How to Design Concrete Structures Using Eurocode 2 Beams

CONTENTS

• PROPERTIES OF CONCRETE FOR USE IN EUROCODE 2
  • [1] Compressive Strength
  • [2] Tensile Strength
• STEEL AND REINFORCEMENT
• SELF WEIGHT OF COMMON CONSTRUCTION MATERIALS
• GENERAL ARRANGEMENT (G.A.)
  • My style of General Arrangement (GA)
  • MEMBER DATA
• SECTION ANALYSIS AND DESIGN FORMULAE
  • [1] Analysis of singly reinforced section
  • [2] Analysis of doubly reinforced sections
• ANALYSIS AND DESIGN OF THE FLOOR SLABS
  • LOAD ANALYSIS
    • Permanent Loads on Slab
    • Variable Loads on Slab
  • PANEL 1
  • Checking for Deflection
  • PANEL 2
  • PANEL 3
  • PANEL 4
  • PANEL 5
  • PANEL 6
  • PANEL 7
  • PANEL 8
  • PANEL 9
  • PANEL 10 (1st floor Kitchen Balcony)
• ANALYSIS AND DESIGN OF THE FLOOR BEAMS
  • LOADING OF BEAMS
    • SPAN 1 – 3
    • SPAN 3 – 5
    • SPAN 5 – 7
    • SPAN A – C
    • SPAN C – D
    • SPAN D – F
  • BEAM ANALYSIS
    • SPAN 1 – 7
    • Using Claperon's 3-Moment Equation
  • STRUCTURAL DESIGN
    • SPAN 1 -3 (L-beam)
    • Checking for deflection
    • Top reinforcements (Hogging moment)
    • SPAN A-B (L – Beam)
    • SPAN C-D
    • SPAN D – F (Rectangular beam)
• ANALYSIS AND DESIGN OF THE COLUMNS
  • LOAD ANALYSIS
    • Load from Roof
    • Column Axial Loads (ground floor columns)
    • Column Self Weight
    • COLUMN A1
    • COLUMN A3
    • COLUMN A5
    • COLUMN A7
    • COLUMN C1
    • COLUMN C3
    • COLUMN B5
    • COLUMN B7
    • COLUMN B8
    • COLUMN D1
    • COLUMN D2
    • COLUMN D4
    • COLUMN E5
    • COLUMN E8
    • COLUMN F1
    • COLUMN F2
    • COLUMN F4
    • COLUMN G4
    • COLUMN G6
    • COLUMN G8
  • COLUMN ANALYSIS FOR MOMENT (ULS)
  • STRUCTURAL DESIGN
    • COLUMN E5
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In order to fully present the simplified processes involved in the design of a simple reinforced concrete residential building, the case study of the building with the approach elevation shown below has been considered for full structural analysis and design. The building is a relatively small one required to fit into a 10m x 15m plot of land.

PROPERTIES OF CONCRETE FOR USE IN EUROCODE 2

In the design of concrete structures, engineers have the flexibility to specify a particular concrete type(s) aimed at meeting the specific performance requirements for their project. For instance, where calculated deflections exceed serviceability limits, the designer can achieve lower deflections by increasing the class of concrete and the associated modulus of elasticity, rather than by resizing members.

[1] Compressive Strength

In BS EN 206-1: Concrete – Specification, performance, production and conformity, compressive strength is expressed as a strength class. BS EN 1992-1-1 uses the characteristic compressive cylinder strength fck (based on 2:1 cylinder) as the basis of design calculations. Normally, 28 days compressive strength is specified, but in some cases, higher days strength can be specified. In the UK, and as well as in Nigeria, the compressive strength is usually tested using cubes (usually 150mm x 150mm) rather than cylinders. Higher strength is obtained for cubes because the test machine platens offer greater lateral restraint due to the lower aspect ratio. In BS EN 206-1, the 2:1 cylinder strength is taken to be about 20% less than the cube strength for normal structural concrete but with higher strength classes, the cylinder strength achieves a higher proportion of the cube strength. To accommodate these differences, the strength class is defined by both the cylinder and the cube strength (for example, C30/37 Ccyl/cube).

The design compressive strength (f_cd ) for concrete according to BS EN 1992-1-1 is given by;

f_cd = (α_cc*f_cK)/γ_c

where;

f_cK  =  characteristic cylinder compressive strength of concrete at 28 days

γ_c  =  partial (safety) factor for concrete

α_cc  =  a coefficient taking account of long-term effects on the compressive strength (which is reduced under sustained load) and unfavourable effects resulting from the way the load is applied.

[2] Tensile Strength

In design, tensile strength is used in both serviceability and ultimate limit state calculations. For example in considerations of cracking, deflection, shear, punching shear, bond and Anchorage, in the evaluation of the cracking moment for prestressed elements, etc.

The design tensile strength of concrete,f_ctd , according to BS EN 1992-1-1 is taken as:

f_ctd = (α_ct*f_{ctk 0.05})/γ_c

f_{ctk 0.05}  = characteristic tensile strength of concrete at 28 days

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.

STEEL AND REINFORCEMENT

Reinforcement for concrete generally consists of deformed steel bars or welded steel mesh fabric. Normal reinforcement relies entirely upon the alkaline environment provided by a durable concrete cover for its protection against corrosion. Under special circumstances, galvanised, epoxy-coated or stainless steel can be used.

Idealised and design stress-strain diagrams for reinforcing steel

The grade of reinforcement steel denotes the specified characteristic yield stress (fyk). This is obtained by dividing the characteristic yield load by the nominal cross-section area of the bar. For products without pronounced yield stress, the 0.2% proof stress f0.2k is substituted as the yield stress. The idealised and design stress-strain diagrams for reinforcing steel (for tension and compression) according to the stress-strain diagram above.

In general, ductility is inversely related to yield stress. Therefore, in applications where ductility is critical, it is important to ensure that the actual yield strength does not exceed the specified value by a large margin. Ductility is defined using the strain at maximum load (εuk) and the ratio between the maximum and the yield strengths (ft/fyk). Ductility is an essential property if the advantage is to be taken off the plastic behaviour of structures. The greater the ductility, the greater the elongation in axially loaded members, and the greater the rotation capacity in members subjected to flexure. In members where the ultimate strength is governed by the yielding of reinforcement (shallow members with a low percentage of steel), the higher the value of strain at ultimate load, the greater the ductility.

According to Annex C of EN 1992-1-1:2004 (E), the design rules in the diagram above are applicable to steel reinforcement with characteristic yield strength in the range 400 – 600 N/mm2. Details of the actual yield strength of steel available in the UK for the reinforcement of concrete can be found in BS 4449: 2005. This document indicates that steel reinforcement will now be manufactured in three grades, all of 500 N/mm2 characteristic yield strength, but with differing ductility.

SELF WEIGHT OF COMMON CONSTRUCTION MATERIALS

The self-weight of some commonly encountered construction materials are stated HERE

GENERAL ARRANGEMENT (G.A.)

