Building Structures and Technology
Building Structures and Technology
Building Structures and Technology
Second Edition
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Acknowledgments :
Cover Photo:
Reflections in the façade of the John Hancock Building, Boston USA (Geoff Taplin 2000)
The John Hancock building (60 stories) has one of the most infamous facades in the world.
( http://www.pubs.asce.org/ceonline/0600feat.html )
LECTURE NOTES
TABLE OF CONTENTS
1. General
1.1 Purpose:
1. $ cost
2. $ cost
3. $ cost
4. spacing of supports (walls, columns)
5. serviceability (ie stiffness and vibration)
6. adaptability to future changes of use
………plus safety
by material
• floor slab – concrete
• beams – concrete or steel
by method of construction
• steel beams with cast-in-situ slab
• steel beams with precast slab
• cast-in-situ concrete beams with cast-in-situ slab
• precast beams with precast slab
The following notes provide some ‘rules-of-thumb’ for the preliminary sizing of slabs
and beams. The thickness of the slab, or the depth of the beam, is given as a span-to-
depth ratio, l/d.
note that d is the effective depth – ie you must add cover thickness to the concrete (say
40 mm)
eg,
l/d =30 and slab span = 6 metres υ thickness = 200+40 = 240 mm
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 6
Lecture notes
one-way slab
span
steel column
metal
decking
(before
concrete
is cast)
steel beam
• a slab may span one-way onto secondary beams, which in turn span onto primary
beams,
• if the slab is supported by beams on 4 edges, but the aspect ratio (ie longer
supported length divided by shorter supported length) > 2, treat as a one-way slab
• if the beams are at very close centres, the floor system becomes a ribbed slab,
• ribbed slabs are require a lot of labour to build the formwork for the slab
• if wide and shallow concrete beams support the slab, it is called a band beam floor
system
• band beam systems have simple formwork, and are economical with labour
• precast systems are proprietary systems – ie they are products which you buy from a
manufacturer
When the slab is supported by beams or walls on four sides it is a two-way slab.
two-way slab
• if the beams (in two directions) are at very close centres, the floor system becomes a
waffle slab
• waffle slabs require a lot of labour to build the formwork for the slab
We now have some ‘rules-of thumb’ for sizing one-way and two-way slabs, but what
about the size of the beams that support them?
• if the beam is simply supported (ie one span only, or no moment connection
between spans), l/d = 12
• if the beam is continuous (ie 2 or more spans), l/d = 15
• if the beam is a cantilever, l/d = 6
• because there are no beams of walls, flat slabs tend to have large deflections, and for
this reason they are not as popular as they were 10 years ago
flat plate
3. Prestressed concrete
Prestressing refers to the practice of placing high tensile wires in the concrete, and
stretching them once the concrete has hardened.
Prestressed concrete will not be dealt with in this subject, but it is commonly used in
floor slabs and beams, and results in thinner slabs and beams and/or longer spans.
• cost depends upon local practices, but in developed economies easier formwork
leads to lower costs
• prestressed slabs do not adapt well to future changes of use, because of the critical
nature of the prestressing wires
And note, many products in the building industry work on a module of 600 mm (eg
ceiling systems, brick and blockwork). Where possible try to use floor spans which are
multiples of 600 mm, eg
6.0 m – a useful shorter span
7.2 m – a very common span
8.4 m – quite a common span
10.2 m – a long span
But remember, very often other considerations will prevent a 600 mm module being
adopted.
TOPIC 3: FACADES
1. What is a façade?
The facade is the walls of a building
2. Classifying facades
Facades can be either load bearing or non load bearing. Non load bearing facades
require a separate structural frame to support the loads.
•concrete •concrete
•masonry (brick and block) •masonry (brick and block)
•timber •timber
•metal decking
•glass
•GRC and GRP
3. A facade tour
Views of a range of façade types will be shown in the lecture.
4. Common issues:
• weatherproofing
• allowing for movement
• examples of defects
4.1 Weatherproofing
• impermeable facades
o glass
o plastic
o metal sheeting
• low permeability facades
o thick concrete or masonry
• cavity wall construction
o single leaf concrete or masonry
Impermeable facades
5. Examples of defects
Views of a range of façade defects will be shown in the lecture
TOPIC 4: LOADING
1. Introduction
It is a very important step in the total design process to determine the design loads for
the structure. Typical loads for a building are dead load, live load, wind load and
earthquake load. Special consideration is sometimes given to impact and fatigue that
may occur in vehicles, cranes or machinery.
2. Load Combination
We must consider combinations of various loads that can be imposed on the structure.
As the number of loads included in the combination increases, it is customary to
introduce a “live load combination factor” (ψ c), which takes into account the
improbability of the maximum value of each load occurring simultaneously. Load
combinations are given in Australian Standard 1170.1 –1989. Please also pay attention
to the direction of loads.
Different load factors are used for strength limit state design and for serviceability limit
state design. Typical examples are given below:
1.25G + 1.5 Q
or
1.25G + Wu + ψ cQ
or
1.25G + 1.6Feq + ψ cQ
where
G = dead load
Q = live load
Wu = wind load
Feq = earthquake load
Ws
or
ψ sQ
or
G + Ws
or
G + ψ sQ
where
Ws = wind load for serviceability limit state
Department of Civil Engineering, Monash University
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Unit CIV3221 : Building structures and technology 18
Lecture notes
Roof:
Case Q (kPa)
if non-trafficable 0.25
If trafficable 3
Wind direction
Wind speed
Structure height
Department of Civil Engineering, Monash University
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Unit CIV3221 : Building structures and technology 19
Lecture notes
Structure shape
Structure component (wall or roof)
Region (which city)
Topographic condition
serviceability multiplying factor = 0.6 for region A, 0.4 for regions B and C, 0.35 for
region D
pd = p’ x B1 x B2 x B3 x B4
B1 to B4 are factors
p’ = basic pressure (for wall and roof)
5.3 Factors
B1 = regional factor
Region B1
A = normal 1.0
B= intermediate 1.5
C= tropical cyclone 2.3
D = severe tropical cyclone 3.3
B3 = topographic factor
For flat area, B3 = 1.0
Terminology:
External pressure
Internal pressure
Windward sections
Leeward sections
Positive pressure and negative pressure
6. Earthquake Load
Conventional earthquake (seismic) design procedures replace the dynamic earthquake
loads with equivalent static loads. The earthquake loads which are stipulated are
recognised to be much less than the maximum loads possible from a very severe
earthquake. However, AS1170.4-1993 attempts to stipulate earthquake loads large
enough to prevent structural damage and minimise other damage in moderate
earthquakes which occasionally occur, and to avoid collapse or serious damage in
severe earthquakes which seldom occur.
The earthquake design category (A, B, C, D, or E) can be determined using the above
three items – see Table 2.6, AS1170.4-1993.
