# SPECIFICATIONS, LOADS, AND METHODS OF DESIGN

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1 CHAPTER Structural Steel Design LRFD Method Third Edition SPECIFICATIONS, LOADS, AND METHODS OF DESIGN A. J. Clark School of Engineering Department of Civil and Environmental Engineering Part II Structural Steel Design and Analysis 2b FALL 2002 By Dr. Ibrahim. Assakkaf ENCE Introduction to Structural Design Department of Civil and Environmental Engineering University of Maryland, College Park CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 1 The load factors are usually amplifying factors that are used in LRFD design equation to increase the loads. The purpose of increasing the loads is to account for the uncertainties involved in estimating the magnitudes of dead and/or live loads. How close (%) could you estimate the worst wind or snow load that will ever be applied to a particular building? 1

4 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 6 Example 1 (cont d) w u Span of Beam, L L 8 ft 8 ft CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 7 Example 1 (cont d) Computing factored loads and noting that D and L are the only loads to be supported, therefore using Eqs. 1 to 7 result in: 1. U = 1.4( D + F ) = 1.4( ) = 637 lb/ft 2. U = 1.2( D + F + T ) + 1.6( L + H ) + 0.5( Lr or S or R) = 1.2( ) + 1.6( ) + 0.5( 0 or 0 or 0) = 1570 lb/ft 3. U = 1.2 = 1.2 Controls (largest) D + 1.6( Lr or S or R) + ( 0.5L or 0.8W ) ( 455) + 1.6( 0) + 0.5( 640) = 866 lb/ft 4

5 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 8 Example 1 (cont d) 4. U = 1.2 = 1.2 D + 1.6W + 0.5L + 0.5( Lr or S or R) ( 455) + 1.6( 0) + 0.5( 640) + 0.5( 0) = 866 lb/ft 5. U = 1.2D + 1.0E + 0.5L + 0.2S = 1.2 ( 455) + 1.0( 0) + 0.5( 640) + 0.2( 0) = 866 lb/ft 6. U = 0.9D + 1.6W + 1.6H = 0.9 ( 455) + 1.6( 0) + 1.6( 0) = lb/ft 7. U = 0.9D + 1.0E + 1.6H = 0.9 ( 455) + 1.0( 0) + 1.6( 0) = 409 lb/ft CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 9 Example 2 The various axial loads for a building column have been computed according to the applicable building code with the following results: dead load = 200 k, load from roof = 50 k (roof live load), live load from floor = 250 k, compression wind = 80 k, tensile wind 65 k, compression earthquake = 60 k, and tensile earthquake = 70 k. Determine the critical design load using the combinations provided by Eqs. 1 to 7. 5

6 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 10 Example 2 (cont d) 1. U = U = U = 1.2D + a. b. c. = 1.2 = 665 k U = 1.2 U = 1.2 ( D + F ) = 1.4( ) = 280 k ( D + F + T ) + 1.6( L + H ) + 0.5( Lr or S or R) ( ) + 1.6( ) + 0.5( 50) U = 1.2 Controls (largest) 1.6( Lr or S or R) + ( 0.5L or 0.8W ) ( 200) + 1.6( 50) + 0.5( 250) = 445 k ( 200) + 1.6( 50) + 0.8( 80) = 384 k ( 200) + 1.6( 50) + 0.8( 65) = 268 k CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 11 Example 2 (cont d) 4. U = 1.2D + 1.6W + 0.5L + 0.5( Lr or S or R) a. U = 1.2( 455) + 1.6( 80) + 0.5( 250) + 0.5( 50) = b. U = 1.2( 455) + 1.6( 65) + 0.5( 250) + 0.5( 50) 5. U = 1.2D + 1.0E + 0.5L + 0.2S a. U = 1.2 b. U = 1.2 a. U = 0.9 b. U = 0.9 a. U = 0.9 b. U = 0.9 ( 200) + 1.0( 60) + 0.5( 250) + 0.2( 0) = ( 200) + 1.0( 70) + 0.5( 250) + 0.2( 0) 6. U = 0.9D + 1.6W + 1.6H ( 200) + 1.6( 80) + 1.6( 0) = ( 200) + 1.6( - 65) + 1.6( 0) 7. U = 0.9D + 1.0E + 1.6H 308 k = 76 k ( 200) + 1.0( 60) + 1.6( 0) = 240 k ( 200) + 1.0( 70) + 1.6( 0) = 110 k 518 k = 286 k 425 k = 295 k 6

7 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 12 Strength (or Resistance) Factors Strength factors are usually reduction factors that applied to the strength (stress, force, moment) of the member to account for the uncertainties in material strengths, dimensions, and workmanship. With a resistance factor, the designer attempts to account for imperfection in analysis theory, variation in material properties, and imperfect dimensions. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 13 Strength (or Resistance) Factors This can be accomplished by multiplying the theoretical ultimate strength (also called nominal strength) of each member by a capacity reduction factor φ, which generally less than one. These values are 0.85 for columns, 0.75 or 0.90 for tension members, 0.90 for bending or shear in beams, and so on. Typical reduction factors are provided in Table 1 (Table 2.2, Text). 7