In order to carry out a design properly, the engineer should be able to adequately idealise the structure so as to try and obtain the closest theoretical behaviour of the structure. This interpretation is usually made from standard and well prepared architectural drawing of the proposed building. The architectural drawing enables the engineer to prepare what is normally referred to as the 'general arrangement' of the building, popularly called the 'G.A.'. The G.A. clearly specifies the disposition of the structural elements such as the columns, beams, and the panelling of the floor slabs. The G.A. also contains the labelling of the axes and the members, based on the grid lines. After completing the GA, the engineer makes preliminary sizing of the structural elements which may be governed by past experience or by deflection requirements based on the code of practice. After the sizing, the engineer is faced with the challenge of loading the structure. But let us briefly review how we go about the G.A.

There are no spelt out rules about how to select the appropriate general arrangement of a structure. To me, adequate presentation of the general arrangement has more to with years of design experience. However, I am going to highlight some important guidelines which are very necessary.

[1] Respect the architect's original disposition

For instance, when arranging your columns, do not place columns where the architect has meant to be a free space, nor should the structural elements interrupt the interaction of spaces. Also, your columns and beams should not project out where the architect has intended plane walls or sections etc. Just make your arrangement consistent with the original form of the structure.

[2] Select a stable model

The model or general arrangement you are presenting should be statically stable, and fully representative of the behaviour of the structure.

[3] Buildability and construction consequences

The model you adopt should be buildable. Consider the technical capacities of the local contractors that will execute the design from your model. For instance, in a region where reinforcement bending machines and cranes are not available, why should you provide models that will require the provision of Y40mm bars, or recommend the use of precast elements?

[4] Know the economic and structural consequences

Between more steel and more concrete, which one is more economical? For instance, let us consider an external beam that is 8m long. If there are no openings at the wall panels under the mid-span of the beam, you can comfortably hide an intermediate column there, thereby having two spans 4m each. The original 8m beam will require more steel reinforcements, or deeper sections, or both to satisfy ultimate and serviceability limit state requirements. However, if you introduce a column at the mid-span, there will be a redistribution of stresses, with a hogging moment at the propped mid-span, and hence, generally lesser reinforcements and concrete sections. However, now note that you are going to construct a new column and new isolated base (requiring concrete, reinforcement, and additional excavation cost). Between the two options, which one gave you the most economical solution? That is another influencing factor.

In this article, a small residential building on a 10m x 15m plot of land has been presented for the purpose of analysis and design. The ground floor plan of the building is shown in the diagram below. From the architectural disposition, the building is a two family occupancy arrangement, with each family occupying a floor level. To minimise interruption, the staircase has been placed at the extreme of the left-hand side of the building. The 1st floor of the building is also shown below. It has a balcony at the kitchen, a cantilever sit-out/verandah, and a little balcony by the staircase area. Apart from that, the general arrangement is fairly the same.

My style of General Arrangement (GA)

The architectural drawing of a building can come in many forms. If it comes as a soft copy of a CAD drawing, then the work is made much easier. It is advisable to place the ground floor plan side by side on the graphical user interface window of your CAD program (e.g. AUTOCAD) with the floor plan of the subsequent stories as shown in the diagram below:

Note that architectural drawings come with their own grid lines for different axes. In the current building, we are trying to design here, some details (like gridlines) have been removed from the architectural drawing (Ground & First Floor Plan) for the purpose of clarity of very necessary details like dimension lines. After placing the floor plans side by side on the window, you can start noticing a few different things about the floor plans immediately. For more ideas on how to proceed, copy the floor plans to another location on the window (still leaving the ones you placed side by side). Now, copy the plan of the first floor, and paste it on the ground floor plan, so that the dimensions and axes are matching perfectly.

You can choose a prominent corner of the building as a pick-up point for your copy and paste operation. For more clarity, you can change the colour or thickness (or both) of the ground floor elements and first-floor elements and gridlines, so that you can rightly distinguish between the two.

After you have pasted the first floor on the ground floor, you can now see the interaction of the two floors. All axes that are coincidental will be visible, and all axes that are on the first floor and are not on the ground floor and vice versa will also become visible. At this point, you can also see the outline of block work on the first floor, and this is more critical for the general arrangement of the floor slab. It will properly guide you on the selection of floor beam axis. The block work axis of the ground floor will aid you on the design of the foundation plan. Do not work on the ground floor alone without looking at the first floor – the last thing you will want to happen is to place a column somewhere on the ground floor, and see it popping out through the lobby of the first floor. So carefully make your selections based on matching axes, and fair uniformity. And as I hinted earlier, your arrangement must be consistent with what the architect has in mind. So this is much like art, and you have to use your ingenuity here. (My supervisor during my industrial training once told me that it is one of the hardest things to do in structural design).

As a hint, the next thing to do is to create a rectangular box (say 230 x 230mm) on AUTOCAD and hatch it with any pattern appealing to you (I normally use SOLID). This represents your columns on the floor plan. Now carefully copy this element and start pasting it at the locations where you have decided to place your columns (this usually occurs at intersections between axes). Also, I normally start at the corners of the building; more often than not, columns must be there irrespective of the arrangement. After that, move to the interiors and place your columns as desired. After you are satisfied with what you have done, carefully check the interaction of the arrangement, and make sure that they are reasonable. Areas, where primary and secondary beams will interact, will also become very visible.

To make some points clearer, let us look at a portion of the plan we are considering in this article;

We wish to place columns along axis A. A little consideration will show that we can place columns at points A1, A3, and A5. Also, we can alternatively place columns at points A1, A2, A4, and A5. Without reading further, ponder on that arrangement and see the alternative that you will prefer.

Obviously, A1 and A5 are certain (corner columns). If I should choose to place a column at point A3 (neglecting A2 and A4), these are the implications:

• I will have a larger span for A:1-3 and A:3-5
• I will have a floor beam running along axis 3 (the beam will probably have to run down to axis C before encountering another support unless I extend any other beam to act as a primary beam to it). That's complexity on its own.
• The floor slabs on the bedroom and kitchen will be subjected to block workload from the walls on axis 2 and 4, which some part of it will be subsequently transferred to the beams on axis 3. This is based on the assumption that the ground floor block walls do not carry any load (so I have neglected the effect of the ground floor axis 2 and 4 walls on this assumption)

If I choose the second alternative;

• I will have shorter spans, and three spans instead of 2. The bending moment on Span A:2 – 4 will probably be hogging.
• I will have a wall along axis 3, but by proximity and considering load sharing implication, it will not be critical, and will not affect my designs like the previous alternative.
• My floor beam at axis 2 will stop at axis B, and my floor beam at axis 4 will stop at the wall close to axis B. So I will not have a complex arrangement to deal with.

However, if the drawing comes in the form of paperwork (this is the most typical), then nothing changes in the approach. Place the floor plans side by side, and manually make your decisions, before you go into drafting. It will be better to use tracing paper, place your tracing paper on the ground floor plan and select your column, then place same tracing paper on the first floor to know if any column selected at the ground floor shown at the middle of a room, door or window at the first floor, if this happens, you just need to cancel the column and replace it somewhere within the wall.

