CHAPTER 3 DESIGN OF MULTISTOREY FRAMES Summary:
The traditional approaches to the design of frames are concisely described: - continuous framing with rigid joints and /or simple framing with pinned joints, - the so called “wind moment” method - the “partial strength” approach - rigid-plastic design.
The modern approach to frame design, i.e. semi-continuous framing using semi-rigid joints, is then outlined; how it is to be distinguished from the traditional approaches is explained and the potential benefits (scientific and economic) for its use are raised.
A consistent design process is described in which joint behaviour is accounted for in the global analysis from the outset. It is shown how to modify the simple framing or continuous framing approaches to be more in line with a consistent design process. It is explained that a consistent design process can take different forms which depend on the assumptions about joint behaviour in the global analysis, who is responsible for joint design and/or the degree of collaboration between the parties (designer and fabricator).
Design practices are identified which show how design responsibility is, or can be, shared. It is explained that an understanding of the sharing of design responsibility is essential in order to modify current practice so as to allow a consistent design process to be used.
Objectives: The student should: Understand the different approaches to frame design, both traditional and modern. Have an appreciation of the potential benefits of using a consistent design process which
best accounts for joint behaviour. Understand the consequences that the sharing of design responsibility may have on the
subsequent design. Know how to put a consistent design process into practice. References: [1] Anderson, D., Reading, S.J., The Wind-moment design for unbraced frames, SCI publication P-082, 1991. [2] Anderson D., Colson A., Jaspart J.-P., Connection and frame design for economy, ECCS/TC10 publication N°, 1991 (also published in a number of national journals). [3] Maquoi, R., Chabrolin B., Frame design including joint behaviour, Report EUR 18563 EN, ECSC/European Commission, 1998. [4] ENV 1993-1-1: Design of steel structures: Part 1-1: General rules and rules for buildings. [5] BS5950:Part1 Structural steelwork design of buildings, British Standard Institute.
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CHAPTER 3 DESIGN OF MULTISTOREY FRAMES
1 CLASSIFICATION OF BUILDING FRAMES
For building frame design, it is useful to define various frame systems in order to
simplify models of analysis. For example, in the case of a braced frame, it is not necessary to
separate frame and bracing behavior since both can be analyzed with a single model. On the
other hand, for more complicated three-dimensional structures involving the interaction of
different structural systems, simple models are useful for preliminary design and for checking
computer results. These models should be able to capture the behavior of individual subframes
and their effects on the overall structures.
The remaining of this section attempt to describe what a framed system represents,
define when a framed system can be considered to be braced by another system, what is meant
by a bracing system, and the difference between sway and non-sway frames. Various structural
schemes for tall building construction are also given.
1.1 Rigid Frames
A rigid frame derives its lateral stiffness mainly from the bending rigidity of frame
members inter-connected by rigid joints. The joints shall be designed in such a manner that they
have adequate strength and stiffness and negligible deformation. The deformation must be small
enough to have any significant influence on the distribution of internal forces and moments in
the structure or on the overall frame deformation.
A rigid unbraced frame should be capable of resisting lateral loads without relying on
additional bracing system for stability. The frame, by itself, has to resist all the design forces,
including gravity as well as lateral forces. At the same time, it should have adequate lateral
stiffness against side sway when it is subjected to horizontal wind or earthquake loads. Even
though the detailing of the rigid connections results in a less economic structure, rigid unbraced
frame systems have the following benefits:
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1. rigid connections are more ductile and therefore the structure performs better in load reversal
situation or in earthquakes.
2. from the architectural and functional points of view, it can be advantageous not to have any
triangulated bracing systems or solid wall systems in the building.
1.2 Simple Frames (Pin-connected Frames)
A simple frame referred to a structural system in which the beams and columns are
pinned connected and the system is not capable of resisting any lateral loads. The stability of
the entire structure must be provided by attaching the simple frame to some forms of bracing
systems. The lateral loads are resisted by the bracing systems while the gravity loads are
resisted by both the simple frame and the bracing system.
In most cases, the lateral load response of the bracing system is sufficiently small such
that second-order effects may be neglected for the design of the frames. Thus the simple frames
that are attached to the bracing system may be classified as non-sway frames. Figure 1.1 shows
the principal components - simply frame and bracing system - of such a structure.
There are several reasons of adopting pinned connections in the design of steel
multistorey frames:
1. pin-jointed frames are easier to fabricate and erect. For steel structures, it is more
convenient to join the webs of the members without connecting the flanges.
2. bolted connections are preferred than welded connections, which normally require weld
inspection, weather protection and surface preparation.
3. it is easier to design and analyze a building structure that can be separated into system
resisting vertical loads and system resisting horizontal loads. For example, if all the girders
are simply supported between the columns, the sizing of the simply supported girders and
the columns is a straight-forward task.
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4. it is more cost effective to reduce the horizontal drift by means of bracing systems added to
the simple framing than to use unbraced frame systems with rigid connections.
Actual connections in structures do not always fall within the categories of pinned or
rigid connections. Practical connections are semi-rigid in nature and therefore the pinned and
rigid conditions are only idealizations. Modern design codes such as Eurocode 3 allow the
design of semi-rigid frames using the concept of wind moment design. In wind moment design
the connection is assumed to be capable of transmitting only part of the bending moments (those
due to the wind only). Recent development in the analysis and design of semi-rigid frames can
be obtained from Chen Goto and Liew. (1996). Design guidance is given in Eurocode 3 (1992).
1.3 Bracing Systems
Bracing systems refer to structures that can provide lateral stability to the overall
framework. It may be in the forms of triangulated frames, shear wall/cores, or rigid-jointed
frames. It is common to find bracing systems represented as shown in Figure 1.2. They are
normally located in buildings to accommodate lift shafts and staircases.
In steel structures, it is common to represent a bracing system by a triangulated truss
because, unlike concrete structures where all the joints are naturally continuous, the most
immediate way of making connections between steel members is to hinge one member to the
other. As a result, common steel building structures are designed to have bracing systems in
order to provide side-sway resistance. Therefore bracing can only be obtained by use of
triangulated trusses (Figure 1.2a) or, exceptionally, by a very stiff structure such as shear wall or
core wall (Figure1.2b). The efficiency of a building to resist lateral forces depends on the
location and the types of the bracing systems employed, and the presence or otherwise of shear
walls and cores around lift shafts and stair wells.
1.4 Braced Frames Versus Unbraced Frames
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The main function of a bracing system is to resist lateral forces. Building frame systems
can be separated into vertical load-resistance and horizontal load-resistance systems. In some
cases the vertical load-resistance system also has some capability to resist horizontal forces. It
is necessary, therefore, to identify the two sources of resistance and to compare their behavior
with respect to the horizontal actions. However, this identification is not that obvious since the
bracing is integral within the structure. Some assumptions need to be made in order to define
the two structures for the purpose of comparison.
Figures 1.3 and 1.4 represent the structures that are easy to define, within one system,
two sub-assemblies identifying the bracing system and the system to be braced. For the
structure shown in Figure1.3, there is a clear separation of functions in which the gravity loads
are resist by the hinged subassembly (Frame B) and the horizontal load loads are resisted by the
braced assembly (Frame A). In contrast, for the structure in Fig. 1.4, since the second sub-
assembly (Frame B) is able to resist horizontal actions as well as vertical actions, it is necessary
to assume that practically all the horizontal actions are carried by the first sub-assembly (Frame
A) in order to define this system as braced.
Eurocode 3 (EC3, 1992) gives a clear guidance in defining braced and unbraced frames.
A frame may be classified as braced if its sway resistance is supplied by a bracing system in
which its response to lateral loads is sufficiently stiff for it to be acceptably accurate to assume
all horizontal loads are resisted by the bracing system. The frame can be classified as braced if
the bracing system reduces its horizontal displacement by at least 80 percent.
For the frame shown in Fig. 1.3, the hinged frame (Frame B) has no lateral stiffness, and
Frame A (truss frame) resist all lateral load. In this case, Frame B is considered to be braced by
Frame A. For the frame shown in Fig. 1.4, Frame B may be considered to be a braced frame if
the following deflection criterion is satisfied:
1 0 8
A
B
. (3.1.1)
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5
where
A = lateral deflection calculated from the truss frame (Frame A) alone.
B = lateral deflection calculated from Frame B alone.
Alternatively, the lateral stiffness of frame A under the applied lateral load should be at
least five times larger than that of Frame B:
KA 5 KB (3.1.2)
where
KA = lateral stiffness of Frame A
KB = lateral stiffness of Frame B
The classification of braced and unbraced frame is illustrated in Fig. 1.6.
1.5 Sway Frames Versus Non-Sway Frames
The identification of sway frames and non-sway frames in a building is useful for
evaluating safety of structures against instability. In the design of multi-storey building frame, it
is convenient to isolate the columns from the frame and treat the stability of columns and the
stability of frames as independent problems. For a column in a braced frame it is assumed the
columns are restricted at their ends from horizontal displacements and therefore are only
subjected to end moments and axial loads as transferred from the frame. It is then assumed that
the frame, possibly by means of a bracing system, satisfies global stability checks and that the
global stability of the frame does not affect the column behavior. This gives the commonly
assumed non-sway frame. The design of columns in non-sway frame follows the conventional
interaction equation check approach, and the column effective length may be evaluated based on
the column end restraint conditions. Another reason for defining "sway" and "non-sway
frames" is the need to adopt conventional analysis in which all the internal forces are computed
on the basis of the undeformed geometry of the structure. This assumption is valid if second-
order effects are negligible. When there is an interaction between overall frame stability and
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column stability, it is not possible to isolate the column. The column and the frame have to act
interactively in a "sway" mode. The design of sway frames has to consider the frame
subassemblage or the structure as a whole.
On the basis of the above considerations, a definition can be established for sway and
non-sway frames as: “a frame can be classified as non-sway if its response to in-plane
horizontal forces is sufficiently stiff for it to be acceptably accurate to neglect any additional
internal forces or moments arising from horizontal displacements of its nodes.”
1.6 Sway Frame Classification Based on British Code
A structural frame may be classed as “non-sway” if its sway deformation is sufficiently small for
the resulting secondary forces and moments to be negligible. For clad structures, provided that
the stiffening effect of masonry infill wall panels or diaphragms of profiled steel sheeting is not
explicitly taken into account, this may be assumed to be satisfied if the sway mode elastic critical
load factor cr of the frame, under vertical load only, satisfies:
cr 10
In all other cases the structure or frame should be classed as “sway-sensitive”.
cr may be calculated based on
cr = 200
h
British Code: BS5950:Part 1(2000) provides a procedure to distinguish between sway and
nonsway frames as follows:
1) Apply a set of notional horizontal loads to the frame. These notional forces are to be
taken as 0.5% of the factored dead plus vertical imposed loads and are applied in
isolation, i.e., without the simultaneous application of actual vertical or horizontal
loading.