8 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 14 Strength (or Resistance) Factors Table 1. Typical Resistance (Strength) Factors Type of Loading φ Bearing on the projected areas of pins, web yielding under concentrated loads, slip-resistant bolt shear values Beams on bending and shear, fillet welds with stress parallel to weld axis, groove welds base metal Columns, web crippling, edge distance, and bearing capacity at holes Shear on effective area of full-penetration groove welds, tension normal to the effective area of partial-penetration groove welds. Bolts in tension, plug, or slot welds, fracture in the net section of tension members Bearing on bolts (other than A307) Bearing on concrete foundations CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 15 Reliability The reliability of an engineering system can be defined as the system s ability to fulfill its design functions for a specified period of time. In the context of this course, it refers to the estimated percentage of times that the strength of a member will equal or exceed the maximum loading applied to that member during its estimated life (say 25 years). 8

9 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 16 Reliability Motivation Assume that a designer states that his or her designs are 99.6 percent reliable (this is usually the case obtained with most LRFD design). If we consider the designs of 1000 structures, this does not mean that 4 of the 1000 structures will fall flat on the ground, but rather it means that those structures at some time will be loaded into the plastic range and perhaps the strain hardening range. So excessive deformation and slight damage might occur, but a complete failure. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 17 LRFD In the previous example, it would be desirable to have 100%-reliability. However, this is an impossible goal statistically. There will always be a chance of failure (unreliability), say 2 or 3 %. The goal of the LRFD was to keep this to very small and consistent percentage. 9

10 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 18 LRFD To do this, the resistance or strength R of each member of steel structure as well as the maximum loading Q, expected during the life of the structure, are computed. A structure then is s said to be safe if R Q (1) CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 19 LRFD General Form m φr n γ L i= 1 i ni (2) Where φ = strength reduction factor γ i = load factor for the i th load component out of n components R n = nominal or design strength (stress, moment, force, etc.) L ni = nominal (or design) value for the i th load component out of m components 10

11 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 20 Probability Based-design Approach Versus Deterministic Approach m m R n L i φrn γ ili FS i= 1 i= 1 ASD LRFD According to ASD, one factor of safety (FS) is used that accounts for the entire uncertainty in loads and strength. According to LRFD (probability-based), different partial safety factors for the different load and strength types are used. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 21 LRFD The actual values of R and Q are random variables and it is therefore impossible to say with 100% certainty that R is always equal or greater than Q for a particular structure. No matter how carefully a structure is designed, there will be always some chance that Q exceeds R as shown in Figure 1. 11

12 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 22 LRFD Density Function Figure 1 ( g = R Q) < 0 Area (for g < 0) = Failure probability Load Effect (Q) (L) Strength (R) Origin 0 Random Value CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 23 LRFD Reliability Index β A measure of reliability can be defined by introducing a parameter β, called the reliability index. β can be computed using structural reliability theory and knowledge of the first and second moment statistical characteristics (i.e., mean and COV) for both the strength and load variables. 12

13 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 24 LRFD Reliability Index β (cont d) For two variables and linear performance function, the reliability index b can be defined as the shortest distance from the origin to the failure line as shown in Fig. 2. Mathematically, it can be expressed as µ R µ Q β = (2) 2 2 σ σ µ = mean value of strength or load variable σ = standard deviation of strength or load variable R Q CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 25 Figure 2 Reliability Index β L' Failure Region Design Point Failure Line g = 0 β Survival Region R' The reliability index β is the shortest distance from the origin to the failure surface. 13

15 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 28 Advantages of LRFD LRFD Advantages Provides a more rational approach for new designs and configurations. Provides consistency in reliability. Provides potentially a more economical use of materials. Allows for future changes as a result of gained information in prediction models, and material and load characterization Easier and consistent for code calibration. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 29 Computer Example The computer program INSTEP32 design software can be used to perform the necessary calculations for load combinations provided by Eqs. 1 through 7. This program can also assist you in solving many of the problems presented in the textbook. 15

16 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 30 Computer Example Installation of INSTEP32 1. Insert the CD-ROM in your computer. 2. In Windows Explorer, open the INSTEP32 directory on the CD-ROM. 3. Double-click on Setup. 4. The Setup program will guide you through the installation process. INSTEP32 can be started by clicking the INSTEP32 icon on the Start Menu on the desktop. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 31 Computer Example Example 3: Load Combinations To perform the calculations for load combinations, start INSTEP32 and then select Design Load Combinations from the menu bar. After you have done this, that data entry dialogue shown in the following slides will appear. 16

17 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 32 Computer Example Example 3: Load Combinations (cont d) CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 33 Computer Example Example 3: Load Combinations 17

18 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 34 Computer Example Example 3: Load Combinations Using the computer program INSTEP32, compute the governing factored loads for each of the following: D = 200 k, L r = 50 k, L = 250 k, W = 80 k, and E = 60 k. The input and output of the program is shown in the next slides. Therefore, the critical factored load for design is 665 k. CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 35 Computer Example Example 3: Load Combinations 18

19 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 36 Computer Example Example 3: Load Combinations (cont d) CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 37 Computer Example Example 3: Load Combinations (cont d) Critical Factored load 19

20 CHAPTER 2b. SPECIFICATIONS, LOADS, AND METHODS OF DESIGN Slide No. 38 Computer Example Example 3: Load Combinations (cont d) 20

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