6.5k+

# Precast frame analysis

Authored by: Kim S. Elliott

# Precast Concrete Structures

Print publication date:  November  2016
Online publication date:  November  2016

Print ISBN: 9781498723992
eBook ISBN: 9781315370705

10.1201/9781315370705-4

#### Abstract

This chapter introduces the basic principles and some of the design and analysis procedures, involved in the design of precast concrete skeletal structures, essentially a beam–column framework possibly braced using walls and/or cores, as well as briefly discusses precast portal frames and wall frames. Eurocodes EC0, EC1 and EC2 used to determine the combinations and arrangement of gravity and horizontal loads acting on floors, beams and structures are introduced. The design of reinforced and prestressed concrete elements, connections and structures will follow in later chapters.

#### 3.1  Types of Precast Concrete Structures

This chapter introduces the basic principles and some of the design and analysis procedures, involved in the design of precast concrete skeletal structures, essentially a beam–column framework possibly braced using walls and/or cores, as well as briefly discusses precast portal frames and wall frames. Eurocodes EC0, EC1 and EC2 used to determine the combinations and arrangement of gravity and horizontal loads acting on floors, beams and structures are introduced. The design of reinforced and prestressed concrete elements, connections and structures will follow in later chapters.

Preliminary structural design, which many people refer to as the feasibility stage, is more often a recognition of the type of structural frame that is best suited to the form and function of a building than the structural design itself. The creation of a large ‘open plan’ accommodation giving the widest possible scope for room utilisation clearly calls for a column and slab structure, as shown in Figure 3.1, where internal partitions could be erected to suit any client's needs. The type of structure used in this case is often referred to as ‘skeletal’ – resembling a skeleton of rather small but very strong components of columns, beams, floors, staircases, and sometimes structural (as opposed to partition) walls. Of course, a skeletal structure could be designed in cast in situ concrete and structural steelwork, but here we will consider only the precast concrete version.

The basis for the design of precast skeletal structures has been introduced in Figures 1.11 and 1.13. The major elements (the precast components) in the structure are shown in Figure 3.2. Note that the major connections between beams and floors are designed and constructed as ‘pinned joints’, and therefore the horizontal elements (slabs, staircases, beams) are all simply supported. They need not always be pinned (in seismic zones, the connections are made rigid and very ductile) but in terms of simplicity of design and construction it is still the preferred choice. Vertical elements (walls, columns) may be designed as continuous, but because the beam and slab connections are pinned there is no global frame action and no requirement for a frame stiffness analysis, apart from the distribution of some column moments arising from eccentric beam reactions. The stiff bracing elements such as walls are designed either as a storey height element, bracing each storey in turn, or as a continuous element bracing all floors as tall cantilevers.

In office and retail development, distances between columns and beams are usually in the range of 6–12 m (Figures 1.7 and 3.3) depending on the floor loading, method of stability and intended use. In multi-storey car parks, where the imposed loading (vehicle gross weight <30 kN according to the NA to BS EN 1991-1-1, Table NA.6) is 2.5 kN/m2 it is around 16 m for floor spans × 7.2 m for beams, giving three parking bays between columns (Figure 1.6). The exterior of the frame – the building's weatherproof envelope – could also be a skeletal structure, in which case the spaces between the columns would be clad in brickwork, precast concrete panels, sheeting, etc. Alternatively, the envelope might be constructed in solid precast bearing walls, which dispenses with the need for beams, and is referred to as a ‘wall frame’ (Figure 1.14).

Figure 3.1   Precast skeletal structure showing large unobstructed spaces for the benefit of construction workers and the client.

Figure 3.2   Definitions in a precast skeletal structure.

Examples of residential buildings where a precast wall frame would be the obvious choice are shown in Figures 3.4 through 3.7 – the walls are all load-bearing and they support one-way spanning floor slabs. There is less architectural freedom compared to the skeletal frame, for example walls should (preferably) be arranged on a rectangular grid and of fixed modular distance, usually 300 mm, which is quite important economically. A wall frame may be more economical and may often be faster to build, especially if the external walls are furnished with thermal insulation and a decorative finish at the factory. Figures 1.14 and 1.22 are good examples of this. Distances between walls may be around 6 m for hotels, schools, offices and domestic housing, and 10–15 m in commercial developments. Given this description, wall frames appear to be very simple in concept, but in fact are quite complicated to analyse because the walls have very large in-plane rigidity whilst the connections between walls and floors are more flexible. Differential movement between wall panels and between walls and floors has resulted in major serviceability problems over a 25+ year life, often leading to a breakdown in the weatherproof envelope and the eventual condemnation of buildings, which are structurally adequate.

Figure 3.3   Precast skeletal structure in Portugal. (Courtesy of Ergon, Belgium.)

Figure 3.4   Wall frames are best suited for apartments, hotels, schools, shopping units, as in this example at Rhodes, near Sydney, Australia.

Figure 3.5   Precast wall and slab frames in Kuala Lumpur, Malaysia.

Figure 3.6   Cross-wall system in precast wall frames.

Figure 3.7   Precast wall and slab frame at Strijkijzer, Den Haag, the Netherlands.

The third category of precast building is the ‘portal frame’ used for industrial buildings and warehouses where clear spans of some 25–40 m I-section or T-section prestressed rafters are necessary; Figures 3.8 and 3.9. Although portal frames are nearly always used for single-storey buildings, they may actually be used to form the roof structure to a skeletal frame, and as this book is concerned with multi-storey structures it gives us a reason to mention them. The portal frame looks simple enough and in fact is quite rudimentary in design, providing that the flexural rotations at the end of the main rafters, which we can assume will always cause cracking damage to the bearing ledge, are catered for by inserting a flexible pad (e.g. neoprene) at the bearing. As mentioned before, pinned connections between the rafter and column are the preferred choice – they are easy to design and construct. But the columns must be designed as moment-resisting cantilevers, which might cause a problem in some structures as explained later in Section 3.6.2. A moment-resisting connection is equally possible allowing some moment continuity into the column at the eaves. However, unless the columns are particularly tall, say more than about 8 m, it is not worth the extra effort.

Precast portal frames with flat (or shallow inclination) roof structures comprising prestressed or reinforced beams of 6–8 m span supporting long-span precast folded plate roof elements, spanning around 20 m. This is a popular option for industrial buildings, and in the case of Figure 3.10 used in laboratory buildings. The overhang beam is an option for sun or rain shading.

Table 3.1 reviews the various types of precast structures with respect to their possible applications.

Figure 3.8   Definition of a precast portal frame.

Figure 3.9   Precast portal frame. (Courtesy David Fernandez-Ordoñez, Escuela Técnica Superior de Ingeniería Civil, Madrid, Spain.)

Figure 3.10   Portal frame with folded plate roof units, the University of Sao Carlos, Brazil.

### Table 3.1   Application and types of precast concrete frames

Use of building

Number of storeysa

Interior spans (m)

Skeletal frame

Wall frame

Portal frame

Office

2–0

6–15

2–50

6–15

Retail, shopping complex

2–10

6–10

Cultural

2–10

6–10

Education

2–5

6–10

Car parking

2–10

15–20

2–4

6–8

Hotel

2–30

6–8

Hospital

2–10

6–10

Residential

1–40

4–6

Industrial

1

25–40

Warehouse with office

2–3

6–8 25–40

#### 3.2  Simplified Frame Analysis

One of the most frequently asked questions is … how is a precast concrete structure analysed compared to a monolithic cast in situ one? The first response is to say that a precast concrete structure is not a cast in situ structure cut up into little pieces making it possible to transport and erect. It was mentioned in Chapter 1 that the passage of forces through the prefabricated and assembled components in a precast structure is quite different to a continuous (monolithic) structure. This is certainly true near to connections. It is therefore possible to begin a global analysis by first considering the behaviour of a continuous frame and identifying the positions where suitable connections in a precast frame may be made. A two-dimensional in-the-plane simplification is appropriate in the first instance. This is defined in Figure 3.11 where there are no structural frame components, only simply supported floor units, connecting the 2-D in-plane frames together.

Figure 3.11   2-D simplification of a 3-D skeletal structure.

Figure 3.12 shows the approximate bending moments and deflected shape in a three-storey continuous beam and column frame subject to vertical (gravity) patch loads and horizontal (wind) pressure. Call this frame F1. The beam–column connections have equal strength and stiffness as the members. The stability of F1 is achieved through the combined action of the beams, columns and beam–column connections in bending, shear and axial. This is called an ‘unbraced’ frame. There are points of zero moment (‘contraflexure’) in F1, which depend on the relative intensity of the two load cases. If gravity loads are dominant, beam contraflexure is near to the beam–column connection, typically 0.1 times the span of the beam as shown in Figure 3.13; but if the horizontal load is dominant (more rare), contraflexure is at mid-span, with the final location for combined loading at about 0.15 × span. In the column, contraflexure is always at mid-storey height, and this is a good place to make a pinned (notionally = small moment capacity) connection between two precast columns.

Now, if the strength and stiffness of the connection at the end of the beam are reduced to zero, whilst the column and the foundation are untouched, the resulting moments and deflections in this frame, called F2, are as shown in Figure 3.14. The columns alone achieve the stability of F2 – the beams transfer no moments, only axial forces and shear. The foundations must be moment-resisting (rigid). This is the principle of a pinned jointed unbraced skeletal frame. In taller structures, > three storeys or about 10 m, the large sizes of the columns become impractical and uneconomic leading to bracing. The bracing may be used in the full height, called a ‘fully braced’ frame, or up to or from a certain level, called a ‘partially braced’ frame. The differences are explained in Figure 3.15. The bracing could be located in the upper storeys, providing the columns in the unbraced part below the first floor are sufficiently stable to carry horizontal forces and any second-order moments resulting from slenderness.

