In the construction of any substructure below the ground surface, excavations must be made to the recommended depth before the construction can begin. In the circumstances where deep vertical cuts (say about 3m) are to be made in a soil, provision must be made for the support of the soil to stop it from caving in or collapsing. The design of braced cuts involves the evaluation of the stresses in the members (bracings) used for supporting the cut, and provision of adequate members sizes or sections to avoid structural failure. These are usually classified as temporary works in construction.
Most shallow foundations in cohesive-frictional soils may not need side supports to keep the cut stable. Furthermore, when there is enough space, the sides of the excavation may be sloped, thereby eliminating the need for bracings or support. However, when a vertical excavation exceeds 3m depth, the most common practice is to drive vertical timber planks called sheeting along the sides of the excavation. Horizontal beams called wales are used to hold the sheeting in place, while struts running from side to side of the excavation is used to hold the wales in place.
For deeper cuts of about 5-6m, the use of timber sheeting may become uneconomical and steel sheet pile walls may be used (see figure below). Sheet piles are convenient to use because of their high resistance to stresses. On the other hand, they are relatively lightweight and reuseable when compared with timber.
Typical construction of a braced cut using sheet pile walls
The sheet piles are driven around the boundary of the excavation. As the soil is removed from the enclosure, the wales and struts are inserted. The process continues until the excavation is complete. When the width of the excavation is too wide to permit the economical use of struts across the excavation, tiebacks may be used as shown below.
Typical schematic representation of tieback
Lateral Earth Pressure Distribution in Braced Cuts
To design braced excavations (that is to select the wales, struts, sheet piles etc), the engineer must estimate the lateral pressure that the braced cut will be subjected to. The lateral earth pressure distribution of braced cuts is different from that of conventional retaining walls. This is because the deformation of braced cut walls gradually increases with the depth of the excavation. The deformation is influenced by factors such as type of soil, type of equipment used, depth of excavation, and workmanship.
Due to very little wall yielding at the top of the excavation, the pressure will be very close to at rest at the top. At the bottom of the wall with greater yielding, the lateral earth pressure will be lower than the Rankine active earth pressure. As a result of this, the earth pressure distribution of braced cuts varies significantly with that of retaining walls.
Deformation behaviour of braced cuts and retaining walls
Peck in 1969 presented the pressure distribution diagram of braced cuts. The diagrams were developed from observations and studies obtained from the strut load of Berlin subway cut, Munich subway cut, and New York subway cut. He described the pressure as apparent pressure envelopes which represents the fictitious pressure distributions for estimating the strut loads in a braced cut.
Apparent pressure distribution in braced cuts
(a) For cuts in sand The lateral pressure for cuts in sand is given by;
σ = 0.65γHKa
Where; γ = Soil unit weight H = Height of the cut Ka = Rankine’s active earth pressure
(b) For cuts in soft-medium clay (γH/c > 4)
The lateral pressure is the larger of;
σ = γH[1 – (4c/γH)] or σ = 0.3γH
(c) For cuts in stiff clay (γH/c ≤ 4)
σ = 0.2γH to 0.4γH (average of 0.3γH)
Where c is the undrained cohesion of the soil.
When using the pressure envelops in design, it is very important to keep the following in mind;
The pressure envelope is sometimes referred to as the apparent pressure envelope. The actual pressure distribution depends on the construction sequence and the flexibility of the wall.
The water table is below the bottom of the cut
Sand is assumed to be drained with no pore water pressure
Clay is assumed to be undrained, and pore water pressure is not considered
The excavation depth is greater than about 6 m
Design of the various components of a braced cut
Struts Struts in braced cuts behave like horizontal columns. Hence, the load-carrying capacity depends on the slenderness ratio and flexural buckling capacity. During construction, it is recommended that struts should have a minimum vertical spacing of 3m. In clay soils, the depth of the first strut below the ground surface should be less than the depth of the tensile crack.
To determine the load in the strut, the following procedure can be used;
Draw the pressure envelope of the braced cut and show the strut levels. Assume that the sheet pile walls are hinged at the strut levels except for the top and bottom ones.
Determine the support reaction of the cantilever (for the top and bottom struts), and use simply supported assumption to determine the intermediate struts
Sum up the reactions to determine the load in each strut
Provide and appropriate section for the strut using a suitable design code
Sheet Piles The following steps may be taken in the design of a sheet pile wall;
For each of the section (in plan), determine the maximum bending moment
Pick the highest value of the maximum bending moments (in kNm/m)
Obtain the required section modulus of the sheet pile wall Sx = Mmax/Allowable stress of the sheet pile material
Choose a sheet pile section with a slightly higher section modulus
Wales Wales may be treated as continuous horizontal members if they are properly spliced. Conservatively, they can be assumed to be pinned at the struts.
Design of Braced Cuts – Solved Example
A long trench is excavated in a medium dense sand for the basement of a highrise building. The excavation is supported with sheet pile walls held in place by struts and wales as shown in the figure below. The struts are spaced 4m c/c in plan.
Determine;
(a) The pressure envelope on the wall (b) The loads in the struts (c) The maximum moment for determining an appropriate section for the sheet piles (d) The maximum moment for determining the section for the wales
Solution (a) The pressure envelope on the wall For a braced cut in sand, the apparent pressure envelope is given by;
(c) Bending moment on the sheet pile walls By looking at the shear force diagram, we can confirm that the maximum moment will occur at the cantilever regions and at the points of contraflexure for shear.
For section A-B, the point of contraflexure of the shear force diagram can be obtained using similar triangle;
The maximum moment for the section is 37.25 kNm/m and should be used for the design of the sheet pile wall when factored.
(d) Design moment of the wales We can use the maximum strut load to design the wale.
Mmax = PL2/8
Where L is the spacing of the strut in plan Mmax = (111.746 x 42)/8 = 223.492 kNm/m The design can be carried out accordingly.
Special considerations should be given to braced cuts in layered soils (kindly refer to standard geotechnical engineering textbooks). Furthermore, heave at the bottom of braced cuts in clay soils should also be considered where applicable.
Site managers are saddled with the responsibility of making material requisition and ensuring that construction progresses without too many glitches. One of the major causes of delay in construction is the unavailability of materials on site. Therefore a site engineer is expected to have very quick tips on how to estimate the quantity of materials required to execute a given item of work.
In this article, we are going to provide a basic quantity estimate guide for common construction materials that are used on site. The list is not exhaustive and will be updated from time to time.
(a) Quantity Estimation Guide for Concrete Works
The materials that are required for the production of normal weight concrete are cement, sand, coarse aggregates, and water. For normal strength concrete, the quantity of cement varies with the concrete grade, while the quantity of sand and coarse aggregate remains approximately constant. To estimate the quantity of materials required for concrete works, the first step is to calculate the volume of concrete required for the job.
For slabs; the volume of concrete required (m3) = Area of the floor slab (m2) × thickness of slab (m) For beams; the volume of concrete required (m3) = Depth of beam (m) × width of beam (m) × length of the beam (m) For columns; the volume of concrete required (m3) = length of column (m) × width of column (m) × height of column (m)
For instance, for a (230 x 230)mm column in a duplex, the volume of concrete required is 0.16 m3 (i.e 0.23 m × 0.23 m × 3m = 0.16 m3) assuming that the storey height is 3 m. This volume can be multiplied by the number of columns in the building to obtain the total volume of concrete required for casting all the columns in the building.
After obtaining the volume of concrete required, the quantity of materials required for the production of one cubic metre of concrete can be estimated as follows;
(i) 7 bags (350 kg) of cement is required to produce one cubic metre (1m3)of grade 25 concrete (ii) 1200 kg (1.2 tonnes) of sand is required to produce one cubic metre (1m3) of grade 25 concrete (iii) 1400 kg (1.4 tonnes) of granite is required to produce one cubic metre (1m3) of grade 25 concrete (iv) 175 – 210 litres of water is required to produce one cubic metre (1m3) of grade 25 concrete
The quantities listed above require no additional allowance for waste.
