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Problems of Civil Engineering Consultancy in Nigeria

The wealth creation cycle in construction industry revolves around the political climate, economy, government policies, availability of resources, technology, and skilled manpower. Thousands of engineering graduates leave school in Nigeria each year, and the job opportunities are getting leaner by the day. If I may say, this trend is worse in structural engineering consultancy sector in Nigeria.


In Nigeria currently, construction and real estate companies are recruiting more civil engineering graduates than any other sector. The consultancy sector has been stagnant in terms of job creation recently.  There are a lot of explanations to this trend, and the idea is to keep fresh graduates abreast of the current position of the industry.

Current challenges of the consultancy sector;

(1) Inadequate Regulation
In the classroom, there appears to be a scale of fees for professionals in a building project, but out there in the field, there seems to be no professional standard. For instance in the structural design of buildings, it is not very clear whether we charge on the basis of number of floors, or area of  the building, or a percentage of the estimated total cost of the building. It appears every organisation has its own way of charging for designs in order to balance their book.

Also fresh graduates and students who have acquired design skills do not hesitate in collecting design jobs directly from architects and clients. With all these challenges of no basic control, the consultancy sector has become a haven of negotiation and bargaining power, instead of professional standard. However, it is important to point out that a few Grade ‘A’ consultancy firms still maintain their standards, and those who need them knows where to find them.

(2) Availability of Design Softwares
The availability of commercial softwares have improved speed, accuracy, and output of many consultancy firms. The negative aspect of it is that the demand for man power has significantly reduced. If a consultancy firm lands one design job per month, then they can comfortably make do with just one design engineer depending on the complexity of the jobs. With these softwares guaranteeing speed in output, engineers spend less time in generating working drawings and documents, thereby making organisations realise that they need fewer hands.


(3) In-house Designs
In order to save cost, most construction and real estate companies have resolved to carrying out their designs in-house by employing the services of a structural engineer. As a result of this, fewer jobs get to real consultancy firms, and the sector continues to suffer.

(4) Crowded Industry
The built environment sector in Nigeria is currently crowded with a lot of professionals offering the same services. In our society today, civil engineers are building, builders are building, architects are building, masons are building, and even quantity surveyors are building. In this case, anyone can easily obtain their working drawings from any source at the cheapest rate, work hard on obtaining COREN stamp, and all the way the building goes. If this issue is notaddressed, the consultancy sector will continue to suffer. This has led to many certified professionals into mediocrity in the bid to survive.

civil

(5) Lack of research and creativity
The conventional system of building in Nigeria has remained largely the same over a long period of time. The advances that have been made in materials science and technology rarely reflects in the construction projects executed in Nigeria. Innovation and creativity is what drives any industry, and if there are no new trends, we will remain stuck where we are. Engineers are encouraged to continue exploits in the areas of adaptation, research, sustainability, cost reduction, and improved technology. This is one of the ways consultancy can be revitalised.

(6) Insignificant Public Private Partnership (PPP)
In Nigeria, the construction industry is still largely financed by public funds, apart from the real estate sector that can be said to be shared equally. So for the industry to thrive, there is need for more public Private Partnership (PPP) initiatives to drive the industry forward. It is obvious the government cannot no longer do it alone due to huge capital requirements. This will have significant impact in the consultancy sector.

Eko Atlantic Nigeria1

Finally, I wish to encourage engineers to keep up the acts of hard work and professionalism in all their endeavours. Positive entrepreneurship geared towards service to humanity, research, development, sustainability, and enterprise is highly encouraged. Footprints on the sands of time are not made by sitting down. Nigeria is ours… God bless.


On the Bearing Capacity of Shallow Foundations

Bearing capacity is the maximum load a soil profile can withstand before undergoing excessive deformation and shear failure. It is the most popular and perhaps the most important information needed for the design of shallow foundations. The allowable bearing capacity of soil is used for the proper sizing of shallow foundations so that the load from the superstructure will not exceed the strength of the soil or lead to an excessive settlement.

If a load is applied gradually to a foundation, the settlement will increase. At a certain point when the load equals the bearing capacity of the soil, sudden failure of the soil supporting the foundation will take place. This sudden failure in which the failure surface will extend into the ground is known as ‘general shear failure‘.