Figure 3.12   Deformation and bending moment distribution in a continuous structure subjected to (a) gravity loads and (b) horizontal sway load.

Pinned connections may be formed at other locations. Referring back to frame F1, if the flexural stiffness of the members at the lower end of a column is greater than that at the upper end, the point of contraflexure will be near to the lower (stiffer) end of the column. If the strength and stiffness of the lower end of the column are reduced to zero, whilst the beam and beam–column connections are untouched, the resulting moments and deflections in this frame, called F3, are as shown in Figure 3.16a. The stability of F3 is achieved by the portal frame action of inverted U frames – clearly not a practical solution for factory cast large spans so that this method is used for repetitious site casting. Therefore, a practical solution is to prefabricate a series of L-frames as shown in Figure 3.16b for long-span beams and small-storey height columns in a parking structure. Foundations to F3 may be pinned, although most contractors prefer to use a fixed base for safety and immediate stability.

Figure 3.13   Beam half-joints at 0.1× span close to points of contraflexure in a continuous beam.

The so-called H-frame is a variation on F3. Referring back to frame F1, if pinned connections are made at the points of column contraflexure, structural behaviour is similar to a continuous frame as explained in Figure 3.17. Connections between frames are made at mid-storey height positions. Although in theory the connection is classed as pinned, in reality there will be some need for moment transfer, however small. Therefore, H-frame connections are designed with finite moment capacity, this also gives safety and stability to the H-frames, which by their nature tend to be massive. The foundation to half-storey height ground floor columns must be rigid. The connection at the upper end of the column may be pinned if it is located at a point of contraflexure. If not the connection must possess flexural strength as shown in Figure 3.17, where the H-frame has been used in a number of multi-storey grandstands.

Figure 3.14   Deformation and bending moment distribution in a pinned jointed structure subjected to (a) gravity loads. (b) horizontal sway load.

Figure 3.15   Partially braced structures.

Figure 3.16   Structural systems for (a) portal U-frames and (b) portal L-frames.

#### 3.3.1  Two-dimensional plane frames

The object of analysis of a structure is to determine bending moments, shear and axial forces throughout the structure. Monolithic two-dimensional plane frames are analysed using either rigorous elastic analysis, for example moment distribution or stiffness method, either manually or using a computer program. Moment redistribution may be included in the analysis if appropriate. However, often it is only required to determine the moments and forces in one beam or one column, so codes of practice allow simplified substructuring techniques to be used to obtain these values. Figure 3.18 gives one such substructure, called a ‘subframe’ – refer to (Bhatt et al. 2014) for further details. If the frame is fairly regular, that is spans and loads are within 15% of each other, substructuring gives 90%–95% agreement with full frame analysis.

Figure 3.17   H-frames (a) structural system, (b) deformation and bending moments.

Substructuring is also carried out in precast frame analysis, except that, where pinned connections are used, no moment distribution or redistribution is permitted. Figure 3.19 shows subframes for internal beam and upper and ground floor columns where all beam–column connections are pinned. For rigid connections, refer to Figure 3.18. Horizontal wind loads and sway forces due to imperfections are not considered in subframes because the bending moments due to horizontal loads in an unbraced frame (there are no column moments due to horizontal loads in a braced frame) are additive to those derived from subframes. Elastic analysis is used to determine moments, forces and deflections, but a plastic (ultimate) section analysis is used for the design of the components. Clearly, some inaccuracies must be accepted, but according to ‘Designer's Guide to EN 1992-1-1 and EN 1992-1-2’ (Narayanan and Beeby 2005), a design using the partial safety factors (PSFs) and methodologies in the Eurocodes (design philosophy and materials) is “likely to lead to a structure with a reliability index greater than the target value of 3.8 stated in the code for a 50-year reference period.”

Figure 3.18   Substructuring method for internal beam in a continuous frame.

Figure 3.19   Substructuring methods for internal beam and columns in a pinned jointed frame (a) internal beam. (b) upper floor column and (c) ground floor column.

#### 3.3.2  Design loads on beams and frames

The primary Eurocodes used in the design of precast concrete structures are

• Eurocode 0 ‘Basis of design’ (BS EN 1990 2002).
• Eurocode 1 ‘Actions on Structures – Part 1-1: General Actions – Densities, self-weight, imposed loads for buildings’ (BS EN 1991-1-1 2002) plus other parts dealing with fire, snow, wind, thermal, execution (during construction) and accidental (explosion, impact, etc.) actions.
• Eurocode 2 ‘Design of Concrete Structures – General rules and rules for buildings’ (BS EN 1992-1-1 2004) including coefficients for sway loads due to imperfections (lack of plumb, built-in curvature, etc.).
• Eurocode 3 ‘Design of steel structures. General rules and rules for buildings (BS EN 1993-1-1 2005) where steelwork, steel inserts, welding, etc. is required in the additional parts of EN 1993.

Each pan-European document is accompanied by national annexes (NAs) appropriate to national working practices, regional conditions and established/historical precedence, for example stability ties for robustness are the same as in the British code BS 8110: 1997. This book will refer to the UK NAs to Eurocodes EC0 (NA to BS EN 1990 2002), EC1 (NA to BS EN 1991-1-1 2002), EC2 (NA to BS EN 1992-1-1 2004) and briefly to the NA to Eurocode 3 where steelwork, inserts, welding, etc. is required (NA to BS EN 1993-1-1 2005). Appendix 3A (at the end of this chapter) summarises the content of Eurocodes EC2 Parts 1-1 and 1-2, together with the specific clauses related to precast and prestressed concrete elements in the NA to BS EN 1992-1-1. Reference will also be made to the UK's Published Document PD 6687-1 (PD 6687-1 2010) that gives guidance on some specific items that were not published in the concrete Eurocodes or were in need of additional or noncontradictory additional information. The main items in the PD relating to the design of precast concrete structures are listed in Appendix 3B.

These documents give the magnitude and combinations of loads, loading patterns, and PSFs γf (in BS EN 1990) for gravity and horizontal loads in frames and beams. Four conditions are considered, each with their own values of γf follows

• Serviceability limit state (stress, cracking, deflection, dynamic, fatigue)
• Ultimate limit state (ULS) (strength, buckling)
• Instability limit state (for over-turning)
• Accidental limit state (fire, robustness, progressive collapse)

However, each condition varies depending on the nature of the loads. These are called ‘actions’ in the Eurocodes, and those applicable to the super-structure are as follows:

• Permanent actions: self-weight, dead loads of toppings, finishes, services, permanent walls Gk; prestressing forces P; settlement of supports, sway loads due to imperfection.
• Variable actions: imposed floor live loads; demountable partitions, snow loads Qk; wind loads Wk; temperature effects.
• Accidental actions: fire, impact, loss of support, collapse, explosion, Ad, etc.

Dead, live and wind loads are based on the 95% characteristic value for uniformly distributed load (UDL) known as gk, qk and wk [kN/m2] and for line/beam loads and point loads as Gk, Qk and Wk [kN/m or kN].

The self-weight of plain concrete made with normal-weight aggregates (approx. 2600 kg/m3) is taken as 24 kN/m3, according to BS EN 1991-1-1, Table A.1, unless it is shown by the manufacturer that the characteristic self-weight of elements is different. An additional 1 kN/m3 is made for reinforcement and prestressing tendons, although it is unlikely that tendons will add this amount, for example 10 no. 9.3 mm strands in a 1200 × 150 deep solid slab add only 0.22 kN/m3. The density of wet concrete is taken as 25 kN/m3. The densities or self-weight of other building materials and stored materials in warehouses, etc. are given in BS EN 1991-1-1, Tables A.2 through A.12. Note that the self-weight of masonry units are given in BS EN 771 (BS EN 771 2011) and not in the masonry code (BS EN 1996-1-1 2005).

The design values of actions for each of the limit states depend on the nature of the load (i) to (iii), the use of the floor slabs (e.g. residential, parking, storage) and the number and location of the variable loads. Statistically, it is improbable that all imposed loads will be acting at their characteristic value Qk1, Qk2Qki and at the same time, that is full live loads will not act at all floor levels in a multi-storey building, or live, wind and snow loads will not act at the same time. Exceptions to this obviously apply and the designer must be aware of the certain simultaneous combination, such as full live loads acting on a staircase and landing at the same time, in which case the characteristic load will be taken for both elements.

#### 3.3.2.1  Serviceability limit state

Historically, the ‘characteristic’ imposed (live) load Qk was used in all serviceability calculations of service stresses in prestressed concrete, crack spacing and widths, and short- and long-term deflections using viscoelastic deformations due to creep and other effects such as the relative shrinkage between concrete cast at different times. The Eurocodes consider this too severe for long-term effects of cracking and deflection, and, with the exception of calculating service stresses in prestressed concrete in order to avoid sudden rupture after cracking, reduced values of Qk are permitted as shown in Figure 3.20. This is an illustration of the representative values for the characteristic Qk, combination ψ0Qk, frequent ψ1Qk, and quasi-permanent ψ2Qk values of imposed loading over a period of time, which can of course be extended to the whole life of the structure. In fact, clause A1.4.2 of EN 1990 allows the serviceability criteria to be specified for each project and agreed with the client, but the actual circumstantial definitions recommended to be used with particular serviceability requirements in clause A1.4.2 of the NA to BS EN 1990 are

• for function and damage to structural and non-structural elements (e.g. partition walls, etc.), the characteristic combination, for example stress, strength
• for comfort to user, use of machinery, avoiding ponding of water, etc. the frequent combination
• for appearance of the structure, the quasi-permanent combination, for example deformation, deflections

Figure 3.20   Illustration of variable actions.

