in Figure 1.13 transfer almost all of the beamend moment to the column. The web angles in Figure 1.13 transfer almost all the beamend shear to the column flange and fully ensure that the Y direction displacement at the end of member 1is identical to the Y
2.6 Loads And Environmental Effects
23
direction displacement of joint 4. (Onthe line diagram structure in Figure 1.10,joints 3,4, and 5 lie on the same straight line that is the longitudinal axis of members 6 and 7. Thus,joint 4 is located at the point where the longitudinal axes of members 1,6, and 7 intersect.)Consequently,joint4 is treated as being rigid in the Y direction. However, the top and bottom flange angles in Figure 1.13 are not flexuraLly stiff enough to ensure that the flanges of member 1 always remain completely in contact with the flanges of members 6 and 7. Thus, joint 4 cannot be treated as being completely rigid. Therefore, at the left end of member 1in Figure 1.10,a rotational (spiral)spring is shown to denote that a rotational deformation occurs between joint 4 and the end of member 1.It should be obvious that a semirigid connection is capable of developing more moment than a web connection can develop, but not as much moment as a rigid connection can develop.
In Figure 1.13, the angles are welded to the beam and bolted to the column. M effectivelyis transferred to the top and bottom flange angles. Consequently, due to the action of M, the top flange angle and the web angles flexurallydeform, allowing the top beam flange to translate a finite amount [see Figures 1.13(c) and (d)]. However, the bottom flange angle remains in contact with the column flange. Thus, the gap between the end of the beam and the column flange is trapezoidal after the angle deformationsoccur. The bolts resist V and ensure that the beam end does not translate in the Y direction. 1.6 LOADS AND ENVIRONMENTAL EFFECTS In structural analysis courses, the analytical model and the applied loads are given information, and the focus is on the applicable analysis techniques. In structural design, the loads that are to be applied to the analytical model of the structure must be established by the structural designer. In this country, each state has a building code mandated by law that must be used in the design of an engineered structure. The building code gives minimum design loads that must be used in the design of a building to ensure a desired level of public safety unless the structural designer decides that higher design loads should be used. Coping with building codes and determining the applied loads are topics covered in a structural design course dealing with the design of an entire building. We choose to give only a brief description of loads and environmental effects. However, the terminology used in the discussion conforms to the terminology in the building code definitions for the loads and environmental effects. All loads are treated as being statically applied to the structure, and the load classificationsare dead loads, live loads, and impact loads. Environmental efects due to snow and ice, rain, wind, earthquakes, temperature changes, differential settlement of supports, misfit of members, construction tolerances, soil pressures, and hydrostatic pressures are converted into statically equivalent applied live loads. There are three different types of loads: concentrated loads, line loads, and surface loads. Concentrated loads are applied on a relatively small surface area; examples are wheel loads of cranes, forklifts, and traffic vehicles (particularly on bridges). A Iine loud is confined to a rather narrow strip in the structure;examples are
24
Introduction member weights and partition wall weights. As the name implies, suflace loads are distributed over a large area; examples are the weight of a floor slab or a roof, wind pressure on an exterior wall, and snow on a roof.
1.6.1 Dead Loads
Dead loads do not vary with time in regard to position and weight. Thus, they are not moved once they are in place and, therefore, are called dead loads. A worn floor or roof cover is removed and replaced with a new one in a matter of days. A load that is not there for only an interval of a few days in the 50year life of a structure is considered to be a permanent load and is classified as a dead load. Examples are the weight of the structure; heating and airconditioning ducts; plumbing; electrical conduits, wires, and fixtures; floor and roof covers; and ceilings. Since the weights of the indicated items are provided by their manufacturer, dead loads can be estimated with only a small margin of error. 1.6.2 Live Loads
Gravity loads that vary with time in regard to magnitude and/or position are called livr loads. Examples of live loads are people, furniture, movable equipment, movable partition walls, file cabinets, and stored goods in general. Forklifts and other types of slowmoving vehicles (cranes in an industrial building and traffic vehicles in a parking garage, e.g.) are treated as live loads. An estimated maximum expected value of a live load contains a much larger margin of error than an estimated dead load.
Occupancy Loads for Buildings Building codes specify minimum values that must be used for this classification of loads in the design of a building. Each designer must use a t least the minimum values stated in the applicable building code. Some representative values of uniformly distributed live loads for this classification of loads are 40 psf for apartments, hotel rooms, and school rooms; 50 psf for offices in a professional building; 75 to 100 psf for retail stores; 100 psf for corridors on the exit floor level of public buildings (80 psf for corridors on other floor levels) and for bleachers in a sports arena; 1SO psf for library stacks; and 250 psf for warehouses (floors and loading docks). Traffic Loads f o r Bridges Minimum loads for highway bridges are given in the S t a d a r d Specifications for Highzuay Bridges 1121. Designers usually refer to them as the AASHTO specs since they are published by the American Association of State Highway and Transportation Officials. A lane loading with a roving concentrated load as well as wheel loads for a standardized van and for a semitrailer truck are given in these specifications. Minimum loads for railroad bridges are given in the Speczfications .for Steel Railway Bridges [ 131.Designers usually refer to them as the AREA spccs since they are published by the American Railway Engineering Association. 1.6.3 Roof Loads
I n some of the loading combinations listed in LRFD A4.1 (p. 630), one of the independent loadings is shown as L, or S or R, where L, is roof live load, S is snow load, and li is load due to initial rainwater or ice exclusive of the ponding contribu
1.6 Lunds Arid Eiiviroiirnental Effects
25
tion. Some state building codes give minimum load values that must be used for each of thesevariables. Other state building codes give only a single minimum load value that must be used on the roof. For example, except for counties in either the coastal region or the mountainous region of North Carolina, 20 psf is given as the minimum load value that must be used on the roof. The coastal region counties are subject to hurricane rains and the mountainous region counties are subject to deeper snow accumulations. For each county in these regions, a minimum value greater than 20 psf is listed for either rain or snow.
Snow Loads Snow loads corresponding to a 50year mean recurrence interval are specified in mostbuilding codes. The minimum snow load value that must be used is either listed for each county or shown on a map with varying color shades and corresponding minimum snow load values for a group of counties. A 1 in. snow accumulation on a flat surface weighs about 0.5 psf at mountain elevations and weighs more at lower elevations. Snow loads in the range of 20 to 40 psf are commonly found as the minimum snow load value listed in building codes. If the roof surface is not flat, a reduction factor that is a function of the roof slope may be given to convert the snow load specified for a flat roof to a value for a pitched roof. However, the snow load specified for a pitched roof is given as acting on a horizontal projection of the roof surface. Depending on the profile shape of the roof, the snow depth may not be constant over the entire roof surface. The deepest accumulations can be expected to occur in the roof valleys. Also, snow drifts can occur on a flat roof. If either a flat or sloped roof is below a higher roof on the same building or closeenough toa roof on an adjacent tallerbuilding,snow caneither blow off or slide off the higher roof onto the lower roof. Thus, snow drifts can be expected to occur on some roofs. A structural designer should account for these variations in the snow depth on the roof surface, even if the applicable building code does not explicitly state that such variations must be considered.
Rain or Ice Loads Some building codes group ice loads with snow loads, but LRFD A4.1 (p. 630) groups ice loads with rain. Ice can accumulate on members in an exposed structure (bridges and signs, e.g.). An ice coating on such members increases the structural area exposed to wind. Thus, icing in such cases increases the windinduced loads as well as the gravity direction loads. If the drains for a flat roof become clogged or if rainwater accumulates faster than the drains can remove the water, ponding occurs, causing the roof to sag and to accumulate more water. Thus, rainwater on a flat roof causes more serious problems than snow. A slope of at least 0.25 in./ft is needed on the top surface of a flat roof for rainwater to drain properly. Furthermore, in hurricaneprone regions 120mph winds occur with the heaviest rainfalls, push the rainwater on a flat roof to one side of the roof, and cause ponding. For these conditions, in addition to the primary roof drainage system, a secondary drainage system (scuppers, large holes in parapet walls) located above the primary drainage system can be installed to prevent water from accumulating above a certain level. These roofs are usually designed to resist rainwater loads for the rainwater elevation being at the elevation of the secondary drainage system plus 5 psf.
26
lntroduction
Roof Live Loads Mobile equipment may be used on the roof either during construction or when the roof needs repair. Installation or replacement of an airconditioningunit housed on the roof may require a portable crane to be hoisted to a flat roof and used to lift the unit into place. A flat roof may be used as an outdoor setting for a restaurant or as a helicopter port. These are possible sources of the L, variable in LRFD A4.1 (p. 630). 1.6.4 Wind Loads
Wind on an enclosed building causes a pressure to occur on the windward vertical surface and a suction on the leeward vertical surface. Suction is actually an outward pressurethe atmospheric pressure inside the building is greater than the pressure on the outside of the leeward wall. Wind causes a suction (uplift) on flat (0 5 15") roof surfacesof an enclosed building.On a sloping roof with a meanheight/width10.3and 8 > 15", wind causespressureon the windward slope and a suctionon the leeward slope. Maximum wind speeds vary with geographical location (mountain tops and coastal regions prone to hurricanes may experience 120mphwinds), types of terrain (open, wooded, urban, proximity and shapes of nearby structures),height above the ground, air density, and other factors.Wind speed data are collected by the weather bureau at an elevation of 10 m (32.8 ft) above ground level. Formerly, a recorded wind speed was the speed for a mile of wind flowingpast the recording device.Now, wind speeds are being recorded for a 3secduration gust of wind, which is the familiar type of information given in the local TV weather news. The efects due to wind are converted into an equivalentstaticpressure actingon the structure. Wind pressures based on the maximum wind speed for a B y e a r mean recurrence interval are specifiedin most building codes. A basic wind pressure (function of the mass density of air and the wind velocity) is given in the building code either as a formula or in tabular form (pounds per square foot along the height direction of the building). Wind velocity is least at ground level and increases along the height direction of the building. Shape factors are given for buildings and components of buildings. The basic wind pressure is multiplied by the building shape factor and possibly other given factors to obtain a design wind pressure that is applied to the structure. For example, for an enclosed rectangularshapedbuilding, a shape factor of 1.3 (+0.8on the windward surface and 0.5 on the leeward surface)is not uncommon. The design wind pressure distribution up the side of the building is determined and converted to wind loads acting on the structuralframeworkaccountingfor the way the cladding is supported. In most cases, the wind loads are applied joint loads.Half of the wind load on a wall segment located between two adjacent floor levels and two adjacentcolumn lines goesto the floorslabat the top of this wall segment,and theother half of the load goes to the floor slab at the bottom of thiswall segment. The floor slab surrounds the columns and delivers the wind loads as concentrated loads on the columns at the floor levels (at the joints of the framework). 1.6.5 Earthquake Loads
The effect ofan earthquake on a building is similar to the effect of a footballplayer being clipped. For our purposes, say a clip is a hit around or below the knees and from the blind side. The football player is unaware that he is going to be hit. Consequently,his
1.7 Construction Process
27
feet must go in the direction of the person who hits him, but his upper body does not want to move in that direction until the momentum of his lower body tends to drag the upper body in that direction. An earthquake consists of horizontal and vertical ground motions. The horizontal ground motion effect on a structure is similar to the football player being clipped. It is this type of motion that is converted into an equivalent static loading to simulate the effect of an earthquake on a building. An equivalent static loading (essentially a force F = mu with modification factors accounting for seismic zone, type of occupancy, structural loadresisting characteristic, and soilstructure interactionconditions)is applied at all story levels and in the opposite direction of the ground motion since the foundation of the structure remains stationary in a static analysis. All dynamic loads cannot always be replaced by equivalent static loads, and a dynamic analysis of the structure subjected to timedependent motions induced by an earthquake or rotating machinery should be conducted in such cases. 1.6.6 Impact Loads An impact load is a live load that is increased to account for the dynamic effect associated with a suddenly applied load. Impact loads are applicable for cranes, elevators, reciprocating machinery, and vehicular traffic on highway or railroad bridges. LRFD A4.2 (p. 630) stipulates the percentage of increase in live loads to account for impact. LRFD A4.3 (p. 631) stipulates the horizontal and longitudinal crane forces that must be applied to the crane support beam to account for the effect of moving crane trolleys and lifted loads. Similar longitudinal forces are applied to highway and railway bridges to account for sudden stops of vehicles on a bridge.
1.6.7 Water and Earth Pressure Loads
If a structure has walls (or portions thereof)below the ground level, the active earth pressure must be applied to these walls. If a portion of a structure extendsbelow the water table, water pressure must be applied thereon. Also, water pressure must be applied to dams and flumes. 1.6.8 Induced Loads
The effects due to temperature changes, shrinkage, differential settlement of supports, and misjt of members [l]are also converted into equivalent static loadings. 1.7 CONSTRUCTION PROCESS
If the framework of the structure is made of steel, the constructionprocess involves thefubrication,field erection, and inspection of the erected structural steel. The general contractor chooses the shop to fabricatethe steel and the subcontractorto do the field erection of the steel (insomecases, the generalcontractor erects the steel framework). Field inspection is done by an employee hired by the structural engineer and/or the architect. Field inspection is an integral part of the construction process and the final phase of the design process. Fabrication involves interpreting design drawings and specifications,preparing shop fabrication and field erection drawings, obtaining the material from a steel mill
28
Introduction if the needed material is not in the stockpile,cutting, forming, assembling the material into shippable units, and shipping the fabricated units to the construction site. The fabricator cuts the main members to the correct length, cuts the connection pieces from larger pieces including steel plates, and either punches or drills the holes wherever bolted field connections are specified.A shearing machine is used to cut thin material, and a gas flame torch is used to cut thick material and main members unless extreme precision or a smooth surface is required, in which case the cut is made with a saw. If the design specificationsdo not tolerate as much crookedness in a member as the allowed steel mill tolerances, the fabricator reduces the amount of crookedness by using presses or sometimes by applying heat to localized regions of the member. Bolt holes are made by punching, if possible, or drilling.The holemaking process may cause minute cracks or may make the material brittle in a very narrow rim around the hole. The LRFD B2 (p. 634) requires the structural designer to assume that the bolt hole diameter is 1/16 in. larger than the actual hole in order to account for the material that was “damaged” by the holemaking process. The steel field erection contractor uses ingenuity and experience to devise an erection plan that involves lifting the fabricated units into place with a crane. Without a proper plan, lifting operations may cause compression forces to occur in members of a truss that were designed to resist only tension, for example. Also, improperly lifting a plate girder could cause local buckling to occur. Temporary bracing generally must be provided by the erection contractor to avoid construction failures due to the lack of threedimensional or space frame stability. After permanent bracing designed by the structural designer, the roof, and the walls are in place, the structure has considerably more resistance to wind loads. Consequently, more failures due to wind loads occur during construction due to the lack of an adequately designed temporary bracing scheme by the erection contractor.
1.8 LOAD AND RESISTANCE FACTOR DESIGN A building code for a state is prepared by a committee of experienced structural engineers and is mandated by law to be used in the design of a public building. The state building code defines minimum loads (live, snow, wind) for which the structure must be designed, but the structural designer may use larger loads if they are deemed to be more appropriate. These service condition loads are called nominal loads that are codespecified loads. In the LRFD approach, each nominal load is multiplied by a Ioadfacfor. The factored loads are applied to the structure before performing structural strength analyses needed in the design process. Either an elastic analysis or a plastic analysis due to the factored loads is permitted. LRFD A4.1 (p. 630) requires the following load combinations to be investigated to find the critical combination of factored loads:
1. 1 . 4 0
2. 1.20 + 1.6L + 0.5 (L, or S or R ) 3. 1 . 2 0 + 1.6 (L, or S or R ) + (0.5L or 0.8W) 4. 1 . 2 0 + 1.3W + 0.5L + 0.5 ( L , or S or R ) 5. 1 . 2 0 k 1.OE + 0.5L + 0.2s 6. 0.90 f (1.3W or 1.OE)
1.8 Load And Resistance Factor Dcsigil
29
where The numerical values are load factors. D,L, W, L,, S, and R are nominal loads (codespecified loads).
D
is dead load due to the weight of the structural elements and permanent features on the structure.
L is live load due to occupancy and movable equipment. W is wind load. L, is rooflive load. S is snow load. R is load due to initial rainwater or ice exclusive of the ponding contribution. Crosssectional properties listed in the LRFD Manual for rolled sections are nominal values. Steel mills have + and  tolerances (see LRFD, p. 1188)for the crosssectional dimensions of a rolled shape. The permissible variation in area and weight is +2.5% (see LRFD, p. 1189). For a rolled shape that is used as a tension member with its ends welded to connections, for example, the limit of internal rrsistance (nominal strength) is the crosssectional area times the yield strength of the steel. If bolted connections are used, fracture of the member in the connection region may govern the limit of internal resistance. To account for the uncertainty in the crosssectional area and the steel properties, the nonrinnl strcizgtli (resistance) is multiplied by a resisfance(strength reduction)factor to obtain thedcsip strcwgfhof a tension member. Since a mathematical statement of the design requirement for a tension member is more convenient than words, let: 1. (b = resistancefactor (strength reduction factor) 2. P, = nominal strength (resistance) for a tension member
3.
P,,= required tensile strength (maximum axial tension force obtained from an elastic factored load analysis)
The LRFD Specification requires that (PP,, 2 Pli. Some examples of the strc7ngth reduction factor (resistance factor), (b, are: 1.
qc = 0.85 for axial compression
2. @,= 0.90 for shear
3. (bb = 0.90 for flexure (bending moment)
4. qt = 0.90 for yielding in a tension member 5. q$ = 0.75 for fracture in a tension member
The load and resistance factors in the LRFD Specification were developed using a probabilistic approach to ensure with a reasonable margin of safety that the maximumstrength ofeach memberand eachconnection i n astructure isnot less than the maximum load imposed on each of them. A portion of the margin of safety is in the load factors and the other portion is in the resistance factors.
1 lnfroducfion
In addition to being adequately designed for strength requirements, the structure must perform satisfactorilyunder nominal or service load conditions. Deflections of floor and roof beams must not be excessive. In the direction of wind, relative deflections of the column ends or story drift due to wind load must be controlled. Excessivevibrations cannot be tolerated. Thus, the structural designer must provide a structure that satisfies the owner’s performance requirements and the safety requirement on strength as imposed by the applicable building code and LRFD Specification. After the structure has been adequately designed for strength, the structural designer investigates the performance of the structure under service conditions. In addition to adequate strength, a member and the entire structure must have adequate stifiess for serviceability reasons. Many of the owner’s serviceability requirements can be met by ensuring that deflections do not exceed acceptable limits. Some of the common serviceability problems are [3]: 1. Local damage of nonstructural elements (e.g., windows, ceilings, partitions, walls) occurs due to displacements caused by loads, temperature changes, moisture, shrinkage, and creep. 2. Equipment (e.g., an elevator) does not function normally due to excessive displacements. 3. Drift and/or gravity direction deflections are so noticeable that occupants become alarmed. 4. Extensive nonstructural damage occurs due to a tornado or a hurricane.
5. Structural deterioration occurs due to age and usage (e.g., deterioration of bridges and parking decks due to deicing salt). 6. Motion sickness of the occupants OCCUTS due to excessive vibrations caused by (a) Routine occupant activities (floor vibrations). (b) Lateral vibrations due to the effects of wind or an earthquake.
In Table 1.2,these serviceabilityproblems are categorized as a function of either the gravitydirection deflection or the lateral deflection. It is customary steel design practice to limit the deflection index to: 1. L/360 due to live load on a floor or snow load on a roof when the beam supports a plastered ceiling. 2. L/240 due to live load or snow load if the ceiling is not plastered. 3. h/667 to h/200 for each story due to the effectsof wind or earthquakesonlya range of limiting values can be given for many reasons (type of facade, activity of the occupants, routine design, innovative design, structural designer’s judgment and experience). 4. H/715 to H/250 for entire building height H due to the effects of wind or earthquakescomment in item 3 applies here too.
1.8 Load And Resistance Factor Design
31
The deflection index limits for drift are about the same as the accuracy that can be achieved in the erection of the structure. The largest tolerable deflection due to live load is 0.5% of the member length. Consequently, deflected structure sketches are grossly exaggerated for clarity in textbooks. To aid in the discussion of how the required strength of a member in a structure is determined from a factoredload analysis, we chooseto use the plane frame structure showninFigures 1.14and1.15(seeAppendixA for the results obtained froma factored load analysis). This structure is a roof truss supported by two beamcolumns (members 1 to 4 in Figure 1.15). A beamcolumn is a member that is subjected to axial compression plus bending. Behavior and design of beamcolumns are discussed in Chapter 6.In Figure 1.15, members 1to 4 and the roof truss ends are interconnected to provide resistance due to wind, as well as overall lateral stability of the structure for the gravity direction loads. In Figure 1.15, note the moment springs at the foundation ends of the columns. In the factored load analysis given in Appendix A, we assumed that the moment springs represented a boundary condition of halfwayfixed (G = 2 as explained in Chapter 6) due to gravity loads. To provide resistance due to wind perpendicular to the plane of Figure 1.15,some bracing scheme (seeFigure 1.16for an acceptable scheme) must be devised and designed. For Figure 1.14, the nominal loads are: 1. Dead Builtup roof on metal decking = 8 psf Purlins = 20 lb/ft Truss = 0.15 kips at each interior joint Columns = 40 lb/ft 2. Live (crane loads) 8.0 kips at joints 6 and 18 16.0 kips at joint 12 Table 1.2 Deflection Index and Serviceabilitv Behavior ~
Deflection Index h/1000 h/500 h/300 or L/300
L/200 to L/300 h/200 to h/300 L/100 to L/200 h/100 to h/200 Note:
L = span length of a floor or roof member, h = story height.
~~
~
Typical Serviceability Behavior
No visible cracking of brickwork. No visible cracking of partition walls. Visible architectural damage. Visible cracks in reinforced walls. Visible ceiling and floor damage. Leaks in structural facade. Cracks are visually annoying. Visible damage to partitions and large. plateglass windows. Visible damage to structural finishes. Doors, windows, sliding partitions, and elevators do not function properly.
32
lntroduction
Builtup roof on metal decking
Purlin (roof beam) spans 30 ft, between neighboring trusses
NOTE: This cross section exists at each 30 feet along the length of the structure.
FIGURE 1.14 Cross section of an industrial building.
Encircled numbers are joint numbers; other numbers are member numbers.
FIGURE 1.15 Joint n u m b e r s a n d member n u m b e r s for the structure in Figure 1.14.
1.8 Load A n d Resistance Factor Design
33
3. Snow 20 psf (perpendicular to horizontal surface) 4. Wind
12 psf (pressure) on windward surface
7.5 psf (suction) on leeward surface 11 psf (suction) on roof surface For Figure 1.15, the joint loads are given in Appendix A.
LRFD A4.1 (p. 630) load combinations that must be considered are: 1.4D
1.2D + 1.6L + 0.5 ( L , or S or X) 1.2D + 1.6 ( L , or S or R ) + (0.5L or 0.8W) 1.2D + 1.3W + 0.5L + 0.5 (L, or S or R )
0.9D + 1.3W Discussions of the structure in Figures 1.14 to 1.16 are made in some other chapters of the text. Since the discussions will be related to the required strength of
u m
2
i
(a) Plan view
\
hese members and the top chords of roof trusses form a truss to resist wind.
I
\
FIGURE 1.16 S i d e e l e v a t i o n v i e w a n d p l a n v i e w of b u i l d i n g for F i g u r e 1.14
/
34 lntroduction
connectionsand members, the following examination of displacements for serviceability purposes is presented now. From Appendix A, due to nominal loads: 1. At joint 2 due to 0.9D + W, Ax= 1.024 in. = 0.0853 ft = (h/246),where h = 21 ft.
From item 3 of Table 1.2, a storydrift index of h/246 will be acceptable since h/246 lies in the range of h/667 to h/200. 2. Due to snow plus the crane loads, the vertical deflection at joint 12 is 1.041 in. = 0.08675 ft = (L/692),where L = 60 ft. According to item 2 of Table 1.2, the liveload deflection should not exceed (L/240)= 0.25 ft and 0.087 ft is less than 0.25 ft. Consequently, based on the member properties used for the analysis of Figure 1.15, the truss has more than adequate stiffness for gravity loads. 1.9 STRUCTURAL SAFETY The structuraldesigner must provide a structurethat satisfiesthe owner’s performance requirements and the strength requirements stipulatedby the applicablebuilding code and LRFD Spedcation. Safety, serviceability, and economy are accounted for in designing a structure to fulfill the intended usage during the expected lifetime. A safe structure must perform satisfactorily under the expected loads with little or no damage and without injury to the occupants due to any structural malfunctions. For a properly designed structure,the probabilityof a partial or total collapse due to extreme accidental overloads must be very small. Since forecasting the future always involves some uncertainty, anabsolutelysafestrudureduringitsexpectedlifetimecannotbedesigned. For example, a recordbreakingrainfall,snowfall, windstorm, or earthquake may occur for the locale of the building. Thus, the actual loads on the building may exceed the maximum expected loads used in the design of the structure. The first paragraph on LRFD Commentary, p. 6169, is: The LRFD Specifcation is based on (I) probabilistic models of Zoads and resistance, (2) a calibration of the LRFD criteria to the 1978 edition of the AISC ASD Specificationfor selected members, and (3) the evaluation of the resulting criteria by judgment and past experience aided by comparative design ofice studies of representativestructures. A brief discussion of the probabilistic model and calibration is presented later. However, calibration to fhe 2978AlSCASD Specifcation is related to our discussion of structural safety. Hereafter, the 1978 AISC ASD Specification is referred to as the ASD (Allowable Stress Design) Specification. Consider a plane trusswhose member endsarewelded to gusset plates (jointsin the truss).Structuralsafety of a member in the truss is to be discussed in regard to the ASD and LRFD Specifications.Strength terminology in the LRFDSpeaficationis in terms of forces,but the strength terminology in the ASD Specification is in terms of stresses. We choose to discuss the strength requirements of both specifications in terms of forces. For a tension member in our plane truss, the ASD requirement for strength is Pa 2 P, where
Pu= allowable tension force P, = maximum tension force
1.9 Structural Safety
35
(P, is determined from a truss analysis for the required ASD loading combinations of service loads applied at the truss joints.) Let pY = ABFY
where
Ag = gross crosssectional area
Fy= yield stress of steel If the force in our tension member reaches Py due to an extreme accidental overload, this is classified as a "failure" condition (excessively large deflections certainly will occur even though collapse may not occur). To ensure an adequate margin of safety against this failure condition, the ASD requirement for strength is (Pa = O.6Py)2 P, which can also be written as
( ): P,
=
2 P,
where 1 10 FS (factor of safety ) = =  = 1.67 0.6 6
The ASD Specifications from 1924 to 1989did not give the basis for choosing the indicated FS = 1.67, but this choice can be rationalized as follows. Py = AJY is a nominal value since neither A, nor F, is a perfect parameter. For example, the dimensionsof a steel section can be manufactured only within acceptable tolerances (+ and ). Therefore, we should assume that the minimum failure strength of our truss tension member is less than the nominal value by an amount AP which means that the minimum failure strength is Py AP,. We should assume tKat an extreme accidental overload of AP,occurs that causes the force in our member to be P, + AP,. The failure condition occurs when
(P,+ Us) = (P,  My)
and we obtain
36
lntrodiicfion
For
5 = 0.1
and
PY
APs
= 0.5
ps
we obtain
1.5 F S 1 =+ 0L5   = 1.67 10.1 0.9 Values of FS > 1.67were appropriately chosen by the ASD Specification writers for other failure conditions. Two examples where FS > 1.67 was chosen are:
1.For a tension member with bolted memberend connections, FS = 1/0.5 = 2.00 since the ASD Specification gives P, = 0.5A,FU,where A , is the effective net area and F,, is the specified minimum tensile strength of steel. 2. For elastic column buckling, FS = 23 = 1.92 12 since the ASD Specification gives D
where P,, is the elastic column buckling load. The preceding discussion illustrates that structural design involves a forecast of the actual loads and the actual member strengths by estimating in some way that the chance of high loads and low strengths will occur. The main variables involved in our discussion of structural safety for a proposed steel structure are: 1. Strength
(a) Stressstrain characteristics (b) Crosssectional properties (c) Workmanship in the fabrication shop and in field erection (d) Structural deterioration due to repetitive loads (unloading and reloading) and corrosion, for example, particularly at the connection locations (e) Field inspection and quality control (f) Accuracy of the analysis and design calculations
1.9 Structural Safety
37
2. Loads (a) Magnitude (b) Position (c) Duration (d) Load combinations 3. Consequences of a collapse
(a) Loss of life (b) Property damage (c) Lawsuits and legal fees
The LRFD Specification accounts for the factors that influence strength and loads by using a probabilistic basis (involves probability theory and statistical methods). This probabilistic basis ensures a more consistent margin of safety than was the case in the ASD Specification. In order to discuss how structural safety is achieved in the LRFD Specification, we must define the pertinent LRFD terminology. Each failure condition is referred to as a ”limit state.” A limit state is a condition at which a member, a connection, or the entire structure ceases to fufill the intended function. There are two kinds of limit states: serviceability and strength. Serviceability limit states deal with the functional requirements of the structure and involve the control of deflections, vibrations, and permanent deformations. Examples of strength limit states are yielding of a tension member, fracture of a tension member end, formation of a plastic hinge, formation of a plastic mechanism, overall frame instability, member instability, local buckling, and lateraltorsional buckling. Our discussion of structural safety is related to the strength limit states. Again, consider our truss tension member; yielding of the member is the LRFD limit state chosen for our discussion. If we use a generalized notation, the discussion can be extended to a limit state of bending in a beam, for example. Let R denote the resistance (yield strength P J of our truss tension member. Let Q denote the axial force in the member due to a factored loading combination (for convenience, assume that only dead and live loads are involved). Structural safety is a function of R and Q, which are random variables. R is random due to acceptable tolerances (+ and ), which are necessary in the steel mill (steelmaking and rolling processes), fabrication shop, and field erection. Dead load is random due to the uncertainty in the weight of the nonstructural elements (HVAC, sprinkler systems, electrical and communication systems, insulation, partitions, and ceilings) and the acceptable tolerances in the dimensions of steel members and concrete slab thicknesses, for example. Live load varies from structure to structure within a group of supposedly identical structures and varies as a function of time for each of these structures. Strength test data can be plotted as a histogram or frequency distribution of R (see Figure 1.17).For the data used to plot the histogram, statistical definitions can be used to obtain the mean (average) R , and the standard deviation o,,which is a measure of the dispersion. Probability theory can be used to fit a continuous theoretical curve \probability densifyfunction(PDF) of R ] to the histogram. Similarly, by using field measurement data of dead and live loads, a PDF of Q can be obtained.
38
Infroduction If the PDF of R is normal or Gaussian, then the analytical form of the PDF of R is
where
R,
f
0 =
and the constant 1/ f
=J* x
:1 ( x
j R( x ) dx
 R ,)
'
fR
( x ) dx
i is used so that the normalized PDF of R encloses a unit area:
and (R, + oR), the area under the PDF is 0.6827; that is, the Between (R,  oR) probability of an occurrence within this range is 68.27"/0. The probability of an occurrence between (R,  20R) and (R, + 2oR)is 95.45%. The probability of an occurrence between (R,  3 0 ~ and ) (R, + 3 b R ) is 99.865%. If the PDF of Q is normal or Gaussian, then the previous paragraph is applicable, provided that we replace each R with Q. If R and Q are normal random variables and if we define Z = R  Q as the safety margin, then Z is a normal random variable. The probability of Z < 0 is the probability of failure (achievement of the limit state of
Frequency
I
I
Qm
R < Q is "failure." FIGURE 1.17 Frequency distribution of load Q and resistance R.
R
Q
1.9 Structural Safety
39
Frequency Survival
Failure
FIGURE 1.18 Reliability index b.
yielding) for our truss tension member. The equivalent representation of structural safety shown in Figure 1.18 was used in the LRFD probabilistic model. Figure 1.18 is a schematic plot of the PDF of In ( R / Q )and the shaded area is the probability of failure for our truss tension member. In Figure 1.18, is the distance from the origin to the mean [In (R/Q)Im.Using
where
are the coefficients of variation, we obtain:
and b is called the reliability (safety)index. Thus,in obtaining b for each strength limit state, afirstorder, secondmoment (FOSM) probabilistic model was used. This model only uses the mean values and coefficients of variation for the random variables involved in a particular strength limit state to obtain the reliability index b. For increasing values of P I the probability of failure (achievement of the limit state of yielding) for our truss tension member decreases.
40
Introduction The reliability index pis only a relative measure of safety and must be chosen to give the desired degree of reliability. For each LRFD strength criterion, pwas selected by requiring that the LRFD criterion produce the same design as the corresponding ASD requirement for an "average" design situation. For example, for the LRFD loading combination consisting of gravity loads only, a liveloadtodeadload ratio of 3 was used. This procedure of selecting b is called "calibration" to an existing design criterion (1978 AISC ASD Specification in this case). Therefore, the actual distribution shape of the PDF of R/Q was not required and the resistance factors were chosen such that for D + L or S, targeted values of b = 2.6 for members and b = 4.0 for members were achieved. Thus, for our truss tension member example, the LRFD design requirement for the limits ta te of gross section yielding of the member became
(@pn= 0 . 9 ~2 ~pU ~ ~ ) where
@Pa= design tensile strength P,, = required tensile strength
( P , is the maximum value of the tension member force obtained from a factored load analysis for all required LRFD load combinations.) 1.10 SIGNIFICANT DIGITS AND COMPUTATIONAL PRECISION Estimated dead loads can be revised after all member sizes are finalized. However, estimated live loads and equivalent static loads for the effects due to wind and earthquakes are much more uncertain than the estimated dead loads. Perfection is impossible to achieve in the fabrication and erection procedures of the structure. Also, certain simplifying assumptions have to be made by the analyst to obtain practical solutions. For example, joint sizes are usually assumed to be infinitesimal, whereas they really are finite. Interior joints and boundary joints may be assumed to be either rigid or pinned, whereas they really are somewhere between rigid and pinned. Thus, the final structure is never identical to the one that the structural engineer designed, but the differences between the final structure and the designed structure are within certain tolerable limits. A digit in a measurement is a significant digit if the uncertainty in the digit is less than 10 units. Standard steel mill tolerance for areas and weights is k2.5% variation. Consider a rolled shape in the LRFD Manual with a weight of 100 lb/ft and a crosssectional area of 29.4 in.2. The weight variation tolerance is k0.025(100) = k2.5, and the actual weight lies between 97.5 and 102.5 Ib/ft. Since the third digit in 100 is uncertain by only 5 units, the 100lb/ft value is valid to 3 significant digits. However, the area variation tolerance is &0.025(29.4) = k0.735, and the actual area lies between 28.665 and 30.135 i n . 2 . There are 14.7 units of uncertainty in the third digit of 29.4, and ordinarily only the first 2 digits in the value 29.4 would be significant in engineering calculations, which means that 29.0 would be appropriate. In the LRFD approach, however, the uncertainty in the 29.4 in.2 area is accounted for in the resistance factor and it is appropriate to use the 29.4 in.2value recorded in the LRFD Manual.
Problems
41
Most computers will accept arithmetic constants having an absolute value in the range 1 x 103 to 1 x 1038. Some computers will accept a much wider range. A computer holds numeric values only to a fixed number of digits, usually the equivalent of between 6 and 16 decimal (or base 10) digits. The number of decimal digits held is called the precision of the arithmetic constant. For discussion purposes, suppose that the loads and structural properties are accurate to only 3 significant digits. Most commercially available structural analysis software appropriately use at least 16digit precision in the solution of the set of simultaneous equations. In a multistory building, there may be thousands of equations in a set. If only 3digit precision were used in computerized solutions, the truncation and roundoff errors in the mathematics would in some cases contribute as much uncertainty in the computed results as there is in the structural properties and loads. Reconciliation of the actual number of significant digits in the computed results should be made by the structural designer after all of the computed results are available. We always make electronic calculator computations using the maximum available precision in the same manner that a computer would make them in floatingpoint form. We record our computed results with at least threedigit precision and rounded in the last digit.
PROBLEMS Modify Figure 1.15by deleting the columns (members 1 and 3 ) to obtain a determinate truss supported on a fixed hinge at joint 2 and supported on a roller a t joint 22. Use the nominal loads given in Appendix A for Loadings 1 to 4 a t joints 3 to 21 and 23. For the member specified in each of the following problems, from a truss analysis find the axial force for each nominal load case ( D , L, S, and W). Then, see the factored load combinations required by LRFD(A41) to (A46) on LRFD p. 630. For the member specified in each of the following problems, use the previously found axial forces due to nominal loads and find the axial force for each LRFD load combination. Since a nominal load was not given for E(earthquake), ignore any load combination which is a function of E.
1.1Find the axial force in member 33. Indicate which LRFD load combination gives P, (the maximum axial tension and/or compression). Compare P, to the corresponding P,, given in Appendix A.
1.2 Find the axial force in member 43. Indicate which LRFD load combination gives P , (the maximum axial tension and/or compression). Compare P , to the corresponding P, given in Appendix A. 1.3 Find the axial force in member 10. Indicate which LRFD load combination gives P,, (the maximum axial tension and/or compression). Compare P, to the corresponding P, given in Appendix A.
42
lntroduction 1.4 Find the axial force in member 20. Indicate which LRFD load combination gives P , (the maximum axial tension and/or compression). Compare P, to the corresponding P, given in Appendix A.
CHRPTER
Tension IThmhers
2.1 INTRODUCTION A tension member is designed on the assumption that the member has to provide only axial tensile strength. Cables or guy wires are used as tension members to stabilize wood poles that support telephone and electricity transmission lines. Steel cables (wire ropes) and very slender rods (Lld I 500) have negligible bending stiffness.Thus, the assumption that a cable or very slender rod only provides tensile strength is indeed very reasonable. A tension member in a truss is fastened by welds or bolts at the member ends to either other members or connection plates (gusset plates). Truss members do not necessarily have negligible bending stiffness.Therefore, if the structural designer wants a structure to behave like a truss (alljoints assumed to be pins and no bending occurs in any member), the design details must be chosen such that negligible bending occurs in each member. This means that the design details must (1)provide for all loads except the selfweight of the members to actually occur only at truss joints, and (2) ensure that the joints do not cause appreciable memberend moments to occur. If a structural designer wants a structure to behave in a certain manner, the structural details must be carefully chosen such that the desired structural behavior is closely approximated. The LRFD definition of design strength is a resistancefactor times the nominal strength. Since the resistance factor is less than unity,a resistance factor is a strength reduction factor. Separate LRFD design strength definitionsare given for members, connectors (bolts, welds), and joints (angles, brackets, gusset plates, splice plates, stiffeners).The definitionsof design strength for connection plates, fillet welds, and shear of bolts are given in Chapter 3. In the LRFD approach, the members and connectionsof a structure are designed to have adequate strength to resist the factored loads imposed on the structure. For ,,' where @ is a resistancefactor (strength a tension member, the design strength is @ reduction factor), and P, is the nominal strength (resistance) to be defined in the following discussion. For a tension member, the LRFD design requirementfor strength 43
44
Tension Members is @P,, 2 P,, where P, is the required tensile strength, which is the maximum axial tension force obtained from an elastic factored load analysis. After the structure has been adequately designed for strength, the structural designer investigates the performance of the structure under service conditions. If a tension member is too flexible (does not have enough bending stiffness), (1) special handling in the fabrication shop and in the field erection stages may be necessary, resulting in extra costs; (2) the member may sag excessively due to its own weight; and (3) in a building containing large machines with rotating parts or in a bridge truss exposed to wind, the member may vibrate too much. Thus, in addition to adequate strength, a member and the entire structure must have adequate st#ness for serviceability reasons. Many of the owner’s serviceability requirements can be met by ensuring that deflections do not exceed acceptable limits (see Section 1.8).
2.2 STRENGTH OF A TENSION MEMBER WITH BOLTEDEND CONNECTIONS Consider Figure 2.1, which shows a tension member fastened by bolts to a gusset plate. Along the member at some finite distance from the bolts, all crosssectional fibers of the member on Section 22 in Figure 2.l(d) can attain the yield strength, when the bolts and memberend connection are stronger than the member. If all crosssectional fibers of a member yield in tension, the member elongates excessively, which can precipitate failure somewhere in the structural system, of which the tension member is a part. Consequently, yielding on A, of the member is classified as a failure condition. The memberend connection in Figure 2.1 is called a bearingtype connection since the transfer of axial force in the member to the gusset plate is made by bearing of the bolts at each bolt hole in the member end and gusset plate. Failure due to shear ofthe bolts and bearing at the bolt holes is discussed in Chapter 3, where we find that when failure is due either to shear of the bolts or to bearing at the bolt holes, each bolt in Figure 2.1 is assumed to transfer P , / 3 from the member to the gusset. Bearing at the bolt holes in the member end is shown in Figure 2.l(c), where P denotes the bearing (or pushing) force provided by each bolt. In the connection region of Figure 2.1, the stress distribution due to the applied load is not uniform in the member since some of the crosssectional elements of the member are not bolted to the gusset plate. Hence, a transition region exists in the member from the connection region to some finite distance from the connection where the stress distribution in the member becomes uniform when yielding occurs. Thus, before yielding occurs in the member, the regions of the member end and the gusset end containing the bolt holes usually experience strainhardening, and fracture can occur through the bolt holes either in the member end or in the gusset plate. In Figure 2.1, fracture of the member end occurs on Section 33 that has full P , (or all of P,) on it. Fracture of the gusset end is discussed in Chapter 3. In Figure 2.l(e), the cross section of the member is a pair of angles. Each angle section consists of two elements, and their centerlines form a capital “L”. Sometimes in the LRFD Specification, the elements of an angle section are called legs. When some of the crosssectional elements do not have bolts in them as in Figure 2.1, the force in the elements having no bolts must be shunted to the elements where bolts exist in order for the bolts to transfer their force to the gusset plate through bearing at the bolt holes. Removal of the force from an element that has no bolt in it is made
2.2 Strength of a Tension Member With BoltedEnd Connections
I
I
I I
I
45
1
I
1
1
OC'
(b) Section 11
(a) Member end bolted to gusset plate
P = total bearing force from each bolt 3 P = Pu
(c) FBD of member (pair of angles)
(d) Section 22
(e) Section 33
FIGURE 2.1 Tension member bolted to a gusset plate
through shear to an element that has a bolt in it, and the force removal process is called shear lug. When the shear lag has to occur in too short a distance on the member end, the fracture strength is smaller than when no shear lag exists. The LRFD definition offractureon A, (effectivenet area) accounts for any effect due to shear lag. As shown in Figure 2.2(b), another failure mode of the member end can occur and is called block shear rupture. Note that a displaced block is pushed out of each angle leg that has bolts in it. When block shear rupture occurs, a portion of the resistance is due to shear P,, and the other portion of the resistance is due to tension P,. Fracture occurs on the plane of the block where the larger resistance exists and yielding is conservatively assumed to occur on the other plane of the block. All bolt holes are standard holes except when the structural engineer specifies an oversized hole. In the fabrication shop, standard bolt holes are punched in the
46
Tmsion Members P = bearing force from bolt 3 P = P,
P = bearing force from bolt 6 P = P,
I
I
L
L
Singleangle with bolts only in one leg
Single angle with bolts only in one leg
(a) FBD of member end
P,
= P,
+ Pv
(b) Block Shear Rupture of member end
FIGURE 2.2 Examples of block shear rupture
member, when the material thickness does not exceed the hole diameter. The nominal diameter of a standard bolt hole (LRFD Table J3.3, p. 682) is the bolt diameter plus 1/16in. Instrengthcalculations,theholediameterisdefined(LRFDB2, p. 634) as the nominal diameter of the hole plus 1/16 in. For each cited LRFD Specification,the reader should look in the LRFD Commentary to see if there is any information that may be helpful with the interpretation of that specification. The LRFD design strength definitionsof a tension member with bolted connections are: 1. Yielding on As [see Figures 2.l(c) and (d)] (LRFD D1, p. 644)
$Pn= 0.90FyA, where F, = specified minimum yield stress ( h i ) A,
= gross area of
member (in2)
2. Fracture on A, [see Figures 2.l(c) and (e)] (LRFD D1, p. 644, and B3, p. 634)
(PP, = 0.75FUA,
2.2 Strength of a Tension Member With BoltedEnd Connections
47
where
F, = specified minimum tensile stress (ksi) A, = effective net area of the critical section (in.2) A, = A,U when not all crosssectionalelements transfer some of P, A, = A , when all crosssectionalelements transfer some of P, A , = A,  C(fhd,) = net area of member (in.2)
th = thickness of hole (in.) d, = diameter of hole (in.)
dh = (d + 1/8 in.) for a standard bolt hole d = diameter of bolt (in.) U = ( 1 X / L, ) 5 0.9 = strength reduction factor
L, = length of connection parallel to P, (in.)
X
= connection eccentricity (in.) (see Figure 2.3)
U < 1when all crosssectionalelements of the member do not participate in transferringthe tension forceat the member ends through bearing to the bolts and then the bolts transfer their forces through bearing to the gusset plate. Figure 2.3 (adapted from LRFD Figures CB3.1 and CB3.2) illustrates how the connection eccentricity X is to be computed. For the supposed angle(s) to be used in Figures 2.3 (bd) for computing X in the vertical direction, the implications are that the vertical leg of the supposed angle section is measured from the center of the indicated bolt hole to the free edge of the actual section. X in the vertical direction is measured from the center of the indicatedbolt hole to the horizontal axis through the centroid of the supposed angle section. The strengthreduction factor for shear lag is based on the researchconductedby MunseandChesson[15]and EasterlingandGirowc[l6]. 3. Block shear rupture, abbreviated BSR (LRFD J4.3, pp. 687 and 6228) [see Figure 2.201) for some examples] When Fdnt 2 0.6F$,,, @R,= 0.75(Fdn, + 0.6F/l,) When 0.6F&,,
2FAnt,
@R,= 0.75(0.6Fdn, + F,,A,J where A, = gross area on BSR shear plane(s), in.2 A,, = gross area on BSR tension plane, in.2 A,,, = A ,  [A,,, on BSR shear plane(s)] A,, = Agt [Aholes on BSR tension plane]
48
Tension Members
X = larger of x and All c.g. axes are for gross sections. (b) Connection eccentricities (a) Angle section
Treat as a gross angle section.
X = larger of
x and
(c) C section
I.
(c) W section with bolts in web
X = larger of x and Subdivide on yaxis of symmetry. Treat each half above centerline of hole as a gross angle section.
Subdivide on xaxis of symmetry. Treat each half as a gross WT section.
FIGURE 2.3 Connection eccentricity in shear lag reduction coefficient
2.2 Strength of a Tension Member With BoltedEnd Connections
49
h n p l e 2.1 The A36 steel member in Figure 2.l(a) is a pair of angles, L3.5 x 3 x 0.25, with the long legs fastened to a gusset plate by 0.75in.diameter bolts. Use L, = 2 in., s = 3 in., and 8 = 2 in. The gusset plate thickness is 0.5 in. Find the governing design strength of the tension member for a bearingtype bolted connection.
LRFD, p. 162:
For a single L3.5 x 3 x 0.25,
A = 1.56 in.2
x = 0.785 in.;
y = 1.04 in.
The governing design strength of the tension member is the least $Pn value obtained from: 1. Yielding on A, = 2(1.56) = 3.12 i n 2 [see Figures 2.l(c) and (d)]
$P,, = 0.90F/Ix = 0.90(36)(3.12)= 101 kips 2. Fracture on A, = A,U [see Figures 2.l(c) and (e)]
d,, = d
+ 1/8 in. = 0.75 + 0.125 = 0.875 in. t,, = 2(0.25) = 0.500 in.
d,,t, = 0.875(0.500)= 0.4375 in.2 A,, = A,  d,,t,, = 3.13  0.4375 = 2.69 in.2
From Figure 2.3(b), we see that: (a) the connection eccentricity from the face of the gusset plate isx = 0.785 in. (b) the other connection eccentricity for a supposed long leg = g must be computed.
y, = (leg  g) = (3.5  2) = 1.50 in.
A' = A  y,t = 1.56 (1.50)(1/4) = 1.185 i n 2 t
Y =
Ay  y et ( 1% A'

