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    4-1

    4. Behaviour and Design of Flexural Members (Beams)

    (AS 4100 – Section 5)

    4.1 

    General

    The members subject to bending are referred to as beams or flexural members. They support

    transverse loads through bending and shear actions. The behaviour of flexural members is

    more complicated than compression members. In steel construction, pure flexural members

    do exist, but they are often subject to combined bending and axial load actions. In some cases,

    they can also be subjected to combined bending and torsion, for example, crane girders.

    Undesirable combined loading effects (torsion and other) can be minimised though proper

    detailing of connections to adjacent members. If the members are  predominantly subjected to

    bending action, Section 5 of AS 4100 can be used, and this lecture discusses this case.

    A range of steel sections (UB, UC, welded box and I-girders, RHS, SHS, Angles and

    Channels) can be designed to support transverse loads as beams in building and bridgestructures. They are hot-rolled, welded or cold-formed. The steel sections should have

    adequate bending strength and stiffness (should not deflect too much), and must satisfy

    relevant service and architectural requirements. The UB section is the most efficient beam

    section as the material is located away from the neutral axis. The UB and UC sections are

    most commonly used (1 to 30 m span) because of their structural efficiency and relatively

    easier connections to other members in the structure, while angles and channels are used for

    shorter spans (3 to 6 m). Figure 4.1 shows some examples of flexural members (beams).

    Figure 4.1 Examples of Flexural Members (Beams)

    For longer members subject to heavy loads, compound members made of two or more

    sections can be used (5 to 15 m). However, they increase the cost of fabrication due to

    additional components such as battens and lacing. For larger spans, plate girders (10 to 100

    m), trusses (10 to 100 m) or box girders (15 to 200 m) have to be used.

    4.2 Design Action Effects

    Appropriate design load combinations must be considered to determine the maximum design

    action effects of bending moment (and their distribution along the beam length) and any axial

    loads. For this purpose, simple statics or available computer programs could be used. Elastic

    analyses of beams are commonly used for this purpose. If the members are subjected to

    significant axial loads, they must be designed as beams subjected to combined bending and

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    axial actions (see Section 6 of lecture notes). Otherwise, the following sections can be used in

    which the members are designed to carry bending moment only. The maximum bending stress

    in the beam is calculated using the elastic bending formula, ie. maximum bending stress =

    Maximum moment / Elastic section modulus Z about the appropriate principal axis. The

    bending stress distribution is linear with compressive and tensile stresses in the extreme

    fibres. For unsymmetric sections such as angles, the principal axes are not the regular x and yaxes and have to be determined first.

    4.3 Strength Design of a Flexural Member

    For a member subject to a design bending moment M*, the following limit state requirements

    must be satisfied.

    M* ≤≤≤≤ φφφφ Ms (4.1a) M* ≤≤≤≤ φφφφ Mb (4.1b) 

    where φ  = the capacity reduction factor = 0.9 from Table 3.4 of AS 4100.M

    s = the nominal section moment capacity

    Mb = the nominal member moment capacity

    Bending about the appropriate axis must be checked, ie. x and y axes.

    Major principal axis x Mx* ≤ φMsx  Mx* ≤ φMbx  (4.2a)Minor principal axis y My* ≤ φMsy  My* ≤ φMby  (4.2b)where Mx* and My* are Design Bending Moments about x and y axes

    4.4 Section and Member Capacities of a Beam

    Steel members subject to bending have part of their section (one flange and half the web in an

    I-section) under compression forces. Therefore steel beams also suffer from failures causedby the following buckling modes as in pure compression members (last topic).

    ♦  Local Buckling of Plate Elements (because of Slender Plate Elements – see Figure 4.2(a))♦  Global Buckling of the member (because of Slender Beam – see Figure 4.2 (b))

    Therefore the bending capacity of steel members will depend on these two buckling failure

    modes. AS 4100 design formulae are based on these two buckling failure modes.

    Local buckling : Section capacity formula for φ MsGlobal buckling : Member capacity formula for φ Mb 

    Figure 4.2 Buckling Modes in Beams – (a) Local Buckling

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    BeforeBuckling

    After

    Buckling

     

    Figure 4.2 Buckling Modes in Beams – (b) Global Buckling

    The behaviour of a compact beam is simple if it is either very short or constrained to deflect

    in the plane of the applied load. This is because neither local buckling nor global buckling

    could occur in this case. The applied transverse load/bending moment versus deflection curve

    will be linear up to the first yield point, ie. when the extreme fibres of the beams will start to

    yield. Simple elastic bending theory can be used to determine the stresses and deflections up

    to this point. This corresponds to the yield moment capacity of the beam My = Yield stress f y 

    x Elastic section modulus Z. However, unlike in a pure compression member in which the

    entire cross-sectional area begins to yield at the same time, only the extreme fibres yield in

    beams. Therefore the steel beams can support further load until the entire cross-section yields

    at the point of the largest bending moment. From the first yield point, the load-deflection

    curve is nonlinear and once the entire cross-section yields, the beam deflections increase

    rapidly and a plastic hinge is formed in the beam. This corresponds to the beam’s plastic

    moment capacity Mp = Yield stress f y x Plastic section modulus S based on the simple plastictheory. The plastic section modulus S depends on the cross-section, and can be > 1.5Z.