Floor usage

ψ0

ψ1

ψ2

Domestic, residential, offices

0.7

0.5

0.3

Shopping, congregation

0.7

0.7

0.6

Storage

1.0

0.9

0.8

Traffic area < 3t vehicle weight

0.7

0.7

0.6

Traffic area > 3t vehicle weight

0.7

0.5

0.3

Roof

0.7

0

0

Snow at altitude > 1000 m

0.7

0.5

0.2

Snow at altitude < 1000 m

0.5

0.2

0

Wind pressure

0.5

0.2

0

Source: From NA to BS EN 1990, Table NA.A1.1.

ψ0 is used for load combinations of ultimate strength, ψ1 is used for checking crack widths and decompression stresses for certain durability requirements, ψ2 is used for calculating deflections.

These are according to Expressions 6.14b, 6.15b and 6.16b of EN 1990 as follows.

The design service moment Ms, shear force Vs and end reaction Fs are based on the design service load = characteristic load × set of load factors ψ as follows:

1. ‘Characteristic’ combination according to BS EN 1990, Exp. 6.14b.
3.1
where ψ0 is the characteristic load factor according to the UK NA to BS EN 1991, reproduced here in Table 3.2, for example for domestic usage, ψ0 = 0.7. The symbol “+” means in combination with. For example referring to Figure 3.21, if the stresses at the bottom of a prestressed beam due to prestress = +12.0 N/mm2 (compression), due to dead load = −5.0 N/mm2 (tension), and due to two independent live loads = –8.0 and –4.0 N/mm2, the characteristic stress fb = +12.0 – 5.0 – 8.0 – 0.7 × 4.0 = −3.8 N/mm2 (tension).
2. ‘Frequent’ combination according to BS EN 1990, Exp. 6.15b.
3.2

Figure 3.21   Example of the characteristic, frequent and quasi-permanent combinations of service stresses in a fictitious prestressed concrete section.

where ψ1 and ψ2 are the frequent and quasi-permanent load factors in Table 3.2, for example for domestic usage, ψ1 = 0.5 and ψ2 = 0.3. Continuing the example mentioned earlier, the final frequent stress fb = +12.0 – 5.0 – 0.5 × 8.0 – 0.3 × 4.0 = +1.8 N/mm2 (effectively remains in compression).
3. ‘Quasi-permanent’ combination according to BS EN 1990, Exp. 6.16b
3.3

Continuing the example mentioned earlier would not be meaningful as the ‘quasi-permanent’ combination is used for calculating deflections, but for completeness fb = +3.4 N/mm2 (compression). It is clear from these three examples that the conditions of stress are less onerous with each successive combination, and this is a reflection of the diminishing effect of viscoelastic deformations according to the use of buildings and the effect of specific creep. Note that in Table 3.2 for storage the ψ factors are between 0.8 and 1.0, indicating a higher specific creep.

#### 3.3.2.2  Ultimate limit state

This limit state is known as ‘structure STR’. The design ultimate moment MEd, shear force VEd and end reaction FEd are based on the design ultimate load Ed = characteristic load x set of PSFs γ. Loads are called ‘favourable’ or ‘unfavourable’ in creating the worst possible effects in an element, frame or subframe. The ultimate load combination is according to BS EN 1990, Exp. 6.10, or for STR limit state the least favourable (greater) of Exp. 6.10a and 6.10b, which will always be less than Exp. 6.10. The NA to BS EN 1990, Table NA.A1.2 (B) – Design values of actions (STR) (Set B) give the PSF values. The combinations are

3.4
3.5

with j ≥ 1 and i > 1

where γG,j = 1.35 unfavourable, γG,j = 1.0 favourable, γQ,1 = 1.5, ξ = 0.925

for prestress γP = 0.9 (used for ultimate shear capacity)

NA to BS EN 1990, Table NA.A1.2 (B) notes: “Either expression 6.10, or expression 6.10a together with and 6.10b may be made, as desired. The characteristic values of all permanent actions from one source are multiplied by γG,sup = 1.35, if the total resulting action effect is unfavourable and γG,inf = 1.0, if the total resulting action effect is favourable. For example all actions originating from the self-weight of the structure may be considered as coming from one source; this also applies if different materials are involved. When variable actions are favourable Qk should be taken as 0.” In other words, in frame or subframe analysis 1.35 Gk,j + 1.5 ψ0, Qk (6.10a or b) is carried on one element and 1.0 Gk,j on the adjacent element. Each live load is taken as the ‘dominant’ in turn, with all of the others as ‘accompanying’ in turn, until the maximum combination is found. Therefore, if there are two live loads present there will be three load combinations, that is 6.10a; 6.10b Qk1 dominant; and 6.10b Qk2 dominant.

If the building is domestic with ψ0 = 0.7, use the greater of

• 6.10a 1.35 Gk,j + 1.5 × 0.7 Qk,1 6.10b 0.925 × 1.35 Gk,j + 1.5 Qk,1 + 1.05 Qk2 6.10b 0.925 × 1.35 Gk,j + 1.05 Qk1 + 1.5 Qk2

To satisfy the ULS design, the three load combinations must be used to determine the maximum end reactions, and bending moments and shear forces at all points along the span.

#### 3.3.2.2.1  Effective span

Finally, it is necessary to define the effective span of floor slabs (simply supported, cantilevers) and beams (on dry bearings or mechanical connectors). The clear span of floors ln = distance between beam centres – sum of half breadth of beams. Referring to BS EN 1992-1-1,

3.6

where

• h is the depth of slab (including topping)
• Lb2 and Lb1 are the bearing lengths at either end

For cantilevers, Lb2 is also the width of the support. For continuous elements after completion of the continuity (i.e. stage 2 loading), leff = distance between beam centres. If a bearing medium (pad, plate) is provided, leff is to the centre of the pad, and this is also the case for beams supported on steel inserts, cleats and plates, etc.

Exxample 3.1

Calculate the maximum ultimate end reaction FEd in a simply supported beam of effective span 6.0 m subjected to Gk = 30 kN/m, Qk1 = 20 kN/m, and Qk2 = 40 kN point load at mid-span. ψ0 = 0.7.

Solution

• 6.10a wEd = 1.35 × 30 +(0.7 × 1.5) × 20 = 61.5 kN/m PEd = 1.05 × 40 = 42 kN. Then FEd = 61.5 × 6.0/2 + 42/2 = 205.5 kN. Exp. 6.10b = 1.25 × 30 + 1.5 × 20 = 67.5 kN/m + 42 kN (UDL dominant) Then FEd = 67.5 x 6.0/2 + 42/2 = 223.5 kN. Maximum. Exp. 6.10b = 1.25 × 30 + 1.05 × 20 = 58.5 kN/m + 1.5 × 40 = 60 kN (point load dominant) Then FEd = 58.5 × 6.0/2 + 60/2 = 205.5 kN.

#### 3.3.2.3  Instability limit state

This limit state, known as ‘equilibrium EQU’, is used to check uplift in the back-span of cantilevers, and over-turning of frames including the effect of horizontal wind pressure or other forces. The equilibrium load combination is according to BS EN 1990, Exp. 6.10. The PSFs are given in NA to BS EN 1990, Table NA.A1.2 (A) – Design values of actions (EQU) (Set A) Exp. 6.10 as follows:

3.7

with j ≥ 1 and i > 1

• γG,j = 1.1 (unfavourable) and 0.9 (favourable) γQ = 1.5 (unfavourable) and 1.5 ψ0 accompanying γQ = 0 (favourable) At installation refer to BS EN 1990, Table A2.4(A) as follows: γG,j = 1.05 (unfavourable) and 0.95 (favourable) γQ,i = 1.35 (unfavourable) and 0 (favourable)
Exxample 3.2

Calculate the minimum end reaction FEd in a simply supported beam of effective span 6.0 m with a 3.0 m span overhanging cantilever at one end subjected to Gk = 30 kN/m and Qk = 20 kN/m.

Solution

• 6.10. Main span = 0.9 × 30 = 27 kN/m Exp. 6.10. Cantilever span = 1.1 × 30 + 1.5 × 20 = 63 kN/m. Over-turning moment at cantilever = 63 × 3.02/2 = 283.5 kN Then FEd = 27 × 6.0/2 – 283.5/6.0 = 81.0 – 47.25 = 33.75 kN > 0. No uplift.

#### 3.3.2.4  Accidental limit state

The accidental load combination is according to BS EN 1990, Exp. 6.11. The PSFs are all γ = 1.0. ψ values are given in NA to BS EN 1990, Table NA.A1.3 – Design values of actions for use in accidental combinations of actions, Exp. 6.11 as follows:

3.8

with j ≥ 1 and i > 1

where Ad is the value of the accidental action. ψ1,1 is applied to the dominant action and ψ2,i to the others. However, in accidental situations, it may not be obvious which is which and therefore ψ1 is applied to all.

#### 3.3.3.1  Permanent, variable and wind actions

In frame and subframe analysis without sway, the critical gravity load combinations with their associated PSFs γG and γQ are

1. All spans loaded with the maximum ultimate load wEd,max = γG Gk + γQ Qk (for BS EN 1990, Exp. 6.10a or b).
2. Alternate (‘pattern’) spans loaded with wEd,max on one span and the minimum wEd,min = 1.0 Gk on the adjacent span.

For frame analysis with sway, horizontal loads Wk are combined with gravity Gk and Qk load combinations 6.10a and 6.10b for three situations:

1. Permanent “+” imposed actions (gravity dead + live)
2. Permanent “+” wind actions (gravity dead + wind)
3. Permanent “+” imposed “+” wind actions (all)

### Table 3.3   Partial load factors and safety factors for gravity and horizontal loads

Exp.