2) Carry out a linear elastic analysis and evaluate the individual relative sway deflection
for each storey.
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3) If the actual frame is claded but the analysis is carried out on the bare frame, then in
recognition of the fact that the cladding will substantially reduce deflections, the
condition is reflected and the frame may be considered to be non-sway if
h
2000 where h = storey height
for every storey.
4) All frames not complying with the criteria in (3) are considered to be sway frames.
The classification of sway and nonsway frame is illustrated in Fig. 1.7.
1.7 Classification of Tall Building Frames
A tall building is defined uniquely as a building whose structure creates different
conditions in its design, construction, and use than those for common buildings. From the
structural engineer view point, the selection of appropriate structural systems for tall buildings
must satisfy two important criteria: strength and stiffness. The structural system must be
adequate to resist lateral and gravity loads that cause horizontal shear deformation and
overturning deformation. Other important issues that must be considered in planning the
structural schemes and layout are the requirements for architectural details, building services,
vertical transportation, and fire safety, among others. The efficiency of a structural system is
measured in term of their ability to resist higher lateral load which increases with the height of
the frame (Iyengar et al., 1992). A building can be considered as tall when the effect of lateral
loads are reflected in the design. Lateral deflections of tall building should be limited to prevent
damage to both structural and non-structural elements. The accelerations at the top of the of the
building during frequent windstorms should be kept within acceptable limits to minimize
discomfort to the occupants .
Figure 1.5 shows a chart which defines, in general, the limits to which a particular
system can be used efficiently for multi-storey building projects. The various structural systems
in Fig. 1.5 can be broadly classified into two main types: (1) medium-height buildings with
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shear-type deformation predominant and (2) high-rise cantilever structures such as framed tubes,
diagonal tubes and braced trusses. This classification of system forms is based primarily on
their relative effectiveness in resisting lateral loads. At one end of the spectrum in Fig. 1.5 is the
moment resisting frames, which are efficient for buildings of 20 to 30 stories, and at the other
end is the tubular systems with high cantilever efficiency. Other systems were placed with the
idea that the application of any particular form is economical only over a limited range of
building heights.
An attempt has been made to develop a rigorous methodology for the cataloguing of tall
buildings with respect to their structural systems (Council On Tall Buildings and Urban Habitat,
1995). The classification scheme involves four levels of framing division: primary framing
system, bracing subsystem, floor framing, and configuration and load transfer. While any
cataloguing scheme must address the pre-eminent focus on lateral load resistance, the load-
carrying function of the tall building subsystems is rarely independent. An efficient high-rise
system must engage vertical gravity load resisting elements in the lateral load subsystem in
order to reduce the overall structural premium for resisting lateral loads.
1.8 Lateral load Resistance System
Lateral loads produce transverse shears, over turning moments and lateral sway. The stiffness
and strength demands on the lateral system increase dramatically with height. The shear
increases linearly, the overturning moment as a second power and sway as a fourth power of the
height of the building. Therefore, apart from providing the strength to resist lateral shear and
overturning moments, the dominant design consideration (especially for tall building) is to
develop adequate lateral stiffness to control sway. Several types of lateral systems are described
in the reference by Liew Balendra and Chen (1997).
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Figure 1.1 Braced Multistorey Frame
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Figure 1.2 Common Bracing Systems (a) Vertical Truss System (b) Shear Wall
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Figure 2.4 Mixed frames split into two subassemblies
Figure 1.3 Pinned connected frames splited into two subassemblies
Figure 1.4 Mixed frames splited into two subassemblies
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Figure 1.5 Categorisation of Tall Building Systems.
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Frame is non sway if i < hi /2000 Figure 1.7 Classification of sway frame by BS5950:Part1:2000
A
B
FFrraammee BB iiss bbrraacceedd bbyy FFrraammee AA iiff 8.01B
A
Fig. 1.6 Classification of braced and unbraced frame
1
2
3
4
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Example: Classification of sway frames
The frame shown in Figure 1 consists of steel beams and columns arranged on a 7 m x 7 m grid. The frame has been designed on the basis of "Simple Design" according to BS 5950-1:2000, clause 2.1.2.2. Resistance to sway is provided by two 3.5 m braced bays, one on each 49 m side, as shown in Figure 1. In practice, bracing would also be required parallel to the 28 m side, but this is not considered in this example. Once constructed, the frame will be clad to form an office block, but the stiffening effect of the cladding has not been taken into account in the analysis of the frame. Determine whether the frame is "non-sway" or "sway sensitive" according to BS 5950-1 :2000[1] and, if necessary , calculate the amplification factor kamp.
Fig. 1
Unfactored roof and floor loads Roof: Dead load Wdr = 3.5 kN/m2 Imposed load Wir = 1.0 kN/m2 Floor: Dead load Wdf = 3.5 kN/m2 Imposed load Wif = 6.0 kN/m2
Factored roof and floor loads Consider the following three load combinations: (1) 1.4 dead + 1.6 imposed (2) 1.0 dead + 1.4 wind (dead load resisting overturning due to wind) (3) 1.2 dead + 1.2 imposed + 1.2 wind Gravity loads for load combination 1 Roof: wr' = (3.5 X 1.4) + (1.0 X 1.6) = 6.5 kN/m2 Floor: wf' = (3.5 x 1.4) + (6.0 x 1.6) = 14.5 kN/m2 Gravity loads for load combination 2 Roof: wr' = 3.5 X 1.0 = 3.5 kN/m2 Floor: wf' = 3.5 x 1.0 = 3.5 kN/m2
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Gravity loads for load combination 3 Roof: wr' = (3.5 X 1.2) + (1.0 X 1.2) = 5.4 kN/m2 Floor: wf' = (3.5 x 1.2) + (6.0 x 1.2) = 11.4 kN/m2 Member sizes Roof beam 305 x 127 x 37 UB in grade 8275 Floor beam 406 x 178 x 60 UB in grade 8275 Ground to 2nd floor columns 203 x 203 x 60 UC in grade 8275 2nd floor to roof columns 203 x 203 x 46 UC in grade 8275 Bracing 168.3 x 6.3 CH8 in grade 8275 Sway Stability The sway stability of the structure is assessed by performing an elastic analysis on one of the braced bays, under the action of the notional horizontal forces (NHF), according to the rules in clause 2.4.2 of B8 5950-1:2000[1]. The notional horizontal forces are applied as horizontal point loads at every roof and floor level and are taken as 0.5% of the total factored dead + imposed loads for that level. In this example, since the stability is provided by two braced bays, the notional horizontal forces applied to one bracing system should be taken as half the value calculated for the whole floor or roof. The greatest notional horizontal forces occur in load combination 1 and this case should generally be used when assessing sway stability .However, advantage may be taken of the lower notional horizontal forces in load combinations 2 and 3, if desired. All three combinations are considered below. 1 Load combination 1 (Dead + Imposed) Roof level NHF = 0.005 x 0.5 x 28 x 49 x 6.5 = 22.3 kN Floor level NHF = 0.005 x 0.5 x 28 x 49 x 14.5 = 49.7 kN The result of an elastic analysis on one braced bay (bare frame only) under the action of the notional horizontal forces is shown in Figure 2.
Ground – 1st floor cr
35005.15
200 3.4
2nd floor – 3rd floor cr
30005.36
200 2.8
Therefore, cr 5.15
For clad frames in which the stiffening effect of the cladding has been ignored, the amplifier kamp is given by
3-16
16
5.115.1k
cr
cramp
but kamp 1.0
16.15.115.515.1
15.5kamp
Therefore the forces in the bracing system must be increased by 16% for load combination 1. 2 Load combination 2 (Dead + Wind) Roof level NHF = 0.005 x 0.5 x 28 x 49 x 3.5 = 12 kN Floor level NHF = 0.005 x 0.5 x 28 x 49 x 3.5 = 12 kN The result of an elastic analysis on one braced bay (bare frame only) under the action of the notional horizontal forces is shown in Figure 3.
Fig. 3
Ground – 1st floor 5.170.1200
3500cr
2nd floor – 3rd floor 7.169.0200
3000cr
Therefore, 7.16cr
Since 10cr , the frame is classed as "non-sway" and there is no need to amplify the forces in
the bracing system for load combination 2.
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3 Load combination 3 (Dead + Wind + Imposed) Roof level NHF = 0.005 x 0.5 x 28 x 49 x 5.4 = 18.5 kN Floor level NHF = 0.005 x 0.5 x 28 x 49 x 11.4 = 39.1 kN The result of an elastic analysis on one braced bay (bare frame only) under the action of the notional horizontal forces is shown in Figure 4
Fig. 4
Ground – 1st floor 5.170.1200
3500cr
2nd floor – 3rd floor 82.62.2200
3000cr
Therefore, 82.6cr
Since 10cr , the frame is classified as "sway sensitive" for load combination 3.
For clad frames in which the stiffening effect of the cladding has been ignored, the amplifier
09.15.1)48.615.1(
48.6
5.115.1k
cr
cramp
Therefore, the forces in the bracing system must be increased by 9 % for load combination3.
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1.9 Design Methods for Buildings Three types of design methods are mentioned in BS5950: (1) Simple Design (or Simple Construction). (a) Connections are assumed not to develop significant moments adversely affecting
either the members or the structure as a whole. (b) The beams may be designed as simply supported. (c) The columns are designed to carry axial loads as well as nominal moments from the
reaction shear of the beam, applied at the appropriate eccentricity. (d) Columns must be fully continuous. (e) It is assumed that side sway due to horizontal loading is prevented by inserting bracing
or by utilising shear walls, lift or staircase closures, acting together with shear resistance of the floor slab.
(2) Rigid Design (or Continuous Construction) (a) The connections are assumed to be capable of developing the strength and/or stiffness
required by an analysis assuming full continuity. (b) A full frame analysis may be made using either elastic or plastic methods. The column
design should be carried out based on the full frame analysis. (c) When a plastic analysis is used, the connections should be at least equal in strength and
rigidity to the connected members, and only plastic (Class 1 ) sections should be used. (3) Semi-Rigid Design Practical joints are capable of transmitting some moment and the method takes this partial fixity into consideration. A simpler approach is to assume the restraint moment equal to 10% the free moment in the beam, although other options are also available. In practice, structures are designed to either simple or rigid methods. Semi-rigid design has
seldom found favour with designers.
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3.2 DESIGN FOR SIMPLE CONSTRUCTION
1.0 Introduction In multi-storey construction it is usual to use the same column size and weight through at least 3 storeys to avoid the cost of splicing the columns. The columns are therefore continuous throughout a number of levels However The basis of “Simple Construction” is to assume that that the structure is composed of members connected by nominally pinned joints with resistance to horizontal forces being provided by bracing, shear walls or a lift core. The floor acts as a diaphragm to distribute horizontal load. See Figure 1.