When the soil supporting the foundation is of sand or clay soil of medium compaction, the failure surface of the soil will gradually extend outward from the foundation. When the applied load reaches the bearing capacity of the soil, the foundation movement will be accompanied by sudden jerks, and considerable movement will be required before the failure surface extends into the ground. This is generally referred to as ‘local shear failure‘, and the peak value of the load is not realised in this type of failure.

If the foundation is founded on loose soil, the failure surface will not extend into the ground surface. The load-settlement plot of this interaction will be steep and practically linear. Such failure is referred to as ‘punching shear failure‘.

types of foundation failure
Different types of bearing capacity failure

The methods of determining the ultimate bearing capacity of soils are;

  • General shear failure theory of Terzaghi
  • Theoretical solutions presented by Meyerhof, Hansen, and Vesic
  • Correlations from in-situ tests such as PLT, SPT, and CPT

Information needed for the evaluation of the bearing capacity of soil is obtained from site investigation. Laboratory study of undisturbed samples or in-situ soil tests can be done in order to obtain the shear strength parameters needed for the evaluation of the bearing capacity. Several correlations exist for relating in-situ soil properties from cone penetration test to bearing capacity of the soil.

Soil investigation is therefore one of the most important activities that must be carried out before the commencement of any construction project. In the soil test report, the geotechnical engineer is expected to state the strength of the soil at different layers, and ultimately recommend a suitable foundation. Some of the parameters used in describing the strength of a soil formation for the purposes of estimating the soil bearing capacity are the cohesion and the angle of internal friction of the soil.

In this article, we are going to present an example of how to determine the bearing capacity of soil using the general bearing capacity equation.

Background to the bearing capacity of shallow foundations

Terzaghi in 1943 extended the plastic failure theory of Prandtl to evaluate the bearing capacity for shallow strip footings. After the development of Terzaghi’s bearing capacity equation, several scholars such as Meyerhof (1951 and 1963), Vesic (1973), Hansen (1970), etc worked on this area and refined the solution to what is known as the general bearing capacity equation. This modification allowed for depth factor, shape factor, and inclination factors.

The modified general ultimate bearing capacity equation can be written as;

qu = c’FcsFcdFciNc + qFqsFqdFqiNq + 0.5FγsFγdFγiγBNγ

Where;
Fcs, Fqs, Fγs are shape factors which account for the shearing resistance developed along the surface in soil above the base of the footing
Fcd, Fqd, Fγd are depth factors to determine the bearing capacity of rectangular and circular footings
Fci, Fqi, Fγi are inclination factors to determine the bearing capacity of a footing on which the direction of load application is inclined at a certain angle to the vertical

Solved Example on the determination of bearing capacity

Let us determine the bearing capacity of a simple pad foundation with the following data;

Bearing capacity of shallow foundations

Depth of foundation Df = 0.9 m
Width of foundation B = 1.0 m
Effective cohesion of soil c’ = 12 kN/m2
Angle of internal friction φ’ = 27°
Unit weight of soil = 18.5 kN/m3

The water table is about 9 m below the surface

From table, we can determine the bearing capacity factors;

rtty
Bearing capacity factors culled from Das and Sobhan (2012)

Angle of internal friction φ’ = 27°
Nc = 23.94; Nq = 13.20; Nγ = 14.47

Fcs = 1 + (B/L)(Nq /Nc) = 1 + (1.0/1.0)(13.2/23.94) = 1.551
Fqs = 1 + (B/L)tanφ’  = 1 + (1.0/1.0)tan 27 = 1.509
Fγs = 1 + 0.4(B/L)  = 1 + 0.4(1.0/1.0) = 1.4

Fcd = 1 + 0.4(Df/B)  = 1 + 0.4(0.9/1.0) = 1.36
Fqd = 1 + 2tanφ'(1 – sin φ’)2(Df/B)  = 1 + 2tan27(1 – sin 27)2 (0.9/1.0) = 1.273
Fγd = 1.0