    However, AS 4100 does not allow the use of S values > 1.5 Z. The ratio S/Z is known as the

    Shape Factor. Additional details on the determination of plastic section moduli can be found

    in engineering mechanics and structural analysis textbooks. The presence of strain hardening

    will result in larger plastic moment capacities in practice, however, this additional strength is

    ignored in design. Effects of residual stresses in beams will lead to premature yielding in the

    extreme fibres, however, the plastic moment capacity will not be affected.

    4.4.1 Local Buckling and Section Capacity Formula

    As in columns, the nominal section moment capacity Ms accounts for cross-section yielding 

    and/or local buckling. The nominal section moment capacity is given by:

    Ms = f y Ze  (4.3)

    where f y = yield stress (use smaller f y if web and flange yield stresses are different)

    Ze = Effective Section Modulus

    The effective section modulus depends on whether the section is subjected to elastic or

    inelastic local buckling effects, ie. it depends on the slenderness λe of plate elements (web and

    flange). This plate slenderness is defined as for the case of compression members, ie.

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    λe = (b/t)250

    f y  (4.4)

    where b = the clear width and t = the element thickness

    This plate slenderness is then checked against two limits, the yield limit λey and the plasticitylimit λep  to determine whether elastic or inelastic local buckling will occur in the plateelements. These slenderness limits are given in Table 5.2 of AS 4100. Table 5.2 is reproduced

    here as Table 4.1 for the sake of completeness.

    If all the plate elements have λe values less than the corresponding λep values given in Table5.2 of AS 4100, the section will not be subjected to any local buckling effects (elastic or

    inelastic). Therefore the section will develop its full plastic moment capacity Mp, and is

    called a COMPACT section. The Effective Section Modulus Ze in this case is given by:

    Ze = Lesser of S or 1.5 Z   (4.5)

    Table 4.1 Values of Plate Element Slenderness Limit (λλλλey and λλλλep)

    Plate Element Type Longitud.

    edges

    supported

    Residual

    stresses

    (see Notes)

    Plasticity

    limit

    (λλλλep) 

    Yield

    limit

    (λλλλey) 

    Deformn.

    Limit (λλλλed) 

    SR 10 16 35

    Flat element One HR 9 16 35

    subject to (outstand) LW, CF 8 15 35

    Uniform HW 8 14 35

    compression SR 30 45 90Both HR 30 45 90

    LW, CF 30 40 90

    HW 30 35 90

    Flat element subject to max. SR 10 25 -

    compn. at unsupported

    edge,

    One HR 9 25 -

    zero stress or tension (outstand) LW, CF 8 22 -

    at supported edge HW 8 22 -

    Flat element subject to

    Compression at one edge, Both Any 82 115 -

    Tension at the otherSR 50 120 -

    Circular hollow sections HR, CF 50 120 -

    LW 42 120 -

    HW 42 120 -

    Notes. 1. SR – Stress Relieved

    HR – Hot-Rolled or Hot-Finished

    CF – Cold-Formed

    LW – Lightly Welded longitudinally

    HW – Heavily Welded longitudinally

    2. 

    Welded members whose compressive residual stresses are less than 40 MPa maybe considered to be lightly welded.

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    3. For elements with λe  > λed  noticeable deformations may occur under serviceloading 

    For standard sections use BHP Handbook or AISC design capacity tables;

    For nonstandard sections calculate S and Z, then Ze. The S component for web elements is

    bd2 /4 whereas the S component for flanges can be approximately taken as flange area x the

    distance between the flange and the section centroid.

    If anyone of the plate element’s λe exceeds the plastic limit λep, but not the yield limit λey(λepλey), thesection will be subjected to elastic local buckling. The plate elements exceeding the yield

    limit will buckle locally before the section reaches its first yield moment. The section capacity

    Ms is therefore less than the first yield moment My (see Figure 4.3). In this case, the section is

    called a SLENDER section. The Effective Section Modulus Ze in this case is given by:

       N  o  m   i  n  a   l  s  e  c   t   i  o  n  c  a  p  a  c   i   t  y   M  s

    Plate slenderness b/t250

    f y  

    Mp 

    My 

    Compact Non-compact Slender

    Ms 

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    Ze = Zλ 

    λ 

    ey

    e

      (4.7a)

    for flat plate elements in uniform Compression;

    Examples of this case are the compression flange elements of I-section and Box-section

    subject to major axis bending.

