When to Order the Demolition of a Building

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December 6, 2020

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.

Thank you very much for visiting Structville today.

Hemp and Recycled Aggregate Concrete

By

Ubani Obinna

-

December 5, 2020

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].

hemp fibre — **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.

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).

hemp hurds — **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 Mater 7, 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 Mater 14, 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

How to Design Programme of Work (POW) for Building Construction

By

Mubarak Shittu

-

December 4, 2020

Table of Contents

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.

Structville-programme-of-works Download

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

Design of Pad Foundation

By

Ubani Obinna

-

December 2, 2020

Table of Contents

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;

(a) Analytical (Direct) method
(b) Semi-empirical (Indirect) method
(c) Prescriptive method using presumed bearing resistance (BS 8004)

Using the analytical (direct) method, all limit states should be verified. The ultimate bearing capacity q_ult of a pad foundation should be verified using the expression below;

q_ult = c’N_cs_cd_ci_cg_cb_c+ q’N_qs_qd_qi_qg_qb_q + γ’BN_γs_γd_γi_γg_γb_γ/2

where;
c = cohesion
q = overburden
γ = body-weight
N_i = bearing capacity factors
s_i = shape factors
d_i = depth factors
i_i = inclination factors
g_i = ground inclination factors
b_i = 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 φ_k 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 φ_k = 21°. Note that the safety factor γ_φ is applied to tan φ_k not to φ_k.
Combination 1: γ_φ = 1.0, tan φ_d = tan φ_k/γ_φ= tan 21° = 0.383, φ_d = 21°
Combination 2: γ_φ = 1.25, tan φ_d = tan φ_k/γ_φ = tan 21/1.25 = 0.307, φ_d= 17°

Design values of cohesion
Characteristic value of effective cohesion c‘ = 10 kN/m²
Combination 1: γ_c’ = 1.0, c’_d = c’/γ_c’ = 10 kN/m²
Combination 2: γ_c’ = 1.25, c’_d = c’/γ_c’ = 10/1.25 = 8 kN/m²

Step 2: Calculate design bearing capacity factors. Use the equations in section D.4 of Annex D of Eurocode 7.

i. Overburden factor N_q
N_q = e^{(π × tanφ_d)} × tan²(45 + φ_d/2)
Combination 1: φ_d = 21°, N_q = 7.07
Combination 2: φ_d = 17°, N_q = 4.77

ii. Cohesion factor, N_c
N_c= (N_q − 1) cot φ_d
Combination 1: φ_d = 21°, N_c = 15.84
Combination 2: φ_d = 17°, N_c = 12.28

iii. Body weight factor N_γ
N_γ = 2(N_q − 1) tanφ_d
Combination 1: φ_d = 21°, N_γ = 4.66
Combination 2: φ_d = 17°, N_γ = 2.305

Step 3: Calculate the design shape factors. Use the equations in Annex D of Eurocode 7.

Square base
i. s_q = 1 + sin φ_d
Combination 1: φ_d = 21°, s_q = 1.358
Combination 2: φ_d = 17°, s_q = 1.292

ii. s_c = (s_qN_q − 1)/(N_q − 1)
Combination 1: N_q = 7.07, s_q = 1.358, s_c = 1.419
Combination 2: N_q = 4.77, s_q = 1.292, s_c = 1.369

iii. s_γ = 0.7 (for a square shape)

Step 4: Calculate the overburden pressure, q.
The unit weight of soil is 18 kN/m³ and the safety factor γ_γ = 1
q = 18 × depth of footing = γ_γ× 18 × 0.9 = 16.2 kN/m²

Step 5: Calculate the allowable q_ult:
q_ult = c’N_cs_cd_ci_cg_cb_c+ q’N_qs_qd_qi_qg_qb_q + γ’BN_γs_γd_γi_γg_γb_γ/2

For this calculation, all inclination factors have been ignored.

q_ult = c’N_cs_c + q’N_qs_q + 0.5γ’BN_γs_γ

Combination 1: q_ult = (10 × 15.84 × 1.419) + (16.2 × 7.07 × 1.358) + (0.5 × 18.0 × 4.66 × 1.0 × 0.7) = 409.66 kN/m²
Combination 2: q_ult = (8 × 12.28 × 1.369) + (16.2 × 4.77 × 1.292) + (0.5 × 18.0 × 2.305 × 1.0 × 0.7) = 248 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.