Beneficial

Beneficial

Wind

6.10a

Permanent + imposed

1.35

1.0

ψ0 1.5

0

N/A

Permanent + wind

1.35

1.0

N/A

N/A

ψ0 1.5 = 0.75

All

1.35

1.0

ψ0 1.5

0

ψ0 1.5 = 0.75

6.10b

Permanent + imposed

1.25

1.0

1.5

0

N/A

Permanent + wind

1.25

1.0

N/A

N/A

1.5

All

1.25

1.0

1.5

0

ψ0 1.5 = 0.75

Source: From NA to BS EN 1990, Table NA.A1.2 (B).

Alternative terminology to BS EN 1990: beneficial = favourable, adverse = unfavourable.

For wind load, ψ0 = 0.5 according to Table NA.A1.1 of NA to BS EN 1990 (see Table 3.2).

The fundamental PSF for wind load (notation used here γW) is as NA to BS EN 1990, Table NA.A1.2 (B) – Design values of actions (STR) (Set B) γW = 1.5, and is modified by ψ0 = 0.5 (see Table 3.2) in the same way as for gravity loads. The values for γ and ψ are summarised in Table 3.3.

Exxample 3.3

Calculate the maximum ultimate bending moment MEd at the lower end of the columns of height h = 4.0 m in Figure 3.22. The beam–column connections are pinned, and the foundation is rigid. The distance from the edge of the column to the centre of the beam end reaction is 100 mm. Characteristic beam loading is Gk = 40 kN/m and Qk = 30 kN/m, and the wind pressure equates to a horizontal load Wk = 12 kN. The carry-over moment at the lower end of the column is equal to 50% of the upper end moment due to beam eccentricity. Let ψ0 (gravity load) = 0.7.

Solution

Eccentricity of beam reaction R from the centre of column e = 300/2 + 100 = 250 mm.

• Moment at the lower end of column due to REd, MEd = 0.5 REd e

Figure 3.22   Detail to Example 3.3.

Moment at the lower end of each column due to wind load, MEd = WEd h/2 (because there are two columns)

Ultimate load combinations and moments are summarised in the following table.

wEd (kN/m)

REd (kN)

MEd = 0.5 REde (kNm)

WEd (kN)

MEd = WEd h/2 (kNm)

Total MEd (kNm)

6.10a

P + I

85.5

342.0

42.75

42.75

P + W

54.0

216.0

27.0

9.0

18.0

45.0

All

85.5

342.0

42.75

9.0

18.0

60.75

6.10b

P + I

95.0

380.0

47.5

47.5

P + W

50.0

200.0

25.0

18.0

36.0

61.0

All

95.0

380.0

47.5

9.0

18.0

65.5

Then MEd,max = 65.5 kNm using Exp. 6.10b for all loads (it is interesting to note that according to BS 8110 values for all loads, wult = 1.2 × 70 = 84 kN/m, Mu (gravity) = 42 kNm, Wu = 1.2 × 12 = 14.4 kN, Mu (wind) = 28.8 kNm. Total Mu = 70.8 kNm).

#### 3.3.3.2  Horizontal forces due to imperfections

All buildings, including precast concrete structures built with the greatest practical accuracy, will contain imperfections due to construction methods, errors or natural effects. Some of these are unavoidable, for example over-turning moments due to balconies, hanging façade panels, etc. resulting in horizontal deflection and curvature in columns and walls. The reactions to the precast frame from the inclined staircase shown in Figure 3.23 are not exactly imperfections but demonstrate the point of transferring inclined gravity loads into horizontal forces.

Figure 3.23   Inclined staircase imposing horizontal forces to the structure at Bella Sky Hotel, Denmark. (Courtesy Ramboll, Denmark.)

BS EN 1992-1-1 defines imperfections as ‘possible deviations’ in geometry and ‘positions’ of loads in Section 5.2 and as quantified by code Exp. 5.1 through 5.4 as an ultimate horizontal force Hi = Nθi due the inclination θi according to the height l and number m the elements contributing to the imperfection, recognising the improbability that imperfection will be the same in all elements. The value attributed to θi is related to ‘Class 1 execution deviations’ according to BS EN 13670 (BS EN 13670 2009) and is taken into account at the ULS and accidental design situations, but not at serviceability. Deviations in cross section dimensions are taken into account in material safety factors. Imperfections and deviations should not be included in structural analysis as their effect is additional to first-order bending moments and shear forces, but not when checking deflections.

BS EN 1992-1-1 distinguishes between (a) whole braced or unbraced structures, known as ‘global analysis’, (b) isolated columns in braced or unbraced structures, and (c) floor and roof diaphragm action (see Chapter 8) as follows:

• For structures, the horizontal force Hi is applied to the bracing system, for example cores or shear walls, at each floor level as the horizontal component of the total ultimate gravity load at that floor level. Referring to Figure 3.24a (adapted from BS EN 1992-1-1, Fig. 5.1b).
3.9
3.10
θ0 = 1/200 (NA BS EN 1992-1-1)

Figure 3.24   Examples of the effect of geometric imperfections. (a) Bracing system and (b) isolated column in unbraced structure. (Adapted from BS EN 1992-1-1. 2004, Eurocode 2: Design of Concrete Structures – Part 1-1: General rules and rules for buildings, BSI, London, February 2014, Fig. 5.1a1 and b.)

3.11 $α h =2/3 ≤ ( 2 / √ l ) ≤ 1$
l = total height of braced structure (m)
3.12 $α m = √ 0 . 5 ( 1 + 1 / m )$
m = number of vertical members contributing to the horizontal force on the bracing system.
• For isolated columns, Hi is applied to unbraced columns as an eccentricity ei as shown in Figure 3.24b (adapted from BS EN 1992-1-1, Fig. 5.1a and Exp. 5.2). This is used because columns in a precast structure are statically determinate and there is no moment continuity between the rows of columns.
3.13
where l0 is effective length (m) of the column at the floor level that ei is considered, that is at the second floor level l0 is based on the height to the second floor, etc. αh = 2/3 ≤ (2/√l) ≤ 1, where l = actual length of column at the level that ei is acting (m) and m = αm = 1
• For isolated columns (and minor axis of walls) in a braced structure αh is simplified such that αh = m = αm = 1
3.14 $e i = l 0 / 400$
Exxample 3.4

Calculate the horizontal forces at each roof and floor level and the over-turning moment at the foundation due to imperfection in the braced skeletal structure shown in Figure 3.25. The number of columns in each line is six, and there are five rows of column. There are two sets of shear walls in each of the external rows of columns. The total ultimate gravity load per floor = 15,000 kN and at the roof = 7,000 kN.

Figure 3.25   Detail to Example 3.4 (braced) and Example 3.5 (unbraced).

Solution

• l = 10.5 m, then αh = 2/√10.5 = 0.617 use 2/3 m = 30 columns. αm = √0.5(1 + 1/30) = 0.719 θi = (1/200) × (2/3) × 0.719 = 0.002396 (or 1 in 417) At roof level, Hi,roof = θI N/2 sets of shear walls = 0.002396 × 7000/2 = 8.38 kN per wall At floor level, Hi,floor = 0.002396 × 15,000/2 = 17.97 kN per wall The over-turning moment Mi due to Hi = 8.38 × 10.50 + 17.97 × (7.25 + 4.00) = 290.1 kNm per wall.
Exxample 3.5

Calculate the horizontal forces at each roof and floor level and the over-turning moment at the foundation due to imperfection in one of the internal columns, if the same structure shown in Figure 3.25 is unbraced. The effective length factor for the columns may (in this example) be taken as 2.2. The total ultimate gravity load per column per floor = 900 kN and at the roof = 500 kN.

Solution

• At the roof level. l = 10.50 m, then αh = 2/√10.50 = 0.617 use 2/3. l0 = 2.2 × 10.5 = 23.1 m ei = (1/200) × (2/3) × 23.1/2 = 0.039 m Then Mi due to ei = 500 × 0.039 = 19.5 kNm which equates to Hi = 19.5/10.5 = 1.85 kN At the second floor level. l = 7.25 m, then αh = 2/√7.25 = 0.742. l0 = 2.2 × 7.25 = 15.95 m ei = (1/200) × 0.742 × 15.95/2 = 0.030 m Then Mi due to ei = 900 × 0.030 = 27.0 kNm which equates to Hi = 27.0/7.25 = 3.72 kN At the first floor level. l = 4.00 m, then αh = 2/√4.00 = 1. l0 = 2.2 × 4.00 = 8.80 m ei = (1/200) × 1 × 8.80/2 = 0.022 m Then Mi due to ei = 900 × 0.022 = 19.8 kNm which equates to Hi = 19.8/4.00 = 4.95 kN Total Mi = 19.5 + 27.0 + 19.8 = 66.3 kNm per column.

(Note that Mi for isolated columns is greater per column than if the total Mi for the walls was divided over the total number of columns).

#### 3.3.4  Beam subframe

Figure 3.19a. The subframe consists of the beam to be designed of span L1, and half of the adjacent beams of span L2 and L3. The eccentricity of the beam end reaction from the centroidal axis of the column is e. Alternate pattern loading is used. The height of the column above and below the beam is actually of no consequence to beam. It is assumed that the cross section and flexural stiffness of the column is constant.

3.15
3.16

(Note the shear force in the beam is VEd = wEd,max (L2 – 2e)/2).

3.17

The resulting maximum bending moment in the column is given by

3.18

assuming that R1 < R3 and h1 > h3. Figure 3.26a shows the final moments.

Figure 3.26   Bending moments in a pinned jointed frame for (a) internal beams, (b) upper floor columns. (c) ground floor columns.

#### 3.3.5  Upper floor column subframe

Figure 3.19b. The subframe consists of the column to be designed of height (distance between centres of beam bearing) h2, and half the adjacent columns of heights h1 and h3. Because the column is continuous, the cross section and flexural stiffness EI of each part of the column is considered as shown in the figure. The beams are pattern loaded as mentioned earlier, of span L4/2 and L5/2, and the eccentricity of each beam end reaction from the centroidal axis of the column is e4 and e5, respectively. The moment at the upper end of the designed column is given by

and at the lower end is

3.19

where R4 and R5 are given in Equations 3.2 and 3.3. Figure 3.26b shows the final moments. Note that patch loading produces single curvature in the columns.

