AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
DES140:
Designing for
Lateral-Torsional Stability
in Wood Members
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
Welcome to the Lateral Torsional Stability eCourse.
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1
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Outline
• Lateral-Torsional Buckling Basic Concept
• Design Method
• Examples
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
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 Members
which is a free document from the Free Downloads Library at www.awc.org.
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2
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Limit State – Beam
Deformation
= in-plane deformation
+ out-of-plane deformation
+ twisting
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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.
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3
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Limit State – Beam
Deformation
= in-plane deformation
+ out-of-plane deformation
+ twisting
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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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Equilibrium equations
in-plane moment (X-X)
out-of-plane moment (Y-Y)
twisting moment
M ext
d 2v
= − EI x 2
dx
β M ext
d 2u
= − EI y 2
dz
du
dβ
M ext = GJ
dz
dz
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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, Mcr. 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, …
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Equilibrium equations – coupled
d 2v
= − EI x 2
dx
in-plane moment (X-X)
M ext
out-of-plane moment (Y-Y)
and twisting moment
2
M ext
d 2β
+
β=0
dz 2 EI y GJ
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… the two equations can be combined to form the resulting second-order differential
equation.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Equilibrium equations – boundary conditions
d 2v
= − EI x 2
dx
strong axis bending
M ext
lateral-torsional
- constant cross-section
- constant moment along
unbraced length, Lu
M ocr =
π
Lu
EI y GJ
M
M
Lu
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If a beam has a constant cross-section and a constant induced moment, Mo, along the
unbraced length, L, there is a closed-form solution to the second order differential equation.
By defining a constant k2=Mo2/EIyGJ, 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, Mocr, yields the final form of the equation
shown in the slide. The mathematical details of the development are found on page 21-22
of TR14.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Basic Concept
• Critical Moment – simply supported beam under
constant moment
M ocr =
π
Lu
M
EI y GJ
M
Lu
Mocr = reference case critical
moment
Lu = unbraced length
E = modulus of elasticity
Iy = weak-axis moment of inertia
G = torsional shear modulus
J = torsional constant for the
section
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Here:
Mocr = critical moment for the reference case (simply-supported beam under a constant
moment)
Lu = unbraced length
E = modulus of elasticity
Iy = 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.
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8
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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 Moment (End Moment)
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!!
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In early editions of the NDS, beam stability was addressed using prescriptive bracing
provisions. The effective length approach (see 1.2.2.1.1) 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 )
2. 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 Moment (End Moment) 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Adjustment – Beam Slenderness
• Adjust to allow for flexural rigidities EIx ≅ EIy in
addition to EIx > EIy
γ =1−
M ocr
π
=
Lu
Iy
Ix
EI y GJ
γ
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The closed form Mocr equation was derived for slender beams where the in-plane flexural
rigidity (EIx) is much larger than the out-of-plane flexural rigidity (EIy). 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 Iy by the term γ, where γ
=1-Iy/Ix and Ix is the moment of inertia about the strong-axis.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Adjustment –
Other Loading and Support Conditions
• Requires alternate solution to closed form for Mocr
– See Appendix A.1 for more detail
• Two adjustment methods from reference loading
and support conditions:
– Effective Length approach
– Equivalent Moment factor approach
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The closed-form Mocr equation applies to a single loading condition (simply-supported
beam under a constant moment). Many 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 Mocr 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Effective Length Approach
• Adjusts the unbraced length to an effective length
for specific loading and support conditions
π
M cr =
EI y GJ
le
Mocr = reference case critical
moment
le = effective length
E = modulus of elasticity
Iy = weak-axis moment of inertia
G = torsional shear modulus
J = torsional constant for the
section
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The effective length approach adjusts the reference case critical moment, Mocr, to other
