Propriety of Bending Schedules for Construction Purposes

By

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December 18, 2018

Structural engineers usually issue bar bending schedule (BBS) to clients/contractors after completing the design and detailing of a reinforced concrete structure. Bar bending schedules usually show the length, shape, and quantity of reinforcements needed for a particular phase/element of a project.

The bar bending schedule issued by structural engineers after a design is usually very exact. In other words, engineers provide the exact quantity of steel needed to produce what they have shown in their drawing, without additional considerations. However, the primary challenge faced by contractors on site is the issue of offcuts, since rebars are usually supplied in 12 m length, and the cut length of reinforcements can vary depending on the element. Offcuts are not usually accounted for in the schedule issued by a structural engineer, and practically, not all offcuts are reusable.

In this case, there are usually two major challenges for the contractor;

(1) He can buy more reinforcement than given in the bending schedule to take care of the losses from offcuts. This usually requires additional calculations.

(2) If the contractor purchases reinforcements based on the bar bending schedule only, he risks requesting for supply twice, which will probably double his transportation cost.

Case Study
To make this point clearer, let us consider this practical scenario below. The ground floor of a 3-storey commercial development is 4.0 m high, and the details of the column reinforcement are as shown below;

CT1 – Column Type 1 (230 x 230)mm – 2 Nos
Main bars: 4Y16
Links: Y10-200 c/c

CT2 – Column Type 2 (450 x 230)mm – 4 Nos
Main bars: 6Y16
Links: Y10-200 c/c

CT3 – Column Type 3 (600 x 230)mm – 2 Nos
Main bars: 8Y20
Links: Y10-250 c/c

CT4 – Column Type 4 (450 x 450)mm – 1 No
Main bars: 12Y20
Links: Y10-300 c/c

At a lap length of 45ϕ (45 x diameter of bar), the typical bending schedule prepared by the structural engineer for the main bars of the ground floor to first floor columns is as follows:

From the above bending schedule, the quantity of steel to be purchased (per 12 m length) for the main bars are as follows;

Y16: [242.28/(1.579 x 12)] = 12.78 (say 13 lengths of Y16mm)
Y20: [343.51/(2.466 x 12)] = 11.61 (say 12 lengths of Y20mm)

As you can see, these quantities are easily verifiable from the bending schedule and could have possibly been used in the preparation of the bill. It is the exact quantity required as given in the drawing.

However, let us go to site and see how this will possibly play out:

The total quantity of bar mark 01 that will need to be cut by the iron bender/fitter is 32 pieces (kindly verify), and the cut length of each bar is 4795 mm. What this means is that he can only obtain 2 pieces of bar mark 01 from each 12m length of reinforcement, and an offcut of 2410 mm. This offcut cannot be used anywhere in the column, except perhaps it gets useful in the beams or other places. Since 32 pieces of bar mark 01 are required, he will need 16 lengths of Y16 mm against the 13 lengths calculated in the bending schedule. Of course, there will be 16 pieces of 2410 mm as offcut!

In the same way, the total quantity of bar mark 02 that will need to be prepared by the iron bender is 28 pieces, and the cut length of each bar is 4975 mm. This also means that he can only obtain 2 pieces of bar mark 02 from each 12m length of reinforcement, with an offcut of 2050 mm. In effect, he will need 14 lengths of Y20 mm against the 12 lengths calculated in the bending schedule.

If the contractor did not make this consideration before requesting for supply, he will likely make his request twice, to the detriment of extra cost. Therefore, the effects of offcuts should be verified by the contractor before he can place his order for reinforcement. This is usually very common, especially for someone who has not experienced it before.

Thank you for visiting Structville today, and God bless you.

Do you need help in knowing the exact quantity of reinforcement that you will need to complete your project without needing to go to the market twice? We are here to help you with that. When you send your drawings to us, we will assess all reinforcements and bar marks independently, and tell you where all offcuts can be reused. We achieve this by issuing a special bending schedule and cutting scheme that takes into account re-usage of offcuts for foundations, beams, slabs, and columns. We will also issue a cutting sequence to the iron bender such that wastage can be minimised as practically as possible. This can save you a reasonable amount of money for your projects, and we do this for a very low fee that is based on the quantity of offcuts that we saved for you. Try us today by sending an e-mail to info@structville.com

Plastic Analysis of Frames

By

Ubani Obinna Uzodimma

-

December 16, 2018

The fully plastic moment of a section (M_p) is the maximum moment of resistance of a fully yielded cross-section. The yielded zone due to bending where infinite rotation can take place at a constant plastic moment (M_p) is called a plastic hinge.