    Ze = Z ( λ λ 

    ey

    e

    )2  (4.7b)

    for flat plate elements with maximum compression at an unsupported edge and zero stress or

    tension at the other edge. Examples of this case are I-section subject to minor axis bending

    and inverted T-section subject to major axis bending.

    The effective section modulus can be obtained conservatively by applying reductions to the

    whole section, but a more accurate answer can be obtained if the reductions are applied to the

    section modulus contributed by the slender elements only as appropriate and not to the whole

    section. Example problems in this section illustrate this method.

    Note that BHP’s standard sections are either compact or non-compact (not slender) and their

    Effective section modulus Zex and Zey values are listed in the BHP handbook and AISC design

    capacity tables. Therefore no other calculations are needed. Section capacities of these

    beams in bending are simply obtained by the product f y x Ze.

    Figure 4.3 illustrates the variation in section moment capacity as a function of plate

    slenderness and type of section (compact versus non-compact versus slender).

    For circular hollow sections, the section slenderness

    λs =250

    do 250

    f y  (4.8)

    where do= the outside diameter

    If λs exceeds the section yield slenderness limit λsy Ze = Lesser of Z √(λsy / λs) and Z (2λsy / λs)2

    (4.9)

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    The values of λsy are to be taken as the values of λey given in Table 5.2 of AS 4100 (see Table4.1 in this set of notes).

    For sections with holes that reduce either of the flange areas by not more than

    100 {1-[f y /(0.85f u)]}%, there is little effect on moment capacity and the gross section can be

    used to calculate the elastic and plastic section moduli.

    4.4.2 Example Problems on Local Buckling and Yielding of Beams

     Example Problem No.1

    What is the section moment capacity of 530UB92.4 Grade 300 beams about the major

    principal axis (x)?

     Example Problem No.2

    What is the section moment capacity of 200UC52.2 Grade 300 beams about the minor

    principal axis (y)?

     Example Problem No.3

    Figure 4.4 shows a lightly welded box-girder made of Grade 350 steel. What is its section

    moment capacity about the major principal axis (x)?

    tw = 10 mm f yw = 360 MPa

    tf  = 12 mm f yf = 360 MPa

    Figure 4.4 Welded Box-girder in Example Problem No.3

     Example Problem No.4

    Figure 4.5 shows a lightly welded I-section beam made of Grade 350 steel. What is its section

    360 mm400 mm

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    moment capacity about the major principal axis (x)?

    tw = tf  =10 mm f y = 360 MPa

    Figure 4.5 Welded I-section in Example Problem No.4

     Example Problem No.5

    Figure 4.6 shows a lightly welded plate girder made of Grade 350 steel. What is its section

    moment capacity about the major principal axis (x)?

    tw = 10 mm f yw = 360 MPa

    tf   = 25 mm f yf = 340 MPa

    Figure 4.6 Welded Plate Girder in Example Problem No.5

     Example Problem No.6

    Figure 4.7 shows a welded plate girder made of Grade 350 steel. What is its section moment

    capacity about the major principal axis (x)?

    tw = 10 mm f yw = 360 MPa

    tf  = 15 mm f yf = 350 MPa

    400 mm

    400 mm

    300 mm

    1500 mm

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    Figure 4.7 Welded Plate Girder in Example Problem No.6

     Example Problem No.7

    How do you calculate the section moment capacity of an I-section beam about the minor

    principal axis (y) if its flanges are slender.

     Example Problem No.8

    What is the section moment capacity of a Grade 350 1500 x 10 CHS?

    450 mm

    1500 mm

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    4.4.3 Global Buckling and Member Moment Capacity Mb 

    The Nominal Member Moment Capacity  Mb  accounts for overall flexural-torsional

    buckling (global buckling) of beams. Sections 5.3 to 5.6 of AS 4100 give the design rules

    required to calculate the member capacity of beam segments subject to various restraints and

    loading conditions. A segment in a member subjected to bending is the length betweenadjacent cross-sections, which are fully or partially restrained, or the length between an

    unrestrained end and the adjacent cross-section, which is fully or partially restrained (see

    Figure 4.8). In beam design, each such segment has to be checked individually for possible

    buckling failure as it depends very much on the end restraint and loading conditions.