pressure distribution of pad foundations

(a) Design ultimate bearing pressure
For a concentrically loaded pad foundation, the design earth pressure is given by;

q = P/A_prov

Where;
P = Design axial force of the column = 1.35G_k + 1.5Q_k (kN)
A_prov = 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.

punching shear perimeter of pad foundations

(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.

typical reinforcement works of a pad foundation

Structural Design Example of Pad Foundation

Design a square pad footing for a 250 × 250 mm column carrying a characteristic permanent load G_k of 800 kN and characteristic variable load Q_k of 425 kN. The presumed allowable bearing pressure of the non-aggressive soil is 225 kN/m². f_ck = 30 N/mm²; f_yk = 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 A_prov = 6.25 m²)

Loading at ultimate limit state N_Ed = (1.35 x 800) + (1.5 x 425) = 1717.5 kN

ULS bearing pressure N_Ed/A_prov = 1717.5/6.25 = 274.8 kN/m²

Critical design moment at the face of the column
M_Ed = (274.8 x 1.125²)/2 = 173.89 kNm/m

Effective depth d = 600 – 50 – 16 = 534 mm
k = M_Ed/(bd²f_ck) = (173.89 x 10⁶)/(100 x 534² x 30) = 0.0203
⇒ z = 0.95d = 0.95 x 534 = 507.3 mm
⇒ A_s = M_Ed/0.87f_ykz = (173.89 x 10⁶)/(0.87 x 500 x 507.3) = 788 mm²/m

Provide H16 @ 225 c/c both ways (A_sprov = 893 mm²/m)

Beam shear
Check critical section d away from column face
V_Ed = 274.8 x (1.125 – 0.534) = 162.4 kN/m
v_Ed = 162.4/534 = 0.304 N/mm²

v_{Rd, c} = C_{Rd, c} × k × (100 × ρ₁ × f_ck) ^0.3333
C_{Rd, c} = 0.12
k = 1 + √ (200/d) = 1 + √ (200/534) = 1.611
ρ = 893/(534 × 1000) = 0.00167
v_{Rd, c} = 0.12 × 1.611 × (100 × 0.00167 × 30)^0.333 = 0.33 N/mm²
=> v_Ed (0.304 N/mm²) < v_Rd,c (0.33 N/mm²) beam shear ok

Punching Shear
Punching shear: Basic control perimeter at 2d from face of column
v_Ed = βV_Ed/u_id < v_Rd,c

β = 1,
u_i = (250 x 4 + 534 x 2 x 2 x π) = 7710 mm

V_Ed = load minus net upward force within the area of the control perimeter)
V_Ed = 1717.5 – 274.8 x (0.25² + π x 1.068² + 1.068 x 0.25 x 4) = 422 kN
v_Ed = (422 x 10³)/(7710 x 534) = 0.102 N/mm²
Punching shear is therefore okay

Detailing sketches of pad foundation

What is the likely cause of this failure?

By

Ubani Obinna

-

December 2, 2020

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.

Design of Gabion Walls

By

Ubani Obinna

-

December 1, 2020

Table of Contents

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.

decorative gabion retaining walls — Gabions being used as fence in a home

decorative gabion walls 2 — Gabion wall decoration of a garden

gabion acting as retaining wall — 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).

analysis and design of gabion retaining walls

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; γ_d= 16.0 kN/m³
Friction between gabions; δ_bg.k = 35.0 deg

Loading
Permanent surcharge; ρ_o,G = 5 kN/m²

Soil properties
Slope of retained soil; β = 0.0 deg
Characteristic effective shearing resistance angle; ϕ’_r.k = 30.0 deg
Characteristic saturated density of retained soil; γ_sr = 21.0 kN/m³
Coefficient for wall friction; k_membrane = 0.75
Wall friction angle; δ_r.k= 22.5 deg
Characteristic base friction angle; δ_bb.k= 30.0 deg
Soil bearing capacity; q = 100 kN/m²

The calculation sheet has been made available in PDF format and can be downloaded below;

Gabion-retaining-wall-analysis-design-EN1997-2Download

Thank you for visiting Structville today. God bless you.

ASDIP Announces the Release of STEEL-5 Software

By

Ubani Obinna

-

November 30, 2020

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.”

Detailed information is available about this structural engineering software by visiting ASDIP STEEL. You are also invited to download the Free 15-day Software Trial, or go ahead and Place your Order.

Design of Reinforced Concrete (R.C.) Slabs

By

Ubani Obinna

-

November 30, 2020

Table of Contents

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 M_xx and M_yy and shear forces V_xand V_y but also to twisting moments M_xy on all the four faces.

