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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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4-6
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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4-9
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