Beams are horizontal structural elements used for supporting lateral loads. In conventional reinforced concrete structures, beams usually receive load from the floor slab, but may also be subjected to other loads such as wall load, finishes, services installation, etc. The design and detailing of reinforced concrete beams involves the selection of the proper beam size and the quantity of longitudinal and shear reinforcement that will satisfy ultimate and serviceability limit state requirements. Afterward, it is very important that the beam reinforcements are placed and arranged properly on site, to avoid construction error.
This article presents guides and short notes on how to properly place and arrange beam reinforcements on site. This is very important for students (on industrial training) or fresh graduates that are new to construction site practices.
Typical detailing of a reinforced concrete beam
The technical guides to the detailing and arrangement of beam reinforcements are as follows;
(1) Confirm the formwork dimensions and stability Beam reinforcement placement commences immediately after the carpenters complete the soffit formwork of the floor. At this point, it is important to verify that the formwork dimensions have been done according to the design specification. The checks should include measuring the beam width and the drop of the beam relative to the proposed surface of the slab. The sides of the beam drop should be checked for verticality to avoid having slanted beam sides, and the soffit should be checked for perfect horizontality. It is usually very difficult and depressing to make corrections after the reinforcement has been placed.
Poorly sized beam dimensions can also lead to compromised concrete cover, which can impact the bonding of concrete and reinforcement, the fire rating, and the durability of the building. It is typical to take a bent link (stirrup) and insert it in the beam drop to check if the desired concrete cover has been achieved. It can also be done to confirm that the beam size has been done according to specification. Furthermore, it is important to check the stability and bracing of the formwork to avoid collapse, bulging, or bursting.
(2) Confirm cut length and bending dimensions Before the placement of the rebars commences, the site engineer should confirm that reinforcements have been cut and bent according to the design requirements. Bar bending schedule (BBS) can be used as a check document, provided it is very accurate and synchronised with what has been done on-site. All changes in dimensions or arrangements should be communicated ahead of time.
Cut and bent beam reinforcements at a construction site awaiting tying and placement
(3) Tying of Beam Reinforcements Links and stirrups are used for resisting shear stresses and torsion in a reinforced concrete beam. They are also used for holding the top and bottom reinforcement of beams in place. The correct quantity required should be installed and spaced as recommended in the structural drawing.
Tying and arrangement of beam reinforcement on site
(4) Arrangement of Primary and Secondary Beams For an internal beam (secondary beam), which is supported by another beam on either or both sides, the top bars should rest on the top bars of the primary beam. Also, the bottom bars should rest on the bottom bars of the primary beam as shown in the image below.
(5) Arrangement of Beam-Column Junction at Corners For beams occurring at a column junction at the corner of a building, either beam reinforcements can rest on each other.
Corner beam-column junction
(6) Arrangement of Continuous Beam-Column Junction At a continuous beam-column junction as shown below, the continuous beam with heavier reinforcement will have to be placed on the beam with lighter reinforcement or the beam terminating at that point.
Internal beam-column Junction
(7) Reinforcement Arrangement of Overhang (Cantilever) Beams For cantilever beams, the main bars are the top reinforcements. If a side beam is resting on it (when the cantilever beam is acting as a primary beam) as shown in the image below, the top bars of the secondary beams beam must rest on the top bars of the cantilever beams. In the same vein, the bottoms bars of the side beam (main rebar) must rest on the bottom bars of the cantilever beam.
Well designed and constructed cantilever beam
The arrangement of beams at a site should be checked strictly to conform to the engineer’s design drawings. Failure to check construction details properly may lead to the failure of the member or lack of robustness. We believe that beams are arguably the most common structural members in a building. Hence, attention should be given to its arrangements for robustness and efficient load transfer.
Researchers from the Department of Civil, Environmental, and Geomatic Engineering, ETH Zurich, Switzerland have offered new insights on the failure of Fukae Bridge during the Great Hanshin 1995 earthquake in Kobe, Japan. The magnitude 6.9 earthquake which left many damages behind its wake caused about $100 billion loss in properties, including the collapse of all 18 spans of the elevated Route No. 3 of Hanshin Expressway.
In the catastrophic seismic event, the deck of the bridge which was monolithically connected to 3.1m diameter piers failed and overturned dramatically. However, the massive 17–pile groups supporting the piers survived the earthquake and are still in use, supporting the new bridge. The piles were founded in alluvium sand and gravel formation.