Considering all these consequences, I preferred the second alternative. However, if the building is being located in an area where the soil is so bad that constructing a single footing will be very expensive, we can settle for alternative 1. The final GA I adopted for the whole model is shown diagram below;

MEMBER DATA

Thickness of slab = 150mm

Dimensions of floor beams = 450mm x 230mm

Dimension of roof beams = 250mm x 230mm

Dimensions of columns = 230 x 230mm

SECTION ANALYSIS AND DESIGN FORMULAE

[1]Analysis of singly reinforced section

Analysis of singly reinforced section to EC2

From EC2 singly reinforced concrete stress block;

M_Rd = F_Cz —–(1)

f_cd = (α_cc*f_cK)/γ_c

f_cd = (0.85*f_cK)/1.5 = 0.5667f_cK

Compressive force in concrete = Design stress (fcd) x Area of compression block

F_C = 0.5667f_cK x 0.8xb = 0.4533f_cK xb

z = d – 0.4x —–(2)

Clause 5.6.3 of EC2 limits the depth of the neutral axis to 0.45d for concrete class less than or equal to C50/60. Therefore for an under reinforced section (ductile);

x = 0.45d —–(3)

Combining equation (1), (2) and (3), we obtain the ultimate moment of resistance (M_Rd )

M_Rd = 0.167f_ckbd²  —–(4)

Also from the reinforced concrete stress block, we can obtain the tensile design moment as;

M_Ed = F_ez —–(5)

Where;

F_s = (f_yk/1.15)A_s1  —–(6)

A_s1  = Area of tensile reinforcement required

Substituting equ (6) into (5) and making A_s1  the subject of the formular;

A_s1 = M_Ed/0.87f_ykz —–(7)

The lever arm z in EC2 is given;

From (2), z = d – 0.4x

Therefore, x = 2.5(d – z)

M = 0.453 × fck × b × 2.5(d – z)z

K = M/f_ckbd² —–(8)

K can be considered as the normalised bending resistance

z = d[0.5 + (0.25 – 0.882k)^1/2] —–(9)

[2] Analysis of doubly reinforced sections

If the design ultimate moment is greater than the ultimate moment of resistance i.e.M_Ed  > M_Rd , then compression reinforcement is required. Provided that d₂/x ≤ 0.38 (i.e. compression steel has yielded) where d₂is the depth of the compression steel from the compression face and x = (d − z)/0.4.

The area of compression reinforcement, A_s2 , is given by:

A_s2 = (M_Ed– M_Rd)/0.87f_ykz(d – d₂) —–(10)

Area of tension reinforcement A_s1 , is given by:

A_s1 = (M_Ed/0.87f_ykz) +A_s2  —–(11)

Wherez = d[0.5 + (0.25 – 0.882k)^1/2]

K' = 0.167

ANALYSIS AND DESIGN OF THE FLOOR SLABS

Floor slabs in reinforced concrete construction are normally classified as spanning in one direction or in two directions. The common criteria for judging this classification is the ratio (k) of the long span (Ly) to the short span (Lx), and if the slab is only supported on two opposite sides. When k is greater than 2, the slab is treated as spanning in one direction; however, when k is less than 2, the slab is treated as spanning in two directions. Slabs are normally designed per unit width, hence in design, we normally take b = 1000mm.

For one-way slabs, the bending moments are calculated in the same way as for beams. Continuity across a beam can be treated as fixed support (see analysis of PANELS 6 and 9), if the adjacent slab is spanning in two directions, otherwise it can be analysed like a continuous beam. In detailing, if a slab is assumed to be simply supported at end support, it is advisable to provide reinforcement for a probable negative bending moment due to the monolithic construction of beams and slabs.

Approximate analysis is normally employed in the analysis of two-way slabs. The most popular is the theory of plates utilising the elastic analysis. A slab may be assumed to be freely supported when the slab is not continuous, and the edge bears on a block work. If the slab is monolithic with the beam, then a partial fixity exists. Coefficients for analysis of factored loads on two-way slabs can be found in BS 8110-1:1997 (same with the Eurocodes) and has been extensively used in this article.

It is necessary to point out that the critical span in the analysis of solid slabs is the short span. Therefore, the main reinforcements will lie parallel to the short span, and that is where you check your deflections. For one way slabs, a transverse (secondary) reinforcement of not less than 20% of the main reinforcement should be provided as distribution bars. It is also required that the spacing of the main reinforcements should not be more than 3h or 400mm whichever is less. For secondary reinforcements, this should not exceed 3.5h or 450mm whichever is less (h is the thickness of the slab) (see clause 9.3.1.1 of EN 1992:1-1:2004). In a slab where shear reinforcement must be provided, the thickness must be greater than 200mm. More often than not, it is the control of deflection that influences the area of steel and thickness of slabs than flexural requirements. When a floor slab deflects excessively, the finishes such as tiles may crack, and of course, the occupants of the building will not be happy.

LOAD ANALYSIS

Permanent Loads on Slab

Self weight of slab = 25 KN/m³ × 0.15m = 3.75 KN/m²

Weight of finishes (assume) = 1.2 KN/m²

Partition allowance = 1.5 KN/m²

Total dead load (Gk) = 6.45 KN/m²

Variable Loads on Slab

Leading variable action (Imposed load) Qk₁ = 1.5 KN/m²

Ultimate limit state combination = 1.35GK + 1.5QK

Total load on slab (ULS) = 1.35(6.45) + 1.5(1.5) = 10.9575 KN/m²

PANEL 1

[1]  The floor slab (PANEL 1) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.825/3.362 = 1.055

[2] Moment coefficient for two adjacent edges discontinuous (the dia gram have 2 adjacent edges)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.042

Continuous edge = 0.056

Long Span (α) –Mid-span = 0.034

Continuous edge = 0.045

[3] Design of short span

(1) Mid Span

M = αn(Lx)²  = 0.042 × 10.9575(ULS) × 3.625² = 6.0475 KN.m

M_Ed = 6.0475 KN.m

d = h – Cc – ϕ/2

Assuming ϕ12mm bars will be employed for the construction, Cc is the thickness of the concrete cover

d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)

K = M/f_ckbd²

K = (6.0475 x 10⁶)/(25 x 1000 x 119²) = 0.0171

Since k < 0.167 No compression reinforcement required

z = d[0.5 + (0.25 – 0.882k)^1/2]

z = d[0.5 + (0.25 – 0.882 x 0.0171)^1/2] = 0.95d

A_s1 = M_Ed/0.87f_ykz

A_s1 =  (6.0475 x 10⁶)/(0.87 x 460 x 0.95 x 119) = 133.668 mm²/m

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

To calculate the minimum area of steel required;

fctm = 0.3 × f_ck ^2/3= 0.3 × 25^2/3 = 2.5649 N/mm²

ASmin = 0.26 × (fctm/f_yk) × b × d

= 0.26 × (2.5649/460) ×1000 ×144 = 208.76 mm²/m

Check if ASmin < 0.0013 × b × d (187.2 mm²/m)

Since, ASmin > 187.2 mm², the provided reinforcement is adequate.

