How to Calculate the Lap Length of Reinforcements

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July 29, 2017

Reinforced concrete structures rely on the effective transfer of stress between steel reinforcement bars. Lap splices, where two reinforcement bars overlap within the concrete element, are a common method to achieve this continuity. Lapping of reinforcement is a frequent occurrence in construction due to supply length limitations, construction joints, stage construction, and ease of handling of rebars.

Therefore, determining the appropriate lap length is important for ensuring structural integrity and preventing premature failure. This article discusses how to calculate the lap length for reinforcements in concrete structures.

Factors Influencing Lap Length

Several factors influence the required lap length:

Reinforcement diameter (Φ): Larger diameter bars require longer lap lengths for effective stress transfer.
Concrete grade (f_ck): Higher concrete strength allows for shorter lap lengths due to improved bond characteristics.
Reinforcement yield strength (f_yk): Higher yield strength steel requires longer lap lengths to develop its full capacity.
Concrete cover: The thickness of the concrete cover influences the bond transfer and hence, the anchorage length.
Bond condition: The orientation of the reinforcement (straight or bent) and the concrete cover can lead to a good or poor bond condition which can influence the anchorage length of reinforcements.
Confinement conditions: The presence of stirrups or ties around the lap splice can significantly reduce the required lap length.
Type of Lap Splice: Standard lap splices require longer lengths compared to hooked or welded lap splices.

Methods of Calculating Lap Length

Different design codes and standards provide methods for calculating lap length. Here’s a breakdown of two common approaches:

1. Basic Lap Length Formula:
This simplified method often serves as a preliminary estimate:

Lap Length (L) = 50d

Where:
d = Reinforcement diameter (mm)

This formula assumes standard lap splices in normal concrete conditions.

2. Code-Based Calculations:
Detailed calculations are typically based on specific design codes like Eurocode 2 (EC2) or ACI 318. These codes provide equations and tables that consider various factors influencing lap length:

EC2 Method:
EC2 employs a formula that incorporates the bar diameter, concrete grade, steel yield strength, and a factor accounting for the percentage of bars lapped in one section.

ACI 318 Method:
ACI 318 provides tables with tabulated lap lengths based on bar diameter, concrete strength, and type of splice (standard, hooked, or welded).

Note:

Minimum Lap Length: Codes often specify minimum lap lengths regardless of calculations.
Development Length: Lap length should be greater than or equal to the development length of the bar, which is the minimum length required for the bar to develop its full yield strength in the surrounding concrete.
Longitudinal Spacing of Bars: The spacing between lapped bars within the overlap zone can affect the required lap length.
Lightweight Concrete: Adjustments are often necessary for lap lengths in lightweight concrete due to its lower bond strength.

Anchorage Length

Apart from the lapping of steel reinforcements, reinforcing bars should be well anchored so that the bond forces are safely transmitted to the concrete to avoid longitudinal cracking or spalling. The length of rebar required to achieve adequate transfer of forces is known as the anchorage length. Transverse reinforcement shall be provided if necessary.

Apart from straight bars, other shapes that are specified in the code are;

(a) standard bend,
(b) standard hook, and
(c) standard loop.

The detailing rules and the equivalent anchorage length for each of these standard shapes are defined in EN1992-1-1 Figure 8.1. Types of anchorage are shown in the figure below (Figure 8.1 EC2).

For bent bars, the basic tension anchorage length is measured along the centreline of the bar from the section in question to the end of the bar, where:

l_bd = α₁ α₂ α₃ α₄ α₅ l_b,req ≥ l_b,min ———— (1)

where;
l_b,min is the minimum anchorage length taken as follows:
In tension, the greatest of 0.3l_b,rqd or 10ϕ or 100mm
In compression, the greatest of 0.6l_b,rqd or 10ϕ or 100mm

l_b,rqd is the basic anchorage length given by;
l_b,rqd = (ϕ/4) σ_sd/f_bd ————– (2)

Where;
σ_sd = The design strength in the bar (take 0.87f_yk)
f_bd = The design ultimate bond stress (for ribbed bars = 2.25η₁ η₂ f_ctd)
f_ctd = Design concrete tensile strength
f_ctd = 0.21f_ck^(2/3) for f_ck ≤ 50 N/mm²
η₁ is a coefficient related to the quality of the bond condition and the position of the bar during concreting
η₁ = 1.0 when ‘good’ conditions are obtained and
η₁ = 0.7 for all other cases and for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist
η₂ is related to the bar diameter:
η₂ = 1.0 for φ ≤ 32 mm
η₂ = (132 – φ)/100 for φ > 32 mm

‘Good’ bond conditions are applicable when any of the following conditions are fulfilled:

(a) Vertical bars or almost vertical bars inclined at an angle 45° ≤ α ≤ 90° from the horizontal,
(b) bars that are located up to 250 mm from the bottom of the formwork for elements with height h ≤ 600 mm, or
(c) bars that are located at least 300 mm from the free surface during concreting for elements with height h > 600 mm.

‘Poor’ bond conditions are applicable for all other cases and also for bars in structural elements built with slip-forms, unless it can be shown that ‘good’ bond conditions exist. The different bond regions are shown in the figure above that is reproduced from EN1992-1-1 Figure 8.2.

Description%2Bof%2Bbond%2Bconditins%2BEC2

α₁ is for the effect of the form of the bars assuming adequate cover.
α₂ is for the effect of concrete minimum cover.

α₃ is for the effect of confinement by transverse reinforcement
α₄ is for the influence of one or more welded transverse bars ( φt > 0.6φ) along the design anchorage length l_bd
α₅ is for the effect of the pressure transverse to the plane of splitting along the design anchorage length.

The values of these coefficients can be adequately obtained by following the Table below;

l_0,min = max{0.3α₆ l_b,rqd; 15ϕ; 200}

α₆ = √(ρ₁/25) but between 1.0 and 1.5
where ρ₁ is the % of reinforcement lapped within 0.65l₀ from the centre of the lap

percentage%2Bof%2Blapped%2Bbars%2Beurocode%2B2

Values of α₁, α₂, α₃ and α₅ may be taken as for the calculation of anchorage length but for the calculation of α₃, ΣA_st,min should be taken as 1.0As(σ_sd/f_yd), with As = area of one lapped bar.

SOLVED EXAMPLE FOR ANCHORAGE LENGTH
Calculate the design tension anchorage length of T16mm bar (f_yk = 460 N/mm², concrete cover = 35 mm, Concrete cylinder strength f_ck = 25 N/mm²) for;

(a) When it is a straight bar
(b) When it is bent into any other shape
Assume good bond conditions

Solution
l_bd = α₁ α₂ α₃ α₄ α₅ l_b,req ≥ l_b,min
l_b,rqd = (ϕ/4) σ_sd/f_bd
f_bd =2.25η₁ η₂ f_ctd
η₁ = 1.0 ‘Good’ bond conditions
η₂ = 1.0 bar size ≤ 32

f_ctd = (α_ct f_{ctk 0.05})/γ_c ————— (3)

where;
f_{ctk 0.05} = characteristic tensile strength of concrete at 28 days = 1.8 N/mm² (Table 3.1 EC2)
γ_c = partial (safety) factor for concrete = 1.5
α_ct = coefficient taking account of long-term effects on the tensile strength, this is an NDP with a recommended value of 1.

f_ctd = (1.0 × 1.8)/1.5 = 1.2 N/mm²
f_bd = 2.25 × 1.0 × 1.0 × 1.2 = 2.7 N/mm²

l_b,rqd = (ϕ/4) σ_sd/f_bd
σ_sd = 0.87 × 460 = 400.2 N/mm²>br/> l_b,rqd = (ϕ × 400.2 )/(4 × 2.7) = 37.05ϕ

Therefore;

l_bd = α₁ α₂ α₃ α₄ α₅(37.05ϕ)

(a) For straight bar
α₁ = 1.0
α₂ = 1.0 – 0.15 (C_d – ϕ)/ ϕ
α₂ = 1.0 – 0.15 (35 – 16)/16 = 0.8218
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 1.0 conservative value

l_bd = 0.8218 × (37.05ϕ) = 30.4ϕ = 30.4 × 16 = 486.4 mm
Say 500 mm

(b) For other shape bar
α₁ = 1.0 bC_d; = 35 is ≤ 3ϕ = 3 × 16 = 48
α₂ = 1.0 – 0.15 (Cd – 3ϕ)/ ϕ ≤ 1.0
α₂ = 1.0 – 0.15 (35 – 48)/16 = 1.121 ≤ 1.0
α₃ = 1.0 conservative value with K = 0
α₄ = 1.0 N/A
α₅ = 1.0 conservative value
l_bd = 1.0 × (37.05ϕ) = 37.05ϕ = 37.05 × 16 = 592 mm
Say 600mm

