In our previous article, we were able to evaluate the effects of temperature difference on rectangular tanks. In this article, we are going to evaluate the same effect on a cylindrical tank of the same volume, in order to obtain the internal stresses and displacements in the tank due to temperature differences. This article will serve as a comparison between the response of a rectangular and cylindrical tank to temperature actions.
In our last article, the dimensions of the rectangular tank was observed to be 3m (L) x 3m (B) x 2.5m (H), thereby giving a volume of 22.5 m3. To model an equivalent cylindrical tank of the same height of 2.5 m, the diameter of the tank was obtained as 3.38 m. The other details of the tank are as follows;
Dimensions of columns = 300 mm diameter circular column Dimension of beams = 300 x 500 mm Height of column above ground level = 3 m Diameter of tank = 3.38 m (centre to centre) Height of tank = 2.5 m (centre to centre) Thickness of tank walls and base = 250 mm Support condition = Fixed Temperature inside the tank = 120 oC Temperature outside the tank = 25 oC Maximum hydrostatic pressure from the liquid stored = 25 kPa Modulus of elasticity of concrete = 2.8 x 107 kN/m2 Coefficient of expansion of concrete = 1.0 x 10-5 /oC
Temperature change for axial elongation = Average temperature = (25 + 120)/2 = 72.5 oC Temperature difference = 25 – 120 = -95 oC
When modelled on Staad Pro using the procedure described in the video above, the configuration and results below were obtained.
Fig 1: 3D model of cylindrical water tank
Fig 2: Radial bending moment on the tank shell due to temperature load
Fig 3: Longitudinal bending moment on the tank shell due to temperature load
Fig 4: Twisting bending moment on the tank shell due to temperature load
Fig 5: Radial shear on the tank shell due to temperature load
Fig 6: Longitudinal shear on the tank shell due to temperature load
Fig 7: Hoop tension (membrane) on the tank shell due to temperature load
Fig 8: Longitudinal tension on the tank shell due to temperature load
The differences in internal stresses induced in cylindrical tanks of equal volume and height with the rectangular tank are shown in Table 1.
Table 1: Internal stresses in rectangular and cylindrical tanks due to temperature load
In some factories and industries, tanks are used for the storage of hot liquids which are used in production. In such scenarios, the temperature inside the tank and the temperature in the surrounding may not be the same. It is well known that internal forces are induced in statically indeterminate structures when there is temperature difference as the elements undergo differential thermal expansion/contraction. For simple frames, the internal forces due to temperature difference can be easily obtained using the force method of structural analysis. But for more complex structures like combination of beams and plates, software like Staad Pro can be used for evaluation of temperature difference.
For example, let us consider the reinforced concrete tank with the dimensions shown in Figure 1;
Fig 1: Structural scheme of water tank subjected to temperature difference
Dimensions of columns = 300 x 300 mm Dimension of beams = 300 x 500 mm Height of column above ground level = 3 m Length of tank = Width of tank = 3 m (centre to centre) Height of tank = 2.5 m (centre to centre) Thickness of tank walls and base = 250 mm Support condition = Fixed Temperature inside the tank = 120 oC Temperature outside the tank = 25 oC Maximum hydrostatic pressure from the liquid stored = 25 kPa Modulus of elasticity of concrete = 2.8 x 107 kN/m2 Coefficient of expansion of concrete = 1.0 x 10-5 /oC
The tank has been modelled on Staad Pro (see Figure 2) using the parameters defined above.
Fig 2: Modelling of the tank on Staad Pro
The walls of the tank were subjected to a triangular hydrostatic pressure distribution of 25 kPa. You can check how apply hydrostatic loads on Staad Pro here. The temperature difference action applied to the the tank is shown below.
Temperature change for axial elongation = Average temperature = (25 + 120)/2 = 72.5 oC Temperature difference = 25 – 120 = -95 oC
The application on Staad Pro is shown in Figure 3.
Fig 3: Application of temperature load on Staad Pro
When analysed on Staad Pro, the results shown in Figures 4-8 were obtained for the tank shells at SLS.
Fig 4: Bending moment on the tank shell due to water pressure
Fig 5: Shear stress on the tank shells due to water pressure
Fig 6: Bending moment on the tank shell due to temperature difference
Fig 7: Shear stress on the tank shell due to temperature difference
Fig 8: Displacement of tank shell and frame due to temperature difference
The internal stresses induced in the tank shell due to temperature difference is quite serious and requires detailed attention during design.
