DES140: Designing for Lateral-Torsional Stability in Wood Members

DES140: Designing for Lateral-Torsional Stability in Wood embers Welcome to the Lateral Torsional Stability ecourse. 1

Outline Lateral-Torsional Buckling Basic Concept Design ethod Examples In this ecourse, we ll present the concept of lateral torsional buckling behavior, show how to design for it, and then illustrate the design process with a few examples. Our text will be AF&PA s Technical Report (TR) 14 Design for Lateral-Torsional Stability of Wood embers which is a free document from the Free Downloads Library at www.awc.org.

Basic Concept Limit State Beam Deformation in-plane deformation + out-of-plane deformation + twisting Lateral-torsional buckling is a limit state where beam deformation includes in-plane deformation, out-of-plane deformation and twisting. The load causing lateral instability is called the elastic lateral-torsional buckling load and is influenced by many factors such as loading and support conditions, member cross-section, and unbraced length. An unbraced beam loaded in the strong-axis is free to displace and rotate as shown. 3

Basic Concept Limit State Beam Deformation in-plane deformation + out-of-plane deformation + twisting Design of an unbraced beam requires determination of the minimum moment which induces buckling. When a beam--one that is braced at its ends to prevent twisting at bearing points, but which is otherwise unbraced along its length--is loaded in bending about the strong axis so that lateral torsional buckling is induced, three equilibrium equations may be written for the beam in its displaced position: one equation associated with strong axis bending; one equation associated with weak axis bending, and one equation associated with twisting. 4

Basic Concept Equilibrium equations in-plane moment (X-X) ext EI x d v dx out-of-plane moment (Y-Y) β ext EI y d u dz twisting moment du dz ext GJ dβ dz This displacement and rotation can be represented by three equations of equilibrium which equate the externally applied moment to the internal resisting moment. The in-plane moment equation describes the strong axis bending behavior of a beam and, for small displacements, is independent of the out-of-plane and twisting moment equations which are coupled. Using these equations of equilibrium, a set of differential equations can be developed which, when solved, provides the critical moment, cr. A closed-form solution for the case of a simply supported beam under a constant moment has been widely discussed and derivations can be found in the literature and are provided in Appendix A of the text. We ll briefly describe an approach here. By differentiating twisting moment equation with respect to z and substituting the result into the out-of-plane moment equation, 5

Basic Concept Equilibrium equations coupled in-plane moment (X-X) ext EI x d v dx out-of-plane moment (Y-Y) and twisting moment d β dz + ext EI GJ y β 0 the two equations can be combined to form the resulting second-order differential equation. 6

Basic Concept Equilibrium equations boundary conditions strong axis bending ext EI x d v dx lateral-torsional - constant cross-section - constant moment along unbraced length, L u ocr π L u EI y GJ L u If a beam has a constant cross-section and a constant induced moment, o, along the unbraced length, L, there is a closed-form solution to the second order differential equation. By defining a constant k o /EI y GJ, and rewriting, another equation results which is a second-order linear differential equation that has a general solution. Given that the rotation of the beam at the ends is prevented due to end restraint, these boundary conditions can be expressed as (0) 0 and (L) 0. Using a non-trivial result from (L) 0, substituting the value of k, and solving for the critical moment, ocr, yields the final form of the equation shown in the slide. The mathematical details of the development are found on page 1- of TR14. 7

Basic Concept Critical oment simply supported beam under constant moment ocr π L EI GJ ocr reference case critical y moment u L u unbraced length E modulus of elasticity I y weak-axis moment of inertia G torsional shear modulus J torsional constant for the section L u Here: ocr critical moment for the reference case (simply-supported beam under a constant moment) L u unbraced length E modulus of elasticity y I moment of inertia in the weak-axis (y-direction) G torsional shear modulus J torsional constant for the section Determination of the torsional constant, J, is beyond the scope of TR14. However, equations for determining the value of J for common shapes are provided in the literature [see references] in Appendix A.1. 8

Basic Concept Focus: NDS effective length approach based on unbraced length and prescribed loading conditions does not work well with varying loading conditions NDS prescriptive provisions for sawn lumber are being misapplied to Engineered Wood Products with much higher d/b ratios Tabulated effective lengths in NDS provisions only apply to tabulated load cases; otherwise must assume constant moment over entire length of member ASCE 16 (LRFD) - Equivalent oment (End oment) factor permitted for use with non-rectangular members can provide incorrect results when applied to conditions exceeding the limitations of the method. the equation only applies to members with end-moments and with no reverse curvature. Have to adjust!! In early editions of the NDS, beam stability was addressed using prescriptive bracing provisions. The effective length approach (see 1...1.1) for design of bending members was introduced in the 1977 NDS. Basic load and support cases addressed in the 1977 NDS were as follows: 1. A simply-supported beam with: - uniformly-supported load unbraced along the loaded edge - a concentrated load at midspan unbraced at the point of load - equal end moments ( opposite rotation ). A cantilevered beam with: - uniformly distributed load- unbraced along the loaded edge - concentrated load at the free-end unbraced at the point of load 3. A conservative provision for all other load and support cases But these were limiting and did not work well with varying loading conditions. Later changes added more load cases. In addition, prescriptive NDS provisions intended for sawn lumber were being misapplied to engineered wood products with much higher d/b ratios. Tabulated values in the NDS are for tabulated load cases; and if the load case can t be found for the required application, the designer has to assume constant moment over entire length of member. Also in ASCE 16, the Equivalent oment (End oment) factor permitted for use with non-rectangular members can provide incorrect results when applied to conditions exceeding the limitations of the method. For example, the equation only applies to members with end-moments and with no reverse curvature. Clearly adjustments to the method would be required to increase the flexibility of its application. 9

