Shear walls and frames (comprising of columns and beams) are distinct structural systems that can be used in resisting lateral actions in high-rise buildings. However, as a building goes higher, frames alone become inadequate for lateral stability, hence, the structure can be augmented by shear walls and/or cores. Shear wall-frame interaction for lateral load resistance is complex because shear walls deflect primarily in bending mode, while frames deflect in shear mode.
However, the interaction between shear walls and frames is beneficial for high-rise buildings, since the linkage and stiffness of the floor slab diaphragm and the stabilising elements give better lateral load resistance. Furthermore, since their mode of deflection varies, the frame tends to restrain the shear wall in upper storeys and the shear wall tends to restrain the frame in the lower storeys. This reduces the lateral deflection and improves the overall efficiency of the structural system.
Frame-shear wall interaction
Shear walls are stiffer than columns, hence they take up most of the lateral load. This has sometimes led to the conservative approach of transferring the entire lateral load to the shear walls during design. When the building is very tall, the flexural deformation of the shear wall becomes very pronounced and hence, must be allowed for in the analysis. Shear wall-frame systems have been used successfully in buildings ranging from 10 storeys to 50 storeys.
The difference in behavior under lateral load, in combination with the in-plane rigidity of the floor slabs, causes nonuniform interacting forces to develop when walls and frames are present. This makes the analysis more difficult. If torsion is not considered in the analysis, two simplified manual methods of determining the interaction of frames and shear walls are:
Use of charts given by Khan and Sbarounis (1964) or PCA’S Advanced Engineering Bulletin No. 14, f7), and
Use of Equation (C) by PCA.
In order to use these charts, the structure must be reduced to a single frame and a single wall by the addition of the properties of the separate vertical units. In both references, the stiffnesses (Iw) of all the shear walls are summed to give an equivalent single wall.
In this article, we are going to investigate the effects of shear wall-frame interaction in the resistance of uniformly distributed lateral loads using finite element analysis. A 2D model of a 10-storey and 50-storey building will be used in the analysis.
Model 1A: Analysis of a 10-storey Frame Without Shear Wall
Height of building = 30 m Inter-storey height of building = 3 m Dimension of beams = 400 x 600 mm Dimension of columns = 400 x 400 mm Lateral load = 5 kN/m
Model of a 10 storey frame
When analysed on Staad Pro software, the analysis results are as follows;
Lateral deflection of a 10 storey frame
Bending moment diagram of the 10 storey frame
Summarily, the analysis result of the 10-storey frame without shear walls is as follows;
Deflection Maximum deflection at roof level = 16.989 mm Deflection at 1st floor = 2.393 mm
External Column (windward side) Bending moment = 59.1 kNm Shear force = 39 kN Axial force = 146 kN
Internal Column (windward side) Bending moment = 63.5 kNm Shear force = 39.9 kN Axial force = 72.8 kN
First-floor beams Bending moment = 75 kNm Shear force = 26.4 kN Axial force = 7.68 kN
Model 1B: Analysis of a 10-storey frame with shear wall
Height of building = 30 m Inter-storey height of building = 3 m Dimension of beams = 400 x 600 mm Dimension of columns = 400 x 400 mm Length of shear wall = 2500 mm Thickness of shear wall = 250 mm Lateral load = 5 kN/m
Model of a 10 storey frame with shear wall
Deflection Maximum deflection at roof level = 7.120 mm Deflection at 1st floor = 0.319 mm
Lateral deflection of a shear wall-frame model
External Column (windward side) Bending moment = 13 kNm Shear force = 12.7 kN Axial force = 105 kN
First-floor beams Bending moment = 20.8 kNm Shear force = 8.08 kN Axial force = 18.9 kN
Model 2A: Analysis of a 50-storey building without shear wall
Height of building = 150 m Inter-storey height of building = 3 m Dimension of beams = 400 x 600 mm Dimension of columns = 400 x 400 mm Lateral load = 5 kN/m
Summary of analysis results
Deflection Maximum deflection at roof level = 1516.839 mm Deflection at 1st floor = 13.739 mm
External Column (windward side) Bending moment = 299 kNm Shear force = 171 kN Axial force = 4170 kN
Internal Column (windward side) Bending moment = 340 kNm Shear force = 208 kN Axial force = 1142 kN
First-floor beams Bending moment = 411 kNm Shear force = 145 kN Axial force = 7.94 kN
Model 2B: Analysis of a 50-storey framed building with shear wall
Height of building = 150 m Inter-storey height of building = 3 m Dimension of beams = 400 x 600 mm Dimension of columns = 400 x 400 mm Length of shear wall = 2500 mm Thickness of shear wall = 250 mm Lateral load = 5 kN/m
Deflection Maximum deflection at roof level = 1297.128 mm Deflection at 1st floor = 2.522mm
External Column (windward side) Bending moment = 66.5 kNm Shear force = 38.6 kN Axial force = 3887 kN
First-floor beams Bending moment = 137 kNm Shear force = 53.3 kN Axial force = 46.5 kN
Shear Walls Maximum absolute stress = 15734 kN/m2 Sx = 15604.6 kN/m2 Sy = 4547.11 kN/m2
Discussion of results
Model 1
Element (Action Effect)
Without shear wall
With shear wall
Percentage Difference (%)
Maximum deflection
16.989 mm
7.120 mm
58.09 %
Deflection at first storey
2.393 mm
0.319 mm
86.67 %
Column bending moment
59.1 kNm
13 kNm
78%
Column shear force
39 kN
12.7 kN
67.43%
Column axial force
146 kN
105 kN
28.08%
Beam bending moment
75 kNm
20.8 kNm
72.26%
Beam shear force
26.4 kN
8.08 kN
69.39%
Beam axial force
7.78 kN
18.9 kN
-142.93%
Model 2
Element (Action Effect)
Without shear wall
With shear wall
Percentage Difference (%)
Maximum deflection
1516.8 mm
1297.128 mm
14.482 %
Deflection at first storey
13.739 mm
2.522 mm
81.643 %
Column bending moment
299 kNm
66.5 kNm
77.76 %
Column shear force
171 kN
38.6 kN
77.42 %
Column axial force
4170 kN
3887 kN
6.78 %
Beam bending moment
411 kNm
137 kNm
66.67 %
Beam shear force
145 kN
53.3 kN
63.24 %
Beam axial force
7.94 kN
46.5 kN
-485.642 %
The presence of shear walls reduced the maximum lateral deflection of the 10 storey building by about 58% when compared with the 50 storey building where deflection reduced by just 14.482%. Therefore as a building gets higher, the arrangement of shear walls becomes more and more important. In a study carried out by Aginam et al (2015) at Nnamdi Azikiwe University, Awka, Nigeria, it was observed that shear wall positioning affects the lateral displacement of tall buildings. Shear walls placed externally reduced lateral deflection when compared with shear walls placed internally.
In both models, the column bending moment was reduced by about 78% when the shear wall was introduced in-between the frames. However, column axial force reduced by about 28% in the 10-storey building, while it reduced by just 6.78% in the 50-storey building. Bending moment and shear forces in beams of the framed structures reduced considerably when shear walls were introduced. On the other hand, the presence of shear walls increased the axial force in the horizontal beam members.
References
[1] Aginam C.H., Chidolue C.A., and Ubani O.U. (2015): Effect of Planar Solid shear wall-frame arrangement on the deformation behavior of multi-story frames. IOSR Journal of Mechanical and Civil Engineering 12(1):98-105
[2] Khan, F.R. and Sbarounis (1964): Interaction of Shear Walls and Frames Journal of the Structural Division, Proc.,ASCE 90(3):285-338
The bracing system (stabilising components) of a high-rise building is normally made up of different units: frameworks (beams and columns), shear walls, coupled shear walls, and cores. They all contribute to the overall lateral load resistance of the system, but their contributions can be very different both in weight and in nature, so it is essential for the designer to know their behaviour in order that the optimum bracing system can be produced [1].
Bracing components in a building are assumed to be fully fixed at the base and hinged at the top. Non-stabilising elements are assumed to be hinged at both ends. Since non-stabilising elements must be braced by stabilising elements, they have a negative contribution to load resistance [2].
The different types of stabilising components (bracing elements) in a building are;
Columns and frameworks
Shear walls
Towers and cores
Columns and frameworks
Typical columns and frameworks under construction
A linear structural member which takes vertical compressive loads can generally be called a column. Columns can be constructed of steel, wood, or reinforced concrete depending on the strength and/or the aesthetics required. Columns are found mainly in structures in order to provide support for beams or slabs.
When calculating stability in a structure with columns it is essential to ascertain if the column is stabilising or not [2]. This means that non-stabilising elements have to be held up by the stabilising elements so they have a certain negative effect on the overall stiffness of the system.