0.5Y ', )  ( 1.56)( 1.04)  1.50( 0.25)( 3.5  1.50/ 2 ) = 0.499 in. 1.1875 x = 0.661 in.
X = larger of U = smaller of
8  y ' = 2  0.499 = 1.50 in,
{
l.E/L, 0.9
=11.50/6=0.750
50
Tension Members A, = A,U = 2.69(0.750) = 2.02 in?
@P,= 0.75FJ1, = 0.75(58 ksi)(2.02in.,)
= 87.8 kips
3. BZock shear rupture [see Figure 2.2@),onerowofboltscase]
L, = L, + 2s = 2 + 2(3) = 8 in.
A, = L,t = 8(2)(0.25)= 4.00 i n . 2 A,, = Ap  Ahole
= 4.00  2.5(0.4375)= 2.906 in.'
L, = legg = 3.5  2 = 1.5 in. Ag, = L, t = 1.5(2)(0.25)= 0.75 i n . 2 A,, = Agt Aholes= 0.75  0.5(0.4375) = 0.53125 in.'
FA,,= 58(0.53125)= 30.8 kips 0.6FJn, = 0.6(58)(2.906)= 101.14 kips
@P, = 0.75(O.6F,/ln, + Fflg,) @P, = 0.75[101.14 + 36(0.75)]= 96.1 kips @P,,= 87.8 kips, due to fracture on A,, is the governing design strength of the tension member.
The A36 steel member in Figure 2.4 is a pair of angles, L6 x 4 x 0.375, with the long legs fastened to a gusset plate by 0.75in.diameter bolts. Use L, = 2 in., s = 3 in., g1= 2.25 in., andg, = 2.5 in. The gusset plate thickness is 0.5 in. Find the governing design strength of the tension member for a bearingtype bolted connection.
Solution For a single L6 x 4 x 0.375 y = 1.94 in. x = 0.941 in. A = 3.61 in? The governing design strength of the tension member is the least $P, value obtained
L E D , p. 158:
from: 1. Yielding on Ag
QP, = 0.90F,,Ag= 0.90(36)(7.22)= 233.9 kips 2.
Fracture on A, = A,U
2.2 Strength of a Tension Member With BolfedEnd Connections
51
1
1
4
4
2
4
3
6 P = pU
(a) Member end FBD
(b) Section33
I U P=#pU
P=#pU
(c) FBD Section 11
(d) FBD Section 22
(e)FBD Section 33
FIGURE 2.4 Tension member with two lines of bolts
d, = d
+ 1/8 in.= 0.75 + 0.125 = 0.875 in. th
= 2(0.375) = 0.750 in.
dbth= 0.875(0.750) = 0.656 in.2 A,, = Ag  C(d,t,) = 7.22  2(0.656) = 5.908 in? From Figure 2.3(b), we see that: (a) The connectioneccentriatyfrom the faceof the gusset plate is x = 0.941 in. (b) The other connection eccentricity for a supposed long leg = g must be computed:
ye= (leg  g ) = (6  2.25) = 3.75 in. A' = A  yet = 3.61 (3.75)(0.375)= 2.20 in?
y'=
AYYtt(k05Yt )  (3.61)(1.94) 3.75( 0.375)(6  3.75 / 2 ) = 0.540 in. A' 2.20
X = larger of
x = 0.661 in.
gy'=2.250.540=1.71in.
52
Tension Members
A, = A,,U = 5.908(0.715)= 4.22 in.? $Pn = 0.75F,,AC. = 0.75(58 ksi)(4.22 in.2)= 184 kips 3. Block shear rupture [see Figure 2.2(b), tworowofbolts case]
L,, = L, + 2s = 2 + 2(3) = 8 in. A,, = L,,t = 8(2)(0.375)= 6.00 in.]
A,,,, =A,,,  Aholes= 6.00  2.5(0.656) = 4.688 in.? L, = leg  g, = 6  2.25 = 3.75 in. Ax, = L, t = 3.75(2)(0.375)= 2.81 in.2 A,,, = Ax,  Aholes= 2.81  1.5(0.656)= 1.83 in.2
F,,A,,, = 58(1.83) = 106 kips 0.6F,,An,,= 0.6(58)(4.688)= 163 kips
$pn = 0.75(0.6FUA,,,, + F+Aq,) $Ptj = 0.75[163 + 36(2.81)] = 198 kips
$P,, = 184 kips, due to fracture on A,, is the governing design strength of the tension member. 2.3 EFFECT OF STAGGERED BOLT HOLES ON NET AREA
Figure 2.4(a) is the FBD of a tension member separated from a bearingtype bolted connection. To ensure that the socket can be fitted on each nut of all bolts in a bolt group, LRFD J3.3stipulates that the minimum spacing center to center between two adjacent bolt holes is 2.67d, where d is the nominal diameter of the bolts. To speed up the installation of the nuts, the preferred minimum spacing is 3d. In Figure 2.4(a), s andg, are the parameters for which the preferred minimum spacing is3d. In Example 2.2, a pair of L6 x 4 x 0.375 was used as the tension member in Figure 2.4(a).The usual gages for the long leg of a L6 x 4 x 0.375 are g1= 2.25 in. and g2= 2.5 in. Since g2/3= 2.50/3 = 0.833 in., we cannot used > 0.75 in. if we want to use the preferred minimum spacing of 3d at g,. Suppose that we want to use d = 1in.diameter bolts and the preferred minimum spacing of 3d. Then, as shown in Figure 2.5(a), the bolt holes must be staggered to ensure that C 2 3d and 2s 2 3d. When a staggered bolt hole pattern is used in the connection of a tension member, two or more net sections must be investigated to find the critical net section for fracture on A,. If fracture occurs on the net section shown in Figures
2.3 Effect of Staggered Bolt Holes on Net Area
53
2.5(c) and (d), the fracture surface lies in one plane. However, if fracture occurs on the net section of Figures 2.5(e)and (f),the fracture surface does not lie in one plane. Figures 2.5(d)and (h)have the same net area and the same fracture design strength. However, the internal force in Figure 2.5(h) is smaller than in Figure 2.5(d). Furthermore, the internal force on all net sections at and to the left of Section 33 in Figure 2.5(g)is less than P , .Therefore, the critical net section (where the fracture on A, would occur) is either Figure 2.5(c) or (e).The design strength due to tensile fracture in the net section is the smaller value obtained for the net sections in Figures 2.5(c) and (e). I n Figure 2.5(e), Section 22 shows a potential tensile fracture path across the member, and this path is not a straight line. Such a fracture path is called a staggered path. If a line segment on the staggered path is not perpendicular to P,,, we call that line segment a stagger. As shown in Figure 2.5(e), a stagger has a longitudinal component called s and a transverse component called g. If fracture occurs on Section 22 of Figure 2.5(e), a tensile fracture force exists perpendicular to each line segment of Section 22. On each stagger, a shear force must exist in addition to the tension force. This shear force is required to equilibrate the transverse component of the tension force on the stagger. Therefore, a combined state of failure stress exists on each stagger and we only need the sum of all fracture force components parallel to P,. LRFD 82 (p. 634) gives an empirical definition of the net area based on the research of Cochrane [14], which when multiplied times F,, gives the desired tensile fracture force component parallel to P,. We prefer to show this definition as a formula. Let A , denote the equivalent gross area for a staggered path:
A,v, = A ,
+cL s2t
4g
where
A,?= planar gross section area s = longitudinal component of a stagger
g = transverse component of a stagger Then the empirical definition of the net area given in LRFD B2 for a staggered path can be written as
This definition of A,, is applicable for Figures 2.5(e) and (f).
The A36 steel member in Figure 2.5 is a pair of angles, L6 x 4 x 0.375, with the long legs fastened to a gusset plate by 1in.diameter bolts. Use L , = 2 in., s = 2 in.,g, = 2.25 in., and g, = 2.5 in. The gusset plate thickness is 0.5 in. Find the governing design strength o f the tension member for a bearingtype bolted connection.
54
Tension Members Le
1
+s
*
2s
. ). . 1
r
2s
2s
% (b) Gross section
(a) FBD of the member end
r)l
4
(c) Path 11
1
(d) FBD Path 11
r,2
I
I
I
(e) Path 22
L
(8) Path 33
b
2
3
( f ) FBD Path 22
(h) FBD Path 33
(i) Block Shear Rupture for tension on Path 22
FIGURE 2.5 Staggered bolt configuration in a tension member
2.3 Effect of Staggered Bolt Holes on Net Area
55
Solution LRFD, p. 196: For a pair of L6 x 4 x 0.375, A, = 7.22 in.2 LRFD, p. 159: For L6 x 4 x 0.375, 3 = ( x = 0.941 in.) The governing design strength of the tension member is the least @P,value obtained from: 1. Yielding on A,
$P, = O.90F,,A8= 0.90(36)(7.22)= 234 kips 2. Fracture on A, = A,U
{
U = smaller of
lX/L, 0.9
= 10.941/6 = 0.843
d, = d + 1/8 in. = 1+ 0.125 = 1.125 in. th
dhth =
= 2(0.375) = 0.750 in.
1.125(0.750)= 0.844
We must compute A , for Sections 11and 22 in order to determine the critical net section. Both sections have fullP, on them, but we cannot tell by inspection which section has the smaller A,. Section 11
s=2in.
g = (g2= 2.5 in.)
(
[
A, = A , + C s * t ) =7.22+2 ( 2 ) 2 (2)(0*375)]= 782 k.2 4g 4(2.5) A, = A,  C(dhfh) = 7.82  2(0.844) = 6.13 in.* The critical section forfractureon A, is Section 22:
A, = A,U = 6.13(0.843) = 5.17 i n . 2
@P, = 0.75FdP= 0.75(58 ksi)(5.17
= 225 kips
3. Block shear rupture [see Figure 2.5(i)] L, = L,
+ 2(2s) = 2 + 2(2)(2) = 10 in.
A, = L,t = 10(2)(0.375)= 7.50 in.2 A,, = A ,
 A,,
= 7.50  2.5(0.844) = 5.39 i n . 2
L,=Zegg1=62.25=3.75in.
56
Tension Members
=3.75(2)(0.375)+
( 2 ) 2 (2)(0.375) =3.11 in.’ 4(2.5)
A,, = A,,  AholPs= 3.11  1.5(0.844)= 1.85 in.2 F,A,, = 58(1.85) = 107.3kips 0.6FJn, = 0.6(58)(5.39)= 185.6 kips
@P,,= 0.75(0.6FUA,,,+ FyA,,) = 0.75[185.6 + 36(3.11)]= 223 kips
@P,= 223 kips, due to BSR, is the governing design strength of the tension member.
The A36 steel member in Figure 2.6 is a pair of angles, L6 x 3.5x 0.5, with the long legs fastened to a gusset plate by 0.75in.diameterbolts. Use L, = 2 in., s = 1.5in.,g, = 2.25 in., g, = 2.5 in., and g = 2 in. The gusset plate thickness is 0.5 in. All crosssectional elements of the member contain bolts; therefore, A, = A, , For visualization purposes in computing A,,, one angle is shown flattened out in Figure 2.6(c). The design strength of the tension member is for a bearingtype bolted connection. Is the design satisfactory for P, = 275 kips?
Solution LRFD, p. 158:For a pair of L6 x 3.5 x 0.5, A, = 2(4.50) = 9.00 in.2 1. Yielding on A,
@P, = 0.90F,,AS= 0.90(36)(9.00)= 291.6 kips
(@Pn= 291.6 kips) 2 ( P , = 275) as required 2. Fracture on A , = A,,
d,, = d
+ 1/8 in. = 0.75 + 0.125 = 0.875 in. t,, = 2(0.5)= 1.00 in. d,t,, = 0.875(1.00)= 0.875 in.,
A, = A,  Z(d,,f,,) = 7.22  0.844 = 6.376 in.*
For the net section that passes through only hole B, A,, = A,  C(d,,t,,)= 9.00  0.875 = 8.125 in.2 @Pv= 0.75FUA,= 0.75(58)(8.125)= 353.4 kips
(@Pn = 353.4 kips) 2 (P, = 275) as required
2.3 Ejicect ofstaggered Bolt Holes on Net Area
pair of L6 x 3.5 x 0.5
(b) Gross section
7 P = P, (a) Member end FBD
+
"i
1 f
s=
1.25
3.75
1S O
T
7 P = P,
b
1
(c) FBD of member end flattened out into a plane
pu >
(d) BSR of member end flattened out into a plane FIGURE 2.6 Bolts in all crosssectional elements
57
58
Tension Members For the net section that passes through holes A and C,
A, = Ag  s ( d h f h ) = 9.00  2(0.875) = 7.25 in.'
@Pn= 0.75FJ1, = 0.75(58)(7.25)= 315.4 kips (@P,= 315.4 kips) 2 [(6/7)P, = 6(275)/7 = 235.7 kips] as required For the net section that passes through holes A and B,
$Pn = 0.75FdA,= 0.75(58)(7.225)= 314.3 kips (@Pn= 314.3 kips) 2 (P, = 275) as required For the net section that passes through holes A, B, and C [see Figure 2.6(c)],
@P,= 0.75FJe = 0.75(58)(6.375)= 277.3 kips
($P,
= 277.3 kips) 2 (P, = 275)
as required
3. Block shear rupture [see Figure 2.6(d)]
A , = CL$ = (8 + 6.5)(2)(0.5)= 14.50 in? A,, = A,  Aholes = 14.50  (1.5 + 2.5)(0.875)= 11.0 in.' A,, =
C(Ltt)+C
=(3.75+1.50)(2)(0.5)+
(1.50) (2N05) 5.475 in. 2 4(2.5) ~
A,, =Ast Aholes= 5.475  (1.5 + 0.5)(0.875) = 3.725 in.'
FA,, = 58(3.725)= 216.05 kips 0.6FJ,,
= 0.6(58)(5.39)= 382.8 kips
(bp, = 0.75(0.6F&,+ F,,Agt)
= 0.75I382.8
+ 36(5.475)]= 434.9 kips
($@, = 434.9 kips) 2 (P, = 275) as required
The design is satisfactory for P, = 275.
2.4 Design of a Tension Member With BoltedEnd Connections 2.4
59
DESIGN OF A TENSION MEMBER WITH BOLTEDEND CONNECTIONS For a tension member with bearingtypebolted connections at the member ends, the design requirement for strength is
W n 2 p, and the design strengths qP,,that must be considered are yielding on A,; fracture on A,; BSR; and bearing at the bolt holes. The latter is accounted for when the connection design strengths are computed in Chapter 3. In Examples 2.1 to 2.4,we found in three of the four examples that @Pndue to BSR governed and was only slightly smaller than qP,,due to fractureon A,. Lf BSR governsthe design,we usually can easily make some changes in the connection layout to increase the design strength to a satisfactory level. Consequently, in the design of a tension member with bearingtype bolted connections at the member ends, we recommend the followingitemized approach: 1. Tentatively choose the bolt diameter, number of bolts, spacing center to center between the bolt holes, and the end edge distance. (Note: This step is done by the author for the reader since concepts in Chapter 3 must be used in making the tentative choices.) 2. Assume that @Pndue to BSR does not govern. 3. Using the information given in item 1, pick a trial section that has adequate qP,,for both of the following cases:
a. Yielding on Ag
($Pn = 0.90F,/,) 2 P,
A, 2 Pu/(0.90F,,) b. Fmcture on Ae = AnLl (illustrated for the most general case) (@,, = 0.75FAJ 2 P,
A, 2 Pu/(0.75F,)
Ln U = (1 X / L , ) 5 0.9, X is unknown before the section has been chosen and L, is tentative until all details of the bolt group have been finalized. For the design of tension members, the following U values (from LRFD CB3, p. 6172) are good first estimates: (i) U = 0.85 when longer line of bolts parallel to Puhas 2 three bolts. (ii) U = 0.75 when longer line of bolts parallel to Pyhas only two bolts. 4. Compute @Pndue to BSR to check our assumption. a. If @Pn2 Pu,the trial section is satisfactory and it becomes the chosen section. Exit the design process.
60
Tension Members b. If W,,< P,, increase $P,, by either changing some of the previously chosen values in item 1 (to increase the length of the shear plane, e.g.) or by choosing a section whose elements in which BSR occurs are thicker. Now, compute I/ = ( 1 2 / L , ) I 0.9 and compare it to the assumed U value. If they differ by more than 2% compute @Pndue tofracture on Acand check the strength design requirement for fracture on Ae. If necessary, repeat step 4 until the design is satisfactory. Examples 2.5,2.7, and 2.8 discuss the design of tension members in the “truss region” of the structure shown in Figure 1.15. Stability of this structure in the horizontal direction and resistance due to wind loads are ensured by bending of the columns (members 1 to 4). The trusstocolumn connections at joints 2,3,22, and 23 will necessarily have to account for bending of the columns, which causes some bending in the “truss members” attached to those joints. Consequently, in the plane frame analysis of this structure (see Appendix A), the author did not release the moment at the end of any member. Therefore, for some of the members in the ”truss region” of this structure, the bending effects cannot be ignored. In the final design check of such members, P , and Mu must be accounted for simultaneously as discussed in Chapter 6. In the preliminary design phase (selecting trial member sizes) of such members, we account for Muby using an equivalent P,.
For members 34 and 43 in Figure 1.15, select the lightest available doubleangle section with long legs back to back (see LRFD, p. 198)of A36 steel that can be used in a bearingtype bolted connection. For the connection layout shown in Figure 2.1, the author has determined that the following values are required for the connection design to be satisfactory with standardsize bolt holes: d = 0.75in.diameterbolts; s = 3in. bolt spacing; L, = 1.5in. end edge distance, and each angle thickness t 2 3/16 in. for strength due to bearing at the bolt holes of the member. Note that the length of the connection is L, = L, + 2s = 1.5 +2(3) = 7.5 in. From Appendix A, for members 34 and 43 due to loading 8, we find P , = 66.4 kips (tension)and Mu= 0.16 ftkips. In the final design check of these members, P , and Mu must be accounted for simultaneously as discussed in Chapter 6. In the preliminary design phase of a tensionplusbending member, we account for Mu by using an equivalent P,; in this case, we know (from Chapter 6) that a 10% increase in P, is adequate. Try equivalent PI, = l.lO(66.3) = 72.9 kips. Solutiorr
Assumptions 1. Try U = 0.85 since there are three bolts in the longer line of bolts. 2. @P,[due to BSR does not govern.
For yielding on A,, the design requirement is
2.4 Design of a Tension Member With BoltedEnd Connections
61
(@Pn= 0.90F@,) 2 (equivalent P, = 72.9 kips) 7
L
~
0.9F,,
72.9 = 2.25 in.= 0.9( 36)
Forfracture on A, = A,U, the design requirement is
[@P,= 0.75F, (A,U)] 2 (equivalent P , = 72.9 kips)
A,, 2 [
"
0.75FuU
=
72.9 = 1.97 in.* 0.75( 58)(0.85)
1
dhth= (0.75 + 0.125)t = 0.875t A , = A,  2(0.875)t = A,  1.75t A, 2 (A,,+ 1.75t = 1.97 in.2 + 1.75t) Summary of the design requirements I. t 2 3/16 in. for each angle due to bearing at the bolt holes. 2. A, 2 2.25 in.*for yielding on A,.
3. A, 2 (1.97
+ 1.75t) forfracture on A,.
(a) For t = 3/16 in., we need: A, 2 [1.97 + 1.75(0.1875)= 2.30 in.2]2 2.22 in.2 Since a pair of angles with long legs back to back is to be chosen, pick the trial section from LRFD, pp. 196 to 199. No available section for t = 3/16 in. has All 2 2.30 in2 (b) For t = 1/4 in., we need: A, 2 [1.97 + 1.75(0.25)= 2.41 in.*]2 2.22 in.2
Try a pair of L3 x 2.5 x 1/4, (Ax= 2.63 i n 2 )2 2.41 in.2U = 0.85 was assumed; check this assumption. From Table 2.1 for leg = 3 in., we find g = 1.75 in. From LRFD, pp.1 64 and 165, for a single L3 x 2.5 x 1/4, we find x = 0.661 in. and y = 0.911 in. that are needed to determine the governing connection eccentricity for each angle. From Figure 2.3(b), we see that (1)the connection eccentricity from the face of the gusset plate is x = 0.661 in.; and (2) the other connection eccentricity for a supposed long leg = g must be computed. The yedge distance is ye= (leg g) = (3  1.75)= 1.25 in.
For simplicity in the following computations, a prime is placed on all variables for the supposedangle section. A' = A  yet= 2.63/2 (1.25)(1/4) = 1.0025 in.2
62
Tension Members
y‘=
AY  Y ef(leg0.5Ye

A’
(m)( 0.911)  1.25(0.25)( 3  1.25/ 2) = 0.456 in. 1.oo
X = larger of
U = smaller of
x = 0.661 in.
gy’= 1.750.456 = 1.29in.
1X/L, 0.9
{
= 11.29/7.5 = 0.828
Since (U= 0.828) < (assumed U = 0.85), revise A, required forfracture on A, to A, 2 [1.97(0.85)/0.828+ 1.75(0.25)= 2.46 in.2]
Try a pair of L3 x 2.5 x 1/4, (A,
= 2.63 in?) 2 2.46 in?. This is the original trial section. We cannot find any lighter pair of angles for f = 1 / 4 in. Weight = 9.0 lb/ft.
(c) For t = 5/16 in. and U = 0.85, the Ax requirement is
A, 2 t1.97 + 1.75(0.3125) = 2.52 in.2]2 (2.22 in?).
Try a pair of L2.5 x 2 x 5/16, (A, = 2.62 in?) 2 2.52 in.2 Check our assumption that U = 0.85: ye= (leg g)= (2.5  1.375) = 1.125 in.
A‘ = A yet = 2.62/2 (1.125)(5/16)= 0.958 in?
TABLE 2.1 Usual gages in angle legs, inches
b4 g
Leg
s 81
g2
7
8 1
42
1
3
4 1
3
2
3
3
2
From LRFD, p. 913.
2 21 4 1
2
2
1
4
5
6
3, 1
3
2,
2
13 4
2
1
3
2, 3
14
3
18
3
2 1
18
1
3
1
I4
12
18
14
1
7
_7
3 
8
8
4
1 5 8
2.6 Design of u Tension Member With WeldedEnd Connections
63
~(1.31)(0.809)  1.125(0.3125)( 2.5  1.125/ 2) = o.395 in y ' = Ay  y e t(leg  0 . 5 )~A' 0.958
X = larger of
U = smaller of
x = 0.559 in.
8  IJ ' = 1.375 0.395 = 0.980 in.
{
lX/L, 0.9
= 10.980/7.5=0.869
(U= 0.869) > (assumed U = 0.85). Revise A, required forfiucture on A, to A, 2 [1.97(0.85)/0.869 + 1.75(0.3125)= 2.47 in2]
Try L2.5 x 2 x 5/16, (As = 2.62 2 2.47 in2.This is the original trial section. We cannot find any lighter pair of angles for t = 5/16 in. Weight = 9.0 Ib/ft. Conclusion: Two acceptable sections of equal weight have been found that satisfy the design requirements thus far. Our preference is a pair of L3x 2.5 x 1/4 with long legs backtoback. Now we must compute @', due to BSR for the trial section to check our assumption that BSR does not govern the design selection. See Figure 2.7: dhth
= (0.75 + 0.125)(2)(0.25)= 0.4375 in?
L, = 1.5 + 2(3) = 7.5 in.
A* = (7.5)(2)(0.25)= 3.75 i n . 2 A,, = 3.75  2.5(0.4375) = 2.656 in? A,, = (1.25)(2)(0.25)= 0.625 i n . 2 A,, = 0.625  0.5(0.4375) = 0.406 in?
FA,, = 58(0.406)= 23.55 kips 0.6FAn, = 0.6(58)(2.656)= 92.43 kips #Pn = (0.6F,,An,+ F,,A$) = 0.75[92.43 + 36(0.625)]= 86.2 kips (@,' = 86.2 kips) 2 (P,= 66.3 kips)
As we assumed, BSR does not govern our choice of the section.
Conclusion: Use a pair of L 3 x 2.5 x 1/4; weight = 9.0 lb/ft. 2.5
STRENGTH OF A TENSION MEMBER WITH WELDEDEND CONNECTIONS Consider Figure 2.8, which shows a tension member fastened by fillet welds to a gusset plate. Along the member at some finite distance from the welds, as shown
64
Tension Members pair of L3 x 2 x 0.25
L"
7 1.9 kips
FIGURE 2.7 Block shear rupture of bolted member end
in Section 1.1.4 and Figure 1.4(c), all crosssectional fibers can attain the yield strength, when the welds and gusset plate are stronger than the member. In the region of the memberend connection, an edge on one leg of each angle in Figure 2.8(c) is not welded to the gusset plate. Therefore, the stress distribution due to the applied load is not uniform in the member end. However, the LRFD definitions account for a shear lag effect in the member end when only transverse welds are used at the member end to transfer the force in the member to the gusset plate. A transition region exists from the connection region to some finite distance from the connection where the stress distribution in the member becomes uniform when yielding occurs in the member. Thus, before yielding occurs in the member, the connection region of the member end usually experiences strainhardening, and fracture can possibly occur in the region where strainhardening occurs due to shear lag. The LRFD design strength definitions of a tension member without any holes in it and with filletwelded end connections are: 1. Yielding on A, [see Figure 2.8(b)] (LRFD D1, p. 644) @,' = 0.90FflP,
2. Fracture on A, (LRFD D1 p. 644) @,' = 0.75FJ1,
where Ap,the effective area, is to be determined from: (a) When holes exist in the member (LRFD D1, p. 644), A , = A,, (b) When holes do not exist in the member (LRFD B3, p. 634),
A,= AU where A and U are to be determined from: (i) When the weld group consists only of transverse welds, A
= area
of the directly connected elements
u = 1.0
2.5 Strength of a Tension Member With WeldedEnd Connections
1


65
Gusset plate
on 11
'1 Gusset plate: 5 x 5/8: Fillet welds: 0.25 in.;
Fillet weld
Member: pair ofL3 x 2 x S/16 L , = 4.50 in.; L = I .SO in.; L = 3.OU i n
(a) Member end connection detail
(d) Section 33
(c) Section 22
j"qJJp L =L A =shadedarea +4
(e) End View 44
(t) Block shear rupture of meinhcr end
FIGURE 2.8 Tension member fillet welded to a gusset plate
(ii) When the member is a single plate fastened by only longitudinal welds for ( L , = length of longitudinal weld) I (w= width of plate),
A =A, U = 1.00 when L, 2 2w
U = 0.87 when 2w > L,2 7 . 5 ~ U = 0.75 when 1 . 5>~LL2 70 (iii) When preceding item (i) or (ii) is not applicable,
A = A,
u = (1 X/L,) 5 0.9
66
Tension Members
L, = length of longest longitudinal weld
X = connection eccentricity 3. Blockshear rupfure, abbreviated BSR (LRFDJ4.3, p. 687)[seeFigure 2.8(e)for an example]BSR design strength is a functionof the weld group arrangement in the weldedend connections and the thickness of the block in the member end. Weldedend connections are discussed in Chapter 3.
Sinceall cases we will discuss have no holes in the members, we can simplify the BSR definitions as shown below:
A, = gross area on BSR shear plane(s) (in.2) A, = gross area on BSR tension plane (in?) When F,,A, 2 0.6FfiU,
qP, = O.75(Ffif + 0.6F8,) When 0.6FfiU 2 F,A,,
$P, = 0.75(0.6F,,A, + F 8 , )
EHampte 2.6 A fillet weld group arranged as shown in Figure 2.8(a) is used on each long leg of a pair of L3 x 2 x 0.3125 of A36 steel to fasten this tension member to a 0.625 in. thick gusset plate. Find the governing design strength of the tension member with weldedend connections. Solution
LRFD, p. 198: For a pair of L3 x 2 x 0.3125, A, = 2.93 in.2 The governing design strength of the tension member is the least (PP, value obtained from: 1. Yielding on Ag = 2.93
(PP, = 0.90F& = 0.90(36)(2.93)= 94.9 kips 2. Fracture on A, = A,U
(PP, = 0.75F,,Ap,
U
= smaller of
{
1 X / L , = 1 0.516/ 4.50 = 0.885 0.9
= 0.75(58)(2.93)(0.885)= 114.0 kips
3. Block shear rupfure of the member end [see Figure 2.8(e)]
A, = L, t = 4.5(2)(0.3125)= 2.8125 in.2
2.6 Design of u Tension Member With WeldedEnd Connections
67
A , = L, t = 3(2)(0.3125)= 1.875 in.’ FJt = 58(1.875) = 108.75 kips 0.6FJP, = 0.6(58)(2.8125)= 97.875 kips
4Pn = 0.75(F,,Ar+ 0.6F,,AV)
$Pn= 0.75t108.75 + 0.6(36)(2.8125)]= 127.1 kips For the tension member shown in Figure 2.8, the governing design strength is 4Pn= 94.9 kips. 2.6 DESIGN OF A TENSION MEMBER WITH WELDEDEND CONNECTIONS For a tension member with weldedend connections, the design requirement for strength is W n 2 p,
and the applicable design strengths are yielding on A,, fiacture on A,, and BSR. However, if yielding on A, is not the governingcase, we usually can easily make some changes in the connection layout to increase the other design strengths to a satisfactory level. Consequently, in the design of a tension member with weldedend connections, we recommend the following itemized approach: 1. Assume that yielding on A, governs; pick a trial section that satisfies:
(4Pn = 0.90F,,Ag)2 P, Ag 2 P,/(0.9OFy) 2. Tentatively choose the weld size and weld lengths. (Note:This step is done by
the author for the reader since concepts explained in Chapter 3 must be used in making the tentative choices.) 3. Compute qPndue tofracture on A, and BSR:
(a) If 4Pn2 P , our trial section is satisfactory, it now becomes our chosen section, and we exit the design process. @) If @Pn< P , we must increase $Pn by either changing some of our previously chosen values in item 2 (to increase the length of the shear plane, e.g.) or choosing a section with thicker elements in those elements in which BSR occurs. Repeat step 3 until o w design is satisfactory.
Repeat Example 2.5 except, as shown in Figure 2.9, the long legs of each angle section are to be welded to a gusset plate. The longer longitudinal weld length is 4.50 in. and 3/16in. fillet welds are used. From Example 2.5, for members 34 and 43 in Figure 1.15,
Equivalent P, = 72.9 kips
68
Tension Members
Typical member section
(a) Joint 12 of Figure 1.15
(b) Section 11
FIGURE 2.9 Welded truss joint details for doubleangle members.
Solution Assume that yielding on A, is the governing case; we need: (@,, = 0.90F8,) 2 (equivalent P, = 72.9 kips)
A, 2 (72.9/[0.90(36)]= 2.25 in.*)
Try a pair of L3 x 2 x 1 / 4 : (Ax= 2.38 in2)2 2.25; weight = 8.1 lb/ft. When the final design check is performed in Chapter 6 for members 34 and 43 in Figure 1.15 as a tensionplusbending member with weldedend connections, we will need the governing @Pnwhich is the least of
1. For yielding on A, = 0.90F8, = 0.90(36)(2.38) = 77.1 kips
2. Forfracture on A, = A,U
U = smaller of
{
lX/L, 0.9
= l0.493/4.50 = 0.890
@P,,= 0.75FUA,= 0.75(58)(2.38)(0.890) = 92.1 kips 3. For BSR of the member end (see Figure 2.10) A, = L, t = 4.50(2)(0.25) = 2.25 in.2 A, = L, t = 3.00(2)(0.25) = 1.50 in.2
0.60F&
= 0.6(58)(2.25)= 78.3 kips
F,A, = 58(1.50) = 87.0 kips
2.6 Design o f a Tension Member With WeldedEnd Connections
69
A = shaded area /
(a) End View 22
(b) Block shear rupture of member end
FIGURE 2.10 Block shear rupture of welded member end.
@P,,= 0.75(FUA,+ 0.6F,,AP,)= 0.75[87.0 + 0.6(36)(2.25)]= 101.7kips ($P,, = 102 kips) 2 ( P , = 66.3 kips) The BSR design strength is more than adequate. Tentatively select a pair of L3 x 2 x 1/4 for members 34 and 43 for which the governing $P, = 77.1 kips is applicable in Chapter 6 when this trial selection is checked as a tensionplusbending member with weldedend connections.
txample 2.8 See Figure 1.15 and consider the design of a bottom chord member in the truss portion of thisstructure. Asshown inFigure2.11, the top and bottom chord members of the truss are to be WT sections of A36 steel. Each truss web member is to be a doubleangle section with long legs back to back and fillet welded to the WT chord members at each truss joint. Assume that the same WT section is to be used in Figure 1.15 for members 5 through 14. From Appendix A, for member 10 due to loading 7, we find P , = 114.2 kips (tension) and M u = 1.62 ftkips. In the final design check of these members, P, and M,, must be accounted for simultaneously as discussed in Chapter 6. In the preliminary design phase of a tensionplus bending member, we account for M u by using an equivalent P,; in this case, we know (from Chapter 6) that a 20% increase in P, is adequate. Try Equivalent P, = 1.20(114.2) = 137 kips. Select the lightest available WT7 section of A36 steel for which @P, 2 137 kips. Soltitior1
Assume that yielding on A, governs @P,; the strength design requirement is (@P,= 0.90FPPR) 2 (equivalent P, = 143 kips)
A, t { 137/[0.90(36)] = 4.23 i n 2 1
Fracture on A , is not applicable. BSR is not applicable. However, BSR of the web of the chosen WT is applicable in the
70
Tension Members
r2
4
(c) Section 22
2
(a) Joint 12 of Figure 1.15
t
Ll
n (b) Section 11
FIGURE 2.11 Truss joint details for WT bottom chord member.
design of members 25 to 43 in Figure 1.15. Tentatively select a WT7 x 15 for which:
@P,,= 0.90F,,Ag = 0.90(36)(4.42)= 143 kips @,' = 143 kips is applicable in Chapter 6 when this trial selection is checked as a tensionplusbending member with weldedend connections. 2.7 SINGLEANGLE MEMBERS On pp. 6277 to 6300 of the LRFD Manual, we now find a separate, supplementary specification that deals only with the design of singleangle members. We chose not to discuss singleangle members in this chapter since they usually are subjected to combined bending and axial force, a topic discussed in Chapter 6. 2.8 THREADED RODS
As shown in Figure 1.16, cross braces in roofs and walls may be designed as tension members to resist wind and to provide overall structural stability in a threedimensional sense for gravitytype loads. If the roof slope in Figure 1.14 had been chosen to be greater than about 15", sag rods might be designed as tension members to provide lateral support for the weak axis of the purlins. If sag rods were used in Figure 1.14, they would be perpendicular to the purlins and parallel to the roof surface. They would function in a manner similar to the saddle and stirrups for a horseback rider when the rider stands up in the stirrups. Each end of a sag rod would be threaded and passed through holes punched in each purlin web. A nut would be used on each end of a sag rod for anchorage. Adjacent sag rods would be offset in plan view about 6 in. or less to accommodate installation.
2.9 Stiffness Considerations
71
If rods are chosen as the tension members for cross bracing, a turnbuckle (see LRFD, p. 894) may be used at midlength of the rod in order to take up slack and to pretension the rod. At each end of the rod, clevises (see LRFD, p. 892)or welds may be used to fasten the rod to other structural members. The opening paragraph of LRFD Chapter D (p. 644) states that Sec. J3(p. 679) is applicable for threaded rods. LRFD Table J3.2 (p. 681) gives the tensile design strength of a threaded rod as @,’ = 0.75(0.75F&).
Select threaded rods for the cross braces shown in Figure 1.16 using A36 steel.
Solufion Figure 1.14 gives the nominal wind load on the ends of the building as 12psf pressure on the windward end and 7.5psf suction on the leeward end. The total factored wind load on the building ends is 1.3(0.012+0.0075)[60(26.75)]= 40.7 kips. At least half of this 40.7kip wind load should be applied at the roof level, and the remainder of the wind load is applied at the foundation level. Two pairs of crossbraces are shown in Figure 1.16@).Hence, there are four identical pairs of cross braces that resist wind in the length direction of the building. At any time, only onemember in each pair ofcrossbraces is in tension due to wind. The other member in eachpair of crossbraces is in compressiondue to wind and buckles at a negligiblysmall load.When the wind reverses, the other member in each pair of crossbraces is in tension and resists the reversed wind. Consequently,for each tension member:
L = J(25.5)’ +(30)’ P,
= 39.37ft.
= (40.7/2)(339.37/30)/4= 6.68 kips
and the strength design requirement is [@,, = 0.75(0.75F&)] 2 (P, = 6.68 kips)
’[
P” 6.68 = 0.205 in.’ 0.75(0.75FU) 0.75(0.75)(58)
1
mi2
in.’ 4 d 2 0.453 in. A 1/2 in. diameter rod is acceptable, but no less than a 5/8in.diameter rod is preferred forease of handling during construction.Therefore,use a 5/8in.diameter threaded rod A36 steel. 2 0.205
2.9 STIFFNESS CONSIDERATIONS
After the structure in Figure 1.15 is erected, from the factored load combinations in Appendix A, we find that members 34 and 43 are required to resist a maximum axial tension force of 66.4 kips and a maximum axial compression force of 5.14 kips. In the
72
Tension Members fabrication shop and in shipping the prefabricated truss, a crane is used to lift the prefabricated truss. Unless special lifting procedures are used, members 34 and 43 may be required to resist a larger compressive axial force due to lifting than is required after the structure iserected. The design strength of a compression member is given in Chapter 4. A fundamental parameter in the design strength definition of a pinnedended compression member is the maximum slenderness ratio Llr of the member, where L is the member length and r is the minimum radius of gyration for the cross section of the member. As shown in Appendix B, the definition of radius of gyration for any crosssectional axis x is
rx = where
I,
= moment of inertia about the xaxis
A = gross crosssectional area 1 and r are minimum for the minor principal axis, and are needed to obtain the
maximum slenderness ratio. LRFD B7 (p. 637) states that Llr S 300 is preferable for a tension member except for threaded rods. LRFD Commentary 8 7 (p. 6177) states that the reason for this advisory upper limit on L l r is to provide adequate bending stiffness for ease of handling during fabrication, shipping, and erection. If the tension member will be exposed to wind or perhaps subjected to mechanically induced vibrations, a smaller upper limit on L l r may be needed to prevent excessive vibrations.
For the tension member designed in Example 2.7, compute the maximum slenderness ratio.
Solution From Example 2.7 we find that 1. Member 43 of Figure 1.15 was designed; for this member; L = 90 in. 2. The design choice was a pair of L3 x 2 x 1 / 4 with the long legs separated by and welded to a gusset plate at each member end. For the purposes of this example, assume that thc gusset plate thickness is 3/8 in. See LRFD, p. 198, which shows a sketch of a doubleangle section a n d lists the properties of sections for the Xand Yaxes which are principal axes since the Yaxis is an axis of symmetry. Therefore, for member behavior as a pair of L3 x 2 x 1/4 separated 3/8 in. back to back, the minimum r is the smaller of rr = 0.957 in. and ry = 0.891 in. For behavior as a pair of angles, the m n Y n z u t ? i L / r = (90 in.)/(0.891 in.) = 101, which is much less than the preferred upper limit of 300 for a tension member. However, note that the gusset plate exists only at the member ends. Elsewhere along
Problems
73
the member length, there is a 3/8in. gap between the long legs of the pair of angles. Doubleangle behavior is truly ensured only at the points where the pair of angles is tied together (by the gusset plates a t the member ends). Therefore, we need to determine the individual behavior of each angle. For one L3 x 2 x 1/4, the minimum r is rz = 0.435 in. (from LRFD, p. 165) and the irzaxinzuriz L / r = 90/0.435 = 207, which is less than the preferred upper limit of 300 for a tension member. If we insert a 3/8in. plate between the long legs at the midlength of the member and weld the long legs to this plate, we will ensure doubleangle behavior at midlength of the member as well as at the member ends. Then, the length for singleangle behavior is L = 90/2 = 45 in. and the maximum L / r = 45/0.435 = 103.5 for singleangle behavior. Since 103.5 is very nearly equal to 101 (for doubleangle behavior), we can conclude that we only need to insert and weld a spacer plate at midlength of the member in order to ensure that doubleangle behavior is valid. PROBLEMS 2.1
A36 steel; 7/8in. diameter A325N bolts
s = 2.75 in.
L , = 1.5 in.
g,= 3 in.
y2 = 3 in.
The tension member is a pair of L8 x 6 x 1/2 with the long legs bolted to a 10x 1gusset plate. For the member, compute the design strength due to: 1. Yielding on A, 2. Fracture on A, 3. Block shear rupture Is the design satisfactory for P,, = 298 kips? 2.2
A36 steel; 7/8in. diameter A325N bolts
s = 1.5 in.
L, = 1.5 in.
8, = 2.25 in.
R~= 2.50 in.
The tension member is a pair of L6 x 4 x 1/ 2 with the long legs bolted to a 9 x 1 gusset
L FIGURE P2.1
1
Section 11
74
Tension Members plate. For the member, compute the design strength due to: 1. Yielding on A, 2. Fracture on A, 3. Block shear rupture: Failure mode has a staggered tension path and one shear plane (L, = L, + 6s). Is the design satisfactory for P, = 270 kips?
I
I
I
1
I
I Section 11
FIGURE P2.2
2.3
A36 steel; 1in. diameter A325N bolts S=
3 in.; g = 3 in.; L, = 1.75 in.
The tension member is a C15 x 33.9 bolted to a 16 x 5/8 gusset plate. For the member, compute the design strength due to: 1. Yielding on A, 2. Fracture on A, 3. Block shear rupture Is the design satisfactory for P, = 200 kips?
L,1 FIGURE P2.3
Section 11
Problems A36 steel
2.4
s = 3 in.
75
1in.diameter A325N bolts
g = 3 in.
L, = 1.75 in.
The tension member is a C15 x 33.9 bolted to a 16x 5/8 gusset plate. For the member, compute the design strength due to: 1. Yielding on A8 2. Fracture on A, 3. Block shear rupture
Is the design satisfactory €orP, = 199 kips?
Section 33
1
3
2‘
L 1
Paths forfracture on A
Case 3
FIGURE P2.4
Case 2 Cases for Block Shear Rupture
Case 1
76 Tension Members
A36 steel; 3/4in.diameter A325N bolts
2.5
s = 3 in.
g = 3 in.
L,
=
1.75 in.
A pair of 10 x 3/8 connector plates is used to butt splice the tension member that is a 10 x 3/4 plate. For the member, compute the design strength due to: 1. Yielding on A, 2. Fracture on A, 3. Block shear rupture Is the design satisfactory for P,

= 240 kips?
p,
2@g
OII.
1
Secuon 1  1
FIGURE P 2.5
A36 steel; 3/4 in diameter A490X bolts
2.6
s = 3 in.
g = 3 in.
L, = 1.75in.
A pair of 10 x 3/8 connector plates is used to butt splice the tension member which is a 10 x 3/4 plate. For the member, compute the design strength due to: 1. Yielding on A, 2. Fracture on A, 3. Block shear rupture Is the design satisfactory for P,, = 240 kips?
I: f Section 1  1
FIGURE P 2.6
we
Problcnrs
77
A36 steel; 1indiameter A325N bolts
2.7
s = 3 in.
= 5.5 in.
L 4.77, weld fracture govems.
3. Shear rupture of the gusset alongside two welds:
@R,,/in. = 0.75(0.60FU)f= 0.75(0.60)(58)(0.625)= 16.3 kip~/in. Since 16.3 > [2(4.77)= 9.541, weld fracture govems. For a transverse weld, 1. Weld fracture:
@R,,/in. = (4.77 kips/in.)(1.5) = 7.16 kips/in. 2. Tension rupture of the gusset alongside two welds:
@R,,/in.= 0.75FJ = 0.75(58)(0.625)= 27.2 kip~/in. Since 27.2 > [2(7.16)= 14.31, weld fracture governs.
3.9 Fillet Welds 103 For the fillet weld group, the design strength is C[((PR, /in.)L,] = 2[(4.77)(4.50+ 1.50) + (7.16)(3.00)]= 100.2 kips 3.9.2 Design of Fillet Welds
The maximum fillet weld size that can be deposited in one pass is S, = 5/ 16in. Most of the heat given off in the welding process is absorbed by the parts being joined. Heat in the deposited weld is absorbed faster by the thicker part being joined. Ductility is adversely affected when the weld metal cools too rapidly. Also, the thicker base material is stiffer and provides more restraint to shrinkage of the deposited weld during cooling. To ensure that enough heat is available in the deposited weld for proper fusion of the weld to the base material, a minimum fillet weld size is specified in LRFD Table J2.4 (p. 675) as a function of the thicker part being joined, and L, 2 45, is stipulated in LRFD J2.2b (p. 675). L, > 705, is not permitted for a longitudinal weld. End returns [shown in Figure 3.10(d) and numerically illustrated in Example 3.71 are not required, but they increase the ductility of the connection. When an end return is not provided, the distance from the weld termination point to the end of the connected part must be at least S,. Other limitations and reasons for them are given, respectively, in LRFD J2.2b (p. 675) and CJ2.2b (p. 6220). Only the design of SMAW fillet welds will be illustrated. Figures3.10 (ce) show the possible weld arrangements on the end of a pair of L2.5 x 2 x 5/16 used as a tension member. In Examples 3.6 to 3.8, we will design SMAW fillet welds for each of the weld arrangements to resist P, = 84.9 kips. We will use E70 electrodes and A36 steel for the member and the gusset plate (f = 3/8 in.).Information applicable ineach of the examples is: 1. For thicker part joined, f = 3/8 in. (gusset plate), minimum S, = 3/16 in. 2. f=5/16in. oftheanglesectiongovemsmaximumSW=5/161/16= 1/4in. 3. We choose to use S, = 1/4 in. 4. For a longitudinal weld:
(a) Weld fracture:
(PR,,/in.= 0.75(0.60F,)(0.7O7Sw) = 0.75(0.60)(70)(0.707)(0.25) = 5.57 kips/in.
(b) Shear rupture of the member alongside a weld:
(PR, /in.= 0.75(0.60FU)t= 0.75(0.60)(58)(0.3125)= 8.16 kips/in. Since 8.16 > 5.57, weld fracture mode governs. (c) Shear rupture of the gusset alongside two welds:
(PR,,/in. = 0.75(0.60FU)t= 0.75(0.60)(58)(0.375)= 9.79 kips/in. Since 9.79 < [2(5.57)= 11.141, shear rupture mode governs.
104 Connections for Tension Members
il
Typical member scctiori
(a) Joint side elevation view 3Pair
.
(b) Section I  I
of L2.5 x 2 x 5/16
L =L2
End returns
(Pair of L2.5 x 2 x 5/16)
L = L2
?
3
4
N
(c) Shortest weld arrangement
L
2
(d) Welds with end returns x 2 x 5/16
I .25 in.