Adjustment Beam Slenderness Adjust to allow for flexural rigidities EI x EI y in addition to EI x > EI y γ 1 I I y x ocr π L u EI y γ GJ The closed form ocr equation was derived for slender beams where the in-plane flexural rigidity (EI x ) is much larger than the out-of-plane flexural rigidity (EI y ). Federhofer and Dinnik found this assumption to be conservative for less slender beams. To adjust for this conservatism in less slender beams where the flexural rigidities in each direction are of the same order of magnitude, Kirby and Nethercot suggested dividing I y by the term γ, where γ 1-I y /I x and I x is the moment of inertia about the strong-axis. 10

Adjustment Other Loading and Support Conditions Requires alternate solution to closed form for ocr See Appendix A.1 for more detail Two adjustment methods from reference loading and support conditions: Effective Length approach Equivalent oment factor approach The closed-form ocr equation applies to a single loading condition (simply-supported beam under a constant moment). any other loading and beam support conditions are commonly found in practice for which the magnitude of the critical moment needs to be determined. A closed-form solution is not available for these conditions and differential equations found in the derivation must be solved by other means including solution by numerical methods, infinite series approximation techniques, finite element analysis, or empirically. Solutions for common loading and beam support conditions have been developed and are provided in design specifications and textbooks. Solution techniques used to solve several common loading conditions and approximation techniques used to estimate adjustments for other loading conditions are discussed in more detail in Appendix A. Two methods are used to adjust the closed form ocr equation for different loading and support conditions. The effective length approach and the equivalent moment factor approach. Both of these methods adjust from a reference loading and support condition. 11

Effective Length Approach Adjusts the unbraced length to an effective length for specific loading and support conditions cr π l EI GJ ocr reference case critical y moment e l e effective length E modulus of elasticity I y weak-axis moment of inertia G torsional shear modulus J torsional constant for the section The effective length approach adjusts the reference case critical moment, ocr, to other loading and support conditions by adjusting the unbraced length to an effective length as provided in this equation using the term e l, where: cr critical moment for specific loading and support conditions e l effective length which equals the unbraced length adjusted for specific loading and support conditions 1

Effective Length Approach effective length l e tabulated for specific loading and support conditions as identified in Table 1 of TR14. 13

Equivalent oment Factor Approach Adjusts the reference case to other loading and support conditions using a factor C ocr reference case critical cr b ocr moment C b equivalent moment factor L 0 BD The equivalent moment factor approach adjusts the reference case critical moment, ocr, to other loading and support conditions using an equivalent moment factor, C b, as provided in the equation above, where: C b equivalent moment factor 14

Equivalent oment Factor Approach Origins of C b C b 1 1 1.75 + 1.05 + 0.3.3 1 minimum end moment maximum end moment AISC 1961 Limitation - does not apply to beams bent in reverse curvature 1 L 1 0 BD Prior to 1961, the steel industry used an effective length approach to distinguish between different loading and support conditions. In 1961, the American Institute of Steel Construction (AISC) adopted the equivalent moment factor approach. This approach expanded application of beam stability calculations to a wider set of loading and support conditions. The empirical equation used for design of steel beams when estimating the equivalent moment factor, C b, has changed over the last 40 years. The original equation for estimating C b is shown here, where: 1 minimum end moment maximum end moment This equation provides accurate estimates of C b in cases where the slope of the moment diagram is constant (a straight line) between braced points but does not apply to beams bent in reverse curvature. For this reason, it is only recommended for 1 / ratios which are negative (with single curvature i.e. 1 / < 0) 15

Equivalent oment Factor Approach Origins of C b 1. 5 C b 0.6 0.4 1 minimum end moment maximum end moment 1 ASIC 1993 Limitation - applies only to beams bent in reverse curvature with constant slope of moment diagram between braced points 1 1 L 0 BD The C b equation here was developed for cases where slope of the moment diagram is constant between braced points and better matches the theoretical solution for beams bent in single and reversed curvature. 16

Equivalent oment Factor Approach Origins of C b C b 3 A + 4 1 B max + 3 C + max A absolute value of moment at l u / 4 B absolute value of moment at l u / C absolute value of moment at 3 l u / 4 max absolute value of maximum moment A B C 0 BD Kirby & Nethercot 1979 - applies to beams bent in reverse curvature with better fit of variable slope of moment diagram between braced points l u max To address limitations of the two previous equations and to avoid their misapplication, Kirby and Nethercot proposed the empirical equation shown here which uses a moment-area method to estimate C b. 17