Checking for Deflection

We check for deflection at the short span of slabs

l/d must be < Modification factor

l/d = 3625/119 = 30.46

Modification Factor Bs = 310/σs

σs = (310 x f_ykx A_s1)/(500AS_prov)

= (310 x 460 x 133.668)/(500 x 452) = 19061056.8/226000

σs = 84.34

Bs = 310/84.34 = 3.68 (modification factor can't be more than 2) Bs = 2

Span depth ratio = 2 x 30.46 = 60.92

Since 30.46 < 60.92 (deflection is OK)

(2) Continuous edge

M = αn(Lx)²  = 0.056 × 10.9575(ULS) × 3.625² = 8.0633 KN.m

M_Ed = 8.0633 KN.m

K = 0.0228; La/d = 0.95; A_s1 = 178.2236 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4] Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.034 × 10.9575(ULS) × 3.625² = 4.895 KN.m

M_Ed = 4.895 KN.m

K = 0.0138; La/d = 0.95; A_s1 = 108.1945 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.045 × 10.9575 × 3.625² = 6.479 KN.m

M_Ed = 6.479 KN.m

K = 0.0183; La/d = 0.95; A_s1 = 143.2057 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 2

[1]  The floor slab (PANEL 2) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.205/2.8 = 1.055

[2] Moment coefficient for one short edge discontinuous (name from the span diagram)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.0345

Continuous edge = 0.046

Long Span (α) –Mid-span = 0.028

Continuous edge = 0.037

[3] Design of short span

(1) Mid span

M = αn(Lx)²  = 0.0345 × 10.9575 × 2.8² = 2.9637 KN.m

M_Ed = 2.9637 KN.m

K = 0.0084; La/d = 0.95; A_s1 = 65.5068 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

Permissible L/d = 101.3240; Actual L/d = 23.5394 (Deflection is ok)

(2) Continuous edge

M = αn(Lx)²  = 0.046 × 10.9575 × 2.8² = 3.9517 KN.m

M_Ed = 3.9517 KN.m

K = 0.0111; La/d = 0.95; A_s1 = 86.5489 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4]Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.028 × 10.9575(ULS) × 2.8² = 2.405 KN.m

M_Ed = 2.405 KN.m

K = 0.0068; La/d = 0.95; A_s1 = 53.1579 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.037 × 10.9575 × 2.8² = 3.178 KN.m

M_Ed = 3.178 KN.m

K = 0.0090; La/d = 0.95; A_s1 = 70.2435 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 3

[1] The floor slab (PANEL 3) is spanning in two directions since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.325/3.205 = 1.037

[2] Moment coefficient for one long edge discontinuous (name from the span diagram)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.030

Continuous edge = 0.039

Long Span (α) –Mid-span = 0.028

Continuous edge = 0.037

[3] Design of short span

(1) Mid span

M = αn(Lx)²  = 0.030 × 10.9575 × 3.205² = 3.376 KN.m

M_Ed = 3.376 KN.m

K = 0.0095; La/d = 0.95; A_s1 = 74.6199 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

Permissible L/d = 87.8141; Actual L/d = 26.9328 (Deflection is ok)

(2) Continuous edge

M = αn(Lx)²  = 0.039 × 10.9575 × 3.205² = 4.389 KN.m

M_Ed = 4.389 KN.m

K = 0.0124; La/d = 0.95; A_s1 = 97.0103 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4] Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.028 × 10.9575(ULS) × 3.205² = 3.152 KN.m

M_Ed = 3.152 KN.m

K = 0.0068; La/d = 0.95; A_s1 = 53.1579 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.037 × 10.9575 × 3.205² = 4.1645 KN.m

M_Ed = 4.1645 KN.m

K = 0.0118; La/d = 0.95; A_s1 = 92.0482 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 4

[1]  The floor slab (PANEL 4) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 5.360/4.545 = 1.179

[2] Moment coefficient for one short edge discontinuous (name from the span diagram)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.036

Continuous edge = 0.048

Long Span (α) –Mid-span = 0.028

Continuous edge = 0.037

[3] Design of short span

(1) Mid span

M = αn(Lx)²  = 0.036 × 10.9575 × 4.545² = 8.148 KN.m

M_Ed = 8.148 KN.m

K = 0.0230; La/d = 0.95; A_s1 = 180.0957 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

Permissible L/d = 97.0628; Actual L/d = 38.1933 (Deflection is ok)

(2) Continuous edge

M = αn(Lx)²  = 0.048 × 10.9575 × 4.545² = 10.865 KN.m

M_Ed = 10.865 KN.m

K = 0.0307; La/d = 0.95; A_s1 = 239.9508 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4] Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.028 × 10.9575(ULS) × 4.545² = 6.337 KN.m

M_Ed = 6.337 KN.m

K = 0.0179; La/d = 0.95; A_s1 = 140.0671 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.037 × 10.9575 × 4.545² = 8.3749 KN.m

M_Ed = 8.3749 KN.m

K = 0.0237; La/d = 0.95; A_s1 = 185.1109 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 5

[1]  In this case, we are going to assume (PANEL 5) as a 1.5m long cantilever spanning along the short span. This assumption is incorrect but conservative. It is very usual to increase the live load of balconies, but in this case, let us assume that the same live load is distributed all across the entire floor slab. Let the partition allowance load of 1.5 KN/m2 (which is not possible on balconies in this case) take care of this.

[2] Design of short span

Support moment M_Ed = 12.327 KN.m

K = 0.0348; La/d = 0.95; A_s1 = 272.4644 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

Permissible L/d = 17.1086; Actual L/d = 12.6050 (Deflection is ok)

[3] Shear Design

Support E; V_Ed = 16.436 KN

Using the area of tension reinforcement provided (ASprov = 452 mm2/m)

V_Rdc = 60.4786 KN > V_Ed (Therefore , no shear reinforcement is required)

This is very ok !!!!

PANEL 6

[1] The floor slab (PANEL 6) is spanning in one direction, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is greater than 2.

Hence, K = Ly/Lx = 2.837/1.4 = 2.0264

From our previous designs without much ado,ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 7

[1]  The floor slab (PANEL 7) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 2.8/1.705 = 1.179

[2] Moment coefficient for all edges continuous (name from the span diagram)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.042

Continuous edge = 0.057

Long Span (α) –Mid-span = 0.024

Continuous edge = 0.032

[3] Design of short span

(1) Mid span

M = αn(Lx)²  = 0.042 × 10.9575 × 1.705² = 1.338 KN.m

M_Ed = 1.338 KN.m

K = 0.0038; La/d = 0.95; A_s1 = 29.5739 mm²; ASmin = 172.5217 mm²;

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

Permissible L/d = 101.3240; Actual L/d = 23.5294 (Deflection is ok)

(2) Continuous edge

M = αn(Lx)²  = 0.057 × 10.9575 × 1.705² = 1.8156 KN.m

M_Ed = 1.8156 KN.m

K = 0.0512; La/d = 0.95; A_s1 = 40.1303 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4] Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.024 × 10.9575(ULS) × 1.705² = 0.764 KN.m

M_Ed = 0.764 KN.m

K = 0.00215; La/d = 0.95; A_s1 = 16.886 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.032 × 10.9575 × 1.705² = 1.0932 KN.m

M_Ed = 1.0932 KN.m

K = 0.0313; La/d = 0.95; A_s1 = 24.163 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 8

[1]  The floor slab (PANEL 8) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.715/3.225 = 1.15

[2]  Moment coefficient for 2 edges continuous and 2 edges discontinuous  (name from the span diagram)

NOTE: this value are constant gotten from the code

Short Span (α) –Mid-span = 0.045

Continuous edge = 0.060

Long Span (α) –Mid-span = 0.034

Continuous edge = 0.045

[3] Design of short span

(1) Mid span

M = αn(Lx)²  = 0.045 × 10.9575 × 3.225² = 5.1284 KN.m

M_Ed = 5.1284 KN.m

K = 0.0144; La/d = 0.95; A_s1 = 113.353 mm²; ASmin = 172.5217 mm²;

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

Permissible L/d = 87.8141; Actual L/d = 27.101 (Deflection is ok)

(2) Continuous edge

M = αn(Lx)²  = 0.060 × 10.9575 × 3.225² = 6.8379 KN.m

M_Ed = 6.8379 KN.m

K = 0.0193; La/d = 0.95; A_s1 = 151.138 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

[4] Design of Long span

(1) Mid span

M = αn(Lx)²  = 0.034 × 10.9575(ULS) × 3.225² = 3.8748 KN.m

M_Ed = 3,8748 KN.m

K = 0.0109; La/d = 0.95; A_s1 = 85.644 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

(2) Continuous edge

M = αn(Lx)²  = 0.045 × 10.9575 × 3.225² = 5.1284 KN.m

M_Ed = 5.1284 KN.m

K = 0.0144; La/d = 0.95; A_s1 = 113.353 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 9

[1]The floor slab (PANEL 9) is spanning in one direction, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is greater than 2.