Figure 1: Japan, showing seismicity from 1961 to 1994, location of the 1995 Hanshin-Awaji earthquake, and projected rupture areas of largest historical earthquakes to shake Kobe, which were subduction-zone earthquakes in 1944 and 1946 (Source: https://www.geosociety.org/)
According to the authors, the lessons learned from the Kobe earthquake influenced substantially the seismic practices and codes not only in Japan but also worldwide. However, new insights have been offered on the failure of the bridge as the researchers carried out nonlinear finite element analysis of the bridge. The findings of the study were published in Elsevier -Soils and Foundations Journal.
Previous studies attributed the failure of the bridge deck to inadequate structural design regarding mainly the addition of a prematurely terminated third row of longitudinal reinforcement and insufficient shear capacity due to poor transverse reinforcement. Surprisingly, the failure did not occur at the bottom of the pier (location of maximum bending moment) but 2.5 m above the pilecap, where shear cracking initiated.
Figure 2: Reinforcement details and critical section where the 3rd row of longitudinal rebars was terminated and shear cracking initiated; lower section of a collapsed pier exposed in the Hanshin Expressway earthquake museum (Sakellariadis et al, 2020)
The authors, therefore, re-examined the collapse of the bridge by comparatively assessing the performance of the original foundation, which survived the earthquake and is still in use (for the fully replaced bridge), to that of alternative design concepts, considering nonlinear soil-foundation interaction.
To achieve this, they carried out the static and dynamic response of a single segment of the Fukae bridge employing the FE method using ABAQUS software. Six different foundation configurations were explored, starting with the actual very stiff 17–pile foundation with its large cap and a highly nonlinear rocking shallow footing alternative consisting only of the pile cap. The soil profile was modelled with hexahedral (8-node) elements, adopting a thoroughly validated kinematic hardening model, with a modified pressure-dependent Von Mises failure criterion and associated plastic flow rule.
Key attributes of the FE model of a single segment of the Fukae bridge in Abaqus (Sakellariadis et al, 2020)
To gain deeper insights on the collapse mechanism, critical reinforced concrete (RC) structural members (pier and piles) were modelled and simulated with nonlinear solid elements, employing the Concrete Damaged Plasticity (CDP) model.
The numerical simulation successfully reproduced the shear-dominated failure mode at the longitudinal reinforcement cutoff region. The analysis also confirmed that, despite being highly overdesigned, the pile group foundation experienced limited but non-negligible swaying and rocking during shaking, as a result of which the piles were subjected to tension and combined shear-moment loading.
The resulting stiffness reduction of the cracked under tension piles leads to load redistribution towards the stiffer compressed piles, preventing plastic hinging of the weaker piles (under tension). These findings were found to be consistent with the post-earthquake in-situ testing.
Some of the conclusions inspired by the study are as follows;
Reducing the number of pile rows in the critical direction of seismic loading is promising in improving the seismic performance, but not sufficient to prevent severe damage of the examined pier.
The unconnected piled raft alternative reduces both the structural distress and the settlement.
References L. Sakellariadis, I. Anastasopoulos and G. Gazetas, Fukae bridge collapse (Kobe 1995) revisited: New insights, Soils and Foundations, https://doi.org/10.1016/j.sandf.2020.09.005
Disclaimer The original article cited above is an open-access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) which permits sharing, copying, and redistribution of the material in any medium or format provided appropriate credit is given. It has been presented on www.structville.com in form of a news article.
Building design concepts all over the world are evolving on a daily basis. Architects are coming up with different ways of making buildings appear sleeker, elegant, and more beautiful. While the aesthetic appeal offered by these contemporary buildings are commendable and decorate our environment, they can be a nightmare to structural engineers who have the responsibility of making the buildings safe and stable.
As a modern structural engineer, there is no good reason for not following up on the latest trends in design and construction. All the modern solutions available for analysis, design, and construction of complex-looking buildings should be handy to you. Furthermore, engineers should always follow up on the latest developments in construction materials and techniques by attending workshops, seminars, webinars, expos, conferences, etc that are relevant to their field of practice. That is one of the best ways of staying current and relevant in the industry.
At Structville, we are interested in always driving up curiosity and creativity in the minds of civil engineers. It is not a good idea to ‘over-modify’ the concepts of an architect in the quest for structural stability. Therefore, you should not lack ideas on how to approach any problem that comes to you. That is the major sign of a quality engineer.