loading and support conditions by adjusting the unbraced length to an “effective length” as
provided in this equation using the term le ,
where:
Mcr = critical moment for specific loading and support conditions
le = effective length which equals the unbraced length adjusted for specific loading and
support conditions …
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Effective Length Approach
• effective length
le tabulated for
specific loading
and support
conditions
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… as identified in Table 1 of TR14.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Adjusts the reference case to other loading and
support conditions using a factor
M cr = Cb M ocr
M
M
Mocr = reference case critical
moment
Cb = equivalent moment factor
M
0 BMD
L
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The equivalent moment factor approach adjusts the reference case critical moment, Mocr, to
other loading and support conditions using an equivalent moment factor, Cb, as provided in
the equation above, where:
Cb = equivalent moment factor
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Origins of Cb
2
⎛M ⎞
⎛M ⎞
Cb = 1.75 + 1.05⎜⎜ 1 ⎟⎟ + 0.3⎜⎜ 1 ⎟⎟ ≤ 2.3
⎝ M2 ⎠
⎝ M2 ⎠
M1 = minimum end moment
M2 = maximum end moment
M2
M1
L
AISC 1961
Limitation - does not apply to
beams bent in reverse curvature
M1
M2
0 BMD
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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, Cb, has changed over the last 40 years. The original equation for estimating
Cb is shown here, where:
M1 = minimum end moment
M2 = maximum end moment
This equation provides accurate estimates of Cb 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 M1/M2 ratios which
are negative (with single curvature i.e. M1/M2 < 0)
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Origins of Cb
Cb =
1
⎛M ⎞
0.6 − 0.4⎜⎜ 1 ⎟⎟
⎝ M2 ⎠
M1 = minimum end moment
M2 = maximum end moment
≤ 2 .5
M2
M1
L
ASIC 1993
Limitation - applies only to beams bent in M1
reverse curvature with constant slope
of moment diagram between braced
points
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0 BMD
M2
The Cb 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Origins of Cb
Cb =
12 M max
3M A + 4M B + 3M C + 2 M max
MA = absolute value of moment at lu / 4
MB = absolute value of moment at lu / 2
MC = absolute value of moment at 3 lu / 4
Mmax = absolute value of maximum moment
MA MB MC
0 BMD
lu
Kirby & Nethercot 1979
- applies to beams bent in reverse curvature with better fit of
variable slope of moment diagram between braced points
Mmax
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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 Cb.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Origins of Cb
Cb =
12.5M max
3M A + 4 M B + 3M C + 2.5M max
MA = absolute value of moment at lu / 4
MB = absolute value of moment at lu / 2
MC = absolute value of moment at 3 lu / 4
Mmax = absolute value of maximum moment
MA MB MC
0 BMD
lu
AISC 1997
- applies to beams bent in reverse curvature with better fit of
variable slope of moment diagram between braced points
Mmax
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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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Equivalent Moment Factor Approach
• Comparison of End Moment Equation and 4Moment Equation with theoretical values
Cb
Load Case
Equal End Moments (opposite rotation)
Uniform Load
Single Concentrated Load
Equal End Moments (same rotation)
Mend-min
Mend-max
-1
0
0
1
Theoretical
Value
(Derived)
1.00
1.13
1.35
2.30
End Moment
Equation
(ASCE 16)
1.00
1.75
1.75
2.30
4-Moment
Equation
(TR14)
1.00
1.14
1.32
2.27
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.
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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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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, le
– Load eccentricity factor
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The Mocr 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 Mocr equation for load position relative
to the neutral axis. One method adjusts the effective length. The second method uses a
load eccentricity factor, Ce. Derivation of this factor has been reviewed in the literature
referenced in TR14 and is provided in Appendix A.2 of the document.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Adjustment –
Neutral Axis Load Position
• Load eccentricity factor, Ce
– See Appendix A.2 for derivation
Ce = η2 + 1 − η
kd
η=
2lu
EI y
GJ
Accurate for -0.34 ≤ η ≤ +1.72
(-’ve means below neutral axis)
k = constant based on
loading and
support conditions
d = depth of member
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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 +1.72 (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
1.72 limits Ce to a minimum value of 0.27.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Non-rectangular cross-sections
• Simply supported, uniform moment
• Adjusts for increased torsional resistance
provided by warping of the cross-section
2
⎛ πE ⎞
π
⎟⎟ I y C w
M cr =
EI y GJ + ⎜⎜
lu
l
⎝ u ⎠
Cw = warping constant
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The Mocr 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, Cw, 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Other Beam Forms
• Cantilevers
• Continuous Beams
Need to handle these cases
and varied load patterns!