The behaviour of frameworks under lateral load is complex, mainly because they develop both bending and shear deformations when resisting lateral loads [1]. Due to the complexity of the problem, designers and researchers have made considerable efforts to develop approximate techniques and methods. Perhaps the best and most widespread method is the continuum method which is based on an equivalent medium that replaces the framework [1]. However, due to the availability of numerous commercial computer programs for finite element analysis, this computational challenge has largely been overcome.
Columns and frameworks alone are not suitable for resisting lateral loads once the building exceeds 10 storeys, and must be combined with shear walls and cores.
Shear walls
Typical shear wall under construction
Shear walls, made of reinforced concrete, are used in modern buildings because of their effectiveness in maintaining stability and for the freedom they offer the architect who is designing [3]. Shear walls are stiff in the direction of the wind and can be effectively utilised in resisting the internal stresses due to wind action. A shear wall’s position in a building is often initially decided by the architect.
An architect is trained to design for the building’s function and appearance and not for its stability. Therefore, when a structural engineer is not involved in the first phase of design, it may lead to the shear walls being situated in non-favourable positions [2].
Also, while choosing reinforced concrete walls as partition walls, the architect can be unintentionally gaining stabilising elements. Pierced shear walls or coupled shear walls are described as shear walls with holes. These holes can be windows or doors that are necessary for access or lighting for the building. The force from the horizontal wind load results in shear forces which act within the wall and tension and compression resulting at the ground.
Towers/Cores
Typical core of a building under construction
Towers are rigid cores situated inside tall buildings. Shear walls are often built together to create three-dimensional units. A good example is the U-shaped elevator core but many different shapes exist in building structures. The second moments of area of a reinforced concrete core are normally large and a small number of cores are often sufficient to provide the building with the necessary stiffness to resist lateral loading [1].
Usually, a core will exist with another core or combined with shear walls and/or with columns. The combined effect will give rise to greater resistance to torsion depending on how the units are situated in relation to each other. Ideally, they are situated as far apart as possible for creating a torsional resistance. A disadvantage with using a single core, on its own, is that it is susceptible to torsion and must therefore be heavily dimensioned in order to resist torque [2].
The use of cores is however favourable in that they can be used not only as stabilising units but also as elevator shafts or stairwells. Funnelling of ventilation shafts, water pipes and electric cables can also be hidden within the tower giving the architect more manoeuvrability and the client more effective use of the space provided. Towers which have open cross sections, for example U-shaped or H-shaped, have less resistance to torsion than closed sections and should, in general, be combined with other stabilising components.
References
[1] Zalka K. A. (2013): Structural Analysis of Regular Multi-storey Buildings. CRC Press – Taylor and Francis Group, USA [2] Gustaffson D., and Hehir J. (2005): Stability of tall buildings. M.Sc thesis submitted to the Department of Civil and Environmental Engineering, Chalmers University of Technology, Sweden [3] Aginam C.H., Chidolue C.A., and Ubani O.U. (2015): Effect of Planar Solid shear wall-frame arrangement on the deformation behaviour of multi-story frames. IOSR Journal of Mechanical and Civil Engineering 12(1):98-105
Generically, cement is an inorganic fine powder that when mixed with water undergoes hydration reaction, and forms a paste which becomes the binder that holds aggregates together in concrete. The raw materials from which cement is made are lime, silica, alumina, and iron oxide.
These constituents are crushed and blended in the correct proportions and burnt in a rotary kiln. The resulting product is called clinker. The cooled clinker can be mixed with gypsum and various additional constituents and ground to a fine powder in order to produce different types of cement. The main chemical compounds in cement are calcium silicates and aluminates.
Hydraulic cements set and harden by reacting chemically with water in a process known as hydration reaction. During this reaction, cement combines with water to form a stonelike mass, called paste. When the paste (cement and water) is added to aggregates (sand and gravel, crushed stone, or other granular material) it acts as an adhesive and binds the aggregates together to form concrete, the world’s most versatile and most widely used construction material.
Cement powder in bag
The European standard for cements is BS EN 197-1:2011 Cement –Part 1: Composition, specifications, and conformity criteria for common cements. The Standard defines and gives the specifications of 27 distinct common cements, 7 sulfate resisting common cements, as well as 3 distinct low early strength blast furnace cements and 2 sulfate resisting low early strength blast furnace cements and their constituents.
Standard Designation of Cements
CEM cement designation includes the following information:
i. Cement type (CEM I – CEM V) ii. Strength class (32.5-52.5) iii. Indication of early strength iv. Additional designation SR for sulfate resisting cement v. Additional designation LH for low heat cement
Types of Cements
All types of cement are grouped into five major groups, all of which are mixtures of different proportions of clinker and another major constituent. The five groups are:
CEM I – Portland cement: This comprises mainly ground clinker and up to 5% of minor additional constituents.
CEM II – Portland composite cement: This comprises of seven types which contain clinker and up to 35% of another single constituent.
i. Portland slag cement (CEM II/A-S and CEM II/B-S): This comprises of clinker and blast furnace slag which originates from the rapid cooling of slag obtained by smelting iron ore in a blast furnace. The percentage of the slag varies between 6 and 35%.
ii. Portland silica fume cement (CEM II/A-D): This comprises of clinker and silica fume which originates from the reduction of high purity quartz with coal in an electric arc furnace in the production of silicon and ferrosilicon alloys.
iii. Portland-Pozzolana cement (CEM II/A-P, CEM II/B-P, CEM II/A-Q, CEM II/B-Q): This comprises clinker and natural pozzolana such as volcanic ashes or sedimentary rocks with suitable chemical and mineralogical composition or Natural calcined pozzolana such as materials of volcanic origin, clays, shales or sedimentary rocks activated by thermal treatment.
iv. Portland-fly ash cement (CEM II/A-V, CEM II/B-V, CEM II/A-W, CEM II/B-W): This mixture of clinker and fly ash dust-like particles precipitated from the flue gases from furnaces fired with pulverised coal.
v. Portland burnt shale cement (CEM II/A-T, CEM II/B-T): This consists of clinker and burnt shale, specifically oil shale burnt in a special kiln at 800 °C.
vi. Portland-limestone cement (CEM II/A-L, CEM II/B-L, CEM II/A-LL, CEM II/B-LL).
vii. Portland-composite cement (CEM II/A-M, CEM II/B-M).
CEM III – blast furnace cement (CEM III/A, CEM III/B, CEM III/C): This comprises clinker and a higher percentage (36-95%) of blast furnace slag than that in CEM II/A-S and CEM II/B-S.
CEM IV – pozzolanic cement (CEM IV/A, CEM IV/B): This comprises of clinker and a mixture of silica fume, pozzolanas and fly ash.
CEM V – composite cement (CEM V/A, CEM V/B): This comprises clinker and a higher percentage of blast furnace slag and pozzolana or fly ash.
The clinker contents in different types of cements are given in Table 1. The letters A, B and C designate respectively higher, medium and lower proportion of clinker in the final mixture. However, the percentage of clinker with the designations A, B, C can be different in different types of cement as shown in Table 1.
Cement Type
Clinker Content – A
Clinker Content – B
Clinker Content – C
CEM II
80-94%
65-79%
–
CEM III
35-64%
20-34%
5-19%
CEM IV
65-89%
45-64%
–
CEM V
40-64%
20-38%
–
The second constituent in cement in addition to clinker is designated by the second letter as follows:
S = blast furnace slag D = silica fume P = natural pozzolana Q = natural calcined pozzolana V = siliceous fly ash W = calcareous fly ash (e.g., high lime content fly ash) L or LL = limestone T = burnt shale M = combination of two or more of the above components
Strength Class of Cement
The standard strength of a cement is the compressive strength determined in accordance with EN 196-1 at 28 days and shall conform to the requirements in Table 3 of EN197-1. The three standard classes of cement strength are;
class 32.5
class 42.5, and
class 52.5
Strength Class
28 days Compressive strength (MPa)
Initial setting time (mins)
32.5 N
≥ 32.5, ≤ 52.5
≥ 75
32.5 R
≥ 32.5, ≤ 52.5
≥ 75
42.5 N
≥ 42.5, ≤ 62.5
≥ 60
42.5 R
≥ 42.5, ≤ 62.5
≥ 60
52.5 N
≥ 52.5
≥ 45
52.5 R
≥ 52.5
≥ 45
Sulfate-Resisting Cement
Sulfate resisting cements are used particularly in foundations where the presence of sulfates in the soil which can attack ordinary cements. The sulfate resisting cements have the designation SR and they are produced by controlling the amount of tricalcium aluminate (C3A) in the clinker. The available types are:
i. Sulfate resisting Portland cements CEM I-SR0, CEM I-SR3, CEM I-SR5 which have the percentage of tricalcium aluminate in the clinker less than or equal to 0, 3 and 5% respectively
ii. Sulfate resisting blast furnace cements CEM III/B-SR, CEM III/C-SR (no need for control of C3A content in the clinker).
iii. Sulphate resisting pozzolanic cements CEM IV/A-SR, CEM IV/B-SR (C3A content in the clinker should be less than 9%).