Lift Shaft or stair well
Figure 1 Nominally pinned structure with bracing. This clearly makes the design of beams very much easier as bending moments and shear forces can be found for each beam as though they were simply supported. Design of the columns however is another issue. It is tempting to think that the columns could be designed for pure axial loads but this is not the case. A special set of rules exists in BS 5950 Part 1 for continuous columns in simple construction. The beams at the very least are usually connected on the column face producing some eccentricity of loading and in addition the connection will transfer some moment to the structure however flexible the end connection is. This moment will not affect the design of the beams but it will affect the column. . In order to ensure that the connections behave in an approximation to a pin, care has to be taken to ensure that the connections are not too rigid, and this is generally easily achieved by the use of a minimum number of bolts and flexible plates or angles. See Figure 2. Such connections are designed to carry shear only. 2.0 Section Classification In most cases columns in simple construction will be subject to predominantly axial loading and therefore the most efficient section will be a universal column section. These sections are on the whole not subject to local buckling when S275 steel is used and it is usually sufficient to check that the b/T of the flange is less than 10 and the d/t of the web is less than 40.
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(a) Web Cleats (b) End Plate (c) Fin Plates Figure 2. Typical steel beam to column connections 3.0 Column moments As noted above the connection of the beams to the columns will not only generate a moment in the column due to the eccentricity of the connection but will also generate moments due to the stiffness of the connection. In order to allow for this, the concept of "nominal eccentricities" is introduced These eccentricities are given in clause 4.7.7 and require that for a beam connected to the face of the column the eccentricity should be taken as the actual eccentricity plus 100mm. This eccentricity then generates a nominal moment equal to the beam reaction multiplied by the total eccentricity. (See Figure 3)
Moment = R x (D/2 + 100mm)
R
100mm
D
t/2
Moment = R x (t/2 + 100mm)
R
100mm
t
D/2
D is the depth of the column t is the thickness of the web
Figure 3 Calculation of Notional Moments 3.1 Design Moment
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Having calculated the nominal moment at the a particular level the moment in the column should then be calculated. This is done by assuming that the moment from a beam at any one level will be distributed up and down the column in proportion to the stiffness (I / L) of the columns above and below that level. In order to simplify matters even further, when the ratio of the stiffness is less than 1.5 the code allows these moments to be shared equally. For example consider the column in Figure 4
Figure 4 Column with varying stiffness The stiffness of the column between levels 1 and 2 is equal to I/7 and between levels 2 and 3 is equal to I/3.5 the ratio of the stiffness is therefore
5.125.3
7
7/
5.3/
I
IRatio The moment must therefore be distributed in proportion to the
stiffness
5.3/7/
7/
21
121 II
IMM if I1=I2 then
5.3/17/1
7/121 MM
3
1
7/27/1
7/121 MMM
Similarly
3
232 MM
Note that the shortest (and therefore stiffest) length carries the largest moment. The nominal moments should be assumed to have no effect at the levels above and below the level at which they are applied.
3.5m
7m
Level 3
Level 2
Level 1
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4.0 Effective Length The effective length for calculating the compressive resistance pc will depend on the degree of restraint provided by the incoming beams and the connections. The code leaves this decision to the designer simply giving a table (table 22) to determine what effective lengths to use when the degree of restraint has been decided. Typically the values used are 0.85L or 1.0L depending on the size of the beams connecting to the column where L is the distance between floor levels. Beams which carry more than 90% of their full moment capacity should be taken as being unable to provide rotational restraint and the effective length should therefore be taken as 1.0L
6.6 Slenderness The slenderness of the column should be calculated from the appropriate effective length divided by the radius of gyration. To calculate the compressive strength pc this will be either
x = L
Ex / r
x or
y =L
Ey / r
y :noting
that the effective length about the major axis could be longer than that about the minor axis if the member is restrained on the y-y axis by tie beams, cladding supports etc. Care should always be taken to calculate the value of p
c for both slenderness when the slenderness are almost equal,
as major axis (x-x) buckling is checked using table 24(b) and minor axis (y-y) buckling is checked using table 24(c). To calculate the lateral torsional buckling strength pb the slenderness used is always that about the minor axis. In the case of simple construction clause 4.7.7 allows the value of p
b to be
calculated from LT
= 0.5 L / ry rather than the more complex and accurate value of
LT = u v .
This simpler expression takes into account the values of u, and v as well as the effective length and the shape of the moment diagram. The value of p
b is then obtained from Table 16 in the
usual way. The value of the buckling resistance moment is calculated from Mbs=pbSx. For circular or square hollow sections or rectangular hollow sections where lateral torsional buckling does not take place (i.e. those within the limits of Table 15 then Mbs=Mc 6.0 Interaction Equation for Columns in Simple Frames The single interaction expression is used as follows:
1yy
y
bs
x
c
c
Zp
M
M
M
P
F
Where Fc is the compression due to axial load Pc is the compressive strength M
x is the nominal moment about the major axis
Mbs
is the buckling resistance moment for simple columns M
y is the nominal moment about the minor axis
py is the design strength
Zy is the elastic modulus about the minor axis
It is not necessary to carry out local capacity check.
7.0 Loading and forces
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The loading applied to a structure designed on the basis of simple construction is, except for one important item, the same on any other structure and use should be made of BS 6399 and the load factors in BS 5950 Part1. The single major difference for a structure designed using "simple construction" rather than "continuous construction" is that all the beams may be taken as fully loaded and patterns of loading, to determine the worst bending effects in the column need not be taken into account Theoretically there is no reason why this loading should be neglected in simple construction and despite the fact that the code allows it to be ignored, it would be unwise to do so in situations where pattern loading was bound to occur as part of the function of the structure.(e.g. stacking of paper in a warehouse where one bay was intended to be left empty while the adjacent one is filled). In common with all structures the notional horizontal forces given in clause 2.4.2.4 should be applied at every roof and floor level in combinations which do not contain wind loads. These forces allow for eccentricity of vertical loading due to out of straightness and lack of plumb of the columns etc Generally this load will be less than the wind load but for long narrow structures it may be more severe when considering forces along the length of the building
8.0 Design Summary
1. Determine which section to check – usually the lowest in the continuous run 2. Calculate the maximum beam reactions due to both dead and imposed load at either end
of the length to be checked. 3. Calculate the factored axial load within the length due to all loads. 4. Calculate the moments applied to the column from the beams i.e. M=(e+100mm) 5. Distribute the moments in proportion to the stiffness of the column above and below that
level. If the ration of the stiffness < 1.5 then distribute the moment equally 6. Select the maximum moments about each axis 7. Determine effective lengths for bending and axial load 8. Calculate a slenderness for compression. Usually = LE/ry 9. Determine a compression strength pc and the compressive strength Pc =Agpc 10. Calculate a slenderness for lateral torsional buckling LT = 0.5LE/ry 11. Determine the lateral torsional buckling strength pb and the lateral torsional buckling
resistance moment Mbs
12. Check the length under consideration using 1yy
y
bs
x
c
c
Zp
M
M
M
P
F
9.0 EXAMPLE Check the adequacy of the column shown between levels 1 and 2.
3-24
24
Assumptions
The column size is 203 x 203 x 46 in S275 steel.
The column is continuous and forms part of a structure of simple construction.
The column is effectively pinned at the base.
Beams are fixed to the column by end plates.
Beams connected to the flanges are substantial beams carrying the floor load and are not loaded to more than 90% of their capacity. (Assume an effective length about the x-x axis of 0.85).
Beams connected to the web are small beams carrying lightweight cladding only. (Assume an effective length about the y-y axis of 1.0L).
The axial load shown is derived from level 3 and any subsequent levels but no moments are transmitted down the column to level 2. The load includes self weight of the column above level 2. (Assume self weight between levels 1 and 2 is negligible).
4.7.7
Assumed factored loads
FA = 410 kN
R1 = 40 kN (including self weight)
R2 = 160 kN (including self weight)
R3 = 30 kN (including self weight)
Solution
Axial loading
5000
3000
Level 3
Level 2
Level 1
203
x 20
3 x
46G
rade
S27
5
F ~ axial load from above(including self weight)
AR1
R3
R2
R1
R2
3-25
25
The total axial load
Between levels 1 and 2
Fc = F
A + R
1 + R
2 + R
3 + SW
SW = 1.4 x 5 x 46.1/102 = 3.22 kN F
c = 410 + 40 + 160 + 30 + 3.2 = 643 kN
Between levels 2 and 3
Fc = 410 kN (given)
Moments
Mx = R
2 (D/2 + 100) = 160 310/100
2
2.203
= 32.3 kNm
4.7.7
My = R
1
100
2100
2 2
tR
t
=
100
221
tRR
My = 310/100
2
2.73040
= 1.04 kNm
Stiffness of length 1 to 2 = EI/5
Stiffness of length 2 to 3 = EI/3
The ratio of these = 7.13
5
This is greater than 1.5 so the moments must be distributed in proportion to the stiffness.
4.7.7
Therefore, xxx MMEIEI
EIM 38.0
35
5/)12(
= 12.3 kNm
xxx MMEIEI
EIM 62.0
35
3/)32(
= 20.0 kNm
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26
The moment is distributed in proportion to the stiffness, the shorter (stiffer) member taking the greatest moment.
20
32.3
12.25
Similarly, M
y(2-1) = 0.38 M
y = 0.40 kNm
= 0.62 M
y = 0.64 kNm
Try 203 x 203 x 46 UC (S275 steel)
This size must be estimated from experience.
Depth of section D = 203.2 mm Width of section B = 203.6 mm Web thickness t = 7.2 mm Flange thickness T = 11.0 mm Flange breadth/thickness b/T = 9.25 Web depth/thickness d/t = 22.3 Radius of gyration (major) r
x = 8.82 cm
Radius of gyration (minor) ry = 5.13 cm
Elastic modulus (minor) Zy = 152 cm3
Plastic modulus (major) Sx = 497 cm3
Self weight sw = 46.1 kg/m
Depth of web d = 160.8 mm Cross-sectional area A
g = 58.7 cm2
Design Strength
Flange thickness T = 11 mm < 16 mm Therefore, p
y = 275 N/mm2
Table 9
3-27
27
Section Classification
yp
275 0.1
275
275
Flange b/t = 9.25 Limit for Class 1 section = 9 Limit for Class 2 section = 10 The flange is therefore Class 2
Table 11
Web d/t = 22.3
Limit for a Class 1 section = 11
80
r
r1 = 1
yw
c
pdt
F
Table 11
3.5.5
The highest value of Fc is at the base
F
c = 643 kN
r1
= 02.22752.78.160
10643 3
since r
1 1 Therefore, r
1 = 1
Limit = 11
80
r
= 4011
180
web d/t = 22.3 < 40 Therefore, web class 1
Note that with a UC this will nearly always be the case. It would be conservative to take the section subject to full compression, i.e. r
1 = r
2 = r
3 = 1.0, in which case the limit becomes
40 for all classes of section.