The aftermath of any building collapse is usually horrendous, and the experience is an ugly one for everybody concerned. Human lives might be lost, many might be injured, people will be traumatized, properties will be destroyed, and the environment will become a mess. The media usually responds to such events in a manner that will escalate the woes of the stakeholders involved in the construction of the building. That is why professionals must strive to get it right from start to finish.
From a technical point of view, if the recommended/standard process has been followed in construction, a lotof things will need to go wrong before a building will collapse. There are prominent issues that have been identified as the reasons for building collapse such as inadequate design, poor detailing, faulty construction, use of substandard materials, engaging non-professionals etc. In this article, we are going to focus on a special aspect of design called ‘partial factors of safety’.
The procedures for the design of buildings evolved through the years as researchers gained better understanding of the behaviour of structures and materials. This has particularly led to reductions in the factors of safety applied in the design of buildings, thereby leading to more economical designs. As a matter of fact, it is expected that the factors of safety applied in designs will continue to reduce as we gain better understanding of the behaviour of materials that we are dealing with. These factors of safety are simply associated with the fact of not being ‘so sure’. At one point in time in your life, you must have gone an extra mile in doing something, ‘just in case the worst happens’. This is exactly what the factors of safety in modern codes of practice for design offer us, but they also give us guidance on the extent of extra mile we should go, so that we would not spend too much money unnecessarily. The factors of safety were arrived at after rigourous statistical evaluations.
I have seen a cantilever roof parapet collapse due to poor placement of the reinforcement. That could have been a detailing or construction error but it has nothing to do with factor of safety which is the main topic of discussion here. Before we proceed, let us briefly look at some design principles with emphasis on factors of safety.
Design philosophies The first generally accepted principle in the design of structures is the permissible stress method. In this method;
σmax < σper ————— (1)
where σper is given by σcrit/k. In this method, the coefficient k is assessed with regard to uncertainties in the determination of local load effect σmax and of resistance σper. Therefore, the value of k may ensure with an appropriate level of security, the reliability of the structure. The main insufficiency of this method is perhaps the local verification of reliability (in the elastic range) and the impossibility to consider separately the uncertainties of basic quantities and the uncertainties of computational models for the assessment of action effects and structural resistance. In this method, the probability of failure is controlled by one quantity only, the coefficient k (Holicky, 2009).
The second widely-accepted method of structural design is the method of global safety factor. It is based on the condition;
Xresistance/Xaction > S0 ————— (2)
Accordingly, the calculated safety factor s must be greater than its specified value s0. It is a method which attempts mainly to give a truer picture of the behaviour of elements and their cross-sections, in particular through the aggregate quantities of structural resistance Xresistance and action effect Xaction. As in the case of the permissible stresses method the main insufficiency of this method remains the impossibility to consider the uncertainties of particular basic quantities and theoretical models (Holicky, 2009). The probability of failure can, again, be controlled by one quantity only, i.e. by the global safety factor s.
The last on this list is the partial factor method of design (also called the limit state method), and is currently the most advanced operational method of structural design. Partial factor method of structural design is generally characterized by the inequality shown in Equation (3);
Ed(Fd, fd, ad, θd) < Rd(Fd, fd, ad, θd) ———– (3)
This design concept is deemed satisfactory when the effects of actions Ed are less than or equal to the structural resistance Rd. The basic variables involved in modelling the relationship are the actions (Fd = ψγFFk), material properties (fd = fk/γm), dimensions (ad + Δa) and model uncertainties (θd). The reduction factors (ψ) and partial factors for actions and materials (γF and γm) are used to describe the reliability of the structure.
What defines the safety of a building? According to Section 2 of EN 1990, a structure shall be designed and executed in such a way that it will during its intended life, with appropriate degrees of reliability, and in an economic way,
– sustain all actions and influences likely to occur during execution and use; – remain fit for the use for which it is required.
The last two sentences above generally define what is referred to as ultimate limit state (ULS) and serviceability limit state (SLS), and the two key words to look out for in achieving them are ‘reliability’ and ‘economy’. Ultimate limit states are associated with collapse or other similar forms of structural failure. Serviceability limit states correspond to conditions of normal use (deflections, vibration, cracks, etc.). In general the design should include both safety and serviceability, including durability in both cases. The nature of ultimate limit states is essentially different from the nature of serviceability limit states and should be taken into account in reliability verification. ISO 2394 defined reliability as the ability of a structure to comply with given requirements under specified conditions during the intended life for which it was designed.