In order to find the fully plastic moment of a yielded section, we normally employ the force equilibrium equation by saying that the total force in tension and compression at that section are equal. In frames, the plastic moment can be easily obtained using the kinematic method (upper bound theorem).

When a system of loads is applied to an elastic body, it will deform and show some resistance to deformation, and such a body is called a structure. On the other hand, if no resistance is set up against deformation, such as a body is said to have formed a mechanism. This can occur in structures such as beams and frames.

One of the fundamental conditions for plastic analysis of a framed structure is that the collapse load is reached when a mechanism is formed, in which the number of plastic hinges developed in the structure is sufficient to form a mechanism. The load factor (λ) at rigid plastic collapse is the lowest multiple of the design load which will cause the whole structure or part of it to form a mechanism.

It is important to realise that the number of independent mechanisms in a structure is related to the number of plastic hinge locations and the degree of redundancy of the system.

In beams, identification of critical spans in terms of M_p or load factor can be obtained using the static or kinematic method by considering simple beam mechanisms. But in framed structures, other types of mechanisms such as joint, sway, and gable mechanisms are also considered. Each of these mechanisms can occur independently, but it also possible for a critical collapse mechanism to develop by combination of the independent mechanisms.

Beam and sway mechanism in a framed structure

The purpose of combining mechanisms is to eliminate hinges that exist in the independent mechanism leaving only the minimum number required in the resulting combination to induce collapse.

The steps in the plastic analysis of frame structures are as follows;

Calculate the degree of static indeterminacy of the structure (R_D)
Calculate the number of possible plastic hinges (R_D + 1)
Identify the independent collapse mechanisms and evaluate them
Combine the independent mechanism and eliminate hinges where applicable
Obtain the critical plastic moment Mp which is the highest bending moment in the structure
Check for equilibrium and evaluate the reactive forces
Plot the final plastic moment diagram. No bending moment at any point in the structure should exceed the critical plastic moment.

Solved Example
For the frame loaded as shown below, find the critical M_p value, and draw the collapse moment diagram. The loads are factored.

Degree of static indeterminacy
R_D = (3m + r) – 3n
m (number of members) = 4
r (number of support reactions) = 5
n (number of nodes) = 5
R_D= (3 × 4 + 5) – 3(5) = 2
Therefore the frame is indeterminate to the second order.

A little consideration will show that the possible places where we could have plastic hinges in the structure are at node B, section C, section D (just to the left), section D (just to the right), section D (below) and section F. Nodes A, E, and F are all natural hinges. Therefore, total number of possible hinges = 6

Hence, number of independent collapse mechanisms = 6 – 2 = 4

These independent mechanisms are listed below;
– 2 Beam mechanisms (span B – D and span D – G)
– 1 sway mechanism of the entire frame due to the horizontal load
– 1 joint mechanism

Analysis of the independent mechanisms

Mechanism 1

δ_C = 3 × θ = 3θ
Internal work done due to rotations at internal hinges at B, D, and C
Total internal work done (Wi) = Mpθ + Mpθ + 2Mpθ = 4Mpθ

External work done (We) = 115 × 3θ = 345θ

Let external work done = Internal work done
4Mpθ = 345θ
On solving, M_p = 345/4 = 86.25 kNm

Mechanism 2

δ_F = 3 × θ = 3θ
Internal work done due to rotations at internal hinges at B, and F. No work is done at G because it is a natural hinge.
Total internal work done (Wi) = Mpθ + 2Mpθ = 3Mpθ

External work done (We) = 74 × 3θ = 222θ

Let external work done = Internal work done
3Mpθ = 222θ
On solving, M_p = 222/3 = 74 kNm

Mechanism 3 (Sway)

δ₁ = 4 × θ = 4θ
Internal work done due to rotations at internal hinges at B, and D. No work is done at A, E, and G because they are all natural hinges.

Total internal work done (Wi) = M_pθ + M_pθ = 2Mpθ

External work done (We) = 25 × 4θ = 100θ

Let external work done = Internal work done
2M_pθ = 100θ
On solving, M_p = 100/2 = 50 kNm

Mechanism 4 (Joint Rotation)

Internal work done = M_pθ + M_pθ + M_pθ = 3M_pθ
External work done = 0 (no external work done)

Internal work done = M_pθ + M_pθ + M_pθ = 3M_pθ
External work done = 0 (no external work done)

Let us now consider the possible mechanism combination by preparing the table below

The critical M_p is therefore = 111.167 kNm (This was achieved after other combinations were considered but this was found to be the highest).