The flange is Class 2 and the web is Class 1. The section must therefore be classified as Class 2.
Column Length Level 1 to Level 2
The requirement is that, over this length between the restraints at Level 1 and Level 2, that:
1Zp
M
M
M
P
F
yy
y
bs
x
c
c
Compressive Resistance (Pc)
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28
For Class 1 or Class 2 sections, Pc = A
g p
c. The effective length about the major axis
LEx
= 0.85 L = 0.85 x 5.0m = 4.250m
4.7.4Table 22
The slenderness about the major axis x = 1.48
82.8
425
The compressive strength for = 48.1 and p
y = 275 N/mm2 from Table 24(b)
pcx
= 239 N/mm2 P
cx = 58.7 x 102 x 239 x 10-3 = 1402 kN
Table 23T.24b
The effective length about the minor axis:
LEy
= 1.0L = 1.0 x 5.0m = 5.0m
Table 22
The slenderness about the minor axis
y = 5.97
13.5
500
The compressive strength for = 97.5 and p
y = 275 N/mm2 from Table 24(c)
pcy
= 129 N/mm2 P
cy = 58.7 x 102 x 129 x 10-3 = 757 kN
Table 23
It is clear that, in this case, the compressive resistance about the minor axis (Pcy
) is much less than that about the major axis (P
cx) and the value of Pc to be used in the
interaction expression will be that of Pcy
. This will usually be the case unless the minor axis is restrained at intervals between Level 1 and 2.
Lateral Torsional Buckling (Mbs
)
For a Class 2 section M
b = p
b S
x
LT
= 0.5 L/ry = 7.48
13.5
5005.0
4.3.6.4
4.7.7
For LT
= 48.7 and py = 275 N/mm2
pb = 241 N/mm2 Table 16
Mb = p
b S
x = 241 x 497 x 10-3
= 120 kN
3-29
29
yy
y
bs
x
c
c
Zp
M
M
M
P
F =
3
6
10152275
1039.0
120
25.12
757
643
= 0.85 + 0.1 + 0.009 = 0.96
Therefore Section 1-2 adequate.
Thus the lower section of the column which carries the lowest moment and the highest axial load is adequate.
The column between Levels 2 and 3 has a much lower slenderness and axial load than the length between levels 1 and 2. Therefore, P
c and M
bs will be larger. A
check on this length will usually only be necessary if:
1) The distribution of moments at Level 2 is very uneven, and as a result of this, the upper column attracts more moment.
or
2) The moments generated from the reactions at Level 3 are relatively large, compared to those from the reactions at Level 2.
It can be assumed that neither is true in this case and the section is adequate.
Check Member Capacity using Design Tables For 203 x 203 x 46 UC (S275 steel) – see Table 265 Lex = 4.25m Pcx = 1405kN (hand calculation gives 1402 kN) Ley = 5.0 m Pcy = 760 kN (hand calculation gives 757 kN) For LLT = 5m, Mbs = 120 kNm (hand calculation gives 120kNm)
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30
10.0 Example Figure (a) shows an interior frame of a multi-storey steel building to be designed using simple construction method. The frame is braced against out-of-plane and in-plane sway at each storey level. The floor comprises only primary beams, with flooring and roofing spanning as shown in Fig. (b). The materials used for all steel sections are Grade 43 steel.
Based on the unfactored loading data given below, a) Design the first-storey floor beam and check its ultimate strength and
serviceability limit states. b) Design the columns between the ground and first-storey level, and check their
overall buckling resistance at the factored loading. The column base may be assumed to be pinned, and same column size is to be used for the entire frame.
Floor Beams Uniformly distributed dead Load = 4.5 kN/m2 x 6 m
Uniformly distributed imposed Load = 5.0 kN/m2 x 6 m Roof Beam Uniformly distributed dead Load = 4.0 kN/m2 x 6 m Uniformly distributed imposed Load = 1.5 kN/m2 x 6 m
Figure (a) Figure (b)
6m
4m
4m
4m
roof beam
floor beam
floor beam
floor beam
5m
Lateral Braces
Plane Frame
Pinned
First-storeyColumn
Plane Frame
6m
6m
6m
3-31
31
First-Storey Floor Beam: Assume full lateral restraint Dead load = 4.5 x 6 = 27kN/m Imposed Load = 5 x 6 = 30kN/m Dead Plus Imposed Loading Design ULS: W = 1.4 x 27 + 1.6 x 30 = 85.8kN/m
MWL
KNm
2
8
85 5 36
8386 1
..
FWL
kNv 2
515
2257 5.
Try 457 x 191 x 67 UB S275 Steel Use Design Table Page 196 & 197
kNm405M cx > 386kNm
P kNv 636 > 257.5kN Section is Plastic Unfactored imposed load is 30kN/m Deflection at centre of the beam (assuming simply supported)
5 30 6000
384 205000 29400 108 4
4
4
. mm= Span/ 714 < L/360 OK
Loading on Roof beam Deal load = 6 x 4 = 24 kN/m Imposed Load = 6 x 1.5 = 9 kN/m Dead Plus Imposed Loading Design ULS: W = 1.4 x 24 + 1.6 x 9 = 48kN/m Column design Vertical Load due to roof beam = 48 x 6 / 2 = 144kN
Vertical Load due to each floor beam = FWL
kNv 2
515
2257 5.
Design Axial force at the lower floor column F = 144 + 3 x 257.5 = 916.5kN Effective length L L m mEx Ey 1 0 5 5 0. . (Conservative)
Try 203 x 203 x 60 UC S275 steel Use Design Table Page 263
kN1010Pcy for Ley = 5m
Moment due to eccentricity M kNmx [ . ( . / )] .275 5 209 6 2 100 10 56 43
3-32
32
Design moment distributed to the column below the first storey level
Column stiffness ratio I
I
/
/. .
4
51 25 1 5
M kNmx 56 4
228 2
..
My 0
To find Mb
kNm159M bs for L = 5m
Check the Equation
095.1177.0918.00159
2.28
1010
5.916
Z
M
M
M
A
F
yy
y
bs
x
cg
NG
Try 203 x 203 x 71 UC
kN1190Pcy for Ley = 5m
Moment due to eccentricity M kNmx [ . ( . / )] .275 5 215 8 2 100 10 57 33 Design moment distributed to the column below the first storey level
Column stiffness ratio I
I
/
/. .
4
51 25 1 5
M kNmx 57 3
228 6
..
My 0
To find Mbs kNm198M bs for L = 5m
Check the Equation
918.015.0768.00198
6.28
1190
5.916
Z
M
M
M
A
F
yy
y
bs
x
cg
OK
3-33
33
Homework 4: Simple Construction and Simple Frames 1 Check the overall buckling capacity of the columns shown in the Figure 1a. The columns are made of 203 x 203 x 60 UC in Grade 43 steel. The loads acting on the members are factored loading as shown in Figure 5.1 (a)-(c) (Guide to BS5950, Vol 2, 1986).
6000
3000
1
2
3
(a)
Figure 1
R13 = 60kN
R33 = 60kN
R23 = 162kN
Factored self weight = 28kN (b)At joint 3
R12 = 202kN
R32 = 202kN
R22 = 362kN
Factored Axial Load at the centroid=338kN (c)At joint 2
Pinned
Assumptions: (i) the column is effectively continuous and forms part of a structure of simple construction. (ii) The column is effectively "pinned" at the base. (iii) Connections from R12, R22, R32 and R23 offer partially restraint to the column, whilst the restraints from connections at R21 and R22 can be assumed as pinned. (iv) Moments on the column from the beams are due to nominal eccentricity and not due to partial fixity. 2 A column is 3.5m long and it supports loads as shown in Figure 5.2 and enumerated below: (a) 157.9KN applied axially; derived from construction above. (b) 226.6kN at an eccentricity of 100mm + 1/2 depth of column. (c) 71.4kN (applied twice) at an eccentricity of 100mm + 1/2 web thickness. The two loads in (c) are also 94mm eccentric relative to x-x axis. Check the overall buckling resistance of a 203 x 203 x 46 UC assuming it is effectively held in position at both ends but not restrained in direction at either end (Leech, 1988).
3-34
34
94mm71.4kN
71.4kNBeam 1
Beam 2
Beam 3
157.9kN 226.6kN
Figure 2
x
x
y y
3 A multi-storey building frame has same column size of 254 x 254 x 89UC and constant storey height of 4m. The column under consideration is a typical corner column in the multi-storey building. It has a substantial base so that it is effectively held in both position and direction but the top of the column is effectively held in position only. The loading conditions are shown in Figure 3 and are enumerated below: (a) 1500kN concentric load transmitted from the above and producing an axial force acting at the centroid of the cross section. (b) 200kN reaction from Beam 2 which carries the floor and external wall loads. These loads produce an axial force, and bending moments about both axes in the column. (c) 60kN reaction from Beam 1 which carries the external wall only. This, too, produces bending about both axes. Check the overall buckling resistance of the column (Leech, 1988), assuming Grade 43 steel.
60kNBeam 1
Beam 21500kN
200kN
Figure 3
94mm
94mm
UC 254 x 254 x 89 UC
3-35
35
Q4 Figure Q4 shows an interior frame of a five-storey steel building to be designed using simple construction method. The frame is supporting a water tank - the dead load of which can be simulated by two concrentrated load of 50kN each applied at the top of the frame as shown in Fig. Q4. The frame is braced against out-of-plane and in-plane sway at each storey level. The column between the ground floor and the first floor is braced out-of-plane at the mid-height. The floor comprises only primary beams with precast floor and roof slabs spanning one-way as shown in the plan view of Fig. Q4. The materials used for all steel sections are Grade 43 steel. Based on the unfactored loading data given below, design the columns A and B and the beams using UC and UB sections, and check their resistance at the factored loading. The column bases may be assumed to be pin-supported. Unfactored Loadings: Floor Slab Uniformly distributed dead Load = 4.0 kN/m2
Uniformly distributed imposed Load = 5.0 kN/m2 Roof Slab Uniformly distributed dead Load = 3.5 kN/m2 Uniformly distributed imposed Load = 1.5 kN/m2
Water Tank: Concentrated dead Load P = 50kN
Figure Q4
Plane Frame
6m
4.5m
4.5m One-way precast slab
X Out-of-plane restraints
PlanX 3m
X
P=50kN P=50kN
6m
4m
4m
4m
5m
4m
G
1
2
3
5
4
A
B
3-36
36
3 DESIGN OF CONTINUOUS FRAMES (RIGID CONSTRUCTION)
1 Introduction The majority of multi-storey frames in the UK are designed assuming that the connections are “simple” and do not transmit moments. Resistance to lateral forces is provided by bracing. The columns are designed according to the requirements of clause 4.7.7 in the code with the effective lengths for compression obtained from Table 24. Such a frame is shown in Figure 1 together with typical beam to column connections.