    The segments with full lateral restraint will not be subjected to global buckling effects, and

    therefore, their capacity will not be reduced, ie. Mb = Ms 

    10mm

    min

     

    (a) Continuous lateral restraint

    (b) Intermediate lateral restraints

    Figure 4.8 Segments and types of restraints

    When segments have continuous lateral restraint at the critical flange or continuous lateral

    restraints with both ends fully or partially restrained according to Clause 5.3.2.2 of AS 4100

    (Figure 4.8a), or have intermediate lateral restraints at the critical flange with limited sub-

    segment lengths and both restrained ends (Figure 4.8b) according to Clause 5.3.2.3, or have

    full or partial restraints at both ends with limited lengths according to Clause 5.3.2.4, they canbe considered fully laterally restrained. A simple example for continuous lateral restraint

    Concrete slab

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    occurs in the case when the compression flange of a beam is connected to a concrete slab

    (Figure 4.8a).

    The critical flange at any cross-section is considered the flange that deflects the farther during

    buckling in the absence of any restraint at that section (Clause 5.5). Hence it is the

    compression flange for the segment with both ends restrained whereas it is the top flange forthe segment with one end unrestrained under gravity loads (bottom flange for wind uplift).

    4.4.3.1 Member Capacity of Segments without Full Lateral Restraint

    A beam segment made of a compact section and that is fully laterally restrained would not fail

    until well after the onset of yielding, ie. it will reach the full plastic moment capacity Mp. As

    stated in the last section, full lateral restraint can be provided by connecting the top flange to a

    concrete slab, or by restraints at sufficiently close intervals. In addition, twisting of the beam

    section at the supports must also be prevented (torsional restraint). However, a large

    proportion of beam segments fail before this due to either local buckling of compression plate

    elements or lateral buckling with twisting of the whole beam segment (global buckling known

    as lateral torsional or flexural torsional buckling). The reduction to section capacity due to

    local buckling of compression plate elements has been explained already in Section 4.4.1.

    This section therefore describes the reduction in the member capacity of segments without

    full lateral restraint due to lateral torsional buckling.

    4.4.3.2 Lateral torsional buckling

    Figure 4.2(b) shows the lateral torsional buckling of a beam segment bending about its major

    principal axis. In this buckling mode, the beam buckles out-of-plane by deflecting laterally

    and twisting about the longitudinal axis and thus relieves itself from the stiffer major axis in-plane bending. In other words, in-plane loading leads to buckling failure in the less stiff

    direction. The lateral displacement and twisting, which occur during this global buckling

    mode, can be explained. The compression flange of the beam segment is like a column, and

    thus is prone to buckling about the minor principal (weaker) axis, leading to lateral

    deflections. Twisting of the beam segment occurs because the compression flange displaces

    laterally while the tension flange tends to resist the lateral displacement. The lateral torsional

    buckling depends mainly on the cross-section geometry, unbraced length and end restraints.

    For a simply supported beam made of constant cross-sections such as I- and channel sections

    subjected to uniform bending (constant bending moment), the elastic lateral torsional

    buckling moment Mo is given by the following well known formula.

    Mo = )]()[( 2

    2

    2

    2

    e

    w

    e

     y

    l

     EI GJ 

    l

     EI    π π +   (4.10)

    where

    le  = the distance between the restraints preventing lateral deflection and twisting (but

    allows free rotation in plan), known as effective length.

    EIy = the flexural rigidity about the minor axis

    GJ = the uniform torsional rigidity

    EIw = the warping torsional rigidity

    E, G = Elastic and Shear moduli

    Torsion constant J = Σbt3 /3 for open sections

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    Warping constant Iw = Iy df 2 /4 for an I-section where df  is clear web depth

    Iw =

    48

    23

    w f  f bt b

     (8 -

     x

    w f  f 

     I 

    bt b2

    3) for a channel section

    Iw = 0 for an angle section, a tee-section, or a narrow rectangular section

    The lateral torsional buckling involves essentially two types of deformations, lateral

    deflection of compression flange through minor axis bending and twisting of the beam about

    the longitudinal axis. This is reflected by the buckling formula, which therefore includes the

    flexural rigidity about the minor axis EIy and the two types of torsional rigidities GJ and EIw.

    The commonly used sections (UBs, typical I-sections and channel sections) are open sections

    (easier to connect to other members) with a high Ix to Iy  ratio and have narrow flanges (to

    eliminate local buckling). This means these sections have comparatively low values of these

    rigidities (EIy, GJ and EIw), and are therefore susceptible to lateral torsional buckling. Theyneed adequate lateral/torsional restraints at the supports and at points of concentrated loads.