Ideas on how to support a complex architectural design may not always come immediately, but if you sleep and play over it, bearing in mind the ‘basic’ principles of statics of structures, and having a wide knowledge of different structural schemes, you will eventually find the solution. In practical design, you seldom need to think far or explore complex theories before you get an idea; the basic knowledge of statics of structures is usually sufficient.
I came across an interesting architectural design on Instagram. No credit was given for the design, but the handle that posted the pictures is @academiadeestruturas. I am posting the pictures of the design below for us to explore our ingenious ways of carrying out the structural design of the building.
It is my hope that by the time a lot of professionals have contributed their ideas on how to go about the design, a lot of knowledge must have been passed around. For this structural design challenge, you can post your ideas as a comment or you can support with relevant sketches if necessary. The requirements of the challenge are as follows;
(1) Describe the structural scheme that you will use to support the building (2) Describe the possible load path of the structural scheme (3) Comment on the stability of the selected structural scheme (4) State the most likely material that will work for the design
In a general, let us assume that the floor should be able to withstand a live load of 3.0 kN/m2. No calculations whatsoever are needed.
The pictures of the architectural design are given below;
You are also free to comment or advise on the structural scheme selected by others. However, keep all interactions and use of language civil, professional, and respectful. Every engineering solution is a solution, but some solutions are better than the others. Remember to share this challenge with your friends and colleagues. Thank you, and God bless you.
A lintel is a secondary structural member used to support a wall spanning above an opening in a masonry wall. They are usually categorised as elements of minor structural importance unless the span of the opening is very large. Lintels are usually horizontal members with a rectangular cross-section and are often made of reinforced concrete or steel. Other materials such as stone, brick, reinforced brick, and timber can also be used.
Lintels are important in a building, and the details can be included in the structural drawing in the form of lintel schedule to show the rebar specification based on the size of openings. Such a schedule can aid the quantity surveyor to capture the cost of the lintels in the bill of quantity.
In Nigeria, the most popular material used for lintel is reinforced concrete, which may be cast-in-situ or precast. Cast-in-situ lintels can be constructed to span around the entire length of walls in the building, and are often referred to chained lintels. On the other hand, they can be constructed to span across the opening concerned without extending far beyond the opening. Precast lintels are often used to span across the opening concerned only.
Chaining of lintels in a building
Precast lintels always come in handy for residential buildings and are mostly employed by value engineers. The major benefits include cost and time saving for the project or contractor.
Precast concrete lintel
Essentially, the idea behind thin precast lintels is that the provision of top reinforcements and hanger bars can be neglected for lightly loaded one-span members in bending. Reinforcements are provided at the bottom of the lintel to take up all the flexural stresses. The conventional cast-in-situ lintel, however, has an edge over precast lintel if the building will be subjected to remodeling or if the lintels spans between reinforced concrete columns.
Advantages of Precast Lintels
The advantages of using precast lintels in construction are as follows;
Significant saving in reinforcement is obtained by using offcuts from other elements
It more sustainable because of the reduction in cement and aggregates requirements
Quality control in the production of the lintel is enhanced
It also saves and/or reduces the cost of formwork
It reduces the cost of labour
It is relatively easy to construct
Saves time on the construction program since it can be produced and stored ahead of schedule.
Disadvantages of Precast Lintels
A lifting mechanism may be required to put the lintel in place
The robustness of the wall/element is reduced
The bond strength of the wall and the lintel is weaker when compared with cast-in-situ
Specifications for Precast Lintels
(1)Sizing: The width of a lintel should be equal to the width of the wall they are supporting or sitting on, unless otherwise specified. The depth of a lintel should be a minimum of 100 mm, and the length depends on the span of opening. An end bearing length of at least 200 mm beyond the face of the opening should be provided. When the span of a lintel exceeds 2m, a more serious consideration should be given.
(2)Concrete: A minimum concrete grade of 15 MPa should be used for the construction of lintels, unless otherwise specified. The quality of the cement, aggregate, and water should be the minimum requirement used for the construction of major reinforced concrete elements.
(3) Reinforcements: A 2 or 3 number of high yield 12mm diameter bars (2Y12 or 3Y12) placed at the bottom of the lintel is usually sufficient. However, the minimum area of reinforcement required for the section provided should not be violated (0.13% of the cross-sectional area). Top reinforcements (hanger bars) with links (stirrups) may or may not be provided depending on the depth adopted. When links are provided, the minimum area and spacing required should be provided.