Cl 4.7.7 Table 24
(a) Web Cleats (b) End Plate (c) Fin Plates
Figure 1 Examples of simple design- assumes that the connections do not transmit moments
In some cases however the presence of bracing may not be acceptable, and the horizontal forces must be resisted by the stiffness of the members and the connections.
Extended end plate connection
Figure 2 Examples of continuous design - assumes the connections do transmit moments
Rigid Frame – members are connected by rigid joints
3-37
37
2. Loading
2.1 Gravity loads
In order to ensure that the worst load effects are considered it is necessary to consider both full vertical loading, factored dead plus imposed, and various patterns of imposed loading (see Figure 3). It is not necessary to change the load factor on the dead load (1.4) unless it is resisting overturning, in which case it should be taken as 1.0
Cl 5.1.2
Maximum beam span moments Maximum beam support moments
Maximum single curvature bending Maximum double curvature bendingin columns in columns
Figure 3 Pattern Loading
2.2 Notional horizontal forces In load cases which do not involve horizontal loads (i.e. load combination 1 of clause 2.4.1.2) notional horizontal forces should be applied to structures to allow for the effect of practical imperfections such as lack of verticality of columns. They should be taken as equal to 0.5% of the factored (dead plus imposed) load at each level. They should be taken as acting in any one direction at a time and should be used in combination with the full vertical loading as indicated in Figure 4 below.
Cl2.4.2.4
3-38
38
0.5% of (D+I)
0.5% of (D+I)
0.5% of (D+I)
0.5% of (D+I)
Figure 4 Notional horizontal forces
Notional horizontal forces should NOT: a) be applied when considering overturning b) be applied when considering pattern loading c) be combined with applied horizontal loads d) be combined with temperature effects e) be taken to contribute to the net reactions at the foundations.
Factored dead load
1% of DL4
1% of DL3
1% of DL2
1% of DL1
Wind load or
Wind load or
Wind load or
Wind load or
Greater of DL4
DL3
DL2
DL1
DL 1-4 are the total dead load at each floor level
Figure 5 Horizontal Forces for use with load combinations 2 and 3
Resistance to horizontal forces may be provided in a number of ways as follows: a) triangulated bracing members
b) moment resisting joints and frame action
c) cantilever columns, shear walls, staircase and lift shaft enclosures
d) or a combination of these.
3-39
39
3. Analysis 3.1 Determination of moments and forces The elastic analysis of rigid frames is usually dealt with by using one of the many commercial software packages now available for elastic analysis of frames. Most software will carry out a first order structural analysis; i.e. the software will analyse the frame in its undeflected shape. Deflections will be calculated but moments and forces will be determined ignoring any second order effects due to sway (see below). BS 5950:1 addresses the way in which the sway effects can be incorporated in design and also ensures the stability of the frame. This lecture will deal with these requirements as they appear in BS 5950:1 which are, for the most part, contained in Section 5, Annex E and Annex F.
Section 5 Annex E Annex F
3.2 In plane stability In order to understand the concept of in-plane stability as presented in BS 5950:1, it is important to consider two basic concepts of structural behaviour. The first is second order effects and the second is instability.
3.3 Second order effects Second order effects are, those effects due to sway of the frame. The sway will cause eccentricity of vertical loading, which will generate second order moments due to eccentrically applied axial load in the members. These second order moments are commonly referred to as being due to the "P-" effect; i.e. an axial load P applied at an eccentricity (see Figure 6). Two very important concepts should be recognized: P effects arise due to horizontal loading and from any effect such as lack of
verticality or out of straightness of the column, which could cause the frame to move.
P effects do not necessarily cause instability. What is required is a method of
determining whether the P effects are significant and, if so, whether or not they will they cause instability.
P
P
F
Figure 6 P- effects Figure 7 Instability
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40
3.4 Instability The concept of instability can be understood by consideration of an axially loaded vertical slender cantilever strut as shown in Figure 7. At low axial loads,P, a small disturbing force,F, will cause the cantilever to deflect by a finite amount and, when the disturbing force is removed, the cantilever will return to its original shape. As the axial load is increased, the small disturbing force will cause the cantilever to deflect by a larger amount due to increasing P- effects. Eventually the resistance to the disturbing force will vanish and the cantilever will deflect in an uncontrolled manner. The value of the load P which would cause this failure, in a member of infinite material strength and with no imperfections, is known as the elastic critical load. The ratio between this load and the factored load on the frame is known as the elastic critical load factor cr. The load at which a pin-ended member would become unstable in this manner is known as the Euler load PE =
2EI/L
2
The load at which a cantilever would become unstable is given by Pcrit = 0.25
2EI/L
2
= 2EI/4L
2 =
2EI/(2L)
2 .
Hence the effective length LE for a cantilever is 2L. For sway frames, when
cr is greater than 10 the frame is usually judged to be stable and
the second order effects can be neglected. A value of cr
less than 4.6 generally indicates a potentially unstable frame which requires careful second order analysis.
4. Continuous multi-storey frames Consider the frame shown in Figure 8. In addition to the normal linear elastic deflection two other features are potentially significant. Due to the height of the structure there will be considerable axial load in
structure which implies that stability effects should be carefully considered. As the structure distorts, the horizontal deflection will become significant and
the P effects will be large generating significant second order moments.
P
P
Figure 8 Deformation of a continuous multistory frame
3-41
41
5. Classification of frames The code deals with multi-storey frames by first classifying them as independently braced or unbraced and as sway or non-sway frames. A frame may be regarded as braced if the bracing system carries the horizontal loading and unbraced if this loading is carried by bending action in the columns. According to the classification used in BS5950, frames are regarded as non-sway if the P- effects are deemed to be small enough to be ignored and, conversely, they are regarded as sway sensitive if the P- effects are significant in the design.
Cl2.4.2.6
5.1 Independently braced frames Where sway stability is provided by an independent system of resistance to horizontal forces, it may be treated as non-sway if:
a) the stabilizing system has a spring stiffness at least four times larger than the total spring stiffness of all the frames to which it gives horizontal support (i.e. the supporting system reduces horizontal displacements by at least 80%)
and b) the stabilizing system is designed to resist all the horizontal loads applied
including the notional horizontal forces.
Figure 9 below shows the distortion of a continuous multistory frame with independent bracing system.
Cl5.1.4
Figure 9 Independently braced frame
5.2 Non-sway frames
For clad structures where the stiffening effect of masonry wall panels or other cladding is not taken into account: the frame can be classed as a non-sway frame if the elastic critical load factor of the frame in the sway mode cr, under vertical load only, satisfies cr > 10 For multi-storey frames with moment resisting joints and without sloping members cr can be taken as the smallest value of
cr = 200
h
where h is the storey height and is determined for each storey from application of the notional horizontal forces using a linear elastic analysis - as shown in figure 10.
Cl 2.4.2.6
3-42
42
Figure 10 Appliplication of notional load for frame classification 5.3 Sway frames By default, any frame which does not satisfy the above conditions is a sway frame. However, if cr is less than 4, the frame will be very sensitive to sway and the design will require a second order analysis. It is likely that the serviceability limit state for deflection may be difficult to satisfy.
6 Elastic design of continuous multi-storey frames 6.1 Design of independently braced frames
Independently braced frames as defined above and shown in Figure 9 should be designed:
to resist gravity loads (load combination 1). the non-sway mode effective length of the columns should be
obtained using Annex E. pattern loading should be used to determine the most severe moments and
forces. Sub-frames may be used to reduce the number of load cases to be considered. the stabilizing system must be designed to resist all the horizontal loads applied
including the notional horizontal forces.
Cl5.6.2
6.2 Design of non-sway frames Non-sway as defined above, and shown in Figure 11 below, should be designed:
to resist gravity loads (load combination 1) the non-sway mode effective length of the columns should be obtained using
Annex E. pattern loading should be used to determine the most severe moments and forces. sub frames may be used to reduce the number of load cases the frame should be checked for combined vertical and horizontal loads without
pattern loading.
Cl 5.6.3
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
1
2
3
4
3-43
43
Figure 11 Non-sway continuous frame
6.3 Design of sway sensitive frames Sway sensitive frames, such as is shown in Figure 12, should be designed as follows:
Check in the non-sway mode i.e. design to resist gravity loads (load combination1) as for independently braced frames without taking account of sway. (without notional horizontal forces, but with pattern loading).
Check in the sway mode for gravity load (i.e. load combination 1) plus the notional horizontal forces without any pattern loading.
Check in the sway mode for combined vertical an horizontal loads (i.e. load combinations 2 and 3), without pattern loading.
Provided that cr is greater than 4, the sway should be allowed for by using one of the following methods:
a) Effective length method. In this method, sway mode effective lengths from Figure E2 of Annex E are used for the columns.
b) Amplified sway method. The sway moments should be multiplied by the
amplification factor kamp and the internal forces adjusted to maintain equilibrium with the applied loads. In this method non-sway mode in-plane effective lengths from Figure E1 of Annex E should be used for the columns
If cr is less than 4.0 then the frame is potentially quite sensitive to stability effects and a full second order elastic analysis should be used.
Cl 5.6.4 Annex E Cl2.4.2.7 Annex E
Figure 12 Sway sensitive frame
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44
7. Amplification factor and sway moments "Sway-sensitive" frames. All structures that are not classed as "non-sway" (including those in which the stiffening effect of masonry infill wall panels or diaphragms or profiled steel sheeting is explicitly taken into account see 2.4.2.5) should be classed as "sway-sensitive". Except where plastic analysis is used, provided that cr is not less than 4.0 the secondary forces and moments should be allowed for as follows:
Cl 2.4.2.7 2.4.2.8 2.4.2.5
a) if the resistance to horizontal forces is provided by moment-resisting joints or by cantilever columns, instability effects may be allowed for either by using sway mode in-plane effective lengths (see Annex E) for the columns and designing the beams to remain elastic under the factored loads, or alternatively by using the method given in (b); b) by multiplying the sway effects by the amplification factor kamp determined from the following: 1) for clad structures, provided that the stiffening effect of masonry infill wall panels or diaphragms of profiled steel sheeting is not explicitly taken into account:
kamp = 5.115.1 cr
cr
but greater than or equal to 1.0
Annex E Cl2.4.2.8 Cl2.4.2.5
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45
2) for unclad frames, or for clad structures in which the stiffening effect of masonry infill wall panels or diaphragms of profiled steel sheeting is explicitly taken into account:
kamp = 1cr
cr
but greater than or equal to 1.0
If cr is less than 4.0, then the frame is potentially unstable and a full second order elastic analysis should be used. If plastic analysis is used, reference should be made to 5.7 for multi-storey frames. Sway effects. The distortions of a frame may be divided into two components; 1) those which arise from sway with zero rotation of every joint and 2) those which arise a result of joint rotations without any sway of the frame. The sway effects are the moments, forces and shears which arise as a result of sway deformations. In the case of a symmetrical frame, with symmetrical vertical loads, the sway effects can correctly be taken as comprising the forces and moments in the frame due to the horizontal loads. In every other case, the sway effects may be found by using one of the following alternatives:
a) Deducting the non-sway effects: 1) Analyse the frame under the actual restraint conditions. 2) Add horizontal restraints at each floor or roof level to prevent sway, then analyse the frame again. 3) Obtain the sway effects by deducting the second set of forces and moments from the first set. In 1) moments are due to sway + non-sway distortions. In 2) moments are due to non-sway distortions. In 3) moments are thus only due to sway distortions.