(b) Quantity Estimation Guide for Reinforcement Works
The quantity of reinforcements required can vary considerably depending on the structural design output of the structure. Therefore, it is important to refer to the reinforcement detailing and bar bending schedule for quantity estimation. It is also important to note that the bar bending schedule issued by structural engineers does not take into account offcuts and wastage. It is therefore, important to consider the propriety of bar bending schedule for construction purposes. More often than not, contractors prepare their own bar bending schedule (taking offcuts into account) before placing an order.
Reinforcements are usually quantified by weight (kg or tonnes), which can be converted to length (reinforcements are usually supplied per 12 m length). The unit weight of different sizes of reinforcements are as follows;
Diameter of bar (mm)
8
10
12
16
20
25
Unit weight (kg/m)
0.395
0.616
0.888
1.579
2.466
3.854
Reinforcement Quantity Estimation in the Floor Slab of a Duplex
In a floor slab of about 4-5 bedrooms duplex (200 – 250 m2 of floor slab area), the reinforcement requirement is about 2.5 – 4.0 tonnes of Y12 mm (main reinforcements) and 0.3 to 0.6 tonnes of Y10 mm as distribution bars. The exact quantity is determined from the design drawings.
In the absence of design drawings and an idea is needed about the probable cost of reinforcement for a floor slab, the following relationship can be used for slabs of moderate span;
12.5 kg of Y12mm bottom reinforcement is needed per square metre (1 m2) of slab
20% to 30% of the weight of Y12mm (in kg) provided at the bottom is needed at the top (hogging areas)
10% to 15% of the weight of Y12mm (in kg) provided at the bottom is needed as Y10mm (distribution bar of hogging reinforcements)
About 12 kg of binding wire is required to tie 1 tonne of reinforcement
Example Estimate the quantity of reinforcement needed to construct a 250 m2 of moderately spanned floor slab in a duplex.
Solution Quantity of bottom reinforcement required = 12.5 kg × 250 = 3125 kg = 3.125 tonnes of Y12 mm Quantity of top reinforcement required (using 25% of bottom reinforcement) = 0.25 × 3125 = 781 kg = 0.781 tonnes of Y12 mm Quantity of distribution bars required (using 12% of bottom reinforcement) = 0.12 × 3125 = 375 kg = 0.375 tonnes of Y10 mm The total quantity of reinforcement required = 4.281 tonnes Quantity of binding wire required = 12 × 4.281 = 52 kg
Note that this quantity can considerably vary based on the design result. Therefore, the quantity presented here is just a rough idea, and the reinforcements are assumed to be spaced at 150 – 175 mm c/c. This may however be an uneconomical design, but we need a rough idea of the cost, right? When the reinforcement is spaced at 200 mm c/c, 10.7 kg per square metre can be assumed for the bottom reinforcements.
Reinforcement Quantity Estimation in the Beams of a Duplex
In the absence of design results, the quantity of reinforcement required in beams of moderate spans (say equal to or less than 4m) in a duplex can be estimated as follows;
7.9 – 9.5 kg of Y16mm per linear metre of a beam
2.5 kg of Y8mm per linear metre of a beam as links/stirrups
3.55 kg of Y12 mm per linear metre of a lintel
2.5 kg of Y8mm per linear metre of a lintel as links/stirrups
Please note that the quantity stated above is not guaranteed to satisfy structural requirements. Consult a structural engineer for this purpose.
Example Estimate the quantity of floor beam reinforcement required in a moderate duplex if the total length of all the floor beams in the building is 130 m.
Solution Quantity of Y16 mm bars required = 9.5 × 130 = 1235 kg = 1.235 tonnes of Y16 mm Quantity of links reinforcement required = 2.5 × 130 = 325 kg = 0.325 tonnes of Y8 mm
Reinforcement Quantity Estimation in the Columns of a Duplex
In the absence of design results, the following can be used to estimate the quantity of reinforcement in columns of 230 x 230 mm dimensions in a duplex (one suspended floor only).
25 kg of Y16mm per 3m high column
5.33 kg of Y8mm per 3m column as links/stirrups
Note that this quantity can considerably vary based on the design result.
Example Estimate the quantity of reinforcement required to construct 25 number of 230 x 230 columns in the ground floor of a duplex. All spans in the building are moderate.
Solution Quantity of Y16mm required = 25 × 25 = 625 kg (0.625 tonnes of Y16 mm) Quantity of Y8mm required = 25 × 5.33 = 133.25 (0.133 tonnes of Y8 mm)
12 pieces of 1″ x 12″ x 12′ plank formwork is required to cover an area of ten square metres (10 m2)
4 pieces of marine board (1.2 m x 2.4m) is required to cover an area of ten square metres (10 m2) 1 piece of 2″ x 3″ x 12′ softwood is required per square meter (1 m2) of slab
(f) Miscellaneous
The weight of one bag of cement is 50 kg The volume of one bag of cement = 0.034 m3 The volume of a standard builder’s wheelbarrow = 0.065 m3 The volume of one standard head pan is = 0.0175 m3
These quantities are true when the headpan/wheelbarrow is unheaped
The demolition of a concrete structure may be done for a lot of reasons, irrespective of the cost and environmental implications. It is not always an easy easy decision to make, but sometimes, it just has to be done. It is therefore important to know the right conditions when the demolition of a building is necessary or important. The owner, consultant, or government agencies can order the demolition of a concrete structure due to the following reasons;
(1) Violation of building regulations Buildings or parts of buildings that are constructed without approval may be marked for demolition by the government or town planning agencies. These include buildings that violate regulations such as setbacks, airspace, etc.
Such demolitions may be in part or in whole. On the other hand, buildings that are built over canals, drainage, high voltage transmission lines, are also usually marked for demolition. Constructing an industrial building in a residential area without approval and vice versa can lead to a demolition order too. They are generally described as illegal structures.
(2) Court/government order The court or government can give orders for the demolition of a building due to legal reasons or otherwise. For instance, during the 2020 COVID-19 pandemic, a governor in Nigeria ordered and supervised the demolition of a hotel for violating lockdown orders. Sometimes, orders are given for the outright demolition of buildings that are used to perpetrate crimes such as criminal hideouts, kidnapper’s den, etc.
(3) Construction defects The client, consultant, or supervising agency can order the demolition of parts or the whole of a construction work due to gross defects. If the defects are deemed too serious that it will compromise the integrity of the finished structure, the demolition of the defective member(s) may be ordered.
(4) Marked as unsafe Distressed or old buildings that are deemed to pose a safety risk can be marked for demolition by regulatory agencies.
(5) Encroachment into the right of way of other developments When structures encroach into the right of way of new major developments such as highways, railways, airports, etc, they can be marked for demolition. However, for buildings that have approval, appropriate compensation is usually provided for the owners of the building. For instance, it has been reported that the proposed Fourth Mainland Bridge in Lagos, Nigeria, will affect about 800 houses.
(6) Property owner’s desire to get rid of a structure A property owner can decide to demolish an existing structure in order to erect a new one or use the space for another thing.
Generally, the demolition of a reinforced concrete structure can be a very challenging and expensive task, and oftentimes, approval may be needed from the local government before the commencement of the operation. Due to the cost and environmental implications of demolishing a building, it should always be a last resort after all other options have been explored. There are different methods of demolishing a building, but that is outside the scope of this article.
While an imminent demolition of a building is possible due to the six reasons given above, we are going to focus on the number (3), which directly affects structural engineers.
Demolition of a Building for Structural Reasons
Structural engineers are often invited to approve, evaluate, or assess existing buildings or buildings under construction. For an existing building or on-going construction to be deemed fit for human habitation, there must be a minimum safety standard for all parts and the whole of the building.
The latest design standard used all over the world for reinforced concrete design is the limit state design principle. In limit state design, a limit of performance is defined for a structure for it to be deemed acceptable. Generally, a structure is expected to satisfy the ultimate and serviceability limit state requirements.
During design, the limit state concept is achieved by factoring the loads and materials to be used in the construction. The concept also takes into account all imperfections and geometric tolerances. The generality of the concept is that the effects of the design actions should be less than the resistance of the structure. This defines the accepted limit state.