Compression anchorage (α₁ = α₂ = α₃ = α₄ = α₅ = 1.0)
l_bd = 37.05ϕ

For poor bond conditions
Anchorage for ‘Poor’ bond conditions = ‘Good’/0.7

	Good bond conditions		Poor bond conditions
Concrete class	• Straight bars in tension / compression • Other-than-straight bars in compression	• Other-than-straight bars in tension and adequate cover c_d > 3Φ	• Straight bars in tension/compression • Other-than-straight bars in compression	• Other-than-straight bars in tension and adequate cover c_d > 3Φ
C16/20	55Φ	39Φ	78Φ	55Φ
C20/25	47Φ	33Φ	67Φ	47Φ
C25/30	41Φ	29Φ	58Φ	41Φ
C30/37	36Φ	26Φ	52Φ	36Φ
C35/45	33Φ	23Φ	47Φ	33Φ
C40/50	30Φ	21Φ	43Φ	30Φ

Example on the calculation of lap length of 4X16mm bars of a column in a multi-storey building
Since the bars are in compression,
α₁ α₂ α₃ α₅= 1.0
As calculated above, l_bd = 37.05ϕ

Let us say that over 50% of reinforcement is lapped within 0.65l₀ from the centre of the lap
Hence, we will take α₆= 1.5

Lap length therefore = 1.5 × 37.05ϕ = 55.57ϕ = 55.57 × 16 = 889.2mm
Say 900 mm

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How to Calculate the Number of Blocks Required to Complete a 3 Bedroom Flat

By

Ubani Obinna Uzodimma

-

July 26, 2017

Table of Contents

Sandcrete blocks are very popular construction materials for the construction of walls in buildings. They can serve as load-bearing elements or as mere partition panels. There are different types and sizes of sandcrete blocks that are employed in building construction. The size of the block (dimensions) can be a factor in determining the number of blocks required to successfully complete a building.

In order to properly know the cost of constructing a building, it is very important to know the number of blocks that will be able to complete the building. The number of blocks required to construct a building is largely dependent on the type of block, size of the building, area of walls to be built, etc.

on going bungalow construction — **Figure 1:** On-going bungalow construction

In this article, we are going to present a very simple approach to determining the number of blocks required to complete a three-bedroom flat. The floor plan of a three-bedroom flat is shown below, which can comfortably sit on a plot of land of 450 m². The building consists of a sitting room, three bedrooms (all en-suite), a visitor’s toilet, a dining, a kitchen, a store, and sit-out areas. It is required to estimate the quantity of 6 inches blocks (without holes) required to complete the project.

FLOOR%2BPLAN — **Figure 2:** Floor plan of a 3-bedroom flat

The data from the building is as shown below;

DOORS
Door type 1 (D1) = 1 number (Dimension = 2100 × 1200)
Door type 2 (D2) = 6 numbers (Dimension = 2100 × 900)
Door type 3 (D3) = 4 numbers (Dimension = 2100 × 700)

WINDOWS
Window Type 1 (W1) = 12 numbers (Dimension = 1200 × 1200)
Window Type 2 (W2) = 4 numbers (Dimension = 600 × 600)

Storey height (height of the building from floor to ceiling) = 3000 mm

Foundation details of the bungalow are given below.

Foundation%2Bto%2BDPC — **Figure 3:** Foundation details of the building

A section through the wall of the building is given below;

Section%2BThrough%2Bwall — **Figure 4:** Section of the wall

We are going to carry out the calculation in three main phases of construction of bungalow which are;

Foundation to DPC
DPC to lintel level, and
Lintel level to overhead level.

Estimation of the number blocks

The simplest approach is to estimate the number of blocks required in a metre square of wall.

The planar dimensions of a standard block in Nigeria is (450mm × 225mm).

6%2Binches%2Bblock%2Bwithout%2Bholes — **Figure 5**: Typical dimensions of 6 inches blocks in Nigeria

The planar area of 1 block = (0.225 × 0.45) = 0.10125 m²
Therefore one metre square of a wall will contain;
1/0.10125 = 9.876

Therefore, one metre square of wall will contain approximately 10 blocks.

PHASE 1: FOUNDATION TO DPC

At this stage, the total length of the walls of the building and the height of the walls at the substructure level are considered. There are no openings, and the entire foundation excavation length is considered. For the building we are considering, the foundation layout is as shown below;

FOUNDATION%2BPLAN — **Figure 6:** Foundation layout of a three-bedroom flat

Using any method of your choice, you can calculate the total length of the walls at the foundation level.
This is approximately equal to 100256 mm = 100.256m from my calculations.

The height of the wall from foundation to DPC (ignoring thickness of mortar) as shown in Figure 3 = 900mm = 0.9m

Therefore, the total area of wall = 100.256m × 0.9m = 90.2304m²
If 10 blocks are required for 1m2 of a wall, therefore (10 × 90.2304) = 903 blocks is required for 90.2304m² of wall.

The number of blocks required to raise the building from foundation to DPC = 903 blocks (disregarding damages/waste).

PHASE 2: DPC TO LINTEL LEVEL

In this phase, we will consider the effects of openings in the buildings which includes the doors, windows, and other openings as specified in the drawing. These openings are seen in the verandah areas and some of the internal spaces in the building. The same process of calculating the total length of walls is also adopted, after which we minus the area of the openings.

This process is shown below;

Total Length of walls (excluding areas with no walls like verandahs) = 87420 mm = 87.42m

Height of wall = 2100 = 2.1m
Therefore area of wall = (87.42 × 2.1) = 183.582 m²
Total area of all doors from building data = 19.74 m²
Total area of all windows from building data = 18.72 m²
Therefore the net area of walls = 183.582 m² – 19.74 m² – 18.72 m² = 145.122 m²
Hence, the total number of blocks required to take the building from DPC to lintel level = 145.122 × 10 = 1451.22 = 1452 blocks (disregarding wastes)

PHASE 3: LINTEL LEVEL TO OVERHEAD LEVEL

Here, we are going to assume that the building will be chained at the lintel level (this means that the entire level of the building will be cast and reinforced with steel at the lintel level. This offers the advantage of helping the building behave as a unit, and helps in reducing cracking in the building).

Obviously, the length of the wall in this case will not consider openings, so we are going to make use of 100.256m as calculated from the foundation level.

Hence, the length of walls = 100.256m
Height of wall = 675mm = 0.675m

Total area of walls = (100.256 × 0.675) = 67.6728 m²
Hence, the total number of blocks required to take the building from the lintel level to the roof level = 67.6728 × 10 = 676.728 = 677 blocks.

Ultimately, the total number of blocks required to complete the construction of a three-bedrooms flat = 903 + 1452 + 677 = 3032 blocks (disregarding wastage).

BB96016D 458D 4C07 A5AB C2A7BBC128F0 — **Figure 8:** Bungalow with completed block work

To consider wastage, you can make a 20% increase, so that the number of blocks to be moulded or bought = 1.2 × 3032 = 3638 blocks. However, when you are moulding and can guarantee the quality of blocks to be moulded, 10% increase will be satisfactory for the job.

READ ALSO IN THIS BLOG;

How to estimate the quantity of materials required for moulding blocks.

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How to Estimate the Quantity of Sand and Cement Required for Moulding Blocks

By

Ubani Obinna Uzodimma

-

July 26, 2017

In Nigeria, the 9 inches hollow block (450mm x 225mm x 225mm) is normally used for the construction of blockwork of buildings exceeding one storey. The length of the block is 450mm (18 inches), the height is 225mm (9 inches), and the width is 225mm (9 inches). Contractors and builders are often faced with the challenge of either buying blocks for construction or moulding it on their own. When the conditions are right, builders will prefer to mould their own blocks, especially for the purpose of controlling the quality of the blocks, and to reduce the cost of transportation.

Let us assume that you have decided to mould your own blocks, and you have been faced with the challenge of estimating the quantity of cement and sand to purchase that will satisfy the construction requirement. This article will give you a guide on how to estimate the quantity of sand and cement required for moulding blocks in Nigeria. I am going to work through the steps, so that you will be able to make calculations just in case you are using any other size of block.