Note that by virtue of the combination, internal hinges at B, D₂, and D₃ were eliminated. So the critical collapse mechanism is as given below;

Note that the implication of our analysis above is that points C, F and D₁ are plastic hinges, with a full plastic moment of 111.167 kNm. While C and F are sagging, D₁ is hogging. No other bending moment in the structure should exceed this value. Once a higher moment is discovered, it means that the critical collapse moment was not appropriately obtained.

∑M_F^R = 0 (anti clockwise positive)
3V_G = 111.167
V_G = 37.056 kN

∑M_C^L = 0 (clockwise positive)
3V_A – 4H_A= 111.167 ———– (1)

∑M_D^L = 0 (clockwise positive)
6V_A – 4H_A– (115 × 3) = -111.67
6V_A – 4H_A= 233.833 ———– (2)

Solving (1) and (2) simultaneously;
V_A = 40.721 kN
H_A = 2.623 kN

∑M_A = 0 (clockwise positive)
12V_G + 6V_E– (74 × 9) – (115 × 3) – (25 × 4) = 0
6V_E= 666.328
V_E = 111.055 kN

∑F_x = 0
H_E= 25 + 2.623 = 27.623 kN

Bending Moment
You can verify that having obtained the support reactions, the bending moment diagram can easily be plotted without strenuous calculations.

Thank you for visiting Structville today, and God bless you.

How to Analyse Retaining Walls for Trapezoidal Load

By

Ubani Obinna Uzodimma

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December 15, 2018

In the analysis of retaining walls subjected to earth pressure, it is very common to observe trapezoidal load distribution on the walls. Normally, earth pressure on a retaining wall is assumed to adopt a triangular load distribution, but due to surcharge which is usually assumed to act on the ground surface, the top of the wall experiences some degree of lateral pressure.

When the top of a retaining wall is free and the base assumed to be fixed, the retaining wall can be analysed easily using the three equations of equilibrium by modelling the wall as a cantilever. The load cases of surcharge, earth pressure, and ground water can be analysed separately, and the results added algebraically. When the wall top of a wall is propped, a statically indeterminate behaviour is assumed, and the wall can be analysed using the theory of plates and shells.

Numerous approaches such as finite difference method, finite element analysis, classical theory, etc can be adopted to solve problems of such nature. Anchor (1992) adopted a convenient approach of replacing the surcharge pressure with equivalent soil pressure. This was done by adding the maximum soil pressure and surcharge pressure algebraically, and assuming an equivalent triangular distribution. Another approach recommended by Reynolds et al (2008) is to separate the load cases, analyse them separately, and sum the results together. For walls spanning in two directions, this method is accurate for edge bending moments, but approximate for span moments.

In this post, we are going to explore the behaviour of the walls of a tank subjected to earth pressure and surcharge pressure (find the section of the tank above). We are going to investigate the methods described above, and compare them with finite element analysis using Staad Pro software.

Let us consider a propped cantilever wall loaded as shown below. The pressure at the top of the wall represents the surcharge pressure at ULS, while the pressure at the bottom of the wall represents the combined pressure of surcharge and retained earth at ULS.

TRAPEZOIDAL%2BLOAD%2BON%2BRETAINING%2BWALL

(1) By plate theory

We are going to analyse the walls using coefficients that can be readily obtained from charts and tables. We will separate the trapezoidal load into its triangular and rectangular components as shown below, and analyse them separately;

Using Moment Coefficient from Table
Note that these coefficients contain allowance for torsion at the edge, and are picked from Reynolds et al (2008).

Vertical Span
Maximum negative moment at base (triangular) = 0.048 × 70.599 × 4.3² = 62.658 kNm/m
Maximum negative moment at base (rectangular) = 0.073 × 5.625 × 4.3²= 7.592 kNm/m
Total Moment negative base moment = 62.658 + 7.592 = 70.250 kNm

Maximum positive moment (triangular) = 0.0185 × 70.599 × 4.3² = 24.149 kNm/m
Maximum positive moment (rectangular) = 0.055 × 5.625 × 4.3² = 5.720 kNm/m
Total approximate mid-span moment = 24.149 + 5.720 = 29.869 kNm/m

Horizontal Span
Maximum negative moment at the edge (triangular) = 0.033 × 70.599 × 4.3² = 43.077 kNm/m
Maximum negative moment at the edge (rectangular) = 0.032 × 5.625 × 6.4² = 7.3728 kNm/m
Total Moment negative edge moment = 43.077 + 7.3728 = 50.449 kNm

Maximum positive moment (triangular) = 0.011 × 70.599 × 4.3² = 14.359 kNm/m
Maximum positive moment (rectangular) = 0.024 × 5.625 × 6.4² = 5.5296 kNm/m
Total positive moment = 19.886 kNm/m

(2) By Equivalent Pressure Method

We are going to analyse the pressure load on the wall by summing up the pressure due to retained earth and surcharge and analysing them as triangular distribution.