#### 3.3.6  Ground floor column subframe

Figure 3.19c The subframe consists of the column to be designed of height (distance between the centre of first floor beam bearing and 50 mm below top of foundation (see section 9.4)) h1, and half the adjacent column of height h2. All other details are as before. If the foundation is rigid (moment resisting), the moment at the upper end of the designed column is given by Equation 3.5 with appropriate notation. The carry-over moment at the lower end is equal to 50% of the upper end moment. If the foundation is pinned, the upper end moment is given by

3.20 $M c o l , u p p e r = ( R 4 e 4 − R 55 e 5 ) 0.75 E I 1 h 1 0.75 E I 1 h 1 + E I 2 h 2$

and the lower end moment is zero. Figure 3.26c shows the final moments. Patch loads produce single curvature in the columns.

Exxample 3.6

Determine, using substructuring techniques, the bending moments in the beam X and columns Y and Z identified in Figure 3.27. The beam–column connections are pinned, and the foundation is rigid. The distance from the edge of the column to the centre of the beam end reaction is 100 mm. Characteristic beam loading is Gk = 40 kN/m and Qk = 30 kN/m.

Solution

wEd,max = max{1.35 × 40 + 1.05 × 30; 1.25 × 40 + 1.5 × 30} = max {85.5; 95.0} = 95.0 kN/m; wEd,min = 40 kN/m.

Beam subframe

e = 450/2 + 100 = 325 mm

• Equation 3.1. M1 = 95.0 × (8.000 – 2 × 0.325)2/8 = 641.5 kNm

Column Y subframe

Beam end reactions R1 = 95.0 × 8.000/2 = 380.0 kN; R2 = 40 × 6.000/2 = 240.0 kN

• e1 = e2 = 300/2 + 100 = 250 mm

Figure 3.27   Detail to Example 3.6.

• but EI1/h1 = EI2/h2
• Equation 3.5. At upper and lower ends, Mcol = (380.0 – 240.0) × 0.250 × 0.5 = 17.5 kNm

Column Z subframe

Beam end reactions as before. e1 = e2 = 450/2 + 100 = 325 mm

Given that E is constant

Equation 3.5. At upper end, Mcol,upper = (380 – 240) × 0.325 × 451/(451 + 211) = 31.0 kNm

At lower end, Mcol,lower = 50% × 31.0 = 15.5 kNm.

#### 3.4  Connection Design

Connections form the vital part of precast concrete design and construction. They alone can dictate the type of precast frame, the limitations of that frame, and the erection progress. It is said that in a load-bearing wall frame the rigidity of the connections can be as little as 1/100 of the rigidity of the wall panels −200 N/mm2 per mm length for concrete panels versus 2.7–15.0 N/mm2 per mm length for joints (Straman 1990). Moreover, the deformity of the bedding joint, that is the invisible interface where the panel is wet bedded onto a mortar, between upper and lower wall panels can be 10 times greater than that of the panel.

The previous paragraph contained the words connections and joints to describe very similar things. Connections are sometimes called ‘joints’ – the terminology is loose and often interposed. The definition adopted in this book is as follows:

• Connection: is the total construction between two (or more) connected components: it includes a part of the precast component itself and may comprise several joints.
• Joint: is the part of a connection at individual boundaries between two elements (the elements can be precast components, in situ concrete, mortar bedding, mastic sealant, etc.)

For example in the beam–column assembly shown in Figure 3.28, a bearing joint is made between the beam and column corbel, a shear joint is made between the dowel and the angle, and a bolted joint is made between the angle and column. When the assembly is completed by the use of in situ mortar/grout, the entire construction is called a connection. This is because the overall behaviour of the assembly includes the behaviour of the precast components plus all of the interface joints between them. Engineers prove the capacity of the entire connection by assessing the behaviour of the individual joints.

Structurally, joints are required to transfer all types of forces – the most common of these being not only compression and shear, but also tension, bending and occasionally torsion. The combinations of forces at a connection can be resolved into components of compressive, tensile and shear stress, and these can be assessed according to limit state design. Steel (or other materials) inserts may be included if the concrete stresses are greater than permissible values. The effects of localised stress concentrations near to inserts and geometric discontinuities can be assessed and proven at individual joints. However, connection design is much more important than that because of the sensitivity of connection behaviour to manufacturing tolerances, erection methods and workmanship.

Figure 3.28   Moment and shear transfer at a bearing corbel.

It is necessary to determine the force paths through connections in order to be able to check the adequacy of the various joints within. Compared with cast in situ construction, there are a number of forces which are unique to precast connections, namely frictional forces due to relative movement causes by shrinkage, etc pretensioning stresses in the concrete and steel, handling and self-weight stresses. In the example shown in Figure 3.29, a reinforced concrete column and corbel support a pretensioned concrete beam. The figure shows that there are 10 different force vectors in this connection as follows:

1. A: diagonal compression strut in corbel
2. B: horizontal component reaction to force at A
3. C: vertical component reaction to force at A
4. D: internal diagonal resultant to forces B and C
5. E: diagonal compression strut in beam
6. F & G: horizontal component reaction to force E
7. H: tension field reaction to forces E & F
8. J: horizontal friction force caused by the relative movement of beam and corbel
9. K: horizontal membrane reaction to beam rotation due to eccentric prestressing

The structural behaviour of the frame can be controlled by the appropriate design of connections. In achieving the various structural systems in Section 3.3.2, it may be necessary to design and construct either/both rigid and/or pinned connections. Rigid monolithic connections can only truly be made at the time of casting, although it is possible to site cast connections that have been shown to behave as monolithic, for example cast in situ filling of prefabricated soffit beams before and after casting as shown in Figure 3.30a and b. The advantages lost to in situ concreting work (cold climates in particular), the delayed maturity, the increase in structural cross section, and the reliance on correct workmanship, etc. detract this solution in favour of bolted or welded mechanical devices. Rigid connections may be made at the foundation where there is less restriction on space as shown in Figure 3.31. In very simple terms, a bending moment is generated by the provision of a force couple in rigid embedment, that is no slippage when the force is generated. Pinned connections are designed by an absence of this couple, although many connectors designed in this way inadvertently contain a force couple, giving rise to spurious moments which often cause cracking in a region of flexural tension.

Figure 3.29   Force paths in beam to column corbel connection.

To gain an overview of the various types, Figure 3.32 and Table 3.4 show the locations, classification and basic construction of connections in a precast structure.

In theory, no connection is fully rigid or pinned – they all behave in a semi-rigid manner, especially after the onset of flexural cracking. Using a ‘beam-line’ analysis, Figure 3.33, we can assess the structural classification of a connection. Although the beam-line approach was developed for structural steelwork in c1936, research carried out since 1990 has shown that the method is appropriate to precast connections (Elliott et al. 1998, Elliott et al. 2003a,b, Ferriera et al. 2003, Elliott and Jolly 2013).

The moment–rotation (M-θ) diagram in Figure 3.33 is constructed by considering the two extremes in the right hand part of Figure 3.33. The hogging moment of resistance of the beam at the support is given by MRd > wL2/12, and the rotation of a pin ended beam subjected to a UDL of w is θ = wL3/24EI. The gradient of the beam line is 2EI/L. The M-θ plot for plots 1 and 2 give the monolithic and pinned connections, respectively. In reality, the behaviour of a connection in precast concrete will follow plots 3, 4 or 5, etc. If the M-θ plot for the connection fails to pass through the beam-line, that is plot 5, the connection is deemed not to possess sufficient ductility and should be considered in design as ‘pinned’. Furthermore, its inherent stiffness (given by the gradient of the M-θ plot) is ignored. Conversely, if the M-θ behaviour follows plot 3 (the gradient must lie in the shaded zone and the failure takes place outside the shaded zone), the effect of the connection will not differ from a monolithic by more than 5%.

Figure 3.30   Cast in situ concrete topping over precast soffit beams forms a fully rigid connection. (a) Prepared for casting and (b) as cast and awaiting structural topping to also cover the slabs.

#### 3.5  Stabilising Methods

Structural stability and safety are necessary considerations at all times during the erection of precast concrete frames. The structural components will not form a stabilising system until the connections are completed – in some cases, this can involve several hours of maturity of cast in situ concrete/grout joints, and several days if structural cast in situ toppings are used to transfer horizontal forces. A stabilising system must comprise two things as shown in Figure 3.34:

Figure 3.31   Precast column to pocket connection.

Figure 3.32   Types of connections in a precast structure.

1. A horizontal system, often called a ‘floor diaphragm’ because it is extremely thin in relation to its plan area
2. A vertical system in which the reactions from the horizontal system are transferred to the ground (or other substructure)

### Table 3.4   Types of connections in precast frames

Connection type

Location in Figure 3.32

Classification

Method of jointing

1

Pinned

Dowel

2

Rigid

Dowel plus continuity top steel

3

Pinned

Dowel

4

Rigid

Bolts (couple)

Column splice

5

Pinned

Bolts/dowel

Rigid

Bars in grouted sleeve (couple)

Steel shoes

Beam – column face

6

Pinned

Bolts

Welded plates

Notched plates

Dowels

Beam – column corbel

7

Pinned

Dowel

Beam – column corbel

8

Rigid

Dowel plus continuity top steel

Beam – beam

9

Pinned

Bolts

Dowels

Slab – beam

10

Pinned

Tie bars

Slab – wall

11

Pinned

Tie bars

Column – foundation

12

Pinned

Bolts

Column – cast in situ beam or retaining wall

13

Pinned or rigida

Bolts Rebars in grouted sleeve

Figure 3.33   Definition of moment–rotation characteristics.