Initial data
Number of blocks required = 3000 pieces
Recommended production = 1 bag of cement to produce 35 blocks

Step 1: Calculate the volume of the block

For the block size shown above;
Volume of block without holes = (0.45 × 0.225 × 0.225) = 0.02278125 m³
Volume of holes = 2(0.225 × 0.125 × 0.15) = 0.0084375 m³
Therefore volume of the block = 0.02278 – 0.0084375 = 0.0143 m³

Step 2: Calculate the volume of the 35 blocks
If the volume of 1 block is 0.0143, the volume of 35 blocks = (0.0143 × 35) = 0.502 m³

Step 3: Calculate the volume sand required for the 35 blocks
The volume of 1 bag of cement is 34.72 litres = 0.03472m³.
This is obtained by knowing that the mass of 1 bag of cement = 50kg, and the density = 1440 kg/m³

The volume of a standard builder’s wheelbarrow is 0.065 m³ (unheaped). We assume that approximately 2 bags of cement (4 head pans of cement) will fill one builder’s wheelbarrow. Now, we can estimate the number of wheelbarrow trips of sand that the moulder should provide in order to make 35 blocks from one bag of cement.

Total volume of 35 blocks required = 0.502 m³

Let the number of wheel barrow trips of sand be x

Hence, (volume of 1 bag of cement) + (Total volume of sand) = Volume of 35 blocks
Hence, 0.03472 + x(0.065) = 0.502 m³
On solving, x = 7.1889 unheaped wheelbarrow trips of sand
The volume of sand required to make 35 blocks = 7.1889 × 0.065 = 0.46728 m³

However, note that this leads to a mix ratio of about 1:14 which will not be very good when subjected to a compressive strength test. The ideal mix ratio for blocks for proper strength should be about 1:10 or 1 bag to 25 blocks.

Step 4: Calculate the total volume of materials required
We can therefore estimate the quantity of materials to be purchased;

If 1 bag of cement is needed for 35 blocks, therefore 86 bags of cement is needed to mould 3000 blocks (gotten by 3000/35)

If 0.46728 m³ of sand is required to make 35 blocks, therefore, 41 m³ of sand (about 67.60 tonnes assuming the density of dry sand = 1600 kg/m³) is needed to make 3000 blocks.

For a 5 tonne tipper of 3.8 m³ capacity, we have to order for 11 trips of sharp sand.

You have to be very sure of the size of the tipper that will be delivering the sand to you in order to make your estimates more reasonable. On the other hand, you can make the considerations for variations between the weight of materials when wet and when dry. However, this will never be critical for block moulding based on experience. But in concrete it can be very considerable.

Therefore summarily, we need 86 bags of cement and 68 tonnes of sand to mould about 3000 pieces of 9 inches blocks with holes (for 1 bag = 35 blocks).

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Design of Slender Columns According to EC2

By

Ubani Obinna Uzodimma

-

July 24, 2017

Loads from slabs and beams are transferred to the foundations through the columns. In typical cases, columns are usually rectangular or circular in shape. Normally, they are usually classified as short or slender depending on their slenderness ratio, and this in turn influences their mode of failure. Slender columns are likely to fail by buckling than by crushing.

Columns are either subjected to axial, uniaxial, or biaxial loads depending on the location and/or loading condition. Eurocode 2 demands that we include the effects of imperfections in structural design of columns. Column design is covered in section 5.8 of EN 1992-1-1:2004 (EC2).

Slender Columns Requirements in EC2

Clause 5.8.2 of EN 1992-1-1 deals with members and structures in which the structural behaviour is significantly influenced by second order effects (e.g. columns, walls, piles, arches and shells). Global second order effects are more likely to occur in structures with a flexible bracing system.

Column design in EC2 generally involves determining the slenderness ratio (λ), of the member and checking if it lies below or above a critical value λ_lim. If the column slenderness ratio lies below (λ_lim), it can simply be designed to resist the axial action and moment obtained from an elastic analysis, but including the effect of geometric imperfections. These are termed first order effects. However, when the column slenderness exceeds the critical value, additional (second order) moments caused by structural deformations can occur and must also be taken into account.

So in general, second order effects may be ignored if the slenderness λ is below a certain value λ_lim.

λ_lim = (20.A.B.C)/√n ——- (1)

Where:
A = 1/(1 + 0.2ϕ_ef) (if ϕ_ef is not known, A = 0.7 may be used)
B = 1+ 2ω (if ω is not known, B = 1.1 may be used)
C = 1.7 – r_m (if r_m is not known, C = 0.7 may be used)
Where; ϕ_ef = effective creep ratio (0.7 may be used)
ω = A_sf_yd / (A_cf_cd); mechanical reinforcement ratio;
A_s is the total area of longitudinal reinforcement
n = N_Ed / (A_cf_cd); relative normal force
r_m = M₀₁/M₀₂; moment ratio
M₀₁, M₀₂ are the first order end moments, |M₀₂| ≥ |M₀₁|

If the end moments M₀₁ and M₀₂ give tension on the same side, r_m should be taken positive (i.e. C ≤ 1.7), otherwise negative (i.e. C > 1.7). For braced members in which the first order moments arise only from or predominantly due to imperfections or transverse loading r_m should be taken as 1.0 (i.e. C = 0.7).

Also, clause 5.8.3.1(2) says that for biaxial bending, the slenderness criterion may be checked separately for each direction. Depending on the outcome of this check, second order effects (a) may be ignored in both directions, (b) should be taken into account in one direction, or (c) should be taken into account in both directions.

Solved Example

Let us consider the structure shown below. The effects of actions on column member BC is as shown below. It is required to design the column using the following data;

f_ck = 25 N/mm², f_yk = 460 N/mm², Concrete cover = 35mm

Second moment of area of beam AB = (0.3 × 0.6³)/12 = 0.0054 m⁴
Stiffness of beam AB (since E is constant) = 4I/L = (4 × 0.0054) / 6 = 0.0036
Second moment of area of column BC = (0.3 × 0.6³)/12 = 0.0054 m⁴
Stiffness of column BC (since E is constant) = 4I/L = (4 × 0.0054)/7.5 = 0.00288

Remember that we will have to reduce the stiffness of the beams by half to account for cracking;

Therefore, k₁ = 0.00288/0.0018 = 1.6

Since the minimum value of k₁ and k₂ is 0.1, adopt k₁ as 1.6. Let us take k₂ as 1.0 for base designed to resist moment.

Take the unrestrained clear height of column as 7000mm

l_o = 0.5 × 7000√[(1 + (1.6/(0.45 + 1.6))) × (1 + (1.0/(0.45+ 1.0)))] = 6071 mm

Radius of gyration i = h/√12 = 600/√12 = 173.205
Slenderness ratio λ = 6071/173.205 = 35.051

Critical Slenderness for the x-direction
λ_lim = (20.A.B.C)/√n
A = 0.7
B = 1.1
C = 1.7 – M₀₁/M₀₂ = 1.7 – (-210/371) = 2.266

n = N_Ed / (A_c f_cd)
N_Ed = 3500 × 10³ N
A_c = 300 × 600= 180000 mm²
f_cd = (α_cc f_ck)/1.5 = (0.85 × 25)/1.5 = 14.167 N/mm²
n = (3500 × 10³) / (180000 × 14.167) = 1.3725

λ_lim = (20 × 0.7 × 1.1 × 2.266 )/√1.3725 = 29.786

Since 29.786 < 35.051, second order effects need to be considered in the design

Design Moments
M_Bot = 210 KNm, M_Top = 371 KNm
e_i is the geometric imperfection = (θ_i l₀/2) = [(1/200) × (6071/2)] = 15.1775 mm
e_iN_Ed = 15.1775 × 10^-3 × 3500 = 53.121 KNm

First order end moment
M₀₁ = M_Bot + e_iN_Ed = -210 + 53.121 = -156.879 KNm
M₀₂ = M_Top + e_iN_Ed = 371 + 53.121 = 424.121 KNm

Equivalent first order moment M_0Ed
M_0Ed = (0.6M₀₂ + 0.4M₀₁) ≥ 0.4M₀₂ = 0.4 × 424.121 = 169.648 KNm
M_0Ed= (0.6 × 424.121 – 0.4 × 156.879) = 191.721 KNm