Can you exhaust your factors of safety? To keep what we showed in equation (3) simple, let us consider BS 8110-1:1997 code of practice (which has been withdrawn), where we normally apply partial factors of safety to loads and materials. At ultimate limit state, we factor the design loads on a building as follows;
P= 1.4gk + 1.6qk ————- (4)
Where gk and qk are the dead load (permanent actions) and imposed loads (variable actions) respectively. A little observation of Equation (4) will show that we are actually increasing the value of the dead load in the building by 40%, and the value of the live load by 60%. If we sum both actions up, we can say that we are designing the building to sustain loads that are almost in excess of 60% of what we anticipate that it will carry in its design life – which is like the collapse load.
In the Eurocodes, the design action of a building at ultimate limit state is given by;
P= 1.35gk + 1.5qk ————- (5)
As can be seen from equations (4) and (5), the partial factors of safety for loads (actions) reduced in the Eurocodes, but the partial factors for materials remained almost the same. For characteristic strength of concrete, the partial factor γc remained 1.5 in both codes of practice, while for steel, the partial factor of 1.15 was also maintained. We should note that the later releases of BS 8110 prior to its withdrawal reduced factor of safety for steel reinforcements from 1.15 (0.87fy) to 1.05 (0.95fy) after it was discovered that the reinforcements manufactured and tested in the laboratories in the UK rarely fell below the minimum strength.
Having taken note of all these partial factors, is it possible for a structure to fail due to actions and materials? The answer still remains yes. In Nigeria, most local contractors do not pay attention to the quality of materials used in construction, and some property owners may misuse the building by overloading it. The most popular design code in Nigeria is still BS 8110, which is widely applied and accepted by the government and other institutions. Let us use the standard to show how the factors of safety can be squandered.
Concrete We have seen cases where the 28 day characteristic strength of concrete used in construction was observed to be about 13 MPa. If grade 25 concrete was used in the design, and partial factor of 1.5 applied, the conservative design strength will be (25/1.5 = 16.67 MPa). In this case, the factor of safety for the concrete is gone.
Steel In most published works available online, we have seen cases where the average characteristic yield strength of reinforcements falls below 400 MPa. If the yield strength of 460 Mpa was used in the design (0.95fy = 0.95 x 460 = 437 MPa), we will discover that the factor of safety for steel is also gone.
Loads The imposed load used in the design of most residential buildings is 1.5 kPa. Sometimes, the usage of a buildings can be abused by say for example, converting a residential building to a place of worship/gathering, or storage house. If a building designed as residential building is used to stack four bags of cement spread all over the floor, we can assume that the slab has been subjected to an imposed load of about 6 kPa. At ultimate limit state, the building was designed to take imposed load of (1.6qk = 1.6 x 1.5 = 2.4 kPa), and with imposed load of 6 kPa, there will be ‘real fire on the mountain’.
Therefore we can see that through poor materials and misuse of building through overloading, the factors of safety used in a design can be exhausted, thereby placing the structure on the risk of imminent collapse. All hands must be on deck to make building collapse a thing of the past in Nigeria.
References Holicky M. (2009): Reliability Analysis for Structural Design (1st Edition). Sun Media Press Stellenbosch
Cement is one of the most popular construction materials in Nigeria, and has been extensively utilised in civil engineering works such as building construction, bridges, transport infrastructures, rigid pavements, etc. It is very important for construction professionals in Nigeria to know the current standards and different types or classes of cement we have, and their various suitability for different construction works.
There are three grades or classes of cement in Nigeria, namely grades 32.5, 42.5, and 52.5. These grades corresponds to the minimum 28th day compressive strength of the cement mortar after curing. They are also referred to as cement strength classes of 32.5 MPa, 42.5 MPa and 52.5 MPa respectively. In terms of strength, the classes of cement are 32.5N, 32.5R, 42.5N, 42.5R, 52.5N, 52.5R. The 32.5 category must have strength between 32.5 N/mm2 and 52.5N/mm2, while for the 42.5 grade must have a strength range between 42.5 N/mm2 and 52.5 N/mm2. The minimum strength of the third category is 52.5 N/mm2.
The standard document for cement specification in Nigeria is the NIS 444-1 Cement – Part 1: Composition, specifications and conformity criteria for common cements. This standard defines 27 products in the family of common cement that are grouped into 5 main types (COREN, 2017). The five groups of cement we have in Nigeria are shown in Table 1.