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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 2. 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 Cb and k. Derivation of Cb 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Other Beam Forms
• Cantilevers
• Continuous Beams
Need to handle these cases
and varied load patterns!
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To clarify how to use the provisions in this report to design continuous beams, it must be
clear that the derivation of the Mocr 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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
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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 (Mutton 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Outline
• Lateral-Torsional Buckling Basic Concept
• Design Method
• Examples
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
Now let’s look at the design process for beams subject to lateral torsion.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
NDS Beam Stability Equations
History
• Pre-1977
– Prescriptive provisions
• 1977 NDS
– Effective length approach
• Simple beam
– Uniformly loaded, unbraced along the loaded edge
– Mid-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 & Madsen
• 1991 NDS
– Wider range of load cases possible, some changed
Hooley & Devall
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
In early editions of the NDS, beam stability was addressed using prescriptive bracing
provisions. The effective length approach (see 1.2.2.1.1) 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 )
2. 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
Madsen. Hooley and Madsen 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 Madsen 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
NDS Beam Stability Equations
History
• 1991 NDS: compare le and Cb approaches
for simply supported rectangular beams with
concentrated load at midspan – unbraced at point of
load:
M cr =
2.40 EI y
le
le = 1.37lu
But for reference condition (simply supported with a
constant moment:
le = 1.84lu
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
In the NDS provisions which use an effective length approach to adjust for different load
and support conditions, the buckling stress, FbE, 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:
le = 1.37 lu, 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 Mocr
equation, the NDS effective length equation is: le = 1.84 lu.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
NDS Beam Stability Equations
History
• 1991 NDS: compare le and Cb approaches
to compare at the reference condition for the equivalent
moment factor:
Cb = 1.0
Thus,
Cb =
1.84lu
le
at
le = 1.84lu
or
le =
1.84lu
Cb
must have a common starting point for both approaches
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The equivalent moment factor approach and the NDS effective length approach can be
compared by recognizing that, for the reference condition,
Cb = 1.0 and le = 1.84 lu. Therefore, Cb = 1.84 lu / le. Solving for le yields the result in the
slide.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
NDS Beam Stability Equations
History
•
Table 1
•
NDS 2005 Table 3.3.3
–
inconsistencies arise
– assumes the beam is
always loaded on the
compression edge
– non-identified loading and
support cases: most
severe length assumption
applies (ie. 1.84 lu < le <
2.06 lu)
– Hooley & Duvall
assumption of load case
interpolation between
uniform loading and single
point loading is
inconsistent with
derivation of the equations
and Cb
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Table 1 compares effective lengths derived using Equations 7 and 12 in TR14 and effective
lengths tabulated in the 2005 NDS. It should be noted that some of the NDS effective
length equations are inconsistent with estimates from Equations 7 and 12. 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 lu <
le < 2.06 lu). 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.
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30
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Strong-Axis Moment Capacity, M’
– Compression edge fully supported along length, ends prevented from
rotation
M =M
'
*
M* = moment resistance for strong
axis bending multiplied by all applicable
factors except Cfu, CV, and CL
– Compression edge not laterally braced over full length or any portion
M ' = CL M *
CL = beam stability factor
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When a bending member is bent about its strong axis and the moment of inertia about the
strong axis (Ix) exceeds the moment of inertia about the weak axis (Iy), 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, M ', about the strong-axis of a member
laterally braced as noted above,
where:
M' = adjusted moment resistance
M* = moment resistance for strong axis bending multiplied by all applicable factors except
Cfu, CV, and CL.
The adjusted moment resistance about the strong axis of members that are not laterally
braced for their full length or a portion thereof is M’ = CL M*
where:
CL = beam stability factor.
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31
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Beam Stability Factor, CL
α
1 + αb
⎛ 1 + αb ⎞
CL =
− ⎜
⎟ − b
1.90
⎝ 1.90 ⎠ 0.95
2
αb =
M cr
M*
Mcr = critical buckling moment
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The beam stability factor, CL, shall be calculated as shown here. When the volume effect
factor, CV, is less than 1.0, the lesser of CL or CV shall be used to determine the adjusted
moment resistance.