(member end FBD)
PI
(2.5  y) = I .69 in ,;/ = 84.0 kips
T
J  0.809 in p* (e) Balanced weld arrangernentc.g. of weld forces and member coincides
FIGURE 3.10 Welded truss joint details for doubleangle members
5. For a transverse weld, (a) Weld fracture:
@R,?/in. = (5.57 kips/in.)(l.5) = 8.35 kips/in. (b) Tension rupture of the gusset alongside two welds:
@R,,/in. = 0.75FJ = 0.75(58)(0.375)= 16.3 kips/in. Since 16.3 < [2(8.35)= 16.71, tension rupture mode governs.
Note: This failure mode is not listed in LRFD Table J2.5, but it can govern as shown when Appendix J2.4 is used to obtain the design weld strengths.
3.9 Fillet Welds 105 6. Other design strength values for the design in Figure 3.10(a)are: (a) For yielding on A8 of the member, $Pn= 84.9 kips. (b) Forfractureon Aeofthemember, Lc> 2.19 in. is required for GP"284.9kips. (c) For BSR of the gusset, LD2 3.00 in. is required for (pP,,2 84.9 kips.
EHample 3.6 For the weld arrangement shown in Figure3.10(c),note that L, = 2.50 in. is given and L, = L, is required. Design the welds for P,, = 84.9 kips using the applicable information in the last paragraph preceding this example.
Solution From items 5(b)and 4(c)of the list preceding this example, the design strength of the fillet weld group is C[($R, /in.)L,,] = (16.3)(2.50)+ (9.79)(L2+ L3) = 40.75 kips + 19.58L2
The strength design requirement is
c[(@R,, /in.)L,,l 2 p, (40.75 kips + 19.58LJ 2 84.9 kips
L, 2 2.25 in L, 2 ( L , 2 3.00 in.) for the required BSR strength [see item 6c] Use L,
= L, = 3.00 in.
for each L2.5 x 2 x 5/16.
For the arrangement shown in Figure 3.10(d),note that there are end returns and L, = L,. Using the minimum permissible length for each end return and the applicable information preceding Example 3.6, design the welds for P, = 84.9 kips. Solution Let L,, = length of an end return. From LRFD J2.2b, at the free end of a tension member: 1. L,, 2 [2S, = 2(0.25) = 0.5 in.] is required. 2. If we want the end return to be fully effective in our weld strength calculations, L,, 2 [4S, = 4(0.25) = 1.00 in.] is required. 3. When L,, < 4S,, then S, = L,,/4 (or, when rupture of the base material governs, onefourth of the design strength for rupture of the base material) must be used in the weld strength calculations.
106 Connectionsfor Tension Members For each end return, choose L, = 0.5 in. Since the end returns are transverse welds, (Le, = 0.5 in.) < [4S, = 4(0.25) = 1.00 in.], and item (5b) in the itemized list preceding Example 3.6 is applicable, the total usable design strength for the end returns is $Rn /in.= 2(16.3 k/in.)(0.5 in.)/4 = 4.08 kips Also, item (4c) in the itemized list preceding Example 3.6 is applicable. The design strength of the fillet weld group is C[(@R,/in.)L,] = 4.08 kips + (9.79)(L, + L3) = 4.08 kips + 19.58L,
The strength design requirement is
W $ R , /in.)L,12 p , (4.08 kips + 19.58LJ 2 84.9 kips
L, 2 4.13 in. L, 2 (L, 2 3.00 in.) for the required BSR strength [see item 6c] From LRFD J2.2b, we find that L, I [70S, = 70(0.25) = 17.5 in.] is required. Use L, = L, = 4.25 in. with 0.5in. end retums for each L2.5 x 2 x 5/16.
Eaample 3.8 The weld lengths L, and L, in a balancedweld arrangement are chosen such that the centroid of the weld forces coincides with P, which is the axial force in the member. For the balancedweld arrangement shown in Figure 3.10(e),note that L, > L,, and use the applicable information preceding Example 3.6 to design the welds for P, = 84.9 kips.
Solution From items 501) and 4(c) of the list preceding Example 3.6, the design strength of the fillet weld group is
P, = (16.3)(2.50)= 40.75 kips P, = 9.79L, P3 = 9.79L3 The strength design requirement is (PI + P,
+ P3) 2 ( P , = 84.9 kips) [40.75 + 9.79(L2+ L,)] 2 84.9 (L, + L3)]2 4.51 in.
At P, in Figure 3.10(e),Zh4 = 0 gwes
3.10 Connecting Elements in a Welded Connection 107 2.5P2+ 1.25P1= 1.69(PU) 2.5(9.79L2)+ 1.25(40.75)= 1.69(84.9) L, = 3.78 in. L, 2 (L, 2 3.00 in.) for the required BSR strength [see item 6(c)] L, 2 (4.51  3.78 = 0.73 in.) is required. L, 2 [4S, = 4(0.25) = 1.00 in.] also is required.
Use L, = 3.75 in. and L, = 1.00 in. on each L2.5 x 2 x 5/16. 3.10 CONNECTING ELEMENTS IN A WELDED CONNECTION A gusset plate (a memberend connector plate) may have an irregular shape [see Figure 3.10(a)]to accommodate the fastening of several member ends at a joint. The gusset plate shape in Figure 3.9(a) is the simplest form of a gusset plate. More than one connector plate or more than one connecting element may be used. Figure 3.8(c)shows examples of block shear rupture of the connecting element at the end of members 10 and 39. Block shear rupture is a tearing failure mode that can occur along the perimeter of welds. As shown in Figures 3.8(c)and (g), a block of materialcanbe tornout of theconnection.AsshowninFigure3.11,iffracturestarts to occur on the tension plane of the block, yielding simultaneously occurs on the shear plane@)of the block. Fracture occurs on the block plane(s) that provides the larger possible fracture force. In Figure 3.8(a), the connecting elements can fail by yielding on their gross sections.For the middle plate, block shear rupture can occur as depicted for members 10 and 39 in Figure 3.8(c). In Figure 3.6, the connecting elements can fail by yielding on their gross sections or by shear rupture between the pair of dotted lines along path aa. If width w1 of the middle plate is larger than implied, block shear rupture can occur for the middle plate as depicted for members 10 and 39 in Figure 3.8(c). In Figure 3.7, the connectingelements can fail by yielding on their gross sections bb and cc. If the width w,of the middle plate is large, block shear rupture can occur for the middle plate as depicted for members 10 and 39 in Figure 3.8(c). The LRFD design strength definitions for failure of the connectingelement(s)in a welded connection are: 1. Yielding on As of the connecting element@)[LRFD J5.2(a),p. 688] (see Figure 3.9 Section 33 for an example)
$Rn= 0.90FYA, where F , = yield strength Ag = gross area of the plate = tpbp t, = thickness of
the plate
b, = width of the plate perpendicular to P, 2. Block shear rupture (LRFDJ4.3, p. 687)
108 Connectionsfor Tension Members
(see Figure 3.8(c) for examples) For welded connections, there are no holes; therefore, let
A, = gross area on BSR shear plane(s) A, = gross area on BSR tension plane When F,,A,L 0.6FUA,,
$Rn= 0.75(FUA,+ 0.6F,,A,) When 0.6 F,,A,,>F,,A,
@Rn= 0.75(0.6F,A7,+ F,,A,) 3. Stwur rripfure strength (LRFD J4.1, p. 687). [see path aa on the plates in Figures 3.6(a) and (b) for an example]
$Rll = 0.75(0.6F,AV) and the parameters are as previously defined for block shear rupture.
hample 3.9 Find the design strengths of the 5 x 5/8 gusset plate of A36 steel in Figure 3.9(a).See Figure 3.11 for the BSR failure mode of this gusset plate.
Soliition 1. Yielding on A,? = 5(0.625) = 3.125 in.2
$R,l = 0.9FflP,= 0.9(36)(3.125)= 101.25 kips 2. Block shear rupture
F,,A, = (58 ksi)(4.25 in.)(0.625 in.) = 154.06 kips 0.60F,,4, = 0.60(58 ksi)(6.75in.)(0.625in.) = 146.81 kips
$Rn = 0.75[FI,A, + 0.6F,,A,] @Rn= 0.75[354.06 + 0.6(36)(6.75)(0.625)]= 183.9 kips For the design shown in Figure 3.9(a), a summary of our computed design strengths is:
1. For the gusset plate (from Example 3.9)
(a) (PR,,= 101.25 kips due to yielding on As (b) (PR,,= 183.9 kips due to BSR 2. For the weld group (from Example 3.5, (PR,,= 100.2 kips) 3. For the member (from Example 2.6)
Problems
109
Yield plane
Fracture plane Gusset plate I
i 7 
(a) Tensile fracture and shear yielding possibility
t.
PI
Lv
*I
Fracture plane
I \ Yield plane
Gusset plate I
(b) Tensile yielding and shew fracture possibility For the 5/16 by 5 in. gusset plate shown in Figure 3.Y(a). L, = 3.50 + ( 5  3.50)/2 = 4.25 in. L,, = 6.75 in.
Figure 3.11 Block shear rupture possibilities for a gusset plate. = 94.9 kips due to yielding on Ax (a) @Pn (b) @Pn = 114 kips due to fracture on A(, (c) #Pn = 127 kips due to BSR
The governing design strength is 94.9 kips, wtuch is the least of the design strengths, and the LRFD strength design requirement is (@Ptl= 94.9 kips) 2 P,.
PROBLEMS 3.1 A36 steel; 7/8in.diameter A325N bolts
s = 2.75 in.
L, = 1.5 in.
g,= 3 in.
g, = 3 in.
w,= 3 in.
The tension member is a pair of L8 x 6 x 1/2 with the long legs bolted to a 10 x 1 gusset plate. Compute the design strength for: 1. Bolt shear 2. Bearing at the bolt holes 3. The gusset plate (yielding on A,; fracture on Atl)
Is the design satisfactory for P , = 298 kips?
110 Connections for Tension Members
L
1
Section 11
FIGURE P3.1
3.2 A36 steel; 7/8in.diameter A325N bolts s = 1.5 in.; L, = 1.5 in.; g1= 2.25 in.; g, = 2.50 in.; W, = 2.25 in. The tension member is a pair of L6 x 4 x 1/2 with the long legs bolted to a 9 x 1 gusset plate. Compute the design strength for:
1. Bolt shear 2. Bearing at the bolt holes 3. The gusset plate (yieldingon A,; fracture on A, for each path with fullP, on it)
Is the design satisfactory for P, = 270 kips?
I I e
I
 1
Section 11
FIGURE P3.2
3.3 A36 steel; 1in.diameterA325N bolts
s=3in.; g = 3 i n . ; L,= 1.75in.; we=3.5in. The tension member is a C15 x 33.9 bolted to a 16 x 5/8 gusset plate. Compute the design strength for: 1. Bolt shear 2. Bearing at the bolt holes 3. The gusset plate (yielding on A,; fracture on A,)
Is the design satisfactory for P, = 200 kips?
Problems
PU
4
I
I
1
I
.
111

Lt?
I I
1
Section 11
FIGURE P3.3
3.4 A36 steel; 1in.diameter A325N bolts
s = 3 in. g = 3 in. L, = 1.75 in. W, = 3.5 in. The tension member is a C15 x 33.9 bolted to a 16 x 5/8 gusset plate. Compute the design strength for:
1. Bolt shear 2. Bearing at the bolt holes 3. The gusset plate (yielding on Ag;fracture on A,,)
Is the design satisfactory for P, = 200 kips?
Section 1 1
FIGURE P3.4
3.5 A36 steel; 3/4 in diameter A325N bolts
s = 3in. g = 3 in. L, = 1.75 in. A pair of 10 x 3/8 connector plates is used to butt splice the tension member which is a 10 x 3/4 plate. Compute the design strength for: 1. Bolt shear 2. Bearing at the bolt holes
112 Connections for Tension Members 3. The splice plates (yielding on A,; fracture on A,)
Is the design satisfactory for P, = 240 kips?
3.6 A36 steel; 3/4in.diameter A490X bolts
L, = 1.75 in. s = 3 in. g = 3 in. A pair of 10 x 3/8 connector plates is used to butt splice the tension member which is a 10 x 3/4 plate. Compute the design strength for: 1. Bolt shear 2. Bearing at the bolt holes 3. The splice plates (yielding on As;fracture on A, for each path with full P, on it )
Is the design satisfactory for P, = 240 kips?
3.7 A36 steel; 1in. diameter A325N bolts 1.75 in.; g = 5.5 in Each flange of the member(W8 x 31) is bolted to an 8 x 5/8 gusset plate at the member end. Compute the design strength for: s = 3 in.
L,
=
1. Bolt shear 2. Bearing at the bolt holes 3. The gusset plate (yielding on A,; fracture on A,)
Is the design satisfactory for P, = 280 kips?
Problem 113
Block shear rupture of each member tlangc
P,
Section 1  1
FIGURE P3.7
3.8 A572 Grade 50 steel; 1in. diameter A490X bolts
s = 3 in.; g = 7 in.; L , = 7.75 in. The tension member is a pair of C15 x 33.9 bolted to a pair of 10 x 5/8 gusset plates. Compute the design strength for: 1. Bolt shear 2. Bearing a t the bolt holes 3. The gusset plates (yielding on A,; fracture on All; block shear rupture)
Is the design satisfactory for PI,= 495 kips?
w asp,,
k
.I Section 1  I
FIGURE P3.8
114 Connections for Tension Members
3.9 A36 steel; 7/8in.diameter A325N bolts
s = 1.50 in.; L, = 1.5 in.; g1 = 2.25 in.;g, = 2.50 in.; g = 2.50 in. The tension member is a pair of L6 x 4 x 1/2 with all legs bolted to 3/4 in. thick splice plates. Compute the design strength for: 1. Bolt shear 2. Bearing at the bolt holes
Is the design satisfactory for P, = 308 kips? Design the splice plates. That is, specify the minimum acceptable combinationof widths for the vertical and horizontal plates such that the pair of plates is satisfactory for P, = 308 kips due to yielding on A,; fracture on A,, ;and block shear rupture.
2s I
1
‘ I
I
I
I
I44kd (a) Side elevation view
,1
g
g
(b) Section along path 123
FIGURE P3.9
3.10 A572 Grade 50 steel; E70 electrodes; 1/4in.fillet welds; w,= 1.5 in.
L, = 6 in. = overlap length of member end on the gusset plate
The tension member is a pair of L5 x 3 x 5/ 16 with the long legs welded to a 5/8 in. thick gusset plate. Compute the design strength for: 1)fracture of the fillet welds 2) the gusset plate (yielding on A8)
Is the design satisfactory for P, = 208 kips?
Problems 115
JL
PU
m
b
l (b) Section 11
(a) Connection details
FIGURE P3.10
3.11 A36 steel; E70 electrodes; 1/4in. fillet welds
In the truss joint shown, the vertical member is a pair of W x 3 x 5/16 with the long legs welded to the web of a WT9 x 23 which is the horizontal member. L , = (d  k  S,) = overlapping length of the L5 x 3 x 5/16 member end on the web of the WT9 x 23. S, = size of the fillet weld; d and k are properties of the WT9 x 23. Compute the design strength for: 1)the welds 2) block shear rupture of the connector plate (web of WT9 x 23).
Is the design satisfactory for P,
= 150 kips?
(b) Section 11
1
2 (a) Truss joint detail
FIGURE P3.11
7(c) Section 22
116 Connections for Tension Members 3.12 A36 steel; E70 electrodes; 5/16 inch fillet welds; P, = 300 kips
A tension member (C15 x 33.9) is welded to a t x 16.5 in. gusset plate. Compute the minimum acceptable thickness for the gusset plate due to yielding on A,. Use the preferred increment for thickness on LRFD pl133 and choose the actual minimum acceptable thickness that can be used. Each end return of the weld group is 0.25d, where d = depth of C15 x 33.9 section. Find the minimum value of L, that satisfies the design requirement for the welds. 1
I +L1
I
I
1I 1

pu
Section 11
FIGURE P3.12
3.13 A36 steel; E70 electrodes; 3/8in. fillet welds; L, = 5.25; P, = 270 kips
A pair of L6 x 4 x 7/16 with long legs welded to a w x 3/4 in. gusset plate serves as a tension member. Compute the minimum acceptable value of w for the gusset plate due to yielding on A Assume that the preferred increment for width is 1/ 8 in. 8' and choose the actual minimum acceptable w that can be used. Compute the design strength for: 1. Fracture of the fillet welds 2. The gusset plate (yielding on A,; block shear rupture)
(a) Connection details FIGURE P3.13
(b) Section 1  1
Problems 117 3.14 A36 steel; E70 electrodes; use the maximum acceptable fillet weld size
A tension member (C15x 33.9) is to be buttspliced for P,, = 323 kips with a t x 14 x ( 2 4 + 0.5 in.) splice plate. Choose the minimum acceptable splice plate thickness (see LRFD, p.1133). Fillet welds are located as shown in the cross section and on the back side of the web of the C15 x 33.9 at the ends of the splice plate. Find the minimum acceptable value of L,. Ignore block shear rupture. 1
I4 I I
I
I
b
*
L,
L1
:+ I I
pu
E
Columns 4.1 INTRODUCTION
A member subjected only to an axial compressive force is called a column. Practically speaking, it is impossible for a member to be subjected only to an axial compressive force. When a lab test of such a member is conducted, locating the centroid of the member's cross section at the member ends in order to apply the axial force concentrically cannot be done perfectly. Also, the member is not perfectly straight. Consequently, the initial crookedness increases as the axial compression force is applied. Hence, the member is subjected to a combination of bending and axial compression; such a member is called a beamcolumn. However, as we will see in Chapter 6, the design strength definition of a beamcolumn is an interaction equation that contains a term due only to column action and another term due only to bending action. This chapter deals with column action only. Chapter 5 deals with bending action only. In a truss analysis, only an axial tension or an axial compression force is assumed to exist in each member. Also, there are other situations where the designer deems that bending is negligible and considers only column action in the design of a compression member. Consequently, for practical reasons as well as for the subsequent treatment of beamcolumns, we need to consider column action as a separate topic. Figure 4.1 shows a compression member bolted to a gusset plate. Examination of the FBD in Figure 4.l(c) shows that the maximum compressive force in the net section for a bearingtypebolted connectionis always less than the axial compressive force in the gross section. Therefore, if the only holes in a compression member are at the member ends for a bolted connection, the net section is not involved in the design strength definition of a column. As shown in Figure 4.2, when the axial compressive force in a pinnedended member reaches a certain value called the critical load, or buckling load, the member buckles. If the column is a W section, the buckled shape is a half sine wave in the XZ118
4.2 Elastic Euler Buckling Of Columns 119
ac,
rGusset plate 1
I
(d) Section 11
0 (a) Member end bolted to gusset plate
@I (b) Section 22
P = total bearing force from each bolt 3 P = Pu
38
(c) FBD of member end (pair of angles)
(e) Section 33
FIGURE 4.1 Compression member bolted to a gusset plate. plane. This type of buckling is called column buckling. If the flange and web elements of the W section are not properly configured, 'local buckling of these crosssectional elements can occur before column buckling occurs.Therefore,both types of buckling must be discussed in this chapter. For the W sections usually chosen to serve as a column, the flange and web elements have been configured such that local buckling does not occur before column buckling occurs. The simplest type of column buckling isfzexurul buckling, which denotes that the member bends about one of the principal axes when the column buckles. No twisting of the cross section occurs for flexural buckling. If the cross section twists when column buckling occurs, t h s is called flexuraltorsional buckling. 4.2
ELASTIC EULER BUCKLING OF COLUMNS As shown in Figure 4.3, the boundary conditionsof a column significantly influence the buckled shape, which can be used to compute the bucklingload. The chord length between the points of inflection on the buckled shape is called the effective length KL, where L is the member length. For an isolated column, values of K are shown in Figure 4.3 for various boundary conditions. In Section 4.5, we will discuss how to determine K for a column that is integrallyconnected to other members in a structure. For simplicity,we choose to use a pinnedended column [case(d)of Figure4.31in our initial discussion. For a pinnedended column, note that K =1and KL = L.
120
Columns
j
Buckled shapc neh;;::
I
is reached
f
+ H = O
r
IP (a) Prior to loading
y
t (b) Section 11
r
(c) Buckled FBD
I
M
~
=Pr
IP (d) FBDat z
FIGURE 4.2 Elastic pinnedended column
The first theoretical buckling load solution was published in 1744 by Leonhard Euler [4] for a flagpole column [case (a)of Figure4.31. In 1759,Euler [5]published the solution for a pinnedended column. In his solutions, Euler assumed that the member was elastic, prismatic, and perfectly straight before the axial compressive load was applied. As the axial load was slowly applied, he assumed that the member remained elastic and perfectly straight until the value of the axial load reached the critical load, or buckling loud. Then, he reasoned that the member had reached a state of critical equilibrium and buckled into an assumed shape that was dependent on the boundary conditions at the member ends. Consequently, as Euler chose to do, we refer to the value of the axial compressive load at which the member buckles a s the critical load. The prismatic, pinnedended member in Figure 4.2(a)is assumed to be perfectly straight before the axial compressive load, P, is slowly applied. Also, the member is assumed to be weightless and elastic. When P reaches the critical load value P,,the member bows into a bent configuration, and the cross section does not twist when bowing occurs. Bowing or bending occurs about the crosssectional axis of least resistance, which is the minor principal axis y. Note that the member deflects in the
4.2 Elastic Euler Buckling Of Columns 121
Per‘ I+
K=2.I (2.0)
K = 2.0 (2.0)
(a)
(b) Remarks: I . x marks a point of inflection at which M = 0 2. K = values are recommended for design (theoretical values of K )
FIGURE 4.3 Effective length of isolated columns.
direction of the major principal axisx, which is perpendicular to the bending axis. At an arbitrary distance z from the origin of the member, the bending moment in the buckled member is My =Px (4.1) which can be substituted into the moment curvature relation
to obtain the governing differential equation
For mathematical convenience, let c2 =PI(El,); then Eq. (4.3) becomes dzx+c2x
dz
=
= (J
(4.4)
The solution of Eq. (4.4) is x = A sin cz+B cos cz
(4.5)
122 Columns
At z = 0, the boundary condition is x = 0, which requires B = 0. At z = L, the boundary condition is x = 0 = A sin(cL), which has a nontrivial solution only when CL = nz, where n is an integer number, and
For a column pinned on both ends as shown in Figure 4.2, n = 1 is applicableand gives the least buckling load, which is
Pm =
z2El, L2
(4.7)
For mathematical convenience, let
where
As = gross area of the cross section
r,, = radius of gyration about y axis By substituting ly = A r: into Eq. (4.7)and then dividing both sides of the resulting equation by Ag ,we ottain the critical stress:
Note that we could not find a value for the coefficient A in Eq. (4.5). We only determined that A must be greater than zero. Therefore, the buckled shape in Figure 4.2 is a half sine wave with an indeterminate amplitude, A > 0. 4.3 EFFECT OF INITIAL CROOKEDNESS ON COLUMN BUCKLING
Perfectly straight members cannot be manufactured. As shown in Figure 4.4, each rolled steel sectionhas an initial curvature upon arrival at the fabrication shop. Note that e is the maximum deviation from a straight line connectingthe member ends and L is the member length.If e > L/lOOO, some of the crookednessmust be removed since e 5 L/lOOO is required for the member to be acceptable.For a rolled steel section, the average value ofe = L/1500 (see LRFD Commentary E2, p. 6192). The behavior of an initially crooked member subjected to axial compression is shown in Figure 4.5, where the maximum outofstraightness was assumed to occur at midheight of the member. L/r is a fundamental parameter in Eq. (4.8),which is the critical stress definition for an initially straight, pinnedended member. L/r is called the slenderness ratio, in which L is the member length and r is the radius of gyration for the axis about which bending would occur for Euler buckling (initially perfectly straight member). The effect of initial crookedness on the critical load is greatest in the range 50 < L/r < 135 for a column of A36 steel. Based on experimental and theoretical investigations in which the cross section did not twist when buckling
4.3 Effect Of Initial Crookedness On Column Buckling 123
(b) Section 11
(a) Camber (member is bent about xaxis)
+w X
Y
(d) Section 22
(c) Sweep (member is bent about yaxis)
FIGURE 4.4 Initial crookedness:camber and sweep. occurred, the LRFD Specification writers chose the following definitions for an elastic, prismatic, pinnedended column with an outofstraightness of e = L/1500:
P, =
0.877 a E l
L2
(4.9)
(4.10)
If the column is not pinnedended, then pc, =
0.877 a E l
w2
(4.11)
(4.12)
which can be written as
F, = ( T 0.877) Fy
(4.13)
where (4.14)
Ac is computed for the principal axis having the larger slendernessratio, KL/r, and L
124
Columns
Euler buckling occurs pcr

(Initially perfectly straight member) x = 0 prior to buckling
t P (a) Initially crooked member
X
(b) Load vs. deflection for e = 0
LRFD assumptions: (1) e = U15OO
(2)
9, = 0.877 (Euler T.,.)
(3) Column design strength
 4 e It (c) Load vs. deflection for initially crooked member
FIGURE 4.5 Initially crooked, elastic, pinnedended column.
is the distance between braced points for each principal axis. When Ac 2 1.5,Eq. (4.13) is applicable. For a column that is elastic when buckling occurs, the LRFD design requirement is
@Pn
= p,,
where
@ (r, = 3.70 in.) and (KL), = (KL), = 20 ft = 240 in., we know that ( K L / r).,, > ( K L / r ) , and the cross section bends about the yaxis when column buckling occurs. Therefore,
130 Columns
2: = 0.529
)
Fcr = (0.658 It F, = (0.6580.529 )( 36 ksi ) = 28.85 ksi &P,, = 0.85AZm= 0.85(26.5)(28.85)= 650 kips
An A36 steel W14 x 90 section is used as an axially loaded compression member and the effective length is KL = 20 ft for both principal axes. Use the LRFD column design table on p. 320 to find the column design strength.
Solution Enter LRFD p. 320 at (KL), = 20 ft for a W14 x 90 and F = 36 hi.Find CpP,, = 650 kips. In Example 4.1 we found qP,, = @Pny = 650 kips, w h i x agrees with the value in the LRFD column design table. Therefore,Example 4.1 shows how an entry in the LRFD column design table for W sections was obtained.
An A36 steel W14 x 90 section is used as an axially loaded compressionmember with (KL), = 20 ft and (a), = 10 ft. For this section, we showed in Example 4.1 that local buckling does not govern @,.’ Therefore, find the column design strength.
Solution From LRFD p. 130, for a W14 x 90: A = 26.5in.
d = 14.02in.
OSb, k = 1.375in.
= 10.2 tf
  25.9
tw
Since (KLIr), > (KLIr),,, the crosssectionbends about the x axis when column buckling occurs.
4.5 Efective Length 131
F, = (0.658)': F, = ( 0.658)o'92 (36) = 33.2 ksi (PP, = (PPm= O.85ARc, (PP, = 0.85(26.5)(33.2)= 748 kips
An A36 steel W14 x 90 section is used as an axially loaded compression member with (KL), = 20 ft and (KL), = 10 ft. Use the LRFD column table to find the column design
strength, which is governed by the larger of (KL), and (KL),/ (r,/rJ .
Solution Enter LRFD p. 320 for a W14 x 90 and Fy= 36 ksi. At (KL), = 10ft, find @Pny = 767 kips. Column design strength values for (KL), are not given, but they can be easily obtained from the given information. At the bottom of the LRFD column table, find rJry=1.66foraW14x90.Enter thetableatUU),/(r,/ry) =20/1.66= 12.05ft;uselinear interpolation to find
(PP,,
= 749  0.05 (749738)= 748.45 kips
(PP, = 748 kips
(smaller of @P,, and @Pny )
wluch agrees with the solution obtained in Example 4.3.
The followingare given P, = 300 kips; (KL),= 20 ft (KL), = 10ft and A36 steel. For each nominal depth Listed, find the lightest W section that satisfies the LRFD specifications for axial compression:
1. W14 2. w12 3. w10 4. W8
Solution The design requirement is (PP, 2 (P, = 300 kips). We startby selectinga section that satisfiesthe design requirementfor the yaxis. For this selected section we use its r,/r,ratio Listed at the bottom of the column table and compute (U),/(r,/r,). If (KL),/(r,/rJ > (KL), ,column buckling occurs with the section bending about the x axis, and we must enter the column table with an assumed value for (KL),/ (r,/ry) in order to choose a section that satisfies the design requirement. Assume that r,/rywiU be the same as it was for the section selected for
132
Columns the yaxis. If rJryfor the selected section differssignificantly from the assumed value, compute a revised value of (KL)J(r,/rJ to use in making the next selection. 1. Select the lightest W14 that satisfies the design requirement. For F,, = 36 ksi, enter LRFD p. 321 at ( K L ) , = 10 ft and find the least value that exceeds 300 kips. If yaxis bending governs the column design strength, we find that a W14 x 43 = 312) L 3001 is the lightest choice.
For a W14 x 43, rJry = 3.08 is found at the bottom of the table and is used to compute the following assumed value for entering the table to make the selection that satisfies the design requirement for the x axis:
(KL)J (rJrJ = 20/3.08 = 6.49 ft
[$P,=
W14 x 43
= 350)] >
= 312) 2 300
This is the lightest W14 that satisfies the design requirement. Note that the assumed value of rx/r, = 3.08 used in entering the table at 6.49 ft was the same as the r,/r,value for the section selected for the xaxis. That is, the assumption made was correct. 2. Select the lightest W12 that satisfies the design requirement. For F, = 36 ksi, enter LRFD p. 325 at ( K L ) , = 10 ft and find W12 x 45
($P,l, = 330) 2 300
Enter at (KL),/(rJry) = 20/2.65 = 7.55 ft and find W12 x 45
($Pnx = 360) > ($P,,
=
330) 1300
Note that when the LRFDcolumn table is being used to determine the column design strength, @Pnis governed by the larger of ( K L ) J (rJrJ and (KL),,. 3. Select the lightest W10 that satisfies the design requirement. For F,, = 36 ksi, enter LRFD p. 327 at (KL), = 10 ft and find W10 x 45
( @',,y
= 337) 2 300
Enter at ( K L ) , / ( r x / r J = 20/2.15 = 9.30 ft and find W10 x 45
($P,l, = 346) > (q+d',,,, = 337) 2 300
4. Select the lightest W8 that satisfies the design requirement. For F, = 36 ksi, enter LRFD p. 328 at (KL),, = 10 f t and find W8 x 48
Enter at ( K L ) x / ( r x / r J = 20/1.74 W8 x 48,
((JP~,,, = 362) 2 300 =
11.5 ft and find
(@P,lx= 342) 2 300
Note: For W8 x 40, (q%',ly = 298) = 300, but ( ( J P ,=~280) ~ < 300. W8 x 40 does not satisfy the design requirement. The lightest W8 choice is W8 x 48,
( I P P ,=~ ~362) > (@Pny = 342) 2 (P,, = 300 kips)
4.5 Efective Length 133
Figure 4.8 is an unbraced plane frame. Due to sidesway buckling, all members bend about their xaxis. At the support joints, use the recommended G values given in the last paragraph on LRFD p. 6186. Use Eqs. (4.27)and (4.28) to find (KL), for columns 1to 5. Also find 4P,,, for members 1to 5.
Solution W12 x 120 (Ag= 35.3 in?; I, = 1070 in?;Y, = 5.51 in.) W30 x 173 (I, = 8200 in?) W30 x 116 (I, = 4930 h.4) From the last paragraph on LRFD p. 6186,
GI= 10 G, = 10
G,= 1 At the interior joints of Figure 4.8,
1070 __ 20 = 0.326 4930 30
G, =
1070 1070 +G, = 2o 4930 +30
'
l5 =0.338 8200 40
From Eq. (4.27), the effective length factors for columns 1 to 5 are 1.6G1G, +4(G, +G, )+7.5 = 1.74 G, + G, + 7.5
13.4
Columns
3oft
4oft
4oft
30 Er
All columns ale w12 x 120.
FIGURE 4.8 An unbraced plane frame.
1.6G3G, +4(G, +G, )+7.5 = 1.74 G, +G, +7.5 1.6G4GJ+4(G4+G,)+7.5 = 1.20 G, + G, + 7.5 1.6G6G, +4(G, +G7)+7.5 = 1.24 G, +G, +7.5 l.6G7G8+4(G, +G8)+7.5 = 1.10 G, + G, + 7.5
On LRFD p. 323 for a W12 x 120 and Fy = 36 hi, for each of the columns 1 to 5, enter at (HJ,/(r,/rJ and find $Pm: [(KL)/(r/rJ],= 1.74(20)/1.76= 19.77 ft
($Pm),= 799 kips
[(a)/ (rflJI2 = 1.74(20)/1.76 = 19.77 ft ($Pnr)2= 799 kips
[(KL)J(Y&,,,)], = 1.20(15)/1.76 = 10.23 ft
($P& = 996 kips [(KL)J(rJrJ], = 1.24(20)/1.76 = 14.09 A ($Pnr), = 927 kips [(KL)J(rJr,J5 = 1.10(15)/1.76 = 9.375 ft
(@Pm)5 = 994 kips
4.6 Local Buckling of the CrossSectional Elements 135
In Eq. (4.28),all members are assumed to be elastic, and a point of inflection (M = 0) is assumed to occur at midspan of each girder in the frame. A girder (restraining member) is a bending member that is attached to one or more column ends at a particular joint in a frame. Each girder end provides a rotational resistance at the column end(s)when columnbuckling occursfor the frame.At each end of a girder, the girderend rotationalstiffness in Eq. (4.28)is assumed to be 3El/Ox2) = 6EZ/L.To account for inelastic column buckling and the M = 0 point not being at midspan of the elastic girders in an unbraced frame, the definition of the relativejoint sh$?zess parameter G is
(4.29)
z = E,/E
is as defined in Eq.
(4.23)
actual girder end rotational stiffness
Y=
(6EI/ L)8
(4.30)
For example, when the far end of a girder in an unbraced frame is: 1.fixed, then y= (4EI/L)/(6EI/L) = 0.667. 2. hinged, then y= (3EI/L)/(6EI/L) = 0.5.
Eqns (4.29) and (4.23) are also valid for braced frames, for which we must use actual girder  end rotational stiffness Y= ( 2E l / L (4.31) For example, when the far end of a girder in a braced frame is: 1.fixed, then y= (4EI/L)/(2EI/L) = 2. 2. hinged, then y= (3EI/L)/(2EI/L) = 1.5.
Using z= E,/E = 1is conservative,as was done in Eq. (4.28),but the correct y must be used for each girder. 4.6
LOCAL BUCKLING OF THE CROSSSECTIONAL ELEMENTS Suppose an identical amount of materialis made into closed shapes (pipesand tubes) and open sections (W, C, and L). When the crosssectional area is the same for all shapes, the shape with the largest radii of gyration and with compression elements thick enough to prevent local buckling is the most efficient shape for resisting a compression load. Steel pipes (LRFD, p. 336) and structural steel tubes (LRFD, p. 339) are very good shapes for a column cross section, but attaching other members to these shapes can be difficult and expensive. The open sections are less efficient column sections, but attaching beam and girder sections to them is routine. Some examples of stiffened and unstiffened compression elements in column crosssectional shapes are shown in Figure 4.9 to further clarify the definitions for compression elements.
136 C o h m n s
3
(a) Stiffened compression elements
Larger b/t governs
Larger b/t governs (b) Unstiffened (projecting) compression elements
FIGURE 4.9 Crosssectional compression elements.
Examples of the local buckling mode shapes for the flange and for the web of a W sectionare shown in Figure4.10. Also, in Figure4.10, we assumed that the flange and the web buckled independently of each other. Theoretical discussions of elastic buckling ofthinplate elements are available [6,7] and give Eq. (4.32) as the critical stress for elastic buckling of thinplate elements subjected to a uniaxial, uniform, compressive stress [see Figure 4.ll(a)]:
F,, =
kn2E 12(1  v 2 ) ( b / t ) 2
(4.32)
where
k = constant depending on a h and the edge support conditions E = modulus of elasticity v = 0.3 = Poisson's ratio b/t = widthtothickness ratio of the plate a/b = lengthtowidth
ratio of the plate
See Figure 4.11(b)for example values of k. For a/b 2 4, the half wave length of the buckled shape is on the order of the width b. As shown in Figure 4.10 at any cross section: 1. When local buckling ofa W sectionj7atige occurs, the local buckling mode shape is antisymmetric and the web provides some rotational resistance. Parallel to
4.6 Local Buckling of the Crosssectional Elements 137
n
I 
1
Number of half sine waves is a function of d~ and b/fof flange. (a) Flange local buckling of a W section column
(b) Section 1 I
(c) Section 22
Number of half sine waves is a function of d h and l,/l of web (d) Web local buckling of a W section column
FIGURE 4.10 Local buckling modes in a W section column.
the applied compressive stress, one flange edge is free and the other edge (at the junction of the flange and web) can be assumed to be such that k = 0.7 (about midway between hinged and fixed). Each halfof a W sectionflange is an uiistiffened cotitpression element. 2. When focal biickfing ofa W secfbn zveb occurs, the local buckling mode is symmetric and the flanges provide considerable rotational resistance since for column buckling, the structural designer must prevent twisting of the cross section at the member ends and at any intermediate, weak axis, columnbraced points. Therefore, the flanges are restrained at the ends of each unbraced column length, and the torsional resistance of the flanges can be developed in each unbraced column length. At each junction of the web and flanges, the web edge is somewhere between fully fixed and hinged. The web is a stiffened compression element for which we can assume that k = 5.0 (longitudinal edges are about onethird fixed).
138
Columns
F,, =
kn2 E 12 ( l  V 2 ) ( b / t ) 2 t = plate thickness b = plate width a = plate length
I&
Boundary conditions of unloaded edges are shown below
(a) Long plate with loaded edges simply supported
Case 1
Boundary conditions of unloaded edges One edge simply supported, other edge free
k 0.425
One edge fixed against rotation, other edge free
1.277
3
Bothedges simply supported
4.00
4
Oneedgefixed against rotation, other edge simply supported
5.42
5
Both edges fixed against rotation
6.97
k
b
=i

(b) Section 11
FIGURE 4.11 Coefficients of k for Eq. 4.32 (adapted from [25]).
Note the following: 1. When local buckling of the compression elements in a column cross section occurs, these elements continue to resist some more compressive load until a considerable amplitude of the column buckled shape occurs. However, when column buckling occurs, the member cannot resist any more compressive axial load. 2. Inelastic local buckling of plates can occur when either the b/t ratio or the L/t ratio is small enough. Fortunately, we seldom have to deal directly with Eq. (4.32). For frequently encountered situations, experts on plate buckling have chosen realistic k values
4.6 Local Buckling of the CrossSectional Elements 139 based on currently available theoretical and experimental research, satisfactory performance of existing structures, and engineeringjudgment to devise definitions of critical stress for local buckling of column crosssectionalelements. For example, buckling experts used F,/Fy = 0.7 to account for the presence of residual stressesand imperfections in uniformly compressed elements and made the followingchoices of k to obtain the indicated A, expressions on LRFD p. 638: 1. Unstiffened elements:
Single angles: k = 0.45 Flanges: k = 0.7
A,
= 76 /
fi
A, = 95 / A,
Stems of tees: k = 1.28
= 127/
2. Stiffened element
Web of a W section column: k = 5.0
A,
= 253 /
fi
When b/t of each compression element in a column cross section is less than A, on LRFD p. 638, local buckling of a compression element in a column cross section does not occur before column buckling occurs, and the design strength of a column is given by LRFD E2 (p. 647). When local buckling of a compression element in a column cross section occurs and limits the column buckling strength, LRFD B5.3 (p. 637) refers the reader to LRFD Appendix B5.3 (p. 6105) for the reduced design strength definition of a column. Figure 4.12 provides some explanatoryinformation to aid in coping with LRFD Appendix B5.3 when local buckling of a compression element in a column cross section limits the column design strength. The possible conditions that may be encountered are: 1. The column cross section contains only unstiffened elements [see Figure
4.12(a)].Astress reductionfactor Q, [seeLRFD,p.6106;Eqs. (AB51 to6)]must be computed for the unstiffened element having the larger b/t ratio. Q, must be used in computing the critical stress F, due to local buckling [LRFD,p. 6107; either Eq. (AB515) or (AB516)1. Then, the column design strength is computed: @Pfl= 0.85A cr. If flexuraltorsional buckling can occur, Q,must be used in computing t e critical stress F, due to flexuraltorsionalbuckling [LRFD Eq. (AE3l),p. 648]. The smaller of F, due to local buckling and F , due to flexuraltorsional buckling must be used in computing the column design strength: @P,,= O.85AFc,.Example 4.13 illustrates the computations involved in computing $P,,for a column cross section containing only unstiffened elements. 2. The column cross section contains only stiffened elements [see Figure 4.12b1, an area reduction factor Q, must be computed for each stiffened element having b/t > A,. The definition of Q, = A, /A where A, = Ag Z A , and A, = the ineffective area of a stiffened element. h e ineflective areas in Figure 4.12(b) are the crosshatched areas. Q, must be used in computing
aF
140
Columns the critical stress F,, [LRFD, p. 6108; either Eq. (AB515) or (AB516)). Unfortunately, the ineffective areas are a function of F,,, and an iterative procedure must be used to determine Q,. Then, the column design strength is computed: @Pn= 0.85A$,,. 3. The column cross section contains unstiffened and stiffened elements [see Figure 4.12(c)]. Q = Q,Q, must be used in computing the critical stress F,,. Then, the column design strength is computed: @Pn= 0.85A$,,. See items 1 and 2, respectively,for the computationsof Q, and Qa.The governing F,, is the least F, value computed as described in items 1 and 2.
I ' 7 Q = Qs is governed
by larger b/t is a stress reduction factor (see LRFD B5.3a)
(a) Sections have only unstiffened compression elements
A = an ineffective area
Each crosshatchedarea is an ineffective area. Qa is an area reduction factor
(b) Section has only stiffened compression elements
Q,= [LRFD Eqn (AB53) or (AB54)]
A i = an ineffective area
The crosshatched area is an ineffective area. (c) Section has unstiffened and stiffened compression elements
FIGURE 4.12 Slender compression elements.
4.6 Local Buckling of the CrossSectional Elements 141
The objective of this example is to illustrate how to obtain the column design strength when local buckling of un unstfleened compression element occurs and reduces the column buckling strength. For F, = 36 ksi, (KL),= 6 ft, and a WT8 x 13, find $P,.
Solution WT8 x 13 As = 3.84 in?
d = 7.845 in. br = 5.50 in.
t, = 0.250 in. t, = 0.345 in.
Y,
= 2.47 in.
When b/f of the flange element and/or stem element exceeds the applicable ;L, from LRFD Table 85.1 (p. 638), local buckling may limit $Pnx: Oe5
b,  0.5( 5.50) 0.345
(7
FLB does not occur, but stem local buckling may limit $P,. LRFD Appendix B5.3a [item (d) on p. 61071 and B5.3d must be used to compute $Pnx:
Q, =
20 ,000
F,, ( d l t , ) ’

20’ooo =0.564 36(31.38)’
See LRFD B5.3d: Q, = 1.00since our section has only unstiffened elements. LRFD Eq. (AB512):
Q = QSQ,
= Qs (1.00) = Q,
142
Columns
F,, = Q( 0.658)QA:xF, = 0.564( 0.658)0~M”28 ( 3 6 )= 19.80 ksi CpPnx = 0.85A,F, = 0.85(3.84)(19.80)= 64.62 kips If we had incorrectly used LRFD E2 to compute CpP,, we would have obtained #Pnx = 112.4 kips. Since 112.4/64.62 = 1.739, if we had not accounted for stem local buckling, we would have overestimated the design strength by 73.9%.
The objective of this exampleis to illustratehow to obtain the column design strength $PnY for a W21 x 44and Fy= 36 ksi when local buckling of a stiffened compression element occurs and reduces the column buckling strength.
Solution w21 x 44:
A, = 13.0 in.
ry = 1.26 in.
t, = 0.350 in.
h/t, = 53.6
h = 53.6tW= 18.76 in.
From LRFD Table B5.1 (p. 638),
LRFD Appendix B5.3b (item ii) and B5.3~must be used in computing CpP, when WLB (web local buckling) may limit the column design strength. Note that FLB does not occur. The procedure for determining the design strength due to WLB is outlined in this paragraph. First, we must use LRFD Eq. (AB512) to determine a reduced effective width be of the web (the stiffened element). See Figure 4.13 where the crosshatched area is the ineffective area. Then, LRFD Eq. (AB514) is used to determine Q,, which is an area reduction factor. Since Q, = 1.00 [see LRFD B5.3~(item ii)], Q = Q,Q, = LOOQ, = Q, . Next, either LRFD Eqn (AB515)or (AB516)is used to compute the critical stress T,. Finally, @PnY = 0.85A8,, gives the column design
strength due to WLB. Unfortunately,f in LRFD Eq. (AB512)is F,; therefore, the procedure described in the previous paragraph is an iterative procedure. That is, we must assume a value offin order to determine be and at the end of the procedure we find F,,. If the assumed f = F,, we compute (PP,, = 0.85A$,,; otherwise, we assume another value of J determine be, and so forth. Fortunately, LRFD Appendix B.5.3~states that Ag and r are for the actual cross section. For this example, we use Ag = 13.0 in.*and Y,, =1.26in. as tabulated in the steel manual for the W21 x 44 section.
4.6 Local Buckling of the CrossSectional Elements 143 The (KL), value at which WLB ceases to limit $Pnyis obtained from the condition that Q, = 1, which occurs for the value offthat gives b, = h in LRFD Eq. (AB512).
2
f= Then, as shown, we obtain the Assume that AY I 1.5:
[ F,,
(E) = 22.28 ksi 53.6
(a), value at which WLB ceases to govern $Po,.
= (0.658)n:r
]
Fy = (f = 22.28 ksi)
For F, = 36 h i , we obtain
azCy=
1
log (22.28/ 36) =i.im log (0.658)
a,
=i.0707
(KL), =120.3 in. = 10.02 ft For (KL), 2 10.02 ft, WLB does not govern $Pnyand LRFD E2 is applicable for the determination of $Pny.However, for 0 5 (KL), < 10.02 ft, LRFD Appendix B5.3b to B5.3d must be used to determine $Pnyusing an iterative procedure:
X
FIGURE 4.13 W section with a slender compression element.
144 Columns
1. At (KL), = 0, assume that Q = (Q, = 0.95) and find
QA', =0.95(0)2= 0 f = F,,
= Q ( 0.658)Q"b F, = 0.95( 0.658)'"
(36) = 34.2 ksi
2. Usingf= 34.2 ksi and h / t , = 53.6 in LRFD Eq. (AB512),we find
[$ $[
= 45.571
(L, = 5.3 ft) 12.5(240.6) c  2.5( 240.6) + 3( 236.4) = 1.008 + 4( 240.6) + 3( 236.4)
C , m , = 1.008( 250) = 252 @Mpx= 258 ft  kips
@Mnx = smaller of
@MnX = 252 ftkips
3. For column effects, See Figure 6.14. Since the axial compression force in members 1and 2 is not negligible, the braced frame formula or nomograph on LRFD p. 6186 cannot be used to obtain K, for the girder. Be conservative and use K, = 1:
,
8.06
(F)y1.7226 =
[ Kk/r 574IT Ac =


29,000
= 57.14
 o . ~ o 9 3. 2. For W12, 2.20 Iu 53.42. Only for the lightest three sections is u > 3. 3. For W10, 2.12 5 u 5 2.77. 4. For W8, 2.15 5 u 5 2.16.
Therefore, we recommend u= Z,/zy = 3 as the assumed value for the first trial section. If we replace rn, by m, we obtain
+,
+ m ( M u x+ u M , y ) ]
@CPfl
where
which can be estimated as shown here. [(fi),/(rJr,)] I(a),, F , = 36 ksi, and F , = 50 ksi: For cb = 1, Lb = (a),, 1. @P,, can be obtained from the column tables for a given (KL),. 2. @ can & be I obtained ,,, from the beam design charts for Lb =
(a),.
Use Procedure 1 and select the lightest W section of A36 steel for P, = 12.8 kips
L = 30 ft
C, = 1
M , = 241 ftkips
L, = (KL), = 6 ft
(KL), = 30 ft
Solution
1
Equivalent M u , = From LRFD p. 4130,
Cb= 1
plot point: (Lb= 6, Mu, = 251)
W21 x 44
(@Mu= 250) = 251
This is how we chose the W21 x 44 for member 3 in Figure 6.14 for the loads stated in Example 6.7, where we checked a W21 x 44 for strength and found it was acceptable.
6.9 Preliminary Design
279
Use Procedure 1and select the lightest W section of A36 steel for
Mu, = 241 ftkips
P, = 12.8 kips cb = 1.67
L = 30 ft
Lb =
(a), = 15 ft
(a), = 30 fi
Solution Equivalent M u = From LRFD p. 419, W21 x 44
(#Mpx= 258) 2 251
From LRFD p. 4132,
cb = 1.67
plot point (Ly M&b)
=(15,251/1.67 = 150)
W21 x 44 lies to right of and above this point. Choose W21 x 44 as the trial section.
Use Procedure 2 and select the lightest W12 and lightest W14 of A36 steel for P, = 261 kips; Mu,= 96.8 ftkips L = 30 fi;
cb =
1.67; Lb = (a), = 15 ft;
(a), = 30 ft
Solution 1. For W12 and Fy= 36 ksi, LRFD p. 312 for KL = 15, m = 1.55
#Pn2 [P, + mMux= 261 + (1.55)(96.8) = 411 kips] LRFD p. 324
(a), = 15
= 485) 2 412 W12 x 65 (@Pny
[ ( a ) J ( r J r , ) = 30/1.75 = 17.14 ftl > [ (a), = 151 [#P, = #Pm = 460  0.14(14) = 458 kips] 2 412
Try W12 x 65. (This is how we chose the W12 x 65 for Example 6.4.) 2.
For W14 and Fy= 36 ksi LRFD p. 312 for KL = 15, rn = 1.4
#Pn2 [ P , + m M ,
= 261 + (1.4)(96.8) = 397 kips]
280 Members Subject
to Bending and Axial Force
LRFD p. 321:
(KL), = 15
W14 x 61,
(@Pny = 412) 2 397
[(KL)J(r/,) = 30/2.44 = 12.3 ft] < [(KL), = 151 (@Pn= @Pny = 412 kips) 2 397 Try W14 x 61.
Use Procedure 2, LRFD column tables, F, = 36 ksi, and select the lightest W14 and W12 for
P, = 50.9 kips L = 15 ft
M,,, = 186 ftkips
L, = (KL), = 7.5 ft
( K L ) , / ( r J r y ) = 9.51 ft
Solution 1. For W14 and F, = 36 ksi LRFD p. 312 for K L = 9.5 ft, m = 1.5
@Pn2 [P, + mMux= 50.9 + (1.5)(186) = 330 kips] LRFD p. 321 W14 x 48 (@Pnx= 358) 2 330 Try W14 x 48.
2. For W12 and F, = 36 ksi LRFD page 312: For K L = 9.5 ft., rn = 1.7
$Pn2 [P, + mMux= 50.9 + (1.7)(186) = 367 kips] LRFD p. 325: W12 x 50
(@Pn,= 376) 2 367
Try W12 x 50.
PROBLEMS 6.1 In Example 2.7, for A36 steel and a welded connection, a pair of L3 x 2 x 1/4 with long legs back to back was chosen as the trial section for members 34 and 43 in Figure 1.15. Also known from Example 2.7:
P, = 66.3 kips (tension)
and
M u= 0.18(12) = 2.16 inkips
@Pn= 0.90F,,Ap,= 0.90(36)(2.38)= 77.1 kips Does the trial section satisfy the LRFD H1.l strength design requirement? 6.2 See Figure P6.2. For Fy = 36 ksi, q,, = 1.2 kips/ft; PI,= 170 kips; (KL), = L, =
Problem
281
L = 10 ft, assume that yieIding on A, governs (PP, and select the lightest acceptable W section that satisfies the LRFD H1.l strength design requirement.
L
FIGURE P6.1
6.3 Solve Problem 6.2 for F, = 50 ksi.
In Problems 6.4 to 6.10, does the W section indicated in each problem satisfy the LRFD C1 and H1.2 requirements for a beamcolumn? 6.4 See Figure P6.4. W12 x 45, F, = 50 ksi.
q,
P,, = 85 kips
= 1.0 kips/ft
(KL), = L,, = 15 ft
L = 30 f t
L
FIGURE P6.4
6.5 See Figure P6.5. W12 x 120, F, = 50 ksi.
Q,, = 60 kips
P, = 360 kips
( K L ) , = L, = 12 ft
L = 24 ft
&
FIGURE P6.5
6.6 See Figure P6.6. W12 x 90 F, = 36 ksi.
M, = 130 kips
P, = 600 kips
FIGURE P6.6
(KL), = L, = L = 12 ft
282 Members Subject to Bending and Axial Force
6.7 See Figure P6.7. W14 x 61 Fy = 36 ksi.
MI = 170 kips
M, = 60 kips
P, = 90 kips
(KL), = L b = 10 ft
L = 20 ft
(KL),/(TJY,) = 9.51 ft
FIGURE P6.7
6.8 In Example 4.11 for A36 steel, a pair of L3 x 2 x 1/4 with long legs back to back and welded to a 3/80in.thick gusset plate was chosen as the trial section for members 5 to 14 in Figure 1.15 and in Appendix A. From Example 4.11, for these beamcolumns we also know:
@Pn= 135 kips
P, = 122.5kips
M u = 1.91ftkips
Does the trial section satisfy the LRFD H1.2 strength design requirement? 6.9 Select the lightest acceptable WT section of A36 steel that satisfies LRFD H1.2 for members 5 to 14 in Figure 1.15and in Appendix A. 6.10 Solve Problem 6.9 using Fy = 50 ksi steel.
In Problems 6.11 and 6.12, the member in each indicated figure is subjected to bending about the xaxis. Select the lightest acceptableW section from the indicated column section series that satisfies the requirements of LRFD C1 and H1.2 for the stipulated grade of steel. 6.11 For Figure P6.4, W12, F, = 50 ksi, 9, = 0.86 kips/ft; P , =.261 kips; (KL), = L b = 15 ft; L = 30 ft. 6.12 For Figure P6.5, W12, Fy= 50 ksi, Q, = 32 kips; P, = 350 kips; (KL), = L b = 15 ft; L = 30 ft.
6.13 See Figure 6.13. F, = 50 ksi.
L = 15 ft; L, = (KL),= 15 ft Members 1 and 2: W12 x 40 L, = (KL), = 6 ft Member 3: W18 x 35, L = 30 ft; k, accounts for G = 10 at joints 1and 2. Nominal loads are: Dead: w = 1.5kips/ft Live: w = 1 kips/ft Snow: w = 0.6 kips/ft H4 = 1.5 kips H , = 2.4 kips Wind: w = 0.507 kips/ft PDELTA analyses were performed for each of the required LRFD load combinations (p. 630) and the results at the ends of Member 3 are:
Problems
283
1. For 1.213)+ 1.6L + 0.5S,
X3= 13.0 kips
Y3 = 55.5 kips; M3 = 180.5 ftkips; M4= 180.5 ftkips
2. For 1.213)+ 1.6s + 0.5L
X3= 11.45 kips
Y3= 48.9 kips
M3= 159 ftkips
M4= 159 ftkips
Does the design satisfy LRFD H1.2 (p. 660)?
FIGURE P6.13
6.14 See Figure 6.14. Fy = 50 ksi.
Members 1and 2: W14 x 48, L = 15 ft; L, = (KL), = 15 ft Member 3: w21 x 44, L = 36 ft; L b = (a), = 6 ft k, accounts for G = 10 at joints 1and 2. Nominal loads are : Dead: w = 1.5 kips/ft Live: w = 1kips/ft Snow: w = 0.6 kips/ft Wind: w = 0.507 kips/ft H3 = 2.4 kips H4 = 1.5 kips PDELTA analyses were performed for each of the required LRFD load combinations (p. 630) and the results at the ends of member 3 are: 1. For 1.2D + 1.6L + 0.5s
X 3 = 19.5kips
Y3 = 66.6 kips
M3=
270.8 ftkips
M4= 270.8 ftkips
2. For 1.2D + 1.6s + 0.5L
X3= 17.2 kips
Y3 = 58.7 kips
M3 = 239 ftkips
M , = 239 ftkips
Does the design satisfy LRFD H1.2 (p. 660)? 6.15 See Figure 6.15. Fy = 50 ksi.
Members 1and 2: W10 x 33, L = 15 ft; Lb = (KL), = 15 f t Member 3: W10 x 39, L = 15 ft; Lb = (KL), = 15 ft Members 4 and 5: W24 x 62, L = 36 ft; Lb = (a), = 6 ft.
284 Members Subject to Bending and Axial Force
k, accounts for G = 10 at joints 1 to 3. Nominal loads are: Dead: w = 1.5 kips/ft Live: w = 1 kips/ft Snow: w = 0.9 kips/ft H , = 2.4 kips H , = 1.5 kips Wind: w = 0.507 kips/ft PDELTA analyses were performed for each of the required LRFD load combinations (p. 630) and the results at the ends of members 4 and 5 are: 1. For 1.20 + 1.6L + 0.5S,
X,= 5.22 kips
Y,= 55.2 kips
M, = 72.4 ftkips
M, = 580.7 ftkips
X, = 5.22 kips
Y6 = 55.2 kips
M , = 580.7 ftkips
M, = 72.4 ftkips
2. For 1.20 + 1.6s + 0.5L,
X, = 5.07 kips
Y, = 53.6 kips
M, = 70.3 ftkips
M,= 564.1 ftkips
X, = 5.07 kips
Y6= 81.0 kips
M , = 564.1 ftkips
M, = 70.3 ftkips
Does the design satisfy LRFD H1.2 (p. 660)?
FIGURE P6.15
6.16 See Figure 6.16. F, = 50 ksi.
L, = ( K L ) , = 12.5 ft. Members 1 to 4: W14 x 43, L = 12.5 ft; L, = (KL), = 6 ft. Member 5: W21 x 44, L = 30 ft; L, = (KL), = 6 ft. Member 6: W18 x 35, L = 30 ft; k, accounts for G = 2 at joints 1 and 2. Nominal loads are: zu, = 2.1 kips/ft Dead: ws= 2.61 kips/ft Live (reduced): w,= 1.54 kips/ft Snow: w,= 0.9 kips/ft Wind: w6= 0.507 kips/ft
H3 = 3.25 kips H,= 1.62 kips
H4= 2.03 kips H, = 1.02 kips
Problems
285
PDELTA analyses were performed for each of the required LRFD load combinations (p. 630) and the results at the ends of members 5 and 6 are: 1. For 1.20 + 1.6L + 0.5S,
X,= 18.2 kips X,= 31.9 kips
Y3= 83.9 kips
Y5= 44.55 kips
M,= 349.3 ftlups
M4= 349.3 ftkips
M,= 190 ftkips
M, = 190 ftkips
2. For 1.20 + 1.6s + 0.5L,
X,= 25.7 kips
Y, = 58.5 kips
M, = 254 ftkips
M4= 254 ftkips
X, = 33.3 kips
Y5 = 59.4 kips
M,= 241 ftkips
M, = 241 ftkips
Does the design satisfy LRFD H1.2 (p. 660)? 6.17 See Figure 6.17. F,, = 50 ksi. W
H6
UI
H4
P6.16
Members 1 to 4: W10 x 33, L = 12.5 ft Members 5 to 6: W10 x 39, L = 12.5 ft Members 7 to 8: W24 x 62, L = 30 ft Members 9 to 10: W21 x 50, L = 30 ft k, accounts for G = 2 at joints 1 to 3. Nominal loads are:
L, = (KL), = 12.5 ft L, = (KL), = 12.5 f t L, = (KL), = 6 ft L, = (KL), = 6 f t
Dead: w7= w8= 2.61 kips/ft w9= w , =~ 2.1 kips/ft Live (reduced): w7= w8= 1.54 kips/ft Snow: w9= wl0 = 0.9 kips/ft Wind: w9= wl0 = 0.507 kips/ft
H4 = 3.25 kips H , = 2.03 kips H7 = 1.62 kips H9 = 1.02 kips PDELTA analyses were performed for each of the required LRFD load combi
286 Members Subject to Bending and Axial Force
nations (p. 630) and the results at the ends of members 7 to 10 are: 1. For 1.20 + 1.6L + 0.5s
X,= 8.88 kips X,= 8.88 kips
Y4 = 71.1 kips
M4= 157.5ftkips
M5 = 541.6 ftkips
X, = 14.87 kips
Y5= 96.7 kips Y7= 38.1 kips
M7 = 89.6 ftkips
M 8= 282.7 ftkips
X,= 14.87 kips
Y8 = 51.0 kips
M8= 282.7 ftkips
Mg = 89.6 ftkips
M5= 541.6 ftkips M , = 157.5 ftkips
2. For 1.20 + 1.65 + 0.5L
X, = 11.0 kips
Y4= 50.4 kips
M,= 124.0 ftkips M , = 369.3 ftkips X,= 11.0 kips Y, = 66.7 kips M,= 369.3 ftkips M, = 124.0 ftkips X,= 14.9 kips Y,= 49.9 kips M7= 102.7ftkips M, = 386.8 ftkips X,= 14.9 kips Y,= 68.9 kips M, = 386.8 ftkips M, = 102.7 ftkips 3. For 1.20 + 1.6s + 0.8W
X4= 6.95 kips Y4 = 39.5 kips M4= 83.5 ftkips M , = 309.1 ftkips X,= 8.34 kips Y,= 52.3 kips M , = 280.5 ftkips M , = 121.2 ftkips X,= 14.5 kips Y7zz 44.5 kips M7= 86.3 ftkips M , = 351.7 ftkips X,= 13.4 kips Y,= 61.7 kips M , = 344.6 ftkips M , = 94.7 ftkips Does the design satisfy LRFD H1.2 (p. 660)? w
FIGURE P6.17
W
CHAPTER
Bracing Requirements
7.1 INTRODUCTION This chapter discussesbracing requirements and provides conservative guidelines adapted from Winter [18], McGuire [19], and Galambos [20] for the preliminary design of the bracing. The following types of bracing are considered:
1. Diagonal bracing in a braced frame (LRFD C1 and C2.1) (a) Cross braces (each truss diagonal is designed as a tension member) (b) K braces (one Ktruss diagonal is in compression; other diagonal is in tension) 2. Weakaxis column braces (LRFDB4) 3. Compression flange braces of a beam (to prevent lateraltorsional buckling) 7.2
STABILITY OF A BRACED FRAME In the following discussion, we assume that lateral stability for the braced frame direction of a structure is to be provided by diagonal bracing in vertical, cantilever, plane trusses (see Figures 7.1 to 7.5). The structure in Figure 7.1 is composed of unbraced frames in the Ydirection and braced frames in the Xdirection. Only a twobay by threebay structure was chosen to simpllry the graphical presentation. The flat roof slab for this structure is assumed to be rigid inplane and adequately fastened to the roof framing members such that the joints at the top end of all columns translate the same amount when the structure is subjected to wind independently in each of the Xand Ydirections. Therefore, each of the three diagonal braces provides onethird of the total bracing required for the sum of the weakaxis column buckling loads and the total factored wind load in the X direction. If a structure has more bays in the Xdirection than shown in Figure 7.1, the structural designer probably will choose to locate some of the required diagonal bracing in two or more bays (see Figure 7.2). 287
288
Bracing Requirements
A36 steel; All columns: W14 x 48;All girders (Y direction): W24 x 55 (a) Roof framing plan
15 ft
Y
(b) Section ll(unbrdced frame)
z
L
&
L=3Oft
&
L
(c) Section 22 (braced frame)
Figure 7.1 Onestory building.
The lateral bracing in the Ydirection of Figure 7.1 must provide adequate stiffness to control drift and adequate strength to prevent overturning for each of the required loading combinations [LRFD Eqs. (A41) to (A46)]. For simplicity in the discussion, lateral bracing is discussed only for each of the loading combinations defined by LRFD Eqs. (A43) and (A44). The size of the diagonal braces can be chosen using the preliminary design guidelines given here. Then, accounting for secondorder effects (LRFD C1 and C2.1), a structural analysis of the vertical,
4 spaces at L = 4L
Figure 7.2 Bracing in more than one bay.
I
7.2 Stability ofa Braced Frame 289 Diagonal cross braces are to be designed as tension members. Only one diagonal is active at a time.
mhyft
mO=h/L
(a) Details of the ideal braced bay in Figure 7.1
Maximum aCCeptdbk outofplumb is ug = h i 5 0 0
X