Beam and plate element — **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 (L_y) to shorter span (L_x) greater than 2.

One way slab arrangement — **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).

Typical detailing of a one-way simply supported reinforced concrete slab — **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 l_n is the clear span (distance between faces of supports), the effective span l_effis given by;

l_eff= l_n + a₁ + a₂

effective length of one way slabs 1 — **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 (γ_GG_k+ γ_QQ_k) other spans carrying only γ_GG_k.
Any two adjacent spans carrying (γ_GG_k + γ_QQ_k). All other spans carrying only γ_GG_k

Steps in the design of 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 M_Ed and shear forces V_Ed
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 = M_Ed/(f_ckbd²)
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;
A_s1 = M_Ed/(0.87f_ykz)

(a) Concrete cover
The nominal concrete cover in slabs is expected to satisfy the requirement;

C_nom = C_min + ∆_c,dev

C_min 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;

A_s,min = (0.26f_ctm/f_yk)b_td but not less than 0.0013bd

where b_t = width (for slab design 1000 mm), d = effective depth

The maximum area of steel provided should not exceed 0.04A_c.

(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.

detailing of continuous one way slab — **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 (A_s,req) from calculation is 523 mm²/m, you can provide H12@200 c/c (Area of steel provided A_s,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.

simplified curtailment rules of floor slabs — **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 (v_Ed = V_Ed/bd) must be checked against the shear capacity of an unreinforced section (V_Rc,d). This is given by;

V_Rd,c = [C_Rd,c k(100ρ₁ f_ck )^1/3 + k₁.σ_cp] ≥ (V_min + k₁.σ_cp)

Where;
C_Rd,c = 0.18/γ_c
k = 1 + √(200/d) < 0.02 (d in mm);
ρ₁ = A_s1/bd < 0.02 (In which A_s1 is the area of tensile reinforcement which extends ≥ (l_bd + d) beyond the section considered)
V_min= 0.035k^(3/2)f_ck^0.5
K₁ = 0.15; σ_cp = N_Ed/A_c < 0.2f_cd
(Where N_Ed is the axial force at the section, A_c = cross sectional area of the concrete), f_cd = 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 × β_s

The limiting basic span/ effective depth ratio is given by;

L/d = K [11 + 1.5√(f_ck)ρ₀/ρ + 3.2√(f_ck) (ρ₀/ρ – 1)^1.5] if ρ ≤ ρ₀

L/d = K [11 + 1.5√(f_ck) ρ₀/(ρ – ρ’) + 1/12 √(f_ck) (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 √(f_ck)
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

β_s= (500A_s,prov)/(f_ykA_s,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:

M_sx = bending moment in strips with span l_x = M_sx = α_sxql_x²
M_sy = bending moment in strips with span l_y = M_sy = α_syql_x²

where l_x is the shorter span of the panel, l_y is the longer span of the panel and q is the design ultimate load per unit area. Values of α_sx and α_sy are given in Table 3.13 of BS 8110-1:1997 for different ratios of l_y and l_x, where l_y 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:

M_sx= β_sxql_x²
M_sy = β_syql_x²

where;
M_sxis the maximum design moment either over supports or at midspan on strips with span l_x
M_sy is the maximum design moment either over supports or at midspan on strips with span l_y
q is the design ultimate load per unit area,
l_x is the shorter span, and
l_y is the longer span

The coefficients can be obtained from Table 3.14 of BS 8110-1:1997.

Design Example of 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
f_ck = 25 N/mm²
f_yk= 460 N/mm²

General arrangement of building — **Fig 7**: General arrangement of a floor slab

Load Analysis of 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 (g_k) = 6.45 kN/m²

Variable Load on slab
Leading variable action (Imposed load) q_k1 = 1.5 kN/m²

Total load on slab (ULS) = 1.35g_k + 1.5q_k= 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 (l_y) to the shorter side (l_x) 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
M_Ed = αnl_x²= 0.042 × 10.9575 × 3.625² = 6.0475 kNm/m
d = h – C_c – ϕ/2

Assuming ϕ12mm bars will be employed for the construction
d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)

k = M_Ed/(f_ckbd²) = (6.0475 × 10⁶)/(25 × 1000 × 119²) = 0.0171

Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.0273)] = 0.95d

A_s1 = M_Ed/(0.87f_ykz) = (6.0475 × 10⁶)/(0.87 × 460 × 0.95 × 119) = 133.668 mm²/m