Table 1: Types of cement grade in Nigeria (COREN, 2017)
From Table 1, it can be seen that there are three grades of cement in Nigeria, namely grades 32.5, 42.5, and 52.5.The strength classes of cement are 32.5N, 32.5R, 42.5N, 42.5R, 52.5N, 52.5R. As described earlier, 32.5 category must have strength between 32.5 N/mm2 and 52.5N/mm2, while for the 42.5 grade must have a strength range between 42.5 N/mm2 and 52.5 N/mm2. The minimum strength of the third category is 52.5 N/mm2 (Joel and Mbapuun, 2016). These are strengths after 28 days. The appendage “N” refers to a class of cement with normal early strength, while “R” Refers to those with high early strength (Oyenuga, 2014).
It is very important to note that the most common type of cement in Nigeria is the Portland Limestone Cement (PLC) and not Ordinary Portland Cement (OPC). As a matter of fact, OPC conforming to CEM I class of cement is not available in Nigeria’s open market. The cement available in the open market of Nigeria is the Portland Limestone Cement designated as CEM II in NIS 444-1 (2003). PLC is a modified OPC which is produced by adding 6 -35 % of limestone to OPC. It has a lower clinker content range of 65 – 94 % compared with OPC’s range of 95 -100% (Joel and Mbapuun, 2016) It has lower carbon footprint than OPC and is deemed more environmentally friendly.
Fig 1: Dangote’s cement plant, Obajana
According to COREN (2017), any specified grade of concrete can be produced using any strength class of cement provided appropriate mix design procedure is followed. However, since the year 2014, Standards Organisation of Nigeria (SON) has maintained that grade 32.5 PLC cement should be limited to plastering works, block making, and light concrete works. For water retaining structures, class N cement (normal early strength) is recommended due to their low drying shrinkage that will positively affect the crackwidth of the concrete (Ubani, 2018). Also, high grade and rapid hardening cements may not be too good for plastering works to avoid plaster cracks that might arise from drying shrinkage, especially given the high temperature in Nigeria.
There are many manufacturers of cement in Nigeria, but according to United Capital’s 2019 report, Dangote Cement Plc is the undisputed largest player in the Nigerian cement market. Using installed capacity as a gauge, Dangote Cement Plc is the clear leader wielding 60.6% of the market share with Nigerian installed capacity of 29.3 MMT. Lafarge Africa Plc (10.5 MMT) and BUA Group (inclusive of CCNN) (8.0 MMT) accounts for 21.8% and 17.6% respectively. There is also a fringe player PureChem Industries Limited based in Ogun State with 900,000 MT (United Capital, 2019).
According to Dangote cement’s website, the premium cement is produced in three grades which are 32.5R, 42.5R and 52.5R. They recommended their grade 32.5R for plastering works, low rise buildings and masonry. The other higher grades can be used for high rise structures and mega infrastructures. All Dangote cements available in the open market are Portland Limestone Cement. However Dangote grade 42.5N is available in the market.
From Lafarge’s website, their premium cement brands are Elephant, UNICEM, Ashaka, Elephant Supaset, and Powermax. The exact strength grade of Elephant and Unicem cement is not explicitly stated in their website, but the grade on bag label is 32.5R. They recommended it for medium strength concrete works, suspended slabs, masonry works (block making, plastering, mortar), and reinforced concrete works. All Lafarge cement products in Nigeria’s open markets are also Portland Limestone Cement.
From the label on Elephant Supaset cement, the strength grade is 42.5N, but the information on their website recommended it for block making, precast elements such as poles, culverts, interlocking stones, etc.
BUA cement is manufactured as grade 42.5R as labelled on their bag. The cement is manufactured as Portland Limestone Cement too.
Purechem Cement is manufactured as grade 32.5R cement, and as Portland Lime Stone Cement.
Having known these few specifications, you can select the most appropriate cement grade for your construction work.
References Council for the Regulation of Engineering in Nigeria (COREN) (2017): Concrete mix design manual. Special Publication No. COREN/2017/016/RC
Joel M., and Mbapuun I. D. (2016): Comparative analysis of the properties of concrete produced with Portland Limestone Cement Grade 32.5N and 42.5R for use in rigid pavement work. Global Journal of Engineering Research (15): 17-25
NIS 444-1 Cement – Part 1: Composition, specifications and conformity criteria for common cements. Standards Organisation of Nigeria.
Oyenuga, V., 2014. Cement not Responsible for Building Collapse in Nigeria. An Editorial in This day Newspapers of 13th May 2014.
Ubani O.U. (2018): Structural Design of Swimming Pools and Underground Water Tanks. Structville Integrated Services Limited, Nigeria.