Cl2.4.2.5 Cl 5.7 Cl2.4.2.8
Step 1 Step 2
Sway moment = Moment in Step 1 – Moment in Step
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46
b) Direct calculation: 1) Analyse the frame with horizontal restraints added at each floor or roof level to prevent sway. 2) Reverse the directions of the horizontal reactions produced at the added horizontal restraints. 3) Apply them as loads to the otherwise unloaded frame under the actual restraint conditions. 4) Adopt the forces and moments from the second analysis (step3) as the sway effects which require magnifying by kamp.
Cl2.4.2.8
8. Annex E (The effective length method) 8.1 Background The effective length of a member is defined as that length which, if it were pin-ended, would behave as the real member with its actual end conditions. For simple columns a number of easily defined (idealised) end conditions exist. Values of effective lengths, which should be adopted in design, are given in Table 22 of the code. These are illustrated in Figure 13 of these notes. In some cases the effective lengths given differ from the theoretical values because of the difficulty of attaining perfect fixity against rotation.
None Position Position Position
Position Position Position Position
Direction Direction Direction
Direction Direction Direction
1.0 L 0.85 L 0.7 L 2.0 L 1.2 L
Position
Restraint
Restraint
Practical L E
Figure 13 Effective lengths LE for simple members
Annex E Cl 4.7.3
R3
R2
R1
Step 2: Sway Moment Step 1
R3
R2
R1
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47
In a rigid frame however, the end conditions are dependant on the stiffness of the members meeting at the joint. 8.2 Charts of BS 5950 Work by Wood
(1) has lead to the development of the column design charts, given in
Annex E of the code, which is much more accurate than the results of the simplified values shown in the table of Figure 13. It also leads to more economic designs. The approach is based on the consideration of a limited frame shown in Figure 14. This consists of the column under consideration plus all the members which frame into it at either end. A fuller description of the derivation of these charts may be found in reference 2.
For non-sway frames, the charts have been derived by considering the joint rotation at each end of the column and then calculating the elastic critical load using stability functions. For sway frames, the additional degrees of freedom allowing sway in each of the three storeys has been included. When using the charts, the values of k
1 and k2 are required. These are the sum of the
stiffnesses (I/L) of the two columns meeting at the upper and lower joint respectively divided by the sum of the stiffnesses of all the members meeting at that joint.
When using the charts: k
1 the distribution factor at the upper end of the column being designed is
k1 = (K
c + K
u) / (K
c+K
u+KTL +KTR)
k2 the distribution factor at the lower end of the column being designed is
k2 = (K
c + K
L) / (K
c+K
L+KBL +KBR)
The stiffness K for each member is taken as a function of I / L where I is the second moment of area and L is the actual length with the following provisos: For a beam which is not rigidly connected to the column K should be taken as
zero If either end of the column being designed is required to carry more than 90% of
it's moment carrying capacity ( reduced for the presence of axial load) the value of k1 or k2 as appropriate should be taken as 1.0. This is because the plasticity will be reducing the stiffness of the remaining elastic core significantly.
If a beam supports a concrete or composite floor slab, whether in a sway or a non-sway frame, its K value should be taken as I / L provided that the beam does not carry axial force other than that due to sharing wind or notional loads between the columns
For a beam which carries more than 90% of its moment capacity, a pin should be inserted at that location; once again because of the loss of stiffness due to yielding.
For other conditions, the appropriate values of K are given in Tables E1, E2 and E3 of the code. Table E1 is reproduced as Figure 14a whilst Tables E2 and E3 are summarised in Figure 14b.
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48
0.2
0.4
0.6
0.8
1.0
0.2 0.4 0.6 0.8 1.0 0
0.5
0.6
0.7
0.9 0.8
1.0
a) Non sway frames
k 2
k 1
Fixed Pinned
Pinned
0.2
0.4
0.6
0.8
1.0
0.2 0.4 0.6 0.8 1.0 0
1.0
k 2
k 1
Fixed Pinned
Pinned
1.2
1.3
1.4
1.6
2.0 3.0
1.1
b) Sway frames
K TL K TR
K u
K L
K c
K BR K BL
INFINITY
Figure 14 Limited Frame- Effective length ratios for continuous frames
Conservative formulae for the curves
The code gives conservative formulae for both of these sets of curves as follows: For non sway 2
E 1 2 1 2L / L 0.5 0.14 k k 0.055(k k )
For sway frames 0.5
1 2 1 2E
1 2 1 2
1 0.2(k k ) 0.12k kL / L
1 0.8(k k ) 0.6k k
The code also gives more exact formula, derived from the formulation of the stability functions themselves, which do not produce curves but which allow individual points to be determined for specific values of LE/L, k1, and k.
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49
8.3 Base stiffness and strength The fixity condition assumed at the base of a column has developed for historical reasons. Even today, it is common to assume a base to be either pinned or fixed. The code guidance on this topic has been clarified in the 2000 version. Designers are reminded that the behaviour is influenced by the detailing of the steelwork at the base of the column, its attachment to the concrete base, the interaction between this base and the surrounding soil, as well as the compressibility of the ground itself. The code states that, in the absence of detailed knowledge of the foundation stiffness, the following assumptions may be made.
Cl 5.1.3
8.3.1 Nominally rigid base If the column is connected to a suitable foundation then: In an elastic global analysis, the base stiffness may be assumed to be equal to the stiffness of the column in limit state calculations. When determining deflections under serviceability conditions the base may be assumed to be rigid. This recognizes the fact that the base is less stiff at high loading where yielding of some of the components including the ground may have taken place. In a plastic global analysis, the moment capacity of the base may take any value between zero and Mp of the column member provided that the foundation is designed for that moment together with any co-existing axial load. In an elastic-plastic global analysis, the assumed base stiffness should be consistent with the assumed base capacity but should not be greater than that of the column.
Cl5.1.3.2
8.3.2 Actual pinned base If the base has a pin or rocker plate its stiffness should be taken as zero.
Cl5.1.3.1
8.3.3 Nominally pinned base If the base is assumed to be pinned for the design of the foundation, then this same assumption should be made for the global analysis to calculate the distribution of forces and moments for the frame under the ultimate limit state. However, (i) when checking frame stability, or determining effective lengths, the base stiffness may be taken as 10% of the column stiffness. (ii) when calculating deflections at serviceability conditions, the base stiffness may be taken as 20% of the column stiffness.
Cl5.1.3.3
8.3.4 Nominally semi-rigid base A nominal base stiffness of up to 20% of the column stiffness may be assumed in elastic global analysis provided that, for consistency, the base is also designed for the resulting moments and forces.
Cl5.1.3.4
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50
9 Frames with partial sway bracing Annex E contains charts, similar to those already noted for non-sway and sway frames, which facilitate the determination of effective lengths for frames which are partially braced by infill wall panels (Figures E4 and E5 of the code). These two charts are for two relative stiffnessess kp of the panels to the stiffness of the bare frame equal to 1 or 2. The value of kp is given by: kp = h
2Sp/(80EKc) but 2
where: E is the elastic modulus for steel h is the storey height cis the sum of I/h for all the columns in that storey Sp is the sum of the stiffness of every panel in the storey measured as the horizontal force required to produce unit horizontal deflection. Sp is given by (0.6(h/b) ) t Ep/[1+(h/b)
2]
2
h/b is the ratio of storey height to panel width t is the panel thickness Ep
is the modulus of elasticity of the panel material For a value of kp between 0 and 1 or 1 and 2, interpolation between the effective length ratio values obtained from the charts may be used.
Annex E
10. Beam stiffness values 10.1 Beam does not support a floor slab The charts (E1 to E5) were derived using the assumption that the ends of the beams remote from the column under consideration are encastre. This is not however always the case and guidance on modified beam stiffness is given in table E2 some typical values are given below for beams which do not support a concrete or composite floor slab. for non-sway frames, where it would be expected that single curvature bending
will occur in the beams K should be taken as 0.5(I/L). For a sway frame, where the beams will be in double curvature bending, the
beam stiffness should be taken as 1.5(I/L). Further values are given in table E2
Annex E
Non S way Frame S way Frame
Figure 15 Critical buckling modes
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51
10.2 Buckled mode shape The value of the beam stiffness to be used in calculating the restraint coefficients k1 and k2 depend on both the deformed shape of the frame in the critical mode and also on the presence or otherwise of an axial load in the beam. In the sub-frames used to develop the charts it was assumed that the remote ends of all beams were prevented from rotating. Clearly this is not the case in practice and in recognition of this factors are introduced which modify the basic I/L values. The modifications required depend upon whether or not the frame supports concrete floor slabs. For frames with concrete floor slabs the factors are given in Table E1 of the code and reproduced as Figure 16 below.
Loading condition For the beam
Non-sway mode Sway mode
Beam directly supporting concrete floor or roof slab
1.0 ( I/L ) 1.0 ( I/L )
Other beams supporting direct loads
0.75 ( I/L ) 1.0 ( I/L )
Beams with end moments only
0.5 ( I/L ) 1.5 ( I/L )
Figure 16 Stiffness coefficients Kb of beams with floor slabs For frames which do not support concrete floor slabs, the factors are given in Table E2 of the code and which may be determined as the values in bold type in Figure 17. 10.3 Beams carrying axial load It is the presence of axial loads in the columns which decrease the frame stiffness. The presence of an axial load in a beam will also decrease the stiffness of that member and, if the axial load is significant, it should be taken into account when determining the value of the beams stiffness. For frames carrying concrete floor slabs axial loads due to notional horizontal forces and those arising from the sharing of wind loading may be regarded as insignificant. For other conditions Table E3 of the code may be used which is reproduced as Figure 17 below. The term in the brackets represents the reduction ratio leading to a conservative approximation for the beam stiffness. Rotational restraint at far end of beam Beam stiffness coefficient Kb Fixed at far end 1.0 (I/L ) { 1 – 0.4 (Pc/PE)} Pinned at far end 0.75 (I/L ) { 1 – 1.0 (Pc/PE)} Rotation as at near end (double curvature) 1.5 (I/L ) { 1 – 0.2 (Pc/PE)} Rotation equal and opposite to that at near end (single curvature)
0.5 (I/L ) { 1 – 1.0 (Pc/PE)}
Pc is the axial load in the beam. If Pc = 0 then the final term in the bracket is unity. Remember that PE = 2EI /L2. Figure 17 Stiffness coefficients for beams including the effects of an axial load Pc
Annex E
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52
11 Effective lengths derived using critical load For all frames, the in-plane effective lengths can be obtained from the elastic critical load factor as follows
ccrE F
EIL
2
where cr is the elastic critical load factor of the whole structure calculated using all the vertical loads on that structure including the load supported by those parts which do not contribute to the sway resistance in that plane. Fc is the axial load in the member under factored design loading. cr may be determined by computer analysis or for continuous multi-storey frames by the sway method described in appendix F of the code which states that
cr = max200
1
Annex E
where
max is the maximum value of h
Lu for each storey (see Figure 18)
u is the notional deflection at the top of the storey L is the deflection at the bottom of the storey and h is the storey height.