Low Early Strength Cement
These are CEM III blast furnace cements. Three classes of early strength are available with the designations N, R and L respectively signifying normal, ordinary, high and low early strength.
Examples of Cement Designation
CEM II/A-S 42.5 N This indicates Portland composite cement (indicated by CEM II), with high proportion of clinker (indicated by letter A) and the second constituent is slag (indicated by letter S) and the strength class is 42.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 42.5 MPa) and it gains normal early strength (indicated by letter N).
CEM III/B 32.5 N This indicates blast furnace cement (indicated by CEM III); with medium proportion of clinker (indicated by letter B) and the strength class is 32.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 32.5 MPa) and it gains normal early strength (indicated by letter N).
CEM I 42.5 R-SR3 This indicates Portland cement (indicated by CEM I), the strength class is 42.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 42.5 MPa) and it gains high early strength (indicated by letter R) and is sulfate resisting with C3A content in the clinker less than 3%.
CEM III-C 32.5 L – LH/SR This indicates blast furnace cement (indicated by CEM III), the strength class is 32.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 32.5 MPa) and it gains low early strength (indicated by letter L) and is sulfate resisting (indicated by letters SR) and is of low heat of hydration (indicated by LH).
Common Types of Cements
Of different types of cement, the most common ones are mainly six:
i. CEM I ii. CEM II/B-S (containing 65-79% of clinker and 21-35% of blast furnace slag) iii. CEM II/B-V (containing 65-79% of clinker and 21-35% of siliceous fly ash) iv. CEM II/A-LL (containing 80-94% of clinker and 6-20% of limestone) v. CEM III/A (containing 35-64% of clinker and 36-65% of other constituents) vi. CEM III/B (containing 20-34% of clinker and 66-80% of other constituents)
Columns are vertical or inclined compression members used for transferring superstructure load to the foundation. The structural design of reinforced concrete (R.C.) columns involves the provision of adequate compression reinforcement and member size to guaranty the stability of the structure. In typical cases, columns are usually rectangular, square, or circular in shape. Other sections such as elliptical, octagonal, etc are also possible.
Columns are usually classified as short or slender depending on their slenderness ratio, and this in turn influences their mode of failure. In framed structures, columns are either subjected to axial, uniaxial, or biaxial loads depending on their location and/or loading condition. EN 1992-1-1:2004 (Eurocode 2) demands that we include the effects of imperfections in the structural design of columns. The structural design of reinforced concrete columns is covered in section 5.8 of EC2.
When columns are not properly designed, they can fail by;
crushing
buckling
shear, or
by the combination of any of the above
The steps in the design of reinforced concrete columns are;
Determine design life
Assess actions on the column
Determine which combinations of actions apply
Assess durability requirements and determine concrete strength
Check cover requirements for appropriate fire resistance period
Calculate min. cover for durability, fire, and bond requirements
Analyse structure to obtain critical moments and axial forces
Check slenderness
Determine area of reinforcement required
Check spacing of bars
Braced and Unbraced Columns
Lateral stability in braced reinforced concrete structures is provided by shear walls, lift shafts, and stairwells. Therefore all lateral actions are transferred and resisted by the stiff stabilising members. In a braced column the axial load and the bending moments at the ends of a column arise from the vertical loads acting on the beams. The horizontal loads do not affect the forces or deformation of the column. The columns do not contribute to the overall horizontal stability of the structure.
In unbraced structures, resistance to lateral forces is provided by bending in the columns and beams in that plane. The column ends can deflect laterally. In a column in an unbraced structure, the axial force and moments in the column are caused not only by the vertical load on the beams but also by the lateral loads acting on the structure and additional moments due to the axial load being eccentric to the deflected column.
Figure 1: Braced and unbraced column
However, most concrete buildings are designed as braced structures. Unbraced structures are rare and are used only if there is a need for uninterrupted floor space.
Loads/Actions on Columns
The major action effects on columns are compressive axial force, bending moment, and shear force. In the manual design of reinforced concrete columns, the design axial force can be obtained using the tributary area method or by summing up the support reactions from the beams supported by the column. The self-weight of the column should be included in the calculation of the design axial force.
In the tributary area method, the floor panels supported by the columns are divided into equal parts, and the load from each part transferred to the nearest column. The process is shown in Figure 2.
Figure 2: Tributary area method for column load analysis
For the manual analysis of column design moment, sub-frames can be used to obtain the maximum design moment (see Figure 3). The recommendation is that you should consider the adjacent beams fully fixed, while you reduce their stiffness by half, because it will be an overestimation of the stiffness of the beams to consider all the ends fully fixed. (Details of this can be found in Reynolds and Steedman, 2005, Table 1, and Reynolds, Steedman, and Threlfall, 2008, Table 2.57).
Figure 3: Sub-frame arrangement in the design of columns
Slenderness in the Design of Reinforced Concrete Columns
Clause 5.8.2 of EN 1992-1-1 deals with members and structures in which the structural behaviour is significantly influenced by second-order effects (e.g. columns, walls, piles, arches and shells). Global second-order effects are more likely to occur in structures with a flexible bracing system.
Column design in EC2 generally involves determining the slenderness ratio (λ), of the member and checking if it lies below or above a critical value λlim. If the column slenderness ratio lies below λlim, it can simply be designed to resist the axial action and moment obtained from elastic analysis, but including the effect of geometric imperfections. These are termed first-order effects. However, when the column slenderness exceeds the critical value, additional (second-order) moments caused by structural deformations can occur and must also be taken into account.
So in general, second order effects may be ignored if the slenderness λ is below a certain value λlim.
λlim = (20.A.B.C)/√n ——- (1)
where: A = 1/(1 + 0.2ϕef) (if ϕef is not known, A = 0.7 may be used) B = 1+ 2ω (if ω is not known, B = 1.1 may be used) C = 1.7 – rm (if rm is not known, C = 0.7 may be used)
Where;
ϕef = effective creep ratio (0.7 may be used) ω = Asfyd / (Acfcd); mechanical reinforcement ratio; As is the total area of longitudinal reinforcement n = NEd / (Acfcd); relative normal force rm = M01/M02; moment ratio M01, M02 are the first order end moments, |M02| ≥ |M01|
If the end moments M01 and M02 give tension on the same side, rm should be taken positive (i.e. C ≤ 1.7), otherwise negative (i.e. C > 1.7). For braced members in which the first-order moments arise only from or predominantly due to imperfections or transverse loading rm should be taken as 1.0 (i.e. C = 0.7):
Also, clause 5.8.3.1(2) of EC2 says that for biaxial bending, the slenderness criterion may be checked separately for each direction. Depending on the outcome of this check, second-order effects (a) may be ignored in both directions, (b) should be taken into account in one direction, or (c) should be taken into account in both directions.
Slenderness of isolated members
According to clause 5.8.3.2 of EC2, the slenderness ratio of an isolated member is defined as follows:
λ = l0/i —— (2)
where: l0 is the effective length i is the radius of gyration of the uncracked concrete section (i = h/√12 for rectangular sections)
Effective length of isolated members
From Figure 5.7 of EC2, examples of effective length for isolated members with constant cross-section are given as shown in Figure 4. This gives the examples of different buckling modes and corresponding effective lengths for isolated members.
Figure 4: Effective length of isolated members (Fig 5.7 EC2)
However, for compression members in regular braced frames, the slenderness criterion should be checked with an effective length l0 determined in the following way:
Where; k1, k2 are the relative flexibilities of rotational restraints at ends 1 and 2 respectively. L is the clear height of the column between the end restraints k = 0 is the theoretical limit for rigid rotational restraint, and k = ∞ represents the limit for no restraint at all. Since fully rigid restraint is rare in practice, a minimum value of 0.1 is recommended for k1 and k2.
According to Table 4.15 of Reynolds, Steedman and Threlfall (2008),
“In the above equations, k1 and k2 are the relative flexibilities of rotational restraint at nodes I and 2 respectively. If the stiffness of adjacent columns does not vary significantly (say, difference not exceeding 15% of the higher value), the relative flexibility may be taken as the stiffness of the column under consideration divided by the sum of the stiffness of the beams (or, for an end column, the stiffness of the beam) attached to the column in the appropriate plane of bending. Otherwise, the effective column stiffness should be taken as the sum of the stiffness of the columns above and below the node.
The stiffness of a member is 4EI/L for members fixed at the remote end, and 3EI/L for members pinned at the remote end, where I is the second moment of area of the cross-section allowing for the effect of cracking (for beams, 50% of the value for the uncracked section could be used), and L is the length of the member.