Structural failure is real, and it is the duty of structural engineers to identify all possible modes of failure in a structure, and design against them with appropriate factor of safety. When a building fails, we can identify the most probable cause of the failure by looking at the crack patterns and their location.
By looking at the image above, can you identify the cause of failure of the building? Let us know your answer in the comment section. Thank you very much.
Equivalent horizontal forces (EHF) are not strictly actions, but are forces that are applied to a frame in combination with other actions to model the effect of frame imperfections. Another alternative of doing this is to model the frame out of plumb. According to clause 5.3.2(6) of Eurocode 3, if a frame is sensitive to second order effects, member imperfection must be modelled in the analysis if the member has a moment resisting joint. In this post, we are going to show how to model the effects of imperfection for gravity actions in portal frames.
Determination of EHF According to clause 5.3.2(3) of EC3, for frames sensitive to buckling in a sway mode the effect of imperfections should be allowed for in frame analysis by means of an equivalent imperfection in the form of an initial sway imperfection and individual bow imperfections of members. The imperfections may be determined from:
ϕ = ϕ0 αh αm
Where ϕ0 = 1/200 αh = 2/√h (h is the height of the structure in metres) αm = √[0.5(1+ 1/m)] (where m is the number of columns in the row)
Solved Example Model the effect of imperfection in the frame shown in Figure 1 using the equivalent horizontal force approach Columns – UB 610 x 229 x 125 Rafters – UB 533 x 210 x 92
Fig 1: Portal frame subjected to gravity action
When analysed under the gravity action shown above, the following reactive forces and bending moment diagram was obtained.
Fig 2: Bending moment and support reaction of portal frame under gravity load
For the example above; αh = 2/√h = 2/√7 = 0.755 (h is the height of the structure in metres) αm = √[0.5(1 + 1/m)] (where m is the number of columns in the row) αm = √[0.5(1 + 1/2)]= 0.8660 Therefore; ϕ = (1/200) × 0.755 × 0.8660 = 0.0032
The equivalent horizontal forces are calculated as:
HEHF = ϕVEd
However, sway imperfections may be ignored where HEd ≥ 0.15VEd
Choosing to incorporate the effect of imperfection in our analysis, the equivalent horizontal force is given by; HEHF = 0.0032 × 114.27 = 0.366 kN
We will now model and analyse the frame for load combination 1 with the effects of imperfection included. The analysis will be done with the base pinned.
Fig 3: Portal frame with equivalent horizontal force
The resulting bending moment diagram with the effect of imperfection is given below;
Fig 4: Bending moment and support reaction with the effect of imperfection
A little consideration of the above results will show that the effects of imperfection can be safely ignored in the design of the structure. Thank you for visiting Structville today.
A box girder is formed when two web plates are joined by a common flange at both the top and the bottom. The closed cell which is formed has a much greater torsional stiffness and strength than an open section and this is the main reason why box girder configuration is usually adopted in long span bridges. The torsion and distortion rigidity of box beam is due to the closed section of the box beam.
It has been shown by reseachers that long span bridges with wider decks and eccentric loading on cross-section will suffer in curvature in longitudinal and transverse direction, thereby causing heavy distortion of cross-section. As a result, long span bridges with wide decks should have high torsional rigidity in order to reduce the distortion of cross-section deck to a minimum. This is one of the downsides of T-beam and deck system. Accordingly, box girders are more suitable for larger spans and wider decks, and any eccentric load that will cause high torsional stresses which will be counteracted by the box section (see Figure 1 for typical box girder bridge). However, despite being an efficient cross-section, we should know that that the analysis of such sections are more complicated due combination of flexure, shear, torsion, and distortion. In this post, we are going to evaluate the potentials of Staad Pro software in the analysis of box girder bridge subjected to Load Model 1 of Eurocode 1 Part 2.
Fig 1: Curved box girder bridge
The cross-section of the bridge deck is shown in Figure 2. The following data was used to model the bridge deck on Staad Pro.
Length of bridge = 30 m Support condition = Pinned to the piers Unit weight of concrete = 25 kN/m3 Thickness of sections = 250 mm (web and flanges) Unit weight of asphalt overlay = 22 kN/m3 Thickness of asphalt = 75 mm
Hence, the bridge deck will be subjected to the following actions;
(1) Self weight of girder (2) Weight of asphalt overlay (3) Load Model 1 uniformly distributed load (4) Load Model 1 tandem wheel load (modeled as moving load on Staad Pro) (5) Pedestrian load on the cantilever
In this analysis, self weight of girder and asphalt were modelled as one load case, Load Model 1 (UDL) and pedestrian action were modeled as a separate load case, while Load Model 1 (tandem wheel load) were modelled as another separate load case.