k
(b) Actual erection position
V]
=
(P,
/I
+ P,
u)/L
(c) Final deflected position
I Y LI P = V,/sin O
Figure 7.3 Tensionmember cross braces.
cantilever, and plane trusses containing the diagonal braces must be performed to ensure lateral stability of the structure in the Ydirection. In this lateral stability analysis, axial deformation of all members in the vertical bracing system shall be included (LRFD C2.1). In the preliminary design guidelines given for sizing the diagonal braces, axial deformation of only the diagonal braces is included. As shown in Figure 7.3, during construction the columns cannot be perfectly plumbed and are erected with an acceptable outofplumbness u, I h/500 (see LRFD, p. 6254). The diagonal braces are installed in this outofplumb position and subsequently subjected to nominal loads that increase as construction progresses to completion. 7.2.1 Required Stiffness and Strength of Cross Braces
The derivations for the required stiffness and strength of a typical diagonal brace in Figure 7.3(c) are made for a loading combination that contains a term for wind. The solutions for a loading combination that does not contain a wind term are obtained
290
Bracing Requiremmfs by deleting the term involving the factored wind force P, from the solutions that contain a term for wind. P, is the sum of the column buckling loads to be resisted by one diagonal brace. P, is the total factored load to be resisted by one diagonal brace due to wind on the end of the building and is applied in the X direction. P, and P, are applied to the joint at the top end of the diagonal brace. At reaction 2 in Figure 7.2(c), C M y= 0 gives
v,= P, h L+ P, u where u, = original outofplumbness
u1 = drift that occurs due to P, and P, u = (u,
+ u,) = the final deflected position of the column tops
The force in the brace is
V ,  P,h+P,u Pb =sin6 Lsin6 and the elongation of the brace is (eb =
cos 6) = [(PLEA),= pb/s,]
where S, = (EA/L), is the actual stiffness of the brace. Thus, for this loading combination, the required stiffness is
s, = Pb= eb
s, =
V , /sin 6 ucos6
P,h+P,u  P, + P , h / u uLsin6 cos6 Lsin6 cos6
To be conservative, assume that the maximum outofplumbness u, = h/500 exists and the maximum acceptable value of additional drift u1 occurs. Note that u, is due to factored loads and must be chosen by the structural designer since the LRFD Specification does not give any guidelines on the maximum acceptable drift. For serviceability of a steel framework structure, the drift index is customarily chosen to be in the range of h/667 to h/200. For 1.2D + 1.6L and L / D = 3, this corresponds to 1.5(D + L). To be conservative, we recommend that u = 1.5(h/200)= 3h/400 = h/133 be used in the preliminary design of the brace. Thus, the required stiffness is
['b
=(?),I2
(P,
+ 133 P, )
L S i n 6 COS6
For a tension member, the LRFD strength requirement is
(Lsin6)
7.2 Stability of a Braced Frame
291
Thus, the strength requirement is
Summary The design requirements for each diagonal brace are 1. For stiffness,
2. For strength
The preceding preliminary design guidelines for a onestory building can be applied in each story of a multistory building (see Figure 7.4). Then, in the story for which the diagonal brace is being designed, the definitions of P, and P, are nc
P,
=CP, j=1
where
nc = total number of columns in the story P, = @Pny of the jth column in the story P, = C W j j=i+l
where
ns = total number of stories i = number of the floor level in the story
W j= total factored wind load applied at the jth floor level Consider a brace that is perpendicular to the members being braced. As in the preceding discussion, let P, denote the compressive force being braced. Temporarily, assume that the brace does not have to resist lateral loads due to wind or earthquakes. The longstanding conservative rule of thumb is that the required strength of thisbrace is 0.02Pu.If the brace is inclined to the members being braced as in Figure 7.l(c),0.02Puis the horizontal component of the required strength of the brace. However, the brace must have adequate stzfiess and strength to quallfy as lateral bracing.
292
Bracing Requirements
ith floor member
Figure 7.4 Multistory braced frame.
The diagonal braces in Figure 7.l(a)are to be designed for the bracing configuration in Figure 7.l(c) and the following conditions: F , =36 ksi
F, =58 ksi
tan 6 = h/L = 15/30 = 0.5
h = 15 ft; L = 30 ft; 6 = 26.57";
sin 6 = 0.447
Select the minimum acceptable diameter A36 threaded rod to serve as the diagonal brace in Figure 7.l(b) for the following LRFD load combinations: Loading 1
1.2D + 1.6s + 0.5L which is LRFD (A43)
7.2 Stability ofa Braced Frainr~ 293 Loading 2 1.2D + 1.3W + 0.5L + 0.5s which is LRFD (A44) For all loadings, use P, = CP,, where P,l = nominal strength of each column. In loading 2, assume that P,, = 3.30 kips.
Solutioii Compute P, = XP, for the W14 x 48 columns of A36 steel. For each exterior column,
For each interior column, r
1
Therefore, weakaxis column strength governs. From the LRFD column tables, we find (bP, = @PI,,,.= 270 kips for each column. Each diagonal brace must provide the weakaxis bracing for four columns. Therefore, each diagonal brace must be satisfactorily designed for: Loading 1
P, = CP,l = 4(270/0.85) = 1271 kips Loading 2
P, = 1271 kips
and
P,,, = 3.30 kips
The strength design requirement is
From LRFD Table J3.2 (p. 681), the design tensile strength for an A36 threaded rod is
$JP,~ = 0.75 (0.75Fl,A,,) = 0.75 (0.75)(58ksi)A, = 32.625A,,
where A, = gross area of the threaded rod. Loading 2 governs P,, :
P, = 
h ( PI,, + P, /133) LSinO
(15)(3.30+1271/133) = 14.4 kips (30)(0.447)
To satisfy the strength requirement, we need
294
Bracing Requirements ((Pp,= 32.625Ab) 2 (pb = 14.4 kips) A, 2 (14.4/32.625 = 0.441 in.2) (d2/4) 2 0.441 i n . 2 d 2 0.7496 in. Try d = 3/4 in. threaded rod: (A, = 0.442 in?)2 0.441 in.2 Note: By the 2% ruleofthumb method, P, = 0.02(1271+ ~ . ~ O ) / C6O=S32.1 kips which would require 32.1/14.4 = 2.22 times more strength than we actually need. Check the stiffness design requirement, which is
=(y)b]’[ (I
Pu +133P, )
[’b
s. d c o d )
1
h  l5 = 33.56 ft L , =sm6 0.447
For loading 1,
For loading 2,
p , + 133p , Lsin6 cos6

1271+ 133(3.30) = 143 ft (30)(0.447)(0.894)
Since 382 > 143,we would use a 3/4in.diameter threaded rod as the diagonal brace. For this choice of threaded rod, note that L/d = 33.56(12)/0.75= 537, and for threaded rods we prefer to use d 2 5/8 in. and L/d 5 500, which corresponds to L/r 5 2000. Instead of the threaded rod chosen in Example 7.1, suppose that either a singleangle or doubleangle tension member with fieldwelded end connections is to be selected for the diagonal brace. If the welds are adequately designed such that the tension member govern the design strength, (Pp, is the k i s t of 0.90F&,, 0.75F,/,U, and (PR, for block shear rupture. A, of the diagonal brace must be chosen such that
h ( P , +P, /133) (Lsine)
are satisfied for strength and stiffness. Instead of the threaded rod chosen in Example 7.1, suppose that either a singleangle or doubleangletension member with bolted end connections is to be selected
7.2 Stability of a Braced Frame
295
for the diagonal brace. If the bolts are adequately designed such that the tension member governs the design strength, @Pn is the least of O.90Fflb,0.75F"Z.I (AbAholes), and @Rnfor block shear rupture. A, of the diagonal brace must be chosen such that
h(Pu + P , /133) ( L sine)
are satisfied for strength and stiffness. 7.2.2 Required Stiffness and Strength of K Braces Figure 7.5 shows how the diagonals of a K truss participate to provide lateral bracing for a braced plane frame. The axial force in each diagonal is P,. One diagonal is in tension and the other diagonal is in compression. Thus, the strength requirement of qbP, 2 P, must be satisfied for both diagonals and @Pnof the compression diagonal usually is smaller than (PP, of the tension diagonal. However, when the wind directionis reversed, the original tension diagonalbecomes a compressiondiagonal. Therefore, both diagonals must be identical. Z
L
L=30ft
tan
4
O = hl( L12)
(a) Section 22 (braced frame) of Figure 7. I(a)
V , = ( P , h + P,, u ) l L
e b = ucostl
tan 0 = h / ( L l 2 )
Pb = V,I sin 0
(b) Final deflected position
Figure 7.5 Ktruss braces.
296
Bracing Requirements The derivations for the required stiffness and strength of each diagonal brace in Figure 7.5(b) are similar to those for the cross braces. The design requirements are: 1. For stiffness,
[s,
=(?) 2
( P u +133P,)
( L sin
e
cos
2. For strength,
W"2
e)
1
h ( P , + P , /133)
( L sin e)
The diagonal braces in Figure 7.l(a) are to be designed for the bracing configuration in Figure 7.5 and the following conditions:
F,, =36 ksi
F , =58 ksi
tan 8 = h/ (L/2) = 15/15 = 1
h = 15 ft
8 = 45"
L = 30 ft sin 0 = 0.707
Select the minimum acceptable pair of A36 steel angles that can serve as the diagonal brace in Figure 7.5 for the following LRFD load combinations: Loading 1 1.20 + 1.6s+ 0.5L which is LRFD (A43) Loading 2 1.20 + 1.3W + 0.5L + 0.55 which is LRFD (A44) For all loadings, use P, = CP,, where P,, = nominal strength of each column. In loading 2, assume that P, = 3.30 kips. Select the pair of angles for: 1. Weldedend connections 2. Boltedend connections
Solution
Note that Example 7.2 is Example 7.1 with the following changes: 1. K braces (Figure 7.5) are to be designed instead of X braces [Figure 7.l(c)] 2. A pair of angles is to be selected instead of a tie rod.
Therefore, from Example 7.1, each diagonal brace must be satisfactorily designed for Loading 1
7.2 Stability of a Braced Frame
297
P,, = CP, = 4(270/0.85) = 1271 kips Loading 2
P,
=
1271 kips
P,,, = 3.30 kips
and
The strength design requirement is:
k ( P , +P, /133) (Lsin8)
For loading 1, P, = 0, and ( 1271)(15)/ 133 P, = k ( P,k / 133) = 6.76 kips (Lsin8) (30)(0.707)
For loading 2, P, = 3.30, and
Pb =
(15)(3.30+1271/133) (30)(0.707)
= 9.09 kips
Assume that @Pa2 9.09 kips for the diagonal as a compression member will govern the size. (KL), = (KL),, = 15/0.707 = 21.2 ft. From LRFD p. 370, a pair of L4 x 3 x 0.25 with two intermediate connectors and A = 3.38 in.2is chosen. (@P,= 17 kips) 2 9.09 kips. Check the stiffness design requirement, which is
( p , , +133p, )
[s, = ( L ; 4 ) , ] 2 [
(Lsin8 cos8)
For loading 1,
p,,

Lsin8 cos8
1271 = 84.8 kipdft (30)(0.707)(0.707)
For loading 2,
P, + 133P,  1271 + 133( 3'30) = 114 kips/ft (Lsin8 cose)  (30)(0.707)(0.707) Since 4624 > 114, more than adequate stiffness is provided. Now, we must check the diagonal brace as a tension member: 1. Check the strength of the tension diagonal with weldedend connections:
@FFpx = 0.9(36)(3.38)= 110 kips @F& = 0.75(58)(0.85)(3.38)= 125 kips
298
Bracing Requirements
($P,= 110 kips) 2 17 kips Therefore, the compression design strength governed as assumed. The welds will be designed for $P,= 17kips instead of Pb = 9.09 kips. That is, the design compressive strength of the member will be developed in the design of the welds. For E70 electrodes and 3/16in. fillet welds, the strength of the transverse welds on the ends of the pair of angles is [2(0.75)(0.6)(70)(0.707)(3/16)(4)(1.5) = 50.1 kips] 2 17 kips
Thus, no longitudinalwelds are needed for strength on a tension member and there is no block shear rupture condition to check. However, ductility will be improved by using longitudinal welds with end returns instead of the transverse welds. 2. Check the strength of the tension diagonal with boltedend connections. The bolts will be designed for $P, = 17kips instead of P, = 9.09 kips. For one 5/8in.diameter A325N bolt in double shear, (#R, = 22.1 kips/bolt) 2 17 kips is adequate at each end of the member. Check bearing: 2 (0.25) > 0.375 in.; the 3/8in.thick separator plate governs.
$R, = 0.75(2.MfFU)= 0.75(2.4)(0.625)(0.375)(58) [$R, = 24.5 kips/(bolt location)] 2 17 kips (OK) Check design tension strength:
When there is only one bolt, we use the recommendationon LRFDpage 6172 which is A, = net area of the connecting element in each angle
[@,,Ap, = 0.75(58)(2)(0.25)(4  0.75) = 70.7 kips] 2 17 kips (OK)
Check block shear rupture: The minimum end distance is the larger of 1.125in. and 1.5(0.625)= 0.9375 in Ahole
= 0.5[2(0.25)(0.75)]= 0.1875 in.’
A, = 2 (1.125)(0.25)= 0.5625 in.’ A,, = Ap  Ahole= 0.375 in.’ Agt= 2 (0.25)(1.5)= 0.75 in.’
A,, = A,,  Ahole= 0.5625 in.2 @R,= 0.75(F,,An,+ 0.6F8,) = 0.75[58 + 0.6(36)](0.5625)
(@R,= 33.6 kips) 2 17 kips (OK)
7.3 Weak Axis Stability of a Column
299
7.3 WEAK AXIS STABILITY OF A COLUMN For a pinnedended, W section column (see Figure 7.6), (PP,, governs the column design strength. A perfectly straight and perfectly plumb column does not exist. The LRFD definition for @P,contains the assumption that the initial crookednessis a half sine wave with an amplitude of u, = L/1500 [see Figure 7.6(c)]. The maximum acceptable outofplumbness is U, = L/500 [see Figure 7.6(d)].Figure 7.6(e)shows the buckled configuration of an imperfect, pinnedended, W section column. Let n denote an integer ( n2 2), and h = L/n denote the distance along the column length between laterally braced points. Then (KL), = (h = L/n). Figures 7.7(ce)are for the case n = 2. For a pinnedended column, (KL), = L. For some W sections used as columns, rJr, > 3. The larger of (KL)J (rJrJ and (KL),, governs the column design strength #Pn. In Figure 7.7(c), when
and when elasticbuckling OCCUTS, (PP, for this column is four times (PI', of the pinnedended column in Figure 7.7(a). By bracing the weak axis of a column [see Figures 7.7(c)to 7.91 at a sufficient number of points such that
L
P+l
A
I I
L 2
~
d (a) Perfect column
d
a
(C)
(c) Initially crooked column.
Half sine wave is assumed, u , = L / lo00 (d) Out of plumb column, U, = L 1500 (e)Buckled configuration. (b)Section 11
Figure 7.6 Pinnedended column.
300
Bracing Requirements then qP,,= In Figure 7.7(c),the roller support at midheight of the column is shown to indicate that a weakaxis column brace exists at this location. Using a roller support to indicatecolumnbracedpoints at intermediate locations along the length is customary in textbooks until more specific details are required. In the following discussion, an elastic spring [see Figure 7.7(c)]is used to indicate a weakaxis column brace. The elastic spring may be thought of as an axially loaded member whose axial stiffness is sb
= (PLEA),
where subscript b denotes brace, and P, = axial force in the brace
L, = length of the brace A, = gross area of the brace E, = modulus of elasticity of the brace
‘t IX (b) Section 11
(c) Weakaxis brace at midheight
(a) Pinnedended column
i

‘h
‘h
h=
I
L
hinge
I
2
I
Figure (e) = Figure (d)
AR
(d) Buckled column (e) Used in mathematical derivations
Figure 7.7 One weakaxis brace for a column.
7.3 Weak Axis Stability ofa Column
301
The axial strength of the brace is (f#)Pn)band ($pn)b2 Pb is required, where P, is the force in the brace when weakaxis column buckling occurs. If a brace with adequate stiffness and strength is located at the L/n points along the height of a pinnedended column, then (KL),, = (h = L/n). If h < (KL)Jr,./ry), then we have strengthened the column such that @,' = @.,' The purpose of the following discussion is to provide guidelines on what is adequate stiffness and strength for an intermediate column brace. The braces must be installed such that they prevent lateral translation and twisting. If twisting is not prevented at the braces, the governing column design strength will be due to torsional buckling for (KL),= L. 7.3.1 Bracing Stiffness and Strength Requirements when h = L/2
In Figure 7.7(c), the buckled shape is two half sine waves. There is a point of inflectionon the sine curve at the midheight support. Therefore,the internal bending moment M = 0 at this location. This is important to remember since the following discussion is based on M = 0 at the laterally braced points. Since M = 0 at each braced point, for mathematical and graphical convenience we use pinnedended, straightcolumn segments between the braced points as shown in Figures 7.7(e) to 7.9. In Section7.3.5, we will show how to cope with the situation when the point of inflection on the buckled shape does not occur at a braced point. If less stiffness is provided than that shown in the following derivation, the weaker symmetrical mode shape [Figure 7.7(a)with a dimple at the brace location] will develop. Thus, it is essential that the brace have at least the required stiffness shown here. In Figure 7.7(e),the maximum outofplumbness U, = L/500 and the maximum outofstraightness u, = L/lOOO have been assumed for the weakaxis buckled shape with one intermediate lateral brace Pb.Note that P,, (LRFDnominal column strength for the yaxis) is the buckling load. Also, note that we have chosen to be more conservativeby using u, = L/1000 than the u, = L/1500 assumed in the LRFD column strength definition. For the FBD shown in Figure 7.7(e), from C My = 0 at the bottom support, LR'+U,P,,, = hPb
From C F, = 0,
From C My = 0 at the top end of the bottom column segment as a FBD,
hR=
(",
1
O+U P,,
As recommended by Winter [18], assume that u1 = u, in u = u, Pb = O.oO8Pfly
+ ul;then,
302
Bracing Requirements Pb is the required bracing strength to produce P, for the buckling shape of two half sine waves, each of length h = L/2. Since U,/2 and u, existed when the brace was installed, then Pb =SbUl
s, =8 PLny s b is the required bracing strength to produce Pnyfor the buckling shape of
two half
sine waves, each of length h = L / 2 . The design requirements of an axially loaded member serving as the brace are (@,,)b>(Pb =0.008Pny)
7.3.2 Bracing Stiffness and Strength Requirementswhen h = W3
The buckled shape for Figure 7.8(a) is three half sine waves, and thisantisymmetric mode shape is the one desired. If less stiffness is provided than that shown in the following derivation, the weaker symmetricalmode shape (only one half sine wave with a dimple between the braces) shown in Figure 7.8(b) will develop. Thus, it is essential that the brace have at least the required stiffness shown here. From Z My= 0 at the bottom support,
LR = U,Pny + hP,
= 2hPb
From C F, = 0,
From C My= 0 at the top end of the bottom column segment,
Assume that u1 = u, in u = u,
+ ul; then Pb = 0.018Pny
Since UJ3 and u, existed when the brace was installed, then Pb = s , u ,
7.3 Weak Axis Stability of a Column
TFR
u, = L1500
(a) Desired buckled shape
u, = Lllooo u = U O + u,
d
303
+R b P
ny
(b) Weaker buckled shape
Pb = Sb u1
Figure 7.8 Two weakaxisbraces for a column.
7.3.3 Bracing Stiffness and Strength Requirementswhen h = U4 The buckled shape for Figure 7.9 is four half sine waves, and this antisymmetric mode shape is the one desired. If less stiffness is provided than that shown in the following derivation, the weaker symmetrical mode shape (only one half sine wave with dimples at the brace locations) will develop. Thus, it is essential that the brace have at least the required stiffness shown here. Even though all braces are assumed to have the same stiffnessin Figure 7.9, the theoretical mode shape relation is that the displacement at the center brace is maximum and the displacementat the other two braces is 0.707 times the maximum displacement. These relations are shown in Figure 7.9. From Z M y= 0 at the bottom support, LR’ + U,Pw
From Z F, = 0,
+ 2 hPb = 0.707(h + 3 h )Pb
304
Bracing Requirements
5'  _L 4
u,
= L1500
u, = LllOOo
u = uo+ u u = u,+ u'
I pny Figure 7.9 Three weakaxis braces for a column.
From C My = 0 at the top end of the bottom column segment,
Assume that u1 = u, in u = u,
+ ul; then, Ph = 0.0273 PnY
Since U,/4 and u, existed when the brace was installed, then,
Pb
=SbU*
s, =7.3.4
27.3 Pny
L
Bracing Stiffness and Strength Requirements When h = Lln for Large n The buckled shape is n + 1half sine waves and is antisymmetric. If less stiffness is provided than that shown below, the weaker symmetrical mode shape (only one half sine wave with dimples at the brace locations)will develop. Thus, it is essential that the brace have at least the following required stiffness and strength:
7.3 Weak Axis Stability ofa Column
305
Pb = 0.008 nPnY
'.3.5 When Point of Inflection Does Not Occur at a Braced Point For the structure in Figure 7.10, we showed [l, pp. 6516531 that the M = 0 point is at 0.5179L from the bottom support, whereas the brace is at 0.6L from the bottom support. From C Mu= 0 at the bottom support,
LR ' + U P,, = 0.6 PbL D
R'=0.6Pb ' "Y 500 From C F, = 0,
PY , R = 0.4 Pb + 500 The ordinate of the half sine wave at the braced point is 0 . 9 5 1 where ~ ~ u, is the ordinate at midheight of the column. Formathematicalconvenience, assume that u1 = 0.951~; thus, u = 0.951 (u, + u l ) . From I;My= 0 at the top end of the bottom column segment, 0.5179 LR = (0.998u / 0.951+ 0.5179 U ,) P,,
P, = 0.0101P,,
u,
= L/500
u, =
L/lOOo
u = u,+
u,
Pb = Sb u ,
't
I
X
'1 a 0 r4 m
QI 4
2
'?
Section 11
Figure 7.10 Weakaxis column brace does not occur at M = 0 point.
306
Bracing Requirements Since U,and u, existed when the brace was installed, then pb = sbu,
s, =
10.7Pny
L
If elastic buckling occurs, according to the LRFD column strength definition, Pny=0.877[
K2EI (0.5179L)'
]
)
= 0.877( 3.73n2EI L2
Since0.5179L is very close to 0.5L, we can use the solutionsfor Pband S, obtained in Section 7.3.1 for h = L / 2 and compare the requirements. Again, if elastic buckling occuls when the brace is placed at midheight of the column,
]
= 0.877(":El) 
The strength required for a brace at 0.6L from the bottom end of the column as shown in Figure 7.10 is 0.010(3.73)/ [0.008(4)]= 1.17times the strength required for a brace at midheight of the column. The stiffnessrequired for a brace at 0.6L from the bottom end of the column as shown in Figure 7.10 is 10.7(3.73)/ [8(4)] = 1.25 times the strength required for a brace at midheight of the column.Thus, the reader should not blindly use the stiffness and strength guidelines given in Sections7.3.1 to 7.3.4 if the braces are not placed at h = L/n points along the column length. 7.3.6 Example Problem
See Figure 7.11 for the structurein which we&& column bracing is to be provided. Member 1mustbedesignedasatensionmembertobracethetopendofthefourcolumns. Seesection7.2.1forthedesignrequirementsofthismember.Member2mustbedesigned as a tension member to brace the weak axis of the four columns at midheight and to transfer the tension force in member 1to the foundation support. Select the minimum acceptable diameter A36 threaded rod to serve as the diagonalbraces.
Solution
For each column, @Pn = (PP,, = 364 kips @Pny = 384 kips
7.3 Weak Axis Stability of a Column
I
307
d
(a) Side elevation view (braced frame)
')1
L=30ft
I
4 T
(b) Section 1  1 (an u n b d
frame)
(c) section 22
A36 steel;All columns:W14 x 48;All girders (Ydirection): W24 x 55
Figure 7.11 Example 7.3onestory building.
For Member 1,
P, = 4 (364/0.85) = 1713 kips P, = 2.48 kips (anassumed value for illustration purposes) The strength design requirement is:
h( P, + P, /133) ( L sin 6 ) From LRFD Table J3.2 (p. 681), the design tensile strength for an A36 threaded rod is @Pfl= 0.75 (0.75FJb) (bp,,= 0.75 (0.75)(58 h i ) A, = 32.625Ab
where A, = gross area of the threaded rod. For Member 1,
Loading 2 governs Pb : Pb =
h ( P w +P, /133) L sine
 15( 2.48+1713/133) = 10.9 kips 30( 0.707 )
308
Bracing Requirements To satisfy the strength requirement, we need
(@Pn= 32.625Ab)2 (P, = 10.9 kips) A, 2 (10.9/32.625 = 0.334 in?) (xd2/4)2 0.334 in.2 d 2 0.652 in.
Try d = 3/4 in. threaded rod: (Ab = 0.442 in.2) 2 0.334 i n . 2 Check the stiffness design requirement, which is
['b
=(?),I2[
( P, + 133P, ) (Lsin0 c o s 6 )
h L , =sin0
15 = 21.2 ft 0.707

Since 605 > 136 and [L/d = 21.2(12)/0.75 = 3391 I500, a 3/4in.diameter A36 steel threaded rod is acceptable for member 1. Member 2 Since h = L/2 = 7.5 ft, Section 7.3.1 gives the design requirements for a weakaxis column brace at midheight:
(@Pn),2 ( P b =0.O08Pn,)
[s, =(?),]2> We must add the axial force from member 1 to the strength requirement since member 2 transfers the force in member 1 to the foundation support:
P,
= 10.9 + (0.008)(1713)/0.707 = 30.3 kips
To satisfy the strength requirement, we need
(@Pn= 32.625Ab)2 (P, = 30.3 kips) A, 2 (30.3/32.625 = 0.929 in.2)
(xd2/4)2 0.929 in.2 d L 1.09 in.
Try d = 1.125in. threaded rod: (A, = 0.994 in2)2 0.929 in.2
7.4 Lateral Stability of a Beam Compression Flange
309
Check the stiffness design requirement:
Sb = 29,000(0.994)/21.2= 1360 kips/ft We must add the vertical component of the axial force from member 1to the stiffness requirement since member 2 transfers the force in member 1 to the foundation support:
8 [1713+ 0.707(10.9)]/15= 918 kips/ft Since 1360> 918 and [L/d = 21.2(12)/1.125 = 2261 1500, a 1.125in.diameterthreaded rod is acceptable as the diagonal brace. 7.4
LATERAL STABILITY OF A BEAM COMPRESSION FLANGE For lateral bracing purposes, the compression flange and the top half of the web of a W sectionbeam can conservatively be treated as a column.Thus, the crosssectional area of the equivalent column is A, = A/2, where A is the gross area of the W section used as a beam. Then, the design requirements given in Section 7.3 can be used to design the lateral braces located at intervals of h = L/n along the length of a beam to prevent lateraltorsionalbuckling.
Each flume girder spanning 30 ft in Figure 7.12 is a W24 x 55, F = 36 ksi that was designed assuming lateral braces were to be provided at intervafs of h = L/n = 30/4 = 7.5 ft. The lateral braces are to be provided by a pair of angles as shown in Figures 7.12(b) and (c) in conjunction with the cross braces shown in Figure 7.12(b). Design the bracing members.
Solution We have a simply supported W24 x 55, F,, = 36 ksi that spans 30 ft with lateral = 337 ftkips bracing provided at intervals of 7.5 ft. From the LRFD beam charts, @Mnx for C, = 1 and L b = 7.5 ft:
12.5Mm,, c, = 2.5Mm,, +3M, +4M,
+3M,
c, = 2.5+ 3(55/64) + 412.5 = 1.06 ( 15/16) + 3(63/64) @MnX = smaller of
C , M , = 1.06(337) = 357 @Mpx= 362 ft  kips
@MnX = 357 ftkips Treat QMnx= 357 ftkips as a couple with a lever arm of
a = 0.95d = 0.95(23.57)= 22.4 in. = 1.87 ft
310 Bracing Requiremenfs
t
t
t
Span=30ft I(a) Each flume girder: W24 x 55 I
I
1
Lateral bracing member
+Web
stiffener
Web stiffener
w24 x 55 I
w24 x 55 1
7.5 ft
@) Section 1  I
4 spaces @ 7.5 = 30 ft
3%
.5 ft
Lateral bracing member ( c )Top plan view
I
L Diagonal brace
4 spaces @ 7.5 = 30 ft
(d) Plan view of buckled W24 x 55 flume girder compression flange
Figure 7.12 Example 7.4 lateral braces for two girders.
Then, the axial compressive force in the equivalent column [see Figure 7.12(d)]is
P, = 357/1.87 = 191 kips
The force for which lateral bracing is to be provided is P,, = P,/@ = 191/0.85 = 225 kips From Section 7.3.3 for h = L/4,the design requirements for the brace are Pb = 0.0273P,
s, =
27.3 Pny
L
Pb = 0.0273(225)= 6.14kips
7.4 Lateral Stability ofa Beam Compression Flange
311
From LRFD p. 364, for (KL), = (KL), = 7.5 ft and P , = 6.14 kips, try a pair of L2 x 2 x 1/4 with two intermediate spacers: ($Pn= 10.0 kips) 2 (P, = 6.14 kips)
Check the stiffness requirement:
[S, = 29,000(0.960)/7.5= 3712 kip~/ft]1 [27.3(228)/30= 2071 (OK) Use a pair of L2 x 2 x 1/4 with two intermediatespacers as the lateral brace between the top flanges of the W24 x 55 flume girders. In Figure 7.12(b), each web stiffener serves as the endconnector plate for the pair of L2 x 2 x 1/4 and prevents twisting of the W24 x 55 at each brace location. Choose a pair of angles welded at their ends to the underside of the top flange of each W24 x 55 to serve as the bracing diagonals in Figure 7.12(c).For the end diagonals that must resist an axial compression force: Pd = (3P,)/0.707 = 3(6.22)/0.707= 40.0 kips L = 7.5/0.707 = 10.6 ft
From LRFD p. 364, try a pair of L3.5 x 3 x 1/4:
(@Pfi= 46 kips) 2 (Pb= 40.0 kips) Check the stiffness requirement: [ S , = 29,000(3.13)/7.5/0.707 = 17,120 kip~/ft]2 [27.3(3)(228)/30= 6221 (OK) Use a pair of L3.5 x 3 x 1/4.
GHRPTER
Connections
8.1 INTRODUCTION
The theoretical analysis techniques for a bolted and a welded connectionof the same type usually contain some assumed behavioral features that are very similar. Consequently, we believe that the most appropriate presentation is to discuss bolting and welding for a particular type of connection and then to move on to the discussion of another type of connection. Connectors subjected to concentric shear were discussed in Chapter 2. 8.2 CONNECTORS SUBJECTEDTO ECCENTRIC SHEAR In this type of connection(seeFigure 8.1), the resultant force acting on the connectors does not coincide with the center of gravity of the connectors. The alternate terminology is that the resultant force acting on the connectors is eccentric with respect to the center of gravity of theconnectors. Figure 8.1 shows the following cases of the connectors being subjected to eccentric shear:
1. A bracket plate is fastened to the flange of a column to support a load [see Figures 8.l(a) and (b)].LRFD pp. 125to 1210 contain some tabular information that is useful in the design of a bracket plate. 2. A plate is shopwelded either to the flange or to the web of a column and either bolted or fieldwelded to a beam web at the ends of a simply supported beam [see Figure 8.l(c)]. The connectors on the beam end are subjected to eccentricshear and a negligible moment. However, the groove weld connecting the plate to the column flange is subjected to shear and eccentric tension (see Section 8.5). 3. A shear splice of a beam [see Figure 8.l(d)] is connected only to the ends of each beam web. Therefore, each spliced member end has a shear and a negligible moment that must be transferred through the splice to the adjacent spliced member end. 322
8.2 Connectors Subjected To Eccentric Shear 313
(a) Plate bolted to column flange
(b) Plate welded to column flange
1
I
(d) Beam shear splice
(c) Plate welded to column flange and bolted to beam web
FIGURE 8.1 Connectorssubjected to eccentric shear.
Two methods of analysis for connectors subjected to eccentric shear will be discussed: the ultimate strength method and the elastic method. The ultimate strength method allows more economical fastener groups to be used, provides a more consistent factor of safety,and was used in the preparation of the tabular information in LRFD Tables 818 (p. 840) through 825 (p. 887) for bolt groups and in LRFD Tables 838 (p. 8163)through 845 (p. 8210)for weld groups. However, the ultimate strength method is an iterative procedure. If a structural designer has a connection configuration that does not conform to one of those shown in the LRFD tabular information,the elastic method is simpler and more conservative, but in some cases it is excessively conservative. We choose to describe the methods of analysis for connector groups subjected to an eccentrically applied failure load that is inclined to the vertical axis of the connector groups. 8.2.1 A Bolt Group Subjected to Eccentric Shear
Ultimate Strength Method The following discussionis applicablefor a bearingtype connection and is based on the description given by Crawford and Kulak [21]. Figure 8.2(a) shows the freebody diagram of a bracket plate subjected to an inclined, factored load that is applied at a point whose eccentric location with respect to the center of gravity of the bolt group is denoted by ex and ey.At a point called the
314 Connections
instantaneous center of rotation (ICR),only a pure rotation pof the bracket plate is assumed tooccur.TheICRpointislocatedataninitiallyunknowndistanceofe,,,with respect to the center of gravity of the bolt group. The forces at the holes in the bracket plate are due to bearing of the bolts on the bracket plate. Each bearing force is assumed to be perpendicular to its radius from ICR.For the ith force,
Fi
=en(1e'or,~
)o.55
where
en= least of
I
(a) FBD of bracket plate
[
DSSB DBSBH DSCSB
I
(b)Components of eccentric force
(c)Details for ith bearing force
FIGURE 8.2 Ultimate strength method for eccentric shear of a bolt group
8.2 Connectors Subjected To Eccentric Shear 315 DSSB = design shear strength of a single bolt DBSBH = design bearing strength at a single bolt hole DSCSB = design sliprritical strength of a single bolt (if applicable)
e = 2.718 (base of natural logarithm) ri = radius from ICR to the center of the ith bolt hole
P = 4naJ~m & = 0.34 in. (from LRFD p. 830) T,,
= radius from ICR to the center of the most remote bolt hole
The three equations of statics are
z F i sin 6; P, sin a: = 0 i=l
cFi n
cos6, Pu cos a = 0
i=l
n
[ ( e o , + e x ) cos a+e,sin alp,   ~ T , F =, O i=l
and the parameters are defined in Figure 8.2. In these equations, the known parameters are the bolt spacings (gand s), ex, e ,a, and @rn. We want to find the value of P, from the last equilibrium equation tkat also satisfies the first two equilibrium equations. An iterative procedure must be used to find this value of P,. For an assumed value of eox,the ICR point is located, T,, is determined, and P, ii, F, and 6, are computed. P, is then determined from the last equilibriumequation.If this value of P, satisfiesthe first two equilibrium equations, the assumed value of eOxwas correct. Otherwise, we must assume another value of eoxand repeat the process. Obviously, a computer program needs to be written to conduct the iterative search for the correct value of eox.Let F, and F, respectively, be the absolute value obtained on the lefthand side of the first and second equilibrium equations.Then, we can start by assuming eox= 0.10in. and increase eoxby an increment of 0.02 in. until F, /(P,sin a)I 0.01 and F,/(P, cos a)I0.01, which means we ensure that no more than a 1% discrepancy in equilibrium can occur. This type of procedure was used to obtain the C values given in LRFD Tables 818(p. 840)through 825(p. 887),where C = Pu/(@rn).If we use the LRFD tabular values, the LRFD strengthdesign requirement for the bolt group is
[w n = c(tbr, ) ] 2 pu Note: LRFD p. 832tells us for these tables that: 1. Linear interpolation on any given page is acceptable. 2. Linear interpolation between pages is not acceptable.
316 Connections
Elastic Method The eccentric force components in Figure 8.3(a) are transferred to the center of gravityof thebolt group that necessitates,asshowninFigure8.3(~),that wealsomust have M = e,V, + ep,at the center of gravity of the bolt group. Then, the following assumptions are made: 1. At each bolt hole location due to V , only, there is an upward force of V,/n, where n = total number of bolts. 2. At each bolt hole location due to H , only, in the negative xdirection there is a force of H,/n, where n = total number of bolts. 3. DueonlytoM=e,V, +eP,,theplateisassumed torotateaboutCG (thecenter of gravity of the bolt group) and in the direction of p. A radius is connected from CG to the center of each bolt hole. As shown in Figure 8.3(d),F, due only
(b) Components of eccentric force
t3 (a) FBD of bracket plate
M = e,V,
+ eY H ,
(c) Forces at CG of bolt group
(d) Details for ith bearing force due to M shown in ( c )
FIGURE 8.3 Elastic method for eccentric shear of a bolt group
8.2 Connectors Subjected To Eccentric Shear
317
to M is assumed to be perpendicular to Y,. The moment of F , about CG is r,F,, and this moment is opposite to the direction of M. Also, we assume that F, is proportional to Y,, which can be mathematically written as F, = r , P . Since p is unknown, we can arbitrarily choose p = F,/rl. Then
]:(
Fi = ri
and the equilibrium condition that must be satisfied is
which enables us to obtain
1=1
and
):(
Fi = ri Let
F,, = Ftsin8, F,, = F, C O S ~ ~ , Then, the shear in the most heavily loaded bolt is
F, =/(?+Fr,)’
+(++Fy,)2
The LRFD strengthdesign requirement for the most heavily loaded bolt is @Rri 2 ‘ j
where
@Rn = least of
[t6
DSSB = design shear strength of a single bolt DBSBH = design bearing strength at a single bolt hole DSCSB = design slipcritical strength of a single bolt
(if applicable)
The elastic method is not iterative, but it gives an excessively conservative value of Fi in some cases.
318 Connections
For the bearingtype connection shown in Figure 8.4, we are given the following numerical information: A36 steel bracket plate,g = 3 in., s = 6 in., ex= 5 in., minimum bearing thickness = 0.5 in., and four 7/8in.diameter A325X bolts in single shear. The objectives of this example are to obtain the maximum acceptable value of P, for this bolt group by using: 1. LRFD p. 846 2. The ultimate strength method equations
3. The elastic method equations
Solution 1 From LRFD p. 824, Cpr, = 28.1 kips/bolt = (single shear of one bolt) From LRFD p. 826, Cpr, = (91.4kips/in.)(0.5 in.) = 45.7 kips/(bolt hole) From LRFD p. 846, for g = 3 in., s = 6 in., ex=5 in., and n = 2 bolts in each vertical row of the bolt group, we find C = 2.10,which enables us to compute the LRFD strengthdesign requirement for the bolt group:
[ @R, = C ( Cprn ) = 2.10(28.1) = 59.0 kips] 2 P,
w FIGURE 8.4 Example 8.1
8.2 Connectors Subjected To Eccentric Shear 319
Solution 2
Try e,, = 2.40 in. x1 = 2.40  1.5 = 0.9 x2 = 2.40 + 1.5 = 3.9 Y=
p=== Am rm
y1 = 3
r3 = r1 = 3.1321
y2 = 3 4.9204
r4 = r2 = 4.9204
0.06910
4.9204
F~=&pr, (1e'ori~ F3 = F, = 26.28
F4= F2= 27.58 Ti
= 2[ 3.1231(2628) + 4.9204 ( 27. 5), = 4 t
i=l
2riFi A
p, = i=l
cox + e x
4
FY = C F i cos 6; = 2 i=l
 436*03 = 58.92 kips 2.40 + 5.00
[
26.28( 0.9) 27.58( 3.9) 3.1321 4.9204 +
Vertical equilibrium is only violated by 0.17% and another iteration cycle is not needed. P, = 58.9 kips is very nearly the same as Pu= 59.0 kips found in Solution 1.
Solution 3
Try P, = 59.0 kips: M = e p u = 5(59) = 295 in.kips At each bolt hole due only to P, = 59.0 kips, there is an upward force of 59.0/4 = 14.75 kips. At each bolt hole due only to M = 295 in.kips, all radii are equal and Y:
=x: +y: =(1.5)'+(3)' =11.25 4
x r z = 4(11.25)=45.0 i=l
i=l
320
Connections For i = 1to 4,
Fi = 3.354(6.56) = 22.0 kips
F,, = F, sin
22.0( 3 ) ei = = 19.7 3.3541
The resultant force on the most heavily loaded bolt is
1
F, = ( F X j )
+
+F,,
7/ =
= 31.51
which is 31.51/27.58 = 1.14 times the force in the most heavily loaded bolt by the ultimate strength method. Since the LRFD strength requirement is that @Rfl2 Fp by the elastic method the maximum acceptable value of P, = (28.1/31.51)(59.0)= 52.6 kips. By the ultimate strength method, the maximum acceptable value of P, = 59.0 kips. Therefore, by using the ultimate strength method in this example, we can have 59.0/52.6 = 1.12 times more applied load than we can have according to the elastic method.
For the bearingtype connection shown in Figure 8.5, we are given the following numerical information: A36 steel bracket plate; g = 5.5 in.; s = 3 in.; ex = 15 in.; minimum bearing thickness = 0.5 in.; 7/8in.diameter A325X bolts in single shear; P, = 100 kips; and a = 45". Using LRFD p. 855, find the number of bolts required.
Solution From LRFD p. 824, @, = 28.1 kips/bolt = (single shear of one bolt)
From LRFD p. 826, Cpr, = (91.4 kips/in.)(0.5 in.) = 45.7 kips/(bolt hole) On LRFD p. 855, for g = 5.5 in., s = 3 in., ex = 15 in., and n = 5 bolts in each vertical
row of the bolt group, we need [@, =C(@rfl)=C(28.l)]>(P, =100kips)
C 2 3.56
For n = 5 bolts in each vertical row of the bolt group, we find C = 3.64 and [@R,= 3.64(28.1) = 102 kips] 2 (P, = 100 kips).
8.2 Connecfors Subjected To Eccentric Shear
e
321
ox
FIGURE 8.5 Example 8.2
8.2.2 A Weld Group Subjected to Eccentric Shear
Ultimate Strength Method The following discussion is based on the descriptions given by Butler et al. [22] and Lesik and Kennedy [33]. Figure 8.6(a)shows the freebody diagram of a fillet weld group fastened to a bracket plate subjected to an inclined, factored load that is applied at a point whose eccentric location with respect to the center of gravity of the weld group is denoted by e, and e .At the ICR point, only a pure rotation b of the bracket plate is assumed to occur. d e ICR point is located at an initially unknown distance of eOxwith respect to the center of gravity of the bolt group. The shear force on each of two typical differential elements in the weld group is shown in Figure 8.6(a). Each shear force is assumed to be perpendicular to its radius from ICR. For the ith force,
F,
={Pn =0.75(0.6F,A,)(l.0+0.50~in'~~ 6 ) [ p ( 1 . 9  0 . 9 p ) ] 0 . 3 }
where
F , = weld electrode strength (AJi = (0.707S,L,)i (SJi = weld size (LJ, = weld length chosen for the ith element
6; = angle measured in degrees as shown in Figure 8.6(c)
ri = radius from ICR to center of ith element
322 Connections
(b)Components of eccentric force
(a) FBD of weld group
m ........... ,.... .../_ ..,.,.,
x.:.:::::::?:.
X1
ICR
Each theta angle is measured in degrees from the lengthdirection axis of a fillet weld to the force on the differential weld element. (c) Details for the force on a weld element
FIGURE 8.6 Ultimate strength method for eccentric shear of a weld group T,
= radius from ICR to center of element with minimum AU/ri
[ A . = 17.39(e+6)O.@ s,
Ii 5 2.72(s,
)i
The three equations of statics are n
C FP,~ sina = 0 i=l
n
CF, P, cosa = 0 i=l
[(e, + e x ) cos a+e, sin alp,  $ T i ~ i
=
o
i=l
where Fi and Fyj, respectively, denote the x and y components of Fi. In these equations, the known parameters are the dimensions and properties of each fillet
8.2 Connectors Subjected To Eccentric Shear
323
weld in the weld group, ex, e,,, and a . We want to find the value of P, from the last equilibrium equation that also satisfies the first two equilibrium equations. An iterative procedure must be used to find this value of P,. For an assumed value of e,,, the ICR point is located, Fi is computed for an appropriately chosen (LJi, and P, is then determined from the last equilibrium equation. If thisvalue of P, satisfies the first two equilibrium equations, the assumed value of e,, was correct. Otherwise, we must assume another value of e,, and repeat the process. Obviously, a computer program needs to be written to conduct the iterative search for the correct value of eo,. Let F , and F , respectively, be the absolute value obtained on the lefthand side of the first and second equilibrium equations.Then, we can start by assuming eox= 0.10 in. and increase e,, by an increment of 0.02 in. until F , /(P,sin a)50.005and F Y / (P, cos a) I 0.005. This type of procedure was used to obtain the C values given in LRFD Tables 838(p. 8163) through 845 (p. 8210). If we use the LRFD tabular values, the LRFD strengthdesign requirement for the weld group is
( 4 R n = 16CC,S,L, ) 2 P, where C = tabular value (which includes 4 = 0.75) C, = electrode coefficient from LRFD Table 837 S, = weld size (in.)
L, = length of connection (in.)
P, = required strength of weld group (kips) Note: LRFD p. 8157 tells us for these tables that: 1. Linear interpolation on any given page is acceptable. 2. Linear interpolation between pages is not acceptable. 3. C, accounts for
(a) The C values being done for E70 electrodes. (b) Additional strength reduction factors of 0.9 for E70 and E80 electrodes, and 0.85 for El00 and E l l 0 electrodes;these strength reductions account for the uncertainty of the extrapolation from E70 test results to higher strength electrodes. Elastic Method
Figure 8.7(a) shows a weld group subjected to eccentric shear. For mathematical convenience,eachweld is consideredto be a line coincidentwith the edge of material to be welded. Thus, each weld is treated as having an effective throat thickness of t, = unity.For the weld group in Figure 8.7(a): 1. The area i s A = 2b + d. 2. The center of gravity of the weld group is located at
x=  b 2 A
3. The polar moment of inertia is
324
Connections
H,
"U
/jpu HU
(b) Components of eccentric force X
M = ex V ,
(a) FBD of weld group
+e
Y H,
(c) Forces at CG of weld group
(d) Details for ith force due to M shown in (c)
FIGURE 8.7 Elastic method for eccentric shear of a weld group
IP = I , +I,
=
2 b 3 + 6 b d 2 + d 3 +6Fbd 12
The eccentric force components in Figure 8.7a are transferred to the center of gravity of the weld group that necessitates, as shown in Figure 8.7(c), that we also must have M = e,V, + ep, at the center of gravity of the weld group. Then, the assumptions made are: 1. Due to V , only, there is an upward force per inch of weld length of qY=
V,/A. 2. Due to H , only, in the negative xdirection there is a force per inch of weld length of 9, = H,/A. 3. Due only to M = e,V, + eyH,, the plate is assumed to rotate about the center of gravity of the weld group and in the direction of p.
8.2 Connectors Subjected To Eccentric Shear
325
A radius is connected from CG to the center of each differential weld element. As shown in Figure 8.7(d),Fjdue only to M is assumed to be perpendicular to rj.The moment of Fjabout CG is rjFj,and this moment is opposite to the direction of M . Also, we assume that Fi is proportional to rt, which can be mathematically written as F j = Y I P . Then, at a point whose coordinates are (x, y), the force per inch of weld length in the x and y directions, respectively, is !
9x
MY
= I P
, Mx
9y
= I P
Let qidenotethe maximum value due to the sum of the force components found in items 1 to 3. Then, the shear per inch of weld length in the most heavily loaded differential weld element is
=\/'bx +9:
+ ( 9 y +9$ Note that for thejth differential weld element, 9x and 4: are in the same direction; also, qYand q i are in the same direction. The LRFD strength design requirement is 9I
where
@Fw = 0.75 (0.60FExx) FExx = weld electrode strength t, = 0.707Sw
The elastic method is not iterative, but it gives an excessively conservative value of
q, in some cases.
For the filletwelded connection shown in Figure 8.8, we are given the following numerical information: A36 steel bracket plate; b = 4 in.; d = 8 in.; ex = 12 in.; E70 electrodes (C, = 1);and 5/16in. fillet welds. Obtain the maximum acceptable value of P, for this weld group by using: 1. LRFD p. 8187 2. The elastic method equations.
Solution 1 From LRFD p. 8187, for kL = (b = 4 in.), L = (d = 8 in.),and a L = (ex= 12in.),which means that k = 0.5 and a = 1.5, we find C = 0.706 and
P, = 16CC,SJ = 16(0.706)(1)(5/16)(8)= 28.2 kips
326 Connecfbns
w
FIGURE 8.8 Example8.3
Solution 2 A = 2b + d = 2(4) + 8 = 16 i n . 2
I, I,
=
=
2b3 +6bd2 + d 3 +6Tbd 12
2 ( 4 ) 3 + 6 ( 4 ) ( 8 ) *+ ( 8 ) 3+6(1)(4)(8) = 197.33in. 12
@,t, = 0.75(0.60)(70)(0.707)(5/16) = 6.96 kip~/in. Try P, = 28.2 kips:
M
= ep, = 12(28.2)= 338 in.kips
Due to P, only, there is an upward force per inch of weld length of
qy = P,JA = 28.2/16 = 1.76 kip~/in. At (xi. yi) = (3,4) due only to M = 338 in.kips,
The resultant force on the most heavily loaded point in the weld group is 9 j = d w = 1 / ( 6 . 8 6 ) ~+(1.76+5.156)* =9.74kips/in.
8.3 Bolts Subjected To Tension And Prying Action 327 By the elastic method, the shear/in. at the most heavily loaded point is qj= 9.10 kips/ in. Since the LRFD strength requirement is that $FJ, 2 qj, by the elastic method P, = (6.96/9.74)(28.2)= 20.2 kips
Recall that P, = 28.2 kips for the ultimate strengthmethod for which, in thisexample, we can have 28.2/20.2 = 1.40 times more applied load than we can have according to the elastic method.
For the filletwelded connection shown in Figure 8.8, we are given the following numerical information: A36 steel bracket plate; b = 4 in.; d = 8 in.; ex= 12 in.; E70 electrodes (C, = 1);and P, = 50 kips. Using LRFD p. 8187, find the required weld size.
Solution From LRFD p. 846, for kL = (b = 4 in.), L = (d = 8 in.),aL = (ex= 12in.), which means that k = 0.5 and a = 1.5, we find C = 0.706 and
S, 2 {Pu/(16CC,L)=50/[16(0.706)(1)(8)]= 0.553 in.} Use S, = 9/16 in. 8.3 BOLTS SUBJECTEDTO TENSION AND PRYING ACTION
Figure 8.9 shows a hanger connection in which the bolts are subjected to tension. Also, the top bolt in the angle section of Figure 1.12(c)is subjected to tension. The bolts in the web angle of Figure 1.12(c)are subjected to shear and tension due to a small bending moment (see Section 8.7). As shown in Figures 8.9(c)and (d),each bolt is subjected to a direct tension force T + Q, where T = Pu/4and Q is a prying force.For this type of connection, according to LRFD J3.3: 1. A highstrength bolt must be used. The required bolt strength is T + Q.
2.
Fully tensioned bolts should be used in bolt groups subjected to prying action in order to reduce the deformationsof the connectionparts and the prying forces.The empirical analysis and design procedures given on LRFD pp. 115 to 1111may be used for hanger connections.The minimumfeasiblypossible value of b in Figure 8.9 should be chosen to reduce the prying force. See LRFD p. 1111for an example problem. As shown in Figure 8.10, preferably the hanger should be stiffened to eliminate prying action. If more than one bolt on each side of the hanger web is used, hanger stiffeners should be used.
328
Connections 1*
Stiffener may be required for the bottom flange. See LRFD K 1.8.
Girder
7
Stiffener
+
Hanger
1J
pu (b) Section 1  I
(a) Side elevation view
Q
T+Q
T+Q
Q
ti6 T
2T
(c) Deformed hanger
u t
P, = 4 T
9
Right half of flange
MIKT (d) Flange bending moment diagram
FIGURE 8.9 T section hanger connection
f FIGURE 8.10 Stiffened T section hanger connection
8.4 Bolts Subjected To Tension And Shear
329
8.4 BOLTS SUBJECTEDTO TENSION AND SHEAR Figure 8.11 shows a truss diagonal member used to provide tension bracing for a braced frame.The vertical component of P , subjects the bolt group at the end of the tension member to uniform shear. The horizontal component of P, subjects the bolt group at the end of the tension member to uniform tension. Each bolt in this bolt group is assumed to resist an equal share of each component of P,. LRFD J3.7 and LRFD Table J3.5 (p. 684) give the needed design information for these bolts in a bearingtype connection. LRFD Figure CJ3.1 (p. 6226) shows the interaction curve as three straight lines approximating an ellipse.This figure is the basis for the limiting tensile stress equations given in LRFD Table J3.5. If a slipcritical connection is desired, LRFD J3.8 and LRFD Tables J3.1and J3.6 are applicable.
110 kips
LA x 3 x 0.25
u Column
(a) Side elevation view
U
Lr
(b) Section 1  1
FIGURE 8.11 Bolts subjected to shear and tension
330
Connections
For the bearingtype connection shown in Figure 8.11, determine the number of 7/8in.diameter A325X bolts required.
Solution Try four bolts: 0.707 P,  0.707(110) = 32.3 ksi nA, 4(0.6013)
1
[ F, = 1171.5fD = 1171.5(32.3) = 68.55ksiI I90ksi Since (F, = 68.6) 2 V, = 32.3), four bolts are acceptable if the bearing strength is adequate.
If a slipcritical connection is desired in Example 8.5, are four lin.diameter A325X bolts adequate? Assume standard bolt holes, 0 = 30.6 kips, L = 45.8 kips, and the governing factored loading combination is 1.20 + 1.6L. Therefore, P, = 110kips (the same as in Example 8.5) and 0 + L = 76.4 kips. S o h t ion The service tension force in each of the four bolts is 0.707(0 + L )  0.707( 76.4) = 13.5 kips 4 4 From LRFD Table J3.1, the minimum pretension force is Tb= 51 kips:
T=
0.707 T
0.707 (13.5) = 12.15 0.7854
The limiting shear strength is
1
[ l) (
F, = 1
(17ksi)= 1';f)(17)=12.5ksi
Since (F, = 12.5) 2 Vu= 12.15),four bolts are adequate for a slipcriticalconnection if the bearing strength is adequate. 8.5
CONNECTORS SUBJECTEDTO ECCENTRIC TENSION AND SHEAR Figures 8.12 and 8.13 show examples of beamtocolumn connections in which the connectors are subjected to eccentric tension (due to a bending moment) and shear. Since we preferred to show the eccentric force (reversed beam reaction) in Figures
8.5 Connectors Subjected To Eccentric Tension And Shear pu
e .
331
1
Section 11 (a) Beam seat angle bolted to column flange