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
0.5%(D+I)
1
2
3
4
Figure 18 Sway under notional loads
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53
12. Member Capacity Checks The following two checks are required for columns in multi-storey continuous frames: (i) Local capacity check, and (ii) Overall buckling check. LOCAL CAPACITY CHECK
Check (1) or (2) or (3)
Tension 1cy
y
cx
x
t
t
M
M
M
M
P
F ------------------------- (1)
Compression 1cy
y
cx
x
yg
c
M
M
M
M
pA
F------------------------ (2)
Alternatively where the cross-section is class 1 or class 2
121
z
ry
y
z
rx
x
M
M
M
M -----------------------(3)
OVERALL BUCKLING Tension – check moment capacity and resistance Compression Generally: Check both (4) and (5)
1yy
yy
xy
xx
c
c
Zp
Mm
Zp
Mm
P
F ----------------------------(4)
1yy
yy
b
xLT
cy
c
Zp
Mm
M
Mm
P
F ------------------------(5)
Alternatively for I and H sections with equal flanges check (6), (7) and (8)
15.05.01
cy
yxy
cx
c
cx
xx
cx
c
M
Mm
P
F
M
Mm
P
F ------------------(6)
11
cy
c
cy
yy
b
xLT
yc
c
P
F
M
Mm
M
Mm
P
F -------------------------(7)
3-54
54
1)/1(
)/1(
)/1(
/5.01(
cyccy
cycyy
ccx
cxcxx
PFM
PFMm
PFM
PFMm
cx
--------------(8)
To include the sway effect (or second order effect), either effect length method or amplified sway method may be used as illustrated by the following fomula:
TYPICAL LOAD COMBINATONS USING AMPLIFIED SWAY METHOD Alternatively, The member forces obtained from the structural analysies using based on the following load combinations can be used directly to check the overall buckling of members.
Load Combinations for designing sway frames 1.4 Dead + 1.6 Imposed + Af NHL (Notional horizontal load to be applied in one direction at a time) 1.2 (Dead + Imposed Af Wind) 1.4 (Dead Af Wind) 1.0 Dead Af 1.4 x Wind
Af = moment amplification factor
0.1Z
Mm
M
MmF
yy
yy
b
xx
c
Effective length for sway frame
amplified moment or
P
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55
Example 1: Determine the effective length factors LE/L for the column numbered 1, 2 and 3 of the non-sway frame shown below. Assuming that the column bases are pinned. What are the column corresponding effective length ratios if (a) the floor beams (except the roof beams) carrying moments exceeding 90%
of their moment capacity. (b) the roof beams having pinned supports. (c) the column bases are fixed instead of pinned. Note: all beams support concrete floor slab.
1
2
3
All beams I = 21500cmx4
All columns I = 6090cmx
4
Solution: Use Annex E, Fig. E1
kk k
k k k kc u
c u TL TR1
, kk k
k k k kc L
c L BL BR2
Column 1:
36.0)720/21500360/6090(2
360/6090360/6090K1
36.0)720/21500360/6090(2
360/6090360/6090K 2
L LE / .0 62 Fig. E1 Column 2:
K16090 360
6090 360 21500 7200 36
/
/ /.
53.0720/21500)360/6090(2
360/6090360/6090K 2
L LE / .0 66 Fig. E1 Column 3:
K12 6090 360
2 6090 360 21500 7200 36
( / )
( / / ).
3-56
56
91.0K2 5.1.2.4b L LE / .0 76 Fig. E1 Special Cases (a) if floor beams carrying moments > 90%Mp, then beam stiffness Column 1: k k L LE1 21 0 1 0 1 0 . . / . Column 2: k k L LE1 20 36 1 0 0 78 . . / . Column 3: 97.0L/L90.0k0.1k E21 (b) if roof beams are pinned at their ends Column 2: k k k L LTR E1 210 053 0 083 . . ( ) / . (c) if column bases are fixed. Column 3: k k L LE1 20 36 0 5 0 66 . . / .
3-57
57
Example 2: Determine the effective length factors LE/L for the columns of the rectangular portal frame shown below. Assuming that the beam is not supporting a concrete floor slab and the column bases are (a) fixed, (b) pinned.
I beams cmx 21500 4 I columns cmx 6090 4 Solution:
k kbeam column 21500
71730
6090
60910
(a) base fixed
kk k
k k k kc u
c u TL TR1
Since the beam does not directly support a concrete floor slab, for a sway frame
kI
LTR 1 5. Table E2
k1
10 0
10 0 0 1 5 30
10
550 18
..
k2 0 (if calculations justify infinite restraint) Cl5.1.3 L LE / .1 06 Fig. E.2 (b) Base pinned using Figs. 22 and 24 k asabove1 0 18 . ( ) k if basesareactuallypinned2 1 0 . ( ) 5.1.3.3 L LE / .2 10 Fig. E.2
7.17m
6.09m
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58
Example 3: Estimate the elastic critical buckling load factor for the four storey frame shown below. Find the effective length ratios of all columns if the design load is 500kN in each column, and all members are 4.0m long and I = 2000cm4, E = 20500kN/cm2.
500kN 500kN5kN
0
0
0
Linear elastic AnalysisDeflection in cm
3.273
2.463
1.522
0.6031st
2nd
3rd
4th
Horizontal notional loads = 0.5%(factored gravity loads) Solution: Use Annex F.2. Notional horizontal load in each storey = 0.5%(factored gravity loads). Conduct a linear elastic analysis to give sway deflection . F.2 Compute sway indices for each storey
su L
storeyheight
Storey No. Sway index, s
1 0.00151 2 0.00230 3 0.00236 4 0.00202
Select s max = 0.00236
cr
s
1
200
1
200 0 002362 12
max .. F.2
Axial load in the column is F = 500kN
Euler load of column = 2
2EI
L =
2
220500 2000
4002529
kN
3-59
59
55.1500x12.2
2529
FL
EI
L
L2
cr
2E
E.6
Using limited frame method, for intermediate column Annex E2.1
k k1 21 1
1 1 1 50 57
.
.
L
LE 1 58. Fig. E2
Both the limited frame method and the elastic critical load factor methods give almost the same effective length ratio for the column.
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60
Example 4 Estimate the elastic critical buckling load factor for the four storey frame shown below. Find the effective length of the column AB if the design load is as shown in the figure. The storey height is 3.5m, and the beam length is 3.5m. All members are UC 152x152x37kg/m, E = 20500kN/cm2.
Horizontal notional loads = 0.5%(factored gravity loads) Frame Classification for sway or nonsway Limiting value for uncladded frame h/4000 = 350/4000 = 0.0875cm since interstorey sway deflections due to notional load at all storey level are greater than the limiting value of 0.0875cm, the frame is classified as sway frame. Compute Elastic Buckling Load Factor Use Annex E.6 and F.2 Notional horizontal load in each storey = 0.5%(factored gravity loads). Conduct a linear elastic analysis to give sway deflection . F.2 Compute sway indices for each storey
su L
storey height
(Storey height = 350cm)
3.5m
100kN 100kN 1kN
1kN
1.5kN
1.5kN
Lateral Deflection at the storey level (cm)
1.503
1.305
0.942
0.412
100 100
150 150
150 150
A
B 3.5m
3.5m
3.5m
3.5m
3.5m
3-61
61
Storey No. Sway index, s
1 0.00118 2 0.00151 3 0.00104 4 0.00056
Select s max = 0.00151
crs
1
2001
200 0 00151 3311max . . < 4.0 F.2
Since cr is less than 4.0amplified sway method cannot be used. Column Effective Length
Forces in column AB (from stucturla analysis)
Axial load in column AB, F = 504.7 kN
Euler load of column = 2
2EI
L =
2
2
20500 2200350
3667
kN
Effective length factor for column AB
= 48.17.504x311.3
3667
L
EI
L
L2
cr
2E
Using limited frame method, Annex E.2.1 2 kb = 1.5I/L (double curvature bending for beams) ka = 0.5 (rigidly connected to foundation)
A
B
504.7 kN
504.7 kN
2.7 kNm
5.973 kNm
3-62
62
57.05.1011
11k1
53.1L
LE Fig. E2
For UC 152 x 152 x 37 S 275 stl, use design table Page 265 Local Capacity Check (Simplified formula)
OK0.146.00703.039.085
973.5
1300
7.504
M
M
A
F
cx
x
y
Overall Buckling Check (simplified formula)
0.1Z
Mm
Z
Mm
P
F
yy
yy
xy
xx
c
(In-Plane Buckling )
0.1Z
Mm
M
Mm
P
F
yy
yy
b
xLT
cy
(Lateral Torsional Buckling)
My=0 for major axis bending only Lex=1.48(3.5m)=5.18m Pcx = 899kN Ley = 3.5m Pcy = 667kN Therefore Pc = 667kN (control by weak axis buckling) = -2.7/5.973 = - 0.452 mx = 0.51 Table 26 mLT = 0.45 Table 18
0.180.075
97.5x51.0
667
7.504
Z
Mm
Z
Mm
P
F
yy
yy
xy
xx
c
0.180.0068
973.545.0
667
7.504
Z
Mm
M
Mm
P
F
yy
yy
b
xx
cy
OK
NOTE: AMPLIFIED SWAY METHOD CANNOT BE USED AS cr < 4 Netheless, we proceed with the use of amplided sway moment mehtod
By Amplified Sway Method
3-63
63
Af = 1cr
cr
= 43.1131.3
31.3
Bsaed on original column length Lex=3.5m Pcx = 1110kN In plane Ley = 3.5m Pcy = 668kN out-of-plane = -2.7/5.973 = - 0.452 mLT = 0.45 (Table 18 of BS5950)
mx = 0.51 (Table 26 of BS5950) In Plane Buckling
0.180.075
97.5x51.0x43.1
668
7.504
Z
Mm
P
F
yy
xx
c
Lateral Torsional Buckling (out of plane buckling; no amplification for moment)
0.180.068
973.545.0
668
7.504
M
Mm
P
F
b
xx
cy
OK
Local Capacity Check Do we use amplified moment??