Since we are assuming vertical loads, take Fci = Fqi = Fγi = 1.0

q = (18.5 kN/m3 × 0.9 m) = 16.65 kN/m2

qu = c’FcsFcdFciNc + qFqsFqdFqiNq + 0.5FγsFγdFγiγBNγ
qu = (12 × 23.94 × 1.551 × 1.36 × 1.0) + (16.65 × 1.509 × 1.273 × 1.0 × 13.20) + (0.5 × 1.4 × 1.0 × 1.0 × 18.5 × 1.0 × 14.47) = 1215.55 kN/m2

Using a factor of safety (FOS) of 3.0
qallowable = qu /FOS = 1215.55/3.0 = 405.183 kN/m2

So with this, the allowable bearing capacity of the soil can be stated as 405 kN/m2

Question of the Day

Structville daily questions
From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.


Today’s Question
For the compound frame loaded as shown above, find the bending moment just below point B under the action of the externally applied load. Assume linear elastic response.

Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.

E-mail: info@structville.com
WhatsApp: +2347053638996

You can also visit Structville Research for downloads of civil engineering materials.

STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL
We have this very affordable design manual available…

final%2Bfront%2Bcover

Do you want to preview the book, click PREVIEW
To download full textbook, click HERE

Solution to Questions of the Week (4th week, May, 2018) and Winners

SOLUTIONS

Earlier this week, I started the daily questions program on Structville and I am happy for the kind of the reception the program has received. We are going to summarise the all questions asked last week, provide solution to the questions, and recognise those who participated.


Wednesday 23 May, 2018
We were asked to determine the equation that the two beams shown below have in common in their final state. The question has the conditions of assuming linear elastic response.

Structville%2BQuestion%2BChallenge

Solution
We can see that the difference in the beams is based on their support conditions. The first beam is simply supported, while the second beam is fully fixed at the supports. Under the action of the externally applied uniformly distributed load, the simply supported beam develops a maximum sagging moment of wL2/8 which occurs at the mid-span. However, the shear force at the support remains wL/2.

In the free state, the second beam develops a maximum sagging moment of wL2/8 which occurs at the mid-span, and a hogging fixed end moment of wL2/12 which occurs at the support. Now in the final state (combining free moment and fixed end moment), the maximum sagging moment becomes  wL2/8 wL2/12 = wL2/24 (at the midspan). But the shear force remains wL/2. So the correct answer is D.

Those who got the answer correct are:

  • Ogungbire Adedolapo
  • Palo Tuleja
  • Nitesh Mithapelli

Thursday, 24 May, 2018
We were required to determine the support reaction and bending moment at support C of the compound beam loaded as shown below;

STRUCTVILLE%2BCHALLENGE%2B2

Bending Moment at Support C
The bending moment at support C can be expressly determined by considering the cantilever moment from the concentrated load at the free end of the overhang.

MCR = -2 kN × 2m = -4 kN.m


Support Reaction at Support C
There are two easy ways of obtaining the vertical reaction at support C.
Since the values of the reaction at A and B are already given, we can easily sum up the vertical forces. (upward forces positive, downward forces negative)

Hence
 -2 – (6 × 4) – 2 + 1 + 34.6 – Cy = 0
Cy – 7.6 = 0
Cy = 7.6 kN (downwards)

On the other hand, we can take moment about point G just to the left. But Structville question is designed so that very simple approaches can be used to obtain the answers. So the correct answer is C.

The people that got the answers correct are as follows:

  • Ogungbire Adedolapo
  • Ovie Agbaga
  • Theodoros Gianneas
  • Subramanian Narayanan
  • Peter O.

Friday 25 May, 2018
In this one, we are required to obtain the bending moment just to the left of point E and the horizontal support reaction at support B.

QUESTION%2BOF%2BTHE%2BDAY%2BFRAME

To obtain the moment just to the left of node E, we have to carry out a very simple calculation. The support reaction at support A has already been provided.

Hence;
MGL = (2 × 4) – (2 × 42)/2 = – 8kNm
This is just as simple as that.


The horizontal support reaction can be readily obtained by summing up the horizontal forces. All forces pointing towards the right are taken as positive, while forces pointing towards the left is taken as negative.