Fdet welds and returns.
Note:For clarity, these welds were omitted in the leftmost view.
Section 22
(b) Beam seat angle welded to column flange
FIGURE 8.12 Connectors for beamseat angles 8.12(a) and (b), the beam sitting on the beam seat and the required top angle to provide lateral bracing of the top beam flangewere not shown.Theseomitted details are shown in the figures on LRFD pp. 9128 and 9138. 8.5.1 Weld Groups The design strength of the welds on a beam seat [seeFigure 8.12@)and the figure on LRFD p. 91281is given in LRFDTable 97 (p.9137)for some seat angle sizes.Entries in LRFD Table 97 (p. 9137) were computed by the elastic method for this weld group. In the elastic method for this weld group treated as line elements, the welds are subjected to a shear/in. of D
where ZLw = sum of the line element lengths (including the returns) in the weld group. Also,the weld returns are subjected to a tension/in. of
where = distance from top of weld group to neutral axis I = moment of inertia of weld group e = distance from weld group to beam reaction P,
Note: Examples 8.10 and 8.11 illustrate how to determine the value of e used in computing any entry in LRFD Table 97(p. 9137).
332
Connections
Notes: Column flange stiffeners shown may be required. The welds were omitted in both views for clarity. The required welds are: (1) A fillet weld each side of the beam web. (2) Either a full penetration weld between each flange and the plate or a fillet weld all around each flange. Section 1 1 (a) Plate welded to beam end and bolted to column flange
(b) Gabled frame
(c) Moment splice at the crown of a gabled frame
FIGURE 8.13 Connectors for beamend plates
The design requirement is
[ @FWf ,
= 0.75( 0.6 FExx)(0.707SW) ] 2 ( q =
4.9:)
Alternatively, if we conservativelyignore the weld returns, the special case on LRFD p. 8163 for the ultimate strength method of a weld group can be used to determine the design strength of the welds on a beam seat. For the weld group described in Figure 8.13(a),the procedure recommended on LRFD p. 1024 can be used. In this procedure, if only fillet welds are used, the eccentric tension is assumed to be resisted by the fillet weld group surrounding the beam tension flange and the shear is assumed to be resisted by the fillet weld group on the beam web. Therefore, in this procedure the eccentric tension and shear are uncoupled; that is, they are treated independently.
8.5 Connectors Subjected To Eccentric Tension And Shear
333
8.5.2 Bolt Groups
For the elastic method and the ultimate strength method, respectively, Figures 8.14(a) and 8.14@) schematically show the bolt tension force distributions that without much thought appear reasonable to assume in a bearingtype bolt group subjected to a bending moment and a shear. If either of these bolt tension force distributions is assumed, Figure 8.14 is applicable for Figures 8.12 and 8.13, and LRFDJ3.7is applicable for the bolts since they are subjected to tension and shear. We will show how involved the elastic method and the ultimate strength method are for the bolt tension force distributions assumed in Figure 8.14. Then, other bolt tension forcedistributions that may be assumed in certain cases to simplify the mathematics will be noted. Consider a bolt group that has m vertical lines of bolts and n horizontal rows of bolts. The total number of bolts in the bolt group is nb = mn. For example, in Figure 8.14, m = 2, n = 5, and nb = 10. Other parameters common to both methods are
T>#
k=3
/jj= fc
(a) Elastic method
n=5
7 @
4
Mu 1
FY (b) Ultimate strength method
Note: These assumed bolt group behaviors are shown on a plate welded to a beam end and bolted to a column flange (not shown).
FIGURE 8.14 Bolts subjected to shear and eccentric tension
334
Connections a = depth of compression stress block
b = width of compression stress block di = distance from neutral axis to ith row of bolts in tension Ai = mA, = total area of bolts in ith row of bolts in tension A, = area of one bolt
k = number of first row of bolts above neutral axis (in Figure 8.14, k = 3) Elastic Method The neutral axis is found by taking moments about the bottom edge of the compression zone: n
U
3( a b ) = C ( a + d , ) A i L
i=k
An iterative procedure must be used to determine u, and then the moment of inertia is
a3b I =+xAidf i=k
and the tension force in the ith row of bolts is
T. =
M,Aidi 1
which is valid for i = k to n. The tension stress in each bolt in the ith row is
and must not exceed the tension stress limit F, given by the formulas in LRFD Table J3.5.If we assume that each bolt in the bolt group resists an equal amount of shear, then the bolt shear stressfD is
where
Vu= factored shear force acting on the bolt group nb = total number of bolts in the bolt group
A, = area of one bolt
Ultimate Strength Method Equilibrium of horizontal forces requires that i=k
8.5 Connectors Subjected To Eccentric Tension And Shear
335
where Fy= yield stress of the plate
Ti = mA$, A, = area of one bolt
F, = tension stress limit specified in LRFD Table J3.5 is a function offo. The usual assumption for the bolt shear stressf, is that
where V , = the factored shear force acting on the bolt group. Since k is unknown, a n iterative procedure must be used to determine
and then the design bending strength of the connection is 0.9F,,a2b
(PMn=
+$diTi
2
i=k
The design requirement is that (PM,2 Mu. As we noted prior to the presentation of the elastic method and the ultimate strength method, there are other bolt tension force distributionsthat may be assumed in certain cases to sirnphfy the mathematics. Bolt force distributions that are assumed in the LRFD Manual procedures for beam seats and end plates will be noted. The various bolt group types customarily used for beam seats are shown on LRFD p. 9128.The shear design strength of the bolts for eachbolt group type is given in LRFD Table 96 (p. 9136). In preparing this table, the tension in the bolts due to Mu= P,e was ignored. For example, from LRFD p. 824 for a 0.75in.diameterA325N bolt in single shear, we find thatfv = 36.0 ksi and (PR, = 15.9 kips/bolt. For six bolts, +R, = 6(15.5) = 93.0 kips, which agrees with the shear design strength of 93.0 kips given on LRFD p. 9136 for connection type C (sixbolts). The LRFD Manual does not give a justificationfor ignoringMu= P,e in a bolt group on a beam seat and accounting for Mu= Pue in a weld group on a beam seat. For mend plate welded to an entire sectionof a beam end and bolted to a column flange, the bolt force distributions assumed in the LRFD Manual design procedure are stated in the first paragraph on LRFD p. 1024.The bolts located an equal distance above and below the beam tension flange are assumed to resist tension only. The tension force in each of these bolts is assumed to be equal. (PFt from LRFD Table J3.2 (p. 681) or, alternatively, LRFD 815 (p. 827) is used in choosing a sufficient number of these bolts to develop the beam flange force.
F, = MU dt,
336
Connections where Mu= required factored beam end moment
d = depth of the chosen beam section ?= flange thickness of the chosen beam section The other bolts in the bolt group are assumed to resist only shear. The shear force in each of these bolts is assumed to be equal. LRFD Table ID (p.824) is used in choosing a sufficient number of these bolts to develop the required factored beam end shear V,. 8.6 TRUSS MEMBER CONNECTIONS AND SPLICES An adequate number of bearingtype connection and splice examples for tension members was given in Chapter 2. For compressionmembers in trusses, the memberend and splice end forces must be developed by the connectors. Therefore, the discussion in Chapter 2 for connectorsin tension member ends is also applicable for the connectors in truss compression member ends. If a truss compression member needs to be field spliced, the moment of inertia for each principal axis of the splice connection parts should not be less than those of the compression member being spliced. Also, the buckling strength of the splice connection parts must not be less than the column design strength. Since Chapter 2 contains only bearingtype connections, a discussion of slipcritical connections is given. When highstrength bolts are fully tensioned, the parts being connected are clamped together by the tension force in the bolts. A friction force equal to the clamping force times the coefficient of friction on the clamped surfaces is developed. If the maximum axial force in the member due to service loads does not exceed the available friction force, the connection is classified as being slipresistant. According to LRFD Commentary J3.8 (p. 6226), slippage occurs at about 1.4 to 1.5times the maximum service load. LRFD J3.8(p. 683) and LRFD Tables J3.5 and J3.6 give the information needed for slipcritical connections, which may be designed either at service loads or at factored loads. In LRFD Table J3.6,the nominal slipcritical shear strength of a highstrength bolt is really the nominal friction force per bolt that can be counted on to occur in a slipresistant connection. Slipcritical connections must also satisfy the design requirements as a bearingtype connection.
See Figure 2.6 and Example 2.6 where, due to factored loads, the governing design strength of the connection was found to be 50.6 kips. If we want this connection to be satisfactory as a slipcritical connection, what is the maximum acceptable force due to service loads? Assume that LD = 1.5,where L is live load and D is dead load, and assume the loading combination that governs is 1.20 + 1.6L.
Solution 1.20 + 1.6(1.50) = 50.6 kips 0 = 50.6/3.6 = 14.1 kips L = 1.5(14.1) = 21.2 kips
8.7 Column Base Plates
337
Required maximum service load = D + L = 35.3 kips. See LRFD p. 829. Due to service loads for one 0.75in.diameter A325 bolt in a standardsizedhole and subjected to doubleshear,the maximum acceptableslipcritical shear = 15.0kips/bolt. In Figure 2.6(a),there are three bolts in the connection.Therefore, the maximum acceptableserviceload = 3(15)= 45.0 kips. Since45.0 235.3, the connection satisfies all W’Ddesign requirements when the three bolts are fully tensioned. If the required maximum service load = D + L = 52.1kips, for example, then the number of bolts needed for the connection to qualify as being slipresistantwould be 52.1/15 = 3.47 bolts and four bolts would have to be used. 8.7 COLUMN BASE PLATES Suggested column base plate details are given on LRFD p. 1155. The suggested design procedure on LRFD p. 1157 to 1160 for a column base plate is applicable when the base plate is required to transfer only an axial compression force from a column to a reinforced concrete footing or to a reinforced concrete pedestal (pier). Two example problems are given on LRFD p. 1160and 1164.Suggested anchor bolt details are shown on LRFD pp. 888 to 891. The LRFD Manual does not suggest a design procedure for a column base plate when the column is subjected to an axial compression and bending. For the two eccentricallyloaded column cases shown in Figure 8.15, we recommend the following design procedure based on the ultimate strength method. In Figure 8.15(a),the base plate dimensions are denoted as
B = width H=d+2h’ t = thickness where d = depth of the column section
h’ 2 (w,+ C,) we= minimum edge distance (see LRFD pp. 682,1157) C, = minimum clearance for socket wrench head (see LRFD p. 813) H is estimated prior to entering the base plate design procedure. Consequently, H is known in the base plate design procedure. For welding purposes, we need B L b,+2(S, + 1/16 in.),where b,= flange width of the column section. B, = minimum B required for bearing strength is computed as shown for each case. The required values in inches for B and Hare rounded up to either the next integer or the next even integer depending on the availability of the desired plate size. See LRFD p. 1133for the preferred increment of thickness. Our discussion assumes that the design bearing strength of the concrete is GcP,, =0.6(1.7fc’Al ) = 1.02f,‘AI
which was obtained from LRFD Eq. (J92)for A, 2 4A, where A, = area of base plate resting on the concrete support
A, = area of concrete support on which the base plate sits
338
Connections
p+T+q
Section 22
Section 11 (b)Large eccentricity case
(a) Small eccentricity case
FIGURE 8.15 Eccentricallyloadedcolumn base plates Case 2 ;(e = M Y/ PY)IH/6 For the assumption that a plane section remains plane when subjected to an axial compressionplus bending, when e = H / 6 the neutral axis is located at the left end of the base plate. Therefore, no anchor bolt is required to resist any tension due to M u when e = H / 6 . The depth of the rectangular compression stress block is conservatively assumed to be u = H 2e [seeFigure 8.15(a)].Equilibrium of the vertical forces requires that
[ (bc P p = ( 1.02f,'B,
)( H
 2 e ) ] = P,
where B, = minimum B required for bearing strength
B, =
P"
1.02fc'(H2e) At the face of the most heavily loaded column flange, the design bending strength requirement is
8.7 Column Base Plates
[
$JM
= 0.9 F,
339
(i) ] [1.02f:B h '(:)]
B
2
where B = actual value chosen for baseplate width 2.27fc'B
t2h'
If B is large enough, the critical sectionsare in the vicinity of the column flange tips [see Figure 8.15(a)].Note that the location of these sections is identical to the assumptiononLRFDp.2101,butwechosetousebinsteadofn todenotethedistance from these sections to the edge of the plate. To be conservative, the plate design bending strength is computed only for the length a:
[
I');(
[
@Mn= 0.9FYa
)I
T)( B m
2 1.02fiub( b
which requires that
Case 2: (e = MJPJ > W6 As shown in Figure 8.15@),the anchor bolts on the left end of the base plate are required to resist tension. This case is considerably more complex than case 1. However, the base plate thicknessformulasderived in case 1are applicablefor case 2 and are not repeated here. An additional base plate thickness requirement for plastic bending of the plate due to T, [seeFigure 8.15@)]is
t 2 2.108
iTu(Xiwe
where T, = required tension strength of anchor bolt group. Equilibrium of the vertical forces requires T , = C,  P, where
c, = 1.02fc'Ba in which B and u are unknown. Moment equilibrium requires T, (h:)+P,
(Y)=M,
Substitutionof T, = C,  P, and then substituting for C, we find that 1.02fc'B
340
Connections in which B is unknown and h is approximately known. For example, initially we can assume that h = H  (minimum w e). Alternatively, we can assume that h = H  0.5h'. For a trial value of B 2 [bf+ 2 (S, + 1/16 in.)], we can compute a and T,. Then, we can use the followingprocedure adapted from Shippand Haninger [23]to determine the size and the number of anchor bolts required to produce T,. An identical anchor bolt group will be used on the compression end of the base plate. If the governing loading combination includes wind, an identical anchor bolt group on each end of the base plate will be necessary since the direction of M u reverses when wind reverses. Let n denote the total number of anchor bolts; thus, there are n/2 anchor bolts on each end of the base plate. Each anchor bolt is assumed to resist an equal amount of shear. The design requirement for the anchor bolts that produce T, is
where
qR,, = design tension strength of one anchor bolt (p. 827) C, = shear coefficient (accounts for the effects of various shear failure surfaces) C, = 1.10 when the base plate is embedded in the concrete support and the top surface of the base plate is flush with the support surface
C, = 1.25when the base plate is recessed in grout
C , = 1.85 when the base plate is supported on, but not recessed in, grout. If a structural designer wants to try a particular anchor bolt group having n / 2 bolts of a certain size on each end of the base plate, the required value of B can be computed by using the following formulas:
Mu+P, ( h  $ ) p, + T , B2
pu +Tu 1.02fcra
Select an A36 steel base plate for the following reaction requirements of a W14 x 99 column: P, = 524 kips, M u= 120 ftkips = 1440 in.kips, and V, = 24 kips due to 1.2D + 1.6L + 0.5(L, or S or R). For the support, assume that the concrete grade is 3 ksi.
Solution For a W14 x 99, b,= 14.57in. and d = 14.16 in.
8.7 Column Base Plafes
341
For an anchor bolt diameter d,, the minimum value of h’ = 4.25d,. We need an estimated anchor bolt diameter to begin our design. When case 1 is applicable, the anchor bolt diameter required for strength is zero since T , = 0. However, for erection purposes we would not use less than a 0.75in.diameterfor the column size used in this example. Larger anchor bolt diameters may be required when case 2 is applicable. From LRFD p. 813 for a 0.75in.diameter bolt, the minimum socket wrench head clearance is C, = 1.25 in. From LRFD p. 682 for a 0.75in.diameter bolt, (minimum we) = 1.25 in. However, LRFD Table 113 (p. 1157) indicates the hole diameter in a base plate for a 0.75in.diameter bolt should be 1.3125 in., which is larger than what was assumed on LRFD p. 682, and we find a revised estimate of
(minimumw,)=1.25in.+(1.31251.25)=1.3125in.AssumethatHl[14.16+2(1.25 + 1.3125)= 19.3 in.]. Try H = 20 in. and h‘ = (20  14.16)/2 = 2.92 in. Since (e = M,/P, = 1440/524 = 2.75 in.) I (H/6 = 20/6 = 3.37 in.), Case 1 is applicable and a = H  2e = 20  2(2.75) = 14.5 in. For strength, we need B, =
p, 1.02fc’(H2e)

524 = 11.81 in 1.02(3)(14.5)
Since (b,= 14.57 in.) > 11.8 in., choose B = 16 in., which gives
b = [16  0.8(14.57)]/2 = 2.17 in. The required base plate thckness is
For the investigated loading combination, a 16 x 20 x 1.25 baseplate is needed. If we only want to use an anchor bolt group with n = 4 bolts and a base plate embedded in grout, the required tension design strength of one bolt is
From LRFD p. 827 for a 0.75in.diameter A325 bolt, we find that (@Rn= 29.8 kips/ bolt) 2 7.5 kips/bolt.
Select an A36 steel base plate for the following reaction requirements of a W14 x 99 column: P, = 280 kips, M u = 240 ftkips = 2880 in.kips, and V , = 48 kips due to 1.20 + 1.3W+ 0.5L + O.5(Lror S or R). For the support, assume the concrete grade is 3 ksi. Solution For a W14 x 99, b,= 14.57 in. and d = 14.16 in. Try d, = 0.875 in.; h‘ = 4.25db= 3.72 in.; H 2 [d + 2’ = 14.16 +2(3.72) = 21.6 in.]. In Example 8.8, for a different loading combination, we needed B = 16 in. Try a base plate size of B = 16 in. and H = 22 in., which gives h’ = (22 14.16)/2 = 3.92 in. and b = 2.17 in.
342 Connections
Since (e = MJP, = 2880/280 = 10.29 in.) > (H/6 = 22/6 = 3.67 in.),Case 2 is applicable. Try h = H  2.25 in. = 23  2.25 = 20.75 in:
P,, ( 2 h  H ) + 2 M U 1.02f ,’B u = 20.75  (20.75) 
280( 41.5  23) + 2 (2880) = 6.36 in. 1.02( 3)(16)
T,, = 1.02fJBa  P, = 1.02(3)(16)(6.36) 280 = 31.4 kips If we only want to use an anchor bolt group with n = 4 bolts and a base plate embedded in grout, then the required tension design strength of one bolt is
From LRFD p. 827 for a 7/8in.diameter A325 bolt, we find that
($Rn= 40.6 kips/bolt) 2 430.7 kips/bolt
r
__ 1 , t 2 [ 2 . 1 Tu0 ( 8h , ,w:/,/)y,’801=2’1w( =.2 31.4( 3.92  2.25)’ = 0.64 in.] = 0.64 in. 36(16)
For the investigated loading combination, a 16 x 23 x 1.75 base plate embedded in grout with four headed 7/8in.diameter A325 anchor bolts is adequate. 8.8 COLUMN SPLICES
Suggested column splice details for columns in a multistory building are given on LRFD pp. 1164to 1191.Column splicesare usually occur at 4 ft above the finish floor level, which means that the same W section is used in two stories. An axial compressive forceusually can be totally transferred by bearing of the upper column on the lower column at the splice location. This requires that the column ends at the splice locations be milled to provide a smooth bearing surface. The section depth of the upper column is usually less than the section depth of the lower column. This requires that filler plates or shims be inserted between the splice plates and the upper column flanges. In some cases, as shown in case III on LRFD p. 1175and in case IX on LRFD p. 1189, a butt plate must be welded to the bottom end of the upper column end at a column splice location to provide an adequate bearing seat on the top end of the lower column. The column splice plates must be designed to resist the required factored moment and shear in the column at a splice location. When all the required factored axial compression force in a column cannot be resisted by bearing, the remainder of
8.9 Simple Shear Connections For Beams
343
this force must be resisted by the splice plates and their connectors. All the required factoredaxial tension forcein a column must be resisted by the spliceplates and their COMeCtOrS. 8.9
SIMPLE SHEAR CONNECTIONS FOR BEAMS The connectionsfor this category are designed to transfer only the required factored shear at the end of the beam from the beam end to the column. Definitions of fully restrained (FR) and partially restrained (PR) connection types are given on LRFD p. 625. In the structural analysis of the structureshown in Figure 1.10, joints 2,5,7, and 8 are rigid (nondeformable).FR connections must be designed at these joints. An FR connectionmust have adequatestrength and stiffness to transfer the required shear and bending strengthsat the beam ends to the column without any appreciable change in the angle between each beam and column. At joint 10 in Figure 1.10, the beam end is hinged to the column. A PR connection is designed at joint 10 to transfer only the required beam end shear to the column. At joint 4 in Figure 1.10, there is a rotational spring between the end of member 1and the column.A PR connectionis designed to transfer the required beam end shear and the required rotational spring moment from member 1to the column. When loads are applied to the structure, there is a sigruficantchange in the anglebetween the left end of member 1and the columns. The moment at the end of member 1is on the order of 20 to 60%of the moment that would exist if the left end of member 1were rigidly attached (fully restrained) at joint 4. When the moment capacity of the connected parts is negligible in PR construction, the terminology "simple framing" is often used to describe the type of construction. When no moment is to be transferred by a connection, structural designers refer to the connection as a simple connection, a shear connection, or a nomoment connection.Simpleconnectionsare designed to be flexibleenough to allow the beam end essentiallyto rotate freely as assumed at a hinge in structuralanalysis. Simple shear connections commonly used for beams are web framing angles, beams seats, shear endplate connections,and single plates. Informationfor each of these connections is given in LRFD Part 9. The simple connection choice is often a matter of personal preference, but there are situations when the choice is dictated by framing conditions, fabrication costs, and convenience in field erection. Suggested details for beamtobeam connections are given on LRFD pp. 9179 and 1066.The top flangesofallbeamsinafloorsystemmustbeatthesameelevation. For twomutually perpendicular sets of beams in a floor system, the beam ends in one set must be coped to eliminatethe otherwise interfering flanges (seeLRFD pp. 8226, 1066). Tabular information that is useful in the required checking of block shear rupture at coped beam ends is given on LRFD pp. 8214 to 8224.
8.9.1 Beam W e b Connections Recommended design procedure information for allbolted doubleangle COMWtions and example problems are given on LRFD pp. 911 to 914. Tables of these connectionsare given on LRFD pp. 922 to 987. Although a gage of 3in. is shown in these tables, other gages may be used, provided that the design bearing strength is reduced for a gage in the range of 2.67d to 3d, where d = bolt diameter. As permitted by footnote c in LRFD Table J3.4 (p. 682), the minimum end distance of 1.25 in. is
344
Connections used. These connections are designed on the assumption that the effects due to eccentric shear are negligible. Therefore, each bolt in these connections is assumed to resist only an equal share of the required beam end shear (see LRFD Table 92, p. 922). The design strength of the connection angles is assumed to be governed by shear through the net section. In a bearingtype connection, the design strength of the connection may be governed by bearing on the beam web or on the connection angles. The length of the connection angles should be at least T/2, where T is the flat height of the beam web. If the connectionangle length is less than T, the angles should be attached on the web as close as possible to the compression flange of the beam in order to provide lateral stability for that flange. Recommended design procedure information for onesided allbolted doubleangle connections and example problems are given on LRFD pp. 9161 to 163. A bolt group in each leg of the single angle is subjected to an eccentric shear that must not be ignored. The tabular information can be used to design a bolt group subjected to an eccentric shear. In welded web connections,either a WTsection or a flat plate may be less expensive than a web framing angle and serves the same structural purpose. However, as shown in the figureon LRFD p. 9128,we would install either a top angle or a flat plate to provide lateral support for the compression beam flange. Recommended design procedure information for bolted/welded and allwelded doubleangle connections and example problems are given on LRFD pp. 915 to 920. Tables of these connections are given on LRFD pp. 988 to 990. The weld group on each side of the beam web is subjected to an eccentric shear that is not ignored in the design of these connections.
8.9.2 Unstiffened Beam Seats
Recommended design procedure information for allbolted unstiffened beam seats, example problems, and tables for these connections are given on LRFD pp. 9128 to 9136. Recommended design procedure information for allwelded unstiffened beam seats, example problems, and tables for these connections are given on LRFD pp. 9132,9135, and 9137. A single angle may serve satisfactorily as an unstiffened beam seat when the required factored beam end shear is small. For a beam supported by a seat angle, as shown in the figure on LRFD p. 9128, either a light angle or plate must be attached either to the top flange of the beam or to the upper portion of the beam web to provide lateral support for the compression beam flange. Usually, an L4 x 4 x 0.25 is chosen for this top angle. Since the entries in Table 97 for a 4in. outstanding leg of an angle for a welded beam seat are identical to the corresponding entries in Table 96 for a bolted beam seat, the critical section for bending of the angle leg was assumed to be the same whether a bolted beam seat or a welded beam seat is used. This is consistent with the assumption that the bolt group is subjected only to shear. In Figure 8.16(b),we show the assumed deformation of the seat angle. Only the outstanding leg of the angle is subjected to bending. The inclined bearing forcebetween the beam and the beam seat has a horizontal component that pins the heel of the seat angle against the column flange. Therefore, the following design bending requirement of the outstanding angle leg must be satisfied:
8.9 Simple Shear Connections For Beams
345
Column web
e=
3 N +4
2
( ): = 83 +5N 
e, = e  t +
Heel of seat angle remains in contact with the column web.
t
(b) Assumed behavior
(a) Beam seat details
FIGURE 8.16 Bolted beamseat connection
1
=0.225FyBt2 > ( M u= P u e c ) where
B = length of beam seat angle section t = thickness of beam seat angle section
e, = eccentricity of the beam reaction P , with respect to the critical section e = eccentricity of the beam reaction P, with respect to the heel of the angle
e = 0.75 + N / 2 N = bearing length defined in the next paragraph e, = e  ( t + 0.375) = 0.375 + N / 2  t Note: t + 0.375 is assumed to be representative of the value of k for an angle section. In preparing LRFD Tables 96 and 97, a beam setback of 0.75 in. was assumed and is made in the following discussion. The critical section in bending is at the toe of the fillet, which is assumed to be located at t + 0.375 in. from the heel of the angle. If beam web yielding [LRFD Eq. (K13)] governs the minimum bearing length Nq required on the angle leg,
N,
''
=Fptw
2.5k
346 Connections
If beam web crippling [LRFD Eq. (Kl5)] governs the minimum bearing length N , required on the angle leg,
The actual bearing length used in preparing the LRFD Manual tables was
The bearing length N is the larger of Nv,N,, and N,,,. Assume that the bearing force is uniformly distributed on the bearing length N:
e, = 0.75 + N/2  (f + 0.375) = 0.375 + N/2  f The value of f that satisfies the design bending strength requirement for the angle leg is ~
P, +J[Pu+0.9FyB(0.375+N/2)]Pu f2
0.45 FyB
Compute t for the desired value of B; the usual values of B are 6 and 8 in. Let n be the required number of bolts in a beam seat. The design requirementfor a bearingtype bolt group in a beam seat is that n 2 P J R , where R, is the smaller of the single shear design strength of one bolt and the bearing design strength at one bolt location.
See figures on LRFD p. 9128. Venfy entries LRFD Table 96 for the following: Thebeam seat is LA x 4 x 0.75 x 8 Fy= 36 hi., and a type D bolted connectionisused with three A325N bolts; d = 7/8 in. The beam is a W16 x 50:
t, = 0.380 in.
fr= 0.630 in.
d = 16.26 in.
k = 21/16 in.
Fy = 36 hi.
Solution Bolt shear strength The bolts are assumed to be subjected only to shear. For single shear,
@R,= 3(21.6) = 64.8 kips
which agrees with the tabulated value on LRFD p. 9136. Bending strength of the outstanding leg As noted, N,,, was used for the bearing length N in preparing the LRFD tables. Also, t, = 3/8 in. was used in preparing the LRFD tables:
8.9 Simple Shear Connections For Beams
N, = pu 2 Fryt

347
55 = 2.04 in. 2( 36)(0.375)
For illustration purposes, we will also compute Nq and N,. For the prevention of web yielding,
N,
=
P U
Fptw
2.5k =
36 (0.380)
For the prevention of web crippling:
N,
=( ?)[
51(0.380) 1/36 55(0.630)/ 0.380
= 0.385 in.
l]( %)I5
The bearing length N is the larger of Nq, NW, and N,:
N = N,,, = 2.04 in. For the bending strength requirement of the seat angle, we need:
P, + J [ P , +0.9FyB(0.375+N/2)]P,
t> 55
+ ,/[
0.45 F, B 55 + 0.9( 36)( 8)(0.375+ 2.04/2)]( 55)
t2
0.45( 36)( 8)
= 0.744 in.
Use t = 0.75 in. If we wish to verrfy the design bending strength tabulated in the LRFD Manual, we proceed as follows:
e,
= e  ( t +0.375) = 0.375
N t +2
55.6 N = N m = (PRn = 2.06 in. 2 Fry t 2 ( 36)(0.375)
e, =0.375+2*06 0.75 = 0.655 in. 2
(PR,, = P , =
0.225( 36)(8 ) ( 0.75)’ = 55.6 kips 0.655
and the tabulated bending strength on LRFD p. 9136 is (PR, = 55.6 kips.
348
Connections Alternatively, we could have treated the design bending strength as unknown, which requires the following approach for the determination of the design bending strength:
N
e, = e  ( t + o m ) = o m +  2
t
Substitute N into e, and e, into Muto find
$R,, = P, = b + & G where b=2F,f,,, (f0.375)=2(36)(0.375)(0.750.375)=10.125 ~ = 0 . 9 F , ~ f , F , B t=~0.9(36)(0.375)(36)(0.75)' = 1968.3 Thus, we obtain the design bending strength:
@R,,= P, = 10.125+
y'( 10.125)2+ 1968.3 = 55.6
kips
See the figures on LRFD p. 9128. Verify the entries in LRFD Table 97 for the following. The beam seat is L7 x 4 x 0.75 x 8, F, = 36 ksi. with5/16 in. E7Oxx fillet welds. The beam is a W21 x 62: t, = 0.400 in.
f,= 0.615 in.
d = 20.99 in.
k = 11/8 in.
F, = 36 ksi
Solution Weld design strength The welds are assumed to be subjected to shear and bending. According to LRFD p. 9132, this design strength was computed using the elastic method. Include the length of the weld returns in q, due to shear. Ignore the length of the weld returns in qM due to bending moment (see Figure 8.17): S, = 5/16 in.
B =2S, = weld return length = 0.625 in.
8.9 Simple Shear Connections For Beams
349
B = 2Sw
H
S,=
weld size
FIGURE 8.17 Example 8.11
$R,, = P, = 10.125+
J(10.125)2 + 1968.3 = 55.6 kips
e = 0.75 + N/2 N = N = P U 2F,t,

pu = 0.0370 P, 2(36)(0.375)
[$ (0.6FE,t,) = 0.75(0.6)(70)(0.707SW)] 2 q is required.
Solve for P,, which is renamed as the design strength $Rn:
$Rn= 55.0 kips and the tabulated value on LRFD p. 9137 is $Rn = 53.4 kips. Bending strength of the outstanding leg As noted, N,,,was used for the bearing length N in preparing the LRFD tables. Also, t,, = 0.375 in. was used in preparing the LRFD tables:
N,
= pu

50 = 1.85 in. 2( 36)(0.375)
2 F, t , For illustration purposes, we will also compute Nq and Nw,. For the prevention of web yielding,
N WY
=
P U
F p t 1”
2.5 k =
36 ( 0.400)
350
Connections For the prevention of web crippling,
50  1]( 51( 0.400)21/36*0.615/0.400
s) l5
= 2.35 in.
The bearing length N is the larger of NW,N,, and N,:
N = (N,= 1.85 in.) For the bending strength requirement of the seat angle, we need
Pu +,/[P,, +0.9FyB(0.375+N/2)]Pu f>
f2
50
0.45 FyB
+ 4[ 50 + 0.9( 36)(8)(0.375 + 1.85/ 2)]50 0.45(36)(8)
=0.687 in.
Use f = 0.75 in. The computationsfor the verificationand determination of the design bending strength are identically the same as in Example 8.10. 8.9.3 Stiffened Beam Seats
When the required factoredbeam end shear exceeds the design strengthvalues listed in the unstiffened beam seat table, the beam seat angle is stiffened by either vertical angles or plates to eliminate bending in the outstanding leg of the seat angle. Recommended design procedure information for allbolted stiffened beam seats, example problems, and tables for these connections are given on LRFD pp. 9138 to 9144. Recommended design procedure information for bolted/welded stiffened beam seats, example problems, and tables for these connections are given on LRFD p. 9140,9143, and 9146. The paired stiffener angles shown in the figure on LRFD p. 9138 can be separated to accommodate column gages. A filler or spacer plate should be inserted in the separation gap and stitchbolted as one does in a doubleanglecolumn when the angles are not in contact.
See the figures on LRFD p. 9138. Venfy the entries in LRFD Table 98 for the following.Use A36 steel. The beam seat is a 6 x 10 x 0.375 plate stiffened by a pair of W x 5 x 5/16 x 8.625, and attachedby a type Abolted connection(sixA325Nbolts;d= 7/8 in.)toa W3Ox99beam.
8.9 Simple Shear Connections For Beams
351
Solution Bolt shear strength The bolts are assumed to be subjected only to shear. For single shear,
4Rn= 6(21.6) = 129.6kips which agrees with the tabulated value on LXFD p. 9144.
Pair of angle stiffenem For the bearing strength, LRFD J8.1 (p. 689): #& = 0.75 (1.8FA) The effectivebearing length of the stiffener is the outstanding leg length minus 0.75 in:
Apb=2(5  0.75)(5/16) = 2.66 in.2 #R,, = 0.75(1.8)(36)(2.66) = 129 kips which agrees with the tabulated value on LXFD p. 9144. Check widththickness ratio
When the angle legs are in continuouscontact,
If the angles are separated, a thicker angle is required to satisfy the limiting widththickness ratio.
See the figures on
LRFD p. 9138.Venfy the entries in LRFD Table 99for the
following.use A36 SteeL The beam seat is a 6x 12x 0.375plate stiffenedby a 6x 15x 0.625plate and attached by 5/16 in. met welds (E7OXX) to a W30 x 116 (f, = 0.565 in.) Solution
For the stiffener plate, (t = 0.625 in.) > (f, = 0.565 in.) as required
(t = 0.625 in.) 2 [2(5/16)= 0.625 in.] as required
Check the widththickness ratio:
asrequired
352
Connections
FIGURE 8.18 Example 8.13
Design strength of the S, = 5/16 in. welds: The welds are assumed to be subjected to shear and bending. This design strength was computed using the elastic method (see Figure 8.18):
q =  p=u
p, pu =O.0278Pu 2(L+B)2(15+3)
A
From the top of the weld group to the centroid of the weld group is c=
2( 15)(7.5) =6.25 in. 2(15+3)
!
+15(7.56.25)2 +3(6.25)* =844 W=6in.
Since the only variables on LRFD p. 9145 are W,L, and the weld size, assume that
N = W 0.75 = 60.75 = 5.25 in. e = W   N = 6   5.25 = 3.375 in. 2 2 Mc qM
p, (3.375)(6.25) = 0.0250P" 844
=I= q
=
d
m
[@(0.6FE,f,) = O.75(O.6)(7O)(O.707Sw)]2 q is required
Solve for P,, which is renamed as the design strength @Rn: (PR, = 186 kips
and the tabulated value on LRFD p. 9145 is (PR, = 154 kips.
8.10 Moment Connectionsfor Beams
353
8.9.4 Shear EndPlate Connections
At the end of a beam, as shown in the figure on LRFD p. 991, a plate can be shopwelded to the beam web and fieldbolted to a column flange to serve as the means to transfer the required factored beam end shear to the column. This type of connection is classified as a shear endplate connection. Recommended design procedure information for these connections is given on LRFD pp. 991 to 9127. The end plate should be attached on the beam web as close as possible to the compression flange of the beam in order to provide lateral stability for that flange. 8.9.5 Bracket Plates On LRFD p. 125, there are twofiguresof bracket plates. Design information for these
bracket plates is given on LRFD pp. 125to 1210. LRFD Table 121gives the critical net section indicated in the figure for that table. For the bracket plate, the design bending requirement for the extreme fiber in tension is
( +Mn = O.9FyS,, ) 2 P,e where
S,, = net section modulus of the critical section e = eccentricity of the beam reaction P, with respect to the critical section Examples 8.1 and 8.2 discussed the design strength of a bolt group for a bracket plate. For design purposes (to select the required number of bolts), the tables on LRFD pp. 840 to 887 can be used. 8.10 MOMENT CONNECTIONS FOR BEAMS
The connectionsfor thiscategory are designed to transfer the required factored shear and moment at the beam end to the supporting member. 8.10.1 BeamtoBeam Connections and Splices
Suggested design details for moment connections and splices of a beam are shown on LRFDpp. 1057and 1066.In a floor system, one may desire that the lighter loaded set of intersecting beams be continuous at their supports (the heavier loaded set of intersectingbeams, which are usually called girders). As shown on LRFD pp. 1057 and 1066, the beam end shears at the girder supports are resisted by either web connections or by beam seats. The continuous beam end moments are transferred either through top and bottom spliceplates or through top splice plates and the beam seats. The moment to be transferred can be treated as a couple whose lever arm is d + tp, where d is the beam depth being connected and fp is the thickness of the connectionplate. The top splice plate is subjected to a tension force of T , = M,/(d + tp)where M uis the requiredmoment at the continuousbeamsupport.If a tensionspliceplate isbolted to the beam flange being C O M & ~ ~ , the chosen splice plate must satisfy LRFD J5.2. For a splice plate that transfers a compression force of C, = M u/(d + t,), the limiting widththickness ratios on LRFD p. 639 forflangecover plates are applicable.
354 Connections
If the continuous beam is designed by assuming that
then, for the compression splice plate we must require that
where b is the plate width between the lines of bolts in the splice plate and t is the thickness of the splice plate. At the splice gap location, the design compressive strength requirement of a compression splice plate is that ( o , P , = 0.85A,Fy ) 2 C, where Ag is the gross area of the splice plate. 8.10.2 BeamtoColumn Connections
Suggested design details for this type of connection are shown on LRFD pp. 1010, 1022, and 1023.Only two of the suggested possibilities (seeFigures 8.19 and 8.20) are mentioned in this text. For the extended endplate connection shown in Figures 8.13 and 8.19, the suggested design procedure informationis given on LRFDpp. 1021 to 1035.There are comprehensive example problems in the LRFD Manual,and we see no need to provide any additional examples. The criteria given on LRFD pp. 1035to 1042 must be used to determine when column web stiffeners aligned with the beam flange ends are required. A paragraph on LRFJ3p. 1025 encourages the avoidanceof such stiffeners. According to research by Curtis and Murray [24], excessive column flange bending at the beam tension flange does not occur when the column flange thickness satisfies the following criterion: t,,
> JT O.9Fyb,
where b, = effective column flange length region subjected to bending (in.)
b, = 2 . 5 for ~ ~fourbolt arrangement (in.) [see Figure 8.19(a)]
b, = 3.5pb+s4 for eightbolt arrangement (in.) [seeFigure 8.19(b)] s4 = 2p, + tpt
The other parameters are defined on LRFD p. 1024.
8.10 MomentConnectionsfClrBemns 355
(a) Fburbolt t
(b) Eightbolt anangement
u section 22
FIGURE 8.19 Moment endplate connection
FIGURE 8.20 Top and bottom plate connection
356
Connections
8.11 KNEE OR CORNER CONNECTIONS LRFD Figure 1029 (p. 1067) shows examples of knee or comer connections frequently used in onestory frames designed using FR connections. Figure 8.21(a)showsa squarekneeconnection.The beam extends to the edge of the exterior column flange and sits on a base plate (crosshatchedelement)welded to the top end of the column. A plate is welded to the end of the beam and to the exterior column flange. As shown in Figure 8.21@), the tension flange forces must be transferred by shear into the beam web. Usually, a pair of beam web stiffeners is required from the points C to D to transfer some of the column flangeforce C, via fillet welds subjected to shear into the web of the girder. If either the design shear strength or the design buckling strength of the beam web is not adequate, a pair of diagonalweb stiffeners must be provided from points A to C.
& L
Section 11
(a) Side elevation view
I T1
f TB