While this can easily be achieved during design, construction defects can compromise the safety of a well-designed building. Furthermore, when a building is exposed to an earthquake, fire, or accidental impact, some level of damages may be observed that will demand the assessment of the structural integrity of the building. It is also important to evaluate the capacity of an existing building when the occupancy class is to be changed from say residential to a storage house.
During construction, defects can come from poor quality of materials or poor workmanship. Some scary construction defects that require serious attention are:
(1) Extreme honeycombs (2) Very poor concrete strength (3) Inadequate yield strength of reinforcement (4) Inadequate member sizes (5) Improper placement and arrangement of reinforcements (6) Insufficient area of reinforcement (7) Poor concrete cover, etc
When a structural defect is observed in construction, something MUST be done to address it, and the onus lies on the structural engineer to provide guidance on what is to be done. This is because the safety of the occupants of the building during and after construction is not negotiable. Nothing should be left to ‘luck’ or ‘guesses’.
Due to the cost and environmental implications, a consultant should order demolition as a last resort unless nothing else can be done. Even when the member sizes and reinforcements provided are inadequate, all options should be explored before a decision is taken.
Some options to be explored are:
(1) Redesign/assessment of the building based on what is existing When the member sizes and areas of reinforcement provided are not in conformity with the design, the building can be remodelled and redesigned to check the capacity of what is existing. The design parameters such as concrete grade and yield strength should be based on the actual properties of the materials used in the construction. As a result, tests like non-destructive testing of concrete and the reinforcement should be carried out.
The engineer should also inspect the building thoroughly for cracks and evidence of structural distress. This can help him predict what is happening to the building or the nature of stresses causing the distress.
When the assessment is done, the structural engineer will then have a basis for making a decision. If the construction is still ongoing, the load path or the structural scheme can be modified accordingly. The other options listed below can be explored too. The cost of this assessment should be decided between the property owner and the contractor.
(2) Introducing additional supports/members to change the load distribution Additional supports or members can be introduced into an on-going or existing buildings in order to alter the load/stress distribution in the structure. This should be an informed decision from the structural engineer after understanding the current state of the building, or the effects of the existing structural arrangement.
(3) Introducing steel members and polymer composites Where possible, structural steel members such as stanchions, universal beams, flats, plates, or angles can be introduced to provide external support to failing structural members. This is a form of retrofitting on its own though. According to a 2016 study in Indonesia, steel plates and fasteners can be used as external reinforcements when the rebars in a structure are inadequate. The use of epoxy grouting, steel plates, and bolt fasteners have been observed to increase the flexural capacity of the strengthened members.
Materials like fibre reinforced polymers (FRP) have also found applications in the strengthening of beams and slabs. The usual strengthening method presumes the placing of the sheets bonded on the tensioned side using resins. The sheets are mounted parallel to the long edge of the slabs.
(4) Retrofitting Retrofitting is basically the process of adding or incorporating new features that were not previously included in an existing design. There are different methods of retrofitting reinforced concrete structures such as concrete jacketing, steel jacketing, use of fibre-reinforced polymer, etc.
One way of retrofitting reinforced concrete beams is by concrete jacketing. In this method, an additional layer of concrete is placed around the existing beam, together with additional longitudinal bars and stirrups, to enhance the flexural and/or shear capacities. Other options of retrofitting beams are the use of fibre-reinforced polymer (FRP) wrapping and the use of steel brackets or haunches.
Concrete jacketing of reinforced concrete columns involves adding concrete with longitudinal and transverse reinforcement around the existing columns. This type of strengthening improves the axial and shear strength of columns while the flexural strength of column and strength of the beam-column joints will likely remain the same.
Concrete jacketing involves drilling holes into the concrete members and producing a very workable but strong concrete mixture. The compatibility of deformation between the existing and new concrete, resistance against delamination, and durability should also be carefully looked into.
It is important to note that there is a price to pay for making these kinds of changes. Apart from the increased cost of construction and safety risks, the architectural concept of the building may also be altered.
If a consultant gets to a site, he can order the demolition of the structural member if the following are observed after thorough tests and investigations. This is a case where the options listed above are not feasible:
The reinforcement provided in the structural members is less than the minimum required, with respect to the expected load on the member.
The reinforcements are poorly placed. For example, placing the reinforcements of a cantilever at the bottom (the props should not be removed in the first place).
The member sizes provided are less than the design capacity
The compressive strength of concrete is less than the design strength
The yield strength of the reinforcement is less than the design strength.
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Hemp and Recycled Aggregates Concrete (HRAC) is a sustainable concrete where coarse aggregates are partially replaced with industrial hemp fibers and recycled concrete aggregates (RCA) [1]. In the bid to achieve sustainable materials in construction and also reduce the carbon footprint and environmental impacts of concrete, researchers have been coming up with various ways to beneficially reuse industrial and agricultural wastes as construction materials.
According to researchers from the Department of Civil and Environmental Engineering, American University of Beirut, Lebanon, the use of HRAC in construction has two main benefits which are; savings on natural resources, and reuse/recycling of waste materials [1].
Hemp fiber is a natural material that has been used extensively in producing a lot of things such as rope, clothing, shoes, accessories, etc. It has also been used with other materials to produce plastics and composites for the automobile industry, and also in the production of bricks and mortars in building construction [2]. Hemp fiber has also been used in the production of fibreboard, thermal insulation in homes, and hempcrete. As a matter of fact, hemp fibers are regarded as one of the toughest natural fibers [3].
Fig 1: Hemp fibers (Source [3])
Recycled concrete aggregate (RCA) is obtained by breaking the lumps of old or demolished concrete works and extracting the coarse aggregates. One of the major reasons to use RCA in structural concrete is to have more environmentally friendly construction [4]. The use of RCA on a large scale may help to reduce the effects of the construction on the environment by reusing waste materials and preventing more natural resources from being harvested.
Fig 2: Recycled Concrete Aggregate (Source [4])
Some scholars have researched the use of hemp fiber in the production of concrete. A 2008 study in France showed that treating hemp fiber in alkaline improved the fiber strength and the fiber-matrix adhesion in concrete [5]. Another study by [6] showed that the strength of hemp fibers can be improved with low lignin content and good fiber separation when treated in 10% (by weight) NaOH (Sodium hydroxide) solution. The hemp fibers were observed to have an average tensile strength of 857 MPa and Young’s modulus of 58 GPa.
Li et al [7] carried out a study to determine the properties of hemp fibre reinforced concrete. The study showed that the addition of hemp fibre into the concrete matrix results in a linear reduction in the specific gravity and the water absorption ratio of the matrix. Furthermore, the compressive strength, flexural strength, toughness and toughness indices, specific gravity, and water absorption ratio of hemp fibre reinforced concrete are all correlated with aggregate size parameters, fibre factors and matrix initial mechanical properties.
A 2014 study by [8] showed that hemp fiber has no negative effect on the strength of the concrete in the long term.
Previous researches done on the strength of concrete produced with recycled concrete aggregate showed minor reductions in the various mechanical properties including compressive strength, splitting tensile strength, flexural strength, and modulus of elasticity. According to [4], the density of RCA is lesser than the density of natural aggregates due to residual adhered mortar. A study by [9] showed that the relative density of RCA (in the saturated surface dry condition) is approximately 7–9 % lower than that of natural aggregate.
Research has shown that the natural aggregate in a concrete mix may need to be replaced with about 25-30% of RCA before significant changes can occur in the compressive strength [4]. A 2008 study attributed the reduction in compressive strength of RCA to water higher water adsorption when compared with natural aggregates [10].
In a 2020 study of HRAC by [11], the coarse aggregate content was reduced by 20% of the concrete volume, and 50% of the natural coarse aggregates were replaced by recycled concrete aggregates. The variables of the study included percentage replacement of natural aggregates with RCA (0 or 50%), maximum size aggregate (10 and 20 mm), hemp fiber length (20 and 30 mm), and fiber surface treatment (alkali, silane, and acetyl).
Fig 3: Hemp hurds (Source [12])
From the study, the replacement of 50% of natural aggregate with RCA reduced the tested mechanical properties by 1 to 10% when the maximum size of aggregate was 10 mm and by 4 to 13% when maximum size of aggregate was 20 mm. When fibers were incorporated in the mix and the coarse aggregate content was reduced by 20%, the compressive strength of the concrete reduced by 37% for maximum size aggregate of 10 mm relative to the control mix.