Provide Y12mm @ 250mm c/c BOT (A_Sprov= 452 mm²/m)

To calculate the minimum area of steel required;
f_ctm = 0.3 × f_ck^2⁄3 = 0.3 × 25^2⁄3= 2.5649 N/mm²(Table 3.1 EC2)

A_s,min = 0.26 × f_ctm/f_yk × b × d = 0.26 × 2.5649/460 × 1000 × 144 = 208.76 mm²/m
Check if A_s,min < 0.0013 × b × d (187.2 mm²/m)
Since, A_s,min = 187.2 mm², the provided reinforcement is adequate.

Check for deflection
We check for deflection at the short span of slabs
k = 1.3 for slab simply supported at one end and continuous at the other end

ρ = A_s/bd = 134/(1000 × 119) = 0.001126 < 10^-3√f_ck (0.005)

Since ρ < ρ₀
L/d = K [11 + 1.5√(f_ck ) ρ₀/ρ + 3.2√(f_ck) (ρ₀/ρ – 1)^3⁄2]

L/d = 1.3 [11 + 1.5√(25) × 0.005/0.001126 + 3.2√25 (0.005/0.001126 – 1)^3⁄2] = 1.3(44.3 + 102.106) = 190.327

Modification factor β_s = 310/σ_s
σ_s = (310f_ykA_s,req)/(500A_s,prov) = (310 × 460 × 133.688)/(500 × 452) = 84.341 N/mm²
β_s = 310/84.342 = 3.67 > 2.0 (take 2.0)

Taking the distance between supports as the effective span, L = 3625mm
The allowable span/depth ratio = β_s× 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
V_Ed = βnl_x

Ultimate shear force V_Ed = 0.44 × 10.9575 × 3.625 = 17.477 kN/m

V_Rd,c = [C_Rd,c.k. (100ρ₁ f_ck)^1/3

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/119) = 2.296 > 2.0, therefore, k = 2

ρ₁ = A_s/bd = 452/(1000 × 119) = 0.003798 < 0.02; K1 = 0.15

V_Rd,c = [0.12 × 2 × (100 × 0.003798 × 25)^1/3] × 1000 × 119 = 60476.947 N = 60.476 kN

Since V_Rd,c (60.476 kN) > V_Ed (17.477 kN), no shear reinforcement is required.

Note that shear reinforcements are rarely required in solid slabs supported by beams. Hence no further checks for shear will be done on the slab.

Design of short span

Continuous edge
M_Ed = αnl_x² = 0.056 × 10.9575 × 3.6252 = 8.0633 kNm/m
Using the procedure described above;
k = 0.0228; La/d = 0.95; A_s1 = 178.2236 mm²; A_s,min= 172.5217 mm²
Provide Y12 @ 250mm c/c Top (A_s,prov = 452 mm²)

Design of long span
Mid span
M_Ed = αnl_x² = 0.034 × 10.9575 × 3.6252 = 4.895 kNm/m
k = 0.0138; La/d = 0.95; A_s1 = 108.1945 mm²; A_s,min = 172.5217 mm²;
Provide Y12 @ 250mm c/c N/Bottom (A_s,prov= 452 mm²)

Continuous edge
M_Ed = αnl_x² = 0.045 × 10.9575 × 3.6252 = 6.479 kNm/m
k = 0.0183; La/d = 0.95; A_s1 = 143.2057 mm²; A_s,min = 172.5217 mm²;
Provide Y12 @ 250mm c/c Top (A_s,prov= 452 mm²)

Reinforcement Detailing

detailing of floor slab — **Fig. 8:** Detailing of Panel 1

Cost of Reinforcement Works in Nigeria

By

Ubani Obinna

-

November 21, 2020

Installation of reinforcements is an important item of work in the construction of reinforced concrete structures. The cost of reinforcement works therefore significantly impacts the total contract sum of any construction project. In the preparation of a bill of quantity (BOQ) or bill of engineering measurement and evaluation (BEME), concrete and reinforcement works are listed as separate items of work.

Therefore, a contractor bidding for such a job is expected to provide his rate of buying, cutting, bending, and installing the reinforcements according to the construction drawing in one sum. While concrete is priced per cubic meters (m³), reinforcement is usually priced per kilogram (kg).