    The Universal Columns (UCs) have a higher resistance to lateral torsional buckling because

    of wider flanges, however, their bending capacity about the major axis is lower than UB

    sections. Despite this, many types of construction allow effective bracing/restraints to be

    included so that these efficient beam sections such as UBs are commonly used.

    On the other hand, the sections with larger values for these rigidities will have greater lateral

    torsional buckling moment capacity. For example, closed sections such as RHS, SHS and

    CHS, square and round bars are not susceptible to lateral buckling and require far less

    bracing. Also, the beam segments bent about their weak (minor principal) axis do not fail bylateral torsional buckling provided that the loads are not applied too high above the shear

    centre. In these cases, the beam segments can be designed based on their section moment

    capacity Ms. Similarly if the distance l is decreased by providing intermediate lateral

    restraints, the beam segments will not buckle laterally, and as stated earlier, it can be assumed

    fully braced and designed based on its section moment capacity Ms.

    4.4.3.3 Effective length

    The elastic flexural torsional buckling moment Mo  of a beam segment is affected by the

    effects of cross-sectional distortion, load height and rotational end restraints. It also dependson the restraint conditions of the ends of the segment. The AS 4100 takes these effects into

    account through the use of an effective length le instead of l as defined next.

    le = k t k l k r l (4.11)

    where k t = twist restraint factor

    k l = load height factor

    k r = lateral rotation restraint factor

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     Restraint Conditions at the Segment End Cross-sections

    The AS 4100 classifies the restraint conditions at a cross-section as either Fully restrained (F),

    Partially restrained (P), Laterally restrained (L) or Unrestrained (U). The two important

    deformations that affect lateral torsional buckling are lateral displacement of critical flange

    and twist about the longitudinal axis. Accordingly, the AS 4100 defines the Fully restrainedcross-sections (F) to be those which are effectively prevented from deflecting laterally and

    twisting. The level of torsional restraint required at the cross-section can be reduced if the

    lateral restraint acts on the critical flange. On the other hand, if the lateral restraint is not at

    the critical flange, a full torsional restraint will be required for a Fully restrained condition at

    the cross-section.

    The partially restrained cross-sections (P) are effectively prevented from deflecting laterally,

    but only partially prevented from twisting. The laterally restrained cross-sections (L) are

    effectively restrained from deflecting laterally at the critical flange, but are unrestrained

    against twisting. The unrestrained cross-sections (U) are those, which have no lateral or twist

    restraints. The cross-sections, which are not effectively prevented from deflecting laterally,

    are also considered as Unrestrained even if they have full twist restraint. Similarly if the

    cross-sections are effectively prevented from deflecting laterally only at points away from the

    critical flange, they are also considered unrestrained if they do not have partial or full twist

    restraint. Figures 5.4.2.1, 5.4.2.2, and 5.4.2.4 of AS 4100 give examples of fully restrained,

    partially restrained and laterally restrained cross-sections. An AISC publication entitled

    “Design of unbraced beams” by Trahair et al. (1993) uses 38 real connections to demonstrate

    the classification into the various restraint types. This enables the designer to choose an

    appropriate restraint for the connection under consideration.

    In the design of a beam, the cross-sections have to be classified as fully (F), partially (P) orlaterally (L) restrained or unrestrained (U), based on which the beam is divided into segments

    or sub-segments. A segment is a length between fully or partially restrained cross-sections

    (FF, FP, PP) whereas a sub-segment has one end laterally restrained (L) (FL, PL, LL). In the

    design of beams there is no need to separate the two types of segments.

    Twist restraint factor

    Table 4.2 Twist Restraint Factor kt 

    Restraint arrangement Factor k t 

    FF, FL, LL, FU 1.0FP, PL, PU

    w

    3

    w

    f 1

    n

    ])t2

    t)(

    l

    d[(

    1+  

    PP

    w

    3

    w

    f 1

    n

    ])t2

    t)(

    l

    d(2[

    1+  

    Notes: d1 = clear depth of web

    nw = number of webs

    tf  = thickness of critical flange tw = thickness of webl = segment length

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    This factor takes into the effects of cross-sectional distortion that occurs in beams with deep

    thin webs (see Figure 4.9). The cross-sectional distortion increases the twist rotation and thus

    reduces the buckling capacity. It depends on the web flexibility. Table 4.2 gives the twist

    restraint factor k t as a function of the cross-section geometry, length and restraint conditions

    at segment ends. For most of the standard sections and connection details, this factor is closer

    to unity. Note that it is also unity when both ends are fully or laterally restrained.