Nominal second order moment M₂
Specified concrete cover = 35mm
Diameter of longitudinal steel = 32 mm
Diameter of links = 10 mm

Thus, the effective depth (d) = h – C_nom – ϕ/2 – ϕ_links
d = 600 – 35 – 16 – 10 = 539 mm

1/r₀ = ε_yd/(0.45 d)
ε_yd = f_yd/Es = (460 /1.15) / (200 × 10³) = 0.002
1/r₀ = 0.002/(0.45 × 539) = 8.2457 × 10^-6
β = 0.35 + f_ck/200 – λ/150 (λ is the slenderness ratio)
β = 0.35 + (25/200) – (35.051/150) = 0.2413

Kϕ = 1 + βϕ_ef ≥ 1.0 (ϕ_ef is the effective creep ratio, assume 0.87)
Kϕ = 1 + (0.2413 × 0.87) = 1.2099 ≥ 1.0
Assume K_r = 0.8
1/r = K_r.Kϕ. 1/r₀ = 0.8 × 1.2099 × 8.2457 × 10^-6 = 7.981 × 10^-6
e₂ is the deflection = (1/r) (l₀²) / 10 = 7.981 × 10^-6 × 6071² / 10 = 29.415 mm

M₂ = N_Ed.e₂ = 3500 × 29.415 × 10^-3 = 102.954 KNm

Design Moment M_Ed
M_Ed = maximum of {M_0Ed + M₂; M₀₂; M₀₁ + 0.5M₂}
M_Ed = maximum of {191.721 + 102.954 = 294.675 kNm; 424.121 kNm; -156.879 + (0.5 × -102.954) = -208.356 kNm}

Longitudinal Steel Area
d₂ = C_nom + ϕ/2 + ϕ_links
d₂ = 35 + 16 + 10 = 61 mm

d₂/h = 61/600 = 0.1016
Reading from chart No 2; d₂/h = 0.10;

M_Ed/(f_ck bh²) = (424.121 × 10⁶) / (25 × 300 × 600²) = 0.1571
N_Ed/(f_ck bh) = (3500 × 10³) / (25 × 300 × 600) = 0.777

From the chart, (A_sF_yk)/(bhf_ck) = 0.53
Area of longitudinal steel required (As) = (0.53 × 25 × 300 × 600)/460 = 5185 mm²

Provide 6Y32 + 2Y20 (A_Sprov = 5452 mm²)

A_s,min = (0.1 N_Ed)/f_yd = (0.1 × 3500 × 1000) / 400 = 875mm², 0.002bh = 0.002 × 300 × 600 = 360 mm²
A_s,max = 0.04bh = 0.04 × 300 × 600 = 7200 mm²

Links
Minimum size = 0.25ϕ = 0.25 × 32 = 8mm < 6mm
We are adopting Y10mm as links
Spacing adopted = 300mm less than {b, h, 20ϕ, 400mm} Provide Y10 @ 300 mm links

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Static Analysis of Suspension Bridges: A solved Example

By

Ubani Obinna Uzodimma

-

July 17, 2017

Cables are made from high-strength steel wires that are twisted together. They offer a flexible structural system, which can resist loads only by axial tension. Cables allow engineers to cover very large spans especially in bridges and other suspended structures . Structurally, cables are extremely efficient because they make the most effective use of structural material in that their loads are carried solely through tension through the wire. Therefore, there is no tendency for buckling to occur either from bending or from compressive axial loads.

As load-bearing elements, cables have several unique features. One of them is that vertical loads give rise to horizontal reactions at the support, which, as in case of an arch, is called the thrust. To accommodate the thrust it is necessary to have a supporting structure. It may be a pillar of a bridge, tower, or pylon. Cables are usually employed in suspension bridges, cable-stayed bridges, tower guy wires, roofs, etc.

In this post a simplified static model of suspension bridge is presented for the purpose of structural analysis and design (linear first order analysis). The girder has an internal hinge at point G.

SUSPENSION%2BBRIDGE%2BMANUAL%2BANALYSIS%2BSOLVED%2BEXAMPLE

GEOMETRICAL PROPERTIES

We know that the thrust at the support of the cables = H = Cx = Dx

We can therefore verify that Cy = Dy = Htan⁡α
You can verify that α = tan^-1(4/7) = 29.744°
β = tan^-1(3/7) = 23.198°

SUPPORT REACTIONS

∑M_B = 0
35Ay + 35Cy – 12Cx + 12Dx – (12 × 35²)/2 – (300 × 26) = 0
35Ay + 35Cy – 7350 – 7800 = 0
35Ay + 35(0.5714H) = 15150
35Ay + 20H = 15150 ————– (1)

∑M_G^L = 0
17.5Ay + 17.5Cy – 12Cx – (12 × 17.5²)/2 – (300 × 8.5) + 5H = 0
17.5Ay + 17.5(0.5714H) – 12H + 5H – 1837.5 – 2550 = 0
17.5Ay + 3H = 4387.5 —————– (2)

Solving (1) and (2) simultaneously;
Ay = 172.653 KN; H = 455.537 KN

∑M_A = 0
35By + 35Dy – 12Dx + 12Cx – (12 × 35²)/2 – (300 × 9) = 0
35By + 35(0.5714H) = 10050
35By + 20H = 10050 ———— (3)

∑M_G^R = 0
17.5By + 17.5Dy – 12Dx + 5H – (12 × 17.5²)/2 = 0
17.5By + 17.5(0.5714H) – 12H + 5H = 1837.5
17.5By + 3H = 1837.5 ————– (4)

Solving (3) and (4) simultaneously;
By = 26.939 KN; H = 455.357 KN

Hence, Cy = Dy = H tan⁡α = 455.357 × 0.5714 = 260.195 KN

ANALYSIS OF THE JOINTS OF THE CABLE

Analysis of support C

∑F_x = 0
-455.357 + F_C1 cos29.744 = 0
F_C1 = 455.357/cos⁡29.744 = 524.453 KN

Analysis of joint 1

∑F_x = 0
– F_C1cos⁡α + F₁₂ cos⁡β = 0
-524.453cos29.744 + F₁₂ cos⁡23.198 = 0
F_C1 = (524.453 × cos29.744)/cos⁡23.198 = 495.411 KN

∑F_y = 0
F_C1sin⁡α – F₁₂ sin⁡β – P₁ = 0
(524.453sin29.744) – (495.411sin⁡23.198) – P₁ = 0
P₁ = 65.047 KN

Analysis of joint 2

∑F_y = 0
– F₁₂ sin⁡β – P₂ = 0
– 495.411sin⁡23.198 – P₂ = 0
P₂ = 195.147 KN

Therefore, the equivalent loading on the girder is given below;

INTERNAL STRESSES ON GIRDER

Bending Moment on the girder
M_A = 0
M₁ = (172.635 × 7) – (12 × 7²)/2 = 914.445 KNm
M_E = (172.635 × 9) – (12 × 9²)/2+ (65.047 × 2) = 1197.809 KNm
M₂ = (172.635 × 14) – (12 × 14²)/2 + (65.047 × 7) – (300 × 5) = 196.219 KNm
M_G = (172.635 × 17.5) – (12 × 17.5²)/2 + (65.047 × 10.5) – (300 × 8.5) + (195.147 × 3.5) = 0

Coming from the right
M_B = 0
M₄ = (26.939 × 7) – (12 × 7²)/2 = -105.427 KNm
M₃ = (26.939 × 14) – (12 × 14²)/2 + (65.047 × 7) = -343.525 KNm
M_G = (26.939 × 17.5) – (12 × 17.5₂)/2 + (65.047 × 10.5) + (195.147 × 3.5) = 0

BENDING%2BMOMENT%2BDIAGRAM%2BOF%2BSUSPENSION%2BBRIDGE%2BGIRDER

Shear force on the girder

Q_A = 172.635 KN
Q₁^L = 172.635 – (12 × 7) = 88.635 KN
Q₁^R = 172.635 – (12 × 7) + 65.047 = 153.682 KN
Q_E^L = 172.635 – (12 × 9) + 65.047 = 129.682 KN
Q_E^R = 172.635 – (12 × 9) + 65.047 – 300 = -170.318 KN
Q₂^L = 172.635 – (12 × 14) + 65.047 – 300 = -230.318 KN
Q₂^R = 172.635 – (12 × 14) + 65.047 – 300 + 195.146 = -35.172 KN
Q_G^L = 172.635 – (12 × 17.5) + 65.047 – 300 + 195.146 = -77.172 KN
Q₃^R = 172.635 – (12 × 21) + 65.047 – 300 + 195.146 = -119.172 KN
Q₃^L = 172.635 – (12 × 21) + 65.047 – 300 + 195.146 + 195.146 = 75.974 KN
Q₄^L = 172.635 – (12 × 28) + 65.047 – 300 + 195.146 + 195.146 = -8.026 KN
Q₄^R = 172.635 – (12 × 28) + 65.047 – 300 + 195.146 + 195.146 + 65.047 = 57.021 KN
Q₄^R = 172.635 – (12 × 35) + 65.047 – 300 + 195.146 + 195.146 + 65.047 = -26.939 KN

SHEAR%2BFORCE%2BDIAGRAM%2BOF%2BSUSPENSION%2BBRIDGE%2BGIRDER

Obviously, there are no axial forces in the bridge girder.