Fx is certain percentage of V depending on the gravity load and height of each level.
. ⋅ I ⋅ G g ⋅ a ⋅S
125
V=
T2 / 3 ⋅ R f
hn
Fundamental period: T=
46
hn
Period for the orthogonal direction: T =
58
Fx = C vx ⋅ V
G gx ⋅ h x
If the height of the structure is less than 23 meters: C vx = n
∑G gi ⋅ hi
i =1
7 Example
If V = 125 kN, find the horizontal force at level 3 with the conditions given below.
There are uncertainties in load, fabrication, material and theoretical models. Two
different methods are available to take into account the uncertainties, namely the
working stress design method and the limit state design method.
S
S* ≤
SF
where
S* = Design Stress
Limit state design is a design method in which the performance of a structure is checked
against various limiting conditions at appropriate load levels. The limiting conditions to
be checked in structural steel design are ultimate limit state and serviceability limit
state. Ultimate limit states are those states concerning safety, for example, load-carrying
capacity, overturning, sliding, and fracture due to fatigue or other causes. Serviceability
limit states are those states in which the behavior of the structure under normal
operating conditions is unsatisfactory, and these include excessive deflection, excessive
vibration, and excessive permanent deformation.
In essence, the designer attempts to ensure that the maximum strength of a structure (or
elements of a structure) is greater than the loads that will be imposed upon it, with a
reasonable margin against failure. This is the “ultimate limit state” criterion. In addition,
the designer attempts to ensure that the structure will fulfill its function satisfactorily
where subjected to its service loads. This is the “serviceability limit state” criterion.
Department of Civil Engineering, Monash University
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Unit CIV3221 : Building structures and technology 24
Lecture notes
The ultimate limit state criterion can be illustrated by the Figure shown below. This
figure shows hypothetical frequency distribution curves for the effects of loads on a
structural element and the strength, or resistance, of the structural element. Where the
two curves overlap, shown by the shaded area, the effect of the loads is greater than the
resistance of the element, and the element will fail. The structure must be proportioned
so that the overlap of the two curves is small, and hence the probability of failure
occurring is small enough to be acceptable.
E
R
Frequency
The basic equation for checking the limit state condition is:
S* ≤ Φ ⋅ R n
where
Rn = Nominal Capacity
Φ= Capacity Factor
S* = Design Load (action) Effects
Advantages:
• Consistent Reliability
(different Φ for different cases, calibrated using reliability analysis,
i.e certain reliability index and probability of failure)
• More Economical
(saves material)
2. Methods of Analysis
The following methods are available as shown in the figure below.
λP λP
λW
λ = load factor
∆ = sway
λ
First order elastic
Elastic buckling
λc
Advanced analysis
∆
(b) Load deflection responses
The assumptions made are: geometry of structure remains unchanged and material
properties remain unchanged.
The assumptions made are: geometry of structure keeps changing and material
properties remain unchanged.
The assumptions made are: geometry of structure remains unchanged and material
properties keep changing.
The design action effects shall be determined using a rigid plastic analysis. It shall be
permissible to assume full strength or partial strength connections, provided the
capacities of these are used in the analysis. The rotation capacity at none of the hinges
in the collapse mechanism must be exceeded.
The assumptions made are: only P-δ effect is considered and only elastic buckling load
is obtained.
The assumptions made are: geometry of structure keeps changing and material
properties keep changing. The analysis includes residual stresses, geometrical
imperfections and erection procedures.
For a frame comprising members of compact section with full lateral restraint, an
advanced structural analysis may be carried out provided the analysis can be shown to
accurately model the behaviour of that class of frame. The analysis shall take into
account the relevant material properties, residual stresses, geometrical imperfections,
second order effects, erection procedures and interaction with the foundations. For the
strength limit state, it shall be sufficient to satisfy the section capacity requirements for
the members and the requirements for the connections.
M * = δ b ⋅ M *m
M * = δ s ⋅ M *m
Cm N omb
δb = λm = for each member
1 N*
1−
λm
π2 ⋅ E ⋅ I
N omb =
( k e ⋅ L) 2
The modifying factor (β e) which accounts for the conditions at the far ends of the beams
are given in the table below.
N oms
Cm ∑(L
)
δs = λ ms = for each storey
1 N*
1−
λ ms ∑( L )
Noms = π 2 EI / (keL)2 for each column
Limitation: δ ≤ 1.4
1. Introduction
one-way slab
span
two-way slab
Note that shear capacity is almost never a problem (load intensity is lower than on
beams, and the shear capacity is higher because the section is thinner)
span L
1 metre
pf sy
φM u = φA st fsy d 1 −
17 . fc′
• Shear is not usually a problem (ligatures not required)
• Deflection (serviceability limit state) is often the critical requirement
If the slab is continuous, design a one metre wide strip as a continuous beam – use
matrix analysis to calculate the bending moment diagram,
PL3 L
∆=
48EI
L ü
E - non linear
time dependent
shrinkage affected
E = ρ1.5 × 0.043 f cm
ρ = concrete density
fcm = mean compressive strength
3.1 Shrinkage
and if I restrain one face with steel reinforcing bars, it becomes this concrete:
3.2 Creep
ultimate strain
≈ 2.5
creep strain
3.3 The value of I
To avoid calculating the non-linear time dependent cracking, creep and shrinkage
effects, we can use deemed to comply span to depth ratios.
5 wL4
∆= ×
384 EI
∆ 5 wL3
∴ = ×
L 384 I 3
E 3 bd
bd
I ∆
3 bE
L bd 3 L
∴ =
d 5 w
384
L L
as calculated must be greater than the actual of the beam, if the
d d
SERVICEABILITY LIMIT STATE is to be satisfied.