Fig 2: Cross-section of box girder bridge
The typical combined traffic action on the bridge deck is shown in Figure 3.
Fig 3: Box girder bridge subjected to Load Model 1 of Eurocode
When modelled on Staad Pro, the configuration in Figure 4 was obtained. The rendered 3D view is shown in Figure 5.
Fig 4: Model of box girder bridge on Staad Pro
Fig 5: 3D rendering of the box girder bridge
When the structure was analysed, the following results were obtained;
Fig 6: Transverse bending moment due to self weight of girder and weight of asphalt overlay
Fig 7: Longitudinal bending moment due to self weight of girder and weight of asphalt overlay
Fig 8: Transverse bending moment on girder due to traffic UDLFig 9: Flexural-distortional behaviour of the girder due to traffic UDLFigure 10: Transverse bending on the section (flexural – distortional) due to wheel load at the most critical location
Figure 11: Longitudinal bending on the section (flexural – distortional) due to wheel load at the most critical location
Fig 12: Torsional-distortional action effect due to wheel load at the most onerous section
From the results, it could be seen that uniform loads have barely no effect on the torsional-distortional behaviour of box girders, however, unbalanced wheel loads are critical in the behaviour of such structures. We will carry out more studies on this bridge deck behaviour using Staad Pro. Thank you for stopping by today.
Columns are vertical structural elements used in transmitting floor load to a lower level or to the substructure. As a result, their stability and structural resistance are of major importance in the integrity of a structure. In buildings, columns are usually classified as either slender, short, or intermediate.
Based on the loading they are subjected to, columns can be described as being axially loaded (subjected to axial load only), uniaxially loaded (subjected to axial load and bending moment in one direction), or biaxially loaded (subjected to axial load and bending in two directions).
While the procedure for the design of columns is well known by structural engineers, the process of constructing columns on site to achieve what is in the drawing is another different challenge altogether. The success and accuracy of any building have something to do with the setting out. In this article, we are going to briefly describe the procedures for setting out columns from the foundation and from the floor slab using locally available construction tools.
Let us briefly review this process.
Setting out columns from the foundation
Column starter bars will normally start from the foundation which can be a pad, raft, or pile cap. For local construction works, setting out works will involve establishing the building lines, and marking out column locations on the profile board which must be established all round the building.
Detailed information on how to set out buildings are abundant on the internet. These points marked out on the profile board are used as reference for excavation of the trenches for the foundation. After excavating to the required depth, the foundation should be blinded according to the drawing specifications.
Fig 1: Excavated trench for building foundation
After blinding the footing excavation, lines can be thrown as appropriate, and the proper location of each column can be identified. Therefore, the accuracy of column positioning at the foundation stage depends on the accuracy of the setting-out work. This is why proper setting out is important, especially during the foundation stage. The footing reinforcement and column reinforcement starters can be installed properly and held in place until the concreting of the foundation is done.
Fig 2: Setting out starter bars using lines
Setting out columns from the floor slab
If the foundation has been completed and there is a need to continue the structure from the ground floor or first-floor slab, then setting out will need to be done for the kickers of the columns. Setting out of kickers can be done by surveyors or the site engineer using simple tools such as tape, builder’s square, and lines.
It is usually erroneous to take a corner of a building (edge of a slab) as an absolute reference unless you are sure that the edges of your slab are perfectly straight and aligned in accordance with the drawing. Sometimes after casting the slab, you can experience an error of about ±20 mm with reference to the drawing. If you want to avoid cranking the column rebars, you should deal with such errors immediately.
Fig 3: Column reinforcements with kickers
The fastest way of dealing with this is to divide the floor of the building into two as shown in Figure 4 (red lines). The lines must not be strictly through the centre of the building, but from any convenient point that will enable you to take measurements with ease.
However, you should ensure that the intersection of the lines is at 90 degrees to each other. You can use a laser machine or the 3-4-5 method to ensure that the lines are square. Instead of using any edge of the building as a reference, you should the centrelines you established as a reference to set out the position of the columns. You can rarely go wrong using this method, and another advantage is that if there are errors in the dimensions of the floor slab, the error is shared equally.
Fig 4: Division of floor slab into parts for kicker setting out
Suggestions on improving this article are welcome and will be incorporated.
The characteristic compressive strength of concrete to be used in construction is usually specified in the working drawings, design calculation sheets, and contract documents. The grade of concrete specified is used by bidders/contractors in pricing the job. Therefore, it is expected that it must be achieved on site.