Therefore;
4 kN – 7 kN + 4.875 kN – Bx = 0
1.875kN – Bx = 0
Bx = 1.875 kN

Therefore the correct answer is C.

The people that got the answer correct are;

  • Thaddeus Odinakachi Ekwugha
  • Ogungbire Adedolapo
Winner(s)
From the rules of the exercise, the winner for this week is Ogungbire Adedolapo (he got all three correct !!!). Mr. Adedolapo will receive special academic materials from Structville, and also, we will give him a one month data subscription for any network of his choice. Well done Mr. Adedolapo, and to all others who participated in the exercise. From my own point of view, this is neither a competition nor an exam, but just a way of teaching, learning, and discussing civil engineering on the internet.

My sincere appreciation also goes to all the people who commented on various social platforms. However for your response to be recognised specially, you must post it on this blog. Some people also commented on the blog anonymously. Thank you so much for your contributions. Let us look forward to this weeks questions. God bless.


Structville Question of the Day

QUESTION%2BOF%2BTHE%2BDAY%2BFRAME

Structville daily questions
From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.


Today’s Question
For the compound frame loaded as shown above, find the bending moment just to the left of point E and the horizontal support reaction at support B assuming linear elastic response.

Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.

E-mail: info@structville.com
WhatsApp: +2347053638996

You can also visit Structville Research for downloads of civil engineering materials.

STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL
We have this very affordable design manual available…

final%2Bfront%2Bcover

Do you want to preview the book, click PREVIEW
To download full textbook, click HERE

Design of Large Span Cantilever Structures

Large span cantilevers are delightful to everyone, but very challenging to engineers during design. This is because the engineer is battling with a lot of design factors, and he must guaranty the stability and good performance of the structure.

The most prominent issues in the design of reinforced concrete cantilever structures are;
(1) Excessive deflection and
(2) Balance of moment for stability

In this post, I am going to comment on some common design solutions to large span cantilevers. The simplest solutions are discussed below;

long%2Bspan%2Bcantilever

(1) Controlling deflection
For relatively shorter spans (say less than 1.5m), increasing the depth of the section or increasing the quantity of steel reinforcement looks like an express solution without very serious consequences. However as the span of the cantilever increases, increasing the depth will increase the design load and add to the design challenges. You may have to increase the compressive strength of the concrete (employing high strength concrete), and heavily reinforce the tension and compression zones of the section to assist in resisting deflection. On the other hand, you can try prestressing, and if your architect permits, the use of trusses to support the cantilever will provide a very fast and simple solution.


Furthermore, forming a web kind of design (like introducing grids) will improve the load distribution on the members, offer higher resistance, and ultimately bring about shallower sections. This is one of the most efficient ways of dealing with large span cantilevers. For improved aesthetics, the web can be adequately covered with finishes to make the whole arrangement look like one single section. See the example below where the use of multiple members have been used to approach a relatively large span cantilever problem.

IMG 20180508 092446

(2) Balance of moment for stability
For cantilever structures to stand, the moment generated at the fixed end must be balanced, otherwise equilibrium problem will ensue. If the cantilever structure has no backspan (discontinuous), then the foundation must be used to provide the balance needed, otherwise the structure will topple (EQU problem).  This also means that the column of the cantilever must be designed to resist heavy moment (STR problem). The exercise below has been presented to give you a little idea of what is being discussed. You can attempt the problem and place your solution in the comment section of this post.

Exercise
For the structure loaded as shown below, proportion the dimensions of base to counteract the moment from externally applied load. (Take width of base = 1.2 m, Unit weight of concrete = 24 kN/m3). Make all other necessary assumptions in your design. 
strucr

On the other hand, if the cantilever has a backspan (continuous), then the bending moment is distributed to all members meeting at that node, and column moment is alleviated to some extent. So we really do not need the base to help provide equilibrium, but the backspan of the structure and interaction with other members provide the balance needed. So we have mainly STR problem. The picture below shows a cantilever with a backspan. However, in some cases, the base will be required to assist in maintaining equilibrium.

back

There is a structural design challenge that is currently ongoing, and some Nigerian civil engineering students are carrying out a design on a structure that has many cantilevers that are spanning as long as 3 meters. We are all looking forward to the solutions that they will come up with. So let us anticipate.