1
∆ bE 3
0.045
L L
= for beams (Cl. 8.5.4)
d k 2Fdef
or
1
∆ bE 3
0.053
L L
= for slabs (Cl. 9.3.4, rearranged)
d k 2 Fdef
I 1
(for an uncracked section, =
3 12
= 0.083)
bd
b = width (one metre for a slab)
k2 = 5/384 for a simple span
= 1/185 for an end span
= 1/384 for an interior span
Fdef = load per metre, including allowance for shrinkage and creep
= (1.0 + kcs)g + (ψ s + kcs ψ l)q
A sc
where k cs = 2 − 12
. ≥ 0.8
A st
Asc
Ast
Ly
Bending moments
spanning in the x direction, M*x = β x w Lx 2
spanning in the y direction, M*y = β y w Lx 2
note that it is Lx in both cases
(if Ly /Lx > 2, treat as 1-way spanning, expect M*x = wLx 2 /8, M*y = 0)
To check the span/depth ratio for two way slabs, first rearrange the formula,
1 1 1
∆ bE 1 ∆
3 ∆ 3 3
0.053 bE L bE
L L 0.053 3 L
= = = k4
d k 2 Fdef k 2 Fdef F
def
where,
1
0.053 3
k4 =
k2
1
0.053 3
k4 = = 1.6
5
If simply supported,
384
To allow for two way spanning slabs, the value of k4 is adjusted,
edge condition k4
values of Ly /Lx
1.0 1.25 1.5 2.0
all edges continuous 4.00 3.40 3.10 2.75
1 short edge discontinuous 3.75 3.25 3.00 2.70
1 long edge discontinuous 3.75 2.95 2.65 2.30
2 short edges discontinuous 3.55 3.15 2.90 2.65
2 long edges discontinuous 3.55 2.75 2.25 1.80
2 adjacent edges discontinuous 3.25 2.75 2.50 2.20
1 long edge continuous 3.00 2.55 2.40 2.15
1 short edge continuous 3.00 2.35 2.10 1.75
all edges discontinuous 2.50 2.10 1.90 1.70
TOPIC 8: FOOTINGS
1. Introduction
load = N*
column
footing
pressure = qu
T
Department of Civil Engineering, Monash University
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Unit CIV3221 : Building structures and technology 40
Lecture notes
Under reinforced means that the steel yields before the concrete crushes – ie there is not
too much steel
pf sy
φM u = φA st fsy d 1 −
17 . fc′
2.2 Shear failure
(φ = 0.7 for shear)
Shear forces are carried by the concrete and the steel (ligatures) acting together,
Vu = Vuc + Vus
shear shear
carried by carried by
concrete ligs
d
β 1 = 1.11.6 − 0 ≥ 1.1, is a depth effect
1000
β 2 = 1, unless axial tension is large
β 3 = 1 unless the load is near to the support
If ligs must be designed, they must provide a shear strength of φVus, (found from Vus =
Vu – Vuc).
critical
section for SECTION
bending
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Unit CIV3221 : Building structures and technology 42
Lecture notes
d
B
(D-d)/2
PLAN
The critical section for shear is at ‘d’ from the face of the column (as for a beam)
d [(D-d)/2]-d
V* = pressure x area
D−d
= qu × B × − d
2
Choose D so that you do not need to use ligatures.
a conical shape
‘punches out’ of
the footing
The failure cone is approximated as a prism with vertical sides, and the sides are taken
to be at d/2 from the column face, where d is the effective depth of the footing.
x+d
x
B y y+d
d/2
shear surface = u x d
shear capacity
1. Introduction
Conventional reinforced concrete slabs like this
require a lot of formwork to contain the wet concrete until it has gained strength
2. Design issues
No Australian Standard exists yet.
‘Design of Composite Slabs for Strength’ – BHP Structural Steel, 1998 is the best
reference.
• ultimate strength of the bare steel to carry the wet concrete (not considered here)
• ultimate strength of the reinforced concrete slab
o yielding of the steel (flexure)
o crushing of the concrete (flexure)
o slip between the steel and the concrete (longitudinal slip)
• serviceability deflection of the composite slab (treat as for a normal rc slab)
Design for negative bending (hogging) as a normal reinforced concrete slab – ie ignore
the metal sheeting.
0.85f’C
b compression
NA C
DC
Tsh
tension ysh
cross section strain stress at failure
0.5Tsh
from top of slab to the line of force C =
0.85 f c′b
0.5Tsh
lever arm between C and Tsh = Dc − ysh −
0.85 f c′b
therefore, ultimate moment,
. Tsh
05
φ Mu = φ Tsh Dc − y sh −
0.85 fc′b
where φ = 0.8
Calculation of Tsh
3. The sheeting slips along the concrete/steel interface before either 1 or 2 occurs
(longitudinal slip failure),
Shear between the sheeting and the concrete has two components:
1. friction
• equals vertical reaction at the support (R*, kN) x coefficient of friction (µ)
2. mechanical resistance
• equals area of sheeting x resistance per unit area of sheeting (Hr, kPa)
• is due to the geometry of the sheeting cross section
UDL A
reaction, R* A
(per metre width) x
free body diagram
Hrx + µR
µR
x
interface shear strength = Hrx + µR (per metre width)
So if there is not sufficient distance from the support to the section AA, the cross
section will fail by longitudinal slip, rather than by flexural failure.
In summary,
Tsh = minimum of
Note that the strength of composite slabs is a function of the position in the span, as
well as the properties of the cross section. This is not the case for reinforced concrete or
steel beams.
1. Introduction
Comprise a steel beam and a concrete slab, joined with shear connectors to achieve
composite action between the two elements.
In building construction the concrete slab is often (but not always) a composite slab – ie
cast on metal sheeting
Composite beam comprise a steel beam connected to a concrete slab with shear
connectors, such that the two components act as one beam cross section.
The concrete is in compression and the steel is in tension – this is the best application
for each material.
2. References
Australian Standard AS2327.1-1996 Composite structures Part 1: Simply supported
beams
3. Shear connectors
Shear connectors are an essential component of composite beams.
They can be
• welded stud shear connectors (most common)
• welded channels
• high strength bolts
complete shear connection - the strength of the beam cross section is not limited by the
strength of the shear connection
partial shear connection - the strength of the shear connection limits the section
capacity
5. Design issues
1. ultimate flexural strength of a cross section - complete shear connection
failure is by yielding of the steel beam in tension, or crushing the compression
concrete
2. ultimate flexural strength of a cross section - partial shear connection
failure occurs in the shear connection
3. longitudinal shear failure within the slab
a failure plane develops within the slab
4. ultimate shear strength of a cross section
design for shear as a plain steel beam (ignore the concrete)
Department of Civil Engineering, Monash University
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Unit CIV3221 : Building structures and technology 56
Lecture notes
5. deflection
this is complex, because of the deflection caused by the wet concrete on the bare
steel beam. and because of creep and shrinkage effects
6. vibration
must be checked for composite beams
bcf
Dc
b1
bsf1 section AA
A A
span
this width Lef
of slab is
considered
effective
area A
VAy y
τ=
Ib
so, the shear force per unit length along the shear surface,
VAy
τb =
I
For example, if
τb = 200 kN/m
and if,
we use welded shear studs with dbs = 19 mm and fuc = 410 MPa
f’c = 32 MPa, Ec = 28,600 MPa
then fvs = 93 kN.
We must apply a capacity reduction factor φ = 0.85 to the shear stud strength, fvs., which
reduces it to 79 kN per stud.
So the horizontal spacing of the studs along the beam needs to be,
This uses an ultimate strength approach. The horizontal force that can be transferred at
any cross section equals the total strength of the shear connection between that cross
section and the free end of the beam.
So if the number of studs between a cross section and the end is n, the horizontal force
that can be transferred at that cross section,
F = n × φ × f vs × k n
018
.
where φ = 0.85, and k n = 118
. − allows for the fact that the reliability of
n
the connection increases with increasing number of studs.