The price to be quoted per kg of reinforcement is expected to cover the cost of purchase and supply of the materials to the site, cost of cutting, bending, and installation according to the design specification. It is also expected to cover the contractor’s profit and overhead. The use of bar bending schedule (BBS) can enhance the process of reinforcement handling and evaluation.

Read Also…
How to quote/price for concrete in Nigeria

Therefore, a contractor bidding for a job in reinforced concrete construction is expected to know the basic cost of reinforcement in the market, the cost of transportation, the cost of labour, and the cost of machinery that will be needed to execute the job. Initial market research may be required before the tender is computed and submitted to the client.

In Nigeria, there are different categories of reinforcements that are used for construction works, and each of them is priced differently in the market. As found all over the world, the use of mild steel (f_y = 250 N/mm²) is no longer common, and it is rarely utilised in construction in Nigeria. However, some people still use R6mm (commonly called the ‘quarter rod’) and R8mm as links for columns and beams, but it is rarely specified in construction drawings these days. The categories of reinforcements in the Nigerian markets are;

Local reinforcements (manufactured locally)
TMT reinforcements (manufactured locally)
Foreign reinforcements (imported into the country)

Local reinforcements are manufactured locally by the steel industries in Nigeria and are the cheapest class of reinforcements in the market. They have a reputation of not always meeting the minimum yield strength requirements, and are not usually employed in large-scale or serious projects.

TMT reinforcements are high-quality reinforcements produced from recycled steel through the process of thermomechanical treatment. They usually satisfy the yield strength requirements and are widely used for all kinds of projects in the country, including bridges. They are more expensive than local reinforcements. As at July 2022, while one tonne of reinforcement is sold at about ₦415,000.

Foreign reinforcements are usually imported from Germany, Ukraine, or Russia, and are slightly more expensive than TMT reinforcements. However, the presence of TMT reinforcements is making foreign reinforcements less attractive.

Therefore, careful attention to design details and specifications is needed during the pricing of reinforcements. We are going to show below the process of pricing for reinforcement works in Nigeria. The information that will be needed for the evaluation of the cost of reinforcement works in Nigeria are as follows;

Cost of reinforcement in the market (based on the size and type of reinforcement)
Cost of binding wire
Cost of delivery to site
Cost of labour for cutting, bending, and installation
Cost of machinery (cutting machine and bending machine)

Calculation of the Cost of Reinforcement Works in Nigeria

For example, let us calculate the cost of installing 3650 kg of Y12 reinforcements in the floor slab of a building.

As a rule of thumb, about 12 kg of binding wire is required to tie 1 tonne of reinforcement (this also means that about 0.012 kg of binding wire is required to tie 1 kg of reinforcement). Therefore about 44 kg of binding wire is required for this project.

Cost calculation
Basic cost of reinforcement (including delivery to site) = say ₦430,000 per tonne = ₦430 per kg
Basic cost of binding wire (including delivery to site) = ₦17,000 per 20 kg = ₦ 850 per kg (= ₦10.2 per kg of reinforcement)
Basic cost of labour (cutting, bending, and installation) = ₦25,000 per tonne = ₦25 per kg
Assuming that the use of machinery is not applicable in the project or factored into the cost of labour, total basic cost = ₦430 + ₦10.2 + ₦25 = ₦465.2 per kg

Make 10% allowance for waste and laps = 1.1 x 465.2 = ₦512
Allow 20% for contractor’s profit and overhead = 1.2 x ₦465.2 = ₦ 559 per kg

Therefore the current cost of reinforcement works in Nigeria is ₦559 per kg of reinforcement.

Therefore, the cost of fixing the floor slab reinforcement = ₦559 x 3650 = ₦2,040,350:00 (Two million, forty-six thousand and three hundred and fifty Naira)

For your construction and building design works, contact;

Structville Integrated Services Limited (RC 1491828)
Whatsapp/Call: +2347053638996
E-mail: info@structville.com

Design of Shear Walls

By

Ubani Obinna

-

November 21, 2020

Table of Contents

Shear walls are structural members that provide additional lateral stiffness to a building by resisting shear, moment, and axial forces, which are produced due to gravity and lateral loads. In some design cases, the entire lateral loads coming to a building assumed to be resisted by shear walls alone, especially when it is the major stabilising/bracing component of the building. The design of shear walls involves providing adequate cross-section and reinforcements to resist bending, shear, axial, and twisting forces due to gravity and lateral loads.