To check out the nature of the competition, click HERE.

56d

Check out this beautiful cantilever structure below. Have you designed long span cantilever structures before? Kindly let us know how you approached it.

FGTTT


Structville Question of the Day (Thursday, 24/05/2018)

STRUCTVILLE%2BCHALLENGE%2B2

Structville daily questions
From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.


Today’s Question
For the compound beam loaded as shown above, find the bending moment and support reaction at support C assuming linear elastic response.

Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.

E-mail: info@structville.com
WhatsApp: +2347053638996

You can also visit Structville Research for downloads of civil engineering materials.

STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL
We have this very affordable design manual available…

final%2Bfront%2Bcover

Do you want to preview the book, click PREVIEW
To download full textbook, click HERE

Structville Question of the Day

Structville%2BQuestion%2BChallenge

Structville daily questions
From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.


Today’s Question
For the two beams loaded as shown above, which formular is common to each of them when evaluating the internal forces in the final loaded state? (assume linear elastic response)

Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.

E-mail: info@structville.com
WhatsApp: +2347053638996

You can also visit Structville Research for downloads of civil engineering materials.

STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL
We have this very affordable design manual available…

final%2Bfront%2Bcover

Do you want to preview the book, click PREVIEW
To download full textbook, click HERE

Design and Calculation of Formwork Requirement of Slabs

Formworks are structures used to support and hold fresh concrete in place in order to obtain the desired shape prior to setting, curing, and hardening. Formwork can be temporary (struck from concrete after curing) or permanent. Normally, formwork is designed to support load from fresh concrete (including hydrostatic pressure), their self-weight, live load from working personnel, and load from other equipment.

A typical formwork configuration for floor slabs is shown in the figure below;

Configuration
Formwork for slab

Before construction commences, it is very important that the formwork be well-designed. The design effort required will be determined by the form’s size, complexity, height above the ground, and materials (considering reuses). In all cases, the strength and serviceability of the formwork should be considered in the design. Furthermore, the entire system’s stability and member buckling should be checked.

Formwork for reinforced concrete structures should be designed to safely handle all vertical and lateral loads that may be imposed until the concrete structure can take such loads. The weight of reinforcing steel, fresh concrete, the weight of the forms themselves, and numerous live loads exerted during the construction are all loads on the forms.

Furthermore, wind load can cause lateral forces that must be resisted by the formwork to avoid lateral failure. The formwork system or in-place structure with suitable strength for that purpose should carry vertical and lateral loads to the ground.

Loads and Pressures on Formworks

Unsymmetrical concrete pouring/placement, impact from machine-delivered concrete, uplift, and concentrated loads caused by keeping supplies on the freshly constructed slab must all be considered while designing the forms. There will rarely be precise knowledge about the loads that will be applied to the forms, therefore the designer must make certain safe assumptions that will hold true in most situations. The sections that follow are intended to assist the designer in establishing the loading on which to base form design for typical structural concrete situations.

Lateral Pressure of fresh Concrete on Formwork

The gravity load on a horizontal slab or beam formwork differs from the loads imposed by fresh concrete against wall or column formwork. The freshly placed concrete behaves like a fluid for a short time until the concrete begins to set, causing hydrostatic pressure to act laterally on the vertical forms.

Concrete pressure on formwork is primarily determined by several or all of the following factors:

  1. Rate of placing concrete in forms (R)
  2. The temperature of concrete (T)
  3. Weight or density of concrete (ρ)
  4. Cement type or blend used in the concrete.
  5. Method of consolidating the concrete.
  6. Method of placement of the concrete.
  7. Depth of placement.
  8. Height of form.

Lateral Pressure equations

The American Concrete Institute has spent a lot of time and effort researching and studying form design and construction techniques. The complete hydrostatic lateral pressure, as defined by the following equations, is the maximum pressure on formwork, according to ACI Committee (347).

p = ρgh (kPa) ——— (1)

The set characteristics of a concrete mixture should be understood, and the level of fluid concrete can be calculated using the rate of placement. When more than one placement of concrete is to be made, (h) should be taken as the complete height of the form or the distance between horizontal construction joints for columns or other forms that can be filled quickly before stiffening of the concrete occurs.