For maximum size aggregate of 20mm, compressive strength reduced by 31.1% relative to the control mix. The study recommended that HRAC mixes should not be used in members subjected to direct compression such as columns.
Other mechanical properties such as modulus of elasticity, modulus of rupture, and splitting tensile strength and long term/durability test results from the study can be obtained from the open access article.
References [1] Ghosn S., Hamad B. (2020): Durability Evaluation of Hemp Fibers and Recycled Aggregates Concrete. In Proceedings to the XV International Conference on Durability of Building Materials and Components DBMC 2020, Barcelona [2] https://en.wikipedia.org/wiki/Hemp#Fiber(Assessed on the 5th of December, 2020) [3] https://hempgazette.com/industrial-hemp/hemp-fiber-production/(Assessed on the 5th of December, 2020) [4] McNeil, K., Kang, T.HK (2013): Recycled Concrete Aggregates: A Review. Int J Concr Struct Mater7, 61–69. https://doi.org/10.1007/s40069-013-0032-5 [5] Sedan, D., Pagnoux, C., Smith, A., & Chotard, T. (2008). Mechanical properties of hemp fibre reinforced cement: influence of the fibre/matrix interaction. Journal of the European Ceramic Society,28(1), 183–192. [6] Pickering, K. L., Beckermann, G. W., Alam, S. N., & Foreman, N. J. (2007). Optimising industrial hemp fibre for composites. Composites Part A—Applied Science and Manufacturing,38(2), 461–468. [7] Li, Z., Wang, X., & Wang, L. (2006). Properties of hemp fibre reinforced concrete composites, Composites: Part A Applied Science and Manufacturing, 37(3), 497–505. [8] Awwad, E., Mabsout, M., Hamad, B., Farran, M.T. and Khatib, H. (2012): Studies on fiber-reinforced concrete using industrial hemp fibers. Construction and Building Materials, 35, 710-717 [9] Limbachiya, M. C., Leelawat, T., & Dhir, R. K. (2000): Use of recycled concrete aggregate in high-strength concrete. Materials and Structures, 33, 574–580 [10] Yang, K.-H., Chung, H.-S., & Ashour, A. F. (2008): Influence of type and replacement level of recycled aggregates on concrete properties. ACI Materials Journal, 105(3), 289–296. [11] Ghosn, S., Cherkawi, N. & Hamad, B. (2010): Studies on Hemp and Recycled Aggregate Concrete. Int J Concr Struct Mater14, 54 (2020). https://doi.org/10.1186/s40069-020-00429-6 [12] Novakova P., Sal J. (2019): Use of technical hemp for concrete – Hempcrete. IOP Conf. Series: Materials Science and Engineering 603 (2019) 052095 doi:10.1088/1757-899X/603/5/052095
Featured Image Credit: Novakova P., Sal J. (2019): Use of technical hemp for concrete – Hempcrete. IOP Conf. Series: Materials Science and Engineering 603 (2019) 052095 doi:10.1088/1757-899X/603/5/052095
To ensure adequate planning and delivery of a project, consultants and clients often require that the contractor submit a programme of work (POW) that shows the sequence of activities that will be carried out in order to meet the expected completion date of the project. A construction work programme is a document that shows the series of work items, their relationship, and the time allocated for their execution. They are like the road map to the proposed construction.
Clients, consultants, project managers, and construction site managers also use POW to plan milestones which is an important tool or key performance indicators (KPIs) for milestone payment or achievement.
The POW can be presented in the form of a Gantt chart showing all the tasks/activities, duration, start and finish date and milestones from site mobilisation to commissioning according to the scope of work. It also serves as a control document to monitor the progress of work with respect to the actual work done. A report could be provided weekly or monthly regarding the status of work to the project stakeholder.
The level of details required from the POW is usually defined by the consultant and it desirably kept simple. However, for POW done for in-house usage, resources may be added. POW is submitted every monthly site meeting to show project progress to project stakeholders. However, when it appears that the progress of work is critically lagging, a request for extension of time (EOT) could be initiated quickly.
Tasks/Activities in a Construction Work
The easiest way to identify tasks in construction is to break the construction work into phases. For typical building construction, these phases can be broken down as follows;
Pre-construction activities
Mobilisation and Setting Out
Substructure (foundation) construction
Concrete frame works (ground floor to first floor)
Block work
Finishes, etc
Under preconstruction activities, tasks such as site clearing, obtaining permission to commence construction, host community negotiations, environmental assessments, harmonization of construction drawings, etc can be identified. Timing and duration can also be assigned to these tasks so as to have a defined target. Under mobilisation and setting out works, all unique tasks involved should also be listed according to the peculiarities of the project.
While construction activities are usually specified in an order, it is possible for two different activities to be going on simultaneously. For instance, reinforcements works can commence while setting out operations are still on-going. The setting of blocks on the ground floor can commence during the casting of columns on the second floor, etc.
Therefore all tasks in a project are identified bearing in mind the resources that are available for their successful completion. All machinery, man-power, materials, and funds needed to execute any item of work should be given special consideration during the preparation of the programme of work. For instance, a contractor who intends to use ready mix concrete and pumping of concrete may not have the same programme of work with a contractor who wishes to use manual labour.
After the activities are defined and set up, another important aspect is to establish the relationship among the tasks. The major importance of this is to figure out the critical path and the best sequence of works. There are some activities which when they are not done can stop almost every other activity on site. For instance, during substructure works, excavation should be done quickly and completely because without it being completed, every other activity is affected.
Tasks in a construction work can be connected to each other in four ways. These are;
Finish-to-Start (FS): If the activity can’t start without finishing the predecessor activity, it is called an FS relationship.
Start-to-Start (SS): If for any reason, two or more activities must start together, it is said they have a SS relationship.
Finish-to-Finish (FF): If two or more activities finish together, they have a FF relationship.
Start-to-Finish (SF): If an activity can’t be finished until another activity starts, that means they have an SF relationship.
Once the sequence of all the activities has been identified, the project manager can issue the program of work. For simplification, programme of work can also be broken down and handed over the site manager. An example of a simplified programme of work for casting a first-floor slab is given below;
Duration of Tasks
It is important that an experienced execution team be involved in the planning or at least understand how the planning was done. Many projects have failed to achieve proper success because of an improperly thought out timeline assigned to an activity. There are some documents or magazines where someone can pick the duration of works, but the person has to pay proper attention to the conditions that warranted such duration.
In our own point of view, the duration assigned to any item of work should be based on experience. If a comparable item of work has been completed in the past, the project manager should borrow ideas from the project based on the number of workers deployed, the machinery used, the weather conditions, etc. In the image below, we present a simple POW for the finishes of a building.
Advantages of Programme of Works
It serves as part of KPIs for a project
Work plans are generated from POW
Data gathering
To boost credibility to show understanding of what the scope of the project entails from initiation to closeout.
Special Considerations for Preparation of Programme of Works
Understanding of the contract documents
Bill and drawings
Phasing, milestone, and milestone payment if any
Scope of work and work breakdown structure
Method of construction (client and contractor)
Rate and data for resources allocation to each task/activities
Weather condition, period of the year, and associated risk(s)
Cash flow (client and contractor)
Timeline and cost relationship of task and targets
Logistics.
Sample Programme of Works for a Residential Building
A simple programme for a building to be completed in 3 months has been attached in pdf format (see below for the screenshot of the attached sample of a construction programme of work). You can download the PDF file below.
PS: We can help you manage your civil engineering projects of any kind inclusive of research, design, and construction. We pride ourselves in excellence and creativity. Send an e-mail to info@structville.com
Pad foundations are isolated rectangular, square, or circular slabs that are provided under reinforced concrete columns or column stubs to safely transmit the column load to the ground. They are a type of shallow foundation that is widely used all over the world especially in areas where the soil possesses good bearing capacity. They are also referred to as isolated bases or spread footings. The design of pad foundations involves sizing the base slab to satisfy geotechnical requirements and providing adequate thickness and reinforcements to satisfy structural requirements.