For flat slabs, the beam stiffness should be based on the dimensions of the column strip. At nodes where the beams are considered as nominally simply-supported, and at bases not designed to resist column moments, k should be taken as 10. At bases designed to resist column moments, k may be taken as 1.0”.
Methods of Analysisof Reinforced Concrete Columns
According to clause 5.8.5 (1), the methods of analysis include a general method, based on non-linear second order analysis and the following two simplified methods:
(a) Method based on nominal stiffness (b) Method based on nominal curvature
The method of nominal curvature has been used in this article, which is mainly suitable for isolated members with a constant normal force.
According to clause 5.8.8.2, the design moment is:
MEd = M0Ed + M2 —– (4)
where: M0Ed is the 1st order moment, including the effect of imperfections, M2 is the nominal 2nd order moment
The maximum value of MEd is given by the distributions of M0Ed and M2; the latter may be taken as parabolic or sinusoidal over the effective length.
Note: For statically indeterminate members, M0Ed is determined for the actual boundary conditions, whereas M2 will depend on boundary conditions via the effective length.
By differing first order end moments, M01 and M02 may be replaced by an equivalent first order end moment M0e:
M0e = 0.6M02 + 0.4M01 ≥ 0.4M02 ——–(5)
M01 and M02 should have the same sign if they give tension on the same side, otherwise opposite signs. Furthermore, |M02| ≥ |M01|.
The nominal second order moment M2 in Expression (4) is;
M2 = NEde2 ——- (6)
where: NEd is the design value of axial force e2 is the deflection = (1/r) l02/c
1/r is the curvature l0 is the effective length c is a factor depending on the curvature distribution
For constant cross-section, c = 10 (≈ π2) is normally used. If the first-order moment is constant, a lower value should be considered (8 is a lower limit, corresponding to constant total moment).
Simplified Design Steps when λ < λlim (Arya, 2009)
According to clause 5.8.3.1 of EC2, if the slenderness, λ, is less than λlim, the column should be designed for the applied axial load, NEd, and the moment due to first-order effects, MEd, being numerically equal to the sum of the larger elastic end moment, M02, plus any moment due to geometric imperfection, NEd.ei, as follows:
MEd = M02 + NEd.ei ——— (7)
Where; ei is the geometric imperfection = (θil0/2) θi is the angle of inclination and can be taken as 1/200 for isolated braced columns and l0 is the effective length (clause 5.2(7)).
According to clause 6.1(4) the minimum design eccentricity, e0, is h/30 but not less than 20 mm where h is the depth of the section.
Once NEd and MEd have been determined, the area of longitudinal steel can be calculated by strain compatibility using an iterative procedure.. However, this approach may not be practical for everyday design and therefore The Concrete Centre has produced a series of design charts, similar to those in BS 8110:Part 3, which can be used to determine the area of longitudinal steel.
Curvature
According to clause 5.8.8.3 of EC2, for members with constant symmetrical cross sections (including reinforcement), the following relation may be used:
1/r = Kr.Kϕ.1/r0 ———(8)
where: Kr is a correction factor depending on axial load Kϕ is a factor for taking account of creep 1/r0 = εyd/(0.45d) εyd = fyd/Es d is the effective depth Es is the elastic modulus of steel = 200 kN/mm2
However, if all reinforcement is not concentrated on opposite sides, but part of it is distributed parallel to the plane of bending, d is defined as
d = (h/2) + is ——— (9)
where is is the radius of gyration of the total reinforcement area.
Kr = (nu – n) / (nu – nbal) ≤ 1 ——— (10)
where: n = NEd /(Acfcd), relative axial force NEd is the design value of axial force nu = 1 + ω nbal is the value of n at maximum moment resistance; the value 0.4 may be used ω = Asfyd /(Acfcd)
As is the total area of reinforcement Ac is the area of concrete cross-section
The effect of creep should be taken into account by the following factor:
Kϕ = 1 + βϕef ≥ 1.0
where: ϕef is the effective creep ratio β = 0.35 + fck/200 – λ/150 λ is the slenderness ratio
Simplified Design Steps of Columns when λ > λlim (Arya, 2009)
When λ > λlim, critical conditions may occur at the top, middle or bottom of the column. The values of the design moments at these positions are, respectively (see Figure 5):
(i) M02 (ii) M0Ed + M2 (iii) M01 + 0.5M2
M0Ed is the equivalent first-order moment including the effect of imperfections at about mid-height of the column and may be taken as M0e as follows:
M0e = (0.6M02 + 0.4M01) ≥ 0.4M02 ———(11)
Where; M01 and M02 are the first order end moments including the effect of imperfections acting on the column and M02 is the numerically larger of the elastic end moment acting on the column i.e.
|M02| > |M01|
M2 is the nominal second order moment acting on the column and is given by
M2 = NEd.e2 Where;
NEd is the design axial load at ULS e2 is the deflection = (1/r) (l02)/10
Figure 5: Column design bending moment
Biaxial Bending of Reinforced Concrete Columns
According to clause 5.8.9 of EC2, separate design in each principal direction, disregarding biaxial bending, may be made as a first step. Imperfections need to be taken into account only in the direction where they will have the most unfavourable effect.
No further check is necessary if the slenderness ratios satisfy the following two conditions
λy/λz ≤ 2.0 and λz/λy ≤ 2.0 ——–(12)
and if the relative eccentricities ey/h and ez/b satisfy one the following conditions:
(ey/heq)/(ez/beq) ≤ 0.2 or (ez/beq)/(ey/heq) ≤ 0.2 ——— (13)
where:
b, h are the width and depth of the section beq = iy.√12 and heq = iz.√12 for an equivalent rectangular section λy and λz are the slenderness ratios l0/i with respect to the y- and z-axis respectively iy, iz are the radii of gyration with respect to y- and z-axis respectively ez = MEd,y /NEd; eccentricity along z-axis ey = MEd,z /NEd; eccentricity along y-axis MEd,y is the design moment about y-axis, including the second-order moment MEd,z is the design moment about z-axis, including second order moment NEd is the design value of axial load in the respective load combination
If the condition of Expression (12 and 13) is not fulfilled, biaxial bending should be taken into account including the 2nd order effects in each direction (unless they may be ignored according to clauses 5.8.2 (6) or 5.8.3 of EC2). In the absence of an accurate cross-section design for biaxial bending, the following simplified criterion may be used
(MEdz/MRdz )a + (MEdy/MRdy)a ≤ 1.0 ——–(14)
where: MEd,i is the design moment around the respective axis, including a 2nd order moment. MRd,i is the moment resistance in the respective direction a is the exponent; for circular and elliptical cross-sections: a = 2
For rectangular sections, see Table 1;
Table 1: Values of ‘a’ exponent for rectangular sections
NEd is the design value of axial force NRd = Acfcd + Asfyd, design axial resistance of section.
Where: Ac is the gross area of the concrete section As is the area of longitudinal reinforcement
Design example of reinforced concrete columns
Design a 230 x 230 mm biaxially loaded reinforced concrete column with a clear height of 4050 mm. The forces acting on the column are given below. fck = 25 MPa, fyk = 460 Mpa, Concrete cover = 35 mm
43.098 < 46.359, second order effects need not to be considered in the x-direction
Critical Slenderness for the z-direction
A = 0.7 B = 1.1 C = 1.7 – M_01/M_02 = 1.7 – ((-3.569)/7.138) = 2.2 n = NEd/(Acfcd) = 0.5336
44.044 < 46.381, second order effects need not to be considered in the z – direction
Design Moments (x-direction)
M01 = 6.592 kNm, M02 = 13.185 kNm
e1 is the geometric imperfection = (θi l0/2) = (1/200 × 2862/2) = 7.155 mm Minimum eccentricity e0 = h/30 = 230/30 = 7.667mm. Since this is less than 20mm, take minimum eccentricity = 20mm
e1 is the geometric imperfection = (θi l0/2) = (1/200 × 2865/2) = 7.1625 mm Minimum eccentricity e0 = h/30 = 230/30 = 7.667mm. Since this is less than 20 mm, take minimum eccentricity = 20 mm
Beams are horizontal structural elements designed to carry lateral loads. When they are inclined or slanted, they are referred to as raker beams. Floor beams in a reinforced concrete building are normally designed to resist load from the floor slab, their own self-weight, the weight of the partitions/cladding, the weight of finishes, and other actions as may be applied. The design of a reinforced concrete (R.C.) beam involves the selection of the proper beam size and area of reinforcement to carry the applied load without failing or deflecting excessively.
Under the actions listed above, a horizontal reinforced concrete beam will majorly experience bending moment and shear force. Depending on the loading and orientation, the beam may experience torsion (twisting), as found in curved beams or beams supporting canopy roofs. For raker beams, the presence of axial force can be quite significant in the design.