Typical reinforcement cutting and fixing for a precast lintel
Technical Guide for Construction of a Precast Lintel
(1) Surface preparation: A flat surface is required for casting of a precast lintel. This can be achieved on a well-leveled floor or on a marine plywood placed on a hard surface. The marine board will cover all undulations that may come from the surface, and will give a smooth finished appearance.
(2) Ease of removal: A cement sack or membrane should be spread on the surface to receive the lintel concrete, and to aid in quick removal without sticking with the platform. Alternatively, the surface of the marine board can oiled using an approved material.
(3) Shuttering/Mould Construction: The mould for the precast concrete can be achieved using planks or marine boards joined together. Special moulds made of steel or plastic can also be used where available.
(4) Concrete cover: Provision of nominal concrete cover (to the links) not less than 15mm should be made. Normally, fire rating, bond, and durability (exposure conditions) of the lintel should inform the choice of concrete cover.
(5) Concreting: A concrete mix of strength not less than 15 MPa after 28 days should be carefully poured and consolidated in the form/shuttering prepared. A vibrator is usually not needed for precast lintels as consolidation can be achieved by tamping the concrete or tapping the sides of the form/shuttering. The top surface can be finished using a trowel.
(6) Curing: Remove the forms after 24 hours and cure using any suitable method for at least 7 days before placement.
(7) Handling and Storage: Precast concrete lintels are fragile elements that can get damaged when not properly handled. All forms of impact or dropping from height should be avoided. The elements should be stored on a flat surface and protected until they are due for placement.
Arrangement of temporary forms on a flat surface for precast lintel
A unit of precast lintel of 100mm x 230mm x 1300mm
Placement/Installation of Precast Lintels
Place high-quality cement mortar around the end bearing of the masonry wall that will support the lintel
Lift the precast lintel carefully and place it on the mortar taking note of alignment and level
Apply temporary support to transfer the stresses efficiently.
Recommendation
Architects should make an allowance of using precast lintels in their designs by making window or door opening spans on masonry walls. Though this may not always be feasible due to the presence of columns in important areas.
Low-cost and mass housing schemes can employ precast lintels to save cost and improve on delivery time.
Value Engineers, Contractors, and Consultants should adopt precast lintels in their constructions.
Conclusion
The use of precast lintel is a cost and time-saving alternative in construction. It is also a sustainable practice that can lead to the reuse of waste materials on-site like empty cement bag, old planks/formworks, reinforcement offcuts, and waste concrete materials.
PS: At Structville, we value sustainability and value engineering. Do you want a review and analysis of your drawings and bills to see where you can save cost and time without sacrificing quality? You can reach us at info@structville.com to have a deal with our team of value engineers.
A plate is a flat structural element that has a thickness that is small compared with the lateral dimensions. Plates are bounded by two parallel planes, called faces, and a cylindrical surface called an edge or boundary. The thickness is usually constant but may be variable and is measured normal to the middle surface of the plate. Circular plates are common in many structures such as nozzle covers, end closures in pressure vessels, pump diaphragms, turbine disks, and bulkheads in submarines and airplanes, etc. They are also used as bases or covers of cylindrical water tanks and structures. When they are resting on the ground, they behave as plates on an elastic foundation.
Typical circular plate covering an undergound tank
In the elastic analysis of circular plates, it is convenient to express the governing differential equation in polar coordinates, even though rectangular (Cartesian) coordinates can be used (as applied in this article). This can be readily accomplished by a coordinate transformation and can be found in many pieces of literature.
If the coordinate transformation technique is used, the following geometrical relations between the Cartesian and polar coordinates are applicable;
x = a cosθ; y = a sinθ; x2 + y2 = a2 —- (1) θ= tan-1(y/x)
Referring to the above; ∂r/∂x = x/r =cosθ; ∂r/∂y = y/r =sinθ ∂θ/∂x= -y/r2= -sinθ/r; ∂θ/∂y = x/r2= cosθ/r
The governing differential equation (biharmonic equation) for the elastic deflection of a plate in polar coordinates is given by equation (2);
This is equivalent to the moment-curvature relationship of a thin plate which is given by (3);
∂4w/∂x4 + 2∂4w/∂x2∂y2 + ∂4w/∂y4 = q/D — (3)
Where D is the flexural rigidity of the plate which is given by; D = Eh3/12(1 – ν2) — (4)
h = thickness of the plate E = Modulus of elasticity of the plate material ν = Poisson ratio of the plate material
Circular Plate Clamped at the Edge and Subjected to Uniform Load
Let us consider a circular plate clamped at all edges (fixed) and subjected to a uniform lateral load q. The boundary of the plate is given by y2 + x2 = a2
For a such a clamped circular plate, the slope and deflection at the edges are zero; w = 0; ∂w/∂x = 0; ∂w/∂y = 0 along y2 + x2 = a2
The solution to the deflection equation of the plate is given by;
w = c(x2 + y2 – a2)2 which satisfies the condition given above.