Equivalent oment Factor Approach Origins of C b C b 3 A + 4 1.5 B + 3 max C +.5 max A absolute value of moment at l u / 4 B absolute value of moment at l u / C absolute value of moment at 3 l u / 4 max absolute value of maximum moment A B C 0 BD AISC 1997 - applies to beams bent in reverse curvature with better fit of variable slope of moment diagram between braced points l u max Finally, this modified version of the Kirby and Nethercot equation appeared in AISC s 1993 Load and Resistance Factored Design Specification for Structural Steel Buildings. This equation provides a lower bound across the full range of moment distributions in beam segments between points of bracing. 18

Equivalent oment Factor Approach Comparison of End oment Equation and 4- oment Equation with theoretical values C b Theoretical End oment 4-oment end-min Value Equation Equation Load Case end-max (Derived) (ASCE 16) (TR14) Equal End oments (opposite rotation) -1 1.00 1.00 1.00 Uniform Load 0 1.13 1.75 1.14 Single Concentrated Load 0 1.35 1.75 1.3 Equal End oments (same rotation) 1.30.30.7 In general, the end moment equation is only accurate when loading is through end moments at bracing points. For wood construction, there are few loading conditions that would provide for the proper use of this equation. In general, the end moment equation is only accurate when loading is through end moments at bracing points. For wood construction, there are few loading conditions that would provide for the proper use of the end moment equation. 19

Adjustment Neutral Axis Load Position Adjusts for load position relative to neutral axis Two adjustment methods from reference loading and support conditions: Effective length approach, l e Load eccentricity factor The ocr equation assumes that loading is through the neutral axis of the member. In cases where the load is above the neutral axis at an unbraced location, an additional moment is induced due to displacement of the top of the beam relative to the neutral axis. This additional moment due to torsional eccentricity reduces the critical moment. Similarly, a load applied below the neutral axis increases the critical moment. The design literature referenced in TR14 provides several means of addressing the effects of load position. There are generally two methods used to adjust the ocr equation for load position relative to the neutral axis. One method adjusts the effective length. The second method uses a load eccentricity factor, C e. Derivation of this factor has been reviewed in the literature referenced in TR14 and is provided in Appendix A. of the document. 0

Adjustment Neutral Axis Load Position Load eccentricity factor, C e See Appendix A. for derivation C e η + 1 η kd η l Accurate for -0.34 η +1.7 (- ve means below neutral axis) u EI y GJ k constant based on loading and support conditions d depth of member The eccentricity factor takes the form above. Flint demonstrated numerically, that the equation for η is accurate for a single concentrated load case where η lies between -0.34 and +1.7 (negative values indicate loads applied below the neutral axis and positive values indicate loads applied above the neutral axis). Limiting η to a maximum value of 1.7 limits C e to a minimum value of 0.7. 1

Non-rectangular cross-sections Simply supported, uniform moment Adjusts for increased torsional resistance provided by warping of the cross-section cr π l u EI y GJ πe + l u I y C w C w warping constant The ocr equation under-predicts the critical moment for beams with non-rectangular crosssections, such as I-beams. For more accurate predictions of the critical moment for simply supported nonrectangular beams under uniform moment, an additional term can be added to account for increased torsional resistance provided by warping of the cross-section. Procedures for calculating the warping constant, C w, are beyond the scope of TR14; however, procedures exist in the referenced literature. For many cross-sections, the equation shown here is overly complicated and can be simplified by conservatively ignoring the increased resistance provided by warping.

Other Beam Forms Cantilevers Continuous Beams Need to handle these cases and varied load patterns! Currently, there is a lack of specific lateral-torsional buckling design provisions for cantilevered and continuous beams. For cantilevered beams, direct solutions for a few cases have been developed and are available in the literature referenced in TR14 and have been reprinted in Tables 1 and. However, for other load cases of cantilevered beams which are unbraced at the ends, approximation equations, do not apply. For these cases, designers have used conservative values for C b and k. Derivation of C b and k values for additional cantilevered load cases is beyond the scope of TR14. Provisions in TR14 also apply to continuous beams. However, since there are various levels of bracing along the compression edge of continuous beams and at interior supports, applying beam stability provisions may be confusing. As a result, most continuous beams are designed with additional full-height bracing at points of interior bearing. 3

Other Beam Forms Cantilevers Continuous Beams Need to handle these cases and varied load patterns! To clarify how to use the provisions in this report to design continuous beams, it must be clear that the derivation of the ocr equation assumes that there is no external lateral load applied about the weak axis of the beam through bracing, supports, or otherwise. Also, it assumes that supports at the ends of members are laterally braced. Additional considerations are required for continuous beams. A continuous beam can be assumed to be braced adequately at interior supports if provisions are made to prevent displacement of the tension edge of the beam relative to the compression edge. Such provisions include providing full-height blocking, providing both tension and compression edge sheathing at the interior supports, or having one edge braced by sheathing and the other edge braced by an adequately designed wall. Otherwise, the designer has to use judgment as to whether to consider the beam to have an unbraced length equal to the full length of the member between points of bracing, an unbraced length measured from the inflection point, or an unbraced length that is between these two conditions. The designer s judgment should consider the sheathing stiffness and the method of attachment to the beam. A beam that is about to undergo lateral torsional buckling develops bending moments about its weak axis and these moments can extend beyond inflection points associated with strong axis bending. Further research is being considered to quantify the roles that tension side bracing and inflection points have on the tendency for lateral torsional buckling. 4