Hence, K = Ly/Lx = 3.825/1.285 = 2.9766

From our previous designs without much ado,ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

PANEL 10 (1st floor Kitchen Balcony)

[1]  In this case, we are going to assume (PANEL 10) as a 1.1m long cantilever spanning along the short span. It is very usual to increase the live load of balconies, but in this case, let us assume that the same live load is distributed all across the entire floor slab. Let the partition allowance load of 1.5 KN/m2 (which is not likely on balconies in this case) take care of this.

[2] Design of short span

Support moment M_Ed = 6.629 KN.m

K = 0.0187; La/d = 0.95; A_s1 = 146.5212 mm²; ASmin = 172.5217 mm²;

ProvideY12mm @ 250mm c/c BOT (AS_prov = 452 mm²/m)

Permissible L/d = 27.0197; Actual L/d = 9.2437 (Deflection is ok)

[3] Shear Design

Support E; V_Ed = 12.053 KN

Using the area of tension reinforcement provided (ASprov = 452 mm²/m)

V_Rdc = 60.4786 KN > V_Ed (Therefore , no shear reinforcement is required)

This is very ok !!!!

ANALYSIS AND DESIGN OF THE FLOOR BEAMS

Floor beams in a reinforced concrete framed building normally receive load from the floor slabs, carry their own self-weight, the weight of the blocks or bricks (walls), and the weight of finishes. In typical reinforced concrete buildings, these beams are usually T – beams, L – beams, or rectangular beams. T- beams are usually internal beams, while external beams are usually L – beams. Beams that are not carrying any slab load are more often rectangular sections.

In the manual analysis of floor beams, loads are transferred from slab to beams based on the yield line assumption. This depends on if the slab is spanning in one-way or two-way. Typically, for a two-way slab, the loads are either triangular (for the beam parallel to the short span direction of the slab) or trapezoidal (for the beam parallel to the long span direction of the slab). This can be seen in the diagram below, and the typical load distribution on the beam is shown in the second diagram below. Those who are experienced in manual analysis can verify that such analysis can be tedious especially for continuous beams with variable loading and unequal spans. However, for beams of equal span and similar loading, coefficients for bending moment and shear can be obtained from Chapter 12 of Reynolds and Steedman (2005). However, for the sake of convenience, the load transferred from the slab to the beam can be treated as a uniformly distributed load, and the formulas for the transfer of such loads are given in Chapter 13 of Reynolds and Steedman (2005). These formulas are stated in a table below:

Load transfer from slab to beams based on yield line method

Slab loading on the Long Span (Ly) and short span (Lx)

The formula for Equivalent slab load as UDL (Reynolds and Steedman, 2008)

Floor beam Layout and numbering

LOADING OF BEAMS

It is pertinent to point out the approach that has been adopted in the modelling of the self-weight of the beam. As we already know, the self-weight of the slab has already been accounted for in the transfer of the load from the slab to the beam. So the weight of the beam that we account for is the 'drop' of the beam of the beneath the slab. In this case, if the total depth (h) of the beam is 450mm, and the thickness of the slab (hf) is 150mm, then the depth (drop) to be considered on the beam is;

Hd = 450 – 150mm = 300mm

If the width of the web (b_w) of the beam is 230mm, and the density of concrete (ρ_c) is 25 KN/m3 then the self weight of the beam is given by;

Sw = ρ_c × Hd × b_w = 25 KN/m³× 0.3m × 0.23m = 1.725 KN/m

At ultimate limit state, the factor for unfavourable load combination γ_GK = 1.35

Hence Gk = 1.35 × 1.725 = 2.32875 KN/m

The weight of block work is also taken into account. According to Oyenuga (2009), the unit weight of block work in Nigeria (finishes inclusive) is 3.47 KN/m². When we multiply this with the height of the wall, we can rightly obtain the load of the wall. We also factor it as a dead load with γ_GK = 1.35

SPAN 1 – 3

The floor slabs (PANEL 1) are spanning in two directions since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.825/3.625 = 1.055

We can observe from the general arrangement of the structure that the span is parallel to the long span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

ρ = nLx/2 (1 – 1/3k²)

ρ = 10.9575 x 3.625/2 (1 – 1/3(1.055)²)

ρ = 13.912 KN/m

Self weight of the beam (ULS) = 1.35 × 0.3m × 0.23m × 25 KN/m³ = 2.32875 KN/m

Weight of block work (ULS) = 1.35 × 2.75m × 3.47 KN/m² = 12.882 KN/m

Therefore total load on span 1 – 3 (ULS) = 13.912 + 2.32875 + 12.882 = 29.123 KN/m

SPAN 3 – 5

A little consideration will show that the floor slabs (PANEL 2) are spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.625/2.8 = 1.295

We can observe from the general arrangement of the structure that the span is parallel to the short span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

ρ = nLx/3

ρ = (10.9575 x 2.8)/3 = 10.227 KN/m

Self weight of the beam (ULS) = 1.35 × 0.3m × 0.23m × 25 KN/m³ = 2.32875 KN/m

Weight of block work (ULS) = 1.35 × 2.75m × 3.47 KN/m² = 12.882 KN/m

Therefore total load on span 3 – 5 (ULS) = 10.227 + 2.32875 + 12.882 = 29.123 KN/m

SPAN 5 – 7

The floor slab (PANEL 3) is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.625/3.325 = 1.09

We can observe from the general arrangement of the structure that the span is parallel to the short span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

ρ = nLx/3

ρ = (10.9575 x 3.325)/3 = 12.144 KN/m

Self weight of the beam (ULS) = 1.35 × 0.3m × 0.23m × 25 KN/m³ = 2.32875 KN/m

Weight of block work (ULS) = 1.35 × 2.75m × 3.47 KN/m² = 12.882 KN/m

Therefore total load on span 5 – 7 (ULS) = 12.144 + 2.32875 + 12.882 = 29.123 KN/m

SPAN A – C

You can observe that PANEL 1 is spanning in two directions, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is less than 2.

Hence, K = Ly/Lx = 3.825/3.625 = 1.055

We can observe from the general arrangement of the structure that the span is parallel to the short span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

ρ = nLx/3

ρ = (10.9575 x 3.625)/3 = 13.240 KN/m

Self weight of the beam (ULS) = 1.35 × 0.3m × 0.23m × 25 KN/m³ = 2.32875 KN/m

Weight of block work (ULS) = 1.35 × 2.75m × 3.47 KN/m² = 12.882 KN/m

Therefore total load on span A – C (ULS) = 13.240 + 2.32875 + 12.882 = 28.451 KN/m

SPAN C – D

PANEL 9 is spanning in one direction, since the ratio (k) of the longer side (Ly) to the shorter side (Lx) is greater than 2.