FA
A MMy
f x
cx 10. ??
YES. Cross section capacity should be checked using the second order moment.
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64
Example 5: A 4 storey frame shown in Fig. Q5 is designed to support a water tank with 500 kN factored force applied at the top of the four columns. All members are 203 x 203 UC 60 UC and 4m long. Perform frame classification and evaluate the overall stability of the columns assuming that the self weight of the structure has been included in the loading and horizontal force is negligible. If the loads are increased to 1000kN per column, propose alternative solution without changing the column size. Use S275 steel. 152mmOD x 101mmOD (DIN Standard ST52) Apply notional load of 0.5% the factored gravity load to the top storey as shown NHLx = NHLy = 0.005 x 1000 = 5 kN Storey Horizontal deflection at the storey level
under notional load (mm)Side sway in the y
direction
Side sway in the x direction
1 2.1 6.3 2 5.3 15.253 8.7 24.634 11.6 32.73
Interstorey Drift Maximum interstorey drift in y direction is 3.4mm (8.7mm-5.3mm) Maximum inrstorey drift in x direction is 9.38 mm (24.63 – 15.25) Classifcation for nonsway frame interstorey deflection must be less than h/2000 = 4000/2000 = 2mm The frame is sway frame in both directions.
NHLy
NHLy
x
y
500 kN
500 kN 500 kN 500 kN
NHLx
NHLx
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65
Compute sway indices for each storey Buckling about the minor axis will control the design
su L
storeyheight
Storey No. Sway index, s (for sway in the x direction only)
1 0.001512 0.00230 3 0.002364 0.00202
Using limited frame method Annex E2.1 For top column (4th storey)
57.05.111
11k
4.05.11
1k
bottom
top
44.1L
LE Fig. E.2
For 2nd and 3rd storey column
57.05.111
11kk bottomtop
L
LE 1 58.
For first storey Kbottom = 0; ktop = 0.57, LE/L = 1.28 Fig. E2 For member capacity check use Ley = 1.58 x 4 = 6.3m Design table Page 263 203 x 203 UC 60 Pcy = 733 kN > 500 kN OK If applied load increased to 1000kN, the sway deflection will double under a notional load of 0.5% the gravity load. Pcy = 733 < 10000K Provide bracing to prevent sway buckling in the y direction and allow sway in the
4 m
4 m
4 m
4 m
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66
Annex E2.1
500kN 500kN5kN
0
0
0
Linear elastic AnalysisDeflection in cm
3.273
2.463
1.522
0.6031st
2nd
3rd
4th
Horizontal notional loads = 0.5%(factored gravity loads) Solution: Use Annex F.2. Notional horizontal load in each storey = 0.5%(factored gravity loads). Conduct a linear elastic analysis to obtain sway deflection . F.2 Compute sway indices for each storey
su L
storeyheight
Storey No. Sway index, s
1 0.00151 2 0.00230 3 0.002364 0.00202
Select s max = 0.00236
cr
s
1
200
1
200 0 002362 12
max .. F.2
Axial load in the column is F = 500kN
Euler load of column = 2
2EI
L =
2
220500 2000
4002529
kN
3-67
67
55.1500x12.2
2529
FL
EI
L
L2
cr
2E
E.6
Using limited frame method, for intermediate column Annex E2.1
k k1 21 1
1 1 1 50 57
.
.
L
LE 1 58. Fig. E2
Both the limited frame method and the elastic critical load factor methods give almost the same effective length ratio for the column.
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HOMEWORK 4: CONTINUOUS FRAMES
1 Design the following plane frames using coninuous construction method. The structures are braced against out-of-plane and in-plane sway at the storey level. The materials used for all steel sections are S275 steel. a) Design the floor beam(s) and check for ultimate strength and serviceability limit states. b) Design the columns, and check their overall buckling resistance at the factored loading. The column base may be assumed to be pinned, and same column size is to be used for the entire frame. c) Check the serviceability deflection of the beams.
Figure (a) Floor Loading: Dead Load = 24 kN/m Imposed Load = 9 kN/m
Figure (b)
The unfactored loading data are given below: Floor Beams Distributed dead Load = 27 kN/m
Distributed imposed Load = 30 kN/m Roof Beam Distributed dead Load = 24 kN/m
Distributed imposed Load = 9 kN/m
6m
4m
(a)
Lateral Brace
6m
6m
4m
Lateral Braces
(b)
Floor Beam
Roof Beam
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2 Design the rigid portal frame shown below using S275 Steel. The frame is effectively braced against out-of-plane sway at the floor level. Dead load on floor = 4kN/m2 Imposed load on floor = 5kN/m2 Wind load at the storey level = 10kN Frame spacing = 6m. 3 Design the multi-storey frame shown below assuming rigid construction. The frame is effectively braced against out-of-plane sway at the floor level. 16.6kN
17.0kN
14.2kN
13.0kN
4m
4m
4m
4m
6m 6m 6m 6m
Wind Loads
Frame spacing = 6m
Loads Roof Dead load = 4.0kN/m2 = 24.0kN/m Imposed load = 1.5kN/m2 = 9.0kN/m Floors Dead load = 4.5kN/m2 = 27.0kN/m Imposed load = 5.0kN/m2 = 30.0kN/m Deflection :Limits: Interstorey deflection < Storey height / 400 Maximum frame drift < Frame height / 400
6m
5m
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4 Briefly explain the two approaches for designing columns in continuous frames. (Hint: amplified sway approach and the effective length approach.) (a) Give three typical load combinations that are commonly used for analysing continuous
frames and explain how the moment amplification effects can be included in the analysis.
(b) Explain how the outrigger and belt trusses help in resisting lateral forces in a core braced frame.
(c) A 30-storey building frame is constructed using rigid frame with several internal braced
frames. Explain how the braced frame and unbraced rigid frame help each other in resisting lateral load.
(d) Explain how the tube-in-tube system work as compared with the exterior tube system? Q5 In a case where the framing system consisting of simple framing and a bracing system ( a rigid frame), is it true that the simple frame has to be considered non-sway since under sway the effective length of the columns in the simple framing is infinity? What about the bracing system, can one design the frame as sway if the lateral deflection shown that it exceed the value for the frame to be considered as non-sway? In this case can the simple frame take the side sway as a result of the lateral sway of the bracing system or should it be prevented from swaying since the simple frame is not able to take the side sway. Even though the bracing system is able to reduce the lateral deflection by 80 % but the deflection can still be large to cause side sway. How should we take into account of this sway in the design of simple frame since the joints are not allowed to take substantial moment which could be as result of the sway
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Q6 Figure 6a shows an elevation view of a six-storey continuous frame, which is braced against out-of-plane deflection at the storey level, but free to deflect in the N-S direction as shown in Fig. 6b. a) Perform a check on the frame to determine whether it is a sway or non-sway frame and
determine the moment amplification factor using the amplified sway method. b) Determine the effective length factors for columns A and B using the limited frame
method. c) Check columns A and B in Figure 6a for overall buckling using the load combination:
1.4 Dead Load + 1.6 Imposed Load + Notional horizontal Load (NHL). Member forces are given in Table 6.
d) Propose a lateral-load resisting system for the building and comment on it in view of the design assumptions made.
All steel sections are S275 steel.. Table 6
Member forces in Columns A and B due to dead loads, imposed loads and
notional horizontal load.
Column 1.6 x Imposed Load 1.4 x Dead Load Notional horizontal Load F (kN) Mx (kNm) F (kN) Mx (kNm) F (kN) Mx (kNm)
A 1022 0 1008 0 0 30 B 1300 0 1474 0 0 120
F = Axial compression, Mx = Moment about the major axis
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All beams 457 x 191 x UB 89
19.52
Storey deflection under NHL (mm)
28.14
23.83
4.20
15.21
8.41
3.75m
3.75m
3.75m
5.75m
3.75m
G
1
2
3
5
4
A
B
3.75m
6
254 x 254 x 89UC
356x368x129UC
7m
Fig. 6a Section View of a Six-Storey continuous frame
N S
7m
Fig. 6b Typical floor plan of a Six-Storey frame
Six-storey frame
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Q7 Figure Q7 shows the framing plans for a 20-storey office building. The roofs and floors are cast-in-situ concrete slabs supported on steel beams (non-composite). All the lateral loads are resisted by the internal core. The framework surrounding the core is designed using simple construction method. e) Design the floor beam FB1, as shown in Fig. Q7c, for moment and shear. Check the
beam deflection due to imposed load. f) Design the perimeter column PC1 at the lowest storey as in Fig. 1c assuming simple
construction.
Grade 50 steel should be used. Loading: Roof and floors - dead 6.5kN/m2 ; imposed load 4.0 kN/m2 External curtain wall - 1 kN/m2 Estimated column weight (including concrete encasement) 2.5kN/m Live load reduction may be omitted.
Figure Q7b Plan View
Figure Q7a Elevation View
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Q8
Figures Q7a&b shows the original framing plans for a multi-storey office building in which the internal core is designed mainly for the lateral wind load, and the steel framework surrounding the core is designed to resist the gravity load only. If the original scheme is found to give rise to unacceptable deflection at the top of the building, proposed alternate methods to enhance the lateral stiffness of the building against the wind load.
Figure Q7c Floor layout plan
Figure Q7d Beam to column connections
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Q9 The members of the unbraced rigid-jointed two storey frame shown in Fig. 1 are all of S275 steel. The member properties are shown in Table 1.
(i) By using a structural analysis software (SAP2000, STAAD PRO, etc.), perform elastic analysis to determine the internal force distribution in the members for the frame subject to the factored loads as shown.
(ii) determine the effective length of the columns (iii) determine the amplified first-order design moments M* for the columns; (iv) determine the adequacy of the columns for the actions determined by elastic
analysis using either effective length method or amplified sway method;
Table 1. Member properties Member Section Ix (cm4) A (cm2) py (N/mm2) 1-3, 10-12 UC 203 x 203 x 46 4570 58.7 275 2-11 UB 610 x 229 x 113 87300 144 275 3-12 UB 457 x 191 x 74 33300 94.6 275
5.0
5.0
Fig. 1. Frame and loads (m, kN)
5
10
5 1010 10 10
10 2020 20 20
3.0 3.0 3.0
2
3
1
5 7 9
4 6 8 11
12
10 17
3.0 3.0 3.0
10 5
20 10
14 16
13 15
Simple frame Rigid frame