I 
(b) FBD of flanges and web
FIGURE 8.21 Comer connection
'h
8.22 Knee or Corner Connections
357
Design the squareknee connection in Figure 8.21 for a W24 x 55 beam and a W14 x 82 column. For the W24 x 55 section, the required strengths are V, = 60 kips, P, = 24 kips, and Mu= 320 ftkips. For the W14 x 82 section, the axial compressive design strength is @Pn= 558 kips and the required strengths are V,,= 24 kips, P, = 60 kips, and Mu= 320 ftkips. Use A36 steel and E70 electrodes.
Solution In Figure 8.21(b), assume: 1. The beam flange forces are Mu TD ==
0.95d
p,, 2
c, = Mu +p,, 0.95d
2
320(12) 24   = 160 kips 0.95(23.57) 2 320(12) 24 +=184 0.95(23.57) 2
kips
2. The column flange forces are
Mu c, =+=
0.95d
p,, 2
320(12) 60 += 313 kips 0.95(14.31) 2
Design the beam end plate: The beam flangewidth is 7.01 in. and the column flange width is 10.13in. Choose the end plate width as 8 in. to accommodate fillet welds on the column flange. Since the end plate is in tension, the design strength requirement for yielding on As is [0.9(36)(8t)]2 (TB= 253)
t 2 0.976 in. is required. Use t = 1in. The column flange thickness is 0.855 in. and the endplate thickness is 1 in. The thicker part joined is 1in., for which the minimumacceptableS, = 5/16 in. and the maximum acceptable S, = 15/16 in. Try S,= 5/8 in. The design requirement for the weld group on the column flange is [0.75(0.6)(70)(0.707)(0.625)(8 + X,)]2 (TB= 253)
L, 2 10.2 in. is required for S, = 0.625 in. to develop the column flange force TB= 253 kips. On the end of the beam web, the design requirement for the weld group is [0.75(0.6)(70)(0.707S,)(2)(21)]2 (TB= 253)
358 Connections
For strength, S, 2 0.277 in. is required. Use S, = 5/16 in., which is the minimum acceptable weld size. Check the beam web:
( h/ t , = 54.6)5 (187,/=
= 69.7)
From B to A in Figure 8.21,the design shear strength of the web is
4V,, = 0.9(0.6)(36)(21)(0.395) = 161 kips which is less than the required strength TB= 253 kips. Therefore, on the beam end, there is an excessforceof 253  161 =92kips. Ifthe web buckling strength from LRFD K1.6 in the A to C directionis inadequate, a pair of diagonal web stiffenersmust be designed. From D to A, the design shear strength of the web is +V,,= 0.9(0.6)(36)(11)(0.395) = 84 kips which is less than the required strength TD= 160kips. Therefore, parallel to the top beam flange, there is an excess forceof 160 84 = 76kips.If the web buckling strength from LRFD K1.6in theA to C directionis inadequate,a pair of diagonalweb stiffeners must be designed. The diagonal length of the web in compression is
4
d, =
= 23.7 in.
From LRFD K1.6 @. 694),the web buckling design strength is
[
+R,,=0.9 4100(0'395)3 23.7
*]
= 57.6 kips
and the required strength is
P,,= J(92)' +(76)2 = 119 kips c P,, a pair of diagonalweb stiffenersmust be designed for the region from Since +R,,
A to C for P,, = 119 kips.
In Figure 8.20,the ends of the diagonal stiffener need to be mitered to fit the corners in bearing since the diagonal compressionforce exists at points A and C. The length of the diagonal stiffener is
L, =.\1(23.5720.505)' +[14.322(0.855)]' =25.85 in. From LRFDK1.9 (p. 696),KL = 0.75 (25.85)= 19.4in.is the effective length of the pair of diagonal web stiffeners to be installed between points A and C. Try a pair of 3.25 x 23.75 x t plates: b =325 = 8.67 s; t 0.375
1
= 15.8
as required.
Sincethe ends of the diagonalstiffenerare mitered to fit the cornersinbearing, LRFD
8.22 Knee or Comer Connections 359
J8.1 (p. 689) requires that [& = 0.75(1.8)(36AJ] 2 (119  57.6 = 61.4 kips)
[A, = 2( 3 . 2 50395 +2% l5)t]>1.26 in.* t 2 0.251 in. is required.
Use t = 0.25 in.
From LRPD K1.9 @. 696), the effective area of the diagonal web stiffener is A, = 2(3.25)(0.25)+ (25)(0.395)(0.395)= 5.53 in?
The moment of inertia for column buckling perpendicular to the web plane is
I, =
0.25[2 ( 3.25) + 0.395 J = 6.83ine4 12
($1,
=
0.75 ( 25.85) = 17.5 1.11
From L,RFD E2 (p. 639):
ac = i7.5,/3ZEi% = 0.617
a:
= 0.380
[qPn= 0.85(5.53)(0.658°3ao)(36) = 144kipI 2 ( P , = 119kips) Therefore, a pair of 3.25 x 0.25 x 23.75 plates is adequate. Minimum fillet welds along the stiffenermust be provided to prevent buckling of each individual web stiffener plate as a column in the direction of the plane of the beam web. The thicker part joined is 0.395 in. and the minimum fillet weld size is 3/16 in. Check the beam web at point C in Figure 8.21(b).Since the diagonal stiffener is required, we would automaticallyprovide a pair of web stiffenersfrom C to D.From the FBD of this stiffener in Figure 8.20(b), the required compressive strength of the stiffener from C to D is P, = 313  (92 + 60)= 162 kips LRFDJ8.l @. 689) requires [@,, = 0.75(1.8)(36Ab]2 161kips
[A, = 2( 3.25+*G 0395 15)t] 2 3.31 in.2 t 2 0.660in.
US^ t = 0.75 in.
360 Connections
From C to D, the design shear strength of the beam web is
$Vn= 0.6(36)(23.57)(0.395)= 181 kips ($Vn= 181 kips) 2 (P, = 161 kips) Since this design strength is so close to the required strength, we would extend the stiffener to the toe of the top web fillet. Therefore, the strength requirement of the weld group (4 welds) is 4 [0.75(0.6)(70)(0.707S,)(21)]
2 (P, = 161)
The weld leg size required for strength is S, 2 0.086 in., and the minimum acceptable S, = 0.25 in. for the 0.75in. thicker part joined. Use S, = 0.25 in. PROBLEMS 8.1 Select an A36 steel base plate for the following reaction requirements of a W14 x 82 column: P , = 400 kips, M u= 120 ftkips, and V, = 30 kips. For the support, assume that the concrete grade is 3 ksi. 8.2 Select an A36 steel base plate for the following reaction requirements of a W14 x 82 column: P, = 200 kips, M u= 300 ftkips, and V, = 60 kips. For the support, assume that the concrete grade is 3 hi.
8.3 A pair of LA x 3.5 x 0.5 is fillet welded to a column web (t, = 0.510 in.) and bolted to a beam end (t, = 0.510 in.).Five 0.75in.diameter A325N bolts in a bearingtype connection with standard size bolt holes are in the 3.5in. angle legs and the beam web. The pair of angles, the beam, and the column are A36 steel.
Use LRFD Table 93 (p. 988) and determine the required weld size for the weld group. Venfy the design strength tabulated for weld group B. This design strength was computed using the elastic method. Include the length of the weld returns in qo due to shear. Ignore the length of the weld returns in 9,,, due to bending moment (see Figure P8.3): B = 2Sw
k4
S,,, = weld size FIGURE PS.3
Problems 361
PU
PU
==
41,
A
2(L+2SW)
S, = weld leg size
2S, = weld retum length
e = 2.25 in. (from LRFD, p. 8271)
q = d d +q; [t#1(0.6F,tJ
= 0.75(0.6)(70)(0.707Sw)] 2 q is required
Solve for P,, which is renamed as the design strength @Kn. 8.4 Solve Problem 8.3 for 7/8in.diameter A325N bolts. 8.5 Solve Problem 8.3 for 1in.diameter A325N bolts. 8.6 Solve Problem 8.3 for A307 bolts.
8.7Using LRFDTables92 and 93, selecta pair of anglesbolted to beam web and welded to column flange. The beam is a W24 x 55 and the column is a W14 x 82. The weld group on the pair of angles is also on the column web. Use A36 steel for the angles, the beam, and the column. Use 0.75in.diameter A325N bolts in a bearingtype connection. The required shear strength of the beam end is 100 kips. 8.8 Solve Problem 8.7 for A307 bolts. 8.9 As shown on LRFD p. 989, a pair of L3 x 3x 5/16 is filletwelded to a column web and to a beam end. The pair of angles, the beam, and the column are A36 steel. Venfy the entries given in LRFD Table 94, using:
(a) LRFD Table 842 for weld group A. (b) The elastic method for weld group B. Include the length of the weld returns in ql,due to shear. Ignore the length of the weld returns in qM due to bending moment (see Figure P8.9): p, 9 == A
P U
2(L+2SW)
S, = weld leg size
2S, = weld retum Iength Mc q M == I
(Pue)(L/2)3(P,e) 2 ( ~ ~ / 1 2 ) L~ e=3in.
362
Connections
S= ,
weld size
FIGURE pS.9
q=dqO’+q;
[+(0.6F,te)
= O.75(O.6)(7O)(O.707Sw)] 2 q is required
Solve for P,, which is renamed as the design strength @,. 8.10 If the web thickness of the beam and the column in Problem 8.9 is t, = 0.510 in.,do the weld sizes for Problem 8.9 satisfy LRFD J2.2b(p. 675)? 8.11 Use LRFD Table 94to select a pair of angles and the weld sizes to be used for the following conditions. The beam is a W24 x 55 and the column is a W14 x 82. Weld group B is on the column web. Use A36 steel for the angles, the beam, and the column. The connection must develop the design shear strength of the beam end. 8.12 Determine the block shear rupture strength of the connection in Example 91(p. 916)by using:
(a) LRFD equations for BSR @) LRFD Tables 847 and 848 (pp. 8215to 8224) 8.13 Determine the block shear rupture strength of the connection in Example 92@. 918)by using:
(a) LRFD equations for BSR (b) LRFD Tables 847and 848 (pp. 8215to 8224). 8.14 See the figure on LRFD p. 991.Verify the entries in Table 95for the endplate connection designed in Example 96 (p. 992).
8.15 Repeat Example 96 and Problem 8.14 for Fy= 36 ksi. 8.16 See LRFD pp. 9161 and 9162.Venfy the entries in Table 911 for the connection designed in Example 913(p. 9163). 8.17 See LRFD p. 9163.Verify the entries in Table 912for the connection designed in Example 914(p. 9166).
Problems 363 8.18 For the bearingtypeconnection shown in Figure P8.18,we are given the followingnumerical information:A36 steelbracket plate and column,g = 5.5 in., s = 3 in., ex= 6 in., column flange thickness = 0.855 in., 0.75in.dia.meter A325X bolts in single shear, and P,, = 50 kips. Use TXFD p. 852to determine the number of bolts required. Use LRFD pp. 125 to 1210 to determine the required bracket plate thickness for M = PJ?.
I
I I
I
*
FIGURE P8.18
8.19 For the bearingtype connection shown in Figure P8.18,we are given the followingnumerical information: A36 steel bracket plate and column, g = 8 in., s = 3in., ex= 14in., column flange thickness = 0.710 in., 7/8in.diameter A325X bolts in single shear, and P,, = 75 kips. Use TXFD p. 852to determine the number of bolts required. Use LRFD pp. 125 to 1210 to determine the required bracket plate thickness for M = PJ?. 8.20 For the welded connection shown in Figure P8.20, we are given the following numerical informatio~~ A36 steel bracket plate and column, b = 4 in., d = 10 in., e, = 12 in., column flange thickness = 0.855 in., E70 electrodes for the fillet
FIGURE P8.20
364 Connections
welds, and P, = 50 kips. Use LRFD p. 8187 to determine the required weld size. Determine the required bracket plate thickness such that M/S, 2 0.9Fy,where
M = P,e S, = 1,= elastic section modulus of the bracket plate
d/2
Also, the cantilevered plate length divided by the thickness must not exceed
A, = 9 5 / J F y 8.21 For the welded connection shown in Figure P8.20, we are given the following numerical information: A36 steel bracket plate and column, b = 12in., d = 14 in., ex = 14 in., column flange thickness = 0.710 in., E70 electrodes for the fillet welds, and P, = 100 kips. Use LRFD p. 8187 to determine the required weld size. Determine the required bracket plate thickness such that M / S , 2 0.9Fy,where
M = P,e
s, =I x  section modulus of the bracket plate  elastic d/2
Also, the cantilevered plate length divided by the thickness must not exceed
A, = 9 5 / E 8.22 Design the beam splice shown in Figure P8.22 Use A36 steel for the pair of splice plates and the W24 x 55 beam. The required shear strength of the beam ends is V, = 121kips. Use 0.75in.4iameter A325X bolts in a bearingtype connection and E70 electrodes. Determine the required number of bolts, the weld size, and the size of each splice plate.
uu I
I
FIGURE P8.22
8.23 Design the beam splice shown in Figure P8.22. Use Fy= 50 ksi steel for the pair of splice plates and the W24 x 55 beam. The required shear strength of the beam ends is V , = 167 kips. Use lin.diameter A325X bolts in a bearingtype connection and ESO electrodes. Determine the required number of bolts, the weld size, and the size of each splice plate. 8.24 Solve Example 8.5 for the tension member inclined at tan a = 0.5.
ProbIems 365 8.25 Solve Example 8.6 for the tension member inclined at tan a = 0.5. 8.26 Design a beam splice as shown in LRFD Figure 1020 (p. 1057).Use A36 steel for the splice plates and the W24 x 55 beam. The connectionmust develop $MpI = 330 ftkips and @V,,= 121 kips. Use 7/8in.diameter A325X bolts in a bearingtype connection and E70 electrodes. Determine the required number of bolts, the weld size, and the size of each splice plate. The girder is a W30 x 90.
8.27 Design an allweldedbeam splicesimilar to the welded half of LRFD Figure 1020 (p. 1057). Use A36 steel for the splice plates and the W24 x 55 beam. The connection must develop $Mpx= 362 ftkips and $V,,= 121 kips. Use E70 electrodes. Determinethe required weld sizes, the beam seat size, and the size of the spliceplate. The girder is a W36 x 135. 8.28 Solve LRFD Example 102 (p. 1016) for Mu= 362 ftkips, a factored beam end shear of 66 kips, and a factored column web shear of 36 kips. Also, LRFD Example 106 Solution A is applicable.
8.29 Solve Problem 8.28 for a W24 x 55 beam framed to a W14 x 82 column. M u = 503 ftkips. The factored beam shear is 92 kips. The factored column web shear is
50 kips. Use F,, = 50 ksi steel for the connection parts. Use lin.diameterA325X bolts and E80 electrodes.
GHAPTER
Plate Girders 9.1 INTRODUCTION In LRFD Chapter G (p. 658),we find that flexural members are classified as either beams or plate girders. Some of the characteristics of beams and plate girders need to be illustrated. Therefore, we choose to state the characteristicsof beams and plate girders in separate figures. The characteristics of a beam are given in Figure 9.l(d), where Fd is the yield strength of the flanges.Infmediafe web stiffeners,as shown in Figure 9.2(a)for a plate girder, are not allowed in the definition of a beam. However, bearing web stiffenersas shown in Figure 9.l(a) may be required by LRFD K1.9 (p. 696). A beam may be a builtup hybrid section (F,of the web and flangesmay be different).LRFD Appendix G (p. 6203) states that inelastic web buckling in a hybrid flexural member is dependent on the flange strain. ThecharacteristicsofaplufegirderaregiveninFigure9.2(c),whereFy,istheyield strength of the flanges. Bearing web stiffeners as shown in Figure 9.2(a) may be required by LRFD K1.9 (p. 696). Intermediate web stiffeners may be required by LRFD F2.3 (p. 6113) or, when tmsionfie2d action is utilized (seeFigure 9.4),by LRFD G4 (p. 6125). Intermediate web stiffeners [see Figure 9.2(a)]are not required when either of the following conditions is satisfied: 1. h l t , S 4 1 8 I G 2. @Vn2 V"
where FF = yield strength of the web V,, = required shear strength
@Vn= design shear strength given in LRFD F2.2 (p. 656) h and f, are defined in Figure 9.l(b). 366
9.2 lntroduction 367
ection 11 is typical cross section except at supports and concentrated load points. (a) Side elevation view
y 4
 _d
2
hs
 d
2 Welded section
Rolled section (W section)
(b) Section 11 when cross section is doubly symmetric
(d) Beam characteristicsare: Rolled or welded section.
1.5:
h
h Maximu=tw
h Maximm=
2000
&14,000
where
Fd = yield strength of the flanges u = clear distance between intermediate web stiffeners h and t, are defined in Figure 9.2(b).
368 Plate Girders
Intermediate web stiffeners (if required)
/
A
(a) Side elevation view of left half span length
L 2
lzJI$
X
2 Doubly symmetric section
Singly symmetric section
(b) Section 1  1 when LRFD G4 stiffeners are not used
(c) Plate Girder characteristics are: Rolled or welded section. Either or both of the following: h >IW
970
&
Intermediate web stiffeners are required by LRFD F2.3 or G4.
FIGURE 9.2 Plate girder.
As shown in Figure 9.2(b), the most commonly used builtup section for a plate girder is composed of two flange plates filletwelded to a web plate. Someother cross sections that have been used for a plate girder are shown in Figure 9.3. Any of the sectionsin Figures9.2 and 9.3may be a hybrid section;that is, for example, the web(s) may be Fy = 36 ksi steel and the flanges may be Fy= 50 ksi steel. In a multistory office or hotel building, large columnfree areas (ballroom, dining room, lobby) may be needed or desirable. Either a builtup plate girder or a truss may be the economical choice to span these columnfree rooms and to support the columns in the stories above these rooms. Students want design rules without any exceptions to be given for them to use in deciding the most economical type of flexural member for a specific design situation. Such rules are impossible for us to give, but the following guidelines may be helpful: 1. For rolled beams, an economical depth is L/20 to L/50, where L is the span length. See LRFD Table 42 (p.430).
9. I
Introduction
369
Notes: Groove welds at each end of the plate are not shown. Usually the grade of steel for the plate is chosen to be less than the grade of steel for the WT sections to create a hybrid section. For example: Plate : Fy = 36ksi WT Fr = 50ksi
(a) Rolled builtup section
r
T
A
(c) Box section
(d) Delta section
FIGURE 9.3 Plate girder cross sections.
2. For builtup beams and plate girders, an economical depth is L/15 to L / 8 . Simple spans of 70 to 150 feet are typical for plate girders in buildings and highway bridges. Shorter simple spans are economical in unusual cases and railway bridges. Longer continuous spans of 90 to400 feet are feasible,but for the longer spans the section depth varies from a maximum at the supports to a minimum at midspan. 3. For parallel chord trusses, an economical depth is L/12 to L / 8 .
A builtup plate girder usually has the following advantages when compared to a truss: 1. Lower fabrication cost. 2. Fewer field erection problems, but a larger crane capacity may be required. 3. Fewer critical points in the member at which design requirements may govern. In a truss, if an overload occurs in any member or connection, the result may be a disaster. However, if an overload occurs only at one point in
a plate girder, some of the neighboring material may help prevent a disaster. We restrict the discussion to the case where the required strengths are determined from an elastic factored analysis. In the LRFD Specification, there are two
370 Plate Girders
design procedures for plate girders. We choose to restrict the following discussion to the case for which the cross sections shown in Figures 9.l(b) to (c) and 9.2(b) are applicable. The two design procedures for this case are referred to as: 1. Conventional Design Method A brief descriptionof the plate girder behavior for this design method is given in Section 9.2. See Section 5.10 and Example 5.8 for the design of bearing web stiffeners. 2. Tension Field Design Method A brief descriptionof the plate girder behavior for thisdesign method is given in Section 9.3. See Section 5.10 and Example 5.8 for the design of bearing web stiffeners.The Tension Field Method is not applicable for (a) Webtapered girders (b) Hybrid girders (c) The memberend panels in a nonhybrid girder (d) Any panels of a nonhybrid girder for which
According to a statement on LRFD p. 6249, plate girders usually are more economicalwhen they are designed by the Conventional Method. The reason cited for thisis that if a thicker web is used, fewer intermediateweb stiffeners are required (possibly none are required)by the ConventionalMethod than for the Tension Field Method. Consequently, considerably less fabrication time is required when the number of intermediate web stiffeners can be held to a minimum or to none. By making the web thicker (adds more steel cost), which requires fewer intermediate web stiffeners, skilled labor costs can be signhcantly reduced, and the overall fabrication cost may be less than for a minimumgirder weight designby the Tension Field Method. 9.2 CONVENTIONAL DESIGN METHOD LRFDF2 (p. 653)is applicablefor shear. LRFDF2.3 is applicablefor the intermediate
(transverse) stiffeners. For bending, when LRFD F1.l to F1.2 are not applicable, LRFD G (p. 658)invokes: 1. LRFD Appendix F1 (p. 6111) when h / t I 970 /
JFsd
2. LRFDG2(p.6122)when h / t , >970/&
Except for having to design intermediate web stiffeners (LRFD F2.3), this method involves the familiar design requirements for a beam when h / t , I 970 / &However, the determinationof the design shear strength (LRFD F2) is more complex than for a beam, and the design bending strength may have to be determinedfrom LRFD Appendix F1 for which there are no availabledesign aids similar to the C, = 1beam charts. When h / t ,> 970 / &previously undiscsussed definitions must be used to determine the design bending strength (LRFD, Appendix G2). Since there are no available design aids similar to the C, = 1beam charts for
9.2 Conventional Design Method
371
LRFDAppendicesF1 andG2, the designprocess is iterative (a trial section is checked and, if necessary, revised, checked, and so on until the design requirements are satisfied). However, on LRFD pp. 4183 to 4185, some tabular information is provided as a guide for selecting a trial section whose depth is in the range of 45 to 92 in. Also, the tables on LRFD pp. 6155 to 6158 may be useful in choosing the intermediatestiffener spacingssuch that the design requirement for shear is satisfied. 9.2.1 Design Strength Definitions
Thedefinitionsthatfollowarenotvalidfor a webtapered girder for which the reader should see LRFD Appendix F3 (p. 6118). For a plate girder subjected to bending about the strong axis only, the design requirements are 1. W n r w A x 2. &Vn2 v, where W , and eVVnare the design bending and shear strengths, Mu, and V, the required bending and shear strengths For the Conventional Design Method, the design strength definitions are: 1. Design bending strength &Mn,
05bf 65 When and tf is defined in LRFD Appendix F1 (pp. 6111).
FY
@4,
h When >tw
970
Ef
&M, is defined in LRFD Appendix G2 (p. 6122), which states that the governing definition is the smaller value obtained from: (a) Tensionflange yield:
w,
=O.9(S,,Wy, )
(b) Compressionflangebuckling:
4Mnx = 0*9(sxc where
RF'G R e Fcr
)
372
Plate Girders
R, =
IZ+a, ( 3 m  m 3 )
for hybrid girders
12 + Za,
Re = 1 for nonhybrid girders a, = ratio of web area to compression flange area
h, and t, are defined in Figure 9.2(b)
s,
= &/yc
s,, = IJy,
and
I, = moment of inertia for strongaxis bending (in.4) y, = distance from xaxis to the extreme fiber in tension (in.)
Fyt = yield stress of tension flange (ksi) yc= distance from xaxis to the extreme fiber in compression (in.) F,, = critical compression flange stress (ksi) (see definitions in next paragraph)
m = F,/F,, In the following equations and conditional relations, the slenderness parameters A, Ap, A,, and C,, are as defined subsequently for Iateraltorsional buckling (LTB) andflange local buckling (FLB).F,, must be computed for LTB and FLB; the smaller F,, governs: (a) When d 5 A,,, Fcr
=
FyJ
(b) When A, < A I A r ,
(c) When d > A,, f
F,,
L PG
=
d2
For LTB,
L, = laterally unbraced length of the compression flange (in.) rT = radius of gyration of a T section comprised of the compression flange and onethird of the compression web area
A, = 756
K
9.2 Conventional Design Method = 286,OOOC,
C,
C, = a bending parameter defined in LRFD Eq. (Fl3), p. 653 For FLB:
A=
0.5b, f
f
65
L p =
.JFyt
A, = C,
230
Jm
= 26,2OOk,
k, =
4
dh/t,
0.35 I k, 2 0.763 2. Design shear strength @,V,
From LRFD F2 (p. 656), we find: (a) If intermediate stiffeners are not used, t,.e design requirements are h I260
h
andeither
418
I
G
tw
fW
or $V,, 2 V ,
where @Vnis defined in LRFD F2 (pp. 656 and 6113). When h 418 2
1IFr.
web shear yielding governs the design shear strength: #V,, =0.90(0.6FFA,) When 418  3 or a/h > (260/h/fuS2,
When 
[mE
(0,982)’
= 10.18
= 2 3 4 , / F = 124.51
A , =(56)(7/16) = 24.5 in.’
@vn=0.9[ 26,400A ~
hltw )
k, ]=o.9[ 26,400( 24.5)(10.18) (128)
‘
I
= 362kips
($Vn= 362 kips) 2 (V, = 360 kips) as required for a/h = 0.982 and a = 55 in. Use a = 55 in.
The builtup W57 x 18 x 206 section on LRFD page 4184 is chosen as a trial section and needs to be investigated for the following conditions: Lb
= 20 ft
Cb= 1.15
Fyw = Fyr’
36 ksi
1. Find @Vnwhen intermediate web stiffeners are not used. 2. Find@M,.
384
Plate Girders Solution 1 For an unstiffened web in a plate girder, from LRFD F2.2 (p. 6113) we find that

~
(7/16) 56
t,
1
= 128 5 260 as required
FromLRFDp.4184andfootnoteconpage4185,@V,,= 177.6kipswhenintermediate web stiffeners are not used. Solution 2 From LRFD p. 4184, for a W57 x 18 x 206: bf= 18 in. t, = 7/16 in.
tf= 1in.
h, = 56 in.
Z, = 1370 in?
S, = 1230 in.3
For a nonhybrid member, since (Z, / S, =1370/1230 = 1.11)< 1.5,
@Mpx= 0.9F,,ZX=0.9 (36)(1370)= 44,388 in.kips = 3699 ftkips which is applicable for all three limit states. a. For FLB,
1
$Ma,= (@Mp,= 3699 ftkips) b. For WLB:
From LRFD p. 6123, for a nonhybrid member we find that R, = 1.0:
@Mr,= 0.9RJy,S, = 0.9 (1.0)(36)(1230)= 39,852 in.kips = 3321 ftkips
@Mnx = 3699  ( 3699  3321 ) 128 Io7 = 3555 ft  kips (162107) c. For LTB, C, = 1.15 and L, = 20 ft. Also, we will need the following section
properties: 1,=2
[l(:i)']+
56(7/16)3 = 972.4 in.4 12
___
L
J
9.2
c,
=
4c
t,b; ( h + t , 1’ 1 ( ~ (56+1)* ) ~ = 789,507in.‘ 24 24
w
1 2 = smaller of
A, =
Conventional Design Method
4(789,507)( 972.4 11,200(13.56) 1230
I
( F ,  F , ) = (3616.5) Fp =36ksi
1’
= 0.213
= 19.5 ksi
s J m= FL
= 151.6
(a, = 50.0) < ( A = 59.85) I ( ar = 151.6) (PM,, =: 0.9 F, S, = 0.9 (19.5)(1230)= 21,587 inkips = 1799 ftkips
59.85  50.0
151.6  50.0
= 4042 ft  kips > (PMpx
(PMm =((PMpx=3699 ftkips) Summary for the determination of (PM,,: a. ForFLB:
(PM, = 3699 ftkips
b. For WLB: (PMm= 3555 ftkips
385
386 Plate Girders
c. ForLTB: @Mm= 3699 ftkips
The governing value is @Mu= 3555 ftkips.
The builtup section used in Example 9.4 is chosen as a trial section and is investigated for the following conditions: Lb=60ft
Cb=l
F,=Fd=36ksi
Find @Mm. Solution See Example 9.4 for the FLB and WLB solutions. From Example 9.4,
ilp = 50.0, X,= 932.3, s, = 1230 in.3 4~~ = 1799 fikips
X, = 0.213,
ry = 4.01 in.,
a, = 151.6,
@M~, = 3699 ftkips
c. ForLTB,
From LRFD pp. 6111 and 6114, we find that @MnX = (0.9M, = 0.9S,F,
F, = C , X ,
a
fi
l+
1
)
= LO(932.3)fi
179.6
= 14.4 ksi
2( 179.6)
4Mm = [ @M, = 0.9( 1230)( 14.4) = 15,941 in. kips = 1328 ft  kips] Summary for the determination of @Mm: a. For FLB @M, = 3699 ftkips
b. For WLB:
@M,, = 3555 ftkips c. ForLTI3:
@M,
= 1328 ftkips
The goveming value is @M, = 1328 ftkips
9.3 Tension Field Design Method
387
9.3 TENSION FIELD DESIGN METHOD
LRFD Appendix G (p. 6122)must be used for shear and the intermediate stiffeners. None of the design strength definitions in LRFD Appendix G have been previously discussed in thistext, and they are more complex than those in LRFD Appendix F1. The design procedure for this method is also iterative as described in Section 9.2 for the Conventional Design Method. For the TensionField Method, a structurallyefficientweb chosen to barely satisfy the design requirement for shear is so slender that web buckling occurs in each panel between the intermediateweb stiffeners.However, the post buckling strength of the web can be utilized if the intermediateweb stiffenersare properly designedto keep the flanges in their origrnal location as described in the next paragraph. Consider Figure 9.4, which shows a simply supported plate girder subjected to a factored loading that is symmetric with respect to midspan. The maximum shear occurs at the supportsand the maximum moment occurs at midspan. In the vicinity of the maximum shear, shear web buckling occurs in the diagonal compression direction in each panel between the intermediateweb stiffeners.The length direction of the ripples due to shear web buckling is parallel to the long directionof the shaded areas in Figure 9.4. After shear web buckling occurs, the following analogy is valid when the intermediate web stiffenersare properly designed. The intermediate web stiffenersare analogousto the vertical compressionmembers in a truss. The diagonal tension region (shaded in Figure 9.4) of each panel between the intermediate web stiffeners is analogous to a diagonal tension member in a truss. In the vicinity of maximum moment,flexurul web buckling occurs in the top region of each panel between the intermediate web stiffeners. The design bending strength of the plate girder cross section is affected by flexural web buckling. As shown in Figure 9.4, the compression flange line is curved due to the factored loading. Curvature of the compression flange causes compression in the web, and the compression is perpendicular to the flange. Therefore, in the region of maximum moment, the post buckling strength of the web must be sufficient when assisted by the intermediate web stiffeners to prevent local buckling of the compression flange in the plane of the web direction. In the region of maximum moment, the intermediate web stiffeners are analogous to the vertical compression members in a truss.
Notes: Bearing stiffeners are shown at the reactions and the concentrated loads. Intexmediate stiffenersare shown between the bearing stiffeners. The shaded areas of the web are "diagonaltension members." FIGURE 9.4 Tensionfield action in a plate girder.
388
Plate Girders
9.3.1 Design Strength Definitions According to LRFD Appendix G3 (p. 6103), tension field action is not permitted for: 1. Hybrid and webtapered girders. 2. The end panels in a nonhybrid girder. 3. The panels of a nonhybrid girder for which a/h > 3 or a h > (260 / h / t w ) 2 . This restriction is to prevent a girder from being too flexible perpendicular to the web for handling purposes during fabrication, shipping, and field erection.
For these cases, the Conventional Design Method must be used. For a plate girder subjected to bending about the strong axis only, the design requirements are: 1. 4 @ n x 2 Mux qvvn2 vu 3. When intermediate stiffeners are required and when 2.
0.9 V,, 0.9 M,,, 0.60 I 5 1.00 and 0.75 I ___ I 1.00 VU MU
the following flexureshear interaction equation must be satisfied at each stiffener location: M U X +0.625 VU 51.375
w n
w n x
where &Mnx and &Vn, respectively, are the design bending and shear strengths, and I$
= 0.9
Mu,= required bending strength
V u= required shear strength
For the Tension Field Design Method, the design strength definitions are: 1. Design bending strength &Mnx The design bending strength definitions are the same as those given for the Conventional Design Method in Section 9.2.1, and they are not repeated here. 2. Design shear strength $vV,,From LRFD G3 (p. 6103), we find that: (a) When
web shear yielding governs the design shear strength: @V,,= 0.90( 0.6FpA,
)
9.3 Tension Field Design Method
389
(b) When
we have
k , =5+
5
(alh)’
Cu= ratio of ”critical” web stress, according to linear buckling theory, to the shear yield stress of the web. C, is determined as follows. When
inelastic web buckling governs the design shear strength for which:
When
elastic web buckling governs the design shear strength for which:
c, =
44,000 k”
I*
( h / t w F, Note: In the memberend panels, the Conventional Design Method must be used to determine @Vn. 9.3.2 Intermediate Stiffener Requirements
Intermediate stiffeners are not required when
where
c, = or
187J5/Fy,l h/tw
390 Plate Girders
In order to satisfy LRFD G1 (p. 6122), stiffenersmay be required regardless of any other LRFD Specification. For such cases, Appendix F2.3 is applicable for the design of the stiffeners. When tension field action is utilized, the design requirements for intermediate stiffeners are I,, 2 a t : j
0.15Dhtw(1C,)18t~ v u
+Vn j = 0.5 when u/h 2 7 otherwise j = 
2.5
(
where
a = clear distance between intermediate web stiffeners (in.) h and t, are defined in Figure 9.2(b) Ist = moment of inertia about an axis in the web center for a pair of stiffenersor about the face in contact with the web plate for a single stiffener (in.4)
A,, = stiffener area (in.*) FFt = yield stress of the stiffeners (hi) D = 1when a pair of stiffeners is used D = 1.8 when a single angle stiffener is used D = 2.4 when a single plate stiffener is used C, and +Vnare as defined in item 2b of Section 9.3.1
V , = required shear at the stiffener location
The intermediate stiffenersare loadbearing and in compression when tension field action is utilized. These stiffeners must be designed such that the following requirements from LRFD Table B5.1 (p. 638) are satisfied:
[ &] [ &]
b< & = 
for plate web stiffeners
b < t
for angle web stiffeners
t
A,=
9.3 Tension Field Design Method
391
Note: When the Conventional Design Method must be used to determinefVfl,LRFD F2.3 (p. 6113) is applicable for the design of the intermediate stiffeners. 9.3.3 Design Examples
LRFD Example 410 (p.4168)is an example of the TensionField Design Method and illustrates: 1. Design of a nonhybrid girder: Fu = 50 ksi. 2.
3. Determination of the design bending strength when lateral braces for the compression flange are provided only at the supports and at the two
concentrated load points: (a) For the two end regions: L, = 17 ft and C, = 1.67. (b) For the center region: Lb = 14 ft and C, = 1. 4. Design of the intermediate stiffeners.
5. In the end panels, the Conventional Design Method is required and is used, but for the other panels the Tension Field Design Method is applicable and is used.
The general design procedure for the Tension Field Method when Lb I Lp is illustrated in LRFD Example410 (p.4168).Therefore,in the followingexampleswe choose to illustrate some strength calculations for a trial section.
Change thewebof thesectioninvestigatedinExample9.4toa56x 1/4plate.Thenew S, = 1135in.3,and Z , = 1222 in?Using section properties are: A, = 56(1/4) = 14.0 the Tension Field Design Method wherever applicable, perform the following design tasks: 1. In the end panels (at the supports of a simply supported plate girder), the Tension Field Design Method is not permitted. Using the Conventional Design Method, find the required a h such that +Vfl= 177.6 kips, which is the same design shear strengthin Example9.4 when intermediateweb stiffeners
were not used. 2. In all other panels the Tension Field Design Method is permitted since
392 Plate Girders
Find the required a/h for an interior panel such that @V,,2 (V, = 160 kips) and design a singleplate intermediate stiffener. 3. Find @Mflx.
Solution 2 To find a h of the end panels, enter LRFD page 6155 with: h _  224 t
W
which indicates that we should try a/h = 0.48; a = 0.48(56)= 26.88 in. Try a = 27 in. and a/h = 27/56 = 0.482 in. for the determination of @V,,in LRFD F2.3 (p. 6113). C
4 J
J
k, =5+=5+
(alh)’
(t [ = ,241,
9vn =0.9[
26400 A k ~
hltw
(0.482)
’
= 26.52
2 3 4 E = 234/%
= 201)
]=w[26,400( 14.0)(26.52)
)
(224)’
I
= 175.8kips
V , = 177.6 kips exceeds @V,,= 175.8 kips by 1%for a = 27 in. and a/h = 27/56 = 0.482 in. Use a = 26 in. and a/h = 26/56 = 0.464 in. for which (@Vn = 187 kips) 2 (V, = 177.6 kips) as required.
Solution 2 To find a/h for an interior panel in which @Vfl 2 (V, = 160 kips) is required for tension field action, enter LRFD p. 6157 with h = 224 tw
Aw
14.0
= 11.4 ksi
which indicates that we should try a/h = 1.3;a = 1.3(56)= 72.8 in. Try a = 73 in. for the determination of @Vfl in LRFD Appendix G3 (p. 6124), which is applicable since =  = 1.3036) I
[(3)’ (z)’ h/t,
5 (n/h)’
k, =5+=5+
=
= 1.347123.0
= 7.94
(1.3036)2
9.3 Tension Field Design Method
44,00Ok,
c, = F,
( h / t,

44,000(7.94) = o.1934 36( 224)2
)2
q5Vn =0.9(0.6AwF,)
393
I
C,
[
J
+ 1.154
1 0.1934
@Vn= 0.9( 0.6)(14.0)(36) 0.1934 +
= 170 kips
1.154
Usea = 73 in. and a/h = 73/56 = 1.3036in. for which (q5Vn= 170 kips) 2 (V, = 160 kips) as required. Design a singleplate intermediate stiffener. When tension field action is utilized, the design requirements for intermediate stiffeners are
160 0.15(2.4)(56)(0.25)(10.1934)18(0.25)2 170
Try 6 x 1/2 plate. (As,= 3.00) 2 2.70 in.2and (b/t = 12.0) 5 15.8 as required. Check required I,,. Since ( a h = 1.3036)2 1,j = 0.5.
I,, 2 [ a t 5 j = 73( 0.25)3 (0.5) = 0.570in. 0.5( 6 )
1 is required
!
= 36.0 2 0.570 in.4 as required
Use a single 6 x 1 / 2 plate as the intermediate web stiffener.
Solution 3 a. For tension flange yield, q5M "* = 0.9 ( S,, R, Fyt) = 0.9 ( 1135)(1.0)( 36) = 36,774 in  kips = 3064 ft  kips b. F,, for FLB,
394
Plate Girders F, =Fyr c. F,forLTB,
C, = 1.15
and
Lb=20ft
.I
Also, we will nee the followIg sectionproperties a, T sectioncomposed of the compressionflange andAJ3, where A,= tw(h,/2)= 0.25(56/2) = 7.00 in?: A, 7.00 A, = A , +=18.0+=20.3 3 3 1(1Q3 (56.0 / 6)(0.2$ + 12 12
I, = I y =
= 486 in.
F, =Fsd d. For compression flange buckling,
A, 14.0 = 0.778 a , =A,
R,, =1
18.0
[ E) ]
0.778 2241200+ 300 (0.77'8)
~ 0 . 9 6 6 21.0
,,
@Mnz= 0.9 ( S, R R eF, ) = 0.9 ( 1135)(0.966)(1.0 )( 36) = 35,518 in  kips = 2960 ft  kips Summary for the determination of
@Mwby the Tension Field Design Method:
a. For tensionflange yield:
@M,, = 3064 ftkips d. For compressionflangebuckling: $Mu = 2960 ftkips
The governing value is @Mm= 2960 ftkips. Since Mu was not given in the problem statement, we are unable to perform the jlexureshear interaction check required by UZED G5 (p. 6125). The solutionfor thisexamplewas obtained for the purpose of comparisonto the solution in Example 9.4 by the Conventional Design Method for the builtup W57 X
9.3 Tension Field Design Method
395
18 x 206 section on LRFD p. 4184 chosen as a trial section and investigated for the following conditions: L b = 20 ft, = 1.15, and Fvw = Fd = 36 ksi. The solutions obtained in Example 9.4 were:
cb
1. $Vn = 177.6 kips when intermediate web stiffeners were not used. 2. Summary for the determination of +Mu: a. For FLB: $M, = 3699 ftkips
b. For WLB $Mnx= 3555 ftkips c. For LTB:
$M, = 3699 ftkips
The governing value is #Mnx= 3555 ftkips.
A comparison of the results for Examples 9.4 and 9.6 is as follows: 1. In Example 9.4, t, = 7/16 in. and the Conventional Design Method was applicable.In Example9.6, t, = 1/4 in. and the Tension Field Design Method
was applicable except for the web in the end panels. 2. In Example 9.4, the web plate weight = (7/16)(56)(490/144) = 83.4 lb/ft. The bearing stiffeners are assumed to be identical in both examples. In Example 9.6, the web plate weight = (1/4)(56)(490/144) = 47.6 lb/ft. The weight of one intermediate stiffener (56 x 6 x 0.5 plate) is 48 lb. For a girder span of 80 ft, 12 intermediate stiffenersare required, which is 12(48/80) = 7.2 lb/ft due to these stiffeners in Example 9.6. According to a queried fabricator, due to fabrication labor this 7.2 lb/ft costs about 4.8 times as much as an equivalent weight in the thicker web of the member in Example 9.4. Thus, for the member in Example 9.6, the total effective web weight accounting for the difference in fabrication costs is about 47.6 + 4.8(7.2) = 82.2 lb/ft, which is 1.2 lb/ft lighter (1.4%lighter) than the web in Example 9.4.
3. The design bending strength for the member in Example 9.6 divided by the design bending strength for the member in Example 9.4 is (2960 ftkips)/ (3555 ftkips) = 0.833.Therefore,the member in Example 9.6 is 17.7%weaker in bending than the member in Example 9.4. The member designed by the Tension Field Method is not the preferred choice for the comparison design examples.
The builtup section used in Example 9.6 is chosen as a trial section and is investigated for the followingconditions: L, = 60 ft, Cb= 1, and Fp = Fyr=36 h i . Find #JM, for compression flange buckling due to LTB.
396
Plate Girders Solution
A,
{
Fcr
300 300 JFyt = j=g = 50.0
=
)]
=1.0(36)[12(147*250*0 12650.0
I
R,, =1
}
= 13.0 ksi I ( F , = 36 ksi)
ar
1200 + 300 a ,
L
X,,
= smaller of
@Mnx = 0.9(S,R,,R,Fc,
0.778 1 1200+ 300( 0.778) 1.o
970
(,,,m
= 1.024
) = 0.9(1135)(1.0)(1.0)(13.0)= 13,280 in. kips = 1107 ft  h p s
Summary for the determination of @Mnx by the Tension Field Design Method a. For tension flange yield: @Mnx = 3064 ftkips
d. For compressionflange buckling: @Mnx = 1107 ftkips The governing value is 4Mnx= 1107 ftkips.
For comparison purposes, the summary from the solution in Example 9.5 for the determination of @Mnx by the Conventional Design Method is given here: = 3699 ftkips a. For FLB: @Mnx = 3555 ftkips b. For WLB: @Mnx c. For LTB: QM,, = 1328 ftkips
The governing value is @Mnx = 1328 ftkips. The member designed by the Tension Field Method is 16.6% weaker, more expensive to fabricate, and is not the preferred choice for the comparison design examples.
Problems
397
PROBLEMS 9.1 Is the section investigated in Example 9.1 satisfactory to use for the factored loading shown in Figure W.1? If the section is satisfactory for shear and moment, design fulldepth web stiffeners at the unframed member ends and at the concentrated load points. See Section 5.10 and Example 5.8 for the design of these stiffeners. 5 1.2 kips
5 1.2 kips
0.192 Wft
+ Lb= 20 ft
5 I .2 kips
Lb =20ft
Lb = 2 0 f t
Lb= 20 ft
FIGURE P9.1
9.2 Is the section investigated in Example 9.2 satisfactory to use for the factored loading shown in Figure P9.2? If the section is satisfactory for shear and moment, design fulldepth web stiffeners at the unframed member ends and at the concentrated load points. See Section 5.10 and Example 5.8 for the design of these stiffeners. 5 1.2 kips
5 1.2 kips
5 1.2kips
0.675 Wft
A
A Lb=20ft
Lb = 20 ft
Lb= 20 ft
Lb= 20 ft
FIGURE P9.2
9.3 The section investigatedin Example 9.4 is to be used for the factored loading shown in Figure P9.3. Find the maximum permissible value of the uniformly distributed load 4,. 19.1 kips
I* Lb=20ft 4 Lb= 20 ft
41 .O kips
Lb= 20 ft
19.1 kips
Lb= 20 ft
FIGURE P9.3
9.4 For LRFD Example 410 (p.4168),find the minimum web thickness that can be used in the Conventional Design Method without any intermediate web stiffeners. Make a comparison of these two solutions as we did at the end of Example 9.6 (see items 2 and 3).
398 Plate
Girders
9.5 For LRFD Example 4 1 3 (p. 4180),try a girder section with 1.25 x 18 flange plates and a 0.25 x 50 web plate. Use the Tension Field Design Method wherever applicable. Design the intermediate stiffeners. Make a comparison of these two solutions as we did at the end of Example 9.6 (see items 2 and 3). 9.6 A plate girder that spans 60 ftbetween two columnsin a multistorybuilding is to be designed. This girder is one of a series over an assembly room where the columns were deleted within the assembly room. The overall girder depth cannot exceed 73.5 in. Service loads on the girder are as follows: live = 0.84 kips/ft; dead = 2.35 kips/ft (includesan estimate for the girder weight). Concentrated loads (from the columnsabove the girder)at 20 ft from each end of the girder are as follows:live = 110 kips; dead = 31 kips. Use the Conventional Design Method without any intermediate web stiffeners and F = 36 ksi for all steel. Assume that the governing LRFD loading combination is 1.d+ 1.6L.
9.7 SolveProblem 9.6 using Fy= 50ksi for the flangesand Fy= 36 ksi for the web.
9.8 Solve LRFD Example 412 (p. 4176) by the Tension Field Design Method using Fp = F,,= 50 ksi and a web thickness of 7/16 in. Try a 28 x 2.5 in. flange plate. Try a = (30 in., 5 @ 42 in., 2 @ 120 in.).Design the intermediateweb stiffeners using a single plate for each stiffener.Make a comparison of these two solutions as we did at the end of Example 9.6 (see items 2 and 3). 9.9 Solve Problem 9.6 by the Tension Field Design Method using a web plate thickness of 1/4 in. 9.10 Solve Problem 9.7 by the Tension Field Design Method using a web plate thickness of 1/4 in.
CHRPTER
10 CompositeMembers
10.1 INTRODUCTION In this chapter, a composite member is defined as consisting of a rolled or a builtup structural steel shape that is either filled with concrete, encased by reinforced concrete, or structurally connected to a reinforced concrete slab (see Figures 10.1 to 10.3). Composite members are constructed such that the structural steel shape and the concrete act together to resist axial compression and/or bending. Most likely the reader has seen a highway bridge of composite steelconcrete construction at the various stages of construction. The girders are structural steel members with shear studs (steel connectors)welded to the top flangeof each girder at discrete intervals along the length direction of the bridge, as shown in Figures 10.2(a)and 10.3. A shear stud is shaped like a bolt with a cylindrical head, but the shear stud doesnot have threads. After the steelpdershavebeenereded, a concrete slabis poured on the top surfaceof the girders [seeFigure 10.2(a)]or on a coldformed steel deck supported by the girders (see Figure 10.3), and the concrete encases the shear studs. When the concretehas cured, the steel girders and the concrete slab are
(a) Concrete filled
I
1
FIGURE 10.1 Compositecolumn sections. 399
400
Composite Members
Shear stud
I L
1 Reinforcing steel Reinforced concrete slab
(a) Steel beam interactive with and supprting a concrete slab
(b) Concreteencased steel beam
FIGURE 10.2 Composite beam sections. interconnectedat discrete intervalsby the shear studs. At the interconnected points, the concrete slab is prevented from slipping relative to the top surface of the steel girders.The steel connectorsare subjectedto shear at the concretesteel construction interface, which explainswhy the steel connectorsare called shear studs. Loads that occur on the bridge surface after the concreteslab has cured are resisted by flexural action of the composite steelconcrete construction. There are more LRFD specifications for composite flexural members than composite compressionmembers. Consequently,we chose the following discussion sequence: composite columns, composite beams, and composite beam columns.
10.2 COMPOSITE COLUMNS As shown in Figure 10.1,composite columnsconsistof rolled or builtup steel shapes that are either filled with concrete or encased by reinforced concrete. An axial compressionload is applied at the top end of the columnby whatever the column was designed to support. The bottom end of the column sits on a surface capable of providing the bearing forces from the steel and concrete in the composite column. Since the bottom end of the column is supported, the steel and concrete composinga compositecolumn act together to resist the axial compressiveload. Note that in most cases there is not any need for steel connectorsin a compositecolumn in order for the steel and concrete to act together in resisting an axial compressive load.
10.2 Composite Cohrnns
401
Coldformed steel deck
(a) Ribs parallel to beam
1
L Rib of coldformed steel deck
@) Ribs perpendicular to beam
(c) Section 11
FIGURE 10.3 Coldformed steel deck and composite beam sections A steel pipe or tube filled with concrete is the most efficient type of composite column. The perimeter steel provides stiffnessand confinement of the concrete core, which resists compression and prevents local inward buckling of the steel encasement. The triaxally confined concrete core in some cases is capable of resisting a compressive strength in excess offc’ which is the 28day compressive strength of a concrete cylinder. This type of composite column has the toughness and ductility needed for earthquakeresistant structures. When a structural steel shape is encased by concrete [see Figure lO.l(b)], a longitudinal steel reinforcing bar is located in each corner of the encasing concrete. Lateral ties are wrapped around the longitudinal bars at sufficiently close intervals along the member length. These Ushaped ties stabilizethe longitudinal bars during construction and prevent outward local buckling of the longitudinal bars when an axial compression load is applied after the concrete has cured. Prior to failure of the composite column, the reinforced concrete encasing the steel shape prevents local buckling of the compression elements in the steel shape. The minimum acceptable
402 Composite Members
amount of longitudinal and transverse reinforcement in the concrete encasement is approximately the same as the minimum amount specified for a tied reinforced concrete column. The behavior of this type of composite column is similar to the behavior of a tied reinforced conrete column. At a uniform compressive strain of 0.002, the encasing concretestarts to become unsound, and spalling is likely to occur. A section analysis for an eccentrically loaded composite column may be performed with the assumption that strains vary linearly across the section with a maximum acceptable compressive strain of 0.003 in the concrete. Perhaps the most popular application of concreteencased steel shapes is the perimeter columns in tubetype highrise buildings. In highrise construction, the steel shape is placed many stories prior to the placement of the encasing concrete. Therefore, the steel shape alone resists the axial compressive load until the encasing concrete has been placed and has cured sufficiently to resist some of the load. Temporary lateral bracing of the steel shapes may be required during construction until the encasing concrete has cured sufficiently [ 0.75ffaccording to LRFD Commentary I1 (p. 6203)]. Concrete subjected to compression creeps (the shortening deformation continues to increase with respect to time without any increase in the compressive stress), but steel does not creep. When composite columns are used, a carefully controlled construction process must be maintained to ensure that the floors remain level (within the allowable tolerance limits). This is particularly true when some of the columns within a story are not composite columns since the rate of shortening is different for composite and noncomposite columns. Chapter 10 in the Guide to Stability Design Criteriafor Metal Structures [25] provides the reader with a discussion on the behavior of composite columns, laboratory test results, and analytical studies. 10.2.1 Limitations
LRFD I2 (p. 661) gives the following restrictions on composite columns: 1. A, 2 0.04Agis required, where A, is the crosssectionalarea of structural steel and Ag is the gross crosssectional area. Otherwise, the member is to be designed as a reinforced concretecolumnin accordancewith ACI 31892R [lo]. 2. Concrete encasing a steel core must be reinforced with longitudinal bars and lateral ties. Spacing of the lateral ties must not exceed twothirds of the least dimension of the composite cross section. A minimum clear cover of 1.5 in. outside all reinforcement is required for fire and corrosion protection. The minimumcrosssectionalarea of the lateral ties and longitudinal bars is 0.007 in.Z/in. of bar spacing.Longitudinalbars provided only for the attachmentof the lateral ties do not have to be continuous at the floor levels. However, any loadcarrying longitudinal bars in the member must be continuous at the floor levels. 3. For normal weight concrete, 8 ksi 2 f,’2 3 ksi is required, where f,’is the specified compressive strength. For lightweight concrete, f f 2 4 ksi is required. The reasons for these restrictions are explained in LRFD Commentary I2 (p. 6203). 4. For the structural and reinforcing steel, FyS 55 ksi is required in the strength calculations. This stress level corresponds to a limit compressive strain of
20.2 Composite Columns 403 0.0018. The encasing concrete starts to become unsound at a uniform compressive strain of about 0.002 and subsequently spalling is likely to occur. Since Fy= 60 ksi reinforcing steel is coIILmonly used, it should be noted that for this grade and higher grades of steel, the analyst must use Fy= 55 ksi in the strength calculations. 5. To ensurethat structuralsteeltubes and pipes filled with concrete yield prior to the occurrence of local buckling, the following thicknesses are required
d vis required for each face width b in a tube. (b) t 2 D d vis required for a pipe wit5 an outside diameter of D. (a) t 2 b
These values are the same as those given in the ACI Code. 6. When concrete encases a steel core consisting of two or more steel shapes, lacing, tieplates, or batten platesmust be used to interconnect the steelshapes to prevent buckling of each shape prior to the hardening of the concrete encasement. 7. At connections, the transfer of load to concrete must be by dired bearing to prevent overloadingeither the structural steel section or the concrete. When the supporting concrete area exceeds the loaded area on all sides, the design strength of the concrete is
(P,P, 5 0.6(1.7fc'A, ) where A, is the loaded area and the limit shownis due to bearing. Note: When the supporting concrete area is identical to the loaded area, the LRED specificationsdo not give the limiting design strength of the concrete due to bearing. For this case, we would use (P, P, 5 0.6 (0.85fc'A
)
which is based on Section 10.15.1 of the ACI Code [lo]. 10.22 Column Design Strength
The design compressivestrengthof a compositecolumn is as defined in LRFDE2 (p. 647) with the following modifications: 1. Ag in LRFD E2 is replaced with A, = gross area of structural steel shape. 2 r in LRFD E2 is replaced with rm, which is the larger of: (a) The radius of gyration of the steelshapeabout the flexuralbuckling axis. (b) 0.3times the overall thickness of the composite cross section in the plane of buckling. 3. Fyand E in LRFD €2are replaced, respectively, with F,,,,, = F , + c , F , ( A , / A , ) + c , ( A , / A , )
E m = E + c3E, (A, / A , ) where A, = crosssectional area of concrete, (in.2)
404
CompositeMembers A, = crosssectional area of longitudinal reinforcing bars, (in2) A, = crosssectional area of the structural steel shape, (in?) E = 29,000 ksi E , = w *.’ w = unit weight of concrete, (pcf) f,’= concrete strength, (ksi) Fy = specified minimum yield stress of structural steel shape, (ksi) Fyr = specified minimum yield stress of longitudinal reinforcing bars, (ksi) c1 = 1.0, cz = 0.85, c3 = 0.4 for concretefilled steel shapes c1 = 0.7,
c2 = 0.6, c3 = 0.2 for concreteencased steel shapes
Using these definitions, we find that the design compressive strength of a composite column is @, P,, = 0.85A, F,,
where
) Fmy
F, = (0.658
when A, > 1.5
0.877
a
when A, I1.5
=
k