Longitudinal reinforcement is used to resist bending moment, and also enhance the shear force capacity of a beam. Stirrups (links) are used for resisting any excess shear force and torsion (where applicable). In deep beams or beams subjected to torsion, sidebars can be used to enhance the torsion capacity and also prevent cracking. The depth (d), width (b), and disposition of reinforcements define the load-carrying capacity of a beam and forms the essence of their design.
Figure 1: Section of a reinforced concrete beam
In typical reinforced concrete buildings, floor beams can be categorised into;
T – beams,
L – beams, or
rectangular beams
Figure 2: Types of beams in a framed concrete building
T-beams and L-beams are generally referred to as flanged beams. T- beams are usually internal beams, while external beams (perimeter beams) are usually L – beams. Beams that are not carrying any slab load are more often rectangular beams. Beams in a reinforced concrete building can also be described in terms of their support condition such as simply supported, cantilever beams, or continuous beams.
The steps in the design of a reinforced concrete beam are as follows;
(a) Preliminary sizing of members (b) Estimation of design load and actions (c) Structural analysis of the beam (d) Selection of concrete cover (e) Flexural design (bending moment resistance) (f) Curtailment and anchorage (g) Shear design (h) Check for deflection (i) Check for cracking (j) Provide detailing sketches
Preliminary sizing of beams
Preliminary sizing of a beam can be influenced in a lot of ways. However, deflection criteria can be used as a starting point in the analysis, even though experience is the best. Long span beams will require deeper sections, and at the same time, the magnitude of loading will also have an effect on the sizing.
Sometimes, architectural constraints can limit your options in selecting the depth and width of reinforced concrete beams. If the headroom of the building is low, you cannot afford very deep sections unless the beams are directly aligned with the partitions. The width of the block walls can also constraint the width of the beam. This is however different if a suspended ceiling will be used to conceal the beams and give the soffit of the floor a uniform look.
A general guide to the size of a beam may be obtained from the basic span-to-effective depth ratio from Table 7.4N of the BS EN 1992-1-1. The following values are can be used as a guide;
Overall depth of beam ≈ span/15. Width of beam ≈ (0.4 to 0.6) × depth.
The width may have to be very much greater in some cases, especially when a larger width is needed to reduce the shear stress in the beam. As said earlier, the size is generally chosen from experience. Many design guides are available which assist in design.
Efective flange width of beams
Clause 5.3.2.1 of EN 1992-1-1:2004 covers the effective flange width of beams for all limit states. For T-beams, the effective flange width, over which uniform conditions of stress can be assumed, depends on the web and flange dimensions, the type of loading, the span, the support conditions, and the transverse reinforcement. The effective width of the flange should be based on the distance lo between points of zero moment, which is shown in Figure 3.
Figure 3: Assumed points of zero moment for beams (Figure 5.2 of EN 1992-1-1:2004 (E))
The flange width for T-beams and L-beams can be derived as shown below. The notations are shown in figure 4.
beff =Σbeff,i + bw ≤ b
where; beff,i = 0.2bi + 0.1lo ≤ 0.2lo and beff,i ≤ bi
Figure 4: Effective flange width parameters ((Figure 5.3 of EN 1992-1-1:2004 (E))
Estimation of design loads
In the manual analysis of floor beams, loads are transferred from slab to beams based on the yield line assumption. However, finite element analysis will use a numerical approach to transfer loads from the slab to the beam. The magnitude of load transferred depends on if the slab is spanning in one-way or two-way. Typically, for a two-way slab, the loads are either triangular (for the beam parallel to the short span direction of the slab) or trapezoidal (for the beam parallel to the long span direction of the slab) as shown in Figure 5.
Figure 5: Load transfer from slab to beam
Typical load distribution from slab to beam is shown in the figure above. The analysis of trapezoidal and triangular loads on beams can be tedious especially for continuous beams with variable loading and unequal spans.
However, for beams of equal span and uniform loading, coefficients for bending moment and shear can be obtained from Chapter 12 of Reynolds and Steedman (2005). For the sake of convenience, the load transferred from the slab to the beam can be approximated as a uniformly distributed load, and the formulas for the transfer of such loads are given in Chapter 13 of Reynolds and Steedman (2005). They are presented in Table 1.
Table 1: Equivalent UDL load transferred from slab to beam
Type of slab
UDL transferred to beam on the long Span (kN/m)
UDL transferred to beam on the short Span (kN/m)
One-way slab
nlx/2
nlx/5
Two-way slab
nlx/2(1 – 0.333k2)
nlx/3
Where; n = ultimate pressure load on the slab = 1.35gk + 1.5qk lx = length of short span ly = length of long span k = ly/lx
Beams in a building can also be subjected to other loads and the typical values are;
Self-weight – Depends on the dimensions of the beam = (unit weight of concrete × width × depth)
Demountable lightweight partitions = 1 kN/m2
Block work with plaster = 3.5 kN/m2
Structural Analysis of Reinforced Concrete Beams
The primary purpose of structural analysis in building structures is to establish the distribution of internal forces and moments over the whole or part of a structure and to identify the critical design conditions at all sections. The geometry is commonly idealised by considering the structure to be made up of linear elements and plane two-dimensional elements.
Linear elastic analysis (with or without redistribution) or plastic can be carried depending on the one suitable for the problem. However for most buildings, linear elastic analysis is very adequate.
For the ultimate limit state only, the moments derived from elastic analysis may be redistributed (up to a maximum of 30%) provided that the resulting distribution of moments remains in equilibrium with the applied loads and subject to certain limits and design criteria (e.g. limitations of depth to neutral axis).
Different methods can be used for the elastic analysis of statically indeterminate beams such as;
Moment distribution method
Clapeyron’s theorem of three moments
Force method
Slope-deflection method
Stiffness method
Some coefficients are also provided in the code of practice for the evaluation of the bending moment and shear force in continuous beams.
Concrete Cover
An adequate concrete cover should be provided in reinforced concrete beams for the following reasons;
■ safe transmission of bond forces ■ durability ■ fire resistance
The minimum cover to ensure adequate bond should not be less than the bar diameter, or equivalent bar diameter for bundled bars, unless the aggregate size is over 32 mm. Under normal conditions, concrete cover of 35mm to 40 mm is usually adequate for beams.
Flexural Design of Reinforced Concrete beams
The stress block of a singly reinforced beam section (Eurocode 2) is shown in Figure 6;
Figure 6: Rectangular stress block Eurocode 2
From EC2 singly reinforced concrete stress block, the moment resistance capacity of the beam MRd is given by;
Compressive force in concrete = Design stress (fcd) x Area of compression block Fc = 0.5667fck × 0.8 x b = 0.4533bfck
From the stress block distribution;
z = d – 0.4x —— (2)
Clause 5.6.3 of EC2 limits the depth of the neutral axis to 0.45d for concrete class less than or equal to C50/60. Therefore for an under reinforced section (ductile failure);
x = 0.45d —— (3)
Combining equation (1), (2) and (3), we obtain the ultimate moment of resistance (MRd)
MRd = 0.167fckbd2 ———- (4)
Also from the reinforced concrete stress block, we can obtain the tensile design moment as;
MEd = Fsz ——- (5)
Where; Fs = fyk/1.15As1 ——-(6) As1 = Area of tensile reinforcement required
Substituting equation (6) into (5) and making As1 the subject of the formula;
As1 = MEd/(0.87fykz) ——(7)
The lever arm z in EC2 is given from equation (2), z = d – 0.4x Therefore, x = 2.5(d – z) M = 0.453 × fck × b × 2.5(d – z)z
Let k = M/(fck bd2) k can be considered as the normalised bending resistance
Rearranging; z = d[0.5 + √(0.25 – k/1.134)] —— (8) z = d[0.5 + √(0.25 – 0.882k)]
where ; k = MEd/(fckbd2) —– (9)
Anchorage of Reinforced Concrete Beams
Reinforcing bars should be well anchored so that the bond forces are safely transmitted to the concrete avoiding longitudinal cracking or spalling. Transverse reinforcement shall be provided if necessary. Types of anchorage are shown in Figure 7 below (Figure 8.1 EC2).