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Plastic Collapse Analysis of Propped Cantilever Beams

By

Ubani Obinna Uzodimma

-

July 13, 2017

We all know that for a propped cantilever beam, there are two possible locations of plastic hinges – which are at the span (point of maximum moment) and at the fixed support. For the propped cantilever loaded as shown above, the degree is static indeterminacy is 1.

Since the number of possible location of plastic hinges is 2, therefore the number of independent mechanisms is;

P = 2 – 1 = 1 (which is a beam mechanism)

As a result, we are going to carry out our elastic analysis in two stages, so as to determine the load factor at which the beam will form a mechanism and completely collapse. We will start by assuming a value of unity for our load factor (i.e. λ =1.0)

STAGE 1

It is very easy to verify that for the beam loaded as shown above, the fixed end moment at support A = 3PL/16 = (3 × 267 × 8) / 16 = 400.5 KNm

It is also easy to verify that the vertical support reaction at support C (R_C) = 5P/16 = (5 × 267)/16 = 83.4375 KNm

Therefore, the maximum span moment at point B (M_B) = 83.4375 × 4 = 333.75 KNm
The bending moment is as given below.

We can also obtain the vertical deflection at point B by quickly placing a unit load at point B on a basic system of the structure (could be a cantilever or a simply supported beam). Then by combining the shapes from the two states of loading using Vereschagin’s rule, we can obtain the deflection at point B. This is given below.

EIδB = 1/6 × 4 [2(400.5) – 333.75] × 4 = 1246
δ_B = (1246/EI) metres

Therefore, for support A to become plastic, the load factor λA₁= M_P/M_E = 500.625/400.5 = 1.25
Also for section B to become plastic, the load factor λB₁ = M_P/M_E = 500.625/333.75 = 1.5

It is therefore obvious that the first plastic hinge will develop at support A.

Therefore the load at failure of support A = 1.25 × 267 = 333.75 KN
λA₁M_A = 1.25 × 400.5 = 500.625 KN.m
λA₁M_B = 1.25 × 333.75 = 417.1875 KN.m

Deflection of beam at failure of support A = δ_B= (1.25 × 1246)/EI= (1557.5/EI metres)

STAGE 2
Now, support A is assumed to have failed (formed a plastic hinge). We now model it as a real hinge and carry out another elastic analysis.

From statics, the maximum moment of the structure is given by;

PL/4 = ( 267 × 8)/4 = 534 KNm

For section B to become plastic and form a hinge;
λB₂ = (M_P– λA₁M_B)/M_E = (500.625 – 417.1875) / 534 = 0.15625

Therefore, the total load factor at collapse (λ) = λA₁ + λB₂ = 1.25 + 0.15625 = 1.40625

Also, the load at complete collapse of the beam = 1.40625 × 267 = 375.468 KN

The deflection at collapse = δ_B= (1.40625 × 1246) / EI= (1752.1875/EI) metres

Verification using the static method

From geometrical relations, you can observe that δ_B = 4θ

Internal work done due to rotations of the structure at full plastic moment = M_Pθ + M_P(2θ) = 3M_Pθ (the rotation at section C will not count because it is a natural hinge).

External work done by the collapse load = 375.468 × 4θ = 1501.871θ

But External work done = Internal work done;
Therefore, 3M_P θ = 1501.871θ

Therefore, M_P = 1501.871/3 = 500.624 KNm

This shows that the load factor we obtained from our analysis is correct.

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Design of Doubly Reinforced Beams | Worked Example

By

Ubani Obinna Uzodimma

-

July 11, 2017

Doubly reinforced beams are reinforced concrete beams that incorporate both compression steel (in the compression zone) and tension steel (in the tension zone). This design is typically employed when the required moment capacity of a beam exceeds that which can be provided by singly reinforced beams (those with only tension steel).

In singly reinforced concrete beams without moment redistribution, k = M_Ed/f_ckbd² is limited to k’ = 0.167. Therefore, once the value of k > k’, the beam must be designed as a doubly reinforced beam. In doubly reinforced beams, compression steel reinforcements are required because the concrete section (in compression) by itself cannot develop the required moment of resistance.

However, when there is a moment redistribution in the section, k > k’ where k’ is dependent on the moment redistribution ratio (δ) and is given by:

k’ = 0.6δ – 0.18δ² – 0.21 where δ ≤ 1.0

In all cases, it is recommended that k’ be limited to 0.167 to ensure ductile failure. It is important to understand that the redistribution of moments derived from elastic analysis of a reinforced concrete structure accounts for the plastic behaviour of the material as it approaches its ultimate limit state.

To facilitate this plastic response, concrete sections must be designed to permit the formation of plastic hinges upon the yielding of tensile reinforcement. This approach ensures a ductile structure characterized by a gradual failure at the ultimate limit state, rather than a sudden, catastrophic collapse due to compressive failure of the concrete.

To guarantee the formation of plastic hinges and the necessary section rotation for ductile behaviour, Eurocode 2 imposes limitations on the maximum neutral axis depth, x_bal. These restrictions ensure that the tensile steel experiences high strains and this promotes the desired plastic response.

The permissible value of x_bal is determined based on the desired degree of moment redistribution. According to Eurocode 2, the limit is x_bal≤ 0.8(δ – 0.44)d for f_ck ≤ C50. However, this limit was modified to x_bal as x_bal ≤ (δ – 0.4)d in the UK Annex to the EC2.

Design of Doubly Reinforced Beam

To ensure a ductile response and prevent sudden compressive failure of the concrete, it is required that the neutral axis depth should not exceed 0.45 times the effective depth (0.45d). This guideline is typically adopted in the design of sections incorporating compression reinforcement (doubly reinforced beam sections).

Rectangular stress block for doubly reinforced beam

In the design of reinforced concrete beams, if the design ultimate moment is greater than the ultimate moment of resistance i.e. M_Ed > M_lim, then compression reinforcement is required. Provided that d₂/x ≤ 0.38 (i.e. compression steel has yielded) where d₂ is the depth of the compression steel from the compression face and x = (d − z)/0.4

The area of steel in compression, A_s2, is given by:
A_s1 = (M_Ed – M_lim) / (f_sc(d – d₂)) —- (1)
f_sc= E_sε_sc ≤ 0.87f_yk
ε_sc = steel compressive strain = 0.0035(x – d’)/x
E_s = Modulus of elasticity of steel = 200000 N/mm²

Area of tension reinforcement:
A_s1 = M_lim / (0.87f_ykz) + A_s2 (f_sc/0.87f_yk) —- (2)

Where:
M_Ed = Design bending moment of the section
M_lim= 0.167f_ckbd²
z = d[0.5+ √(0.25 – 0.882k’)]
where
k’ = 0.167

Design Example of a Doubly Reinforced Beam

To show how this is done, let us consider the flexural design of support A of the beam loaded as shown below. The depth of the beam is 600 mm, and the width is 400 mm.
f_ck= 35 N/mm²; f_yk= 500 N/mm²; concrete cover = 40mm

DOUBLY%2BREINFORCED%2BBEAM%2B%2528QUESTION%2529

DOUBLY%2BREINFORCED%2BBEAM%2B%2528BMD%2529

Top reinforcement (Hogging moment)
Support A (No moment redistribution)
Effective depth (d) = 600 – 40 – 16 – 10 = 534 mm
d₂= 40 + 8 + 10 = 58 mm