Vertical Span
Maximum negative moment at the base base = 0.048 × 76.224 × 4.3² = 67.65 kNm/m
Maximum positive moment at the span = 0.0185 × 76.224 × 4.3² = 26.073 kNm/m

Horizontal Span
Maximum negative moment at the edge = 0.033 × 76.224 × 4.3² = 46.509 kNm/m
Maximum positive moment (triangular) = 0.011 × 76.224 × 4.3² = 15.503 kNm/m

(3) By Finite Element Analysis Using Staad Pro

When modelled and analysed on Staad Pro;

We will have to allow for torsion at the edge moments.

By using Staad Pro;

Vertical Span
Maximum negative moment at base = 58.9 kNm/m (allowing for torsion = 71.7 kNm/m)
Maximum positive moment (mid span) = 31.2 kNm/m

Horizontal Span
Maximum negative moment at edge = 40 kNm/m (allowing for torsion = 52.8 kNm/m)
Maximum positive moment = 18 kNm/m

Twisting moment = 12.8 kNm/m

Let us compare the results from the three methods investigated;

From the above table, it can be seen that the result from the use of coefficient and FEA are very close and comparable, while the result using equivalent approach is giving lower bound results. Therefore, in the absence of computer programs, I recommend the use of coefficients from Table for analysis of retaining walls subjected to trapezoidal loads.

This post is an excerpt from the book;

‘Structural Design of Swimming Pools and Underground Water Tanks‘ by Ubani Obinna. Published by Structville Integrated Services Limited (2018). ISBN 978-978-969-541-6

To obtain this publication in PDF for NGN 2,000, click HERE
For more information, send a Whatsapp message to;
+2347053638996

References
Anchor R.D. (1992): Design of Liquid Containing Structures. Published by Edward Arnolds, UK. ISBN 0-340-54527-5

Reynolds C.E., Steedman J.C., Threlfall A.J. (2008): Reynolds Reinforced Concrete Design Handbook. Spon Press, Taylor and Francis Group (11th Edition)

Ubani O.U. (2018): Structural Design of Swimming Pools and Underground Water Tanks. Structville Integrated Services Limited (2018). ISBN 978-978-969-541-6

Structural Design of Swimming Pools and Underground Water Tanks

By

Ubani Obinna Uzodimma

-

November 27, 2018

I spent a large part of the last few months developing the contents of this booklet on ‘Structural Design of Swimming Pools and Underground Water Tanks (According to the Eurocodes)’. Water is necessary for survival of mankind, but in one way or another, water is relatively scarce. We all know that rain does not fall continuously, and for water to be available for usage in homes, it will have to be fetched/pumped from the stream, or harvested during rainfall, or dug up from the ground. As a result, survival instincts made man to create different means of storing water in order to face the periods of scarcity.

One of the alternatives of storing water is the use of reinforced concrete structures which may be buried under the ground, supported on the ground surface, or elevated above the ground surface. Civil engineers have been at the forefront of designing these infrastructures in order to meet accepted performance criteria. One of the profound requirements of water tanks is water tightness, which can be related to the serviceability limit state requirement of cracking, amongst other factors.

The publication highlighted above focuses on the use of reinforced concrete for construction of underground water tanks and swimming pools. All the factors usually considered in the design of underground water retaining structures such as geotechnical analysis, modelling, loading, structural analysis, and structural design were all presented in an objective manner to the reader. Also, the use of Staad Pro Software was explored in detail on how it can be used to model and analyse tanks and swimming pools.

I am going to use the screenshots below to show you some of the contents, with hope that you will find it interesting. The images are arranged in no specific order.

Structural Design of Swimming Pools and Underground Water Tanks 52

The publication also paid serious attention to how water tightness can achieved on site through careful and standard construction practices. In addition to this 167 pages booklet, I also developed excel spreadsheets that can help make calculation of crackwidths faster, and Staad Pro video tutorials, peradventure you want to master the use of the software. Kindly see screenshot below;

For this reason, there are three different packages that can be purchased to match different levels of interest in the content.