There are different ways of achieving grade 25 concrete on site, and it vary based on each contractor’s experience. This guide is from a personal site experience. If there is no special requirement from the consultants, contractors are usually free to use any acceptable way to achieve the specified grade of concrete.
Normally, trial mixes should be done before the commencement of any project, but experienced contractors are already familiar with the mixes that they have done over and over again unless there is a new consideration to be made. The aim of this article is to show you how you can achieve grade 25 concrete based on site experience only without any technical calculations.
The applicability of this procedure is for low scale – low budget construction where there is no batching plant and sophisticated machines/equipment. If you follow this simple procedure, your concrete cubes for grade 25 concrete will exceed 17 MPa and 25 MPa after 7 and 28 days crushing tests respectively.
To get started, you will need to have the following available:
(1) Trustworthy supervisors (2) Functional concrete mixer (3) Functional concrete vibrator/poker (4) Headpans, 25 litre buckets, and shovels (5) All materials needed for concrete production – gravel, sand, cement (grade 42.5R), and water (6) Concrete cube moulds
Fig 1: Concrete mixer
Fig 2: Concrete vibrator/poker
Steps for Achieving Grade 25 Concrete On-site
Having gotten all materials ready, you can follow the steps outlined below:
(1) Recommended mix ratio To achieve grade 25 concrete in local construction, I recommend a mix ratio of 1:2½:3½. What this means is that for 1 bag of cement, you add 5 headpans of sand, and 7 headpans of granite. The coarse aggregate that has been verified for this mix proportion is 3/4″ granite (19 mm size aggregate). The sand should be properly graded, and the cement should be sound with a strength grade of 42.5N or 42.5R.
Note: The popular 1:2:4 mix ratio sums up to 7 (that is 1 + 2 + 4 = 7), and the recommended 1:2½:3½ mix ratio also sums up to 7. It has been discovered on site that for 3/4″ granite in a 1:2:4 mix ratio, the quantity of granite usually appears excessive at about 0.4 – 0.6 water/cement ratio, thereby giving workmen difficulty in mixing, placing, and consolidating the concrete.
This leads to the addition of excess water which compromises the strength of the concrete. While this can be taken care of by the use of superplasticizer admixture, low-budget constructions do not normally factor admixtures into the cost. Therefore, by slightly adjusting the sand and gravel content from 2:4 to 2½:3½ respectively, a better workable mix is achieved without any serious consequences.
(2) Water Content For the mix ratio described above, the quantity of water to be added must not exceed 25 litres. If it exceeds 25 litres, you have deviated from the provisions of this guide for obtaining grade 25 concrete. On-site, you can achieve this by pouring one bucket of water only into the concrete mixer bunker.
There might be grumblings about the workability of the mix from the workmen, but stick to the plan as it will not affect your finished work. If you have a very tight reinforcement arrangement that you need very workable concrete, you might reconsider using this guide. The bucket size recommended for this is the 25 litres empty paint bucket shown in Figure 3. If you get the mix ratio right and add excess water, the test result will fail.
Fig 3: 25 Litres empty paint bucket
(3) Do not heap the headpans The granite and the sand must not be heaped if you are to achieve grade 25 concrete using this guide. The sand and granite must be made to flush with the top surface of the headpan as shown in Figure 4. If the sand and granite are heaped you have deviated from the provisions of this guide.
Fig 4: Unheaped headpan of sharp sand
(4) Supervision To achieve steps 1, 2, and 3 above, you need strict supervision and monitoring. First of all, you need a supervisor at the loading points, who must ensure that the granite and sand are not heaped. It is good to have different monitors for sand and granite.
To achieve this, after filling each head pan with sand/gravel, it must be levelled using a shovel or piece of wood before being carried to the bunker. Secondly, you need another supervisor at the bunker to ensure that the correct quantity of sand, granite, and water is being poured in for one bag of cement.
Fig 5: Supervision of concreting during local construction
(5) Placement and consolidation The concrete produced from the mixture should be transported without delay to the required point where it is poured and consolidated using a vibrator/poker. The curing process should also begin as soon as it sets.
Fig 6: Placement and Consolidation of concrete for a floor slab
(6) Collection of samples for testing The 150mm x 150mm steel moulds to be used for sample collection should be oiled, and the concrete poured in three layers. Each layer should be subjected to 25 tampings using at least 16 mm diameter reinforcement off-cut. You can do well by hitting the sides of the mould during compaction to ensure proper consolidation.