For an element to be described as a reinforced concrete wall, the length to thickness ratio should be equal to or greater than 4 (Clause 9.6.1, Eurocode 2). A shear wall is said to be short when the height-to-width ratio is less than or equal to one, while it is said to be slender when the height-to-width ratio is greater than or equal to four. For slender walls, bending deformation is dominant, while in short walls, shear deformation is dominant. When the height-to-width ratio is in-between one and four, then the shear wall undergoes both shear and bending deformations.

shear walls and columns in a building — Shear walls and columns in a building

Forces Acting on Shear Walls

Shear walls are designed to resist bending moment, shear, axial, and uplift forces, especially when they are subjected to lateral actions. The lateral forces acting in the plane of a shear wall attempts to lift up one end of the wall and push the other end down. Uplift forces are greater on tall walls and less on low walls. Shear walls resist the shear force parallel to the plane of the wall by cantilever action.

Axial forces in a Shear wall
The axial load in a wall may be calculated assuming the beams and slabs transmitting the loads to it are simply supported.

Transverse moments
For continuous construction, transverse moments can be calculated using elastic analysis. The eccentricity is not to be less than h/30 or 20 mm where h is the wall thickness.

In-plane moments
Moments in the plane of a single shear wall can be calculated from statics. When several walls resist forces, the proportion allocated to each wall should be in proportion to its stiffness.

Location/Placement of Shear of Shear Walls

In a tall building where the shear wall is used for lateral stability, it should be located on each level of the structure. Preferably, shear walls of equal length should be placed symmetrically on all four exterior walls of the building to form an effective box structure. When the shear walls in the exterior frame could not provide sufficient strength and stiffness, the shear walls should be added to the interior frame as well.

If the shear wall is placed at the interior frame of a building, it attracts and resists higher internal forces but may not be too effective in reducing the maximum lateral deflection of the building. However, if a shear wall is placed at the ends of a building, the lateral deflection is reduced considerably when compared to the later.

Architectural disposition in a building may not always give the room for optimum placement of shear walls for good structural performance. In some areas such as lift areas or window/door openings, coupled shear walls may have to be used. However, the best position or arrangement of shear walls in a building is a symmetrical arrangement whereby the shear centre (centre of rotation) will coincide with the centre of gravity of the building in order to reduce or eliminate torsion (twisting) due to lateral loads.

pierced shear wall — Coupled shear wall in a building

Functions of Shear Walls

Shear walls must provide the necessary lateral strength to resist horizontal wind and earthquake forces.
They provide resistance against sliding through connections.
They also provide lateral stiffness to prevent the roof or floor above from excessive side sway.
When the shear walls are stiff enough, they prevent floor and roof framing members from moving off their supports.
Buildings that are sufficiently stiff usually suffer less structural damage due to the presence of shear walls.

Structural Action of a Shear Wall

In a shear wall, the primary mode of deformation is mainly due to flexure but not shear. The mode of deformation of shear walls is such that it has maximum slope at the top and least at the bottom which is the flexural mode shape. The structural action of a shear wall resembles that of a cantilever beam.

Displaced shape of a shear wall under lateral load — Typical deflected shape of a shear wall

Structural Design of Shear Walls

The amount of reinforcement needed for proper detailing of a reinforced concrete wall may be derived using strut-and-tie model. If a wall is however subjected predominantly to out-of-plane bending, the design rules and guidelines for slabs apply. In Eurocode 2, the requirements for the design of columns and shear walls are not so different except in the following areas;

■ The requirements for fire resistance
■ Bending will be critical about the weak axis
■ Rules for spacing and quantity of reinforcement

Reinforcement Detailing of Shear Walls

(a) Minimum and maximum area of vertical reinforcement
According to clause 9.6.2 of Eurocode 2, the minimum and maximum amounts of reinforcement required for a reinforced concrete wall are 0.002A_c and 0.04A_c outside lap locations respectively. It is further stated that where minimum reinforcement controls design, half of this area should be located on each face. The distance between two adjacent vertical bars should not exceed three times the wall thickness or 400 mm, whichever is lesser.

(b) Area of horizontal reinforcement
According to clause 9.6.3 of Eurocode 2, horizontal reinforcement should be provided at each face and should have a minimum area of 25% of the vertical reinforcement or 0.001A_c, whichever is greater. The spacing between two adjacent horizontal bars should not be greater than 400 mm.