Hence, K = Ly/Lx = 3.825/1.285 = 2.9766

We can observe from the general arrangement of the structure that the span is parallel to the short span direction of the slab. Therefore, the equivalent load that is transferred from the slab to the beam can be represented by;

ρ = nLx/5

ρ = (10.9575 x 1.285)/5 = 2.816 KN/m

Load from cantilever slab at the other side = 10.9575 × 0.7m = 7.67025 KN/m

Self weight of the beam (ULS) = 1.35 × 0.3m × 0.23m × 25 KN/m³ = 2.32875 KN/m

No block workload

Therefore total load on span C – D (ULS) = 2.816 + 2.32875 + 7.67025 = 12.815 KN/m

SPAN D – F

A little consideration will show that the area adjacent to the beam has no slab (staircase area). Therefore no load from slab will be transferred.

Self weight of the beam (ULS) = 1.35 × 0.45m × 0.23m × 25 KN/m³ = 3.493 KN/m

Weight of block work (ULS) = 1.35 × 2.75m × 3.47 KN/m² = 12.882 KN/m

Therefore total load on span D – F (ULS) = 3.493 + 12.882 = 16.475 KN/m

BEAM ANALYSIS

SPAN 1 – 7

Using Claperon's 3-Moment Equation

M_A(L_AB) + 2M_B(L_AB + L_BC) + M_C (L_BC) = wl³/4 or (Pa(L² – a²))/L

Analyzing Span 1 – 7

M_A = 0, M_D = 0

M_A(3.825) + 2M_B(3.825 + 2.80) + M_C (3.325) = 29.123 x 3.825³/4 + 25.437 x 2.8³/4

13.25M_B + 3.325M_C = 407.446 + 139.598

13.25M_B + 3.325M_C = 547. 044 —–(1)

M_B(2.80) + 2M_C(2.80 + 3.325) + M_D (3.325) = 25.437 x 2.8³/4 + 27.352 x 3.325³/4

2.8M_B + 12.25M_C = 139.598 + 251.365

2.8M_B + 12.25M_C = 390.923 —–(2)

Solving Equation 1 and 2 Simultaneously

M_B  = 35.303 KNm

M_C  = 23.842 KNm

Calculating the Free Span Moment (M)

Span 1 – 3 = wl²/8 = (29.128 x 3.825²)/8 =  53.27 KNm

Span 3 – 5 = wl²/8 = (25.437 x 2.8²)/8 =  24.93 KNm

Span 3 – 5 = wl²/8 = (27.345 x 3.325²)/8 =  37.79 KNm

Maximum Moment From Each of the Span will now be;

M_max = M – ((M_A + M_B)/2)

Span 1 – 3= 53.27 – (0 + 35.303)/2 = 35.62 KNm

Span 3 – 5= 24.93 – (35.303 + 23.842)/2 = -4.64 KNm

Span 5 – 7= 37.79 – (23.842 + 0)/2 = 25.869 KNm

Calculating the Shear Force

V₁₂  = (29.123 x 3.825)/2 + (0 – 35.303)/3.825 = 46.47 KN

V₂₁  = (29.123 x 3.825)/2 + (35.303 – 0)/3.825 = 64.93 KN

V₂₃  = (25.437 x 2.8)/2 + (35.303 – 23.842)/2.8 = 39.70 KN

V₃₂  = (25.437 x 2.8)/2 + (23.842 – 35.303)/2.8 = 31.51 KN

V₃₄  = (27.343 x 3.325)/2 + (23.842 – 0)/3.325 =  52.63 KN

V₄₃  = (27.343 x 3.325)/2 + (0 – 23.842)/3.325 =  38.29 KN

SPAN A -F

STRUCTURAL DESIGN

SPAN 1 -3 (L-beam)

MEd = 35.62 KNm

Effective depth (d) = 450 – 35 − (16/2) − 8 = 399mm

Beff = 0.2(3395/2) + 0.1(0.85 x 3825) + 230 = 895mm

K = MEd/f_ckbeffd²

K = (35.62 x 10 ⁶ )/(25 x 895 x 399²) = 0.01

Since k < 0.167 No compression reinforcement required

z = d[0.5 + (0.25 – 0.882k)^1/2]

z = d[0.5 + (0.25 – 0.882 x 0.01)^1/2] = 0.95d

A_s1 = M_Ed/0.87f_ykz

A_s1 =  (35.62 x 10⁶)/(0.87 x 460 x 0.95 x 399) = 241.667 mm²/m

Provide 2Y16mm  BOT (AS_prov = 402 mm²/m)

To calculate the minimum area of steel required;

fctm = 0.3 × f_ck ^2/3= 0.3 × 25^2/3 = 2.5649 N/mm²

ASmin = 0.26 × (fctm/f_yk) × b × d

= 0.26 × (2.5649/460) ×230 ×399 = 133.04 mm²

Check if ASmin < 0.0013 × b × d (119.301 mm²)

Since, ASmin > 119.301 mm², the provided reinforcement is adequate.

Checking for deflection

We check for deflection at the short span of slabs

l/d must be < Modification factor

l/d = 3825/399 = 9.59

Modification Factor Bs = 310/σs

σs = (310 x f_ykx A_s1)/(500AS_prov)

= (310 x 460 x 241.667)/(500 x 452)

σs = 171.451

Bs = 310/171.451 = 1.808

Span depth ratio = 1.8 x 171.451 = 308.61

Since 9.59 < 308.61 (deflection is OK)

Use 2Y16mm for the entire bottom span

Top reinforcements (Hogging moment)

MEd = 35.303 KNm

Effective depth (d) = 450 – 35 − (16/2) − 8 = 399mm

Beff = 0.2(3395/2) + 0.1(0.15 x 2800) + 230 = 668.876mm

K = MEd/f_ckbwd²

K = (35.303 x 10 ⁶ )/(25 x 230 x 399²) = 0.0396

Since k < 0.167 No compression reinforcement required

z = d[0.5 + (0.25 – 0.882k)^1/2]

z = d[0.5 + (0.25 – 0.882 x 0.0396)^1/2] = 0.95d

A_s1 = M_Ed/0.87f_ykz

A_s1 =  (35.303 x 10⁶)/(0.87 x 460 x 0.95 x 399) = 239.268 mm²/m

Provide 2Y16mm  BOT (AS_prov = 402 mm²/m)

SPAN A-B (L – Beam)

MEd = 31.73 KNm

Beff = 0.2(3395/2) + 0.1(0.85 x 3625) + 230 = 898mm

K = 0.00089; La/d = 0.95; A_s1 = 209.168 mm²; ASmin = 133.0447 mm²;

Provide 2Y16mm  BOT (AS_prov = 402 mm²/m)

Permissible L/d = 319.896; Actual L/d = 9.0852 Deflection is OK !!!!!