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32
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Critical Buckling Moment, Mcr
πCb Ce
M cr =
C fix lu
E y' 05 I y Gt' 05 J
γ
lu = unbraced length
Iy = weak-axis moment of inertia
J = torsional constant for the
section
Cb = equivalent moment factor
for different loading and
support conditions
Ce = load eccentricity factor for
top-loaded beams
Cfix = 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
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The critical buckling moment, Mcr, shall be calculated as shown here.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Critical Buckling Moment, Mcr – rectangular beams
M cr =
1.3Cb Ce E y' 05 I y
lu
lu = unbraced length
Iy = weak-axis moment of inertia
Cb = equivalent moment factor
for different loading and
support conditions
Ce = load eccentricity factor for
top-loaded beams
E’y05 = adjusted modulus of
elasticity for bending about
the weak axis at the fifth
percentile
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For rectangular bending members, the critical buckling moment shall be permitted to be
calculated as shown here (see TR14 Appendix C.1 for derivation).
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34
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Equivalent Moment Factor, Cb
Cb =
12.5M max
3M A + 4 M B + 3M c + 2.5M max
MA = absolute value of moment at lu / 4
MB = absolute value of moment at lu / 2
MC = absolute value of moment at 3 lu / 4
Mmax = absolute value of maximum moment
Cb = 1.0 (conservatively) all loads & support conditions
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
The equivalent moment factor, Cb, for different loading and support conditions can be taken
from TR14 Table 2 or, for beam segments between points of bracing, calculated as shown
here. For all loading and support conditions, Cb is permitted to be conservatively taken as
1.0.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Equivalent Moment
Factor, Cb
– Tabulated values
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Here is Table 2.
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36
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Load Eccentricity Factor, Ce
Ce = η2 + 1 − η
kd
η=
2lu
E y' 05 I y
Gt' 05 Jγ
Rectangular sections:
η=
k = constant based on loading
and support conditions
(Table 2) or 1.72
conservatively
Ce > 0.27 minimum
1.3kd
lu
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When a bending member is loaded from the tension-side or through the neutral axis,
Ce=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, Ce=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 1.72.
The minimum value of Ce need not be taken as less than 0.27.
For rectangular bending members, η shall be permitted to be calculated as shown (see
TR14 Appendix C.2 for derivation).
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• In a general nut shell:
M ' = CL M *
M * = Fbx* S x
α
1 + αb
⎛ 1 + αb ⎞
− ⎜
⎟ − b
1.90
1
.
90
0
.95
⎠
⎝
2
CL =
αb =
M cr
M*
M cr =
Fbx* = Fbx C D C M Ct C fu Ci
1.3Cb Ce E y' 05 I y
lu
Cb =
12.5M max
3M A + 4 M B + 3M c + 2.5M max
Ce = η2 + 1 − η
η=
E y' 05 =
1.3kd
lu
E y (1 − 1.645COVE )( AdjustShearFreeE )
SafetyFactor
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
So, in a nutshell, here is the design process for lateral torsional buckling of wooden
bending elements.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Cantilevered Beams
– See Table 2; otherwise Cb = 1.0 and k = 1.00
• Continuous Beams
– Cb 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
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
For cantilevered beams, direct solutions for two cases have been developed and are
provided in Table 2; however, for other load cases of cantilevered beams which are
unbraced at the ends, the Mcr equation does not apply. For these cases, designers are
required to use conservative values of Cb = 1.0 and k = 1.00.
For continuous beams, values for Cb 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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Lateral-Torsional Buckling
TR 14 Design Method
• Comparison with Experimental Data
– 1961 Hooley & Madsen 33 beams, varying b/d ratios, loading
conditions, and support conditions
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In 1961, Hooley and Madsen 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 Madsen data was analyzed. A comparison of the predicted
bending strength with the observed bending strength is plotted above.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Outline
• Lateral-Torsional Buckling Basic Concept
• Design Method
• Examples
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
Let’s now apply the design process we’ve learned to a series of worked design examples.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Outline
• Lateral-Torsional Buckling Basic Concept
• Design Method
• Examples
– 1 Beams braced at intermediate points
– 2 Continuous span beams
– 3 Cantilevered ceiling joists
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We’ll look at three problems that will help demonstrate the range of applications of the
design process.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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.