The dimensions of a pad foundation should not be too small so as to cause excessive settlement or bearing capacity failure of the soil. As a matter of fact, allowable bearing capacity is normally used to control settlement during the design of a pad foundation, hence it is treated as a serviceability limit state parameter. The width of a pad foundation is expected not to be less than 1000 mm, and the thickness not less than 150 mm.
Geotechnical Design of Pad Foundation
The geotechnical design of a pad foundation can be carried out according to the requirements of EN 1997-1:2004 (Eurocode 7). Eurocode 7 gives three approaches for the geotechnical design of foundations and they are as follows;
Design Approach 1 (DA1): In this approach, partial factors are applied to actions and to ground strength parameters. Design Approach 2 (DA2): In this approach, partial factors are applied to actions or to the effects of actions and to ground resistances. Design Approach 3 (DA3): In this approach, partial factors are applied to actions or to the effects of actions from the structure and to ground strength parameters.
The three approaches can give very different results when applied in design. However, the UK National Annex to Eurocode 7 permits only Design Approach 1 (DA1). For the design of pad foundation using Design Approach 1, three limit states with their appropriate load combination shall be satisfied for the structure. These limit states are;
EQU: Loss of equilibrium of the structure STR: Internal failure or excessive deformation of the structure itself GEO: Failure due to excessive deformation of the ground supporting the structure UPL: Failure due to uplift of the foundation due to water pressure HYD: Failure due to hydraulic gradient
In the design of pad foundation using DA1, there are two sets of limit state combinations for STR and GEO limit states. Combination 1 is normally used for the structural design of the foundation, while combination 2 is normally used for sizing the foundation. The partial factors for the limit states are given in the table below;
The partial factors for EQU, UPL, and HYD are given in the Table below. They can also be used in the uplift verification of all kind of buried structures.
The partial factors for soil properties is given in the table below;
It should be noted that pad foundations fall under category 2 structures which means that they are conventional structures that are founded on non-difficult grounds. They offer no exceptional geotechnical risk. As a result, routine procedures for field and laboratory testing for design and execution may be used. Geotechnical design of pad foundations can be done by geotechnical or structural engineers. However, the geotechnical design of category 3 structures with abnormal risk can be done by geotechnical engineers only.
The design of pad foundations can be done using any of the following methods;
Using the analytical (direct) method, all limit states should be verified. The ultimate bearing capacity qult of a pad foundation should be verified using the expression below;
where; c = cohesion q = overburden γ = body-weight Ni = bearing capacity factors si = shape factors di = depth factors ii = inclination factors gi = ground inclination factors bi = base inclination factors
Example on the calculation of the bearing capacity of a pad foundation using Design Approach 1 (DA1)
Calculate the bearing capacity of a 1m x 1m pad foundation founded at 0.9 m below a lateritic soil deposit. The characteristic angle of shearing resistance φk of the soil is 21° while the effective cohesion c’ is 10 kN/m2. The water table is 8m below the ground surface, and the unit weight of the soil is 18 kN/m3.
Solution
As the footing rests on a cohesive-frictional soil, the relevant material property is the angle of shearing resistance, φ and the effective cohesion c’.
Design values of angle of shearing resistance Characteristic value φk = 21°. Note that the safety factor γφ is applied to tan φk not to φk. Combination 1: γφ = 1.0, tan φd = tan φk/γφ= tan 21° = 0.383, φd = 21° Combination 2: γφ = 1.25, tan φd = tan φk/γφ = tan 21/1.25 = 0.307, φd = 17°
Step 4: Calculate the overburden pressure, q. The unit weight of soil is 18 kN/m3 and the safety factor γγ = 1 q = 18 × depth of footing = γγ × 18 × 0.9 = 16.2 kN/m2
For sizing of the foundation, Combination 2 allowable bearing capacity should be used (no other factor of safety is to be applied).
For the semi-empirical (indirect) method, a commonly recognized semi-empirical method such as bearing resistance estimation using pressuremeter test should be used. The use of experience and testing to determine SLS parameters that will also satisfy ULS requirements is generally done. An example can be found in Annex E of EN 1997-1:2004.
When using the prescriptive method, a presumed bearing resistance from BS 8004 should be used. When such a method is applied, the design result should be evaluated on the basis of comparable experience.
Example on the Structural Design of Pad Foundation
In the structural design of pad foundations, the reaction under an axially loaded column base may be assumed to be uniformly distributed if the load is concentric without any bending moment. Otherwise, the pressure distribution may be assumed to be varying linearly across the base as shown below.
(a) Design ultimate bearing pressure For a concentrically loaded pad foundation, the design earth pressure is given by;
q = P/Aprov
Where; P = Design axial force of the column = 1.35Gk + 1.5Qk (kN) Aprov = Base area provided for the footing (m2)
(b)
(b) Bending The critical section for bending is at the face of the column on a pad footing or the wall in a strip footing. The moment is taken on a section passing completely across a pad footing and is due to the ultimate loads on one side of the section. No redistribution of moments should be made.
(c) Beam Shear The vertical shear force is the sum of the loads acting outside the section considered. Shear stress is checked at a distance d from the face of the column. It is normal practice to make the base sufficiently deep so that shear reinforcement is not required. The depth of the base is often controlled by the design for shear.
(d) Punching Shear Rules for checking for punching shear resistance are given in section 6.4 of EN 1992-1-1:2004. The punching shear force is the sum of the loads outside the periphery of the critical section. Two punching shear checks should be carried out – at the column perimeter and at between d – 2d from the face of the column.
Structural Design Example of Pad Foundation
Design a square pad footing for a 250 × 250 mm column carrying a characteristic permanent load Gk of 800 kN and characteristic variable load Qk of 425 kN. The presumed allowable bearing pressure of the non-aggressive soil is 225 kN/m2. fck = 30 N/mm2; fyk = 500 N/mm2; Concrete cover = 50 mm
Pad foundation is a category 2 structure, and this design is to be done using prescriptive methods:
Let 10% of the service load account for the self-weight of the pad foundation. Base area A = 1.1(800 + 425)/225 = 5.99 m2 Minimum dimensions of footing = √5.99 = 2.447m
Adopt a square foundation of 2500 mm x 2500m x (600 mm) trial depth (Area provided Aprov = 6.25 m2)
Loading at ultimate limit state NEd = (1.35 x 800) + (1.5 x 425) = 1717.5 kN
Critical design moment at the face of the column MEd = (274.8 x 1.1252)/2 = 173.89 kNm/m
Effective depth d = 600 – 50 – 16 = 534 mm k = MEd/(bd2fck) = (173.89 x 106)/(100 x 5342 x 30) = 0.0203 ⇒ z = 0.95d = 0.95 x 534 = 507.3 mm ⇒ As = MEd/0.87fykz = (173.89 x 106)/(0.87 x 500 x 507.3) = 788 mm2/m
Provide H16 @ 225 c/c both ways (Asprov = 893 mm2/m)
Beam shear Check critical section d away from column face VEd = 274.8 x (1.125 – 0.534) = 162.4 kN/m vEd = 162.4/534 = 0.304 N/mm2
vRd, c = CRd, c × k × (100 × ρ1 × fck) 0.3333 CRd, c = 0.12 k = 1 + √ (200/d) = 1 + √ (200/534) = 1.611 ρ = 893/(534 × 1000) = 0.00167 vRd, c = 0.12 × 1.611 × (100 × 0.00167 × 30)0.333 = 0.33 N/mm2 => vEd (0.304 N/mm2) < vRd,c (0.33 N/mm2) beam shear ok
Punching Shear Punching shear: Basic control perimeter at 2d from face of column vEd = βVEd/uid < vRd,c
β = 1, ui = (250 x 4 + 534 x 2 x 2 x π) = 7710 mm
VEd = load minus net upward force within the area of the control perimeter) VEd = 1717.5 – 274.8 x (0.252 + π x 1.0682 + 1.068 x 0.25 x 4) = 422 kN vEd = (422 x 103)/(7710 x 534) = 0.102 N/mm2 Punching shear is therefore okay
Using the only data available from this picture (which is the crack pattern), what is likely caused failure of this structure?