Beam Bracing Assumptions: member ends at bearing are braced to prevent rotation (twisting) effective bracing must prevent twisting of the beam crosssection Bracing Types Lateral prevent lateral displacement of the beam by holding the compression edge in position (i.e. sheathing) relative to the tension edge Torsional prevent rotation of the cross-section (i.e. cross-bridging, fulldepth blocking) Bracing design currently beyond scope of TR14 Derivation of equations in TR14 are based on the assumption that member ends at bearing points are braced to prevent rotation (twisting), even in members otherwise considered to be unbraced. In addition, effective beam bracing must prevent displacement of the beam compression edge relative to the beam tension edge (i.e. twisting of the beam). There are two types of beam bracing, lateral and torsional. Lateral bracing, such as sheathing attached to the compression edge of a beam, is used to minimize or prevent lateral displacement of the beam, thereby reducing lateral displacement of the compression edge of the beam relative to the tension edge. Torsional bracing, such as cross-bridging or blocking, is used to minimize or prevent rotation of the cross-section. Both bracing types can be designed to effectively control twisting of the beam; however, steel bracing systems that provide combined lateral and torsional bracing, such as a concrete slab attached to a steel beam compression edge with shear studs, have been shown to be more effective than either lateral or torsional bracing acting alone (utton and Trahair) (Tong and Chen). Design of lateral and/or torsional bracing for wood beams is currently beyond the scope of TR14; however, further research is being considered to quantify and develop design provisions for lateral and torsional bracing systems. Guidance for steel bracing systems is provided in Yura and Galambos. There are four general types of lateral bracing systems; continuous, discrete, relative, and lean-on. The effectiveness and design of lateral bracing is a function of bracing type, brace spacing, beam cross-section and induced moment. Lateral bracing is most effective when attached to the compression edge and has little effect when bracing the neutral axis (Yura). There are two general types of torsional bracing; continuous and discrete. The effectiveness and design of torsional bracing is significantly affected by the cross-sectional resistance to twisting. 5

Outline Lateral-Torsional Buckling Basic Concept Design ethod Examples Now let s look at the design process for beams subject to lateral torsion. 6

NDS Beam Stability Equations History Pre-1977 Prescriptive provisions 1977 NDS Effective length approach Simple beam Uniformly loaded, unbraced along the loaded edge id-span concentrated load, unbraced at load point Equal end moments in opposite rotation Cantilevered beam Uniformly loaded, unbraced along the loaded edge Free-end concentrated load, unbraced at load point Conservative provision All other load and support cases method derivations appear in Appendix B Hooley & adsen 1991 NDS Wider range of load cases possible, some changed Hooley & Devall In early editions of the NDS, beam stability was addressed using prescriptive bracing provisions. The effective length approach (see 1...1.1) for design of bending members was introduced in the 1977 NDS. Basic load and support cases addressed in the 1977 NDS were as follows: 1. A simply-supported beam with: - uniformly-supported load unbraced along the loaded edge - a concentrated load at midspan unbraced at the point of load - equal end moments ( opposite rotation ). A cantilevered beam with: - uniformly distributed load- unbraced along the loaded edge - concentrated load at the free-end unbraced at the point of load 3. A conservative provision for all other load and support cases Provisions for these load and support conditions were derived from a report by Hooley and adsen. Hooley and adsen used a study by Flint to derive simple effective length equations to adjust the unbraced lengths for the above mentioned load cases and loads applied at the top of beams. In addition, Hooley and adsen recommended that the unbraced length be increased 15% to account for imperfect torsional restraint at supports. Derivation of these provisions and comparisons are presented in TR14 Appendix B. In 1991, the NDS provisions were expanded to cover a wider range of load cases based on a study by Hooley and Devall. In addition, some load cases were changed. 7

NDS Beam Stability Equations History 1991 NDS: compare l e and C b approaches for simply supported rectangular beams with concentrated load at midspan unbraced at point of load: cr.40ei y le 1. 37lu le But for reference condition (simply supported with a constant moment: le 1. 84l u In the NDS provisions which use an effective length approach to adjust for different load and support conditions, the buckling stress, F be, is calculated using a modified form of the equation which has been simplified for rectangular beams (see TR14 Appendix B). The equation was derived for a simply-supported beam with a concentrated load at midspan - unbraced at point of load, as provided in the equation in this slide. Additional loading conditions were derived and the effective length equations were adjusted to this reference condition, where: e l 1.37 u l, for a simply-supported beam with a concentrated load at midspan - unbraced at point of load For the reference condition (simply-supported beam with a constant moment) in the ocr equation, the NDS effective length equation is: l e 1.84 l u. 8