M_Ed = 761.24 kNm

But since the flange is in tension, we use the beam width to calculate the value of k (this applies to all support hogging moments)
k = M_Ed / (f_ckb_wd²) = (761.24 × 10⁶) / (35 × 400 × 534²) = 0.1906
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d = 0.82 × 534 = 437.88 mm
x = (d − z)/0.4 = (534 − 437.88)/0.4 = 240.3 mm

d₂/x = 58/240.3 = 0.241 < 0.38 (therefore reinforcement bar has yielded and f_sc= 0.87f_yk)

Since k < 0.167, compression is required
Area of compression reinforcement A_S2 = (M_Ed – M_lim) / (0.87f_yk (d – d₂))

M_lim = 0.167f_ckbd² = (0.167 × 35 × 400 × 534²) × 10^-6 = 666.69 kNm

A_S2 = ((761.24 – 666.69) × 10⁶) / (0.87 × 500 × (534 – 58)) = 457 mm²
Provide 3H16 Bottom (A_sprov = 603 mm²)

Area of tension reinforcement A_s1 = M_Rd / (0.87f_yk z) + A_S2
Where z = d[0.5+ √(0.25 – 0.882K’)]
K’ = 0.167
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d

A_s1 = M_Rd / (0.87f_yk z) + A_S2 = ( 666.69 × 10⁶) / (0.87 × 500 × 0.82 × 534) + 457 mm² = 3957 mm²

Provide 3H32mm + 4H25 mm Top (A_Sprov = 4376 mm²)

Comparison of Shear Design to EC2 and BS 8110

By

Ubani Obinna Uzodimma

-

July 11, 2017

Eurocode 2 uses the variable strut inclination method for shear design. It is slightly more complex than the procedure in BS 8110 but can result in savings in the amount of shear reinforcement required.

Like BS 8110, Eurocode 2 models shear behaviour using the truss analogy in which the concrete acts as the diagonal struts, the stirrups act as the vertical ties, the tension reinforcement forms the bottom chord, and the compression steel/ concrete forms the top chord. Whereas in BS 8110 the strut angle has a fixed value of 45°, in EC2 it can vary between 21.8° and 45° and it is this feature which is responsible for possible reductions in the volume of shear reinforcement.

The concrete strut angle changes with respect to the magnitude of the shear force applied. The angle decreases with a reduction in shear stress, with a minimum angle of 21.8° (where cot = 2.5) which happens to be the most frequently used design strut angle.

Shear design in Eurocode 2 often involves comparing the design shear stress (v_Ed) to the concrete shear resistance (v_Rd,c). If v_Ed< v_Rd,c then in the case of one-way spanning slabs no shear reinforcement is required; in the case of beams nominal shear reinforcement is required. v_Rd,cis dependent on the amount of design tensile reinforcement and the design concrete strength.

If v_Ed < v_Rd,max (ie v_{Rd,cot θ = 2.5}), then the shear reinforcement may be calculated as:
A_sw/s ≥ v_Ed/(0.87 × f_yk × z × cot 2.5)

If v_Ed > v_Rd,max (ie v_{Rd,cot θ = 2.5}), then checks against v_{Rd,cot θ = 1.0}and determination of θ are required.
If v_Ed> v_{Rd,cot θ = 1.0} then a larger section or greater concrete strength is required.

Where applied shear stresses are higher, one has to check against the maximum allowable shear stress at the maximum strut angle of θ = 45º (e.g. 5.28 MPa for a C30/37 concrete when cot θ = 1.0) and where necessary determine the actual intermediate strut angle, θ.

Solved Example for Shear Design

For the beam loaded as shown below and disregarding load factors, we are going to obtain the shear reinforcement required for section A. The shear force at the centreline of support has been adopted for this design, and the area of tension steel provided at this section A_s1 = 4825 mm².

Design Data:
Concrete Strength = 35 N/mm²
Grade of steel = 460 N/mm²
Width of beam = 400mm
Depth of beam = 600mm
Effective depth = 543mm
Area of tension steel provided at section A_s1 = 4825 mm²

Shear Design to EC2

Support A; V_Ed = 500.46 kN

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

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/543) = 1.606 > 2.0, therefore, k = 1.606
V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.606)^1.5 × 35^0.5 = 0.421 N/mm²
ρ₁ = A_s/bd = 4825/(400 × 543) = 0.022 > 0.02; Therefore take 0.02

σ_cp = N_Ed/Ac < 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.) Take N_Ed = 0

V_Rd,c = [0.12 × 1.606 (100 × 0.02 × 35 )^(1/3)] 400 × 543 = 172511.992 N = 172.511 kN

Since V_Rd,c (172.511 kN) < V_Ed (500.46 kN), shear reinforcement is required.
The compression capacity of the compression strut (V_Rd,max) assuming θ = 21.8° (cot θ = 2.5)

V_Rd,max = (b_w.z.v₁.f_cd) / (cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc f_ck) / γ_c = (0.85 × 35) / 1.5 = 19.833 N/mm²
Let z = 0.9d

V_Rd,max= [(400 × 0.9 × 543 × 0.516 × 19.833) / (2.5 + 0.4)] × 10^-3 = 689.83 kN

Since V_Rd,c < V_Ed < V_Rd,max
Hence A_sw / S = V_Ed / (0.87.F_yk.z.cot θ) = 500460 / (0.87 × 460 × 0.9 × 543 × 2.5 ) = 1.0235

Minimum shear reinforcement;
A_sw / S = ρ_w,min × b_w × sinα (α = 90° for vertical links)
ρ_w,min = (0.08 × √(F_ck)) / F_yk = (0.08 × √35) / 460 = 0.001028
A_sw/S_min = 0.001028 × 400 × 1 = 0.411
Maximum spacing of shear links = 0.75d = 0.75 × 543 = 407.25

Provide X10mm @ 150mm c/c as shear links (A_sw/S = 1.0467) Ok!!!!

Shear Design to BS 8110-1:1997

Design of support A
Ultimate shear force at the centerline of support
V = 500.46 kN

Using the shear force at the centreline of support;
Shear stress v = V/(bd ) = (500.46 × 10³) / (400 × 543) = 2.304 N/mm²

v(2.304 N/mm² ) < 0.8√f_cu (4.732 N/mm² ). Hence, the dimensions of the cross-section are adequate for shear.

Concrete resistance shear stress
v_c = 0.632 × (100A_s/bd)^1/3 (400/d)^1/4
(100A_s/bd) = (100 × 4825) / (400 × 543) = 2.221 < 3 (See Table 3.8 BS 8110-1;1997)
(400/d)^1/4 = (400/543)^1/4 = 0.926; But for members with shear reinforcement, this value should not be less than 1. Therefore take the value as 1.0

v_c = 0.632 × (2.221)^1/3 × 1.0 = 0.8245 N/mm²
For f_cu = 35 N/mm², v_c = 0.8245 × (35/25)^1/3 = 0.922 N/mm²

Let us check;
(V_c + 0.4)1.322 N/mm² < v(2.304 N/mm²) < 0.8√f_cu (4.732 N/mm²)

Therefore, provide shear reinforcement links.
Let us try 2 legs of Y10mm bars (Area of steel provided = 157 mm²

A_sv/Sv = (0.95 × F_yv) / (b_v (v – v_c)) = [400 × (2.304 – 0.922)] / (0.95 × 460) = 1.264

Maximum spacing = 0.75d = 0.75 × 543 = 407.25 mm
Provide Y10mm @ 100 mm c/c links as shear reinforcement (A_sv/Sv = 1.57)

We can therefore see that disregarding load factor and flexural design requirements, EC2 is more economical than BS 8110 in shear design, and in this case study by about 19%.

Free MATLAB Code for Flexural Design of R.C. Sections According to EC2

By

Ubani Obinna Uzodimma

-

July 5, 2017

In this post, I am going to present an extremely simplified MATLAB code for carrying out flexural design of reinforced concrete beams (rectangular or flanged) according to Eurocode 2. Before you can use this code, you must carry out your analysis and obtain your design forces and moments. You can just copy the code and paste on your MATLAB script. There are no external functions to be called for the program.