rmn Column design strength tables are provided as a design aid on LRFD pp. 573 to 5142 for some configurations of composite columns. For other choices of structural steel shapes, or concretegrades, or longitudinaland lateral reinforcingbars and their arrangements, the preceding formulas are applicable.
On LRFD p. 5129 for a ST14 x 10x 0.375 filled with concrete and (KL), = 30 ft, verify the @P, = 548 kips entry. S o h tion Check the wall thickness:
(t=0.375 ‘ “ , ‘ [ € ~ ~ = 1 4 ~ ~ = 0 . in.] 3 2 2asrequired For the composite section: Ag = A, + A,
A, = area of concrete
10.2 Composite Columns
405
From LRFD p. 1127, ST14 x 10 x 0.375, A, = 17.1 in.2 and r, = 4.08 in. According to LRFD p. 1120, the outside comer radius of a structural tube was assumed to be two times the tube thickness in computing the listed section properties. The area to be deducted for the concrete due to the four comers is
ACOlll,, = ( 4  ~ ) 2t)’ ( = ( 4  ~ ) [ 2 ( 0 . 3 7 5 ) ] ’=0.483 in.’
A , = bhA,o,,,,  A , = 10(14)
 0.483  17.1 = 122.4
in.
A , = A , + A , = 122.4+17.1= 139.5 in.’ (ASIA,= 17.1/139.5 = 0.123) 2 0.04 as required Determine F,, and Em:
c2 = 0.85
c1 = 1
c3 = 0.4
A, =122.4 = 7.16 A,
F my
17.1
+ CZfC’A
= Fy = 46 + 0.85 (3.5)(7.16) = 67.3 ksi
A,
E, E m = E+ c
= w1.5
3 E J
3 2 6 7 ksi
= 29000+0.4(3267)(7.16) = 38,357 ksi
A, rmY
K L / rm
f i = 1451.5a
= ry (ST14xlOxO.375) = 4.08 in.
360/ 4.08
n
Jz
@Pny= 0.85 (0.658

acy2 F,A,
 1.18
a;
= 1.382
)
@Pny= 0.85(0.658)’.382(67.2)(17.1) = 548 kips This agrees with the @P,,= 548 kips entry on LRFD p. 5129.
For the compositeST14x 10x 0.375 column in Example 10.1, determine the minimum acceptable area of a bearing plate located on the concrete at the top of the column.
406
CompositeMembers Solution From LRFD p. 5129: Composite ST14 x 10 x 0.375
(KL), = 30 ft, @Pn= 548 kips From LRFD p. 345, Bare ST14 x 10 x 0.375
(KL), = 30 ft, @Pm = 396 kips. @Pm= @Pn @Pm= 547396 = 151 kips 151 = 42.3 in.2 1.7@,fc' 1.7(0.6)( 3.5) is required since (AB= 42.3 in.2)c (A, = 122.6 in?) For a plate withh/b = 1.4 (ratioof tube dimensions), choose b = 5.5 in. and h = 7.75 in., which give (A, = 42.6 in.2)2 42.3 in.2as required.
Venfy the @Pn= 1420 kips entry at (KL), = 30 ft on LRFD p. 581 for a W12 x 120 (Fy = 36 ksi) encased by concrete (ff = 3.5 ksi) such that the composite column dimensions are b = h = 20 in.
Solution Check the lateral reinforcement (no. 3 at 13 in. on center) requirements: Maximum tie spacing = 0.667(20) = 13.3 in. (Tie spacing = 13 in.) I 13.3in. as required. For one no. 3 bar,
(A, = 0.11 ins2)2 [0.007(13)= 0.091 in?] as required Check the longitudinal reinforcement (four #9) requirements: Clear cover = 1.5 in. is required.
Bar spacing = thickness  2 (clear cover + d,)  d, Bar spacing = 20 2(1.5+ 0.375) 1.128= 15.1 in.
For one no. 9 bar, (A, = 1.00 in.*)2 [0.007(15.1)= 0.106 in?] as required
For the composite section, Ag = bh = 20(20) = 400 i n . 2 From LRFD p. 138 for a bare W12 x 120: A, = 35.3 in? and
ry = 3.13 in.
10.2 Composite Cohmns
407
For four no. 9 bars, A, = 4(1.00)= 4.00 in? A, = Ag (A, + A,.) = 400  (35.3 + 4.00) = 361 in.,
(AJA, = 35.3/400 = 0.0883) 2 0.04 as required
Determine F, and Em:
c1 = 0.7
C, = 0.6
c3 = 0.2
A, 361   = 10.2 A,
35.3
A,  400 A, 35.3
 = 0.113
= 36+0.7(55)(0.113)+0.6(3.5)(10.2)= 61.8 ksi
E , = wI5
Em
= E+2
2f35 = 3267
= 145’.5
hi
c3EcA = 29OOO+0.2(3267)(10.2)= 35,665 ksi As
rT = larger of
(ry of W12 x 120) =3.13in. 0.3b = 0.3( 20) = 6.00 in. rmy= 6.00 in.
@Pny= 0.85( 0.658
a2 cy Fmy A,
)
@Pny= 0.85( 0.658)0.632 (61.8)( 35.3) = 1423 kips This agrees very well with the @Pn= 1420 kips entry on LRFD p. 581. Nofe: Before the encasing concrete hardens, @Pn= 538 kips (from LRFD, p. 323).
On LRFD p. 578 for a W14 x 68 (Fy= 36 h i ) encased by concrete(fi = 3.5 h i ) such that the composite column dimensions are b = 18 in.and h = 20 in., 1. Verify rm/rmy = 1.22. 2. For (KL), = 30 ft and (ZUJy = 15 ft, determine @Pn.
408 Composite Members
Solution ( rx of W14 x 68) = 6.01 in.
rmX= larger of
0.3 h = 0.3 (22) = 6.60 in.
rmx= 6.60 in. ( ry of W14 x 68) = 2.46 in.
rmY= larger of
0.3b = 0.3(18) = 5.40 in. Y,,,
= 5.40 in.
rmx  6.60  = 1.22 rmy
1.22
5.40
> ( K L ) , =15 ft
Enter LRFD p. 578 at 24.6 ft for F,, = 36 ksi and find
qP,, = ($Pnx = 1088 kips) 10.3 COMPOSITE BEAMS WITH SHEAR CONNECTORS The most common case of a composite beam is a steel shape that supports and interacts with a concrete slab as shown in Figures 10.2(a) and 10.3. Either a solid reinforced concrete slab [Figure 10.2(a)]or a concrete slab poured on a steelribbed metal deck (Figure 10.3) can be used. In either case, the shear studs shown or some other type of shear connectors are essential to ensure composite beam action. If the steelribbed metal deck in Figure 10.3has embossments on the upper surface to bond the deck to the concrete slab, the steel deck is classified as a composite deck, and the steel deck can be used in the reinforcement requirement for the concrete slab. 10.3.1 Composite Construction
In shored construction, temporary shores are used to help the structural steel members support the poured concrete. Temporary shores are gravity direction supports located beneath the bottom flange of the steel shape at discrete intervals along the beam length and between the permanent beam supports. After the concrete has cured sufficiently, the temporary shores are removed. For shored construction, composite beam action supports all loads (dead weight of the structure and loads applied on the top surface of the concrete slab). In unshored construction, temporary shores are not used, and only the structural steel members support the freshly poured concrete. After the concrete has cured sufficiently [0.75f; is specified in LRFD 13.4 (p. 665)], composite beam action supports the loads applied on the top surface of the concrete slab. Therefore, the structural steel members must be adequately designed to support all factored loads that exist before the concrete has cured sufficiently. When an elastic stress distribution
20.3 Composite Beams With Shear Connectors
409
is required, superposition must be used in the member section analysis. In stage 1the structural steel members support all factored permanent loads before the concrete has cured sufficiently.In stage 2, composite beam action supports the factored loads applied after the concrete has cured sufficiently. When a piastic stress distribution is permitted and used, load tests have shown that the factored loads can be assumed to be resisted by composite beam action for unshored construction.
10.3.2 Effective Concrete Flange Width The behavior of a concrete slab compositelyconnected by shear studs to steel beams may be conceptually described as follows. A uniform gravity direction load on the top surface of the slab causes: 1. Compression forces in the gravity direction to occur between the slab and the steel beams. 2. Longitudinal shear forces to occur in the shear studs at the concretesteel interface.
Along their length direction and at the top surface of their flanges, the steel beams are subjected to a gravity direction distributed load and to longitudinal shear forces at the shear stud locations.Therefore, each steelbeam is subjected to an eccentrically applied axial compressive load and a lateral load. These loads are distributed along the member length. The concreteslab transfers the gravitydirection load via an interactivecompression forceto the steel beams. Also, the concreteslab is subjected to longitudinalshear forces at the bottom surface of the slab and distributed along the length direction centerlines of the steel beams. Due to the eccentric longitudinal shear forces, the inplane compressive stress distribution in the concrete slab is not uniform. Along the lines where the eccentric longitudinal shear forces are applied, the compressive stress is maximum. Midway between these lines of eccentric longitudinal shear forces, the compressive stress is minimum. If the distance between these lines of eccentric longitudinal shear forces is increased, the maximum compressive stress divided by the minimum compressive stress also increases. To simplify the crosssectionalanalysis of a compositebeam, the compressivestress distributiondescribed is assumed to be constant across an effective slab width attached at the shear studs to a steel beam. The cross section of each composite beam in Figures 10.2(a)and 10.3 consists of the structural steel shape and the effective flange width of the concrete slab, which are interactively connected by the shear studs, which transfer the longitudinal shear forces. Three criteria are given in LRFD 13.1 for the determination of the effective concrete slab width. On each side of the beam centerline, the effective concrete slab width is the least of the following values: 1. b, = L/8, where L =beam span length. 2. b, = S/2, where S = distance to the centerline of the adjacent beam.
3. b, = distance to the edge of the slab.
For a compositebeam on the interior of the concrete slab [see Figure 10.4(a)],the effective flange width is b = b,, + b,R.
410 CompositeMembers b
4 4
blL
(a) ~n interior composite member
b blR
w

Edge of slab
(b) An exterior composite member
FIGURE 10.4 Effective slab width. For a composite beam on the exterior edge of the concrete slab [see Figure 10.4(b)], the effective flange width is b = b, + blR. Examples involvingthe determination of the effectiveflange width are given at the end of Section 10.3.4. 10.3.3 Shear Design Strength
Only the shear design strength of the web for the structuralsteel shape is usable, and LRFD F2 (p. 656) is applicablefor the determination of $Vn, which acts parallel to the gravity direction on a compositebeam cross section. The shear perpendicular to the gravity direction is named horizontal shear in the LRFD Specification. In the behavioraldiscussionwe used the terminology longitudinal shearinstead of horizontal shear. Horizontal shear requirements are given in the next section. 10.3.4 Shear Connectors
LRFD 15.1 (p. 667) defines the material requirements for shear studs and channel shear connectors.The length of a shear stud must be at least four stud diameters.For
10.3 Composite Beams With Shear Connectors
411
other types of shear connectors, LRFD I6 is applicable. The horizontal shear force at the interface between the steel beam and the concrete slab must be transferred by shear connectors and is defined as follows:
I. In regions ofpositive moment (top surface of the concrete slab is in compression), the total horizontal shear between the maximum moment point and the zero moment point is the least of (a) 0.85 fc' Act which is the maximum possible compressive force in the effectivewidth of the concrete slab. (b) AsFYwhich is the maximum possible tensile force in the structural steel shape. When a hybrid structural steel shape is used, ZAsF is applicable, and the sum is made for allelementsin the cross sectionof &e steelshape. (c) ZQ, which is the sum of nominal strengthsof the shear connectorsin the indicated region. 2. In regions ofnegative moment (top surface of the concrete slab is in tension),the total horizontal shear between the maximum moment point and the zero moment point is the smaller of (a) A F which is the maximum possible tensile force in the longitudinal 9. remforcing steel. (b) ZQ,,which is the sum of nominal strengthsof the shear connectorsin the indicated region. When either item lc or 2b governs the definition of the total horizontal shear, this behavior is classified as partiul composite action. Otherwise, the behavior is classified asfull composite action. For a shear stud embedded in a solid concrete slab, the nominal strength of one shear connector is
where A, = crosssectional area of a shear stud, (in?)
F, = minimum specified tensile strength of a shear stud, (hi) For a channel shear connector embedded in a solid concrete slab, the nominal strength of one shear connector is Qn = 0.3 ( t , +0.5tw ) L c where
fr = flange thickness of channel shear connector, (in.) 1, = web thickness of channel shear connector, (in.)
L, = length of channel shear connector, (in.) When full composite action is desired, the minimum required number of shear connectors in each region between a maximum moment point and an adjacent zero moment point is the total horizontal shear in each region divided by Q,. When partial composite action is desired, let N = trial number of shear connectors in a region between a maximum moment point and an adjacent zero moment
412 Composite Members
point. The total horizontal shear transferred in the indicated region = XQ, = NQ,. Then, as described later in Section 10.3.5 and as illustrated in Example 10.12, the design bending strength (PM, for partial composite action must be calculated to determine if the design requirement (PM, 2 M u is satisfied. Shear connectors may be uniformly spaced within each region. However, the number of shear connectors placed between each concentrated load point and the nearest zero moment point must be sufficient to develop M u needed at each concentrated load point. The following restrictions on the placement and spacing of shear connectors are imposed by LRFD 15.6: 1. Minimum lateral concrete cover is 1in. except for shear connectors installed in the ribs of a steel deck. 2. d I 2.5tf, where d = stud diameter and fr= thickness of beam flange to which studs are welded. However, d > 2.5ffispermissible for studs located over the web in a solid slab. 3. Minimum longitudinal spacing of studs is 6d in solid slabs and 4d in the ribs of a steel deck. 4. Minimum transverse spacing of studs is 4d in all cases. However, see LRFD Figure C15.1, p. 6215, for special provisions when the studs are staggered. 5. Maximum spacing of studs is eight times the total slab thickness.
There are special provisions for shear studs embedded in concrete on a formed steel deck (see LRFD Figure C13.3, p. 6211): 1. The usual practice is to fieldweld the studs through the steel deck to the beam
flange. When the studs are shopwelded to the beam flange, holes must be made in the steel deck at the stud locations. 2. Maximum stud diameter is 0.75 in. 3. When the ribs in the steel deck are perpendicular to the steel beam: (a) The steel deck must be anchored to all supporting members at a spacing I 16 in. Welds at the studs and puddle welds (or other devices)elsewhere are acceptable anchorages. (b) Concrete below the top of the steel deck must be ignored in calculating 4. (c) Longitudinal spacing of shear studs I 36 in. (d) Nominal strength of one shear connector is
where A,
= crosssectional area of a
shear stud, (in2)
F , = minimum specified tensile strength of a shear stud, (ksi) 0.85 w
N,= number of shear studs in one rib at a beam intersection
20.3 Composite Beams With Shear Connectors 413
(N,I3 must be used in the R, formula; however, more than three studs may be installed.) w,, h , and H, are defined in LRFD Figure C13.3 p. 6211. H, I(h,+ 3) must be used in the computations. 4. When the ribs in the metal deck are parallel to the steel beam: (a) Asshowninthelast figureof LRFDFigureC13.3~.6211,ataribthedeck may be cut longitudinally and separated to form a concretehaunch over the steel beam. (b) Concretebelow the top of the steel deck can be included in calculatingAc. (c) When h, 2 1.5 in., wr 2 2 in. is required for the first stud in the transverse direction; for Nstuds where N 2 2, w r2 [2in. + 4 (Nl)d] is required where d = stud diameter, (in.) (d) When wr/hr< 1.5, the nominal strength of one shear connector is
where A, = crosssectional area of a shear stud, (in.2)
F, = minimum specified tensile strength of a shear stud, (ksi)
[
R,, =
y(
2  1 . 0 ) ] 51.0
w,h, and H, are defined in LRFD Figure C13.3 p. 6211. H, 5 (h,+ 3)must be used in the computations.
Figure 10.5 shows the cross section of a fully composite structural system consisting of a solid 5in.thick concrete slab connected via shear studs to the top flange of steel beams. Beams: W16 x 31 A, = 9.12 in.2 fr = 0.440 in. F,, = 36 ksi.
L = 30 ft = simply supported span length, S = 10 ft = transverse spacing,
Concrete: fc’ = 3.5 ksi
Jfr = 145
E, = w ’ . ~
w = 145 pcf 1s
f i = 3267
ksi.
Shear studs: d = 0.75 in.
H, = 4d = 3 in.
F, = 60 ksi
For full composite action, determine: 1. b = effective concrete slab width for an interior steel beam.
414 Composite Members Solid 5 in. thick slab
L
I4
s= loft
’
L A r 
A
S=10ft
s= loft
4
All W sections are interior beams.
FIGURE 10.5 Example 10.5.
2. V, = horizontal shear force that must be transferred. Assume that the interior
composite beam is subjected to a uniformly distributed factored load. 3. Minimum required number of shear studs.
Solution 2 The effective slab width on each side of the beam centerline is the smaller of
L 30(12) ==45 8 8
in.
Therefore, the effective slab width, b = 2(45) = 90 in.
Solution 2 In a positive moment region, the horizontal shear force that must be transferred for full composite action is V , = smaller of
0.85fc’Ac = 0.85(3.5)(90)(5)= 1339 kips Asfy = 9.12( 36) = 328 kips V,,= 328 kips
Solution 3 Check the stud diameter. Unless each stud is welded directly over the beam web, d I [2.5tf=2.5(0.440) = 1.10 in.] is required (d = 0.75 in.) S 1.10 in. as required
For one 0.75indiameter shear stud, the nominal shear strength is
(Q,
= 0.5A,,
m)A,F,
A,, = ~ ( d / 2=)~~( 0 . 7 5/)4~= 0.4418 in.’
20.3 Composite Beams With Shear Connectors 415
A g , = 0.4418(60) = 26.5 kips
[0.5( 0.4418),/
= 23.61 5 26.5
Q,= 23.6 kips/stud Between the maximum positive moment point and each adjacent zero moment point, the minimum required number of shear studs is
N = V,/Q, = 328/23.6 = 13.9 Use 14 shear studs between midspan and each simple support. The minimum total number of shear studs required is 2(14)= 28.For only one shear stud directly over the beam web at each location, Stud spacing = 30(12)/28 = 12.9 in. Minimum spacing = 6d = 6(0.75) = 4.5 in. Maximum spacing = 8tslab= 8(5) = 40 in. For placement convenience, use stud spacing = 12 in. and 30 studs.
Solve Example 10.5 for the exterior beam configuration shown in Figure 10.6.
Solution 1 The effective slab width on the interior side of the beam centerline is the smaller of
L 30(12)  45 in. =8 8
s lO(12) 60 in. =2
2
The effective slab width on the exterior side of the beam centerline is the smaller of
Edge distance = 1 f t = 12 in. Therefore, the effective slab width is b = 12 + 45 = 57 in.
Solution 2 In a positive moment region, the horizontalshear forcewhich must be transferredfor full composite action is
416
Composite Members Solid 5 in. thick slab
1 ft (edge distance)
FIGURE 10.6 Example 10.6.
V , = smaller of
0.85ffA, = 0.85(3.5)(57)(5) = 848 kips Asfy = 9.12(36) = 328 kips Vh= 328 kips
Solution 3 Same as in Example 10.5.
Solve Example 10.5with the following modifications: The 5in. total slab thickness consists of a 2in. concrete topping on a 3in. formed steel deck, with the deck ribs spanning perpendicular to the steel beam as shown in Figure 10.7
H,= 4.5 in. = shear stud height h, = 3 in. = deck rib height w,= (4.5 + 7.5)/2 = 6.00 in. = average rib width t, = 2 in. = thickness of concrete slab above the steel deck
Solution 1 Check the special requirements when the concrete slab is poured on a formed steel deck: (h, = 3 in.) < 3 in. as required (t, = 2 in.) 2 2 in. as required (w, = 6.00 in.)2 2.0 in. as required (d = 0.75 in.) 5 0.75 in. as required
(H, = 4.5 in.) 1 (h, + 1.5 in. = 3 + 1.5 = 4.5 in.) as r quir d The solution is the same as in Example 10.5. Solution 2 The concrete below the top of the steel deck must be ignored in calculating A,. In a positive moment region, the horizontal shear force that must be transferred for full
10.3 Composite Beams With Shear Connectors 417
I
I
h , = 3in.

I
I
7.5 in.
 I 
I 4.5 in. I
4.5 in.
1.5 in.
1.5 in.
t L
W16x31 Ribs of metal deck are perpendicularto length centerline of W 16 x 3 1,
FIGURE 10.7 Example 10.7.
composite action is V , = smaller of
0.85fc'Ac = 0.85( 3.5)( 90)(2 ) = 459 kips Asfy = 9.12(36) = 328 kips
V, = 328 kips Solution 3 For one O.7!jin.diametershear stud, the nominal shear strength is
(Q,= 0.5A,
m)
5L F ,
A 2 , = 0.4418(60) = 26.5 kips N , = number of shear studs located in each rib
Try Nr = 2 since (from Example 10.5)stud spacing c 12 in. is probable for N, = 1 and R, < 1:
[
)
]
R , = 0'85wr ( s  l . O ) = x0'85(6) (?4*5 1.0 = 0.601 5 1.0 hr hr
JN,
[ Q, = 0 . 6 0 1 ( 0 . 5 ) ( 0 . 4 4 1 8 ) , /=~ 14.2 kips] I26.5 kips Q, = 14.2 kips/stud
Between the maximum positive moment point and each adjacent zero moment point, the minimum required number of shear studs is
418 Composite Members
N = Vh/Qn= 328/14.2 = 23.1 Use 24 shear studs (12 pairs) between midspan and each simple support. The total number of shear studs required is 2(24) = 48: Longitudinal stud spacing = 30(12)/24 = 12.9 in.
Minimum spacing = 12 in. (rib spacing) Maximum spacing = St,,
= 8(5) = 40 in.
For placement convenience, use 60 studs and longitudinal spacing = 12 in:
Minimum lateral spacing = 4d = 4(0.75)= 3 in. For each pair of shear studs, use lateral spacing = 3 in. Check the stud diameter. Since each pair of shear studs cannot be welded directly over the beam web, (d = 0.75 in.) I [2.5$= 2.5(0.440)= 1.10 in.] as required
10.3.5 Flexural Design Strength For a composite beam with shear connectors, there are two design bending strength definitions. One definition is for the positivemoment region and the other definition is for the negative moment region.
Positive Moment Region In the positive moment region of a fully composite beam with shear connectors, the design bending strength QM,is determined as follows: 1. When h / t I 6 4 0 / E,Q = 0.85 and M, is calculated from the plastic stress
distributionon the composite section.For convenience, this calculation of M, is named the plastic section analysis. The assumptions made in performing a fully composite plastic section analysis are [see LRFD Figure C13.1 (p. 6207)]: (a) Concrete tensile strength is zero. A uniform compressivestress of 0.85fc' is applicable for the concrete compression zone. (b) A uniform stress of F is applicablein the tension and compressionzones of the structural steel shape. The resultant of the forces in the compression zone is equal to the resultant of the forcesin the tension zone. This is an equilibriumrequirement and is not an assumption. 2. When h / t
> 640
/E,Cp= 0.9,and M, iscalculated from the superposition
of elastic stress distributionson the composite section. For convenience, this calculation of M, is named the elastic section analysis. The effects of shoring must be considered in computingM,. The assumptions made in performing an elastic section analysis are:
10.3 Composite Beams With Shear Connectors
419
(a) Strain is proportional to the distance from the neutral axis. (b) Steel stress is E times steel strain, but cannot exceed Fy. For a hybrid structural steel shape, strain in the web may exceed the yield strain; however, at such locations,the stress that must be used in the calculations is FyU. (c) Concrete tensile strength is zero. Concrete compressive stress is Eptimes concrete compressive strain, but cannot exceed 0.85f: The resultant of the forces in the compression zone is equal to the resultant of the forces in the tension zone. This is an equilibrium requirement and is not an assumption.
Negative Moment Region In the negative moment region of a fully composite beam with shear connectors, the design bending strength @M,can be determined from either of the following definitions: 1. Only the structural steel shape can be used to determine @M,in accordance with LRFD F1 (p. 652). 2. @Mncan be determined using @ = 0.85 and M, calculated for the composite section from a fully composite plastic section analysis [see LRFD Figure C13.1 (p. 62031, provided that:
(a) For the structural steel shape, 0.5b, / tf I 65
/K, h / t
uI
I
640
/E,
and L,S Lpfor the flange not in contact with the concrete slab or the metal deck. (b) Adequately designed shear connectors exist in this region. (c) Longitudinal reinforcement in the effective width of the slab is adequately anchored such that ArFYcan be fully developed.
For the fully composite section in Example 10.5 and shored construction, determine
@Mm* Solution
From Example 10.5, we have Effective slab width, b=90in.
Solid slab thickness, t=5in.
A = 3.5 ksi 0.85fc’A, = 0.85 (3.5)(90)(5)= kips W16 x 31: AS=9.12in.2
tW=0.275in.
tf=O.44Oin.
d=15.88in.
420
Composite Members Fy = 36 h i
A S Y= 9.12(36) = 328.32 kips
(PMpx= 146 ftkips
Lp = 4.9 ft.
(hit, = 51.6) I ( 6 4 O / E = 640/* When h / t ,
L@ is p.ermissible. For these Lb regions, the flexural design strength +Mnxmust be deterrmned as for beams analyzed by elastic methods (see Chapters 5,6, and 9). When design by plastic analysis is used, the following LRFD Specification requirements must be satisfied: 1. LRFD A5.1 (p. 631)The steel must exhibit a plastic plateau on the stressstrain curve; consequently, FyI 6 5 ksi must be used. 2. LRFD B5.2 (p. 636)Compression elements in the section must have a widththickness ratio 5 4 [see LRFD Table B5.1 (p. 632)]. 3. LRFD C2 (p. 642)The axial force in a column caused by factored gravity plus factored horizontal loads shall not exceed 0.85+&Fyin a braced frame nor 0.75+p,Fyin an unbraced frame. 4. LRFD E1.2 (p. 647)The column design strength must be governed by inelastic column buckling; that is, the requirement is that
5. LRFD F1.2d (p. 655)The compression flange must be laterally braced at
each plastic hinge location, except at the last plastic hinge to form, and such that Lb I L@, where Lb = laterally unbraced length of the compression flange LpI = limiting value of Lb for plastic design
For an Ishaped member with the compression flange larger than the tension flange and bending about the major axis,
For solid rectangular bars and symmetric box beams bending about their major axis,
442 Plastic Analysis and Design
There is no limit on Lb for members with circular or square cross sectionsor for any member bending about its minor axis. In the L b region of the last plastic hinge to form, and regions not adjacent to a plastic hinge, qMm must be determined from LFRD F1.2 to F1.4. 6. LRFD H1.2 (p. 66O)M, shall be determined in accordance with LRFD C1 which stipulates that secondorder (PA) effeds shall be considered in the design of frames and the requirements stated in item 3 above must be satisfied. 7. LRFD I1 @. 661)qMm of compositemembersshallbe determined from the plastic stress distributions specified in LRFD 13. The followingdiscussionpertainsto item 6in the preceding list of requirements. In our plastic analyses of onestory frames, we use a firstorderplastic analysis since that has been the traditional approach. LRFD C1 requires secondorder (PA)effects to be consideredin the design of frames,but does not give an approximateprocedure to determine Mu based on a firstorder plastic analysis. For onestory frames, we recommend the following approach to account for secondorder effects. Case2
For a member subjected to axial compression and bending in a PCM for which sidesway does not occur M u = BlMpnn
where
Mp.. = maximum moment in each Lb region from the plastic collapse mechanism
C, = 0.85 for a member subjected to transverse loads
Otherwise, C, = 0.6  0.4M1/M2
M1/M2= ratio of smaller to larger moments at the ends of an
region
Ml/M2 is negative when the member is bent in single curvature in an L, region
Otherwise, Ml/M2 is positive P, = axial compression force obtained from the plastic collapse mechanism Pel = z2EZ/(KL)2 Z and KL are for the axis of bending K I 1for braced frame members
22.2 PfasticHinge
483
B, accounts forthe local secondorder PGeffect. There is not any globalsecondorder PA effectwhen sidesway does not occur. Case 2
For a member subjected to axial compression and bending in a PCM for which sidesway occurs,
Mu = B 2 M p where
Mp.. = maximum moment in each L, region from the plastic collapse mechanism
ZPu= required axial strength of all columns in a story
ZPa = L: [n2EI/(KL)2]of all columns in a story I and KL are for the axis of bending K > 1for unbraced frame members
B, accounts for the PA effect when sidesway occurs. Since we cannot separately determine the portion of Mp.. due only to the lateral loads and the portion of M p due only to the gravity loads, Mp..due to all loads is amplified by B,. This approach overcompensatesfor the PA effect, but Case 2 usually does not govern for onestory frames. For multistory frames, we recommend that M ube obtained from a secondorder plastic analysis that directly accounts for the PA effect. 11.2 PLASTIC HINGE In Figure 11.1, the simply supported beam is subjected to a concentrated load at midspan. For convenience in the following discussion, assume that the member weight is negligible. For strongaxisbending,the load W, causes the plastic moment Mpxto occur at the maximum moment location.When the load W,has been applied, the beam has no more resistance to bending at the maximum moment point until strainhardening starts to occur (see Figure 5.4). Accounting for strainhardening is not easy or necessary. It is conservative to ignore the extra bending strength due to strainhardening. Thus,in plastic analysis, the plastic moment is accepted as being the maximum possible moment. Also, for a simple plastic theory, the momentcurvature relationship (see Figure 5.4) is idealized as linearly elastic up to M p and plastic thereafter. Using this idealized momentcurvature relationship is permissible for a plastic analysis to determine the required bending strength since the presence of residual stresses does not affect the plastic moment (seeFigure 5.6). A plastic hinge is a zone of yielding due to flexure in a structural member. The plastic hinge length $is the beam region length in which the moment exceeds the yield moment. As shown in the following discussion, Lph is dependent on the geometry of the cross section and the loading configuration. In Figure 11.1, the moment diagram due to a single concentrated load is linear. Since Mpx = Fyzx
444 Plastic Analysis and Design

Plastic hinge
W U
0.5L
0.5L
(a) Xaxis bending of a W section beam
(b) Yaxis bending of a W section beam
FIGURE 11.1 Plastic hinge.
then where Z, = plastic section modulus
S, = elastic section modulus For W sectionsused asbeams, the shape factor Z,/S,ranges from 1.10to 1.18and the average is 1.14. For some W sections used as columns, Z,/S, ranges as high as 1.23. From similar triangles on the moment diagram in Figure 11.1, Mpx L
 M , F& LPh
11.3 Plastic Collapse Mechanism
445
For Zx/Sx= 1.14,
L,,, = 0.1228L = L / 8 Similarly,for weakaxis bending,
For a uniformly distributed load,
Lph= 0.35L for strongaxis bending Lp,, = 0.5771,for weakaxis bending For a simple plastic theory, the momentcurvature relationship is idealized as linearly elastic up to M pand plastic thereafter, all of the plastic rotation is assumed to occur at the plastic hinges, and the length of each plastic hinge is assumed to be zero. This means that the idealized loaddeflection curve for the member and loading shown in Figure 11.1is linearly elastic up to W,,and is plastic thereafter. At a plastic hinge location, the member behaves as though it were hinged with a constant restraining moment M,. For the member and loading shown in Figure 11.1, a mechanism (geometrically unstable structure) consisting of the real hinges at the member ends and the plastic hinge at midspan forms when W,causes M, to occur at midspan. Since the real hinges cannot resist any moment and the plastic hinge cannot resist any additional moment afterM , occurs, the structuredoes not have any bending strength to resist any load after W,, has been applied. Therefore, for the member and loading shown in Figure 11.1, W,, is the plastic collapse load. 11.3
PLASTIC COLLAPSE MECHANISM In a continuousbeam, the maximum moment points occur at the supports and at a point of zero shear between the supports. In regard to the strength concepts of plastic analysis, the behavior of an interior span of a continuousbeam is the same as for a fixedended beam. Thus, for graphical convenience, we choose to discuss the behavior of a fixedended beam (see Figure 11.2). For the idealized momentcurvaturerelationship (linearlyelastic up toMpand plastic thereafter),plastic hinges form at each support when the applied load value is w p h r which is obtained from wp,,L2/12 = Mpx.As shown in Figure 11.3,when the applied load is increased from w,,, to the plastic collapse load w,,, the member behavior is elastic between the supports, and the memberend moments remain constant at their maximum value of M We px' as see that the first plastic hinge forms at the supports (maximum moment pomt defined by elastic behavior). Then, redistribution of moment occurs until a plastic hinge forms at the zero shear point between the supports when the applied load value is w,,.Each beam segment between plastic hinges is able to move without any increasein load. A systemof suchbeam segmentsis called a prustic corlapse mechanism, which is an unstable structure until strain hardening occurs at one or more of the plastic hinge locations. In Figure 11.2@),the deflected shape of the PCM (p2astic
446 Plastic Analysis and Design
G
3
wL2
wL2
12
12
0.5L
I
0.5L
4
(a) Elastic behavior
r De.fl&
I
shapc of plastic collapse mechanism
0.5L
0.5L
4
(b) Behavior aftex formation of the plastic collapse mechanism occus
FIGURE 11.2 Moment redistribution.
colhpse mechanism)is shown as straight beam segmentsbetween the plastic hinges. When the PCM forms,eachbeamsegment isbent due to Mp..but the furtherbehavior (motion)of the PCM doesnot involve any additionalbending of any beam segment, and the PCM behavior is identical to a hkipsage of straight bars connecting real hinges. This fact is very useful in performing a plastic analysis by a method that involves the motion of the PCM due to an imposed virtual displacement. The types of independent mechanisms are shown in F i p 11.4and discussed here: 1. Beam Mechanism This mechanism can form in any span of a continuousbeam and, as shown in Figure 11.4, in any member of a frame. For a member subjected to a uniformly distributed load, there is only one possible beam mechanism and only three possible plastic hinge locations (onlytwo possible plastic hinge locations if one member end is a real hinge). For a beam subjected to concentrated loads at n locations between the member ends,
11.3 Plastic Collapse Mechanism
I,
I
I
A,
Aah
447
,
I
Apn
Midspandeflectim
FIGURE 113 Idealized loaddeflection curve for beam in Figure 11.2.
there are n possible beam mechanisms and n + 2 possible plastic hinge locations (only n + 1possible plastic hinge locations if one member end is a real hinge). 2. Panel Mechanism This mechanism can occur due to lateral loads. 3. Joint Mechanism This mechanism can occur at the junction of three or more members. 4. Gab& Mechanism This mechanism is characteristic of gabled frames and involvesspreadingof thecolumntopswithrespecttothecolumnbases.Various combinationsof the independentmechanismscanbe made to form a c o m p ite mechanism. Examples of some compositemechanismsare shownin Figure 11.5.
For a structure that is indeterminate to the nth degree, let the number of
448
Plastic Analysis and Design
(a) Beam mechanisms
(b) Panel mechanism
AfAa (c) Joint mechanism
(d) Gable mechanisms
FIGURE 11.4 Independentmechanisms.
redundants be denoted by NR = n. The number of independent mechanismsNlM for such a structure is
NIM = NPPH  NR where NPPH = number of possible plastic hinge locations. 11.4 EQUILIBRIUM METHOD OF ANALYSIS
The equilibrium mefhod of plastic analysis is useful for solving beam mechanisms in continuousbeams and frames. This method is also useful for solving a plane frame for which there is only one redundant. In this method the objective is to find an equilibriummoment diagram in whichM, IM, and Mu=M, at a sufficientnumber of locations to produce a PCM. In the preceding notation, M uis a moment diagram value due to factored loads and M, is the required plastic hinge strength.The steps in the analysis procedure are:
21.4 Equilibrium Method ofAnalysis
449
FIGURE 11.5 Composite mechanism.
1. Select the redundant(s). For a continuous beam, the moments at the supports are chosen as the redundants. 2. Draw the moment diagram for the factored loads.
3. Draw the moment diagram for the redundant(s). 4. Assume that a plastic hinge forms at a sufficient number of locations to
produce a PCM. 5. Solve the moment equilibrium equation for M,. 6. Check to see if MuI Mu,. Our purpose in the analysis is to determine Mu,for the governing plastic collapse mechanism. A value of Mu,obtained by the equilibrium method is a lower bound value. Therefore,the governingvalue of Mupisthe maximum of the required strength values.
450
Plastic Analysis and Design A beam mechanism is simpler than a composite mechanism for a frame. Therefore, we begin our discussion for the equibilibrium method and continuous beams.
Use Fy= 36 ksi and the equilibriummethod of plastic analysis.In Figure 11.6the same W section is to be used for both spans. Select the lightest acceptable W section assuming lateral braces can only be provided as stated in Figure 11.6 for the following conditions: Dead load W, = 11.9 kips w = 0.533 kips/ft W, = 7.10 kips
(W,, W, and w contain an estimate accounting for the member weight.) Live load W, = 22.8 kips
W,= 34.2 kips
w = 2.10 kips/ft
The governing LRFD loading combination is 1.20 + 1.6L.
Solution The factored loads are
W,, = 1.2(7.10)+ 1.6(22.8)= 45.0 kips
W, W,
= 1.2(11.9)+ 1.6(34.2)= 69.0 kips
= 1.2(0.533)+ 1.6(2.10)= 4.00 kips/ft
In Figure 11.7(a),each real hinge location and each possible plastic hinge location is numbered for conveniencein the following discussion.The number of independent mechanisms NZM are
NIM = N P P H  N R = 5  2 = 3 The three independent mechanisms are shown in Figure 11.7(bd). The redundants are chosen as the support moments at points 4 and 6. For span 1,there is only one redundant Mu which is the moment at point 4. As shown in Figure 11.8(a),M4 causes tension in the top fiber of the section at the support and this is a
4
L , = 3 0 ft

4 @ ( L b =I0 f t ) = 4 0 ft L, =4Oft
FIGURE 11.6 Example 11.1:structure and loading.
e
11.4 Equilibrium Method of Analysis 45.0 kips
69.0 kips 4.00 wft
f 1
I
I
2
3
f I
+
5
4
451

6
Comments: 1. Point 1 is a real hinge location. 2. Points 2 to 6 are possible plastic hinge locations. (a) Governingfactored loading: 1.20 + 1.6L
2 (b) PCM 124
(c) PCM 134
3 4.00 wfi
(d) PCM 456
5
FIGURE 11.7 Example 11.1:plastic collapse mechanism.
negative moment. For convenience, we let the moment vector show the correct direction and M4 is the magnitude of this vector. Hence, M 4 is an absolute value. As shown in Figure 11.8@),the moment diagram is drawn by parts for the factoredloads and the redundant applied on a simplysupportedbeam. The factored loads cause a positive moment and the redundant causes a negative moment. For PCM 124,the equilibrium requirements at the possible plastic hinge locations are: 1. M4 = MUp where M , = required plastic hinge strength. 2. ( M 2 = 5 3 0  31M , )
=M
, Since M4 = M , we obtain 4 M , = 530 ftkips 
3 M , = 397.5 ftkips
452
Plastic Analysis and Design 45.0 kips 10 ft
t v,
69.0 kips
10 ft
2
3
?V4 = 6 1 + M4 =53 M4 30 30 Note: M4 is the magnitude of the indicated moment vector.
(a) Loading and member end forces
5
Note: M4 is an absolute value.
3
(b) Factored moment diagram
PCM 124
2
PCM 134
3
(c) Possible plastic collapse mechanisms
FIGURE 11.8 Example 11.1: span 1
2 3. M , =610M, 3
2 =610(397.5)=345ftkips 3
Since
(M3= 345 ftkips) I (Mup= 397.5 ftkips) no moment ordinate exceeds Mu,= 397.5 ftkips and PCM 124 governs for span 1. We could have started the solution for span 1 by assuming that PCM 134 governs and for illustration purposes this is done. For PCM 134, the equilibrium requirements at the possible plastic hinge locations are: 1. M, = Mu, M,=610M 2 ,)=MUY 3
11.4 Equilibrium Method of Analysis
453
Since M , = Mu,, we obtain
5M u , = 610 ftkips 3 Mu,= 366 ftkips 1 3. M , = 530M, 3
1 3
= 530(366)=408 ftkips
Since
( M 2 = 408 ftkips) > (Mup= 366 ftkips) the governing PCM must contain a plastic hinge at point 2. Mu, values obtained by the equilibrium method are lower bound values. Therefore, the required plastic hinge strength in span 1 cannot be less than the maximum computed value, which is Mu, = 397.5 ftkips. Also we know that for the correct value of M u ,all ordinates on theM diagram are less than or equal toM,,. Both these important facts were illustrated in the above calculations. w , =4.w kipdft
M4Gk t
Span = 40 ft
80 kips Note: M4 = M , = is the magnitude of the indicated moment vector.
(a) Loading and member end forces
1
M4
7 ( M 4 + M d
(c) Factored moment diagram
(d) Composite factored moment diagram
FIGURE 11.9 Example 11.1: span 2
454
Plastic Analysis and Design As shown in Figure 11.9(a),for span 2 there is only one possible PCM, and there are two redundants (M4and Md. However, since the structure and loading for this span are symmetric with respect to midspan, M4 = M, and as illustrated in Figure 11.9(d),we can easily find M y = 400 ftkips. Since the same W section1s to be used in both spans, span 2 governs and Mup= 400 ftkips. Therefore, the Mu diagram is shown in Figure 11.10.At each ordinate on the M u diagram, the LRFD design requirement for strongaxis bending is
4Mm 2 Mu where i$Mmis the designbending strength for strongaxis bending. Therefore, at the plastic hinge locations, we must require 4Mpx 2 M , where
$MPx= 0.9F&. = design plastic hinge strength Mup= required plastic hinge strength Using LRFD p. 418 for Fv = 36 h i , we find that W24 x 62
(#Mpx= 413 ftkips) 2 (Mup= 400 ftkips)
is the lightest W section that satisfies the design bending requirement at the plastic hinge locations, when the lateral bracing requirements are satisfied. Using the information given in Figure 11.10, check the lateral bracing requirements. Note that Figure 11.10 shows the Mu diagram associated with the governing plastic collapse mechanism. This is the moment diagram that should be used in checking the lateral bracing requirements. Therefore, in the following Lpl formula, we have shown the definitions as they apply to the M , diagram associated with the governing plastic collapse mechanism. On the left side of the plastic hinge at the center support, we must require that
Lpd2 (Lb = 10 ft)
145.0 kips 169.0 kips
4.00 wft
4 @ ( L * =10 f t ) = 4 0 ft
mftk
FIGURE 11.10 Example 11.1: final moment diagram.
m!ik
11.4 Equilibrium Method of Analysis
455
where
Fy= specified minimumyield strength of the compression flange (ksi)
M,= smaller moment at the ends of the L b region (ftkips) M2 = larger moment at the ends of the yY = radius of
region (ftkips)
gyration about weak axis (ft.)
(M,/M,) is positive when moments cause reverse curvature
Mu,= required plastic hinge strength (ftkips) Therefore,
Fy= 36 ksi MI = 343 ftkips Mu, = 400 ftkips yY = 1.38 in. = 0.115
ft
M , / M , , = 343/400 = 0.8575
L , = [ 3 6 0 0 + 2 2 0 0 ( 0 . 8 5 7 5 ) ] ~= 17.53 ft (36)
(Lpd = 17.53 ft) 2 ( L b = 10 ft) as required On the right side of the plastic hinge at the center support, we must require that Lpd 2 ( L b = 10 ft)
Since M , / M , = 200/400 = 0.5,
L , =[3600+2200(0.5)]=15.0 ( 0 115)
ft
(36)
(Lpd= 15.0 ft) 2 ( L b = 10 ft) as required O n the left side ofthe plastic hinge at the right support, we must require that Lpd 2 ( L b = 10 ft)
Since M , / M , = 200/400 = 0.5, (0 115) L , = [ 3600 + Z O O ( 0.5) J = 15.0 ft
(36)
(Lpd = 15.0 ft) 2 ( L b = 10 ft) as required
For the other L b regions, the design bending requirement is 2 Mu
w ? l X
456
Plasfic Analysis and Design where
@Mnx= smaller of
MI =
[
@Mpx
(@MnX for C, =1) 12.5M + 4 M , +3M,
c, = 2.5Mm,, +3M, where
M,,, = absolute value of maximum M in the L b region MA= absolute value of M at quarter point in the L, region M, = absolute value of M at middle point in the Lb region M, = absolute value of M at threequarter point in the L b region For the L, = 10 ft region at the left support,
12.5(397) = 1.67 c, = 2.5( 397) + 3(99.25) + 4( 198.5)+ 3( 297.75) and from LRFD p. 4128 we find that
[ C,M,
= 1.67(355) = 592
ft  kips] > ( @MpX = 413 ft  kips)
Therefore, (@MnX = 413 ftkips) 2 (Mu= 397 ftkips) as required. For the L, = 5 ft region at the 45.0 kip load,
(Lb= 5 ft) I (Lp = 5.8 ft) Therefore, (@Mnx = 413 ftkips) 2 (Mu= 397 ftkips) as required. For each L, = 10 ft region at the last plastic hinge to form (at midspan of span 2),
12.5(400) = 1.13 c, = 2.5( 400) + 3( 287.5) + 4( 350) + 3( 387.5) From LRFD p. 4128, we find that
[C , M ,
= 1.13(355) = 401.2 ft  kips]
Mu, > 771 ftkips. Since 810/771= 1.05, if we choose to say that Mu, = 810 ftkips, our Mu, is not more than 5% larger than the correct Mu,. In addition to having found a very good estimate of MU,, we also have found that plastic hinges numbered 3 and 4 in Figure 11.39(a) (the correct PCM for loading 1) are located x = 40  5.80 (40/42.72) = 34.57 ft from the eaves. Assume that x = 34.5 ft in Figure 11.39(a)and compute Mu,. The mechanism relation is
u2 = (34.5/40)15a = 208 The virtual work equation is
M u , [ 2( 8 + a ) + 2 (a ) ]= 2.65[ (8069.0)+L( 2 69.0)]v4 Substitution of a = 1.5468and v4 = 34.5a gives Mu, = 786 ftkips, which lies in the previously established range of (810 ftkips > Mu,> 771 ftkips) The preceding calculations for a uniformly distributed load take less time to perform than the case of concentratedloads at the purlin locationson the roof girder. Purlins are spaced at about 5 or 6 ft along the roof girder axis. We have shown for a
Ic
4
+ + + + + + + 2.65 Mft
*
X
4
X
4
40 ft
40 ft
(a) Governing PCM for loading I
I..
b b