Figure 7: Different methods of anchorage (Fig 8.1 EN 1992-1-1:2004)
For bent bars, the basic tension anchorage length is measured along the centreline of the bar from the section in question to the end of the bar, where:
lbd = α1α2α3α4α5lb,req ≥ lb,min
where; lb,min is the minimum anchorage length taken as follows: In tension, the greatest of 0.3lb,rqd or 10ϕ or 100mm In compression, the greatest of 0.6lb,rqd or 10ϕ or 100mm
lb,rqd is the basic anchorage length given by, lb,rqd = (ϕ/4)σsd/fbd
Where; σsd = The design strength in the bar (take 0.87fyk) fbd = The design ultimate bond stress (for ribbed bars = 2.25η1η2fctd) fctd = Design concrete tensile strength = 0.21fck2/3 for fck ≤ 50 N/mm2
η1 is a coefficient related to the quality of the bond condition and the position of the bar during concreting η1 = 1.0 when ‘good’ conditions are obtained and η1 = 0.7 for all other cases and for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist
η2 is related to the bar diameter: η2 = 1.0 for φ ≤ 32 mm η2 = (132 – φ)/100 for φ > 32 mm
α1 is for the effect of the form of the bars assuming adequate cover α2 is for the effect of concrete minimum cover α3 is for the effect of confinement by transverse reinforcement α4 is for the influence of one or more welded transverse bars ( φt > 0.6φ) along the design anchorage length lbd α5 is for the effect of the pressure transverse to the plane of splitting along the design anchorage length
Curtailment of Reinforced Concrete Beams
Sufficient reinforcement should be provided at all sections to resist the envelope of the acting tensile force, including the effect of inclined cracks in webs and flanges.
Figure 8: Illustration of the curtailment of longitudinal reinforcement, taking into account the effect of inclined cracks and the resistance of reinforcement within anchorage lengths (Fig 8.7 EN 1992-1-1:2004)
For members with shear reinforcement the additional tensile force, ΔFtd, should be calculated according to clause 6.2.3 (7). For members without shear reinforcement ΔFtd may be estimated by shifting the moment curve adistance al = d according to clause 6.2.2 (5). This “shift rule” may also be used as an alternative for members with shear reinforcement, where:
al = z (cotθ – cotα)/2 = 0.5z cotθ for vertical shear links z = lever arm, θ = angle of compression strut al = 1.125d when cotθ = 2.5 and 0.45d when cot θ = 1
Shear Design of Reinforced Concrete Beams
Figure 9: Shear force model EC2
In EC2, the concrete resistance shear stress without shear reinforcement is given by;
Where; CRd,c = 0.18/γc k = 1 + √(200/d) < 0.02 (d in mm); ρ1 = As1/bd < 0.02 (In which As1 is the area of tensile reinforcement which extends ≥ (lbd + d) beyond the section considered) Vmin = 0.035k(3/2)fck0.5 K1 = 0.15; σcp = NEd/Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the concrete), fcd = design compressive strength of the concrete).
The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5) VRd,max = (bw.z.v1.fcd )/(cotθ + tanθ) ———- (11)
v1 = 0.6(1 – fck/250) Let z = 0.9d
If VRd,c < VEd < VRd,max Asw/S = VEd/(0.87fykz cotθ)
If VRd,max > VEd, calculate, the compression strut angle. θ = 0.5sin-1 [(VRd,max/bwd))/(0.153fck (1 – fck/250)] ———- (12)
A beam should not deflect excessively under service load. Excessive deflection of beams can cause damages and cracking to partitions and finishes. It can also impair the appearance of the building and cause great concern to the occupants of the building.
The selection of limits to deflection which will ensure that the structure will be able to fulfill its required function is a complex process and it is not possible for a code to specify simple limits which will meet all requirements and still be economical. Limits are suggested in the code but these are for general guidance only; it remains the responsibility of the designer to check whether these are appropriate for the particular case considered or whether some other limits should be used.
For deemed to satisfy basic span/effective depth (limiting to depth/250);
Actual L/d must be ≤ Limiting L/d × βs
The limiting basic span/ effective depth ratio is given by;
L/d = K [11 + 1.5√(fck)ρ0/ρ + 3.2√(fck) (ρ0/ρ – 1)1.5] if ρ ≤ ρ0 ——- (14)
L/d = K [11 + 1.5√(fck) ρ0/(ρ – ρ’) + 1/12 √(fck) (ρ0/ρ)0.5 ] if ρ > ρ0 —— (15)
Where; L/d is the limiting span/depth ratio K = Factor to take into account different structural systems ρ0 = reference reinforcement ratio = 10-3 √(fck) ρ = Tension reinforcement ratio to resist moment due to design load ρ’ = Compression reinforcement ratio
The value of K depends on the structural configuration of the member and relates the basic span/depth ratio of reinforced concrete members. This is given in the table 2;
Table 2: Basic span/effective depth ratio of different structural systems
Structural System
K
Highly stressed ρ = 1.5%
Lightly stressed ρ = 0.5%
Simply supported beams
1.0
14
20
End span of interior beams
1.3
18
26
Interior span of continuous beams
1.5
20
30
Cantilever beams
0.4
6
8
βs = (500As,prov)/(fykAs,req) —–(16)
For flanged sections with b/bw ≥ 3, the basic ratios for rectangular sections should be multiplied by 0.8. For values of b/bw < 3, the basic ratios for rectangular sections should be multiplied by (11 – b/bw)/10. For beams with spans exceeding 7 m, which support partitions liable to be damaged by excessive deflections, the basic ratio should be multiplied by 7/span.
Check for Cracking in Reinforced Concrete Beams
According to clause 7.3.1 of EN 1992-1-1:2004, cracking is normal in reinforced concrete structures subject to bending, shear, torsion or tension resulting from either direct loading or restraint or imposed deformations. However, cracking shall be limited to an extent that will not impair the proper functioning or durability of the structure or cause its appearance to be unacceptable.
Crack width in beams may be calculated according to clause 7.3.4 of EN 1992-1-1:2004. A simplified alternative is to limit the bar size or spacing according to clause 7.3.3. If crack control is required, a minimum amount of bonded reinforcement is required to control cracking in areas where tension is expected. The amount may be estimated from equilibrium between the tensile force in section just before cracking and the tensile force in reinforcement at yielding or at a lower stress if necessary to limit the crack width.
Table 3: Reinforcement detailing of reinforced concrete beams
Specific member issue
Value
Minimum area of steel required (As,min)
0.26(fctm/fyk)btd ≥ 0.0013bd
Maximum areas of steel required (As,max)
0.04bd
Minimum spacing of main bars
max{dg + 5mm, ϕ, 20mm}
Minimum area of shear links (Asw, min)
(0.08bwS√fck)/fyk
Maximum spacing of links
0.75d
In the most simplified manner, the detailing guide for beams is as shown below. However, in the most practical terms for simple residential buildings with moderate spans, reinforcements are rarely curtailed (except at the supports hogging areas) for ease in construction, especially where manual iron bending is employed. Note that to satisfy anchorage requirements, take the bob length for beams as 15 (15 x diameter of reinforcement).
Figure 10: Simplified detailing of a reinforced concrete beam
Design Example of a Reinforced Concrete Beam
A continuous beam in a residential building is loaded as shown below. The beam is an L-beam with effective flange width of 895 mm. The depth and width is 450 x 230 mm. Concrete cover = 35mm, fck = 25 MPa, fyk = 460 MPa
The bending moment and shear force diagram due to the applied load is shown below;
The minimum area of steel required; fctm = 0.3 × fck2⁄3 = 0.3 × 252⁄3 = 2.5649 N/mm2 (Table 3.1 EC2)
As,min = 0.26 × fctm/fyk × b × d = 0.26 × (2.5649/460) × 230 × 399 = 133.04 mm2 Check if As,min < 0.0013 × b × d (119.301 mm2) Since, As,min = 168.587 mm2, the provided reinforcement is adequate.
Check for deflection at the span K = 1.3 for simply supported at one end and continuous at the other end ρ = As/bd = 402/(895 × 399) = 0.0011257 < 10-3√25
Since the beam is flanged, check the ratio of b/bw = 895/230 = 3.89 Since b/bw is greater than 3, multiply the allowable L/d by 0.8
The allowable span/depth ratio = 0.8 × βs × 190.4123 = 0.8 × 1.808 × 146.471 = 275.412
Taking the distance between supports as the effective span, L = 3825 mm Actual deflection L/d = 3825/399 = 9.5864 Since 275.412 > 9.5864, deflection is deemed to satisfy Use 2Y16mm for the entire bottom span.
Top reinforcements (Hogging moment)
Support 3 MEd = 36.296 KNm
Since flange is in tension, we use the beam width to calculate the value of k (this applies to all support hogging moments)
k = MEd/(fckbw d2) = (36.296 × 106)/(25 × 230 × 3992) = 0.0396 Since k < 0.167 No compression reinforcement required z = 0.95d
Shear Design Using the maximum shear force for all the spans Support A; VEd = 65.19 KN VRd,c = [CRd,c.k. (100ρ1 fck)1/3 + k1.σcp]bw.d ≥ (Vmin + k1.σcp)bw.d
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/399) = 1.708 > 2.0, therefore, k = 1.708
σcp = NEd/Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the concrete), fcd = design compressive strength of the concrete.) Take NEd = 0
VRd,c = [0.12 × 1.708(100 × 0.00438 × 25 )1/3] 230 × 399 = 41767.763 N = 41.767 KN Since VRd,c (41.767) < VEd (65.19 KN), shear reinforcement is required.