SPAN C-D

MEd = 32.90 KNm (Hogging)

Beff = Bw = 230mm, Effective depth (d) = 450 – 35 – 8 – 8 = 399mm

K = 0.0359; La/d = 0.95; A_s1 = 216.881 mm²; ASmin = 133.0447 mm²;

Provide 2Y16mm  BOT (AS_prov = 402 mm²/m)

Also allow 2Y16 to run through the Bottom ((AS_prov = 402 mm²)

SPAN D – F (Rectangular beam)

MEd = 20.32 KNm (Sagging)

Beff = Bw = 230mm, Effective depth (d) = 450 – 35 – 8 – 8 = 399mm

K = 0.0222; La/d = 0.95; A_s1 = 133.952 mm²; ASmin = 133.0447 mm²;

Provide 2Y16mm  BOT (AS_prov = 402 mm²/m)

Permissible L/d = 86.5580; Actual L/d = 9.3108 Deflection is OK !!!!!

Follow the Same Procedure in solving the remaining beams

ANALYSIS AND DESIGN OF THE COLUMNS

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. 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 the structural design of columns.

LOAD ANALYSIS

Load from Roof

To design the columns we have in our floor plan, we have to start the load analysis from the roof. The proposed roof beam layout is shown in the diagram below, with each roof beam numbered.

Dead load from roof trusses and concrete fascia on external roof beams = 4 KN/m (assumed)

Dead load on internal roof beam from trusses and services = 2 KN/m (assumed)

Self weight of roof beams = 25 × 0.25 × 0.23 = 1.4375 KN/m

Live load on roof = 0.85 KN/m

At ultimate limit state for external roof beams;

Ed = 1.35gk + 1.5qk = 1.35(5.4375) + 1.5(0.85) = 8.615 KN/m

At ultimate limit state for internal roof beams;

Ed = 1.35gk + 1.5qk = 1.35(3.4375) + 1.5(0.85) = 5.915 KN/m

Column Axial Loads (ground floor columns)

Due to the small values of axial load that will be involved, we are going to neglect the design of first floor columns, and provide 4Y12 all through. We will transfer their loads (which will be calculated) to the ground floor columns, where we will carry out the design. The column axial loads have been obtained by summing up the reactions from all the beams supported by the columns, including the self weight of the column.

Let us use column A1 as an example. At the roof level, the column is supporting beam No 2 (Support Reaction V1 = 13.27 KN) and Beam No 3 (Support Reaction VA = 12.99 KN). At the first floor level (see Analysis and Design of Beam No 1 and 2), the column is supporting Beam No 1 (Support Reaction V1 = 41.38 KN), and Beam No 2 (Support Reaction VA = 42.49 KN). Therefore the summation of all these loads gives the axial load transferred from the beams. For intermediate supports, note that the summation of the shear forces at the support gives the total support reaction (neglect the signs and use absolute value).

The summary of roof beam support reactions is given in Table below:

Column Self Weight

Self weight of 1st Floor Columns (ULS) = 1.35 × 25 KN/m³ × 0.23m × 0.23m × 2.75m = 4.909 KN

Self weight of Ground Floor Columns (ULS) = 1.35 × 25 KN/m³ × 0.23m × 0.23m × 4.050m = 7.231 KN

Total Column Self weight (1st floor and Ground floor) = 12.14 KN

COLUMN A1

Total Columns Self weight = 12.14 KN

Load from roof beams = 13.27 + 12.99 = 26.26 KN

Load from floor beams = 46.21 + 42.49 = 88.70 KN

Total = 127.13 KN

COLUMN A3

Total Columns Self weight = 12.14 KN

Load from roof beams = 35.41 + 11.46 = 46.87 KN

Load from floor beams = 105.33 + 60.85 = 166.18 KN

Total = 225.19 KN

COLUMN A5

Total Columns Self weight = 12.14 KN

Load from roof beams = 17.19 + 5.70 = 22.89 KN

Load from floor beams = 83.64 + 37.91 = 121.55 KN

Total = 156.58 KN

COLUMN A7

Total Columns Self weight = 12.14 KN

Load from roof beams = 43.15 + 9.48 = 52.63 KN

Load from floor beams = 38.26 + 62.45 = 100.71 KN

Total = 165.48 KN

COLUMN C1

Total Columns Self weight = 12.14 KN

Load from roof beams = 27.78 + 11.31 = 39.09 KN

Load from floor beams = 80.80 + 69.16 = 149.96 KN

Total = 201.187 KN

COLUMN C3

Total Columns Self weight = 12.14 KN

Load from roof beams = 11.31 + 16.38 = 27.69 KN

Load from floor beams = 69.16 + 152.23 = 221.39 KN

Total = 261.217 KN

COLUMN B5

Total Columns Self weight = 12.14 KN

Load from roof beams = 29.35 + 19.55 = 48.90 KN

Load from floor beams = 116.74 + 196.89 = 313.63 KN

Total = 374.64 KN

COLUMN B7

Total Columns Self weight = 12.14 KN

Load from roof beams = 9.48 + 26.53 = 48.90 KN

Load from floor beams = 86.04 + 54.03 = 313.63 KN

Total = 188.217 KN

COLUMN B8

Total Columns Self weight = 12.14 KN

Load from roof beams = 15.83 + 12.12 = 27.95 KN

Load from floor beams = 15.44 + 65.61 = 81.05 KN

Total = 121.137 KN

COLUMN D1

Total Columns Self weight = 12.14 KN

Load from roof beams = 21.89 + 2.98 = 24.87 KN

Load from floor beams = 35.33 + 25.43 = 60.76 KN

Total = 141.93 KN

COLUMN D2

Total Columns Self weight = 12.14 KN

Load from roof beams = 20.21 + 10.99 = 31.20 KN

Load from floor beams = 89.35 + 52.82 = 142.17 KN

Total = 178.277 KN

COLUMN D4

Total Columns Self weight = 12.14 KN

Load from roof beams = 7.72 + 7.56 = 15.28 KN

Load from floor beams = 51.97 + 47.21 = 99.18 KN

Total = 126.597 KN

COLUMN E5

Total Columns Self weight = 12.14 KN

Load from roof beams = 10.78 + 42.23 = 53.01 KN

Load from floor beams = 262.51 + 72.23 = 334.74 KN

Total = 399.887 KN

COLUMN E8

Total Columns Self weight = 12.14 KN

Load from roof beams = 40.70 + 13.73 = 54.43 KN

Load from floor beams = 101.67 + 65.61 = 167.28 KN

Total = 233.847 KN

COLUMN F1

Total Columns Self weight = 12.14 KN

Load from roof beams = 11.64 + 4.33 = 15.97 KN

Load from floor beams = 25.87 + 3.98 = 29.85 KN

Total = 57.957 KN

COLUMN F2

Total Columns Self weight = 12.14 KN

Load from roof beams = 10.99 + 29.45 = 40.44 KN

Load from floor beams = 79.98 + 52.82 = 132.80 KN

Total = 185.377 KN

COLUMN F4

Total Columns Self weight = 12.14 KN

Load from roof beams = 10.88 + 11.24 = 22.12 KN

Load from floor beams = 2.90 + 35.83 = 38.73 KN

Total = 72.987 KN

COLUMN G4

Total Columns Self weight = 1.35 × 25 × 0.23 × 0.23 × 7.05 = 12.586 KN

Load from roof beams = 12.76 – 0.36 = 12.40 KN

Total = 24.986 KN

COLUMN G6

Total Columns Self weight = 1.35 × 25 × 0.23 × 0.23 × 7.05 = 12.586 KN

Load from roof beams = 42.96 + 5.0 = 47.960 KN

Total = 60.546 KN

COLUMN G8

Total Columns Self weight = 1.35 × 25 × 0.23 × 0.23 × 7.05 = 12.586 KN

Load from roof beams = 13.01 – 2.82 = 10.19 KN

Total = 22.776 KN

COLUMN ANALYSIS FOR MOMENT (ULS)

For the manual analysis of column design moment, we are going to employ the use of sub-frames, and solve for one unknown rotation at each joint using the displacement method. The answers we are going to get here will vary from the answers from computer methods due to the nature of the loading that will be employed. We will maximize column moment by loading the frames as shown in the diagrams below. The recommendation is that you should consider the adjacent beams fully fixed, while you reduce their stiffness by half because we are overestimating the stiffness of the beams by considering all the ends fully fixed.