When working with mixtures using newly introduced admixtures that increase set time or increase slump characteristics, such as self-consolidating concrete, Eq. (1) should be used until the effect on formwork pressure is understood by measurement.

For concrete having a slump of 175 mm or less and placed with normal internal vibration to a depth of 1.2 m or less, formwork can be designed for a lateral pressure as follows:

For columns:
For determining pressure of concrete on formwork ACI 347 defines a column as a vertical structural member with no plan dimensions greater than 2m. For concrete with a slump (175 mm):

pmax = CwCc [7.2 + (785R/T + 17.8)] ——— (2)

With a minimum of 30 Cw kPa, but in no case greater than ρgh.

For walls: For determining pressure of concrete on formwork ACI 347 defines a wall as a vertical structural member with at least one plan dimension greater than 2m. Two equations are provided for wall form pressure:

(a) With a rate of placement of less than 2.1 m/hr and a placement height not exceeding 4.2 m :
pmax = CwCc [7.2 + (785R/T + 17.8)] ——— (3)

With a minimum of 30Cw kPa, but in no case greater than ρgh.

(b) with a placement rate less than 2.1 m/hr where placement height exceeds 4.2 m, and for all walls with a placement rate of 2.1 to 4.5 m/hr:
pmax = CwCc [7.2 + 1156/(T + 17.8) + 244/(T + 17.7)] ——— (4)
With a minimum of 30Cw kPa, but in no case greater than ρgh.

Where;
Cw = Unit weight coefficient which depends on the unit weight of the concrete
Cc = Chemistry coefficient which depends on the type of cementitious materials

Alternatively, a method based on appropriate experimental data can be used to determine the lateral pressure used for form design.

The unit weight coefficient Cw can be calculated using the Table below;

Density of concreteUnit weight coefficient (Cw )
Less than 2240 kg/m3 Cw = 0.5[1 + ρc/2320] but not less than 0.8
2240 – 2400 kg/m3 Cw = 1.0
More than 2400 kg/m3 Cw = ρc /2320

The value of the chemistry coefficient Cc can be picked from the Table below;

Type of cementChemistry Coefficient Cc
Types I, II, and III without retarders1.0
Types I, II, and III with retarders 1.2
Other types or blends containing less than 70% slag or 40% fly ash without retarders 1.2
Other types or blends containing less than 70% slag or 40% fly ash with retarders 1.4
Other types or blends containing more than 70% slag or 40% fly ash 1.4

In the UK, the formula for calculating the pressure on formworks according to CIRIA Report 108 is given by the equation below, which must not be greater than the hydrostatic pressure.

Pmax = [C1√R + C2K √(H1 – C1√R)]γ

Where:
Pmax = Maximum lateral pressure against formwork (kPa)
R = Rate of placement (m/h)
C1 = Coefficient for the size and shape of the formwork (1 for walls).
C2 = Coefficient for the constituent materials of the concrete (0.3 – 0.6).
γ = Specific weight of concrete (kN/m3).
H1 = Vertical form height (m).
K = Temperature coefficient K = (36/T + 16)2

Vertical loads on Formwork

In addition to lateral pressure, vertical loads are also imposed on formwork. Vertical loads consist of dead and live loads. The weight of formwork, the weight of the reinforcement and freshly placed concrete is dead load. The live load includes the weight of the workers, equipment, material storage, runways, and impact.

Vertical loads assumed for shoring and reshoring design for multistory construction should include all loads transmitted from the floors above as dictated by the proposed construction schedule.

The majority of all formwork involves concrete weighting 22 – 24 kN/m3. Minor variations in this weight are not significant, and for the majority of cases, 24 kN/m3 including the weight of reinforcing steel is commonly assumed for design. Formwork weights vary from as little as 0.15 to 0.7 kN/m2. When the formwork weight is small in relation to the weight of the concrete plus live load, it is frequently, neglected.