The structure is a cantilever and we expect the maximum bending moment and shear force to occur at the support (usually taken as the centreline of the column). Predictions can be made on the cause of the failure based on the nature of the crack…
Kindly let us know your view in the comment section. Thank you and God bless you.
Gabion walls are made up of row upon row of orthogonal cages or baskets (gabions) which are filled with rock fragments/cobbles and tied together. This arrangement forms a block of gravity structure that is able to withstand lateral pressure on it. Furthermore, the permeability of the rock fragments and the flexibility of gabion cages make them particularly suitable for use at sites which are liable to become saturated and where the foundation is composed of relatively compressible materials.
Gabion walls are relatively simple to construct. Where suitable rock is readily available, the use of gabion walls is particularly attractive for reasons of economy and speed of construction. In more recent times, gabion walls are generally used as decorative walls in compounds or gardens and are aesthetically pleasing. However, gabion walls can be designed as retaining structures.
Gabions being used as fence in a home
Gabion wall decoration of a garden
Gabions acting as a retaining wall
When gabions are to be used as retaining structures, they should be considered as gravity retaining walls. There is currently no universally accepted method for designing the individual gabion units. The individual units are placed on each other to form a stable gabion wall. The basic shape of a gabion wall is trapezoidal, but the front and rear faces may be straight or stepped.
It is recommended that back batter be provided for walls higher than about 3 m to improve stability and even out ground bearing pressures. A variety of cage sizes can be produced using suitable materials to suit the terrain. Gabion units are normally in modules of 2 m x 1 m x 1 m.
However, the design method for verification and appropriate functionality of a gabion wall can be carried out at the ultimate limit state analysis (ULS), verifying the following issues;
Soil Bearing: The base pressure applied by the wall must not exceed the ultimate bearing capacity of the supporting soil.
Sliding Resistance: There must be an adequate factor of safety for sliding between the base of the wall and the underlying soil due to the lateral earth pressure. Active earth pressure conditions should be assumed.
Overturning stability: The overturning of the wall due to horizontal earth pressure forces when the retained soil mass become unstable (active failure) should satisfy the required conditions.
Internal stability verification: For each layer of gabion a bearing capacity and sliding resistance verification should be made
Global Stability
Design Considerations of Gabion Walls
The maximum pressure transferred from the gabion wall to the ground should be compared with the safe bearing capacity of the soil. This issue depends on the type of gabion foundation also, which for this case it is considered a non-stiff footing (the gabion is not placed on top of concrete).
Since gabion retaining walls consist of several layers, the internal stability verification should verify that these layers are not collapsing, neither sliding nor the bearing capacity of lower layers lesser than required. These conditions must be satisfied for each layer starting from the top layer to the bottom layer.
The overturning stability of the gabion retaining wall ensures that the soil retained behind the gabion wall is not causing lateral collapse of the wall. For this analysis, it is assumed the gabion wall is rigid body, with no sliding occurrence or partial breakdown between layers. The stabilising force is the weight of the gabion wall, while the destabilising force is the lateral pressure from the retained material or surcharge.
Limit state checks should be carried out at selected planes through the gabion wall, ignoring the resistance contributed by the cage material and the connections between the cages. For stepped walls, stability checks should be carried out at each major change in section shape.
In order to limit deformation, gabion walls should be proportioned in such a way that the resultant force acts within the middle third of the wall’s cross-section. The mobilised angle of wall friction, δ, used in the design should not exceed ϕ’/2, where ϕ’ is the angle of shearing resistance of the compacted backfill. In order for the assumption of δ = ϕ’/2 to hold, the gabion infill must be placed in such manner to achieve a dense mass which will not settle relative to the backfill after construction. Otherwise δ should be assumed to be zero. For a wall to be founded on relatively compressible materials, δ should also be assumed to be zero.
Materials for construction of gabion retaining walls
(1) Gabion Baskets Gabion baskets can be made from a range of materials. Nylon, polypropylene and polyethylene grids have been used. They have the advantage of being lightweight and corrosion-resistant. However, these materials are susceptible to attack by fire and ultraviolet light. A material widely used in the commercial production of gabions is steel wire-mesh, of which there are two types, hexagonal woven and square welded.
The wires used for the wire-mesh in gabion baskets should be mild steel wire with a minimum tensile strength of 350 N/mm2. For permanent applications, the wires should be at least 2.7 mm in diameter and galvanized. Where it is suspected that the infill or retained materials or groundwater are aggressive, the wire mesh should be protected with PVC coating.
(2) Infill Material Rock used for filling gabions should be sound, clean and well-graded. The maximum size of the rock should not exceed two-thirds the depth of the gabion to be filled or 300 mm, whichever is less. The preferred size is 150 mm to 300 mm. The smallest dimension of the rock should at least be twice the largest dimension of the mesh aperture.
(3) Backfill, Filter, and Drainage Materials For a partially-submerged gabion wall, a free-draining granular backfill should be provided so that water pressure will not build up behind the wall when the water level in front of the wall is lowered.
Drainage Considerations for Gabion Walls
A geotextile filter should be provided behind the rear face of the gabion wall to prevent the migration of fines from the backfill into the coarse rock infill. Drainage layers at the rear face are normally not warranted. However, a drainage layer of adequate permeability should be provided at the base of the wall to guard against erosion of the foundation material.
The high permeability of the gabion units will permit direct infiltration through the body of the wall at times of heavy rainfall. In order to minimize the possibility of saturation and erosion of the foundation material under a non-submerged gabion wall, it is good practice to provide a blinding layer with adequate drainage provisions at the level of the foundation. For submerged gabion walls, appropriate measures should be incorporated to prevent scouring and erosion of the foundation.
Design Example of Gabion Walls
In the example below, a 3 m high gabion retaining wall is to be designed according to the requirements of Eurocode 7 (EN 1997-1).
The design parameters and loadings are as follows;
Wall geometry Width of gabion 1; w1 = 2000 mm Height of gabion 1; h1 = 1000 mm Width of gabion 2; w2 = 1700 mm Height of gabion 2; h2 = 1000 mm Step to front face between courses 1 and 2; s2 = 150 mm Width of gabion 3; w3 = 1400 mm Height of gabion 3; h3 = 1000 mm Step to front face between courses 2 and 3; s3 = 150 mm Wall inclination; ε = 0 deg
Gabion properties Unit weight of fill; γd = 16.0 kN/m3 Friction between gabions; δbg.k = 35.0 deg
ASDIP Structural Engineering Software company has announced the release of STEEL-5, the latest version of ASDIP STEEL software module. The Florida, USA, based structural engineering software company has been operational since the year 1992 and offers different modules for civil engineering designs such as reinforced concrete, steel, foundation, and retaining walls. The STEEL-5 is the latest version of the steel package which offers advances and improvements from the previous versions.
ASDIP STEEL software is utilised by professional engineers for the design of steel members and connections. It works with any operating system, and assists structural engineers to transparently design, analyze, check and optimize structural design work. According to the company’s official website, the following are some of the most important features and benefits included in version 5 of ASDIP STEEL;
The new version includes the following five modules:
Base Plate / Anchorage Design – Any combination of axial vertical and horizontal loads, and biaxial moments. It includes uplift and partial bearing analysis. Design of anchor rods and shear lugs. Graphical generation of the breakout areas.
Steel Column Design– Either sway or non-sway columns. Second-order moments analysis to account for slenderness. Multiple load types and load cases. Results sorted by load combination. Graphical generation of the interaction diagram.
Steel / Composite Beam Design – Up to five spans and two overhangs. Multiple load types and load cases. Separate Construction and Final loads analysis. Graphical generation of the shear, and moment diagrams.
Shear Connection Design – Single angle, double angle, Shear Plate, and Tee connections. Multiple options to specify different conditions of the connection elements. Check of the limit states. Graphical generation of the connection in different views.
Moment Connection Design – Flange-Plated, and Welded Flange moment connections. Any combination of vertical loads and moments. Check of the limit states for moment and shear. Graphical generation of the connection in different views.