% ANALYSIS AND DESIGN OF RECTANGULAR AND FLANGED BEAMS PER EUROCODE 2
% THIS PROGRAM WILL CALCULATE THE AREA OF REINFORCEMENT REQUIRED, AND ALSO DO THE DEFLECTION VERIFICATION USING THE DEEMED TO SATISFY RULES OF EC2

clc
disp(‘THIS PROFORMA WAS WRITTEN BY O.U.R UBANI’)
disp(‘DOWNLOADED FROM WWW.STRUCTVILLE.COM’)
disp(‘ANALYSIS AND DESIGN OF BEAMS PER EUROCODE 2’)

% INPUT MATERIAL PROPERTIES
Fck = input(‘Enter the grade of concrete (N/mm^2)Fck = ‘);
Fyk = input(‘Enter the yield strength of steel (N/mm^2)Fyk = ‘);

% INPUT DESIGN MOMENT
MEd = input(‘Enter the ultimate design moment (KNm) MEd = ‘);

% INPUT SECTION AND DESIGN PROPERTIES
h = input(‘Enter the depth of beam (mm)h = ‘);
b = input(‘Enter the effective flange width of the beam (mm)b = ‘);
bw = input(‘Enter the beam width (mm)bw = ‘);
Cc = input(‘Enter concrete cover (mm) = ‘);
dr = input(‘Enter the diameter of reinforcement (mm) = ‘);
dl = input(‘Enter the diameter of links = ‘);

% CALCULATION OF EFFECTIVE DEPTH
disp(‘Effective depth d (mm)’)
d = h-Cc-(dr/2)-dl % Effective depth
disp(‘Depth of reinforcement from the face of concrete’)
do = Cc+dl+(dr/2)

% ANALYSIS FOR INTERNAL STRESSES
ko = 0.167;
k = (MEd*10^6)/(Fck*b*d^2)
if k>ko
disp(‘Since k > ko, Compression reinforcement is required’)
Mcd = (Fck*b*d^2*(k-ko))*10^(-6)
As2 = (Mcd*10^6)/(0.87*Fyk*(d-do))
z = 0.5*d*(1+sqrt(1-3.53*ko))
As1 = ((ko*Fck*b*d^2)/(0.87*Fyk*z))+ As2
else
disp(‘Since k < ko, No Compression reinforcement is required’)
disp(‘Lever arm (la)’)
la = 0.5+sqrt(0.25-0.882*k)
if la>0.95
disp(‘Since la > 0.95,’)
la = 0.95
else
la = 0.5+sqrt(0.25-0.882*k)
end
As1 =(MEd*10^6)/(0.87*Fyk*la*d)
end

% MINIMUM AREA OF STEEL REQUIRED
fctm = 0.3*(Fck^(2/3)) %MEAN TENSILE STRENGTH OF CONCRETE (TABLE 3.1 EC2)
ASmin = 0.26*(fctm/Fyk)*bw*d
if ASmin < 0.0013*bw*d
ASmin = 0.0013*bw*d
end
if As1<ASmin
As1 = ASmin
else
disp(‘Since As1 > Asmin, provide As1 which is the area of steel required’)
end

Asprov1 = input(‘Enter area of tension steel provided (mm^2) = ‘);
Asprov2 = input(‘Enter area of compression steel provided if any (mm^2) = ‘);

% CHECK FOR DEFLECTION

disp(‘DO YOU WANT TO CHECK FOR DEFLECTION?’)

G = input(‘ENTER (1) FOR YES OR (0) FOR NO = ‘);

if G == 1;

disp(‘CHECK FOR DEFLECTION’)

disp(‘BASIC SPAN/EFFECTIVE DEPTH RATIO (K)’)

disp(‘CANTILEVER = 0.4’)

disp(‘SIMPLY SUPPORTED = 1.0’)

disp(‘SIMPLY SUPPORTED AND FIXED AT ONE END = 1.3’)

disp(‘FIXED AT BOTH ENDS = 1.5’)

K = input(‘Enter the selected value of k = ‘);

L = input(‘Enter the deflection critical length of the member (mm) = ‘);

disp(‘Fs is the stress in tensile reinforcement under service loading’)

Fs = (310*Fyk*As1)/(500*Asprov1)

disp(‘Bs = 310/Fs’)

Bs = 310/Fs

disp(‘P is the reinforcement ratio in the section (Asprov1)/(b*d)’)

P = (Asprov1)/(b*d)

disp(‘Po = sqrt(Fck)/1000’)

Po = sqrt(Fck)/1000

P1 = (Asprov2)/(b*d)

if

P <= Po

I_deflection = K*(11+(1.5*sqrt(Fck)*(Po/P)) + 3.2*sqrt(Fck)*((Po/P)-1)^1.5)

else

I_deflection = K*(11+(1.5*sqrt(Fck)*(Po/(P-P1))) + 3.2*sqrt(Fck)*((Po/P)-1)^1.5)

end

CV = input(‘Is the beam is flanged? (1) for YES and (0) for NO = ‘)

if CV == 1

disp(‘S is the ratio of beff/bw’)

S = b/bw

bn = (11-(b/bw))/10

if S >= 3.0;

K_deflection = Bs*0.8* I_deflection

else

K_deflection = Bs*bn* I_deflection

end

if L>7000

Limiting_deflection = (7000/L)*K_deflection

else

Limiting_deflection = K_deflection

end

else

Limiting_deflection = Bs*I_deflection

end

Actual_deflection = L/d

if

Limiting_deflection > Actual_deflection

disp(‘Deflection is satisfactory’)

else

disp(‘DEFLECTION IS NOT SATISFACTORY !!!!! Increase depth of beam, or increase area of steel, or both. Then rerun proforma’)

end

2.0 MATLAB CODE FOR SHEAR DESIGN ACCORDING TO EC2

% SHEAR DESIGN IN EUROCODE 2
clc
disp(‘SHEAR DESIGN ACCORDING TO EC2′)
disp(‘THIS PROFORMA WAS WRITTEN BY O.U.R. UBANI’)

% MATERIALS PROPERTIES
Fck = input(‘Enter the grade of concrete (N/mm^2) = ‘);
Fyk = input(‘Enter the yield strength of steel (N/mm^2) = ‘);

VEd = input(‘Enter the value of shear force at ULS (KN) = ‘);

% SECTION PROPERTIES
h = input(‘Enter the depth of beam (mm) = ‘);
bw = input(‘Enter the beam width (mm) = ‘);
Cc = input(‘Enter concrete cover (mm) = ‘);
dr = input(‘Enter the diameter of reinforcement (mm) = ‘);
dl = input(‘Enter the diameter of links = ‘)
d = h-Cc-(dr/2)-dl % Effective depth
do = Cc+dl+(dr/2)

% CALCULATION OF THE SHEAR CAPACITY OF THE SECTION WITH NO SHEAR REINFORCEMENT

Asprov1= input(‘Enter the area of steel provided in the shear zone (mm^2) = ‘) % The reinforcement must exceed the design anchorage length by at least the effective depth
Crd = 0.12
k1= 1+sqrt(200/d)
if k1>2
k1=2
end
disp(‘Reinforcement ratio’)
P1 = (Asprov1/(bw*d))
if P1>0.02
P1 = 0.02
end