Sub-frames and loading configuration

Moment of inertia of flanged beam I_B = 0.00307648 m⁴(See Appendix 3)

Moment of Inertia of column I_C = 0.0002332 m⁴

I_B/I_C =  13.19; Hence, I_B = 13.19I_C

So to follow the recommendation, we are going to reduce the stiffness of the beam by half, such that I_B = 6.595I_C

STRUCTURAL DESIGN

Column E5 is the column that is subjected to the highest loading.

COLUMN E5

N _Ed  = 399.887 KN

Elastic Moments

X – direction: M₀₁ = 13.185 KNm; M₀₂= – 6.592 KNm

Z – direction: M₀₁ = 7.138 KNm; M₀₂= – 3.569 KNm

Clear column height = 4500 – 450 = 4050 mm

To calculate the effective height of the column (lo)

Let us now calculate the relative stiffnesses in the planes of bending

In the x-direction;

Beam 1: Axis E: 4 – 5 (Flange width) = 0.2(1200/2 + 2607/2) + (0.1 × 0.85 × 1400) + 230 = 728mm

Beam 2: Axis E: 5 – 8 (Flange width) = 0.2(1200/2 + 4315/2) + (0.1 × 0.85 × 5360) + 230 = 1237mm

Second moment of area of beam 1 = 0.0028628 m⁴

Stiffness of beam 1 (since E is constant) = 4I/L = 0.0081794

Second moment of area of beam 2 = 0.0034116 m⁴

Stiffness of beam 2 (since E is constant) = 4I/L = 0.0025460

Second moment of area of column = 0.0002332 m⁴

Stiffness of lower column = 0.00020728

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

K₁ = 0.00020728/(0.001273 + 0.0040897) = 0.0386

Since the minimum value of K₁ and K₂ is 0.1, adopt K₁as 0.1. Let us take K₂ as 1.0 for base designed to resist moment

Io = 2861.5mm

Compare with BS 8110's 0.75L = 0.75 × 4050 = 3037.5 mm

In the z-direction;

Beam 1, Axis 5: B-E

Flange width = 1246 mm

Second Moment of Area = 0.0034191 m⁴

Stiffness of beam = 0.003009

Stiffness of lower column = 0.00020728

K₁ = 0.00020728/0.0015045 = 0.1377

Io = 2924.34mm

Radius of gyration i = h/√12 = 230/√12 = 66.395

λ_x = 2861.5/66.395 = 43.098

λ_y = 2924.34/66.395 = 44.044

Critical Slenderness for the x-direction

λ_lim= 20ABC/√n

A = 0.7, B = 1.1,

C = 1.7 − M₀₁ / M₀₂= 2.199

n = N _Ed  /A_c f_cd

N _Ed = 399.887 × 10³ N, A_c= 230 × 230 = 52900 mm²

n = 0.5336

λ_lim= 46.359

43.098 < 46.359, second order effects need not to be considered in the x-direction

Critical Slenderness for the z-direction

λ_lim= 20ABC/√n

A = 0.7, B = 1.1,

C = 1.7 − M₀₁ / M₀₂= 2.2

n = N _Ed  /A_c f_cd

N _Ed = 399.887 × 10³ N, A_c= 230 × 230 = 52900 mm²

n = 0.5336

f_cd= (0.85 x 25)/1.5

λ_lim= 46.381

44.044 < 46.381, second order effects need not to be considered in the z-direction

Design Moments (x-direction)

M₀₁ = 6.592 KNm, M₀₂ = 13.185 KNm

e1 is the geometric imperfection = (θi x lo/2) = (1/200 x 2862/2) = 7.155 mm

Minimum eccentricity e0 = h/30 = 230/30 = 7.667mm.  Since this is less than 20mm, take minimum eccentricity = 20mm

Minimum design moment = e0NE_d = 20 × 10^-3 × 399.87 = 7.9974 KNm

First order end moment M₀₂ = M_top + eiNE_d

eiNE_d = 7.155 × 10-3 × 399.87 = 2.861 KNm

M₀₂ = M_top + eiNE_d= 13.185 + 2.861 = 16.046KNm

Longitudinal Steel Area

d₂ = C_nom + ϕ/2 + ϕlinks = 35 + 8 + 8 = 51mm

d₂ /h = 51/230 = 0.2217

Reading from chart No 1; d₂ /h = 0.2;

From the chart, A_sF_yk/bhf_ck= 0.15

Area of longitudinal steel required (As) = (0.15 x 25 x 230 x 230)/460 = 431.25 mm²

As,max = 0.04bh = 2116 mm²

Provide 4Y16mm (Asprov = 804 mm²) Ok

Links

Minimum size = 0.25Φ = 0.25 × 16 = 4mm < 6mm

We are adopting Y8mm as links

Spacing adopted = 200mm less than {b, h, 20Φ, 400mm}

Design Moments (z-direction)

M₀₁ = 3.569 KNm, M₀₂ = 7.138 KNm

e1 is the geometric imperfection = (θi x lo/2) = (1/200 x 2865/2) = 7.1625 mm

Minimum eccentricity e0 = h/30 = 230/30 = 7.667mm.  Since this is less than 20mm, take minimum eccentricity = 20mm

Minimum design moment = e0NE_d = 20 × 10^-3 × 399.87 = 7.9974 KNm

First order end moment M₀₂ = M_top + eiNE_d

eiNE_d = 7.1625 × 10^-3 × 399.87 = 2.864 KNm

M₀₂ = M_top + eiNE_d= 7.138 + 2.864 = 10.002KNm

Longitudinal Steel Area

d₂ = C_nom + ϕ/2 + ϕlinks = 35 + 8 + 8 = 51mm

d₂ /h = 51/230 = 0.2217

Reading from chart No 1; d₂ /h = 0.2;

From the chart, A_sF_yk/bhf_ck= 0.05

Area of longitudinal steel required (As) = (0.05 x 25 x 230 x 230)/460 = 143.75 mm²

As,max = 0.04bh = 2116 mm²

Provide 4Y16mm (Asprov = 804 mm²) Ok

Links

Minimum size = 0.25Φ = 0.25 × 16 = 4mm < 6mm

We are adopting Y8mm as links

Spacing adopted = 200mm less than {b, h, 20Φ, 400mm}

NOTE: This calculation is just for a column, you use the same procedure in solving the remaining columns.

How to Design Concrete Structures Using Eurocode 2 Beams

Source: https://engineeringbasic.com/structural-analysis-and-design-of-residential-buildings-using-manual-calculations-according-to-eurocode-2/

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