Using this method, the studs can be equally spaced if the beam has a UDL.
NA in slab dc
Fcc
dsr
decking
Fst
rib
If the neutral axis in is the slab, the steel beam is all yielded, but not all of the concrete
has crushed.
Fcc = Fst
where,
Fst = A sf y
Fcc = 0.85f c′ bcf × d c
so solve for dc.
d
φM bc = 0.9 × Fcc d sr − c
2
potential longitudinal
shear failure surface
transverse reinforcing to
prevent longitudinal
shear failure
1. Section Classification
Three typical load deflection curves are shown in the figure below. Three types of
sections are used in AS4100 to classify the behaviour, namely the compact section, non-
compact section and slender section.
Ideal behaviour
Mp
My
Compact
Non-Compact including
plastic
PL
M0 = design
4
Slender
The compact section must not only develop a moment resistance equal to the plastic
capacity of the member but maintain this resistance through relatively large inelastic
deformations. This will enable the complete structure to redistribute bending moments
and reach the load-carrying capacity anticipated on the basis of a plastic analysis. The
width-to-thickness ratio below which a certain rotation capacity (R defined in the figure
below) can be achieved is called the plasticity width-to-thickness limit. The
corresponding plate slenderness is called the plate plasticity slenderness limit (λep).
M
Mp
1.0
K
Kp
The width-to-thickness limits for non-compact sections are less restrictive than those for
compact sections. The plates composing the cross-section should be capable of allowing
the member to develop a moment resistance equal to the yield moment (My ), and in this
condition the stress in the extreme fibre will be equal to the yield stress fy . In general the
plate will behave elastically at this stage although some deterioration due to the large
compressive residual stresses in the flange tips may be expected.
For slender sections the plates composing the section is not capable to achieve the yield
moment. The width-to-thickness ratio beyond which the yield moment can not be
achieved is called the yield width-to-thickness limit. The corresponding plate
slenderness is called the plate yield slenderness limit (λey).
The concept of compact, non-compact and slender sections are adopted in the
Australian Standard for Steel Structures AS4100-1998 and the American Institute of
Steel Construction LRFD Specification –1993. A similar approach has been adopted in
Eurocode 3 –1992, British Standard BS5950 Part 1- 1990, Japanese Standard AIJ-1990
and the Canadian Standard CSA-S16.1-M89 where a section can be classified from
class 1 to class 4, ie. from a plastic (deformation) section to a slender (buckling) section.
Compact or Class 1 sections can form a plastic hinge with the rotation capacity required
for plastic design. Slender or Class 4 sections can not reach first yield moment due to
local buckling effect. Non-compact sections include Class 2 and Class 3. A Class 2
section can develop the fully plastic moment, but have limited rotation capacity. A
Class 3 section can reach the first yield moment, but local buckling prevents the
development of the fully plastic moment.
2. Section Capacity
Ms = Φ Ze fy
Capacity Factor Φ = 0.90
fy = Yield Stress
Ze = Effective Section Modulus
λ sy − λ s
Ze = Z + ( ) ⋅ ( Z c − Z)
λ sy − λ sp
where Zc = Min{S , 1.5Z}, Z is the elastic section modulus, S is the plastic section
modulus.
Ze Compact
Non-compact
Zc
Slender
Z
λs p λs y λs
λsy can be determined by using stub column tests (see Lecture notes on Steel Columns).
λsp can be determined by using pure bending tests. Plot rotation capacity (R) versus
slenderness λs .
Example 1.
A rotation requirement of R = 4 is adopted in developing the b/t limits in AS4100 for
compact sections. Table 1 gives results of a series of tests carried out for C350 SHS
(square hollow sections) under pure bending. Determine the b/t limit for compact
section for C350 SHS based on the test results. What is the plate slenderness limit (λp )
corresponding to the experimentally determined b/t limit? The yield stress is taken as
350 MPa.
b σy
[Hint: plate slenderness λ = ( ) ⋅ ]
t 250
Table 1Test Results
b/t Rotation (b/t) limit
capacity
(R)
29.1 2.6 40
24.7 8.5 30
b/t ratio
22.2 9.5
23.9 7.2 20
23.4 7.0
10
19.2 11.0
22.7 8.4 0
21.8 7.8 0 2 4 6 8 10 12 14 16
27.3 3.8
Rotation Capacity (R)
b σy 350
Solution: for R=4, lower bound (b/t)limit = 25, λ p = ( ) lim it ⋅ = 25 ⋅ = 30
t 250 250
3. Effective Length
See Section 8.3 and Section 8.4 of CIV 2222 Steel Framed Structures Lecture Notes
Le = k t ⋅ k l ⋅ k r ⋅ L
4. Member Capacity
4.1 Behaviour
It has been assumed so far that the strength of the beam is dependent on the local
buckling of its plate elements. In most cases this assumption is valid. However, if the
beam is laterally unsupported the strength may be governed instead by lateral buckling
of the complete member.
A plot of the relationship between the applied moment (M) and the resulting mid-span
deflection (∆) for a member of length (L) is shown in the figure below. The member,
shown in the insert of the figure, is subjected to end moments producing a uniform
bending moment distribution over the length of the span. Lateral supports are assumed
to be present at the ends of the member so that the laterally unbraced length is equal to
the span.
Mp
My
C A B
M M M
D
∆
L
At low values of M, the member will respond elastically. However, as the moment is
increased yielding will occur due to the strains produced by the applied moment and the
residual strains in the cross-section. Further increase in the applied moment will result
in general yielding over the cross-section as the moment approaches Mp . The movement
of the cross-section during the loading process can be shown in the figure below. As the
member is loaded the cross-section movers vertically from its initial position. At some
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 66
Lecture notes
stage of loading, however, the cross-section may twist and bend about its weak axis, ie.
lateral buckling has occurred. Lateral buckling may occur at any stage during the
loading history, e.g. after the member has reached Mp as shown by curves A and B
above, between My and Mp as shown by curve C and even at moments below My as
shown by curve D. The lateral buckling capacity of the member depends on its unbraced
length and on a variety of cross-sectional properties.
Position
before
loading
Position Position
before after
buckling buckling
4.2 Capacity
The resistance of the member to lateral bending depends on the weak axis bending
stiffness of the cross-section (EIy ). The resistance to a twisting motion can be broken
into two portions. One portion is termed the St Venant resistance and is a function of the
stiffness term, GJ where G represents the shear modulus or the modulus of torsional
regidity and J is the St Venant torsional constant for the section. The second portion of
resistance to twisting is the warping resistance that is developed by cross-bending of the
flanges. For rectangular hollow sections the warping of the section may be ignored.
If full lateral restraint (FLR) is provided to a beam the member capacity of the beam is
the same as the section capacity. The length below which the section capacity can be
achieved is called FLR (Full Lateral Restraint) length.