Let f(x, y) = x2 + y2 – a2 The partial derivatives are given by;
Solved Examples on the Analysis of Circular Plates
A circular plate of thickness (150 mm) and diameter 10m is subjected to a uniform lateral load of 10 kN/m2. Calculate the deflection and bending moment at the centre of the plate assuming; (a) Simply supported conditions (b) Clamped edge conditions (v = 0.2; E = 2.1 x 107 kN/m2)
Solution The flexural rigidity of the plate is given by; D = Eh3/12(1 – ν2) = (2.1 × 107× 0.153)/12(1 – 0.22) = 6152.34 kNm
(ii) Maximum bending moment Mmax = qa2/16 × (3 + v) =(-10 × 52)/16 × (3 + 0.2) = 50 kNm
When analysed using finite element analysis on Staad Pro;
The maximum deflection at the centre of the plate was observed to be 68.749 mm as shown in the table below. This comparable with 68.78 mm obtained using classical theory.
The bending moment obtained using finite element analysis was observed to be 49.9 kNm/m as shown below. A value of 50 kNm was obtained using classical theory.
(b) Assuming clamped edge conditions; (i) Maximum deflection wmax = -qa4/64D = (-10 × 54)/(64 × 6152.34) = 0.0158 m = 15.87 mm
(ii) Bending moment at the center Mmax = q/16 × (1 + v)a2 = 10/16 × (1 + 0.2) × 52 = 18.75 kNm
(iii) Bending moment at the edge Mr = -qa2/8 = (-10 x 52)/8 = -31.25 kNm
When analysed on Staad Pro;
The maximum deflection at the centre of the plate was observed to be 15.964 mm (compared with the value of 15.87 mm obtained using classical theory)
The bending moment values from the finite element analysis result are shown below;
On a construction site, there are different quality control/quality assurance systems to ensure the efficient delivery of a project. One of these is the Bar Bending Schedule (BBS). A bar bending schedule is a document showing the list of structural members, bar mark, type of reinforcement, size of rebar, number of rebars for each member, cutting length, total length, shape, and location/spacing/position of all reinforcements in the working drawing.
Bar bending schedule ensures that reinforcement cutting, bending, and placement are carried out in the most efficient manner on site. It also guides against the excessive waste of reinforcement by minimising the number of useless offcuts. The document is prepared in a manner such that the reinforcement requirements and specification can be recognised and applied without confusion on site. It is issued for different structural elements such as reinforced concrete beams, columns, slabs, foundation, staircase, etc.
In Nigeria, the table showing lists of standard shape codes given in BS8666 doesn’t necessarily conform to the style used in Nigerian construction sites. Some modifications are usually done to avoid confusion, mistakes, and to reduce the workload/effort of the fitters (iron benders). It is very typical for detailers to sketch the actual bending shape on the document instead of referring to the standard shape code.
The importance of bar bending schedule on a construction site cannot be overemphasized. Some of these includes:
To serve as a control document for Clerk of work, Inspectors, Detailers, and Structural Engineers on reinforcement
For quantification of materials by the quantity surveyor during pre-contract and post-contract operations
To assist steel bender and fixer
To fast track construction and supervision;
To plan, detect inconsistencies, and save costs in the handling of reinforcements.
Preparation of Bar Bending Schedule
Preparation of bar bending schedule should not be a haphazard operation. It should begin with the appropriate study of the reinforcement detailing drawings. All inconsistencies or errors (such as repeated bar marks for different types of bars) discovered in the drawing should be reported to the detailer, because as practically as possible, the details of the structural drawing and the bar bending schedule should be synchronised.
The major check lists before the commencement of BBS includes but not limited to the following:
Check the concrete cover to each face
Check the lapping length for tensions and compression bars of each structural elements
Check the direction of bend of each bar
Group each structural unit and floor by floor. For example, all ground floor elements must be scheduled before proceeding to next floor and all footing must finish before proceeding to starter columns.
Bar mark must start from 01 and increasing upward consecutively
Good knowledge of detailing, and considerations affecting the practical construction.