(c) Provision of links
If the compression reinforcement in the wall exceeds 0.02A_c, links must be provided through the wall thickness in accordance with the rules for columns in clause 9.5.3 which are:

The diameter of the transverse reinforcement should not be less than 6 mm or one-quarter of the diameter of the largest longitudinal bar whichever is greater. The maximum spacing is to be S_{cl, max}

S_{cl, max} is the minimum of;

20 times the diameter of the smallest longitudinal bar
The lesser dimension of the wall i.e. the thickness
400 mm

The maximum spacing should be reduced by a factor of 0.6 in the following cases;

In sections within a distance equal to 4 × thickness of wall above or below a beam or slab.
Near lapped joints, if the diameter of the longitudinal bar is greater than 14 mm. A minimum of three bars evenly placed in the lap length is required.

Where the main reinforcement (i.e., vertical bars) is placed nearest to the wall faces, transverse reinforcement should be provided in the form of links with 4 per m²of the wall area.

Design Example of Shear Walls

Design a 225 mm thick shear wall of 3.6 m height at the ground floor of the building. The shear wall is carrying a 200 mm thick slab on the first floor of the building. The action effects on the shear wall are as follows;

Vertical loads
Dead Load G_k = 300 kN/m
Live load Q_k = 55 kN/m

Vertical load due to in-plane bending and wind W_k = ±650 kN/m
Vertical load due to in-plane bending and imperfections G_kH = ±60 kN/m

Maximum moment out-of-plane, floor imposed load as leading action M = 40 kN/m @ ULS
Maximum moment out-of-plane, floor imposed load as accompanying action M = 35 kN/m @ ULS

f_ck = C25/30; f_yk = 500 N/mm²

Check slenderness of wall at ground floor

Effective length, l₀ = 0.75 × (3600 – 200) = 2550 mm (Table C16)
λ = 3.46 × l₀/h = 3.46 × 2550/225 = 39.213 (Cl. 5.8.3.2(1))

Limiting slenderness, λ_lim = 20 ABC/n^0.5 (Cl. 5.8.3.1(1))
where;
A = 0.7
B = 1.1
C = 1.7 – r_m

where;
r_m = M₀₁/M₀₂ = say = –0.25
C = 1.7 – (-0.25) = 1.95
n = N_Ed/A_cf_d

where;
N_Ed (assuming wind is the leading variable action) = 1.35G_k + 1.5Q_k1 + 1.5Ψ₀Q_ki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m
A_cf_d = (225 × 1000) × (0.85 × 30/1.5) = 3825 kN
Therefore, n = 1518.75/3825 = 0.397
λ_lim = 20 ×( 0.7 × 1.1 × 1.95)/0.397^0.5 = 47.66

Therefore as λ < λ_lim wall is not slender and therefore no secondary moments.

Combinations of actions
(a) At ULS, for maximum axial load, W_k is leading variable action
N_Ed = 1.35G_k + 1.5Q_k1 + 1.5Ψ₀Q_ki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m

M_Ed = M + e_iN_Ed ≥ e₀N_Ed (Cl. 5.8.8.2(1), 6.1.4)

where
M = moment from 1st order analysis = 35 kNm/m
e_i = l₀/400 = 2550/400 = 6.375 mm (Cl. 5.2(7), 5.2(9))
e₀ = h/30 ≥ 20 mm = 20 mm (Cl. 6.1.4)

M_Ed= 35 + (0.006375 × 1518.75) ≥ 0.020 × 1518.75
M_Ed = 35 + 9.68 ≥ 30.375 = 44.68 kNm/m

(b) At ULS, for minimum axial load, W_kis leading variable action
N_Ed= (1.0 × 300) – (1.35 × 60) – (1.5 × 650) + (0 × 55) = –756 kN/m (tension)

M_Ed = 35 + (0.006375 × 756) ≥ 0.020 × 756
= 35 + 4.82 ≥ 15.12 = 39.82 kNm/m

(c) At ULS, for maximum out of plane bending assuming Q_kis leading variable action

N_Ed= 1.35(300 + 60) + (1.5 × 55) + (1.5 × 0.5 × 650) = 1056 kN/m

M_Ed = 40 + (0.006375 × 1056) ≥ 0.020 × 1056
M_Ed = 40 + 6.732 ≥ 21.12 = 46.732 kNm/m

or

N_Ed = (1.0 × 300) – (1.35 × 60) – (0 × 55) – (1.5 × 0.5 × 650) = -268.5 kN/m (tension)
M_Ed = 40 + (0.006375 × 268.5) = 41.71 kNm/m

Design load cases
Consolidate (c) into (a) and (b) to consider two load cases:

N_Ed = 1518.75 kN/m
M_Ed= 46.732 kNm/m (out of plane)

and

N_Ed = –756 kN/m (tension)
M_Ed = 46.732 kNm/m (out of plane)

Design: cover above ground
c_nom = c_min + ∆c_dev

where;
c_min = max[c_min,b; c_min,dur] (Exp. (4.1))

where
c_min,b = diameter of bar = 20 mm vertical or 10 mm lacers
c_min,dur= for XC1 = 15 mm
∆c_dev = 10 mm
c_nom= 15 + 10 = 25 mm to lacers (35 mm to vertical bars)

Design using charts
For compressive load:
d₂/h = (25 + 10 + 16/2)/225 = 0.19

Interpolate between charts for d₂/h = 0.15 and d₂/h = 0.2 as shown below

Column interaction chart — Rectangular column interaction chart for d₂/h = 0.15

column intreaction chart 2 — *Rectangular column interaction chart for d*₂*/h = 0.2*

N_Ed/bhf_ck = (1518.75 × 10³)/(225 × 1000 × 30) = 0.225
M_Ed/bh²f_ck = (46.732 × 10⁶)/(225² × 1000 × 30) = 0.030

Gives:A_sf_yk/bhf_ck = 0
Therefore, minimum area of reinforcement required

A_s,min = 0.002A_c (Cl. 9.6.2 & NA)
A_s,min = 0.002 × 225 × 1000 = 450 mm²/m
A_s,min = 450 mm²/m = 225 mm²/m each face
maximum spacing = 400 mm c/c, minimum diameter of rebar = 12 mm diameter

Try H12 @ 300 c/c on each face

For tensile load and moment
Working from first principles, referring to the figure below and ignoring contribution from concrete in tension,

stresses and strain in a shear wall subjected to bending and tension — Stresses and strains in wall subject to tension and out of plane moment

N_Ed= (σ_st1 + σ_st2) × A_s/2
and M_Ed = (σ_st1 – σ_st2) × A_s/2 × (d – d₂)

So σ_st1 + σ_st2 = 2N_Ed/A_s
and σ_st1 – σ_st2 = 2M_Ed/[(d – d₂)A_s]

2σ_st1 = 2N_Ed/A_s + 2M_Ed/[(d – d₂)A_s]
As = (N_Ed/σ_st1) + M_Ed/(d – d₂)σ_st1

σ_st1 = f_yk/γ_s = 500/1.15 = 434.8

A_s = (756 × 10³/434.8) + (46.732 × 10⁶)/[(182 – 43) × 434.8] = 1738.7 + 773 = 2512 mm²

σ_st2 = 2N_Ed/A_s – σ_st1 = 601.9 – 434.8 = 167 MPa

By inspection all concrete is in tension zone and may be ignored.
Use 7H16 @ 175 c/c on both sides for at least 1 m each end of wall (A_sprov = 2814 mm²).

Horizontal reinforcement
A_s,hmin = 0.001A_c or 25% A_s,vert (Cl. 9.6.3(1) & NA)
= 225 mm² or (0.25 × 2814) = 704 mm²/m
This therefore requires 352 mm²/m each side
Use H10 @ 200 (393 mm²/m) both sides

Links
Check 0.02A_c = 0.02 × 225 × 1000 = 4500 mm²> 2814 mm²
Therefore, links not required.
.

Demolition of a Building for Structural Reasons

Tasks/Activities in a Construction Work

Duration of Tasks

Advantages of Programme of Works

Special Considerations for Preparation of Programme of Works

Sample Programme of Works for a Residential Building

Geotechnical Design of Pad Foundation

Example on the calculation of the bearing capacity of a pad foundation using Design Approach 1 (DA1)

Example on the Structural Design of Pad Foundation

Structural Design Example of Pad Foundation

Detailing sketches of pad foundation

Design Considerations of Gabion Walls

Materials for construction of gabion retaining walls

Drainage Considerations for Gabion Walls

Design Example of Gabion Walls

Types of Reinforced Concrete Slabs

One-way Spanning Solid Slabs

Steps in the design of a one-way slab

Two-way Spanning Slabs

Design Example of a two-way slab

Calculation of the Cost of Reinforcement Works in Nigeria

Forces Acting on Shear Walls

Location/Placement of Shear of Shear Walls

Functions of Shear Walls

Structural Action of a Shear Wall

Structural Design of Shear Walls

Reinforcement Detailing of Shear Walls

Design Example of Shear Walls

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