The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5) VRd,max = (bw.z.v1.fcd)/(cotθ + tanθ) V1 = 0.6(1 – fck/250) = 0.6(1 – 25/250) = 0.54 fcd = (αcc fck)/γc = (0.85 × 25)/1.5 = 14.167 N/mm2 Let z = 0.9d
Since 0.200 > 0.18144, adopt 0.200 as the minimum shear reinforcement Maximum spacing of shear links = 0.75d = 0.75 × 399= 299.25 Provide Y8mm @ 250mm c/c as shear links (Asw/S = 0.4021) Ok
Concrete is arguably the most widely used construction material in the world. It is produced from a mixture of cement, sand, gravel, and water through a process known as hydration reaction. In its fresh state, concrete can be poured into different moulds and forms to achieve the desired shape. This is one of the reasons why it is an attractive construction material.
In its hardened state, concrete is very good in compression, but weak in tension. In order to augment this inherent weakness of concrete in tension, steel reinforcement is usually introduced to take up the tensile stresses. Any structure made up of steel reinforcement embedded in concrete to form a load resisting composite is known as a reinforced concrete structure. The process of specifying the member sizes of concrete and the area of steel required to ensure good performance of a structure under load is known as reinforced concrete design.
The key to the good performance of reinforced concrete structures lies in the complementary action of concrete and steel. This composite but complementary action is highlighted in the Table below;
Property
Concrete
Steel
Tensile strength
Poor
Good
Compressive strength
Good
Good (but slender members will buckle)
Shear strength
Fair
Good
Durability
Good
Fair (will corrode if unprotected)
Fire resistance
Good
Poor (will lose strength at elevated temperature)
By looking at the table above, you can see that all the desirable properties listed will be achieved if the two materials are combined. The structural design of reinforced concrete structures aims at taking advantage of the different but complementary characteristics of concrete and steel. Some of the basic theoretical assumptions that are made in design are as follows;
Concrete’s resistance to tension is zero (not practically true, the tensile strength of concrete is about 10% of its compressive strength, but this strength is usually ignored in ultimate limit state design)
The bond between steel and concrete is perfect
Based on these assumptions, all tensile stresses in a structure are transferred to the reinforcements during design. These tensile stresses are transferred by the bond between the concrete and the reinforcement. Perfect bond assumption demands that the strain in the reinforcement is identical to the strain in the adjacent concrete (compatibility of strains). Furthermore, the coefficients of thermal expansion for steel and for concrete are of the order of 10 x 10-6 per ℃ and 7-12 x 10-6 per ℃ respectively. These values are sufficiently close that problems with bond seldom arise from differential expansion between the two materials over normal temperature ranges.
Practically, if the bond between reinforcement and steel is not adequate. the reinforcing bars will slip within the concrete and there will be no composite action. Adequate bonding is ensured by detailing the structure such that the reinforcement is properly anchored in the concrete. Reinforcement bars are also ribbed in order to facilitate bonding with concrete.
Reinforcement bars are ribbed to enhance bonding with concrete
It is normal for cracking to occur in concrete when it subjected to tensile or flexural stress. This cracking, however, does not mean that the structure is not safe provided it is adequately reinforced that the crack width is kept to a minimum. If the crack width is excessive, there may be serviceability and/or durability issues (corrosion of reinforcement) in the structure.
Furthermore, when the compressive or shearing forces exceed the strength of the concrete, then steel reinforcement must again be provided to supplement the load-carrying capacity of the concrete. For example, compression reinforcement is generally required in a column, where it takes the form of vertical bars spaced near the perimeter. To prevent these bars buckling, steel binders are used to assist the restraint provided by the surrounding concrete.
Reinforced concrete has a lot of applications in construction, and has been applied in a lot of structures worldwide – bridges, industries, residential buildings, highrise buildings, swimming pools, retaining walls, highways (rigid pavement) etc. The design of any reinforced concrete structure should start with the understanding and behaviour of the structure to be designed under load. The designer will need to specify the load path (how the load will be transferred from the superstructure to the foundation).
For instance, to design a building, the structure can be broken down into the following elements. This is what is called the general arrangement of the building.
Beams: horizontal members carrying lateral loads
Slab: horizontal plate elements carrying lateral load
Columns: vertical members carrying primarily axial load but generally subjected to axial load and moment
Walls: vertical plate elements resisting vertical, lateral or in-plane loads
Bases and foundations: pads or strips supported directly on the ground that spread the loads from columns or walls so that they can be supported by the ground without excessive settlement. Alternatively, the bases may be supported on piles.
Knowledge of reinforced concrete design starts from knowing how to design the separate elements listed above. However, it is important to recognize the function of the element in the complete structure and that the complete structure or part of it needs to be analysed in order to obtain actions for design.
Designers are expected to follow a generally accepted code of practice in their design and detailing. This is to enable a quick check and understanding of the design by other engineers. Some codes of practice used in the design of concrete structures across the world are;
EN 1992-1-1:2004 – Eurocode 2: Design of concrete structures – Part 1-1: General rules and rules for buildings (European Union) BS 8110-1:1997 – Design of reinforced concrete structures – Rules and general rules for buildings ACI 318-19: Building Code Requirements for Structural Concrete and Commentary IS 456-2000: Plain and Reinforced Concrete – Code of Practice (Indian Standards) CSA A23.3:2014 – Design of concrete structures (Canadian Standards Association) AS 3600:2018 – Concrete structures (Standards Australia)
To improve the safety and serviceability of super-tall buildings in strong winds, aerodynamic optimization of building shapes is considered to be the most efficient approach. Aerodynamic optimization is aimed at solving the problem from the source in contrast to structural optimization which is aimed at increasing the structural resistance against winds [1]. Aerodynamics is concerned with the study of air in motion and how it interacts with solid objects around it.
According to Tanaka et al. [2], the free form architecture being expressed in the modern construction of tall buildings today do not only have the advantage of reflecting the architect’s idea, but also have the advantage of reducing the effects of wind on the structure due to aerodynamic effects. Some aerodynamic modifications in architectural design are one of the effective design approaches which can significantly reduce the effect of the lateral wind force and thus, the building motion [3, 4].
The modifications usually employed to improve aerodynamic response of tall buildings are [5];
tapered cross-section
setback
sculptured top
modifications to corner geometry, and
addition of openings through building
Some aerodynamic optimization of tall buildings [4]
By changing the flow pattern around the building due to aerodynamic modifications of the building shape, (i.e. an appropriate choice of building form), wind response can be moderated when compared to the original building shape. As far as wind loading and resulting motions are concerned for tall and slender buildings, the shape is critical and a governing factor in the architectural design. Wind tunnel test is the most popular method of evaluating the aerodynamic behaviour of a high-rise building.
Distribution of mean vertical velocity around a square and helical building form [2]
Improving the aerodynamic performance of a tall building can be achieved by local and global shape mitigations. Local shape mitigations, such as corner mitigations, have a considerable effect on structural and architectural design, while global shape mitigations have a minor effect on structural and architectural design [6]. Essentially, the precise selection of the outer shape details of a building can result in a significant reduction in forces and motions caused by wind.
Aerodynamic forms in general reduce the along-wind response as well as across-wind vibration of the buildings caused by vortex-shedding by “confusing” the wind (i.e., by interrupting vortex-shedding and the boundary layer around the façade and causing mild turbulence there).
Absolute towers, MISSISSAUGA, CANADA
While irregular forms pose challenges to structural engineers for developing the structural framework, they can be advantageous in reducing wind load effects and building responses [7]. Examples employed in contemporary tall buildings are chamfered or rounded corners, streamlined forms, tapered forms, openings through a building, and notches.
The Shanghai World Financial Center and the Kingdom Center in Riyadh (see below) employ a large through-building opening at the top combined with a tapered form. The proposed Guangzhou Pearl River Tower’s funnel form facades catch natural wind not only to reduce the building motion but also to generate energy using wind. Due to the nature of the strategy which manipulates building masses and forms, this approach blends fittingly with architectural aesthetics.