Figure 3.34   Stabilising systems in braced frames.

The horizontal system is considered in detail in Chapter 8 where reference is also made to the many code regulations on this topic. When subjected to horizontal wind or lack-of-plumb forces, the floor slab acts as a deep beam and is subjected to bending moments Mh and shear forces Vh (h being the subscript used for horizontal diaphragms). The basic design method is shown in Figure 3.35. The design is a three-stage approach:

1. The floor diaphragm is analysed as a long, deep beam which is supported by a number of shear walls, shear cores, deep columns (wind posts), or other kinds of bracing such as steel cross bracing. Figure 3.35a.
2. If there are only two supports (bracing), the analysis is statically determinate and MEd,h and VEd,h may be calculated directly. If there are more than two supports, irrespective of where they are positioned, the analysis is statically indeterminate. The support reactions must first be found by a technique which considers the relative stiffness and position of each support, and the horizontal (e.g. wind load) pressure distribution. The derivation is given in Section 8.1 after which MEd,h and VEd,h may be calculated.
3. The area of reinforcement required to resist MEd,h and VEd,h is determined as follows:
3.21

where

• 0.8 B is the assumed lever arm between the compression zone and the tie steel (the assumption is known to be conservative)
• fyk/γm is the design stress in the tie steel with γm = 1.15

High tensile rebar with fyk = 500 N/mm2 or standard helical strand with fpk = 1770 N/mm2 is used (super strand or Dyform tend to be too stiff to handle) – the reasons are given in Section 8.4.

3.22

where µ is the coefficient of friction as given in BS EN 1992-1-1, clause 6.2.5(2). Hollow core slabs are considered as being untreated and smooth, then µ = 0.6 with no special, that is ex-factory, edge preparation (see Section 8.2). For ex-steel mould with smooth surfaces, µ = 0.5.

• 4. The tie steel Ashd must be placed everywhere moments occur. The tie steel Asvh must be placed only where the shear force is greater than a certain value. This is found by checking that the interface shear stress vEdi = VEd,h/B (D – 30 mm) does not exceed vRdi ≤ 0.15 N/mm2 for smooth and rough surfaces (as in the case of machine cast hollow core units and as-cast precast planks) or vRdi ≤ 0.10 N/mm2 for very smooth surfaces cast against steel moulds, according to BS EN 1992-1-1, clause 10.9.3(12). (The reason for the deduction of 30 mm is explained in Section 8.4.1.)

Figure 3.35   Diaphragm floor action. (a) Deep beam analogy. (b) Reinforced structural topping in double-tee floors. (c) Perimeter reinforcement in hollow core floors.

Diaphragms may be reinforced in several ways. In Figure 3.35b, a reinforced cast in situ topping transfers all horizontal forces to the vertical system – the precast floor plays no part but for restraining the topping against buckling. In Figure 3.35c, there is no cast in situ topping. Perimeter and internal tie steel resists the chord forces resulting from horizontal moments. Coupling bars are inserted into the ends of the floor units, and together with the perimeter steel provides the means for shear friction generated in the concrete-filled longitudinal joints between the units.

Exxample 3.7

Determine the shear wall reactions and diaphragm reinforcement in the floor shown in Figure 3.36a. The precast units are 150 mm deep hollow cored and have an ex-factory edge finish. The characteristic wind pressure on the floor wk = 3 kN/m. Tie steel is high tensile ribbed bar fyk = 500 N/mm2. Suggest some reinforcement details.

Figure 3.36   Detail to Example 3.7. (a) Plan view of floor diaphragm, (b) Wind loading diagram and shear force diagram in the floor diaphragm.

Solution

NA to BS EN 1990, Table NA.A1.2 (B) – Design values of actions (STR) (Set B) γW = 1.5

• Design ultimate wind load = 1.5 × 3.0 = 4.5 kN/m. From Figure 3.36b, support reaction R1 = 50.62 kN; R2 = 84.38 kN. Shear span from LHS (distance to zero shear and hence point of maximum moment) = 50.62/4.5 = 11.25 m. MEd,h,max = 50.62 × 11.25/2 = 284.74 kNm; VEd,h,max = 57.38 kN at RHS of 24 m span. Ashd = 284.74 × 106/0.8 × 5000 × (500/1.15) = 164 mm2. Use 2 no. H12 bars (226). Interface shear stress vEdi = 57.38 × 103/5000 × (150 – 30) = 0.096 N/mm2 < 0.15 N/mm2 allowed. No shear reinforcement needed.

Vertical stabilising systems are dictated by the necessary actions of the structural system, that is skeletal, wall or portal frame. Column effective lengths depend on the type and direction of the bracing. However, there is a broad classification as the structure is

1. Unbraced frame, Figure 3.37, where horizontal force resistance is provided either by moment resisting frame action, cantilever action of columns, or cantilever action of wind posts (deep columns)
2. Braced frame, Figure 3.38, where horizontal force resistance is provided either by cantilever action of walls or cores, in-plane panel action of shear walls or cores, infill walls, cross bracing, etc. or
3. Partially braced frame, Figure 3.39, which is some combination of (1) and (2)

Figure 3.37   Alternative sway mechanisms and resulting column effective length factors.

Figure 3.38   Alternative full-height bracing mechanisms and resulting column effective length factors.

The type of stabilising system may be different in other directions. The floor plan arrangement and the availability of shear walls/cores will dictate the solution. The simplest case is a long narrow rectangular plan where, as shown in Figure 3.40a, shear walls brace the frame in the y direction only, the x direction being unbraced. In other layouts, shown for example in Figure 3.40b, it is nearly always possible to find bracing positions. Precast skeletal frames of three or more storeys in height are mostly braced or partially braced. This is to avoid having to use deep columns to cater for sway deflections, which give rise to large second-order bending moments. Section 6.2.6 refers in more detail.

Figure 3.39   Alternative partial height bracing mechanisms and resulting column effective length factors.

Figure 3.40   Positions of shear walls and cores in alternative floor plan layouts. (a) Positions of shear walls, (b) positions of shear cores or walls around stairs and lift shafts. (c) bracing methods and positions in partially braced irregular or non-symmetrical buildings.

It is not wise to use different stabilising systems acting in the same direction in different parts of a structure. The relative stiffness of the braced part is likely to be much greater than in the unbraced part, giving rise to torsional effects due to the large eccentricity between the centre of external pressure and the centroid of the stabilising system, as explained in Figure 3.40c. The different stabilising systems should be structurally isolated – Figure 3.40d.

In calculating the position of the centroid of a stabilising system, the stiffness of each component of thickness t and length L is given by Ecm,long I, where Ecm,long = long-term Young's modulus (usually taken as 15 kN/mm2) and I = tL3/12. First moments of stiffness are used to calculate the centroid as explained in Example 3.8.

Exxample 3.8

Propose stabilising systems for the five-storey skeletal frame shown in Figure 3.41a. The beam–column connections are all pinned, and the columns should be the minimum possible cross section to cater for gravity loads. Wind loading may be assumed to be uniform over the entire façade. Use only shear walls for bracing.

Hint: the grid dimensions around the stairwell may be taken as 4 m × 3 m, and at the lift shaft 3 m × 3 m.

Solution

A braced frame is required up to the fourth floor, after which a one-storey unbraced frame may be used. It would not otherwise be possible to satisfy the requirement of minimum column sizes for gravity loads. To avoid torsional effects (see Figure 3.40c), the centroid of the stabilising system should be as close as possible to the centre of external pressure, that is at x ≈ 24 m and y ≈ 16 m. It is necessary to first consider the two orthogonal directions.

Stability in y-direction

The centroid of the stability walls x′ ≈ 24 m.

Figure 3.41   Detail to Example 3.8 (dimensions in metres). (a) Plan view and cross-section of framed structure, (b) Plan view showing positions of shear walls.

Select walls as shown in Figure 3.41b. On the assumption that the material and construction of all walls is the same, Young's modulus and thickness of wall are common to all walls and need not be used in the calculation.

Centroid of stiffness $x ′ = ( 4 3 × 0 ) + ( 4 3 × 3 ) + ( 3 3 × 3 3 ) + ( 3 3 × 3 6 ) + ( 4 3 × 4 5 ) + ( 4 3 × 4 8 ) ( 4 × 4 3 ) + ( 2 × 3 3 ) = 25.0 m$

, which is sufficiently close to the required point to eliminate significant torsional effects.

Stability in x-direction

The centroid of the stability walls y′ ≈ 32/2 = 16 m.

Centroid of stiffness $y ′ = ( 3 3 × 0 ) + ( 3 3 × 1 6 ) + ( 3 3 × 3 2 ) 3 × 3 3 = 16.0 m$

, which is at the correct point.

#### 3.6  Comparison of Standard Designs to BS 8110 and Eurocodes

To assist the transition between the British code BS 8110:1997 and the Eurocodes EC0, EC1 and EC2, the following precast reinforced and prestressed concrete elements are designed:

1. Reinforced concrete rectangular beam
2. Reinforced concrete rectangular column
3. Prestressed concrete solid floor slab

Clauses and tables in the codes are indicated within ‘{}’.

#### 3.6.1  Reinforced concrete rectangular beam

A 600 mm deep × 300 mm wide r.c. beam carries uniformly distributed dead and live loading of 40 and 30 kN/m over a simply supported clear span of 5.85 m with bearing lengths of 150 mm. The beam carries office loading. The exposure is internal, and the fire resistance is 60 min. The design life is 50 years.