PACKAGE 1 (STANDARD) COST: NGN 3,000

(1) 167 pages booklet on Structural Analysis of Swimming Pools and Underground Water Tanks (in PDF)

PACKAGE 2 (ADVANCED) COST: NGN 4,500

(1) 167 pages booklet on Structural Analysis of Swimming Pools and Underground Water Tanks (in PDF)

(2) Staad Pro Video Tutorial on Modelling and Analysis of Underground Water Tanks

PACKAGE 3 (SOPHISTICATED) COST: NGN 6,000

(1) 167 pages booklet on Structural Analysis of Swimming Pools and Underground Water Tanks (in PDF)

(2) Staad Pro Video Tutorial on Modelling and Analysis of Underground Water Tanks

(3) MS Excel Spreadsheet for design of slabs, and calculation of crackwidths

All purchases will be delivered via mail within 24 hours of purchase.

To purchase any of the packages, click HERE

Thank you, and God bless you.

For more information;

Call: +2348060307054

WhatsApp: +2347053638996

E-mail: rankiesubani@gmail.com

How will you describe this system of forces?

By

Ubani Obinna Uzodimma

-

August 24, 2018

What is the best description that you will give to this system of forces?

(a) Coplanar parallel system of concurrent forces
(b) Non-coplanar parallel system of non-concurrent forces
(c) Coplanar non-parallel system of concurrent forces
(d) Non-coplanar non-parallel system of non-concurrent forces

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

Are you looking for where you can make free downloads of publications? Visit Structville Research for all your free downloads.

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

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

Is This Ultimate or Serviceability Limit State Failure?

By

Ubani Obinna Uzodimma

-

August 21, 2018

Kindly look at the image carefully, and lend your professional opinion if the failure of the building will be categorized under ultimate or serviceability limit state. By posting and discussing your opinion on the comment section, I am very certain that knowledge and deeper understanding of this topic will be enhanced.

Thank you very much.

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

Are you looking for where you can make free downloads of publications? Visit Structville Research for all your free downloads.

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

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

Aspects of Modelling of Shear Walls

By

Ubani Obinna Uzodimma

-

August 18, 2018

Shear walls are structural elements usually employed in tall buildings to assist in resisting lateral loads. Shear walls can be solid or pierced (coupled), depending on their location in the building. In the design of tall buildings, structural engineers normally throw the entire lateral load (say wind action) to the shear walls, which means that the columns will not be relied on for lateral stability. In a more practical scenario however, the shear walls and columns interact in resisting lateral loads, which can be taken into account.

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In this post, we are going to review how shear walls can be modelled on Staad Pro software and consider the various accuracy levels with each method.

Let us consider the shear wall shown below;

Thickness of the wall = 250 mm
The wall is subjected to a lateral load of 300 kN at the top.

Model 1: Equivalent frame model
In this model, we represent the shear wall as an equivalent frame with representative stiffness for the various elements.

shear%2Bwall%2Bequivalent%2Bframe%2Bmodel

On modelling this Staad Pro, we obtained the result shown below;

The maximum horizontal deflection was obtained to be 158.376 mm.

Model 2: Plate modelling (Finite Element Mesh)

In this model, the shear wall is modelled as a full plate with approximate mesh size of (4 x 4) per square metre. The deflection due to the lateral load is given below;

The maximum horizontal deflection was obtained to be 156.588 mm.

It is important also to compare the deflection obtained by considering different mesh sizes in order to compare the accuracy of utilizing the equivalent frame model. Thank you for visiting Structville today, and God bless.

Buckling of Columns

By

Ubani Obinna Uzodimma

-

August 16, 2018

When structural members are subjected to compressive forces, the members may fail before the compressive resistance (A.f_y) is reached. This premature failure is usually caused by secondary bending effects such as imperfections, the eccentricity of loading, asymmetry of the cross-section etc. In such cases, the failure mode is normally buckling.

The buckling of columns is usually associated with a sudden large displacement (bending) on the column following a slight increase or modification of an existing compressive load. The load at which the buckling of columns occur is usually less than the compressive resistance of the column, and it is mainly influenced by the length, cross-section, and boundary conditions of the column.

This is unlike when the member is subjected to tensile forces, where the member will generally fail when the stress in the cross-section exceeds the ultimate strength of the material. Members subjected to tensile forces are inherently stable.