ACI committee 347 recommends that both vertical supports and horizontal framing components of formwork should be designed for a minimum live load of 2.4 kN/m2 of horizontal projection to provide for the weight of workmen, runways, screeds and other equipment. When motorized carts are used, the minimum should be 3.6 kN/m2. Regardless of slab thickness, the minimum design load for combined dead and live loads should not be less than 4.8 kN/m2 or 6.0 kN/m2 if motorized carts are used.

Horizontal loads on Formwork

Horizontal loads include the assumed value of load due to wind, dumping of concrete, inclined placement of concrete, cable tensions and equipment. The impact of wind increases with height. Horizontal loads should be not less than 1.5 kN/m of floor edge or 2% of total dead load on the form.

Bracing should be provided to withstand the side sway effects which occur when concrete is placed unsymmetrical on a slab form. Wall form bracing should be designed to meet the minimum wind load requirements of the local building code with adjustments for shorter recurrence intervals.

For wall forms exposed to the elements, the minimum wind design load should not be less than (0.72 kN/m2). Bracing for wall forms should be designed for a horizontal load of at least 1.5 kN/m of wall length, applied at the top. Wall forms of unusual height or exposure should be given special consideration.

Formwork Requirement for Suspended Slabs

In Nigeria, the standard given below works for normally proportioned reinforced concrete slabs.

Sheathing – 20 mm thick marine plywood
Joists – 200mm deep wooden H-beams spaced at 600 mm c/c
Stringers – 200mm deep wooden H-beams spaced at 1000 mm c/c
Shores – Steel props spaced at 1000 mm c/c

For low-cost construction, the following can be used;

Sheathing – 25 mm thick wooden planks
Joists – 2″ x 3″ wood (50 mm x 75 mm) spaced at 400mm c/c
Stringers – 2″ x 4″ wood (50 mm x 100 mm) spaced at 600 mm c/c
Shores – Bamboo wood spaced at 600 mm c/c

IMG 2099

Now having known how to arrange formwork for floor slab, the next phase is to determine how to place your order of materials. We will use a typical example to show how it is done.

The first-floor plan of a building is shown below. We are to determine the formwork requirement of the floor slab. We will be neglecting the floor beams in this calculation.

1st%2Bfloor%2Bplan

On studying the drawing,
Length of longer side of building = 16.120 m
Length of shorter side of building = 12.530 m
Therefore, gross area of building =  201.9836 m2

Area of openings (lift and staircase) = 18.21 m2

Therefore net area of slab = 201.9836 – 18.21 = 183.7736 m2

Marine Plywood Requirement
Area of each marine plywood = 2.4m x 1.2m = 2.88 m2

Number of marine plywood required = 183.7736/2.88 = 63.8
Therefore, supply 64 pieces of (2.4m x 1.2m) marine plywood (no allowance for wastage)

If plank were to be used;
Area of one plank = 3.6m x 0.3m = 1.08 m2
Therefore, supply 183.7736/1.08 = 171 pcs of (1″ x 12″ x 12′ plank)

Floor Joist Requirement
Length of wooden H-beam = We normally have variety of 3.9 m or 2.9 m
Considering the longer side of the building;

Number of H-beam required per line = 16.12/3.9 = 4.13 No
We can therefore say, provide 4 No of 3.9m H-beam and 1 No of 2.9m H-beam per line (there will be projections though, and appropriate considerations should be made on site)

Spacing = 600mm

Therefore number of lines required = 12.53/0.6 + 1 = 22 lines

Hence provide;
3.9m H-beam = 4 x 22 = 88 pieces
2.9m H-beam = 1 x 22 = 22 pieces

If it were to be that 2″ x 3″ wood will be used;
Supply length of 2″ x 3″ wood in Nigeria = 3.6m

Number of 2″ x 3″ wood required per line = 16.12/3.6 = 4.47 pcs
At a spacing of 400 mm, we have 12.53/0.4 + 1 = 33 lines

Therefore number of 2″ x 3″ wood required for floor joists = 4.47 x 33 = 148 pieces

Stringer Requirement
We will also be using wooden H-beams for stringers. This will run parallel to the shorter side of the building;