    Before

    bucklingAfter

    buckling

     

    Figure 4.9 Distortion of Beams Figure 4.10 Loads Acting on the Top Flange

     Load height factor kl  

    The loads can be transferred to a beam via its top flange (ex. loading/secondary beam located

    on top flange), the middle (ex. secondary beam connected to the web) or its bottom flange

    (ex. crane girder). The loads acting at the top flange will move with it when it buckles as

    shown in Figure 4.10 This creates additional torques and twist rotations, which reduce the

    buckling capacity. The AS 4100 allows for this effect through the use of a load height factor

    k l. This factor is equal to one when the load acts at the shear centre or below. Table 4.3 gives

    the load height factor k l as a function of the load height position and restraint conditions at

    segment ends. It is also equal to one when the load acts at the segment end of type F, P or L.

    The factor can be as high as 2 for segments of the type FU, PU (one end free) as shown in the

    table. Therefore, this factor needs to be carefully determined.

    Table 4.3 Load Height Factor kl

    Load height positionLongitudinalposition of the load

    Restraint arrangementShear Centre Top Flange

    FF, FP, FL, PP, PL, LL 1.0 1.4Within segment

    FU, PU 1.0 2.0

    FF, FP, FL, PP, PL, LL 1.0 1.0At segment end

    FU, PU 1.0 2.0

     Lateral rotation restraint factor k r 

    This factor takes into the improvement to lateral buckling due to the lateral rotation restraint

    at the segment ends. This is the restraint to the beam’s lateral rotation (not the in-plane

    rotation) when the beam is loaded with vertical transverse loads. The AS 4100 allows for this

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    effect through the use of a lateral rotation restraint factor k r. Table 4.4 gives the lateral

    rotation restraint factor k r as a function of the restraint conditions at segment ends. The AISC

    publication on the Design of unbraced beams by Trahair et al. (1993) recommends the use of

    k r = 1 because lateral rotation restraints are not dependable.

    Table 4.4 Lateral Rotation Restraint Factor kr

    Restraint arrangement Ends with lateral

    rotation restraints

    Factor k r

    FU, PU Any 1.0

    FF, FP, FL, PP, PL, LL None 1.0

    FF, FP, PP One 0.85

    FF, FP, PP Both 0.70

    4.4.3.4 

    Member Capacity

    The elastic flexural torsional buckling moment Mo given in Section 4.4.3.2 is not the ultimate

    strength of real beams. The real beams have initial geometric imperfections and residual

    stresses, and therefore have reduced capacity as shown in Figure 4.11. As seen in the figure,

    the ultimate strength of very slender beams (large le) is approximately equal to the elastic

    flexural torsional buckling capacity whereas the ultimate strength of short beams is limited by

    their section capacity Ms. Other beams have an ultimate strength less than both Mo  and Ms 

    and this strength is reduced depending on the imperfections (initial bow and twist, residual

    stresses and other).

    0 0.5 1.0 1.5 2.0 2.5

    0.5

    1.0Elastic

    Buckling

    Curve

    Slenderness

       N  o  n   d   i  m  e  n  s   i  o  n  a   l  c  a  p  a  c   i   t  y   M   b   /   M  s

     Figure 4.11 Moment Capacity of Real Beams

    The AS 4100 therefore recommends the following equation to calculate the member capacity

    of a beam segment.

    φ Mb = αm αs φ Ms  ≤ φ Ms  (4.12)

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    where αs = a slenderness reduction factor that relates the elastic buckling moment to member

    capacity = 0.6 [ )(]3)[( 2o

    s

    o

    s

     M 

     M 

     M 

     M −+ ] 

    αm = a moment modification factor

    Ms = the nominal section moment capacity

    The elastic flexural torsional buckling moment Mo and the member capacity αs φMs are forbeam segments subject to uniform bending. However, the real beams usually have transverse

    loads and are thus subjected to varying bending moment distribution. Therefore a moment

    modification factor αm is added to the member capacity equation. From a buckling point ofview, the worst case is when a beam is subjected to uniform bending. Therefore in most cases

    the αm factor is greater than 1. The AS 4100 states that the αm factor can be calculated as oneof the following.

    •  Conservatively taken as 1.0• 

    From Table 5.6.1 of AS 4600

    •  Using αm =2*

    4

    2*

    3

    2*

    2

    *

    )()()[(

    7.1

     M  M  M 

     M m

    ++≤ 2.5 (4.13)

    where Mm* = maximum design bending moment in the segment

    M2*, M4* = design bending moments at the quarter points of the segment

    M3* = design bending moment at the midpoint of the segment

    The following gives the αm  values for the most common bending moment distributions. It

    must be noted that the ααααm value is for the beam segment being designed, ie. the segmentbetween restrained cross-sections. Further, the αm values are quite large with the maximumbeing 3.5. This implies that the member capacity is 3.5 times that of uniform bending. Hence,

    the determination of αm values must be done carefully. Note that AISC design capacity tablesassume αm = 1. So the  AISC moment capacity must be multiplied by an appropriate α m value.