In essence, for very short columns or members subjected to tensile forces, the stress is proportional to the load, and the failure load is equal to the yield stress times the area of the cross-section (A.f_y). Commonly, compression members can be classified as short, intermediate, or slender depending on their slenderness ratio. Short elements will fail by crushing, slender elements will fail by excessive lateral deflection, and intermediate elements will fail by the combination of both.

Types of Buckling

Flexural buckling (Euler)
Lateral-torsional buckling
Torsional buckling
Snap-through buckling
Local plate buckling

In this article, we are going to focus on flexural buckling.

In the year 1757, Leonhard Euler developed a theoretical basis for the analysis of premature failure due to buckling. The theory was based on the differential equation of elastic bending of a pin-ended column, which related the applied bending moment to the curvature along the length of the column.

For a pin-ended column, the critical Euler buckling load (K = 1.0) is given by;

P_E = Kπ²EI/L²

Note that L in this case is the effective buckling length which depends on the buckling length between the pinned supports or the points of contraflexure for members with other boundary conditions.

Subsequently, the Perry-Robertson formula was developed to take into account the deficiencies of the Euler method. The formula evolved from the assumption that all practical imperfections could be represented by a hypothetical initial curvature of the column. In the UK, the Perry-Robertson formula was modified and is referred to in BS 5950 as the Perry-Strut formula. As a matter of fact, it also forms the basis of flexural buckling in BS EN 1993 (Eurocode 3).

Worked Example on the Buckling of Columns

In the example below, we are going to show how the failure load of a column changes with an increase in effective length due to the phenomenon of buckling.

Let us investigate the load-carrying capacity of UC 305 x 305 x 158 according to EC3. We will consider the ends of the column to be pinned, and we will start with an initial length of 1.0 m.

Properties of UC 305 × 305 × 158

D = 327.1 mm; B = 311.2 mm; t_f = 25mm; t_w = 15.8mm; r = 15.2mm; d = 246.7mm, b/t_f = 6.22; d/t_w = 15.6; I_zz = 38800 cm⁴; I_yy = 12600 cm⁴; i_y-y = 13.9cm; i_z-z = 7.9 cm, A = 201 cm²

Thickness of flange t_f = 25mm.
Since t_f > 16mm, Design yield strength f_y = 265 N/mm² (Table 3.1 EC3)

Section classification

ε = √(235/f_y ) = √(235/265) = 0.942

We can calculate the outstand of the flange (flange under compression)
C = (b – t_w – 2r) / (2 ) = (327.1 – 15.8 -2(15.2)) / (2 ) = 140.45mm.
We can then verify that C/t_f = 140.45/25 = 5.618
5.618 < 9ε i.e. 5.618 < 8.478.

Therefore the flange is class 1 plastic

Web (Internal compression)
d/t_w = 15.6 < 33ε so that 15.6 < 31.088. Therefore the web is also class 1 plastic

Resistance of the member to uniform compression (clause 6.2.4)
N_C,Rd = (A.F_y)/γ_mo = (201 × 10² × 265) / 1.0 = 5326500 N = 5326.5 kN

Buckling resistance of member (clause 6.3.1)

Since the member is pinned at both ends, the critical buckling length is the same for all axis L_cr = 1000 mm

Slenderness ratio ¯λ = L_cr/(r_i λ₁)
λ₁ = 93.9ε = 93.9 × 0.942 = 88.454

In the major axis
(¯λ_y ) = 1000/(139 × 88.454) = 0.081
In the minor axis
(¯λ_z ) = 1000/(79 × 88.454) = 0.143

Check h/b ratio = 327.1/311.2 = 1.0510 < 1.2, and t_f < 100 mm (Table 6.2 EN 1993-1-1:2005)

Therefore buckling curve b is appropriate for the y-y axis, and buckling curve c for the z-z axis. The imperfection factor for buckling curve b, α = 0.34 and curve c, α = 0.49 (Table 6.1)

Φ = 0.5 [1 + α(¯λ – 0.2) + ¯λ²]

Φ_z = 0.5 [1 + 0.49 (0.143 – 0.2) + 0.143² ] = 0.496
Φ_y = 0.5 [1 + 0.34 (0.081 – 0.2) + 0.081² ] = 0.483

X = 1/(Φ + √(Φ² -¯λ²))
X_z = 1/( 0.496 + √(0.496² – 0.143²)) = 1.02 > 1 (cannot be greater than 1, so take 1.0)
X_y = 1/( 0.483 + √(0.483² – 0.081²)) = 1.04 > 1 (cannot be greater than 1, so take 1.0)