A Linkedin Post by the Founder of ASDIP, Javier Encinas PE, invited structural engineers across the world to download the 15-day trial of the new release in order to check the power and capabilities of the STEEL-5. According to him,
“For some weeks we have announced in advance the coming release of ASDIP STEEL 5, our software for structural steel design. As promised, ASDIP STEEL 5 has been released. In addition to an upgraded layout design and literally dozens of improvements, it includes great new features, such as:
Moment connections (NEW module) AISC 360-16 and ACI 318-19 compliance Custom load combinations Biaxial base plates and anchorage Continuous steel / composite beam
You are invited to download the 15-day free trial and check the software by yourself, hands-on. I have prepared a blog post and a short video with a brief overview of this new version of ASDIP STEEL. Your comments and suggestions are always welcome.”
Reinforced concrete (R.C.) slabs are plate elements used to form the floors of buildings. In a typical reinforced concrete building, reinforcement bars arranged as mats are incorporated into a concrete plate of minimum thickness 125 mm to form a reinforced concrete solid slab. The provision of adequate reinforcement, slab thickness, and proper detailing to satisfy ultimate and serviceability limit state requirements forms the basis of the design of reinforced concrete (R.C.) slab. Satisfying other requirements such as durability, fire resistance, etc are also necessary.
Floor slabs are usually subjected to uniformly distributed loads, partially distributed loads, line loads, or concentrated loads in the transverse direction. A beam is similar to a slab in so many ways but there are fundamental differences in behaviour and stress distribution of the two elements. While a beam is generically a one-dimensional element, a plate is a two-dimensional element. Due to the two-dimensional nature of a slab, it is subjected not only to bending moments Mxx and Myy and shear forces Vxand Vy but also to twisting moments Mxy on all the four faces.
Fig 1: Beam element and slab element
Types of Reinforced Concrete Slabs
A monolithic reinforced concrete slab is essentially a statically indeterminate structure. For a slab of a given shape and support conditions, the distribution of shear forces, bending, and twisting moments in the slab due to externally applied loads cannot be determined easily. The column layout in a monolithic reinforced concrete structure often forms a rectangular grid. Continuous beams may be provided in one direction or two orthogonal directions, to support slabs that may be solid or ribbed in cross-section. Alternatively, slabs can be supported directly on columns to form flat slabs.
Slabs may be simply supported or continuous over one or more supports and are classified according to the method of support as follows:
One-way spanning slab between beams or walls
Two-way spanning slab between the support beams or walls
Flat slabs carried on columns and edge beams or walls with no interior beams
Ribbed slab transferring slab load to beams and walls in one direction
Waffle slab transferring slab load to beams and slabs in two direction
One-way Spanning Solid Slabs
Clause 5.3.1(5) of EN 1992-1-1:2004 suggests that a slab subjected dominantly to uniformly distributed loads may be considered as one-way spanning if either:
It possesses two free (unsupported) and sensibly parallel edges.
It is the central part of a sensibly rectangular slab supported on four edges with a ratio of longer (Ly) to shorter span (Lx) greater than 2.
Fig 2: Typical general arrangement of a one-way slab system
In other words, when beams are provided in one parallel direction only, the slab is a one-way slab. Furthermore, if the longer side of a slab panel exceeds twice the shorter side, the slab is generally designed as a one-way slab, but that does not mean that the slab is transmitting load in one direction only.
One-way slabs may be simply supported or continuous. For one-way slabs supported on two opposite sides, the bending moments are calculated in the same way as for beams. Continuity across a beam is treated as fixed support. In detailing, if a slab is assumed to be simply supported at an end support, it is advisable to provide reinforcement for a probable negative bending moment due to the monolithic construction of beams and slabs (Reynolds and Steedman, 2005).
Fig 3: Typical reinforcement detailing of a simply supported one-way slab
The effective span for one-way slabs is the same as that of beams. If ln is the clear span (distance between faces of supports), the effective span leffis given by;
leff = ln + a1 + a2
Fig 4: Effective span of a simply supported one-way slab
One-way slabs should be designed to resist the most unfavourable arrangement of loads. In clause 5.1.3 of Eurocode 2, the following two loading arrangements are recommended for buildings.
Alternate spans carrying (γGGk + γQQk) other spans carrying only γGGk.
Any two adjacent spans carrying (γGGk + γQQk). All other spans carrying only γGGk
Steps in the designof a one-way slab
The steps in the design of a slab are as follows;
Determine the design life of the structure
Choose a slab thickness determined using deflection requirements, experience, or otherwise
Establish the durability requirements, fire resistance, and adequate concrete cover
Calculate and apply the loads on the slab comprising of the dead and imposed loads
Apply the appropriate load combination
Idealise each slab element and analyse to determine the critical design moments MEd and shear forces VEd
Carry out the flexural design
Check the deflection
Check the shear capacity
Check bar spacing and cracking
For the flexural design of slabs, determine k from;
k = MEd/(fckbd2) If k < 0.167, no compression reinforcement is required, and you can calculate the lever arm; z = d[0.5 + √(0.25 – 0.882k)]
The area of reinforcement required is given by; As1 = MEd/(0.87fykz)
(a) Concrete cover The nominal concrete cover in slabs is expected to satisfy the requirement;
Cnom = Cmin + ∆c,dev
Cmin is expected to satisfy the requirement for durability, fire resistance, and bond, while ∆c,dev is the allowance made for construction deviation (usually 10 mm). The minimum cover for bond should not be less than the bar diameter. For a one-way slab to have a fire rating of one hour (REI 60), the minimum thickness should be 80 mm and the concrete cover (from the surface to the centre of bar) should be minimum of 20 mm. A minimum cover of 15mm + ∆c,dev is adequate for the durability of slabs under exposure class XC1. Therefore under normal circumstances, a concrete cover of 25 mm is usually adequate for floor slabs.
(b) Minimum tension steel The main moment steel spans between supports and over the interior supports of a continuous slab. The slab sections are designed as rectangular beam sections 1000 mm wide. The minimum area of main reinforcement has to satisfy clause 9.2.1.1(1) of Eurocode 2;
As,min = (0.26fctm/fyk)btd but not less than 0.0013bd
where bt = width (for slab design 1000 mm), d = effective depth
The maximum area of steel provided should not exceed 0.04Ac.
(c) Distribution steel The distribution, transverse, or secondary steel runs at right angles to the main moment steel and serves the purpose of tying the slab together and distributing non-uniform loads through the slab. Clause 9.3.1.1(2) states that in the case of one-way slabs, secondary reinforcement of not less than 20 percent of principal reinforcement should be provided. Note that distribution steel is required at the top parallel to the supports of continuous slabs. The main steel is placed nearest to the surface to give the greatest effective depth.
Fig 5: Typical reinforcement detailing of a continuous one-way slab
(d) Slab main reinforcement Slab reinforcement is a mesh and may be formed from two sets of bars placed at right angles. The table below gives bar spacing data in the form of areas of steel per metre width for various bar diameters and spacings. Reinforcement in slabs consists of a large number bars both ways which need to be tied together to form a mat. This is actually an expensive operation (see cost of fixing reinforcement in Nigeria).
It is necessary to point out that the critical span in the analysis of solid slabs is the short span. Therefore, the main reinforcements will lie parallel to the short span, and that is where you check your deflections (see why short spans are critical in floor slabs).
Table 1: Area of reinforcement per metre width of spacing
Spacing
ϕ8 mm
ϕ10 mm
ϕ12 mm
ϕ16 mm
ϕ20 mm
100
502
785
1130
2010
3141
125
402
628
904
1608
2513
150
335
523
753
1340
2094
175
287
448
646
1148
1795
200
251
392
565
1005
1570
225
223
349
502
893
1396
250
201
314
452
804
1256
275
182
285
411
731
1142
300
167
261
376
670
1047
For instance, if the area of reinforcement required (As,req) from calculation is 523 mm2/m, you can provide H12@200 c/c (Area of steel provided As,prov = 565 mm2/m).
(e) Crack Control According to Clause 9.3.1.1(3) of Eurocode 2, if h is the total depth of the slab, then the maximum spacing of reinforcements is normally restricted to;
3h ≤ 400 mm for principal reinforcement
3.5 h ≤ 450 for secondary reinforcement
However, in areas of maximum moment, maximum spacing is restricted to;
2h ≤ 250 mm for principal reinforcement
3 h ≤ 400 mm for secondary reinforcement
For slabs 200 mm thick or greater the bar size and spacing should be limited to control the crack width and reference should be made to section 7.3.3 of the Eurocode 2.