% THIS SECTION IS TO BE CONSIDERED IF THERE IS AXIAL FORCE IN THE SECTION
disp(‘Axial force in the section’)
N = input(‘Enter the value of AXIAL FORCE IF ANY(+VE FOR COMP, AND -VE FOR TENSION)(kN) = ‘)
disp(‘Axial stress in the section (Ds)’)
Ds = (N*1000)/(bw*h)
if Ds > (0.2*0.85*Fck)/1.5
Ds = (0.2*0.85*Fck)/1.5
end
k2 = 0.15;
disp(‘Minimum shear stress in the section (N/mm^2)’)
Vmin = 0.035*k1^(1.5)*sqrt(Fck)
disp(‘Concrete resistance shear stress (VRd) (N/mm^2)’)
VRdQ = ((Crd*k1)*(100*P1*Fck)^(1/3)*(bw*d))/1000 % SHEAR FORCE CONTRIBUTION
VRdN = ((k2*Ds)*(bw*d))/1000 % AXIAL FORCE CONTRIBUTION
VRd = VRdQ + VRdN % TOTAL SHEAR RESISTANCE
if VRd < ((Vmin+(k2*Ds))*(b*d))/1000
VRd = ((Vmin+(k2*Ds))*(b*d))/1000
end
if VRd>VEd
disp(‘Since VRd > VEd’)
disp(‘NO SHEAR REINFORCEMENT REQUIRED, PROVIDE NOMINAL LINKS’)
else
if VRd<VEd
disp(‘Since VRd < VEd’)
disp(‘SHEAR REINFORCEMENT REQUIRED, CALCULATE COMPRESSION STRUT CAPACITY’)
disp(‘Assume strut angle (thetha) = 21.8 deg, cot(thetha)= 2.5’)
thetha = 21.8
V1 = 0.6*(1-(Fck/250))
disp(‘Design compressive strength of concrete (fcd) (N/mm^2)’)
fcd = 0.567*Fck
z = 0.9*d
disp(‘Maximum capacity of compression strut (KN)’)
VRDmax = ((b*z*V1*fcd)/(2.9))/1000
if VRDmax > VEd
disp(‘Since VRDmax > VEd’)
disp(‘OK! Calculate diameter and spacing of links’)
ASMINlinks_to_spacing = ((0.08*sqrt(Fck))/Fyk)*b
ASlinks_to_spacing = (VEd*1000)/(z*0.87*Fyk*2.5)
else
disp(‘Since VRDmax > VEd, it means we need a higher strut angle (beta)’)
disp(‘By calculating the strut angle’)
disp(‘Shear stress at the section (N/mm^2)’)
v = (VEd*1000)/(b*d)
beta = (0.5)*asind((v)/(0.153*Fck*(1-(Fck/250))))
if beta>45
disp(‘Since beta < 45 deg, SECTION INADEQUATE FOR SHEAR, INCREASE DEPTH !!!!!’)
else
disp(‘The ratio of Area of steel/spacing of links’)
ASlinks_to_spacing = (VEd*1000)/(z*0.87*Fyk*cotd(beta))
end
end
end
Area_of_legs = input(‘Enter the area of number of legs selected = ‘)
Spacing = input(‘Enter the spacing = ‘)
disp(‘CHECK’)
T = Area_of_legs/Spacing
if T > ASlinks_to_spacing
disp(‘Shear reinforcement is ok’)
else
disp(‘Increase area of steel, or reduce spacing’)
end
end

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ALSO IN THIS BLOG
Analysis and Design of Continuous R.C. Raker Beam for Stadium using EC2
Two Approaches to the Calculation of Deflection in EC2; Choose a Faster Approach for Yourself

Two Approaches to the Calculation of Deflection in Eurocode 2

By

Ubani Obinna Uzodimma

-

July 5, 2017

The verification of deflection is an important serviceability check in reinforced concrete design. Most designs codes adopt the ‘deemed to satisfy’ rules which limit the deflection to an acceptable depth using Basic Span/Effective depth ratio. This is generally more frequently employed than carrying out the rigorous exact calculation of deflection which is however not
very reliable.

Eurocode 2 has its own set of deemed to satisfy rules, and there are two approaches to calculation of deflection which yields the same result. It is not like that they are two different formulars, but they are just one formular expressed in two different ways. This post shows the two approaches, so that you can select the one you deem fast enough for carrying out your manual calculations.

According to Clause 7.4.2 of EN 1992-1-1, we can verify deflections in the absence of exact calculations using deemed to satisfy basic span/effective depth (limiting deflection to depth/250);

Lateral Deflection of Multi-Storey Buildings Under the Action of Wind Load

(a) BY USING THE FIRST APPROACH
Actual L/d 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]^(3⁄2) ] if ρ ≤ ρ₀ ———– (1)

L/d = K [11 + 1.5√(f_ck) ρ₀/(ρ – ρ’) + 1/12 √(f_ck) (ρ₀/ρ)^(1⁄2) ] if ρ > ρ₀ ———— (2)
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 below;

Table%2Bof%2Bbasic%2Bspan%2Bto%2Beffective%2Bdepth%2Bratio%2BEC2

β_s = (500 As_prov)/(F_yk As_req) ———- (3)

(b) BY USING THE SECOND APPROACH
L/d ≤ Basic ratio × α_s × β_s
Where the basic ratio = 20K

Modification factor α_s which depends on the concrete strength f_ck and reinforcement percentages, is given by:

For 100A_s/bd < 0.1f_ck^0.5;
α_s = 0.55 + 0.0075f_ck/(100A_s/bd) + 0.005f_ck^0.5[ f_ck^0.5/ (100A_s/bd) – 10]^1.5 —————- (4)

For 100A_s/bd ≥ 0.1f_ck^0.5;
α_s = 0.55 + 0.0075f_ck/[100(A_s – A_s’)/bd) + 0.013f_ck^0.25 (100A_s’/bd)]^0.5 (for doubly-reinforced sections) —————- (6)

α_s = 0.55 + 0.0075f_ck/(100A_s/bd) (for singly-reinforced sections) —————- (7)

Modification factor β_s = 310/σ_s —————- (8)
Where:
σ_s is the stress in the tension reinforcement under the characteristic loading. It will normally be conservative to assume;
β_s = (500/f_yk)(A_s,prov/A_s,req) for values of (A_s,prov/A_s,req)≤ 1.5 —————- (9)

Important hints on the calculation of deflection using both approaches;

(1) For flanged sections with b/b_w ≥ 3, the basic ratios for rectangular sections should be multiplied by 0.8. For values of b/b_w < 3, the basic ratios for rectangular sections should be multiplied by (11 – b/b_w)/10

(2) The ratio should be based on the shorter span for two-way spanning slabs, and the longer span for flat slabs.

(3) For beams and slabs, other than flat slabs, with spans exceeding 7 m, which support partitions liable to be damaged by excessive deflections, the basic ratio should be multiplied by 7/span. For flat slabs, where one or both spans exceeds 8.5 m, which support partitions liable to be damaged by excessive deflections, the basic ratio should be multiplied by 8.5/span.

SOLVED EXAMPLE

For the example above we check the deflection for span D-E.
f_ck = 35 N/mm²; f_yk = 460 N/mm²; Effective depth = 840mm; Effective width of flange (B_eff) = 1650mm; A_s,req = 1850 mm²; A_s,prov = 2101 mm²
BY APPROACH 1

ρ = A_s,prov /B_effd = 2101 / (1650 × 840) = 0.0015158;
ρ₀ = reference reinforcement ratio = 10^-3√(f_ck) = 10^-3√(35) = 0.005916
Since if ρ ≤ ρ₀;

L/d = K [11 + 1.5√(f_ck) ρ₀/ρ + 3.2√(f_ck) (ρ₀ / ρ – 1)^(3⁄2)
k = 1.3 (One end continuous)

L/d = 1.3 [11 + 1.5√(35) × (0.005916/0.001516) + 3.2√(35) × [(0.005916 / 0.001516) – 1]^(3⁄2)
L/d = 1.3[11 + 34.630 + 93.608] = 181.00

β_s = (500 As_prov)/(F_yk As_req) = (500 × 2101) / (460 × 1850 ) = 1.234

Check b/b_w = 1650/300 = 5.5
Since b/b_w ≥ 3, multiply L/d by 0.8
Since span is greater than 7m, also multiply by 7/span

Therefore limiting L/d = 1.234 × 0.8 × (7/8) × 181.04 = 156.382
Actual L/d = 8000/840 = 9.523

Since Actual L/d < Limiting L/d, deflection is satisfactory.

BY APPROACH 2

100A_s/bd = (100 × 2101) / (1650 × 840) = 0.1516
Check 0.1f_ck^0.5 = 0.1 × (35)^0.5 = 0.5916

Since For 100A_s/bd < 0.1f_ck^0.5

α_s = 0.55 + 0.0075f_ck/(100A_s/bd) + 0.005f_ck^0.5[ f_ck^0.5/ (100A_s/bd) – 10]^1.5
α_s = 0.55 + [(0.0075 × 35) /(0.1516)] + [0.005 × (35)^0.5 × [(35^0.5/0.1516) – 10]^1.5 = 0.55 + 1.7315 + 4.625 = 6.9065

Basic ratio = 20k = 20 × 1.3 = 26

β_s = (500/f_yk)(A_s,prov/A_s,req) = (500/460) × (2101/1850) = 1.234

Check b/b_w = 1650/300 = 5.5
Since b/b_w ≥ 3, multiply L/d by 0.8
Since span is greater than 7m, also multiply by 7/span

Therefore limiting L/d = 1.234 × 0.8 × (7/8) × 26 × 6.9065 = 155.111
Actual L/d = 8000/840 = 9.523

You can verify that the two methods can be used, with the second method being more approximate.

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