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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 28 feet. The design loads are wdead = 75 plf and
wsnow = 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, lu, can be assumed to be 48 inches.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Given Data
Ridge beam span
Ridge beam unbraced
length
Design Loads
Dead
Snow
l
lu
= 28 feet
= 48 inches
wdead = 75 plf
wsnow = 300 plf
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
Given: The span of the ridge beam, l , is 28 feet. The design loads are wdead = 75 plf and
wsnow = 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, lu, can be assumed to be 48 inches.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Calculate maximum demand moment
M max
( wdead + wlive )l 2 (375)(28 2 )
=
=
= 36,750 ft-lb
8
8
• Select section
2 ½” x 21” 24F-V4 Douglas fir glulam beam
Fbx =
Ey =
2,400 psi
1,600,000 psi
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
First, calculate the maximum demand moment; then pick a glulam beam that might work.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Adjust Material Data - 24F-V4 Douglas fir
Fbx* = Fbx CD CM Ct Cfu Ci
= (2,400)(1.15)(1.0)(1.0)(1.0)(1.0) = 2,760 psi
E’y05 = Ey [1-1.645 COVE](Adjust to shear-free E)/(Safety Factor)
= (1,600,000)[1-1.645 (0.11)] (1.05) / (1.66)
= (1,600,000) / 1.93
= 829,000 psi
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From the glulam grade material properties, determine the adjusted material properties for
strength.
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46
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Geometric Properties - 2 ½” x 21” 24F-V4 Glulam
Sx =
bd 2 ( 2.5)(212 )
=
= 183.8 in3
6
6
Ix =
bd 3 (2.5)(213 )
=
= 1929
12
12
Iy =
db 3 (21)(2.53 )
=
= 27.34 in4
12
12
1/ x
1/ x
0.1
0.1
1/ x
⎛ 21 ⎞ ⎛ 12 ⎞ ⎛ 5.125 ⎞
CV = ⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎝ l ⎠ ⎝d ⎠ ⎝ b ⎠
⎛ 21 ⎞ ⎛ 12 ⎞ ⎛ 5.125 ⎞
=⎜ ⎟ ⎜ ⎟ ⎜
⎟
⎝ 28 ⎠ ⎝ 21 ⎠ ⎝ 2.5 ⎠
in4
where x = 10
0.1
= 0.9871
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Calculate all the geometric section properties, including the volume effect factor.
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47
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Calculate buckling moment capacity
Cb = 1.00
Ce = 1.00
M cr =
=
1.3Cb Ce E y' 05 I y
lu
1.3(1.0)(1.0)(829,000)(27.34)
48
= 613,800 in-lb or 51,150 ft-lb
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Now determine the buckling moment capacity Mcr. Note for Cb, (assume 6 roof joists, from
Table 2), and for Ce, assume braced at point of load.
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48
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - ASD
• Calculate allowable bending moment capacity
M * = Fbx* S x = (2,760)(183.8) = 507,300 in-lb or 42,250 ft-lb
αb =
M cr 51,150
=
= 1.21
M * 42,250
α
1 + αb
1 + 1.21
⎛ 1 + αb ⎞
⎛ 1 + 1.21 ⎞ 1.21
− ⎜
CL =
− ⎜
= 0.88
⎟ − b =
⎟ −
1.90
1.90
⎝ 1.90 ⎠ 0.95
⎝ 1.90 ⎠ 0.95
2
2
M’ = M* (lesser of CV or CL) = (42,250)(0.88) = 37,250 ft-lb
CV = 0.9871
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Now calculate the allowable bending moment capacity of the section.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 1 - Summary
• Structural Check and Demand / Capacity Ratio
M’ ≥ Mmax
37,250 ft-lb ≥ 36,750 ft-lb
M max 36,750
=
= 0.99
M'
37,250
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…and in comparison to demand, it looks good. (and just, with not much to spare!!)