NDS Beam Stability Equations History 1991 NDS: compare l e and C b approaches to compare at the reference condition for the equivalent moment factor: Cb 1.0 at l 1. 84l e u Thus, C b 1.84l l e u or l e 1.84l C b u must have a common starting point for both approaches The equivalent moment factor approach and the NDS effective length approach can be compared by recognizing that, for the reference condition, C b 1.0 and e l 1.84 u l. Therefore, C b 1.84 u l / e l. Solving for e l yields the result in the slide. 9

NDS Beam Stability Equations History Table 1 inconsistencies arise NDS 005 Table 3.3.3 assumes the beam is always loaded on the compression edge non-identified loading and support cases: most severe length assumption applies (ie. 1.84 l u < l e <.06 l u ) Hooley & Duvall assumption of load case interpolation between uniform loading and single point loading is inconsistent with derivation of the equations and C b Table 1 compares effective lengths derived using Equations 7 and 1 in TR14 and effective lengths tabulated in the 005 NDS. It should be noted that some of the NDS effective length equations are inconsistent with estimates from Equations 7 and 1. In cases where a load can be applied on the compression edge of a beam, NDS Table 3.3.3 assumes that the beam is always loaded on the compression edge. For cases not explicitly covered in Table 3.3.3 of the NDS, the most severe effective length assumption applies (i.e. 1.84 l u < l e <.06 l u ). In addition, Hooley and Duvall assumed that effective lengths for beams loaded at multiple points could be estimated by interpolating between the single concentrated load case and the uniform load case. This assumption is inconsistent with assumptions used in the derivation of the equations and the equivalent moment factors. 30

Lateral-Torsional Buckling TR 14 Design ethod Strong-Axis oment Capacity, Compression edge fully supported along length, ends prevented from rotation ' * * moment resistance for strong axis bending multiplied by all applicable factors except C fu, C V, and C L Compression edge not laterally braced over full length or any portion ' * CL C L beam stability factor When a bending member is bent about its strong axis and the moment of inertia about the strong axis (I x ) exceeds the moment of inertia about the weak axis (I y ), the weak axis shall be braced or the moment capacity in the strong axis direction shall be reduced. The compression edge of a bending member shall be braced throughout its length to prevent lateral displacement and the ends at points of bearing shall have lateral support to prevent rotation. The adjusted moment resistance, ', about the strong-axis of a member laterally braced as noted above, where: ' adjusted moment resistance * moment resistance for strong axis bending multiplied by all applicable factors except C fu, C V, and C L. The adjusted moment resistance about the strong axis of members that are not laterally braced for their full length or a portion thereof is C L * where: C L beam stability factor. 31

Lateral-Torsional Buckling TR 14 Design ethod Beam Stability Factor, C L C L 1+ α 1.90 b 1+ α 1.90 b αb 0.95 α b cr * cr critical buckling moment The beam stability factor, C L, shall be calculated as shown here. When the volume effect factor, C V, is less than 1.0, the lesser of C L or C V shall be used to determine the adjusted moment resistance. 3

Lateral-Torsional Buckling TR 14 Design ethod Critical Buckling oment, cr cr πcbc C l fix E u e I γ G l u unbraced length I y weak-axis moment of inertia J torsional constant for the section J ' ' y05 y t 05 C b equivalent moment factor for different loading and support conditions C e load eccentricity factor for top-loaded beams C fix end fixity adjustment E y05 adjusted modulus of elasticity for bending about the weak axis at the fifth percentile G t05 adjusted torsional shear modulus at the fifth percentile The critical buckling moment, cr, shall be calculated as shown here. 33

Lateral-Torsional Buckling TR 14 Design ethod Critical Buckling oment, cr rectangular beams cr 1.3CbC l e u E ' y05 I y C b equivalent moment factor for different loading and support conditions C e load eccentricity factor for top-loaded beams l u unbraced length I y weak-axis moment of inertia E y05 adjusted modulus of elasticity for bending about the weak axis at the fifth percentile For rectangular bending members, the critical buckling moment shall be permitted to be calculated as shown here (see TR14 Appendix C.1 for derivation). 34

Lateral-Torsional Buckling TR 14 Design ethod Equivalent oment Factor, C b C b 3 A 1.5 + 4 + 3 B max c +.5 max A absolute value of moment at l u / 4 B absolute value of moment at l u / C absolute value of moment at 3 l u / 4 max absolute value of maximum moment C b 1.0 (conservatively) all loads & support conditions The equivalent moment factor, C b, for different loading and support conditions can be taken from TR14 Table or, for beam segments between points of bracing, calculated as shown here. For all loading and support conditions, C b is permitted to be conservatively taken as 1.0. 35

Lateral-Torsional Buckling TR 14 Design ethod Equivalent oment Factor, C b Tabulated values Here is Table. 36