The calculation of FLR length has been described in 8.4 of CIV 2222 Steel Framed
Structures Lecture Notes. If full lateral restraint is not provided to a beam there is an
interaction of yielding and buckling. To account the interaction in the presence of
residual stresses and geometric imperfections, AS4100 uses a beam design curve which
combines the beam section moment capacity (Ms) with the elastic buckling moment
(Mo ). The beam design curve is expressed in the form of a slender reduction factor (α s)
given by
Ms Ms
α s = 0.6{ ( )2 + 3 − ( )}
M oa M oa
Mb = Φ α m α s Ms
where α m is called the Moment Modification Factor that accounts for non-uniform
moment as given in Table 5.6.1 of AS4100 for segments fully or partially restrained at
both ends and Table 5.6.2 of AS4100 for segments unrestrained at one end. Moa is the
Elastic buckling moment given by
π 2 EI y π 2 EI w
M oa = ( )( GJ + )
L2e L2e
where
E = Young’s modulus of elasticity
G = shear modulus of elasticity
Iy = second moment of area about the cross-section minor principal y-axis
J = torsion constant for a cross-section
Iw = warping constant for a cross-section
Le = effective length of a laterally unrestrained member
Different approaches were used in determining the beam curves in different codes. Most
of the beam curves were derived from test results of I-sections. A lower bound approach
was adopted in AS4100. A summary of various beam curves are given in the figure
below. Recent research on RHS beams showed that the beam curve in AS4100 is
conservative when applied to RHS beams. An RHS under bending test is shown below.
The proposed beam curve is also shown in the figure below, with the actual expression
give by
M bx = (1056
. − 0.278 λ2 ) ⋅ M px for 0.45 ≤ λ ≤ 1.40
M px
where λ=
M yz
π
M yz = EI y GJ
L
The beam length corresponding to λ of 0.45 is called the plastic buckling length (Lp )
while the beam length corresponding to λ of 1.40 is called the elastic buckling length
(Le) for RHS beams.
1.6
RHS beams (uniform moment)
Test values (maximum)
1.4 ∆Ι )
Test values (at
Basic FEM model (PI and Trahair)
Proposed (PI and Trahair)
1.2 AS4100--1990
This report
1.0
Mbx / Mpx
0.8
0.6
0.4
0.2
0.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0
Non-Dimensional Slenderness λ
Example 2
Determine the plastic buckling length (Lp ) and elastic buckling length (Le) for C350
Cold-Formed RHS 75x25x2.5.
Solution:
Set λp = 0.45,
M px
λp =
M yz
where M yz = π EI y GJ
Lp
therefore
π EI y GJ
Lp = ⋅ λ2p
M px
when E = 200,000 MPa, Iy = 0.0487 x 106 mm4 , G = 80,000 MPa, J = 0.144 x 106 mm4 ,
Mpx = Sx fy = 10.1 x 103 mm3 x 350 MPa = 3,535 x 106 Nmm
Lp = 1906 mm
Set λe = 1.40,
M px
λe =
M yz
π
where M yz = EI y GJ
Le
therefore
π EI y GJ
Le = ⋅ λ2e
M px
when E = 200,000 MPa, Iy = 0.0487 x 106 mm4 , G = 80,000 MPa, J = 0.144 x 106 mm4 ,
Mpx = Sx fy = 10.1 x 103 mm3 x 350 MPa = 3,535 x 106 Nmm
Lp = 18,451 mm
1. Form Factor
The maximum strength of a steel column depends, to a large degree, on the member
length. Steel columns can be normally classified as short, intermediate or long
members. Each range has associated with it a characteristic type of behaviour, and
therefore different techniques must be used to assess the maximum strength. The figure
below shows schematically the relationship between the maximum strength of a column
and its length.
Significant
inelastic action Inelastic action
Material yielding not as significant
Length
A short (stub) column may be defined as a member which can resist a load equal to the
section capacity (N s). The effective cross-sectional area of a section (Ae) will be less
than the gross cross-section (Ag) for cross-sections with very slender plate elements. In
AS4100, the ratio of the effective area to the gross area of the cross-section is termed
the form factor (kf):
kf = Ae/Ag
In other words when the component plates comprising a short column are sufficiently
slender, the cross-section strength will never attain the yield capacity (Ns) due to the
onset of local buckling. The effective width (be) concept is used to approximate the
strength of a plate undergo local buckling (see Section 10.2 of CIV2222 Lecture Notes).
λ ey
b e = b( )≤b
λe
b fy
λe =
t 250
b = clear width
The effective area (Ae) used to calculate the form factor is based on the effective width
of each element, i.e.
Ae = ∑ be ⋅ t
plate_ elements
λey can be determined experimentally from the results of stub column tests as
demonstrated in Example 1, when plotting the ratio Qm (= Pult /Pyield ) versus modified
plate slenderness Sc, which is defined as:
b f y ⋅ 12 ⋅ (1 − ν )
2
b fy 250 ⋅ 12 ⋅ (1 − ν 2 ) λ e
The term Sc can be rewritten as: Sc = ⋅ =
t 250 E ⋅ π2 ⋅ k b 54
Example 1
The ratio Qm (= Pult /Pyield ) is plotted against the modified plate slenderness Sc for C350
and C450 RHS (rectangular hollow sections) based on test results. Determine the yield
slenderness λey.
Qm
k f for RHS
1.0 with b/d = 0.6
0.0
0.0 0.5 1.0 1.5 2.0
Modified Plate Slenderness (S c )
A value of Sc = 0.74 is obtained for the transition, which produces a limit of λey = 40.
2. Section Capacity
The section capacity is defined as
Ns = Φ kf An fy
where Capacity Factor Φ = 0.90, kf is the Form Factor, fy is the Yield Stress and An is
the net area of cross-section. Section 6.2.1. of AS4100 states that for sections with
penetrations or unfilled holes that reduce the section area by less than 100[1-
fy /(0.85fu)]%, the gross area Ag may be used in lieu of the net area An .
All of the BHP 300PLUS UC sections and the Grade 300 WC sections have been
specially tailored so that the gross area is effective in compression (i.e. kf = 1). On the
other hand, the deep web of UB sections means that, for the majority of the BHP-
300PLUS UB sections and all the Grade 300 WB sections, the web is not fully effective
and thus kf < 1.0.
3. Member Capacity
3.1 Behaviour
For longer columns, failure is accompanied by a rapid increase in the lateral deflection.