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50
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2
• Uniformly loaded two-span continuous beam of
equal spans, l
– Find Cb, k, and lu values.
– Compare above values for:
• Top edge unbraced, interior bearing point braced
• Top edge unbraced, interior bearing point unbraced
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Problem Statement: Determine Cb, k, and lu 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.
2. Top edge unbraced, interior bearing point unbraced.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Find Cb
Cb =
12.5M max
3M A + 4M B + 3M C + 2.5M max
Case 1: Top edge unbraced, interior bearing
lu
lu
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: lu = l
• Assume non-derived value of k = 1.72
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Given: The beam has two equal spans of length, l. The value for Cb 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, lu, each
spanning from the end of the beam to the interior support ( lu = l ). Since a value for k has
not been derived, assume k=1.72.
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52
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Case 1: quarter-point moments
= -0.1250 wl 2 at interior bearing
Mmax
= 0.0625 wl 2 at x = 0.25 l
MA
= 0.0625 wl 2 at x = 0.50 l
MB
= 0.0000
at x = 0.75 l
MC
l
MA MB
MC
x
Mmax
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The quarter-point moments for this case are shown above…
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53
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Case 1: Cb
Cb =
12.5(0.1250 wl 2 )
= 2.08
3(0.0625wl 2 ) + 4(0.0625wl 2 ) + 3(0.0000 wl 2 ) + 2.5(0.1250 wl 2 )
l
MA MB
MC
x
Mmax
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
…so working them into the equation for Cb provides the numerical result we need.
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54
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Find Cb
Cb =
12.5M max
3M A + 4M B + 3M C + 2.5M max
Case 2: Top edge unbraced, interior bearing
lu
point unbraced
• Top of beam in compression from end to 0.75l
– unbraced
• Interior bearing point is unbraced
• One unbraced length: lu = 2l
• Assume non-derived value of k = 1.72
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Now let’s look at the second case.
Case 2: 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, lu, which spans between the ends of the beam
( lu = 2l ). Since a value for k has not been derived, assume k=1.72.
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55
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Case 2: quarter-point moments
= -0.1250 wl 2 at interior bearing
Mmax
= 0.0625 wl 2 at x = 0.5 l
MA
= -0.1250 wl 2 at x = l
MB
= 0.0625
at x = 1.5 l
MC
l
l
MA
MC
x
Mmax, MB
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The quarter-point moments for this case are as shown above…
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56
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - ASD
• Case 2: Cb
Cb =
12.5(0.1250 wl 2 )
= 1.32
3(0.0625wl 2 ) + 4(0.1250 wl 2 ) + 3(0.0625wl 2 ) + 2.5(0.1250 wl 2 )
l
l
MA
MC
x
Mmax, MB
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
…and when we work them into the equation for Cb, we get the numerical result for Case 2.
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57
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 2 - Summary
• Uniformly loaded two-span continuous beam of
equal spans, l
Cb
k
lu
Top edge unbraced, interior
bearing point braced
2.08
1.72
l
Top edge unbraced, interior
bearing point unbraced
1.32
1.72
2l
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.
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
Now we can compare the Cb’s for each case.
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58
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
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.
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59
AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Given Data
4 ft
Cantilever span
lu
Joist laterally braced at supporting walls
Load
Concentrated load applied at end of cantilever
Material
#2 Southern pine 2x12 joists @ 2 ft O.C.
4 ft
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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 #2 Southern pine 2x12 joists spaced at 2 feet on
center.