Design the main reinforcement at mid-span and the shear reinforcement at the support and position where nominal links are required. Calculate the crack width and the long-term deflection using the appropriate creep factor and check the span/depth ratio for the area of rebars designed. Use fck/fcu = 32/40, high tensile main bars and links grade fyk/fy = 500 N/mm2, and normal-weight concrete with a 20 mm coarse aggregate.

BS 8110 solution

Eurocodes solution

Durability. BS 8500-1. Table A.4 for 50 years

Exposure XC1. Cover c = 15 + Δc

Exposure XC1. Cnom = 15 + ΔCdev

{7.3} Δc = 5 mm

{4.4.1.3(3)} ΔCdev = 5 mm

Cover to links c = 20 mm

Cover to links = 20 mm

Fire. 1 h

Fire. R60. BS EN 1992-1-2

{Table 3.4} c = 20 mm

{Table 5.5} for b = 300 mm, axis a = 25 mm

Section properties. Let links φ = 8 mm

a = max{20 + 8 = 28 mm; 25 mm}

Assume main bars φ = 32 mm

b = 300 mm; d = 600 – 20 – 8 – 16 = 556 mm

b = 300 mm; d = 600 – 28 – 16 = 556 mm

Flexural design

Self-weight

Self-weight {BS EN 1991-1-1, Table A1.1}

= 0.6 × 0.3 × 24 = 4.32 kN/m

= 0.6 × 0.3 × 25 = 4.5 kN/m

{BS EN 1990, Table A1.2(B) and

Table A1.1} ψ0 = 0.7

wu = 1.4 × 44.32 + 1.6 × 30 = 110.1 kN/m

{Exp. 6.10a; 6.10b} wEd = max{1.35 × 44.5 + 0

0.7 × 1.5 × 30; 1.25 × 44.5 + 1.5 × 30} = 100.7 kNm

Effective span {3.4.1.2}

{5.3.2.2, Fig. 5.4a}

lo = min{5.85 + 0.15; 5.85 + 0.46} = 6.0 m

leff = min{5.85 + 0.15; 5.85 + 0.46} = 6.0 m

M = 110.1 × 6.02/8 = 495.5 kNm

MEd = 100.7 × 6.02/8 = 453.1 kNm

{3.4.4.4} K = 495.5 × 106/40 × 300 × 5562

{3.1.7(3)} K = 453.1 × 106/32 × 300 × 5562

= 0.134 < 0.156 for x/d ≤ 0.5

= 0.153 < 0.206 for x/d ≤ 0.6

z/d = 0.5 + √0.25 – K/0.9 = 0.82 < 0.95

z/d = 0.5 + √0.25 – K/1.133 = 0.84 < 0.95

z = 0.82 × 556 = 455 mm

z = 0.84 × 556 = 467 mm

As = 495.5 × 106/455 × 0.87 × 500 = 2503 mm2

As = 453.1 × 106/467 × 0.87 × 500 = 2230 mm2

Use 2 no. H32 ± 2 no. H25 bars (2,590)

Use 3 no. H32 bars (2,412)

Spacing = (300 – 88 – 114)/3 = 33 mm

Spacing = (300 – 88 – 96)/2 = 58 mm

{3.12.11.1} > 20 + hagg 5 = 25 mm OK

{Table 7.3N} < 100 mm for any value of σs

{Table 7.1, NAD Table NA.4}wk = 0.3 mm

Although compression steel is not required

{Table 3.25} Min As = 0.2% = 360 mm2

{9.2.1.1(1)} Min As = 0.13% = 234 mm2

Use 2 no. H16 (402) at d ′ = 36 mm

Use 2 no. H16 (402) at d ′ = 36 mm

Comments. EC2 requires 11% less area of rebar.

Shear design

{3.4.5.3} lv = 5.85 – 2 × 0.556 = 4.74 m

{6.2.1(8)} lv = 5.85 – 2 × 0.556 = 4.74 m

V = 110.1 × 4.74/2 = 261.0 kN

VEd = 100.7 × 4.74/2 = 238.7 kN

{3.4.5.3} v = 261.0 × 103/300 × 556

{6.3.2. Exp. 6.9}VRd,max = v1 b z fcd 0.5 sin 2θ

= 1.56 N/mm2 < 0.8√fcu = 5.06 N/mm2

v1 = 0.6 (1 − fck/250), z = 0.9d and fcd = fck/1.5

> vc = 0.79 × 1.51/3 × 1.17/1.25 = 0.84 N/mm2

θ = 0.5 sin−1 (238,700/(0.5 × 0.523 × 300 × 501

where 100 As/bvd = 1.5, (fcu/25)1/3 = 1.17

× (32/1.5)) = 8.2° < 22.5° ∴ use cot θ = 2.5

{Table 3.7} v > vc

{6.3.2. Exp. 6.8}

Asv/sv = 300 × (1.56 – 0.84)/(0.87 × 500)

Asw/s = 238,700/501 × 0.87 × 500 × 2.5

= 0.5 mm2/mm = 250 mm2/m/leg

= 0.438 mm2/mm = 219 mm2/m/leg

{3.4.5.5} s < 0.75 × 556 = 417 mm

{9.2.2(6)} s ≤ 0.75d = 417 mm

Use H8 links at 200 mm c/c (250)

Use H8 links at 225 mm c/c (222)

Nominal where v = vc + 0.4 = 1.24 N/mm2

{Exp. 9.5N} Asw,min/s = 0.08 × √32 × 300/500

or where shear force

= 0.272 mm2/m

Vnom = 1.24 × 300 × 556 × 10–3 = 207 kN

VRd,c,min = 238.7 × (0.272/0.438) = 148.0 kN

at 1,530 mm from centre of support.

at 1,120 mm from the centre of support

EC2 requires 13% less area of links, and nominal links to EC2 start at 1.36 times the distance for BS 8110.

Deflection

Short-term Young's modulus

{2.5.4} Es = 200 kN/mm2

{3.2.7(4)} Es = 200 kN/mm2

{Part 2, 7.2} Ec = 20 + 0.2 × 40 = 28 kN/mm2

{Table 3.1} Ecm = 22 (40/10)0.3 = 33.34 kN/mm2

α = 200/28 = 7.14

α = 200/33.34 = 6.00

Long-term Young's modulus

Bottom and sides exposed ho = 2Ac/u = 360,000/(300 + 2 × 600) = 240 mm

{Part 2, Fig. 7.1} φ = 2.45

{Fig. 3.1a} φ(∞,to) = 2.4

{Part 2, 3.6} Ec,long = 28/3.45 = 8.11 kN/mm2

{7.20} Ec,ef = 33.34/3.4 = 9.80 kN/mm2

∴αe = 200/8.11 = 24.64

∴ αe = 200/9.80 = 20.39

αe − 1 (uncracked concrete) = 23.64

αe − 1 = 19.39

Uncracked section properties

Uncracked transformed section

Not required in BS 8110

bh + Σ(αe – 1) As = 234,587 mm2

xu = 300 × 6002/2 + 19.39 × (2,590 × 566 + 402 × 36)/234,587 = 342.3 mm from top

Iu = 8,590 × 106 mm4

Zb = 8,590/257.5 = 33.36 × 106 mm4

{7.4.3} Mcr = 33.36 × 3.02 = 100.8 kNm

{Table 3.1} fctm = 0.3 × 322/3 = 3.02 N/mm2

Ms,QP = (44.5 + 0.3 × 30) × 6.02/8 = 240.7 kNm

> Mcr use partially cracked Ief

Cracked section properties

{Part 2, 3.6} Instantaneous value

Solving first m.o.a. b xc2/2 + (α − 1) As′(xcd′) = α As (dxc)

xc = 202.2 mm

Ix-x,c = 3211 × 106 mm4

Long-term value

Solving first m.o.a. b xc2/2 + (αe×1) As′(xcd′) = αe As (dxc)

xc = 302.0 mm

xc = 279.4 mm

Ix-x,c = 7545 × 106 mm4

Ic = 6,407 × 106 mm4

Curvature 1/rb = M/EI

Effective Ief

{Part 2, 3.3.3) ψ = 0.25 offices

{Exp. 7.18 and 7.19} better presented as

Instantaneous curvature (×10–6 mm−1 units)

Ief = Ic + [(IuIc) 0.5 (Mcr/Ms,QP)2]

wδ = 44.32 + 0.25 × 30 = 51.82 kN/m

Ief = 6407 + [(8,590 – 6,407) × 0.5 × (100.8/240.7)2] = 6599 × 106 mm4

Ms,total = 51.82 × 6.02/8 = 233.2 kNm

1/rb,total = 233.2/(28,000 × 3,211) = 2.59

Ms,Gk = 44.32 × 6.02/8 = 199.4 kNm (dead)

1/rb,Gk = 199.4/(28,000 × 3,211) = 2.22

Long-term curvature

Long-term curvature (×10–6 mm−1 units)

1/rb,Gk = 199.4/(8,116 × 7,545) = 3.26

1/rb = Ms,QP/Ec,ef Ief

1/rb = 3.26 + (2.59 − 2.22) = 3.63

1/rb = 240.7/9,806 × 6,599 = 3.72

{Part 2, 3.7.2} Deflection δ = K lo2 (1/rb)

δ = K leff2 (1/rb)

{Part 2, Table 3.1} K = 0.104

δ = 0.104 × 60002 × 3.63 × 10–6 = 13.6 mm

δ = 0.104 × 6,0002 × 3.72 × 10–6 = 13.9 mm

{3.4.6.3} lo /δ = 440 > 250 OK

{7.4.1(4)} leff /δ = 431 > 250 OK

Span/depth ratio

{3.4.6.1} lo/d = 20 × 0.744 = 14.87

{7.4.2(2)} l/d = 11 + 1.5√32 × 0.432 = 14.67

{Table 3.9} Basic lo/d = 20

where ρo/ρ = 0.432; ρ = 0

{3.4.6.5} MF = 0.55 + [(477 – 332)

ρo = 10–3 √32 = 0.00566

/(120 (0.9 + 5.34))] = 0.744

ρ = 2,230/300 ×566 = 0.0131 > ρo {Exp. 7.16b}

{Table 3.10} fs = (2/3) × 500 × (2,503/2,590)

(Table 7.4N} K = 1

= 332 N/mm2

M/bd2 = 495.5 × 106/300 × 5562 = 5.34

d > 6,000/14.87 = 404 mm < 556 OK

d > 6,000/14.67 = 409 mm < 556 OK

Deflections and span/depth ratios are the same in spite of different approaches in the two codes and that EC2’s Ecm = 1.19 Ec. Changing the creep coefficient φ from 2.4 to 1.5 results in only a 1.3 mm reduction in deflection, showing that φ is not a sensitive parameter.