Therefore N_b,Rd = (X_z A.F_y)/γ_m1 = (1.0 × 201 × 10² × 265) / (1.0 ) = 5326500 N = 5326.5 kN

N_b,Rd = N_C,Rd= 5326.5 kN

This tells us that the column is short and that at a length of 1.0 m, the column will fail by crushing. Buckling is not a significant mode of failure. If we incrementally increase the length of the column by 1.0 m and follow the steps described above, we will arrive at the following results;

Column length (m)	Failure Load (kN)
1	5326
2	3090
3	4700
4	4270
5	3810
6	3330

The graph of the relationship is given below;

Therefore, the load-carrying capacity of compression members reduces with an increase in length. When the height of the pin-ended column was increased from 1m to 6m, a reduction of about 37% was observed in the buckling load. In the design of axially loaded members such as trusses, it is, therefore, advisable to design the members in such a way that the longer members are in tension and the shorter members are in compression.

Thank you for visiting Structville today, and God bless.

Structural Design of Cantilever Slabs

By

Ubani Obinna Uzodimma

-

August 14, 2018

Table of Contents

Cantilever slabs are common features in residential and commercial buildings due to the need to have bigger spaces on upper floors. To achieve this, architects normally extend the slab beyond the ground floor building line, thereby forming a cantilever. Cantilever slabs in reinforced concrete buildings are usually characterised by a reinforced concrete slab projecting from the face of the wall and having a back span that extends into the interior panels of the building.

The design of reinforced concrete cantilever slab is usually governed by serviceability limit state requirements. Typically, the design involves the provision of adequate concrete thickness and reinforcement to prevent excessive deflection and vibration.

During the detailing of reinforced concrete cantilever slabs, the main reinforcements are provided at the top and they extend into the back span by at least 1.5 times the length of the cantilever or 0.3 times the length of the back span, whichever is greater. Furthermore, it is important to provide at least 50% of the reinforcement provided at the top, and at the bottom.

In this article, we are going to show how we can analyse and design cantilever slabs subjected to floor load and block wall load according to the requirements of EN 1992-1-1:2004 (Eurocode 2).

cantilever slab design — Cantilever slab supporting block wall load

Worked Example – Design of Cantilever Slabs

A cantilever slab 200 mm thick is 1.715m long, and it is supporting a blockwork load at 1.0m from the fixed end. Design the slab using the data given below;

Purpose of building – Residential
f_ck = 25 Mpa
f_yk = 460 Mpa
Concrete cover = 25 mm
Height of block wall = 2.75 m
Unit weight of concrete = 25 kN/m³
Unit weight of block with renderings = 3.75 kN/m²

Load Analysis

Self weight of slab = (25 × 0.2) = 5 kN/m²
Finishes (assume) = 1.2 kN/m²
Partition allowance = 1.0 kN/m²
Total characteristic permanent action (pressure load) g_k = 7.2 kN/m²

Permanent action from wall G_k = 3.75 × 2.75 = 10.3125 kN/m

Variable action on slab q_k = 1.5 kN/m²

At ultimate limit state;
n = 1.35g_k + 1.5q_k
n = 1.35(7.2) + 1.5(1.5) = 11.97 kN/m²

Ultimate load from wall = 1.35 x 10.3125 = 13.92 kN/m

Design Forces

M_Ed = (13.92 × 1) + (11.97 × 1.7152)/2 = 31.523 kNm
V_Ed= (13.92) + (11.97 × 1.715) = 34.45 kN

Flexural design

M_Ed = 31.523 kNm
Effective depth (d) = h – C_nom – ϕ/2 – ϕ_links
Assuming ϕ12 mm bars will be employed for the main bars
d = 200 – 25 – 6 = 169 mm

k = M_Ed/(f_ckbd²) = (31.523 × 10⁶)/(25 × 1000 × 169²) = 0.044

Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ]
z = d[0.5+ √(0.25 – 0.882(0.044))] = 0.95d

A_s1 = M_Ed/(0.87f_yk z) = (31.523 × 10⁶)/(0.87 × 460 × 0.95 × 169) = 490 mm²/m
Provide X12@200 c/c TOP (AS_prov = 565 mm²/m)

Check for deflection

ρ = A_s,prov /bd = 565 / (1000 × 169) = 0.0033
ρ₀ = reference reinforcement ratio = 10^-3√(f_ck) = 10^-3√(25) = 0.005
Since if ρ ≤ ρ₀;

L/d = K [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck) (ρ₀ / ρ – 1)^(3⁄2)
k = 0.4 (Cantilevers)

L/d = 0.4 [11 + 1.5√(25) × (0.005/0.0033) + 3.2√(25) × [(0.005 / 0.0033) – 1]^(3⁄2)
L/d = 0.4[11 + 11.363 + 5.9159] = 11.311

β_s = (500 As_prov)/(f_yk As_req) = (500 × 565) / (460 × 490) = 1.253

Therefore limiting L/d = 1.253 × 11.311 = 14.172
Actual L/d = 1715/169 = 10.147

Since Actual L/d (10.147) < Limiting L/d (14.172), deflection is satisfactory.