Kingdom Centre, Riyadh
References
[1] Xie J. (2014): Aerodynamic Optimization of Supertall buildings and its effectiveness assessment. Journal of Wind Engineering and Industrial Aerodynamics (130):88-98 [2] Tanaka H., Tamuro Y., Ohtake K., Nakai M., Kim Y.C., and Bandi E.K. (2013): Aerodynamic and flow characteristics of tall buildings with various unconventional configurations. International Journal of High-Rise Buildings 2(3):213-228 [3] Ilgin E.H. and Gunel M.H. (2007): The role of aerodynamic modifications in the form of tall buildings against wind excitation. METU JFA 24(2):17-25 [4] Elshaer A., and Bitsuamlak G. (2018): Multi-objective aerodynamic optimization of tall buildings openings for wind-induced excitation reduction. Journal of Structural Engineering 144(10):27-38 [5] Kareem, A., Kijewski, T. and Tamura, Y. (1999): Mitigation of Motion of Tall Buildings with Recent Applications. Wind and Structures, 2(3): 201-251. [6] Elshaer A., Bitsuamlak G., and El Damatty A (2016): Aerodynamic shape optimization of tall buildings using twisting and corner modifications. 8th International Colloqium on Bluff Body Dynamics and Applications, Northeastern University, Boston Massachussets, USA [7] Ali M.M., and Moon K.S. (2007): Structural developments in tall buildings: Current trends and future prospects. Architectural Science Review 50(3):205-223
Negative skin friction is a downward drag force exerted on a pile by the soil surrounding it. This is the reverse of the normal skin friction or shaft resistance needed to support piles. If the downward drag force is excessive, it can cause the failure of the pile foundation.
Where applicable, negative skin friction must be allowed when considering the factor of safety on the ultimate load-carrying capacity of a pile. The factor of safety (FOS) where negative skin friction is likely to occur is given by;
Negative skin friction may occur in pile foundations due to the following circumstances;
If a fill of clay soil is placed over a granular soil or completely consolidated soil into which a pile is driven, the consolidation process of the recently placed fill will exert a downward drag of the pile during the process of consolidation.
If a granular fill material is placed over a layer of compressible clay, it will induce the process of consolidation in the clay layer and exert a downward drag on the pile.
Lowering of the water table will increase the vertical effective stress on the soil at any depth which will induce consolidation settlement of the clay. If a pile is located in the clay layer, it will be subjected to a downward drag force.
According to section 7.3.2.2 of EN 1997-1:2004 (Eurocode 7), if ultimate limit state design calculations are carried out with the downdrag load as an action, its value shall be the maximum, which could be generated by the downward movement of the ground relative to the pile.
Furthermore, the calculation of maximum downdrag loads should take account of the shear resistance at the interface between the soil and the pile shaft and downward movement of the ground due to self-weight compression and any surface load around the pile. An upper bound to the downdrag load on a group of piles may be calculated from the weight of the surcharge causing the movement and taking into account any changes in groundwater pressure due to ground-water lowering, consolidation or pile driving.
Computation of negative skin friction on a single pile
The magnitude of negative skin friction (Fn) for a single pile in filled up soil may be taken as;
(a) Cohesionless soil Fn = 0.5K’γf‘Hf2tanδ’
Where; K’ = coefficient of earth pressure = Ko = 1 – sinφ’ γf‘ = Effective unit weight of material causing down drag Hf = Depth of compressible layer causing down drag δ’ = soil-pile angle of friction ≈ 0.5φ’ – 0.6φ’
(b) Cohesive soil Fn = PHfS
Where; P = Perimeter of pile S = Shear strength of soil
Negative skin friction on pile groups When a group of piles passes through a compressible fill, the negative skin friction may be obtained using any of the following methods;
(a) Fng = nFn (b) Fng = sHfPg + γf‘HfAg
Where; n = number of piles in the group γf‘ = Effective unit weight of material causing down drag Pg = Perimeter of the pile group Ag = Cross-sectional area of the pile group within the perimeter Pg S = Shear strength of the soil along the perimeter of the group
where: fctm is the mean value of axial tensile strength of concrete at 28 days, see Table 3.1 of EN 1992-1-1:2004 fyk is the characteristic yield stress of reinforcement steel bt is the mean width of the tension zone; for a T-beam with the flange in compression, only the width of the web is taken into account in calculating the value of bt d is the effective depth of concrete cross-section.
Any section containing less than As,min as given in equation (1) should be treated as an unreinforced section.
The minimum area of reinforcement for different classes of concrete strengths assuming fyk = 500 MPa is given in the Table below;
fck (MPa)
fctm (MPa)
0.26fctm/fyk
25
2.6
0.13%
28
2.8
0.14%
30
2.9
0.15%
32
3.0
0.16%
35
3.2
0.17%
40
3.5
0.18%
45
3.8
0.20%
50
4.2
0.21%
For a T-beam with the flange in tension (such as upstand beams or ground beams), the minimum area of reinforcement is given by equation (2);
As,min = 0.26 (fctm / fyk) Ac,t —– (2)
Where Ac,t is the area of the tension zone above the neutral axis of the reinforcement.
BS 8110-1:1997 made express provisions for flanged beams for different stress states of the web or the flange. The requirements are given in the table below;
Flanged beam in flexure (tension reinforcement)
Definition of Percentage
Minimum Percentage (fy = 500 MPa)
(a) Web in tension
bw/b < 0.4
100As/bwh
0.18%
bw/b ≥ 0.4
100As/bwh
0.13%
(b) Flange in tension
T-beam
100As/bwh
0.26%
L-beam
100As/bwh
0.20%
Maximum area of longitudinal reinforcement The maximum area of tension and compression reinforcement in reinforced concrete beams is;
As,max = 0.04Ac —- (3)
Where; Ac is the area of the cross-section
Shear Reinforcements (links and stirrups) The minimum area of shear reinforcement in beams should be calculated from;
Asw/sbw ≥ ρw,min —– (4)
Where; ρw,min = (0.08√fck)/fyk Asw = Area of shear reinforcement s = spacing of shear reinforcement bw = width of the web
Solved Example For the concrete section shown below, find the minimum area of longitudinal reinforcement required for; (a) When the web is in tension (b) When the flange is in tension fck = 25 MPa, fyk = 500 MPa
(a) When the web is in tension Using Eurocode; As,min = 0.26 (fctm / fyk) btd As,min = 0.26 x (2.6/500) x 250 x 695 = 235 mm2 > 0.0013btd (0.0013 x 250 x 695) = 226 mm2
Using Table 3.25, BS 8110-1:1997; bw/b = 250/1500 = 0.167 < 0.4 As,min = 0.18bwh/100 = (0.18 x 250 x 750)/100 = 338 mm2
(b) When the flange is in tension Using Eurocode; As,min = 0.26 (fctm / fyk) Ac,t Ac,t = (550 x 250) + (1500 x 145) = 355000 mm2 As,min = 0.26 x (2.6/500) x 355000 = 480 mm2
Using Table 3.25, BS 8110-1:1997; As,min = 0.26bwh/100 = (0.26 x 250 x 750)/100 = 488 mm2
The construction of the world’s longest immersed tunnel has officially begun. The Fehmarnbelt Tunnel that will connect Denmark and Germany, is scheduled to be officially opened by 2029. It’s one of Europe’s largest ongoing infrastructure projects, with a budget of more than US$8 million.
The tunnel will have an 18 kilometers extension and will be built across the Fehmarn Belt, a strait between the German island of Fehmarn and the Danish island of Lolland. It will be an alternative to the current ferry service, which takes 45 minutes. Traveling through the tunnel will take seven minutes by train and ten minutes by car.
It will be the longest combined road and rail tunnel in the world, with two double-lane motorways, separated by a service passageway, and two electrified rail tracks. Besides the benefits to passenger trains and cars, it will have a positive impact on the flow of freight trucks and trains.
According to Femern website, the contract for the construction of the 18km tunnel was signed on 30 May 2016. The contract which is worth almost 4 billion Euro was signed between Femern A/S (the Danish state-owned company tasked with designing and planning the link) and international contractors responsible for the establishment of the 18 km Fehmarnbelt tunnel between Rødbyhavn and Puttgarden. The contract with the Dutch consortium, Fehmarn Belt Contractors (FBC) came into force in November 2019. This covers dredging and reclamation.
In May 2020, Femern A/S initiated conditional contracts for the tunnel as well as the portals and ramps, which were signed with the consortium, Femern Link Contractors (FLC). The contracts will be activated with effect from 1 January 2021.
Construction work started in the summer on the Danish side. Work will carry on for a few years in Denmark before moving into German territory. Workers are now building a new harbor in Lolland and in 2021 they will start construction of a factory, both meant to support work on the tunnel. Located behind the port, the factory will have six production lines to assemble the 89 massive concrete sections that will make up the tunnel.
The immersed tunnel and the tunnel factory as well as the portals and ramps
Consortium: Femern Link Contractors (FLC) Contractors: VINCI Construction Grands Projets S.A.S. (France) Per Aarsleff Holding A/S (Denmark) Wayss & Freytag Ingenieurbau AG (Germany) Max Bögl Stiftung & Co. KG (Germany) CFE SA (Belgium) Solétanche-Bachy International S.A.S. (France) BAM Infra B.V. (Holland) BAM International B.V. (Holland) Sub-contractors: Dredging International N.V. (Belgium) Consutants: COWI A/S (Denmark)