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AMERICAN FOREST & PAPER ASSOCIATION
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Example 3 - ASD
• Material Data
#2 Southern pine
From Table 4B of NDS 2001 Supplement:
= 975 psi
Fbx
= 1,600,000 psi
Ey
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From Table 4B of the NDS 2001 Supplement, we glean the material stresses for #2
Southern pine…
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AMERICAN FOREST & PAPER ASSOCIATION
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Engineered and Traditional Wood Products
Example 3 - ASD
• Adjust Material Data
Fbx* = Fbx CD CM Ct Cfu Ci Cr
= (975)(1.0)(1.0)(1.0)(1.0)(1.0)(1.15)
= 1,121 psi
E’y05 = Ey [1-1.645 COVE](Adjust to shear-free E)/(Safety Factor)
= (1,600,000)[1-1.645 (0.25)] (1.03) / (1.66)
= (1,600,000) / 2.74
= 584,500 psi
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… and then adjust the material properties to our design conditions.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Geometric Properties - 2x12 joists @ 2 ft O.C.
bd 2 (1.5)(11.252 )
=
= 31.64 in3
6
6
bd 3 (1.5)(11.253 )
Ix =
=
= 178.0 in4
12
12
Sx =
Iy =
db 3 (11.25)(1.53 )
=
= 3.164 in4
12
12
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Determine the geometric section properties for the joists.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 1: cantilevered joist without end bracing
– Calculate buckling moment capacity
Cb = 1.28 k = 1.0
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Determine the buckling moment capacity using Cb and k from Table 2 for the first case,
cantilevered joists without any end bracing.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 1: cantilevered joist without end bracing
– Calculate buckling moment capacity
η=
1.3kd 1.3(1.0)(11.25)
=
= 0.3047
48
lu
Ce = (η2 + 1) − η = (0.3047 2 + 1) − 0.3047 = 0.7407
M cr =
1.3Cb Ce E y' 05 I y
lu
=
1.3(1.28)(0.7407)(584,494)(3.164)
48
= 47,487 in-lb or 3,957 ft-lb
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The critical moment comes out as shown above. (follow the design process flowchart
shown earlier).
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 1: cantilevered joist without end bracing
– Calculate allowable bending moment capacity
M x* = Fbx* S x = (1,121)(31.64) = 35,468 in-lb or 2,956 ft-lb
αb =
M cr 3,957
=
= 1.339
M x* 2,956
CL =
1 + 1.339
⎛ 1 + 1.339 ⎞ 1.339
= 0.905
− ⎜
⎟ −
1.9
0.95
⎝ 1 .9 ⎠
2
M x' = M x*C L = (2,956)(0.905) = 2,676 ft-lb
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Now determine the allowable bending moment capacity for the joist for the case of
unbraced ends.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 2: cantilevered joist with end bracing
– Calculate buckling moment capacity
Cb = 1.67
Ce = 1.0 assuming
bracing at point
of load
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Let’s repeat the procedure, but this time with braced ends. We use Table 2, noting that the
resulting values for Cb and Ce are different from Case 1.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 2: cantilevered joist with end bracing
– Calculate buckling moment capacity
M cr =
1.3Cb Ce E y' 05 I y
lu
=
1.3(1.67)(1.0)(584,494)(3.164)
48
= 83,644 in-lb or 6,970 ft-lb
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Calculate the critical moment for this case (which is much higher than Case 1)…
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - ASD
• Case 2: cantilevered joist with end bracing
– Calculate allowable bending moment capacity
M x* = Fbx* S x = (1,121)(31.64) = 35,468 in-lb or 2,956 ft-lb
αb =
M cr 6,970
=
= 2.358
M x* 2,956
CL =
1 + 2.358
⎛ 1 + 2.358 ⎞ 2.358
= 0.966
− ⎜
⎟ −
1 .9
0.95
⎝ 1 .9 ⎠
2
M x' = M x*C L = (2,956)(0.966) = 2,857 ft-lb
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… and the allowable bending capacity, which is slightly higher than Case 1.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Example 3 - Summary
• Allowable moment capacities
Unbraced end
M’x
2,676 ft-lb
Braced end
2,857 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.
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Here are the capacity moments in comparison. Braced ends do make a difference.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
Outline
• Lateral-Torsional Buckling Basic Concept
• Design Method
• Examples
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This concludes all the topics of this eCourse.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
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)
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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.
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AMERICAN FOREST & PAPER ASSOCIATION
American Wood Council
Engineered and Traditional Wood Products
TR 14
More detail inside available free from
www.awc.org
Copyright © 2005 American Forest & Paper Association, Inc. All rights reserved.
TR14 is available as a free download from www.awc.org. Look in the Free Download
Library for the report.
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73