    Both ends fully or partially restrained

    1. Uniform moment αm = 1

    2. Central load αm = 1.35

    3. Moment at one end αm = 1.75

    Fig. 4.12 Moment Modification factors ααααm  (a) Both ends fully or partially restrained

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    4. Linear moment variation Calculate end moment ratio βm 

    This figure shows positive βm 

    αm = 1.75+1.05βm+0.3βm2 

    for -1≤βm≤0.6

    αm = 2.5 for 0.6

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    The design rules stated earlier are for the following: segments fully or partially restrained at

    both ends, Open sections with equal flanges, and Segments with constant cross-section. The

    design method for segments with varying cross-section is given in Clause 5.6.1.1 (b) of AS

    4600. Clauses 5.6.1.2 of AS 4600 gives the design method for I-sections with unequal

    flanges. For angle and hollow sections, the same method is used but with Iw = 0 according to

    Clauses 5.6.1.3 and 5.6.1.4, respectively.

    For segments unrestrained at one end, the same method can be used provided the other end

    is fully or partially restrained and laterally continuous or restrained against lateral rotation. In

    this case, appropriate αm  values given in Table 5.6.2 of AS 4600 have to be used. Thefollowing are the αm values for the most common bending moment distributions.

    Unrestrained at one end and the other end fully or partially restrained

    1. Uniform moment αm = 0.25

    2. Point load at the end αm = 1.25

    3.  Uniformly distributed load αm = 2.25

    Figure 4.12 Moment Modification factors ααααm (b) Unrestrained at one end and theother end fully or partially restrained

    The type and arrangement of lateral restraints are very important as they can considerably

    increase the member capacity. Lateral/torsional restraints within the span will be very

    beneficial. For a lateral restraint (bracing) to be effective, it should provide resistance to

    lateral displacement and twist to the critical flange. The lateral restraint/bracing must have

    adequate stiffness to restrain the braced point without moving, but also adequate strength to

    withstand any forces transferred to it by the main beam. In designing the bracing, it is

    assumed to carry 2.5% of the maximum compression force in the flange.

    A beam without intermediate lateral restraints is regarded as a single-segment beam, provided

    that the sections at the supports are adequately restrained against lateral displacement and

    twisting. The addition of one restraint within the span produces a two-segment beam, ie.buckling now takes place between restraints. More than two lateral restraints can also be

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    used. This divides the beam into segments and thus reduces its effective length (increases Mo 

    and thus Mb)

    Uneconomical designs results when slender sections and largely unbraced beams are used.

    Slender sections use only part of the cross-section and thus steel is wasted. Attempts must be

    made to use other sections if possible. Excessive lateral slenderness is measured by αs. If thisvalue is less than 0.7, the design must be considered uneconomical. If possible, the beam

    should be redesigned by introducing more lateral restraints or suitable bracing systems

    (shorter segment length) or by changing the section with increased lateral buckling capacity

    (SHS, CHS, RHS).

    In general, there are two types of beam bracing systems, lateral and torsional. Effective beam

    bracing systems must prevent relative displacement between top and bottom flanges (ie.

    prevents twist). Lateral bracing (restraining members attached to the compression flange) and

    torsional bracing (cross-frame or diaphragms between adjacent beams at intermediate

    locations or continuous bracing by floor systems, metal decks and slabs) can provide this.

    Lateral bracing systems are most effective if they are connected to the critical flange. Some

    bracing systems such as concrete slab attached to the top flange via shear studs can control

    both lateral displacement and twist. Research has shown such combined lateral and torsional

    bracing is more effective, particularly for beams subject to uniform moment. In a common

    example of beams linked together along the span, the beams cannot buckle laterally unless all

    of them buckle. Therefore buckling of an individual beam can occur only between the cross-

    members. If two adjacent members are connected by a properly designed cross-frame or

    diaphragm at intermediate points, those points can be considered as effectively braced (SSRC,

    1998). Both beams can move laterally in the middle, but since they will move equal amounts

    without any twist, the beams can be considered braced at these points. Some designers

    assume the point of contraflexure to be a braced point, however, research has shown this to beincorrect. Figure 4.13 shows some of the bracing systems that are used in steel construction.

    Figure 4.13 Bracing Systems

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    4.4.3.5 Example Problems on Global Buckling of Beams

     Example Problem No.1

    Figure 4.14 shows the roof structure of a service station building. Determine the size of a

    suitable UB section for the main beams B if the wind uplift pressure on the roof is 2.5 kPa.