Lateral-Torsional Buckling TR 14 Design ethod Load Eccentricity Factor, C e C e η + 1 η η kd l u E G ' y05 ' t 05 I y Jγ Rectangular sections: k constant based on loading and support conditions (Table ) or 1.7 conservatively C e > 0.7 minimum 1.3kd η l u When a bending member is loaded from the tension-side or through the neutral axis, C e 1.0 which is conservative for cases where the bending member is loaded from the tension-side. When a bending member is loaded from the compression-side of the neutral axis and is braced at the point of load, C e 1.0. When a bending member is loaded from the compression side of the neutral axis and is unbraced at the point of load, the load eccentricity factor shall be calculated as shown here. For all loading cases, k can conservatively be taken as 1.7. The minimum value of C e need not be taken as less than 0.7. For rectangular bending members, η shall be permitted to be calculated as shown (see TR14 Appendix C. for derivation). 37

Lateral-Torsional Buckling TR 14 Design ethod In a general nut shell: C L ' * CL * F * S bx x 1+ α 1.90 b cr * 1+ αb 1.90 αb 0.95 α b ' 1.3CbCe E y05i y cr lu 1.5 max Cb 3 A + 4 B + 3 c +.5 max C e η + 1 η 1.3kd η l u E y (1 1.645COVE )( AdjustShearFreeE) ' E y05 SafetyFactor F F * bx bx C D C C C t fu C i So, in a nutshell, here is the design process for lateral torsional buckling of wooden bending elements. 38

Lateral-Torsional Buckling TR 14 Design ethod Cantilevered Beams See Table ; otherwise C b 1.0 and k 1.00 Continuous Beams C b and k vary by bracing location Adequate bracing at interior support means: Full height blocking or tension and compression edge bracing at supports are used to prevent displacement of the compression edge relative to the tension edge - Inadequate bracing design as unbraced for full length of beam between points of bearing For cantilevered beams, direct solutions for two cases have been developed and are provided in Table ; however, for other load cases of cantilevered beams which are unbraced at the ends, the cr equation does not apply. For these cases, designers are required to use conservative values of C b 1.0 and k 1.00. For continuous beams, values for C b and k vary depending on the location of bracing. A continuous beam can be assumed to be adequately braced at an interior support if fullheight blocking or tension and compression edge bracing at supports are used to prevent displacement of the tension edge of the beam relative to the compression edge. Otherwise the beam should be designed as unbraced for the full length of the beam between points of bracing. 39

Lateral-Torsional Buckling TR 14 Design ethod Comparison with Experimental Data 1961 Hooley & adsen 33 beams, varying b/d ratios, loading conditions, and support conditions In 1961, Hooley and adsen tested 33 beams with various b/d ratios, loading conditions, and support conditions. Using behavioral equations taken from the design procedure outlined earlier, the Hooley and adsen data was analyzed. A comparison of the predicted bending strength with the observed bending strength is plotted above. 40

Outline Lateral-Torsional Buckling Basic Concept Design ethod Examples Let s now apply the design process we ve learned to a series of worked design examples. 41

Outline Lateral-Torsional Buckling Basic Concept Design ethod Examples 1 Beams braced at intermediate points Continuous span beams 3 Cantilevered ceiling joists We ll look at three problems that will help demonstrate the range of applications of the design process. 4

Example 1 Beams Braced at Intermediate Points Design a glulam roof ridge beam braced with roof joists spaced at 4 feet on center along its length. Problem Statement: Design a glulam beam to be used as a ridge beam braced with roof joists spaced at 4 feet on center along its length. Given: The span of the ridge beam, l, is 8 feet. The design loads are w dead 75 plf and w snow 300 plf. The roof joists are attached to the sides of the ridge beam providing adequate bracing of the beam; therefore, for calculating the beam buckling moment the unbraced length of the beam, l u, can be assumed to be 48 inches. 43

Example 1 - ASD Given Data Ridge beam span l 8 feet Ridge beam unbraced l u 48 inches length Design Loads Dead w dead 75 plf Snow w snow 300 plf Given: The span of the ridge beam, l, is 8 feet. The design loads are w dead 75 plf and w snow 300 plf. The roof joists are attached to the sides of the ridge beam providing adequate bracing of the beam; therefore, for calculating the beam buckling moment the unbraced length of the beam, l u, can be assumed to be 48 inches. 44

Example 1 - ASD Calculate maximum demand moment ( w + w 8 ) l (375)(8 8 dead live max Select section ½ x 1 4F-V4 Douglas fir glulam beam ) 36,750 ft-lb F bx,400 psi E y 1,600,000 psi First, calculate the maximum demand moment; then pick a glulam beam that might work. 45

Example 1 - ASD Adjust aterial Data - 4F-V4 Douglas fir F bx* F bx C D C C t C fu C i (,400)(1.15)(1.0)(1.0)(1.0)(1.0),760 psi E y05 E y [1-1.645 COV E ](Adjust to shear-free E)/(Safety Factor) (1,600,000)[1-1.645 (0.11)] (1.05) / (1.66) (1,600,000) / 1.93 89,000 psi From the glulam grade material properties, determine the adjusted material properties for strength. 46