If the member is extremely slender, the load at which this increased deflection takes
place is not sufficient to significantly yield the member. Thus the maximum load is not
a function of the material strength but rather depends on the bending stiffness of the
member (EI), and its length (L). The failure of a long column may occur at stress levels
below that required to initiate local buckling of slender plate elements. The load
deflection relationship for a long column is schematically shown below.
f y = E εy
N
f =
A
E ( ε y− ε r ) ( F i r s t y i e l d )
fom
Straight
column
Initially crooked
column
u0 u (L/2)
Columns falling into the intermediate range are more complex to analyse but also are
the most common in steel structures. For intermediate length columns, failure is also
characterised by a rapid increase in the lateral deflection, but only after some portions of
the column cross-section have yielded and local buckling of slender plate elements has
occurred. Yielding is initiated first in those portions of the cross-section which have
large compressive residual stresses. The failure in this case is called inelastic instability
and the maximum strength of the column depends not only on the bending stiffness and
length but also on the yield stress of the steel, the distribution of residual stress over the
cross-section, the cross-section slenderness, and the magnitude of the initial
imperfections in columns and component plates of the cross-section. The load deflection
relationship for an intermediate column is schematically shown below.
f y = E εy
N Straight
f =
A column
fom
E( ε y − ε r ) (First yield)
Imperfect
column
u0 u (L/2)
3.2 Capacity
( N s / N om ) + 1 + η 1 2 ⋅ ( N s / N om ) 2
N c = N s ⋅[ ]⋅ 1 − ⋅[ ]
2 ⋅ ( N s / N om ) ( N s / N om ) ( N s / N om ) + 1 + η
where Nom is the elastic buckling load of column and η is the “imperfection parameter”.
The above expression does not consider the influence of residual stresses on the column
strength and behaviour. Since residual stresses are in effect another kind of
imperfection, a simple way of considering them is to empirically adjust the imperfection
parameter η so that the strength prediction shown above is in reasonable agreement with
test results. In principle, this is the approach used in AS4100 since the column strength
equations used therein are based on the Perry-Robertson equation.
AS4100 also places limitations on the actual permissible initial imperfection for
columns (<L/1000 or 3mm). For very slender members, the maximum load carrying
capacity is not greatly reduced by the presence of initial geometric imperfections and
residual stresses. However, for columns of intermediate length the situation is more
serious as it is in this region where the sensitivity to imperfection is greatest. Since the
initial strains corresponding to bending are trigger by initial imperfections, the
imperfection of these two variables (residual strain pattern and magnitude of initial
imperfection) results in a wide scatter in column strengths for intermediate columns.
Nc = Φ α c Ν s = Φ α ckf An fy
where Capacity Factor Φ = 0.90, kf is the form factor, An is the net area of cross-
section, fy is the Yield Stress and α c is the member slenderness reduction factor.
L fy
λn = e ⋅ k f ⋅
r 250
Le = Effective Length of Column
r = rx or ry (Radius of Gyration)
Member Section Constant α b are given in Table 6.3.3 of AS4100, some of the values
are summarised below.
The design provision of AS4100 can be shown in the figure below where the member
compressive strength (N c) divided by the section compressive strength (N s) is plotted
against the slenderness ratio of the member. It can be seen that multiple columns are
used corresponding to different values of α b. The provisions depicted in the figure are
based on the assumption that failure will involve bending about one of the major axes of
the cross-section. This will be the axis associated with the larger slenderness ratio, i.e.
the larger of (Lex/rx ) and (Ley/ry ). In sections having only one axis of symmetry, or in
sections with no axes of symmetry, the possibility also exists that failure will be
accompanied by both bending and twisting of the cross-section and may occur at a
reduced load. Buckling of this kind is called flexural-torsional buckling, and for sections
which may buckle in this manner, the compressive strength should be based on a
consideration of the actual failure modes.
Dimensionless nominal axial force capacity N c / N y
1.0
αb = -1
-0.5
0.8
0
0.5
1
0.6
0.4
0.2
0
0 50 100 150 200 250 300
Example 2
Member capacity
Compare the member capacities of columns with length of 3m, pin - ended, cold -
formed (stress - relieved ) tubes:
C350 (350 MPa) SHS 100x100x3
C350 (450 MPa) SHS 100x100x3
C350 (350 MPa) SHS 100x100x6
C350 (350 MPa) SHS 100x100x6
Solutions:
B − 2t 100 − 2 × 3
b/t = = = 31.33 <33.8
t 3
∴kf =1.0
(b/t)limit = 29.8
B − 2t 100 − 2 × 3
b /t = = = 31.3 > 29.8
t 3
b σy
λe = = 42
t 250
λ ey 40
k f 〈1.0, be = b = 94( ) = 89.5 mm
λe 42
4be t + 4t 2 1110
kf = = = 0.95
4bt + 4t 2 1164
le σy 3000 450
λn = ( ) kf = 0.95 = 99.8
ry 250 39.3 250
α c = 0.60
Comparison
N 450
100 x 100 x3 (350MPa) 100 x 100 x 3 (450MPa)
N 350
Ns 399kN 487kN 22% difference
αc 0.737 0.60
Nc 294kN 292kN -0.7% difference
N 450
100 x 100 x 6 (350MPa) 100 x 100 x 6 (450MPa)
N 350
1. Combined Actions
The following loading and behaviour scenarios are possible for beam-columns:
N*
M *x
M *y
N*
N*
M *y
M *y
N*
N*
M *y
M *x
N*
2. Section Capacity
a) Symbols Used
Actions:
N* = Axial force
M* x , M* y = Bending moment about x, y axes
Capacities:
Ns = Column section capacity
Msx, Msy = Beam section capacity about x, y axes
Reduced capacities:
Mrx , Mry = Reduced beam section capacity
b) Format
The design rules in AS4100 for members subjected to combined bending moments and
axial forces have a two tier approach. For section capacity rules, simple linear
interaction formulae are specified. However, for doubly symmetric compact I-sections
and cold-formed square hollow sections, more advanced interaction rules are specified
as higher tiers in AS4100.
• General formula
N*
M* x ≤ΦMrx = Φ M sx (1 − )
ΦNs
• General formula
N*
M* y ≤ ΦMry = ΦMsy (1 − )
ΦN s
• For special I-sections
N* 2
M* y ≤ ΦMry = Φ1.19 Msy [1 − ( ) ]
ΦN s
• For compact RHS
N*
M y ≤ ΦMry = Φ1.18 Msy (1 −
*
)
ΦN s
• General formula
N* M x* M *y
+ + ≤1
ΦN s Φ M sx ΦM sy
• For special I-sections and compact RHS
M *x r M*y r
( ) +( ) ≤1
Φ M rx Φ M ry
where
N*
γ = 1.4 + ≤ 2.0
φN s
Example 1
1. Determine the design major (x) axis section moment capacity of a 200UC52.2 of
Grade 250 steel which has a design axial compression force of N* = 112 kN.