    Beam

    B

    2m

    Column A

    Elevation

    1.5 m

    Plan View

    Figure 4.14. Service Station Building for Example Problem 1

    11 m 2 m

    2 m11 m

    2 m

    Column APurlins

    Main Beams B

    1.5 m

    8 m

    Metal sheeting Purlins

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     Example Problem No. 2

    Design the beam ABCD shown in Figure 2. The loads shown include load factors. The

    connection and support conditions at A, B and C are as per the AISC publication (Types 1, 11

    and 8, respectively).

    Type 11

    A 3 m B 3 m C 3 m D

    Type 1 Type 8

    Figure 4.16. Beam ABCD for Example Problem 2

    4.5 Serviceability Design of a Beam

    The beams undergo transverse deformations, and must be checked against acceptabledeflection limits. Otherwise, cracking in walls, ceilings and openings located under the

    beams may occur. Deflections under serviceability design loads can be calculated using

    available theoretical formulae or computer analyses. Appendix B of AS 4100 gives

    appropriate deflection limits for beams. The total vertical deflection limit for beams is

    span/250, but span/125 for cantilever beams. Table 1.1 in this set of notes presents the

    recommended values of deflection limits.

    References:

    Trahair, N.S., Hogan, T.J. and Syam, A.A. Design of Unbraced Beams, Steel

    Construction, J. of Australian Institute of Steel Construction, Vol.27, No.1, Feb 1993.

    Structural Stability Research Council (SSRC), Guide to Stability Design Criteria for

    Metal Structures, Ed. By T.V. Galambos, 5th

     Edition, John Wiley & Sons, 1998 

    400 kN

    100 kN

    Bending moment

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    TUTORIAL PROBLEMS ON BEAMS

    Questions on Local Buckling and yielding

    Question 1

    Using the AS4100 limits, which plate elements will buckle locally in the following sections?

    Categorise them into Compact, Non-compact and Slender sections. Assume f y = 250 MPa

    and t = 8 mm. Assume the given centreline dimensions in your calculations.

    Figure 1. Steel Beams

    Question 2

    Determine the Zx, Zy, Sx,, Sy, the shape factors, yielding moment Myield and plastic moment Mp 

    about the major and minor principal axes for the I-section shown in Figure 2. Assume Grade

    250 steel (f y = 250 MPa) and the beam is fully restrained. tf  = 25 mm tw = 16 mm

    (Ans: 1091 x 104  mm4, 633 x 103  mm4; 1289 x 104  mm3, 1017 x 103  mm3, 2727, 3222

    kNm; 158, 254 kNm)

    60 60

    60 60

    270LW

    60 60

    60 60

    960LW

    120 120

    120 120

    1000HW

    250

    720

    HW

    400200

    200 LW

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    275 mm

     

    Figure 2. I-section Beam

    Question 3

    Determine the maximum design moment M* of a lightly welded plate girder of Grade 250

    steel, which has full lateral restraint. tf  = 25 mm tw = 10 mm (Ans. 2908

    kNm)

    300 mm

     

    Figure 3. Lightly Welded Plate Girder

    Questions on Global Buckling and yielding

    Question 4

    Determine the maximum design uniformly distributed load a 460UB67 of Grade 300 steel can

    carry. The beam is simply supported over a span of 5 m, and is fully restrained at the supports

    against lateral deflection and twist rotation, but unrestrained against lateral rotation. Assumethat the load is applied to the top flange. (Ans. 48.6 kNm)

    1170 mm

    1500 mm

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    Question 5

    A simply supported beam with a span of 7 m has a nominal central concentrated live load of

    55 kN on the top flange. The beam is restrained against lateral displacement and twist only at

    the ends, and is free to rotate in plan. Design a suitable UB section. (Ans. 460UB82.1)

    Question 6

    Figure 4 shows the layout of a temporary bridge of 10 m span, used in a construction site. It is

    constructed using a number of 530UB82 steel beams spaced at 2 m with a 100 mm height

    timber deck on top. During construction, this bridge will be subjected to a heavy live load of

    5 kPa of length 5 m as shown in the figure. The density of hardwood timber used can be taken

    as 10 kN/m3.

    a) Determine the adequacy of the 530UB82 to carry the load combination of dead load and

    live load if the timber deck does NOT offer any lateral restraint due to inadequate

    connections between the deck and beams. Your calculations need to include design checksfor bending capacity. (Ans. 94.7 < 182.6 kNm Inadequate)

    b) From the above calculations, how will the beam fail first if tested to failure?

    Figure 4. Bridge Layout