Example 1 - ASD Geometric Properties - ½ x 1 4F-V4 Glulam S x I x I y (.5)(1 ) bd 183.8 in 3 6 6 3 3 (.5)(1 ) bd 199 in 4 1 1 3 3 (1)(.5 ) db 7.34 in 4 1 1 C V 1/ x 1 l 1 8 0.1 1 d 1 1 1/ x 5.15 b 0.1 1/ x 5.15.5 0.1 where x 10 0.9871 Calculate all the geometric section properties, including the volume effect factor. 47

Example 1 - ASD Calculate buckling moment capacity C b C e cr 1.00 1.00 ' 1.3CbC ee y05i y lu 1.3(1.0)(1.0)(89,000)(7.34) 48 613,800 in-lb or 51,150 ft-lb Now determine the buckling moment capacity cr. Note for C b, (assume 6 roof joists, from Table ), and for C e, assume braced at point of load. 48

Example 1 - ASD Calculate allowable bending moment capacity * * F S bx x (,760)(183.8) 507,300 in-lb or 4,50 ft-lb α b cr * 51,150 4,50 1.1 C L 1+ αb 1+ αb αb 1.90 1.90 0.95 1+ 1.1 1.90 1+ 1.1 1.90 1.1 0.95 0.88 * (lesser of C V or C L ) (4,50)(0.88) 37,50 ft-lb C V 0.9871 Now calculate the allowable bending moment capacity of the section. 49

Example 1 - Summary Structural Check and Demand / Capacity Ratio max 37,50 ft-lb 36,750 ft-lb 36,750 37,50 max ' 0.99 and in comparison to demand, it looks good. (and just, with not much to spare!!) 50

Example Uniformly loaded two-span continuous beam of equal spans, l Find C b, k, and l u values. Compare above values for: Top edge unbraced, interior bearing point braced Top edge unbraced, interior bearing point unbraced Problem Statement: Determine C b, k, and u l values for a two-span continuous beam subjected to a uniformly distributed gravity load. Compare these values for the following bracing conditions: 1. Top edge unbraced, interior bearing point braced.. Top edge unbraced, interior bearing point unbraced. 51

Example - ASD Find C b C b 3 A 1.5 + 4 + 3 B max C +.5 max Case 1: Top edge unbraced, interior bearing point braced Top of beam in compression from end to 0.75l unbraced Interior bearing point braced relative to the top tension edge of beam Two unbraced lengths: l u l Assume non-derived value of k 1.7 l u l u Given: The beam has two equal spans of length, l. The value for C b can be calculated for each using the equation above. Let s look now at the first case. Case 1: The top edge of the beam is in compression from the end to the inflection point of the beam (0.75l ) and is unbraced. The interior bearing point is braced relative to the top tension edge of the beam. Therefore, this case has two unbraced lengths, l u, each spanning from the end of the beam to the interior support ( l u l ). Since a value for k has not been derived, assume k1.7. 5

Example - ASD Case 1: quarter-point moments max -0.150 wl at interior bearing A 0.065 wl at x 0.5 l B 0.065 wl at x 0.50 l C 0.0000 at x 0.75 l l A B C x max The quarter-point moments for this case are shown above 53

Example - ASD Case 1: C b C b 3(0.065wl 1.5(0.150wl ) ) + 4(0.065wl ) + 3(0.0000wl ) +.5(0.150wl.08 ) l A B C x max so working them into the equation for C b provides the numerical result we need. 54

Example - ASD Find C b C b 3 A 1.5 + 4 + 3 B max C +.5 max Case : Top edge unbraced, interior bearing point unbraced Top of beam in compression from end to 0.75l unbraced Interior bearing point is unbraced One unbraced length: l u l l u Assume non-derived value of k 1.7 Now let s look at the second case. Case : The top edge of the beam is in compression from the end to the inflection point of the beam (0.75 ) and is unbraced. The interior bearing point is unbraced. Therefore, this case has one unbraced length, u l, which spans between the ends of the beam ( u l l ). Since a value for k has not been derived, assume k1.7. 55

Example - ASD Case : quarter-point moments max -0.150 wl at interior bearing A 0.065 wl at x 0.5 l B -0.150 wl at x l C 0.065 at x 1.5 l l l A C x max, B The quarter-point moments for this case are as shown above 56

Example - ASD Case : C b C b 3(0.065wl 1.5(0.150wl ) ) + 4(0.150wl ) + 3(0.065wl ) +.5(0.150wl 1.3 ) l l A C x max, B and when we work them into the equation for C b, we get the numerical result for Case. 57

Example - Summary Uniformly loaded two-span continuous beam of equal spans, l Top edge unbraced, interior bearing point braced Top edge unbraced, interior bearing point unbraced C b k l u.08 1.7 l 1.3 1.7 l Note: This example assumes no external lateral forces applied. If lateral forces are being resisted by the beam in weak axis direction, a different analysis including effects on beam stability should be considered. Now we can compare the C b s for each case. 58