2. Determine the design minor (y) axis section moment capacity of a 200UC52.2 of
Grade 250 steel which has a design axial compression force of N* = 112 kN
Given:
fy = 250 MPa, capacity factor φ = 0.9, form factor kf = 1.0, compact section
cross-section area An = 6640 mm2 , plastic section modulus Sx = 568,000 mm3 , Sy =
261,000 mm3
Solution:
N*
φMrx = Φ M sx (1 − ) = 0.9 x 142 x [1 – 112/(0.9x1660)] = 118 kNm
ΦNs
Ns = 1660 kN
Ns = 1660 kN
N*
φMry = ΦMsy (1 − ) =0.9 x 65.3 x [1 – 112/(0.9x1660)] = 54.4 kNm
ΦN s
Ns =1660 kN
N* 2
φMry= Φ1.19 Msy [1 − ( ) ] =0.9 x 1.19 x 65.3 x [1–(112/(0.9x1660))2 ]=69.5kNm
ΦN s
φMry =φMsy = 0.9 x 65.3 = 58.8 kNm
3. Member Capacity
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 83
Lecture notes
a) Symbols Used
Actions:
N* = Axial force
M* x , M* y = Bending moment about x, y axes
Capacities:
Mbx = Beam member capacity about x axis
Mbxo = Mbx with α m = 1.0
Ncx = Column member capacity about principal (x) axis with ke = 1.0
Ncy = Column member capacity about y axis
Reduced capacities:
Mi = Reduced in-plane member moment capacity
Mox = Reduced out-of-plane member moment capacity
Formulae are given for the following three cases: beams with FLR, beams without FLR
and beams under biaxial bending.
• General formula
N*
M * x ≤ ΦM i = ΦM sx (1 − )
Φ N cx
• For special I-sections and compact RHS
1 + βm 3 N* 1 + βm 3 N*
M* x ≤ ΦMi = Φ M sx {[1 − ( ) ](1 − ) + 118
. ( ) 1− }
2 ΦN cx 2 φN cx
• General formula
N*
M *x ≤ ΦM ox = Φ M bx 1 −
φN cy
• For special I-sections
N* N*
M ≤ ΦM ox = Φα bc M bxo
*
(1 − )(1 − )
x
φN cy φN oz
GJ + π 2 EI w / L2z
N oz = = Elastic torsional buckling capacity
(I x + I y ) / A
1 1 − βm 1 + βm 3 N*
= +( ) ( 0.4 − 0.23 )
α bc 2 2 φN cy
where
Example 2
Check the in-plane member capacity of the 200UC52.2 beam-column of Grade 250
steel, which is subjected to a combined moment Mx * of 105 kNm and a compression
force of 112 kN. Assume FLR is provided.
Given:
fy = 250 MPa, capacity factor φ = 0.9, form factor kf = 1.0, compact section.
Column length L = 5000 mm, effective length factor ke = 1.0, radius of gyration ry = 89
mm, α b = 0
Solution:
N* 112
Φ M i = Φ M sx (1 − ) = 0.9 ⋅ 142 ⋅ (1 − ) = 116 kNm
Φ N cx 1255
λn = 56
φNc = 1255 kN
1 + βm 3 N* 1 + βm 3 N*
φMi = Φ M sx {[1 − ( ) ](1 − ) + 118
. ( ) 1− }
2 ΦN cx 2 φN cx
112
= 0.9 ⋅ 142 0 + 118
. ⋅ 1− = 144 kNm, φMi ≤φMrx = 139 kNm (from Example 1).
1255
1. Type of Connections
AS4100 allows three forms of construction which relates to the bahaviour of the
connections. It then requires that the design of the connections be such that the structure
is capable of resisting all design actions, calculated by assuming that the connections are
appropriate to the form of construction of the structure or structural part. The design of
the connections is to be consistent with the form of construction assumed. The three
forms of construction are:
(1) Rigid construction – For rigid construction, the connections are assumed to have
sufficient rigidity to hold the original angles between the members unchanged. The joint
deformations must be such that they have no significant influence on the distribution of
the action effects nor on the overall deformation of the frame.
(2) Semi-rigid construction – For semi-rigid construction, the connections may not have
sufficient rigidity to hold the original angles between the members unchanged, but are
required to have the capacity to furnish dependable and known degree of flexural
restraint. The relationship between the degree of flexural restraint and the level of the
load effects is required to be established by methods based on test results. This is
outside the scope of CIV3221.
(3) Simple construction – For simple construction, the connections at the ends of
members are assumed not to develop bending moments. Connections between members
in simple construction must be capable of deforming to provide the required rotation at
the connection. The connections are required to not develop a level of restraining
bending moment which adversely affects any part of the structure. The rotation capacity
of the connection must be provided by the detailing of the connection and must have
been demonstrated experimentally. The connection is then required to be considered as
subject to reaction shear forces acting at an eccentricity appropriate to the connection
detailing.
Typical types of connections for simple construction (also called flexible connections)
are:
• Angle seat
• Bearing pad
• Flexible end plate
• Angle cleat
• Web side plate
• Stiff seat
• Bracing cleat
Typical types of connections for rigid construction (also called rigid connections) are:
This section will only deal with flexible end plate connection (shear connection) and
moment connection (bolted end plate)
Typical flexible end plate shear connections are shown in the figure below. Fabrication
of this type of connection requires close control in cutting the beam to length and
adequate consideration must be given to squaring the beam ends such that both end
plates are parallel and the effect of beam camber does not result in out-of-square end
plates that makes erection and field fit-up difficult. Shims may be required on runs of
beams in a line in order to compensate for mill and shop tolerance. The use of this
connection for two sided beam-to-beam connections should be considered carefully.
Installation of bolts in the end plates can cause difficulties in this case. When unequal
sized beams are used, special coping of the bottom flange of the smaller beam may be
required to prevent it fouling the bolts. Since the end plate is intended to behave
flexibly, damage of the end plate during transport is not normally of concern and may
be rectified on site. For coped beams, the top of the end plate and the bottom of the cope
cut should coincide. Curvature of the end plate due to welding can usually be pulled out
when installing the bolts. Check end plate component width to ensure that it will fit
between fillets of column section when connecting to column web.
Cope length L c
ti (if present)
a ei
Sp
Sp
Sp
a ei
t w
uncoped and single web coped beams (end plate located towards top of beam)
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 88
Lecture notes
L c (If Coped)
t w
ae i a
Sp
Sp
Sp
ae i
ti
uncoped and single web coped beams (end plate located towards bottom of beam)
ti Lc
a ei
a
Sp
t wc
Sp
t w
Sp
Sp
aei
The acting force in the connection is shear force V*. Possible failure modes include
weld failure along the web, bolt failure in shear, end plate component failure in shear,
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 89
Lecture notes
beam web failure in shear at end plate, coped section failure in shear near connection
and coped section failure in bending near connection.
Design is based on determining Vdes, the design capacity of the connection, which is the
minimum of the design capacity {Va, Vb, Vc, Vd, Ve, Vf}. The design requirement is
then Vdes ≥ V*. Each of the design capacity is given below.