Exercise for Students
(1) Provide distribution bars
(2) Verify the section for shear
(3) Check for cracking
(4) Do the detailing sketches

Thank you for visiting Structville today, and God bless.

How to Calculate the Quantity of Mortar (Sand and Cement) for Laying Blocks

By

Ubani Obinna Uzodimma

-

August 9, 2018

Engineers, site managers, and quantity surveyors are always faced with the challenge of specifying as accurately as possible, the quantity of materials needed to execute a specific item of work. In this article, we are going to explain how you can estimate the quantity of mortar (cement and sand) needed to lay blocks per square metre of wall.

It is well known that we need about 10 blocks to build one square metre of wall.

Simple Proof;
Area of wall = 1 m²
Planar area of one standard block (Nigeria) = 450mm x 225mm = 101250 mm²= 0.10125 m²
Therefore, the number of blocks required (disregarding mortar) = 1/0.10125 = 9.87 (say 10 blocks)

Now, how do we estimate the quantity of mortar needed to lay the blocks? First of all, let us look at the dimensions of a typical 9 inches block with holes (the most popular block for building in Nigeria).

From the image of the sandcrete, we can say that the cross-sectional area of the block that receives horizontal mortar is;

A_b = (0.45 x 0.225) – 2(0.15 x 0.125) = 0.0637 m²

Now, let us assume that the mortar will be 25 mm (1 inch) thick.

The typical arrangement of blocks in a one square metre wall is shown below;

arrangement%2Bof%2Bblocks%2Bin%2Ba%2Bone%2Bsquare%2Bmetre%2Bwall

Therefore, we can estimate the volume of mortar required to build one square metre (1 m²) of wall as follows;

Vertical mortar = 8 x (0.025 x 0.225 x 0.275) = 0.0123 m³
Horizontal mortar = 10 x (0.0637 x 0.025) = 0.0159 m³
Total = 0.0282 m³

Therefore the volume of mortar required to lay one square metre (1 m²) of 9-inch block (with holes) can be taken as 0.03 m³ for all practical purposes.

Cement requirement
The typical mix ratio for mortar (laying of blocks) is 1:6
Quantity of cement required = 1/7 x 1440 kg/m³ = 205.71 kg
Making allowance for shrinkage between fresh and wet concrete = 1.54 x 205.71 = 316.79 kg
Quantity of cement required in bags = 316.79/50 = 6.33 bags/m³ of mortar

Therefore, the quantity of cement required to lay 1m² of wall = 0.03 × 6.33 = 0.1899 bags
If 0.1899 bags of cement = 10 blocks
Therefore, 1 bag of cement = 52 blocks

Therefore, 1 bag of cement is required to lay 52 blocks using a 1:6 mix ratio of mortar.

Sand
Quantity of sand required = 6/7 x 1600 kg/m³ = 1371.428 kg
Making allowance for shrinkage between fresh and wet concrete = 1.54 x 1371.428 = 2112 kg
Quantity of sand required in tonnes = 2.112 tonnes/m³ of mortar

Therefore, the quantity of sand for mortar required to lay 1m² of wall = 0.03 × 2.112 = 0.06336 tonnes of sand
If 0.06336 tonnes of sand = 10 blocks
Therefore, 1 tonne of sand = 157 blocks

Therefore, 1 tonne (1000 kg) of sand is required to lay 157 blocks using a 1:6 mix ratio of mortar.

Further Example
To go further, let us assume that we have 150 m² (about 1500 blocks) of wall and want to estimate the quantity of cement and sand needed to lay the blocks.

The volume of mortar required = 0.03 x 150 = 4.5 m³

For 4.5m³ of mortar;
Provide = 4.5 x 6.33 = 29 bags of cement

For 4.5m³ of mortar;
Provide = 4.5 x 2112 = 9504 kg of sharp sand

Therefore you need about 10 tonnes of sharp sand and 29 bags of cement for laying 150m² of 9 inches block wall.

Thank you for visiting Structville today, and God bless.

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