Notations in reinforcement detailing
It is also very important to be familiar with the style used by the draughtsman and its corresponding meaning in the standard detailing practice. For example, in the detailing of a floor slab, the standard method and the corresponding preferred style in Nigeria is as follows:
British Standard Bottom (face): B1 (outer layer) and B2(second layer) Top (face): T1 (outer layer) and T2 (second layer)
Nigerian style B1 = Bottom Bottom (BB) or Bottom (B) B2 = Bottom Top (BT) or Near Bottom (N) T1 = Top Top (TT) or Top (T) T2 =Top Bottom (TB) or Near Top (NT)
It is however common for the notation used in the structural detailing to be given in the design notes.
For example, a main bar of a slab is typically detailed as 12H12-01-250B1 according to standard. However, the common notation used in Nigeria is 12Y12-01-250 BB, which means 12 numbers of high yield deformed bars of 12mm nominal size, at a spacing of 250mm center to center, in the bottom-bottom layer. The bar mark is -01- (Note: BB is the same thing as bottom outer layer). There might not be so much difference in notation based on what is seen above, but it is very important for the detailer to familiarise himself with the loacal standards.
Sections of a Bar Bending Schedule
A typical BBS contains the heading and the schedule section. The heading usually contains information which includes the following:
COMPANY NAME: Provide the name of the company issuing the schedule e.g. Structville Integrated Services Ltd. PROJECT: Provide the title of the project. E.g. Construction of UniAbuja Main Gate JOB NO: This is the number assigned to the job by the company or the client. PREPARED BY: This shows the person that prepared the schedule DRAWING NO: This gives the particular page in the drawing which references the schedule DATE: This shows the date the schedule was prepared with reference to any revised schedule SHEET NO: This shows the page by page documentation/compilation of the schedule
The presentation style of the schedule section of a BBS varies. Different companies always present their schedule in a way that suits them and their workers. Some schedules include the length and number of offcuts expected from a standard length of reinforcement or structural element to guide them in the reuse of reinforcements. For instance, two variants of schedules are shown in the images below.
However, a typical BBS schedule section contains 10 columns in the following order:
[1] MEMBER: This shows the particular member in consideration. E.g base type 1, base type 2, column type 1, column type 2 etc
[2] BAR MARK: This depicts different bar marks present in the drawing. E.g. 01, 02, 03 etc.
[3] TYPE and SIZE: Type relates to whether the specified reinforcement is of high or mild yield, while size depicts the diameter of the bar and typically includes 8, 10, 12, 16, 20, 25, 32, and 40mm bars.
[4] NO. OF MEMBER: This shows the number of times the particular structural elements occur. E.g In the foundation details, how many numbers of Base Type 1 are there?
[5] NO. OF REBAR IN EACH: This means the number of times that particular bar marks occur in that particular member.
[6] TOTAL NO. OF REBAR: This column is the multiplication of columns [5] and [6] to show the cumulative of that particular bar mark.
[7] CUT LENGTH: This column comes next after the previous column. However, in preparing the schedule, the column showing the rebar “shape drawing” has to be done first in order to arrive at the cutting length.
[8] TOTAL LENGTH: This column is the multiplication of columns [6] and [7] to give the overall length of a particular bar mark of a particular element in the entire drawing. The total length is what is used by a Quantity Surveyor to quantify and price. From there, each length or tonnage is gotten as an executive summary. It is also important for data to know how many labourers or man-hours that will be expended on the work.
[9] LOCATION OR POSITION OR SPACING: This is the 9th column on the schedule. Here, if it is a wall, slab or foundation, the layers, and spacing of the bar are shown. If it is a link (stirrup), the spacing is shown and if it is a beam, the position is shown. It is important information for steel fixers.
[10] SHAPE DRAWING: This is the 10th and final column on the schedule. It shows the drawing and the dimensions of all the bends, straights, and the direction each bar is assumed to face after placement. This approach is more detailed and straightforward compared to the main standard where you need a separate sheet of shape code list and separate columns for each dimensions.
For a contractor that uses a fast-track approach, this column is of utmost importance to prefabricate all the shapes and fix all elements ready for placement. E.g all the columns in a story building can be done immediately after mobilization to site. This saves time on the construction program.
Conclusion
The main standard has been domesticated and adapted to be peculiar to the design and construction industry in Nigeria. The Nigerian standard style conforms to BS8666:2005. It is satisfactory and praiseworthy to say that the Nigerian style is comprehensive and easy to understand by a layman. We believe that complex English, Mathematics, Schedule, or Drawing should not be used to confuse sense of good engineering.
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