After filling the moulds and compacting the concrete properly, the surface is leveled and finished smoothly with a trowel. Remove the sample from under the sun (especially in tropics), and cover it with a polyethylene nylon sheet. You can label the sample after about one hour.
Demould the sample the following morning and cure in a clean water tank for the required number of days. 6 or 9 samples might be required for each batch of concreting. However, consultants usually require 7 days and 28 days test results, and an average of 3 cubes should be crushed for each test. In some cases, 14 days’ results might be required.
Figure 7: Preparation of concrete cube samples
If you have followed these procedures properly, you should get satisfactory results from the laboratory for grade 25 concrete. However, note that this does not substitute for proper concrete mix design where it is required. Find below some laboratory test results from the procedure described above. The target characteristic strength after 28 days was 25 MPa. In one of the cases (Figures 10 and 11), regulatory agencies came by themselves and collected samples from the point of casting, and the result was found satisfactory.
Fig 8: 14 days compressive test result of a floor slab (Average 14 days strength = 23.48 MPa)
Fig 9: 28 days test result of the same floor slab (Average 28 days strength = 26.93 MPa)
Fig 10: 7 days test result of a floor beam (Average 7 days strength = 18.46 MPa)
Fig 11: 28 days test result of the same floor beam (Average 28 days strength = 27.14 MPa)
A ribbed slab is a type of reinforced concrete slab in which some of the volume of concrete in the tension zone is removed and replaced with hollow blocks or left as voids. This reduction in the volume of concrete in the tension zone (below the neutral axis) is based on the assumption that the tensile strength of concrete is zero, hence all the tensile stress is borne by the reinforcements in the tension zone. The resulting construction is considerably lighter than a solid cross-section.
This design and construction concept is useful in long-span construction of floors (say spans greater than 5m), where the self-weight becomes excessive when compared to the applied dead and imposed loads, thereby resulting in an uneconomic method of construction. One method of overcoming this problem is to use ribbed slabs which are suitable for longer spans supporting light loading, as in residential or commercial buildings.
Design Example The layout of a floor slab is shown in Figure 1 below. Design the floor to satisfy ultimate and serviceability limit state requirements. (Concrete grade = 30 MPa, Yield strength of reinforcement = 500 MPa, variable action on floor = 2.5 kPa, Fire rating = 1 hour 30 minutes).
Fig 1: Ribbed slab layout
Fig 2: Section through the ribbed slab
Load Analysis For 550 mm c/c rib spacing;
Permanent Actions Weight of topping: 0.050 × 25 × 0.55 = 0.6875 kN/m Weight of ribs: 0.15 × 0.2 × 25 = 0.75 kN/m Weight of finishes: 1.2 × 0.55 = 0.66 kN/m Partition allowance: 1.5 × 0.55 = 0.825 kN/m Self weight of clay hollow pot = 0.65 kN/m Total dead load gk = 3.572 kN/m
Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks Assuming ϕ12 mm bars will be employed for the main bars, and ϕ8 mm bars for the stirrups (links) d = 250 – 25 – (12/2) -8 = 211 mm
k = MEd/(fckbd2) = (21.56 × 106)/(30 × 550 × 2112) = 0.0293 Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)] k = 0.0709 z = d{0.5+ √[0.25 – (0.882 × 0.0293)]} = 0.95d = 200.45 mm
Depth to neutral axis x = 2.5(d – z) = 2.5(211 – 200.45) = 26.375 mm < 1.25hf (62.5 mm)
Therefore, we can design the rib as a rectangular section;
Area of tension reinforcement As1 = MEd/(0.87fyk z) As1 = MEd / (0.87fyk z) = (21.56 × 106) / (0.87 × 500 × 0.95 × 211) = 247.26 mm2
Since VRd,c (23.815 kN) > VEd (17.25 kN), no shear reinforcement is required.
According to clause 6.2.1(4), minimum shear reinforcements may be omitted in ribbed slabs where transverse distribution of loads is possible. But for this design, we will therefore provide minimum shear reinforcement.
Minimum shear reinforcement; Asw / S = ρw,min × bw × sinα (α = 90° for vertical links) ρw,min = (0.08 × √(fck)) / fyk = (0.08 × √30)/500 = 0.000876 Asw/Smin = 0.000876 × 150 × 1 = 0.131 Maximum spacing of shear links = 0.75d = 0.75 × 211 = 158.75 mm Provide H8mm @ 150 mm c/c as shear links.
Slab Topping A142 BRC Mesh can be provided or H8 @ 250mm c/c
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