Example 3 Cantilevered Ceiling Joists Discover the impact of bracing the end of a cantilevered ceiling joist that is unbraced on the compression (bottom) edge. vs Now let s look at our final example. Problem Statement: Determine the impact of bracing the end of a cantilevered ceiling joist that is unbraced on the compression (bottom) edge. 59

Example 3 - ASD Given Data Cantilever span l u 4 ft Joist laterally braced at supporting walls Load Concentrated load applied at end of cantilever aterial # Southern pine x1 joists @ ft O.C. 4 ft Given: A ceiling is designed with a cantilevered span of 4 feet. The beam is laterally braced at the supporting walls. The cantilever is supporting a concentrated load at the end of the cantilever. Design the cantilever with # Southern pine x1 joists spaced at feet on center. 60

Example 3 - ASD aterial Data # Southern pine From Table 4B of NDS 001 Supplement: F bx 975 psi E y 1,600,000 psi From Table 4B of the NDS 001 Supplement, we glean the material stresses for # Southern pine 61

Example 3 - ASD Adjust aterial Data F bx* F bx C D C C t C fu C i C r (975)(1.0)(1.0)(1.0)(1.0)(1.0)(1.15) 1,11 psi E y05 E y [1-1.645 COV E ](Adjust to shear-free E)/(Safety Factor) (1,600,000)[1-1.645 (0.5)] (1.03) / (1.66) (1,600,000) /.74 584,500 psi and then adjust the material properties to our design conditions. 6

Example 3 - ASD Geometric Properties - x1 joists @ ft O.C. S x I x I y bd 6 3 bd 1 (1.5)(11.5 6 ) 31.64 in 3 3 (1.5)(11.5 ) 178.0 in 4 1 3 3 (11.5)(1.5 ) db 3.164 in 4 1 1 Determine the geometric section properties for the joists. 63

Example 3 - ASD Case 1: cantilevered joist without end bracing Calculate buckling moment capacity C b 1.8 k 1.0 Determine the buckling moment capacity using C b and k from Table for the first case, cantilevered joists without any end bracing. 64

Example 3 - ASD Case 1: cantilevered joist without end bracing Calculate buckling moment capacity 1.3kd η l u 1.3(1.0)(11.5) 48 0.3047 C e ( η + 1) η (0.3047 + 1) 0.3047 0.7407 cr 1.3CbC ee l u ' y05 I y 47,487 in-lb or 3,957 ft-lb 1.3(1.8)(0.7407)(584,494)(3.164) 48 The critical moment comes out as shown above. (follow the design process flowchart shown earlier). 65

Example 3 - ASD Case 1: cantilevered joist without end bracing Calculate allowable bending moment capacity α * * x FbxS x b (1,11)(31.64) 3,957,956 cr * x 1.339 35,468 in-lb or,956 ft-lb C L 1+ 1.339 1.9 1+ 1.339 1.9 1.339 0.95 0.905 ' x * xcl (,956)(0.905),676 ft-lb Now determine the allowable bending moment capacity for the joist for the case of unbraced ends. 66

Example 3 - ASD Case : cantilevered joist with end bracing Calculate buckling moment capacity C b 1.67 C e 1.0 assuming bracing at point of load Let s repeat the procedure, but this time with braced ends. We use Table, noting that the resulting values for C b and C e are different from Case 1. 67

Example 3 - ASD Case : cantilevered joist with end bracing Calculate buckling moment capacity cr 1.3CbCe E l u ' y05 I y 1.3(1.67)(1.0)(584,494)(3.164) 48 83,644 in-lb or 6,970 ft-lb Calculate the critical moment for this case (which is much higher than Case 1) 68

Example 3 - ASD Case : cantilevered joist with end bracing Calculate allowable bending moment capacity α * * x FbxS x b (1,11)(31.64) 6,970,956 cr * x.358 35,468 in-lb or,956 ft-lb C L 1+.358 1.9 1+.358 1.9.358 0.95 0.966 ' x * xcl (,956)(0.966),857 ft-lb and the allowable bending capacity, which is slightly higher than Case 1. 69

Example 3 - Summary Allowable moment capacities Unbraced end Braced end x,676 ft-lb,857 ft-lb Note: This example assumes no external lateral forces applied. If lateral forces are being resisted by the beam in weak axis direction, a different analysis including effects on beam stability should be considered. Here are the capacity moments in comparison. Braced ends do make a difference. 70

Outline Lateral-Torsional Buckling Basic Concept Design ethod Examples This concludes all the topics of this ecourse. 71

TR 14 - Appendices A Derivation of Lateral-Torsional Buckling Equations Closed-Form Infinite Series Approximation Strain Energy Approximation All factors B Derivation of 1997 NDS Beam Stability Provisions C Derivation of Rectangular Beam Equations D Graphical Representation of Wood Beam Stability Provisions (for beams loaded through the neutral axis) The Appendices of TR14 are filled with useful information and derivations for all of the material found in the document. It s well worth a read through to gain a more full understanding of all the underlying technology. 7

TR 14 ore detail inside - available free from www.awc.org TR14 is available as a free download from www.awc.org. Look in the Free Download Library for the report. 73