The design capacity of the weld of the end plate to the beam web (Va) is based on the
assumption of vertical shear only on the fillet weld.
The design capacity of the bolts in the end plate (Vb) is based on the assumption of
vertical shear acting at the bolt group centroid. Possible failure modes of bolt shear,
local bearing failure and end plate tearout are considered. For economy, either 4.6/S or
8.8/S bolting category is preferred. 8.8/TB category is uneconomic since it has the same
design capacity as 8.8/S and requires tensioning. The use of 8.8/TF bolting category in
this connection is not recommended since 8.8/TF is designed on a "no-slip" basis. While
this may be desirable in certain restricted instances in order to maintain beam levels, it
also rstricts the horizontal slipping of the end plate, which is an inherent part of the
connection's "flexible" behaviour. This may result in the development of high levels of
restraint moment at the support.
The design capacity of the end plate in shear (Vc) assumes that failure, if it occurs, takes
place on each side of the weld/web interface and that it occurs by shear yielding.
(d) Beam web in shear at end plate Vd = φ (0.6 fyw twb di)
where φ = 0.9
The expression for the shear capacity of the web at the end plate/web interface (Vd) has
been derived by assuming that a near uniform stress distribution applies at the interface
and that therefore, the nominal capacity is given by Clauses 5.11.2 and 5.11.4 of
AS4100.
The expression for Ve and Vf deal with the strength of the section remaining after
coping of the supported member.
The following additional design considerations are worth noting. Flexible end plate
connections will exhibit a wide range of connection flexibility depending on the
connection parameters such as plate thickness, plate depth, bolt category, web thickness.
Rotational flexibility in the connection is required if the connection is to meet the
requirements of AS4100 for simple construction. This is provided by the use of a
relatively thin end plate which deforms out of plane under applied rotation, the use of
snug-tightened bolts which allows the end plate to slip horizontally and detailing a wide
gauge between lines of bolts.
Example 1
See section 4.3.3 of AISC DSC/04-1994 Hogan, T.J. and Thomas, I.R. (1994), Design
of Structural Connections, 4th Edition, Australian Institute of Steel Construction,
Sydney, page 58 to page 59
Typical bolted end plate moment connections are shown in the figure below.
Fabrication of this type of connection requires close control in cutting the beam to
length and adequate consideration must be given to squaring the beam ends such that
end plates at each end are parallel and the effect of any beam camber does not result in
out-of-square end plates which make erection and field fit-up difficult. Shims may be
required to compensate for mill and shop tolerances. 8.8/T (fully tensioned) bolt
category can be used. Holes are 2mm larger than the nominal bolt diameter.
OR
OR
OR
OR
OR
OR
end plate at Knee joint with or without haunch in rigid portal frame
-incoming member inclined to column
or cruciform
The design action effects at the connection could be determined from either a first order
elastic analysis with moment amplification or a second order elastic analysis. Applied
actions at a connection are assumed to be bending moment M*, shear force V* and axial
force N* as shown in the figure below.
N *tm t fb
V*
V *v c N*
x db
M*
N *c m
θ= 0
φ= 90
Type A
y t fb
N *t m db
V* φ N*
x
M*
V *v c
N *c m
Type B
t fb
y
db
N *tm x
V* M *
N*
φ
θ N *c m
V *v c
Type C
(i) The flanges transmit design flange forces due to moment M*, these comprising Ntm *
(tension flange) and Ncm* (compression flange). For the design of the flange and web
welds, the assumption is made that the proportion of the bending moment transmitted
by the web is kmw while the proportion of the bending moment transmitted by the flange
is (1-kmw). The proportion of the bending moment transmitted by the web is given by
lw
k mw =
lw + l f
However for the assessment of the loads on the bolts, and on the end plate and for the
assessment of the necessity for stiffeners and the design of the stiffeners, it is
conventional practice to assume that all the force above and below the neutral axis is
Department of Civil Engineering, Monash University
(File: Name CIV3221 Edition Date: 6:2002)
Unit CIV3221 : Building structures and technology 94
Lecture notes
concentrated at the flanges which is equivalent to assuming that all the bending moment
is transmitted through the flange area.
For the design of bolts and the end plate, it is assumed that the flanges transmitted all of
the design axial force N*, the proportion taken by each being proportional to their
contribution to the total section area. This assumption is made because the bolts, which
must transmit the axial force into the column, are concentrated at the flanges.
• For the flange welds connecting the beam to the end plate:
M*
(1 − k mw ) + N * At
( )
N*fw =
d f − t fb A
Where,
kmw = proportion of the bending moment transmitted by the web (see attached)
At = area of tension flange, A = total area of the section
• For the web welds connecting the beam to the end plate:
Where,
kw = (area of the web)/(total cross-sectional area), kmw as defined above.
M* N*
Total design force in tension flange: N *ft = +
( d b − t fb ) 2
M* N*
Total design force in compression flange: N *fc = ( d − t ) − 2
b fb
Clause 9.1.4 of AS4100 requires that this type of connection be designed for the
following minimum design actions:
Shear force = 40 kN
Axial force = 0
The intention of the AS4100 provision is that connections have a guaranteed minimum
design capacity with some inherent robustness.
Possible failure modes include flange welds failure, web welds failure, bolts failure, end
plate failure and stiffeners failure.
Design capacities are given for each possible failure mode in the connection.
• Flange welds
Check Nfw* < φ Nw = φ ffy bfb tfb for full penetration butt welds
where φ = 0.9 for SP (structural purpose) weld, ffy = yield stress of beam flange, bbf =
width of beam flange, tfb = beam flange thickness.
• Web welds
z + v y ≤ φv w
v*2 *2
Check
N *w 3M *w
v *z = + 2
2Lw Lw
V*
v = *
y
2Lw
where v is strength per unit length, Lw = weld length along web = db – 2tfb
• Bolts
N *ft
Check ≤ φN tf
nt
Vvc*
Check ≤ φVfx orφVfn
n
where nt is the number of bolts in tension, n is the number of bolts in shear, Nt f, Vfx and
Vfn are capacities for a single bolt.
• End plate
Flexure
φ ⋅ f yi ⋅ b i ⋅ t 2i
Check N *ft ≤
a fe
where fyi = yield stress of end plate, bi = width of end plate, ti = end plate thickness, afe
= distance between the bolt center and the top flange of the beam.
Shear
• Stiffeners
N*ft d − t fb
If > b , stiffeners are required.
φ ⋅ f yc ⋅ t wc 2
where fyc = yield stress of column, twc = thickness of column web, db = depth of beam,
tfb = thickness of beam flange.
Example 2
See section 4.8.4 of AISC DSC/04-1994 Hogan, T.J. and Thomas, I.R. (1994), Design
of Structural Connections, 4th Edition, Australian Institute of Steel Construction,
Sydney, page 115 to page 118