FIGURE 11.39 Example 11.10: composite PCM for loading 1.
486
Plastic Analysis and Design uniformly distributed load that the correctlocationsof the plastic hinges, numbered 3 and 4 in Figure 11.39(a), are approximately at a purlin location. Therefore, the solution for a uniformly distributed load is a very good approximation of the solution for the concentrated loads case. For preliminary design purposes, gabled frame solutions based on a uniformly distributed roof loading can be used. We have determinedthat Mup= 786 ftkips for loading 1in Figure 11.37.Usually, the loading combinationthat includes wind does not govern Mupfor a gabled frame. Now we need to show that loading 2 does not govern Mu or we must compute the goveming value of Mup.We prefer to compute M , for &e generalized version of loading 2 and the compositemechanism shown in Figure 11.40. As we will show, the deletion of H, from this solution gives us the exact solution for only a uniformly distributed load.
+ + + + + + +
W"
I'
1
(b) Composite mechanism
FIGURE 11.40 Example 11.10 composite PCM for loading 2.
22.5 Virtual Work Method of Analysis
487
We need to locate the instantaneous center of rotation (ICR)point for the structural segmentconnectingpoints 3 and 6 in Figure 11.40.From similar triangles,
we find that =Y X
hl+R6
L
where
y = h l + ( i ) h , = h l + 2xh L L/2 Solving for R, we obtain
R 6 = 7lL+ 2 h  h 2
1
The horizontal displacement at point 6 is u6 =h l P = R68 which gives the following mechanism relation:
The vertical displacement at point 3 is
v3=xa=(Lx)e which gives the following mechanism relation:
The virtual work equation is
M, [ ( a + e ) + ( e + p ) ] =wYLv3 1 +H,hla 2
Substitution of the mechanism relations gives
M,
=
DifferentiatingM, with respect to x, settingthis result to zero, and solving for x give x=
,,,i /Qpyq} 1+ 1+
1
h2
For o w loading 2 (seeFigure 11.37), H,, = 7.1 kips, w,, = 2.00 kips/ft, L = 80 ft, h, = 20 ft, and h, = 15 ft. Substitution of these values into the preceding formulas gives us x
488
Plastic Analysis and Design = 33.26 ft and Mu, = 624 ftkips.Therefore, loading 1 governs and M , = 786 ftkips. Note that if we set H, = 0 and w,= 2.65 kips/ft in the formuIas obtained for the composite mechanism of loading 2, we obtain x = 34.44 ft and Mu, = 786 ftkips. Therefore, the formulas for x and M , are valid for the case when H,= 0.
See the structure and loadings shown in Figure 11.41. Use Fy = 36 ksi. Assume that the simultaneous occurrence of the beam mechanisms for loading 1 is the governing PCM and determineM, for each member by the equilibriummethod. Use the virtual work method and the relative Mu, values obtained from loading 1 to verify that loading 2 does not govern the Mu, values.
Solution For loading 1and the beam mechanisms [see Figure 11.42(a)],we find, for member 1,
3.12 k/ft
(a) Loading I
2.65 Wft
k
12.74 kips
(b) Loading 2
(c) Real and possible plastic hinge numbers
FIGURE 11.41 Example 11.11: structure and loadings.
11.5 Virtual Work Method of Analysis
489
3.12 Wft

60 ft
30 ft
(a) Loading 1 and PCM
(b) Moment diagram
FIGURE 11.42 Example 11.11: composite PCM for loading 1.
For member 2,
The moment check [see Figure 11.42@)], establishes the indicated Mu, for the columns. If loading 1 (loadingcombination that does not involve wind) governs, we have established that Mu,= 176 ftkips for members 2 and 5, Mu,= 527 ftkips for member 4, and Mu, = 702 ftkips for members 1 and 3. If we choose the least of these values as the base value and express the other values as a function of this base value, we find that the relative Mu,values are Mu, for members 2 and 5,3M, for member 4, and 4M,, for members 1 and 3. Now, we choose to prove that loading 2 does not govern the Mu,values. Any assumed composite mechanism solution for which the required moment diagram ordinates do not exceed the Mu,values obtained for loading 1 is proof that loading 2 does not govern the Mu,values. We choose to investigate the trial PCM shown in Figure 11.43(b).The total internal virtual work is Wint= Mu, (28) + 4MU,8+ 3MU,8+M,,B = 16M,,B
and the total external virtual work is
w,,
[:
= 12.74(15e)+2.65 (60)(30e)
J = 2576.18
From Win,= Wex,,we find that Mu,= 161 ftkips, 3M,, = 483 ftkips, and 4M,,= 644ftkips. When we perform the moment check [seeFigure 11.43(c)],we find that our trial PCM has enabled us to prove that loading 2 does not govern the Mu,values.
490
Plastic Analysis and Design
12.74 kips
30.2 aft 76.3

30 ft
122.4
39.8
(a) hading 2 and the reactions for the trial PCM
e
(b) Trial PCM
(c) Moment diagram
FIGURE 11.43 Example 11.11:compositePCM for loading 2. We have established that loading 1govern and M, = 176 ftkips for members 2 and 5, M, = 527 ftkips for member 4,and M, = 702 ftkips for members 1and 3. We can use the approach illustrated for two bays and a onestory frame when there are two or more bays. As the number of bays increases, a loading combination that involves wind is less likely to govern the relative Mu,values. 11.6 JOINTSIZE In a frame, the plastic hinges do not form at the intersection of the member centerlines as we assumed in the preceding example problems. Consider a onestory rectangular frame (see Figure 11.44).If we use the same W section for members 1and 3, due to a larger axial force in member 3, the plastic hinge at the intersection of members 1and 3 will form at the end of member 3. If the W section chosen for member 3 is not the same as the one chosen for member 1, the plastic hinge will form at the end of the weaker member. At the intersection of members 1,2, and 4,there are three possible plastic hinge locations (one at the end of each
11.6 Joint Size
491
(a) Member numbers, real and possible plastic hinge numbers
(b) Corner (or eave) connection
u (c) Connection at interiorjoint
FIGURE 11.44 Connection details at corner and interior joints.
of the intersecting members). Actually/ as we have described, the plastic hinges form at the interface of the beam and column members. This interface is at the connection locations. Hence, connections are coincident with plastic hinge locations. Connections must be strong enough to develop the required plastic hinge strength for an assumed inelastic rotation capacity of 3 (see footnote c on LRFD p. 639) at the plastic hinge in a member. According to LRFD p. 6171, the “target” reliability index is /3= 3.0 for members and p = 4.5 for connections.The larger value of p= 4.5 means that the connections are expected to be stronger than the members they connect. This means that a plastic hinge at a joint is expected to form outside the connection in one of the interconnected member ends. Therefore, the plastic collapse strength of the frame is somewhat larger than that predicted on the basis of centerline dimensions.Thus, if we ignore the joint size in the plastic analysis, the results obtained are conservative. In a gabled frame (see Figure 11.45),the eave connection usually is haunched in some manner. Sometimes, the crown connection is also haunched. In design, we assume that a plastic hinge will not form within a haunch. Therefore, in the plastic analysis made for the final design considerations,we should assume plastic hinge locations on each end of the haunch. This complicates the plastic analysis computations somewhat, but our final analysis assumptions should be a reasonably good representation of how we intend for the structure to behave. If a beam mechanismgovernsin a plasticanalysisof a frame,we can use the clear The clear span of the beam is the distance span of the beam when we compute Mu,,. between the column flanges. If the column has not already been designed,we use an assumed depth of the column section in computing the clear span and check our assumed depth.
492
Plastic Analysis and Design
A (a) Gabled frame
(b) Eave connection
(c) Haunched eave connections
(d) Crown connections
FIGURE 11.45 Connection details for gable frames.
Problems
493
PROBLEMS 11.1 For the structure and loading in Figure P1l.l, find the correct PCM and MUp. 102 kips

20 ft
'
a
loft
FIGURE P1l.l
11.2 Solve Problem 11.1 with the uniform load replaced by 23.5 kips at each support and 47 kips at the third span points. 11.3 See Appendix D, Figures D.8 and D.9. For five spans of L = 30 ft, w,= 0.832 kips/ft, L, = 0 for the top flange, L, = 6.0 ft wherever needed for the bottom flange, Fy = 36 ksi, select the lightest acceptable W sections for each case and compare the total weights. 11.4 Solve Problem 11.3 for Fy = 50 ksi and I,,= 4.0 ft for bottom flange. 11.5 See Figure P11.5. For the 30ft span, L, = 10 ft. For the 36ft span, L, = 9 ft. Use Fy = 36 ksi and the same W section in both spans. Select the lightest acceptable W section. 24
36
64 kips

k loft
loft
loft
'
18 ft
18 ft
4
FIGURE P11.5
11.6 See Figure P11.6. For the 30ft span, L, = 10 ft. For the 36ft span, L, = 9 ft. Use Fy= 36 ksi. Select the lightest W section that is satisfactory for the 30ft span. Use this W section in the other span and design a pair of cover plates only in the region indicated by a heavy line.
24
l o f t *loft
FIGURE P11.6
36
64 kips
18 ft
18 ft
)I
494
Phstic Analysis and Design 11.7 See Figure P11.7. For the 30ft span, Lb = 10 ft. For the 36ftspan, Lb = 9 ft. Use Fy= 36 ksi. First, selectthe lightestW section that is satisfactory for the 30ft span. Use @M for this W section in the PCM for the other span and select the lightest acceptafie w section for the 36h span. 24
64 kips
36
18 ft
18 ft
4
FIGURE P11.7
11.8 see Figure P11.8. For the 30ft span, Lb = 10ft. For the 36ft span, L b = 12 ft. Use Fy= 36 ksi. First, select the lightestW section that is satisfactory for the 30ft span. Use @M for this W section in the PCM for the other span and select the lightest acceptafie w section for the 36h span. 78
52 kips
26
78
12ft
+
12ft
4
FIGURE P11.8
11.9 !ke Figure P11.9. For the 30ft span, Lb = 10ft. For the 36ft span, L b = 12 ft. Use Fy= 36 ksi. First, selectthe lightestW section that is satisfactory for the 36ft span. Use @M for this W section in the PCM for the other span and select the lightest acceptafie w section for the 30h span. 26
52 kips
12ft
4
12 ft
FIGURE P11.9
11.10 InFigureP1l.lO,useLb=7.5ftandFy=36ksi.Selectthelightestacceptable W section.
+ + + + + + 4.40 wft
b FIGURE Pl1.10
30 ft
 
3oft
Problems
495
11.11 InFigureP11.11,useL,=7.5ftandFy=36hi.Selectthelightestacceptable W section.
+ + + + + + 4.40wft
30 ft
k
30 ft
4
FIGURE Pl1.11
11.12 InFigureP11.12,useL, = L/4ineachspan, the sameW section in all spans, and Fy= 36 hi.Select the lightest acceptableW section that is valid for all spans.
+
4.40wft
+ +
+
30 ft
40ft
30 ft
FIGURE Pl1.12
11.13 In Figure P11.13, use L, = L/4 in each span and F, = 36 hi.First, select the lightest acceptableW section that is valid for the exterior spans. Use CpM for thisW section in the PCM for the interior span and select the lightest acceptabgW section for the 40ftspan.
r
4.40wft I
I
I
I
I
I
i
FIGURE P11.13
11.14 In Figure P11.14, use L, = L/4 in each span and F, = 36 hi.First, select the lightest acceptable W section that is valid for the interior span. Use section in the PCM for the exterior span and select the lightest for the 30ftspans. 4.40wft
FIGURE p11.14
496
Plastic Analysis and Design 11.15 In Figure P11.15, use Fy= 36 h i , L b = 6 ft for the girder, and L b = (KL), = 20 ft for the column. Select the lightest acceptable W section for the girder. Select the lightest acceptable W12 section for the column. Also, select the lightest acceptable W14 section for the column. Use the lighter of the two column selections. 3.12 Wft
(a) Loading 1

1r
+
4
+
4
4
1'
6.35 kips
1r
+ + + +
(a) Loading 1
7.92 kips
4
+
4
+
4
1I
Problems
497
the lightest acceptableW12 section for the column. Also, select the lightest acceptable W14 section for the column. Use the lighter of the two column selections. 11.17 Finish Example 11.10. Use Fy = 36 ksi, eight equal spaces of L, for the girder, and L, = (KL), = 20 ft for the column. Select the lightest acceptable W section for the girder. Select the lightest acceptable W12 section for the column. Also, select the lightest acceptableW14 section for the column. Use the lighter of the two column selections.
11.18 Finish Example 11.11.Use Fy= 36 ksi, L b = 6 ft for the girders,and L, = (KL), = 15ft for the columns. Select the lightest acceptable W section for each girder. Select
the lightest acceptable W12 section for each column. Also, select the lightest acceptable W14 section for each column. Use the lighter column selections. 11.19 In Figure P11.19, use Fy= 36 ksi, five equal spaces of L, for the girder, and L, = (KL) = 20 ft for the column. Assume that the plastic hinges form only in the girders. 8elect the lightest acceptable W section for the girders. Select the lightest acceptable W12 section for the column. Also, select the lightest acceptable W14 section for the column. Use the lighter of the two column selections.
2.65 k/ft
1
(a) Loading I 2.00 Wft
1
(b) Loading 2
FIGURE P11.19
498
Plastic Analysis and Design 11.20 In Figure P11.20, determine the relative values of Mu,,for the trial PCM shown. Compare these values with those that were obtainedin Example 11.11 when the sidewind load was applied in the opposite direction.
(b) Trial PCM FIGURE P11.20
Computer Output for an EHample of an Elastic,
Factored load Analgsis STAAD (structuralanalysisand design),a proprietary computer software program of Research Engineers,Inc., Yorba Linda, CA, was used to obtain a PDELTA analysis of a plane frame (Figures 1.14 and 1.15)from the structure shown in Figure 1.16. The output from STAAD begins immediately after Figure A.1 in which the STAADsign conventionsfor input and output are given. For the user's convenience, STAADplaces a number in front of each line of input and prints these line numbers in the output. The elastic PDELTA analysis was performed for applicable LRFD Load Combinations (A41) to (A46) (p. 630) and accountsfor the secondorder (PA) effects due to joint displacements.Therefore,we do not have to use LRFD Eq.(Cl1) (p. 641) in designing the members of this structure. To minimize the number of output pages, memberend forces are listed for only the governing members and a few other members to indicate the range of axial force and bending moments. ForceY
xtrans joint
A
MomZ
ForceX
joint
forces ~~
X 7
(a) Positive vector directions of global information
(b) Positive vector directions of memberend information
FIGURE A.l Positive directions of global and local forces. 499
500 AppendixA
STAAD PLANE (SEE FIGURES 1.14  1.17) 2. UNIT KIP FEET 3. JOINT COORDINATES 4. 1 0. 0.; 2 0. 21.; 3 0. 25.5 5. 4 6. 21.; 5 6. 26.; 6 12. 21.; 7 12. 26.5 6. a i a . 21.; 9 18. 27.; 10 24. 21.; 11 24. 27.5 7. 12 30. 21.; 13 30. 28.; 14 36. 21.; 15 36. 27.5 a . 16 42. 21.; 17 42. 27.; i a 48. 21.; 19 48. 26.5 9. 20 54. 21.; 21 54. 26.; 22 60. 21.; 23 60. 25.5 24 60. 0. 10. MEMBER INCIDENCES 11. 1 1 2; 2 2 3; 3 24 22; 4 22 23 12. 5 2 4; 6 4 6; 7 6 a; a a 10; 9 10 12 13. 10 12 14; 11 14 16; 12 16 18; 13 i a 20; 14 20 22 14. 15 3 5; 16 5 7; 17 7 9; i a 9 11; 19 11 13 15. 20 13 15; 21 15 17; 22 17 19; 23 19 21; 24 21 23 16. 25 4 5; 26 6 7; 27 a 9; 28 10 11; 29 12 1 3 17. 30 14 15; 31 16 17; 32 i a 19; 33 20 21 1 8 . 34 3 4; 35 5 6; 36 7 8; 37 9 10; 38 11 12 19. 39 12 15; 40 14 17; 41 16 19; 42 18 21; 43 20 23 20. UNIT INCHES 21. MEMBER PROPERTIES IZ 310. 22. 1 TO 4 PRISMATIC AX 1 1 . 8 IZ 20.9 23. 5 TO 14 PRISMATIC AX 5. 24. 15 TO 24 PRISMATIC AX 5.89 IZ 23.3 25. 25 33 PRISMATIC AX 3.13 IZ 3.83 26. 26 TO 32 PRISMATIC AX 2.38 IZ 2.17 27. 34 TO 43 PRISMATIC AX 2.38 IZ 2.17 28. CONSTANTS 29. E 29000. ALL 30. UNIT FEET 31. SUPPORTS 32. 1 24 FIXED BUT KMZ 140.6 UNITS ARE FTKIPS/DEGREE 33. * * 34. VALUE SHOWN IS FOR G = 2 AT THE SUPPORT 35. LOADING 1 36. * DEAD 37. JOINT LOADS 3 8 . 5 7 9 11 15 17 19 21 FY 2.20; 3 23 FY 1.10 13 FY 2.90 39. 4 6 a 10 12 14 16 i a 20 FY 0.20; 2 22 FY 10.90 1.
40.
41. 42. 43. 44. 45.
*
LOADING 2 * LIVE LOAD (CRANES) JOINT LOADS 6 1 8 FY  8 . ; 12 FY 16.
*
Computer Output for an Example of an Elasfic, Factored Load Analysis 501
46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88.
LOADING 3 * ROOF LOAD (SNOW) JOINT LOADS 5 7 9 11 13 15 17 19 21 FY 3.60;
3 23 FY 1.80
* LOADING 4 * WIND LOAD JOINT LOADS 3 FX 0.085 FY 1.; 23 FX 0.085 FY 1. 5 7 9 11 FX 0.170 FY 2.; 15 17 19 21 FX 0.170 FY 2. 13 FY 2. MEMBER LOADS 1 2 UNIFORM GX 0.240; 3 4 UNIFORM GX 0.150 * LOADING 5 * 0.9D + W   TO CHECK DRIFT REPEAT LOAD 1 0.9 4 1.
* LOADING 6 * 1.4D REPEAT LOAD 1 1.4
* LOADING 7 * 1.2D + 1.6L +0.5R REPEAT LOAD 1 1.2 2 1.6 3 0.5
* LOADING 8 * 1.2D + 0.5L +1.6R REPEAT LOAD 1 1.2 2 0.5 3 1.6
* LOADING 9 * 1.2D + 1.6R + 0.8W REPEAT LOAD 1 1.2 3 0.5 4 0.8 * LOADING 10 * 1.2D + 1.3W + 0.5L + 0.5R REPEAT LOAD 1 1.2 4 1.3 2 0.5 3 0 . 5
*
8 9 . LOADING 11
90. * 0.9D + 1.3W 91. REPEAT LOAD
502 Appendix A
1 0.9
92. 93. 94. 95. 96. 97. 98.
4 1.3
* PDELTA ANALYSIS
*
* OUTPUT FOR DRIFT & DEFLECTION CHECKS LOAD LIST 1 TO 5 PRINT JOINT DISPLACEMENTS LIST 2 3 12
JOINT DISPLACEMENTS

JT
XTRANS
YTRANS
ZROTAN
LOAD
(INCHES RADIANS)
2
1 2 3 4 5
0.03821 0.05922 0.05508 1.04229 1.03581
0.01705 0.01178 0.01326 0.00818 0.00714
0.00083 0.00130 0.00119 0.00023 0.00100
3
1 2 3 4 5
0.01814 0.02927 0.02610 1.00881 1.05344
0.01891 0.01421 0.01599 0.00980 0.00718
0.00118 0.00185 0.00170 0.00106 0.00001
12
1 2 3 4 5
0.00000 0.00000 0.00000 1.01927 1.04743
0.35005 0.55002 0.49100 0.27932 0.03551
0.00000 0.00000 0.00000 0.00006 0.00007
99. 100. 101. 102.
* * OUTPUT FOR STRENGTH CHECKS LOAD LIST 6 TO 11 PRINT REACTIONS ALL
SUPPORT REACTIONS  (KIPS FEET) JOINT
LOAD
FORCEX
FORCEY
MOM Z
1
6 7 8 9 10 11
0.88 2.77 2.69 3.53 6.01 7.15
32.41 62.38 64.58 27.84 30.25 6.36
4.31 13.63 13.22 21.41 36.21 40.68
24
6
0.88
32.41
4.31
Computer Output@ an Example of an Elastic, Factored Load Analysis 503 7 8 9 10 11
2 * 77 2.69 4.42 6.92 5.78
62.38 64.58 29.72 33.31 9.31
1 0 3 . PRINT MEMBER FORCES L I S T
M E R END FORCES
M 1
LOAD 6 7 8 9 10 11
2
6 7 8 9 10 11
3
6 7 8 9
JT

(KIPS
AXIAL
13.63 13.22 27.85 44.10 37.48 1 TO 5 25 29 33
FEET)
SHEARY MOMZ
1 2 1 2 1 2 1 2 1 2 1 2
32.41 32.41 62.38 62.38 64.58 64.58 27.84 27.84 30.25 30.25 6.36 6.36
0.88 0.88 2.77 2.77 2.69 2.69 3.53 0.50 6.01 0.54 7.15 0.60
4.31 14.38 13.63 45.49 13.22 44.11 21.41 12.28 36.21 24.45 40.68 41.37
2 3 2 3 2 3 2 3 2 3 2 3
16.48 16.48 47.39 47.39 49.54 49.54 14.49 14.49 16.99 16.99 2.82 2.82
2.09 2.09 6.68 6.68 6.32 6.32 3.06 3.92 5.58 6.98 7.99 9.39
11.59 2.06 37.12 6.06 35.67 6.21 13.18 2.44 24.91 3.19 38.48 0.64
24 22 24 22 24 22 24
32.41 32.41 62.38 62.38 64.58 64.58 29.72
0.88 0.88 2.77 2.77 2.69 2.69 4.42
4.31 14.38 13.63 45.49 13.22 44.11 27.85
1 0 1 4 1 5 20 24 34 39 43
504
AppendixA
10 11
4
6 7 8 9 10 11
5
6 7 8 9 10 11
10
6 7 8 9 10 11
14
6
22 24 22 24 22
29.72 33.31 33.31 9.31 9.31
1.90 6.92 2.82 5.78 1.68
40.67 44.10 62.00 37.48 41.91
22 23 22 23 22 23 22 23 22 23 22 23
2.09 16.48 2.09 16.48 6.68 47.39 6.68 47.39 6.32 49.54 6.32 49.54 7.50 15.66 8.04 15.66 18.90 11.66 12.54 18.90 8.56 1.03 9.44 1.03
11.59 2.06 37.12 6.06 35.67 6.21 36.33 1.31 55.92 1.36 39.38 1.13
2.98 2.98 9.46 9.46 9.01 9.01 2.56 2.56 5.04 5.04 8.59 8.59
0.67 0.67 1.91 1.91 1.96 1.96 0.27 0.27 0.19 0.19 0.62 0.62
2.79 1.26 8.37 3.46 8.44 3.71 0.90 0.70 0.46 0.59 2.90 0.84
12 35.78 35.78 14 12 114.24 14 114.24 12 109.64 14 109.64 12 31.27 31.27 14 12 40.60 40.60 14 5.66 12 14 5.66
0.04 0.04 0.22 0.22 0.01 0.01
0.17 0.48 1.62 0.95 0.81 1.31 0.17 0.40
0.02
0.33 0.04 0.11
20 22
0.67 0.67
1.26 2.79
2 4 2 4 2 4 2 4 2 4 2 4
2.98 2.98
0.03
0.03 0.06 0.06 0.02
0.59
Computer Output for an Example of an Elastic, Factored Load Analysis 505 7 8 9
10 11
15
6 7 8 9
10 11
20
6 7 8 9
10 11
24
6 7 8 9
10
20 22 20 22 20 22 20 22 20 22
9.46 9.46 9.01 9.01 9.40 9.40 14.48 14.48 10.25 10.25
1.91 1.91 1.96 1.96 0.98 0.98 1.33 1.33 0.53 0.53
3.46 8.37 3.71 8.44 1.63 4.34 2.10 6.08 0.62 2.52
3 5 3 5 3 5 3 5 3 5 3 5
15.48 15.48 46.41 46.41 46.87 46.87 18.52 18.52 24.64 24.64 5.18 5.18
0.50 0.50 1.23 1.23 1.29 1.29 0.56 0.56 0.70 0.70 0.11 0.11
2.04 1.20 6.02 3.36 6.15 3.57 2.35 1.28 3.06 1.58 0.53 0.16
13 36.16 1 5 36.16 13 122.47 15 122.47 13 112.48 15 112.48 13 32.43 15 32.43 13 44.75 1 5 44.75 13 4.14 15 4.14
0.11 0.11 0.33 0.33 0.38 0.38 0.10 0.10 0.11 0.11 0.03 0.03
0.04 0.61 0.30 1.91 0.10 1.92 0.03 0.56 0.13 0.70 0.07 0.09
21 23 21 23 21 23 21 23 21 23
0.50 0.50 1.23 1.23 1.29 1.29 0.36 0.36 0.37 0.37
1.20 2.04 3.36 6.02 3.57 6.15 0.89 1.36 0.95 1.46
15.48 15.48 46.41 46.41 46.87 46.87 9.41 9.41 9.77 9.77
506 A p p d i x A
25
11
21 23
9.17 9.17
0.25 0.25
0.46 1.02
6
4 5 4 5 4 5 4 5 4 5 4 5
13.53 13.53 41.33 41.33 41.59 41.59 10.93 10.93 13.07 13.07 3.96 3.96
0.30 0.30 1.09 1.09 1.06 1.06 0.24 0.24 0.30 0.30 0.09 0.09
0.72 0.70 2.28 2* 20 2.21 2.14
12 13 12 13 12 13 12 13 12 13 12 13
1.72 1.72 14.40 14.40 8.69 8.69 1.51 1.51 4.55 4.55 0.64 0.64
0.00 0.00 0.00 0.00 0.00
20 21 20 21 20 21 20 21 20 21 20 21
13.53 13.53 41.33 41.33 41.59 41.59 13.90 13.90 17.92 17.92 0.73 0.73
0.04
3 4
21.88 21.88 66.32 66.32 66.43
0.04 0.04 0.38 0.38 0.37
7 8 9 10 11
29
6 7 8 9 10 11
33
6 7 8 9 10 11
34
6 7
3
8
4 3
0.00 0.00 0.00 0.01 0.01 0.01 0.01 0.30 0.30 1.09 1.09 1.06 1.06 0.32 0.32 0.43 0.43 0.04
0.53
0.57 0.65 0.72 0.27 0.18 0.00 0.00 0.00
0.00 0.00 0.00 0.02 0.01 0.03 0.02 0.02 0.02 0.72 0.70 2.28 2.20 2.21 2.14 0.79 0.71 1.07 0.94 0.14 0.04 0.02 0.05 0.04 0.18 0.06
Computer Outputfor an Example of an Elastic, Factored Load Analysis 507
9 10
11
39
6 7 8 9
10 11
43
6 7 8 9
10 11
104. FINISH
4 3 4 3 4 3 4
66.43 18.23 18.23 22.09 22.09 5.14 5.14
0.37 0.05 0.05 0.08 0.08 0.03 0.03
0.16 0.09 0.03 0.13 0.05 0.11 0.12
12 15 12 15 12 15 12 15 12 15 12 15
0.92 0.92 7.47 7.47 0.30 0.30 0.46 0.46 4.51 4.51 2.53 2.53
0.00 0.00 0.01 0.01 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.02 0.02 0.07 0.07 0.06 0.06 0.00 0.03 0.01 0.04 0.02 0.01
20 23 20 23 20 23 20 23 20 23 20 23
21.88 21.88 66.32 66.32 66.43 66.43 21.82 21.82 27.96 27.96 0.52 0.52
0.04 0.04 0.38 0.38 0.37 0.37 0.01 0.01 0.02 0.02 0.03 0.03
0.05 0.02 0.18 0.04 0.16 0.06 0.12 0.05 0.19 0.10 0.11 0.12
References
1. Smith,J.C., Structural Analysis, New York Harper & Row, 1988. 2. AISC Manual of Steel Construction,Load 6 Resistance Factor Design, Volume I: Structural Members, Specifications,& Codes, 2nd ed.; Volume Ik Connections,2nd ed., Chicago, n. American Institute of Steel Construction,1994. 3. Galambos, T.V., and Ellingwood,B., “ServiceabilityLimit States: Deflection,” Journal of the Structural Division, ASCE, Vol. 112,No. 1,Jan. 1986,pp. 6884. 4. Euler, L., Methodus lnveniendi Lineas Cumus Muximi Minimive Proprietate Guudentes, Appendix: De C u d Elasticis, Lausanne and Geneva, 1744. 5. Euler, L., “Sur la force de colonnes,” Memoires de I’ACademie de Berlin, 1759. 6. Timoshenko, S.P., and Gere,J.M., Theory of Elastic Stability, New York McGrawHill,1961. 7. Bleich, F., Buckling Strength of Metal Structures, New York McGrawHill, 1952. 8. Galambos, T.V., Structural Members and Frames, Englewood Cliffs, NJ: PrenticeHall, 1968. 9. Wood, B.R., Beaulieu, D., and Adams, P.F., ”Column Design by PDelta Method,” Journal ofthe Structural Division, ASCE, Vol. 102,No. ST2,Feb. 1976,pp. 411427. 10. American Concrete Institute, Building Code Requirements@ Reinfirced Concrete, ACI 31&92R, Detroit, MI: 1992. 11. Kavanagh, T.C., “Effective Length of Framed Columns,” Transactions ASCE, Vol. 127,Part II, 1962,pp. 81101. 12. Standard Specirntionsfir Highwuy Bridges, Washington, DC :American Association of State Highway and Transportation Officials (AASHTO), 1989. 13. Sped@tionsjbr Steel Railway Bridges, Chicago, kAmerican Railway Engineering Association, 1992. 14. Cochrane, V.H., ”Rules for Riveted Hole Deductionsin Tension Members,” Engineering NewsRecord (NewYork), Nov. 16,1922,pp. 847848. 15. M m , W.H., and Chesson, Jr., E., “Rveted and Bolted Joints,” Journal of the Structural Division, ASCE, Vol. 89,No. STl,February 1963. 16. Easterling, W.S. and Giroux, L. G., “Shear Lag Effects in Steel Tension Members,” Engineering Journal,AISC,Vol. 30,No. 3 (3rd quarter), 1993,pp. 7789. 17. Yura, J.A., “The Effective Length of Columns in Unbraced Frames,” Engineering 531
532
Journal, AISC, Vol. 8,No. 2 (2nd quarter), 1971,pp. 3742. 18. Winter, G., ”Lateral Bracing of Columns and Beams,” Transactions ASCE, Vol. 125, Part I, 1960,pp. 808845. 19. McGuire, W., Steel Structures, Englewood Cliffs, NJ: PrenticeHall, 1968. 20. Galambos, T.V., “Lateral Support for Tier Building Frames,” Engineering Journal, AISC, Vol. 1, No. 1 (1st quarter), 1964,pp. 1619; Discussion 1, No. 4 (4th quarter), 1964, p. 141. 21. Crawford, S.F., and Kulak, G.L., “Eccentrically Loaded Bolted Connections,” Journal of the Structural Division, ASCE, Vol. 97,No. ST3, March 1971,pp. 765783. 22. Butler, L. J., Pal, S., and Kulak, G.L., “Eccentrically Loaded Weld Connections,” Journal of the Structural Division, ASCE, Vol. 98,No. ST3,May 1972,pp. 9891005. 23. Shipp, J.G., and Haninger, E.R., ”Design of Headed Anchor Bolts,” Engineering Journal, AISC, Vol. 20,No. 2 (2nd quarter), 1983,pp. 59. 24. Curtis, L.E., and Murray, T.M., “Column Flange Strength at Moment EndPlate Connections,” Engineering Journal, AISC,Vol. 26,No. 2 (2nd quarter), 1989,pp. 4150. 25. Galambos, T.V., ed., Guide to Stability Design Criteriafor Metal Structures, 4th ed., Structural Stability Research Council, New York: John Wiley & Sons, 1988. 26. Ravindra, M. K., and Galambos, T. V., “Load and Resistance Factor Design for Steel,” Journal of the Structural Division, ASCE, Vol. 104,No. ST9, Sept. 1978,pp. 13371353. 27. Ellingwood, B.,et al., Development of a Probability Based Load Criterion for American National Standard A58 Building Code Requirements for Minimum Design Loads in Buildings and Other Structures, Special Publication 577,Washington, DC:National Bureau of Standards, June 1980. 28. Beedle, L. S.,Plastic Design of Steel Frames, New York John Wiley & Sons, 1958. 29. Tall, L., ”Residual Stresses in Welded Plates  A Theoretical Study,” Welding Journal, Vol. 43,Jan. 1964. 30. American Welding Society (AWS), Structural Welding CodeSteel(DI.l), AWS, Miami, FL, 1992. 31. Julian, 0.G., and Lawrence, L. S., Notes on J and L Nomgrams for Determination of Effective Lengths, 1959 (unpublished). 32. Dumonteil, P., “Simple Equations for Effective Length Factors,” Engineering Journal, AISC, Vol. 29,No. 3 (3rd quarter), 1992,pp. 111115. 33. Lesik, D. F., and Kennedy, D. J. L., “Ultimate Strength of FilletWelded Connections Loaded in Plane,” Canadian Journal of Civil Engineering, Vol. 17,No. 1, National Research Council of Canada, Ottawa, Canada, 1990. 34. Chen, W. F., and Lui, E. M., Stability Design of Steel Frames, Boca Raton, F L CRC Press, Inc., 1991.
INDEX
Index Terms
Links
A ACI Code
277
Adams, P. F
531
402
Analysis: approximate methods
277
elastic factored load
28
31
43
firstorder
253
258
262
Pdelta method
253
262
plastic
262
440
secondorder
254
262
structural truss
498
262
526
13 118
Analytical model
14
15
46
64
23
Area: effective net ineffective
139
net
52
ASTM
1
3
8
16
B Base plates Beam applied member end moments
181
bearing plates
229
bearing stiffeners
236
bending strength
171
biaxial bending
221
168
236 188
211
This page has been reformatted by Knovel to provide easier navigation.
499
Index Terms
Links
Beam (cont.) block shear rupture
170
bracing
185
builtup
209
camber
180
Cb factor
189
196
charts
192
197
composite
408
compression elements
184
concentrated loads on flanges and web
227
366
connections: moment
353
endplate
330
knee or comer
356
shear
343
bracket plates
353
endplate
353
stiffened seats
350
unstiffened seats
331
web
343
splices connector plates connectors
88
354
344 312
336
342
102 83
coped
170
186
cover plates
210
deflection
169
design flowchart
198
elastic section modulus
172
180
flange local buckling
184
192
flexure formula
172
518
holes in flanges
187
hybrid
179
195
431
203
209
This page has been reformatted by Knovel to provide easier navigation.
353
Index Terms
Links
Beam (cont.) initial crookedness
180
laterally braced
181
lateral support
181
185
lateraltorsional buckling
181
189
local web yielding
232
mechanism
440
momentmature
174
moment redistribution
200
plastic analysis
526
plastic moment
172
plastic section modulus
172
residual stresses
175
183
shear center
146
223
shear strength
170
410
sidesway web buckling
232
sweep
180
205
445
176
torsional constants unsymmetrical bending
223
warping
182
weak axis bending
183
web crippling
229
web local buckling
205
web stiffeners
236
Beamcolumn
8
31
118
biaxial bending
155
258
261
composite
435
design
2
first trial section selection
276
strength
261
primary moment
250
155
261
This page has been reformatted by Knovel to provide easier navigation.
261
Index Terms
Links
Beamcolumn (cont.) secondary moment
250
secondorder effects
248
253
331
344
Beam seats Bearing: failure
88
plates
229
strength
90
stiffeners
236
Bearingtype bolted connections
4
44
bearing strength
44
90
concentric shear
87
eccentric shear
312
minimum distances
87
91
Beaulieu, D.
531
Beedle, L.
440
532
Biaxial bending
155
221
423
519
Bleich, F.
531 66
93
102
Block shear rupture beams
45 170
connector plates
93
102
103
tension members
456
66
70
Bolt holes: staggered
52
standard
45
46
Bolts
44
84
Boundary conditions
18
121
hinge
18
fixed support
18
link
18
prescribed support movement
19
roller
18
138
20
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Boundary conditions (cont.) rotational spring
18
translational spring
18
Braced frame Bracing
33
157
247
124
127
283
beam
309
column
124
127
299
33
248
264
8
12
118
wind
263
287
135
184
288
368
Buckling: column flexuraltorsional
119
lateraltorsional
181
189
205
loads
118
120
127
local
8
12
119
192
203
264
12
120
156
255
257
264
Builtup beams
209
366
Builtup column
148
mode shapes shear
171
story
155
sidesway web
232
Building: code
32 23
multistory
Butler, L. J.
98
321
Camber
123
180
Centroidal axes
508
524
Chen, W., F.
128
532
Chesson, Jr., E.
47
531
Cochrane, V. H.
53
531
C
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Cold bending
6
Column
8
118
action
118
area reduction factor Qa
139
base plates
337
boundary conditions
121
138
buckling
12
36
Euler
119
elastic
119
145
flexural
119
145
flexuraltorsional
119
145
inelastic
125
load
118
120
local
12
135
torsional builtup
118
122
145 148
doubleangle behavior
149
intermediate connections
149
single angle behavior
149
155
composite
400
critical load
118
120
critical stress
122
136
design requirement
124
126
design strength
124
127
155
effective length
119
121
127
initial crookedness
118
122
124
outofstraightness
122
125
127
pinnedended
119
radius of gyration
122
403
residual stresses
125
127
slenderness ratio
122
126
splices
342
140 403
139
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Column (cont.) story design strength strength
155 11
stress reduction factor Qs
139
tangent modulus of elastiaty Et
125
unbraced frame
134
web stiffeners
356
Composite beamcolumns
435
Composite beams
408
124
155
bending strength
418
coldformed steel deck
399
concreteencased
429
deflection
431
effective slab width
409
fully composite section
411
partially composite section
411
plastic neutral axis
420
shear strength
410
shear studs
399
408
shored construction
408
429
unshred construction
408
430
Composite columns
127
401
412
414 425
400
design limitations
402
design strength
403
Compression elements
11
ineffective areas
139
slender
135
127
129
135
184
139
142
stiffened
11
129
135
unstiffened (projecting)
11
129
139
widththicknessratio
11
129
135
184
Computational precision
40
Connecting elements
83
92
102
107
This page has been reformatted by Knovel to provide easier navigation.
Index Terms Connection eccentricity Connections
Links 47 312
beam: moment
353
endplate
330
knee or corner
356
shear
343
bracket plates
312
endplate
353
stiffened seats
350
unstiffened seats
331
web
343
splices
354
353
344
88
312
336
342
93
102
103
170
bearingtype
44
87
312
slipcritical
86
column base plates
337
block shear rupture bolted:
concentric shear
83
corner or knee
354
eccentric shear
312
eccentric tension and shear
330
hanger
327
hunched
332
prying force
328
rigid (fully restrained)
20
21
semirigid (partially restrained)
20
22
122
shear
20
21
343
tension member
83
336
100
321
welded Cover plates
210
Crawford, S. F.
313
531
This page has been reformatted by Knovel to provide easier navigation.
353
Index Terms
Links
Critical load
118
Critical moment
181
Critical net section
120
55
Critical stress
122
136
Curtis, L. E.
354
532
14
169
differential settlement
13
23
drift
30
169
limiting
31
169
live load
31
34
140
D Deflections
precambering prescribed support movement Deflection index Diaphragms Drift
272 169
170 18
19
31 168 30
272
Ductility
1
5
Durability
14
E Easterling Eccentric shear
47 312
elastic analysis: bolt groups
316
weld groups
323
instantaneous center analysis: bolt groups
313
weld groups
321
Eccentric tension and shear
312
Effective length
119
adjustment for inelastic behavior
121
127
135
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Effective length (cont.) bearing stiffeners
237
braced frames
128
factors
121
unbraced frames
128
Effective net area
46
Effective slab width
409
Ellingwood, B..
532
Environmental effects
14
earthquakes
23
rain
23
snow
23
temperature changes
23
wind
23
Equilibrium method
448
Euler, L.
120
64
23
531
F Fabrication
27
Factor: area reduction Qa
139
load
28
resistance
28
safety
35
strength reduction
29
stress reduction Qs
139
Fatigue
5
Field erection
27
Field inspection
27
43
124
8
Fillet welds, see Welds Flange local buckling
11
137
139
184
203 This page has been reformatted by Knovel to provide easier navigation.
192
Index Terms
Links
Flexuraltorsional buckling
119
145
Flexure formula
172
429
518
Floor beams
168
5
16
33
33
157
247
33
248
292
255
257
264
288
134
155
253
261
269
1
118
120
127
155
30
272
Fracture: brittle
5
shear
109
tensile
44
Frames braced diagonal braces multistory unbraced buckling loads drift firstorder analysis
249
gabled
483
497
nolateraltranslation solution
252
265
Pdelta method
253
262
plastic analysis
448
465
479
preliminary design
276
primary moment
269
relative joint stiffness parameter G
128
135
256
secondary effects
248
secondorder analysis
255
sidesway buckling mode
262
sidesway deflection
255
262
Gabled frame
483
497
530
Galambos, T.V.
128
287
531
Gere, J.M.
531
Girder
128
263
530
530
262
G
168
This page has been reformatted by Knovel to provide easier navigation.
287
Index Terms Giroux
Links 47
Girts
168
Grid
17
Gusset plate
93
102
Haninger, E. R
340
532
Hybrid girder
370
118
H
I Ineffective areas
139
Initial crookedness
118
Instability
122
124
128
135
256
18
256
299
14
J Joint: boundary
18
interior
20
numbers
32
relative stiffness parameter G rotational stiffness size
490
support
18
Joists
168
Julian O. G
128
532
Kavanagh, T. C.
128
531
Kennedy
321
532
Kulak, G. L.
313
531
K
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
L Lap splice
88
Lateral support
181
185
Lawrence, L. S.
128
532
Lesik
321
532
Limit states
37
Lintels
168
Loads
23
applied
14
17
concentrated
23
crane
23
31
dead
13
23
24
29
37
30
40
30
40 earthquake
13
23
26
factored combinations
28
33
40
impact
6
27
induced
27
line
23
live
13
24
26
29 40
37 nominal
28
29
31
rain
23
25
29
roof
23
24
26
29
service condition
28
snow
13
25
29
33
traffic
23
24
wind
13
23
26
29
30
34
40 184
192
203
wind suction Local buckling Lui, E.M.
33 12
135
128
532
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
M McGuire, W.
287
Member buckling
12
Member numbers
32
531
Mohr's circle
224
513
Moment of inertia
270
427
Momentcurvature
174
Moment redistribution
200
Munse, W. H.
47
531
Murray, T. M.
354
532
509
N Net area
52
O Outofplumbness
289
299
Outofstraightness
122
125
440
526
127
P Pal, S. Plastic analysis equilibrium method
448
virtual work method
471
Plastic collapse mechanism
445
Plastic design
450
Plastic hinge
440
443
Plastic moment
172
176
Plastic section modulus
172
Plate girder
209
366
bending strength
371
388
definition of
209
366
443
design methods: This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Plate girder (cont.) conventional
370
tension field
366
370
387
hybrid
369
375
380
shear strength
366
373
388
374
389
stiffeners: bearing
366
intermediate
366
Plates: base
337
bearing
229
boundary conditions
138
connector (spacer)
149
local buckling
136
splice tie
88
163 313
336
342
163
Poisson's ratio
5
Precambering
170
Preliminary design
12
136 276
290
Principal axes
513
beam
172
188
191
224
column
120
123
129
149
Probability density function Product of inertia
37 509
Probability of failure
37
38
Purlins
31
168
72
122
R Radius of gyration Ravindra, M. K.
532
Reference axes
508
Relative joint stiffness G
128
135
This page has been reformatted by Knovel to provide easier navigation.
Index Terms Reliability index β Residual stresses
Links 39 8
125
127
Resistance factor φ
28
43
124
Rigid joints
20
180
429
Rolling process Rotational springs
139
4 257
S Section modulus: elastic
172
plastic
172
Semirigid joints
20
22
Serviceability
30
31
37
223
524
problems
30
requirements
30
44
Shear: buckling
171
center
146
concentric
83
connection
20
21
design strength
88
170
373
171
eccentric
312
fracture
109
lag
45
studs, yielding
45
93
340
532
buckling of frames
118
120
127
deflection
270
web buckling
232
Significant digits
40
Single angles
69
149
155
Shipp, J. G.
388
Sidesway: 155
This page has been reformatted by Knovel to provide easier navigation.
257
Index Terms Slipcritical connections
Links 4
Smith, J. C.
531
Soilstructure interaction
257
Spandrel beam
168
86
Splices
88
312
Stability
14
15
Staggered bolt holes
52
Staggered path
53
Standard bolt hole
45
Standard deviation
38
313
336
Steel characteristics: cold bending
8
cooling rate
4
corrosion resistance
1
creep
6
ductility
1
durability
14
hardness
1
heat treatment
3
prior straining
6
strain hardening
2
6
stressstrain relation
2
5
toughness
1
3
uneven cooling
6
8
weldability
7
7 3
5
4 44 6
Steel properties: coefficient of thermal expansion
5
modulus of elasticity
5
Poisson's ratio
5
tangent modulus of elasticity weight
136
125 5
This page has been reformatted by Knovel to provide easier navigation.
342
Index Terms
Links
Steel properties (cont.) ultimate strength
5
yield strength
5
Steels: alloy
1
carbon
1
coldformed
3
high carbon
1
high strength low alloy
1
killed
4
low carbon
1
medium carbon
1
mild carbon
1
quenched
2
rimmed
4
semikilled
4
tempered
2
weathering
3
6
3
3
Stiffeners: bearing
236
366
intermediate
366
374
389
5
14
30
Stiffness Story design strength
155
Strainhdening
2
6
Strength
1
5
design
29
43
fatigue
8
nominal
29
plastic bending required tensile
14
37
43
172 29
43
4
5
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Strength (cont.) ultimate
5
yield
1
4
Strength reduction coefficient U
47
64
Strength reduction factor φ
29
43
124 7
Stress
5
Strest–strain curve
2
5
coupon
2
10
high temperature effects
6
strain hardening
2
7
10
126
member test Stringers
5
125
168
Structural: analysis
13
behavior
13
design
13
economy
30
safety
34
serviceability
34
37
Structure: deflected shape
31
idealized
15
Summation formulas
510
Sweep
123
180
T Tables: bolt shear strength column deflection index
89 130 31
moment of inertia section modulus
192
This page has been reformatted by Knovel to provide easier navigation.
126
Index Terms
Links
Tables (cont.) serviceability problems steels usual gages Tangent modulus of elastiaty Ev Temperature
31 3 62 125 8
Tension members
34
critical section
52
43
design bolted
59
requirements
44
strength
49
welded
67
effective net area
46
net area
47
net section
52
residual stresses
52
59
47
64
8
staggered bolt holes
52
serviceability
72
stiffness considerations
44
71
strength: bearing
44
block shear rupture
45
54
57
52
64 55
63
70 fracture on Ae
45
nominal
43
reduction coefficient U
47
49
reduction factor φ
43
46
required
43
yielding on Ag
44
Threaded rods Timoshenko, S. P.
64
46
70 531
This page has been reformatted by Knovel to provide easier navigation.
66
Index Terms
Links
Torsional buckling
145
Torsional constants
524
Transfer axes formulas
509
Truss
31
33
40
33
134
155
253
12
137
139
142
205
232
356
358
373
389
crippling
229
232
stiffeners
236
355
366
yielding
232
358
388
Weldability
7
Weld eledrodes
4
analysis equivalent beam
15
16
264
270
35
118
270
U Unbraced frame
15 269
Unsymmetrical bending Usual gages
223 62
V Vibrations Virtual work method
14 471
W Web: buckling
yield strength Welding
fillet
97 96
SMAW process Welds
97
96
102
96 97
99
This page has been reformatted by Knovel to provide easier navigation.
Index Terms
Links
Welds (cont.) balanced design
104
design limitations
99
effective area
97
effective throat
97
end returns
101
fracture
100
longitudinal
99
minimum length
101
minimum size
101
separation
98
strength
99
transverse
100
groove
97
plug
97
puddle
97
slot
97
Widththickness ratio
106
11
127
139
Winter, G.
287
301
531
Wood, B. R.
531
Wind load, see Loads
Y Yield moment
172
Yield point
1
Yield strength
1
Yield stress
35
46
Yura, J. A.
156
531
This page has been reformatted by Knovel to provide easier navigation.