Number of H-beam required per line = 12.53/3.9 = 3.21
We can therefore say, provide 3 No of 3.9m H-beam and 1 No of 2.9m H-beam per line

Spacing = 1000mm

Therefore number of lines required = 16.12/1.0 + 1 = 17 lines

Summarily provide;
3.9m H-beam = 3 x 17 = 51 pieces
2.9m H-beam = 1 x 17 = 17 pieces

Shoring Requirement
Steel acrow props will be used for the shoring;
Spacing = 1m c/c

IMG 2097

Number required on the longer side = 16.12/1 + 1 = 17 pcs
Number required on the shorter side = 12.53/1 + 1 = 14 pcs

Acrow props required = 17 x 14 = 238 pieces

Number of props required in the lift and staircase area
Length of lift area and staircase area = 5.055m (6 props)
Width of lift and staircase area = 3.60m (5 props)
Number of props that would have been in lift area = 5 x 6 = 30 pieces

Therefore, the total number of acrow props required = 238 – 30 = 208 pieces

Formwork summary for floor slab;

  • Marine ply wood = 64 pieces
  • 3.9m H-beam = 88 + 51 = 139 pieces
  • 2.9m H-beam = 17 + 22 = 39 pieces
  • Acrow Props = 208 pieces

Note that all these materials can be hired because they are very reusable. This is one of the advantages. For instance, marine plywood can be used 8 times before it gets damaged, and wooden H-beams (joists) are very durable for a long period of time provided they are well handled and protected from long exposure to moisture.

I hope you found this piece of information helpful. Thank you for visiting Structville today and God bless you. Kindly contact info@structville.com and find out how we can be of help to you.


Structural Design of Slabless (Sawtooth) Staircase

Slabless staircases (also called sawtooth staircases) offer aesthetically pleasing alternatives in buildings, and are often a source of wonder to those who do not understand the design and detailing principles underlying their construction.  The processes of determining the design moments in sawtooth staircases have been presented in this post.



According to Reynolds and Steedman (2005), Cusens (1966) showed that if axial shortening is neglected, and the strain energy due to bending only considered, the mid span moment in a slabless staircase is given by the general expression;ms

where;
k0 is the ratio of the stiffness of the tread to the stiffness of the tread
j = number of treads

If j is odd;j%2Bis%2Bodd

If j is even;j%2Bis%2Beven

Design Example
Let us obtain the design forces in a slabless staircase with the following properties;
Thickness of tread and riser = 100mm
Height of riser = 175mm
Width of tread = 300mm
Number of treads = 7
Width of staircase = 1500mm


Load Analysis
Self weight of staircase;
{[(0.3 × 0.1) + (0.175 × 0.1)] / 0.3} × 24 = 3.8 kN/m2
Self weight of finishes = 1.2 kN/m2

At ultimate limit state;
n = 1.35gk + 1.5qk
n = 1.35(5) + 1.5(3) = 11.25 kN/m2

k0 = 175/300 = 0.583chart

Reading from chart, the support moment coefficient can be taken as -0.086
The support bending moment is therefore;
Ms = 0.086 × 11.25 × 2.12 = 4.267 kN.m

j = 7 (odd)
The free bending moment is therefore
Mf = 1/8 × 11.25 × 2.12 × [(72 + 1)/72] = 6.325 kNm

Therefore, the span moment = Mf – Ms = 6.325 – 4.267 = 2.058 kNm

Comparing this answer with finite element analysis from Staad Pro;

MODELON%2BSTAAD%2BPRO

The 3D rendering of the staircase is as shown below;

3D%2BRENDERING

The linear elastic analysis of the structure gave the following results;

Longitudinal Bending Moment

From the result above;
Maximum support moment = 4.11 kNm/m
Mid span moment = 2.84 kN/m
(Kindly compare this answer with the result from manual analysis)

Transverse Bending Moment

My
Twisting Moment
Detailing
The reinforcement detailing for slabless staircase is provided in the form of links. See a sample detailing image below;
FB IMG 1526708726418
That is how far we will go with this post. For more information, kindly contact info@structville.com
Thank you.