(f) Curtailment of bars in slabs Curtailment of bars is done according to the moment envelope. However, clause 9.3.1.2(1) requires that half the calculated span reinforcement must continue up to support. It is further stated that in monolithic construction, where partial fixity occurs along an edge of a slab but is not taken into account, the top reinforcement should be capable of resisting at least 25 percent of the maximum moment in the adjacent span and this reinforcement should extend at least 0.2 times the length of the adjacent span measured from the face of the support.
Fig 6: Typical curtailment rules of solid slabs
The above situation occurs in the case of simply supported slabs or the end support of a continuous slab cast integral with an L-beam which has been taken as a simple support for analysis but the end of the slab might not be permitted to rotate freely as assumed. Hence negative moments may arise and cause cracking.
(g) Shear capacity of slabs Under normal loads, shear stresses are not critical and shear reinforcement is not required in floor slabs. Shear reinforcement is provided in heavily loaded thick slabs but should not be used in slabs less than 200 mm thick (clause 9.3.2 (1)). To check the shear capacity of slabs, the shear stress (vEd = VEd/bd) must be checked against the shear capacity of an unreinforced section (VRc,d). This is given by;
Where; CRd,c = 0.18/γc k = 1 + √(200/d) < 0.02 (d in mm); ρ1 = As1/bd < 0.02 (In which As1 is the area of tensile reinforcement which extends ≥ (lbd + d) beyond the section considered) Vmin = 0.035k(3/2)fck0.5 K1 = 0.15; σcp = NEd/Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the concrete), fcd = design compressive strength of the concrete).
In a one-way slab, the design shear force is calculated from the support reactions or end-shears at the support, while in a two-way slab, they can be obtained from the coefficients in Table 3.15 of BS 8110-1:1997.
(h) Check for deflection The check for deflection is a very important consideration in slab design and usually controls the slab depth. In normal cases, a strip of slab 1 m wide is checked against span-to-effective depth ratios. A slab should not deflect excessively under service load. Excessive deflection of slabs can cause cracking to partitions and finishes.
For deemed to satisfy basic span/effective depth (limiting to depth/250);
Actual L/d of the slab must be ≤ Limiting L/d × βs
The limiting basic span/ effective depth ratio is given by;
L/d = K [11 + 1.5√(fck)ρ0/ρ + 3.2√(fck) (ρ0/ρ – 1)1.5] if ρ ≤ ρ0
L/d = K [11 + 1.5√(fck) ρ0/(ρ – ρ’) + 1/12 √(fck) (ρ0/ρ)0.5 ] if ρ > ρ0
Where; L/d is the limiting span/depth ratio K = Factor to take into account different structural systems ρ0 = reference reinforcement ratio = 10-3 √(fck) ρ = Tension reinforcement ratio to resist moment due to design load ρ’ = Compression reinforcement ratio
The value of K depends on the structural configuration of the member and relates the basic span/depth ratio of reinforced concrete members. This is given in the table 2;
Table 2: Basic span/effective depth ratio of different structural systems
Structural System
K
Highly stressed ρ = 1.5%
Lightly stressed ρ = 0.5%
Simply supported slabs
1.0
14
20
End span of interior slabs
1.3
18
26
Interior span of continuous slabs
1.5
20
30
Flat slab
1.2
17
24
Cantilever slabs
0.4
6
8
βs = (500As,prov)/(fykAs,req)
Two-way Spanning Slabs
Two-way action occurs when a slab is supported on all four sides. If the two dimensions and support conditions are the same, then the load is distributed to all supporting beams equally. In design, a slab is considered to be two-way if the ratio of the longer side to the shorter side is less than two.
For two way slabs, the precise amount and distribution of the load taken by each support, and consequently the magnitude of the bending moments on the slab, are not easily calculated if assumptions resembling real conditions are made. Therefore, approximate analyses are generally used. The method applicable in any particular case depends on the shape of the slab panel, the conditions of restraint at the supports and the type of load.
Two basic methods are commonly used to analyse slabs that span in two directions. They are;
The theory of plates, which is based on elastic analysis, is particularly appropriate to the behaviour under service loads
Yield-line theory, which considers the behaviour of the slab as a collapse condition approaches.
Generally, for rectangular slabs with standard edge conditions and subject to uniformly distributed loads, normally the bending moments are obtained using tabulated coefficients. These coefficients are based on elastic analysis from thin plate theory. The loads used in the analysis are factored to represent the ultimate limit state condition. This is the approach used in BS 8110 for slabs with corners that are not held down (no consideration for torsion). The analysis must take into account the support conditions which can be idealised as fixed, hinged, or free.
For slabs with irregular plan shapes and slabs subject to a combination of point loads and distributed loads, Johansen’s yield line analysis and the Hillerborg strip method provide powerful methods for strength calculations.
Simply supported two-way slabs Where the corners of slabs are free to lift and no provision is made to resist forces at the corners, the maximum moments per unit width are given by the following expressions:
Msx = bending moment in strips with span lx = Msx = αsxqlx2 Msy = bending moment in strips with span ly = Msy = αsyqlx2
where lx is the shorter span of the panel, ly is the longer span of the panel and q is the design ultimate load per unit area. Values of αsx and αsy are given in Table 3.13 of BS 8110-1:1997 for different ratios of ly and lx, where ly is the longer span.
Rectangular panels with restrained edges Where corners of a two-way slab are prevented from lifting and reinforced to resist torsion, the maximum bending moments per unit width are given by the following expressions:
Msx = βsxqlx2 Msy = βsyqlx2
where; Msx is the maximum design moment either over supports or at midspan on strips with span lx Msy is the maximum design moment either over supports or at midspan on strips with span ly q is the design ultimate load per unit area, lx is the shorter span, and ly is the longer span
The coefficients can be obtained from Table 3.14 of BS 8110-1:1997.
Design Exampleof a two-way slab
The general arrangement of the floor plan of a building is shown below. Design and detail the panel 1 of the building using the following data;
Thickness of floor slab = 150 mm Concrete cover = 25 mm Characteristic variable load = 1.5 kN/m2 fck = 25 N/mm2 fyk = 460 N/mm2
Fig 7: General arrangement of a floor slab
Load Analysisof Panel 1
Permanent Loads Self weight of slab = 25 kN/m3 × 0.15m = 3.75 kN/m2 Weight of finishes = 1.2 kN/m2 Partition allowance = 1.5 kN/m2 Total dead load (gk) = 6.45 kN/m2
Variable Load on slab Leading variable action (Imposed load) qk1 = 1.5 kN/m2
Total load on slab (ULS) = 1.35gk + 1.5qk = 1.35(6.45) + 1.5(1.5) = 10.9575 kN/m2
The floor slab (PANEL 1) is spanning in two directions, since the ratio (k) of the longer side (ly) to the shorter side (lx) is less than 2. Hence, k = Ly/Lx = 3.825/3.625 = 1.055 (say 1.1, however it is more appropriate to interpolate from the table).
Moment coefficient for two adjacent edges discontinuous
Short Span Mid-span = 0.042 Continuous edge = 0.056
Long Span Mid-span = 0.034 Continuous edge = 0.045
Design of short span Mid span MEd = αnlx2 = 0.042 × 10.9575 × 3.6252 = 6.0475 kNm/m d = h – Cc – ϕ/2
Assuming ϕ12mm bars will be employed for the construction d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)
Taking the distance between supports as the effective span, L = 3625mm The allowable span/depth ratio = βs × 30.838 = 2.0 × 190.327 = 280.645 Actual deflection L/d = 3625/119 = 30.462 Since 280.645< 30.462 Therefore, deflection is ok.
Shear Verification Maximum shear force coefficient for slab(β) = 0.44 VEd = βnlx
Ultimate shear force VEd = 0.44 × 10.9575 × 3.625 = 17.477 kN/m
VRd,c = [CRd,c.k. (100ρ1 fck)1/3
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/119) = 2.296 > 2.0, therefore, k = 2