Crack width

{Part 2, 3.8.3} Ec/2 = 14 kN/mm2

{7.3.4(1)} wk = sr,max (∊sm − ∊cm)

Spacing = 3 × 28 = 84 mm

{7.3.4(3)} sr,max = 3.4 × 28 + 0.8 × 0.5 × 0.425 × 32/0.075 = 168 mm

cw = 84 × 736 × 10–6 = 0.06 mm

wk = 168 × 913 × 10–6 = 0.15 mm

where ∊m = ∊1 − ∊′ = 804 – 68 = 736 × 10–6

where ∊sm − ∊cm = [206 – (0.4 × 3.02/0.075) ×

1 = Ms (dxc)/Ec/2 Ix-x,c = 334.4 ×

(1 + 6.0 × 0.075)]/200,000 = 913 × 10–6

(556 – 302)/(14,000 × 7545) = 804 x 10–6

but > 0.6 × 206/200,000 = 618 × 10–6

where Ms = 74.32 × 6.02/8 = 334.4 kNm

{7.3.4(2)}ρeff = 2412/300 × 107 = 0.075

ws = 44.32 + 30 = 74.32 kN/m

{7.3.2(3)}Minimum is Ac,eff = (hxc)/3

{Part 2, Eq. 13} ∊′ = 300 × 298 × 302/(3 ×

= (600 – 279.4)/3 = 107 mm

200,000 × 2,590 × 254) = 68 x 10–6

{7.1(2)} fct,eff = fctm = 0.3 × 322/3 = 3.02 N/mm2

σs = αe Ms,QP (dxc)/Ief = 20.39 × 240.7 × (556 − 279.4)/6599 = 206 N/mm2

{3.2.4} cw < cw max = 0.3 mm

{Table NA.4} wk < wmax = 0.3 mm for XC1

Crack width is greater in EC2 due to increased crack spacing, that is 84 and 168 mm in the two codes. The final effective strains are also greater in EC2, possibly because of the assumption in BS 8110 using Ec/2 = 14 kN/mm2 compared to the long-term value in EC2.

#### 3.6.2  Reinforced concrete rectangular column

A two-storey 300 mm × 300 mm edge column supports beams on one side only in an unbraced sway frame as shown in Figure 3.42a. The exposure, fire resistance, design life and beam end reactions are as given in 3.6.1, except that the roof beam may be taken as 60% of the floor beam reactions. The beam reactions act at 80 mm from the face of the column. The flexural stiffness of the beam-to-column connection may (in this exercise) be taken as1/10 of that of the column. The characteristic horizontal wind load is shown in Figure 3.42a.

Design the main reinforcement and specify the shear links. Use fck/fcu = 40/50, fyk/fy = 500 N/mm2, and normal-weight concrete. Effective creep factor φef = 1 (used in EC2). Moment distribution factors at the first floor (upper end of column is pinned, lower end at foundation is fixed) = 4EI/3.5 /(4EI/3.5 + 3EI/3.0) = 54% with 50% carryover (c/o) to the foundation.

Figure 3.42   Detail to code comparison for reinforced concrete column. (a) Frame arrangement, (b) bending moments due to beam eccentricity, horizontal loads (wind and imperfections) and second-order deflections.

BS 8110 solution

Eurocodes solution

Durability is the same as for the beam, ∴ cover to links 20 mm

Fire. 1 h

Fire. R60. BS EN 1992-1-2

{Table 3.4} c = 20 mm

{BS EN 1990, Table A1.1} In fire ψ2 = 0.3 for offices beam load and ψ2 = 0 for wind load

Gk = 44.5 × 6.0/2 = 133.5 kN per beam

Qk = 30 × 6.0/2 = 90 kN per beam

{2.4.2} VEd,fi,max = 133.5 + 0.3 × 90 = 160.5 kN

Per floor plus 60% at roof = 96.3 kN

Self-weight = 0.3 × 0.3 × 25 × 3 = 6.75 kN

NEd,fi at first floor level = 424 kN

Eccentricity of reaction = 150 + 80 = 230 mm

MEd,fi = 160.5 × 0.23 = 36.9 kNm × 54% as distributed = 19.9 kNm

{5.3.2} efi = 19.9/424 = 0.047 m = 47 mm

{5.3.2. Method B, Table 5.2b}as l0 > 3.0 m

e/h = 47/300 = 0.16 < 0.25

Then n = NEd,fi/(0.7(Ac fcd + As fyd))

Try 4 H25 bars = 1963 mm2

n = 424/(0.7 (3002 × 26.67 +1963 × 435) x 10–3

= 0.19 and ω = 1963 × 435/3002 × 26.67 = 0.36

{Table 5.2b} R60 requires bmin/a = 300/25

But 25 mm < c + φlink + φbar/2 = 20 + 8 + 12 = 40 mm ∴ not critical

Section properties.

Assume main bars φ = 25 mm

Assume main bars φ = 20 mm

{3.12.7.1} Links ≥ φ/4 use 8 mm

{9.5.3(1)} Links ≥ φ/4 use 8 mm

b = 300 mm; d = 300 − 20 − 8 − 12 = 260 mm

b = 300 mm; d = 300 − 40 = 260 mm

d/h = 0.87. Use design chart in Figure 3.43a

d/h = 0.87. Use design chart in Figure 3.43b

Self-weight = 0.3 × 0.3 × 24 × 6.5 = 14.0 kN

Self-weight = 0.3 × 0.3 × 25 × 6.5 = 14.6 kN

Effective height factors

{Part 2, clause 2.5} β = 2 + 0.3 αc,min where

{5.8.3.2, Exp. 5.16}

αc = 1.0 for foundation and 10 given for beam

k1 = 0.1 and k2 = 1/(1/10) = 10

β = 2 + 0.3 × 1.0 = 2.3

β = (1 + 0.1/1.1) × (1 + 10/11) = 2.08

Clear first floor l0 = 3500 – 600 = 2,900 mm

le = 2.3 × 2,900 = 6,670 mm. le/h = 22.2

l0 = 2.08 × 2,900 = 6032 mm. λ = l0/i = 69.6

{3.8.1.3} For unbraced le/h > 10 ∴ slender

where i = 300/√12 = 87 mm

{3.8.1.7} le/h < 60 limit

{5.8.3.1(1)} $λ lim = 20 ABC/ n = 33.4$

{3.8.3.1} au = 22.22 × 300 K/2,000 = 73.9K

where A = 1/(1 + 0.2φef) = 1/1.2 = 0.83

$B = 1 + 2 ω = 1.31$

where ω = 0.36

C = 1.7 − rm = 0.7 where rm = +1 for unbraced

n = 502 × 103*/3002 × 26.67 = 0.209

* NEd = 502 kN (see case 1 in the following)

λ >λlim 69.6 > 33.4 ∴ slender

{5.8.8.2(3)} e2 = 19.4 × 10–6 × 6,0322/10 = 71 mm

{5.8.8.3(1)} where 1/r = Kr Kφ fyd/0.45 d Es = 1.05 × 435/0.45 × 260 × 200,000= 19.4 × 10–6 mm−1

{5.8.8.3(3)} Kr Figure 3.43b = 1 (because NEd/fck bh < 0.25 see the following)

{5.8.8.3(4)} Kφ = 1 + (0.35 + fck/250 − λ/150) φef

= 1 + (0.35 + 0.16 − 69.6/150) × 1 = 1.05

Clear height to roof l0 = 6,500 – 400 = 6,100 m

le = 2.3 × 6,100 = 14,030 mm. le/h = 46.7 < 60

l0 = 2.08 × 6,100 = 12,688 mm. λ = l0/i = 146

{3.8.3.1} au = 46.72 × 300 K/2,000 = 327K

e2 = 18.6 × 10–6 × 12,6882/10 = 299 mm

{3.8.3.8}All columns have equal stiffness

where Kφ = 1 and 1/r = 18.6 × 10–6 mm−1

{3.8.5} Column deflection checked as le/h > 30

Horizontal loads and moments per column

Refer to Figure 3.42b

Mwind = (7.0 × 6.5 + 14.0 × 3.5)/3 = 31.5 kNm

Mwk = 31.5 kNm

{3.1.4.2} Total Gk for floor beams + self-weight

{5.2(5)} Imperfection at first floor

Gk,floor = 44.32 × 12 + 3 × 14 = 574 kN

αh = 2/√3.5 = 1.069 use 1, αm = 1

Hfloor = 1.5% Gk = 0.015 × 574 = 8.6 kN

{5.2(7)} ei = (1/200) × 1 × 6.032/2 = 0.015 m

Mi,floor = 302.1* × 0.015 = 4.6 kNm *see the following

At roof Hroof = 0.6 × 8.6 = 5.2 kN

At roof αh = 2 /√6.5 = 0.784, αm = 1

ei = (1/200) × 0.784 × 12.688/2 = 0.025 m

Mi,roof = 181.2* × 0.025 = 4.5 kNm *see the following

MEd,i = 4.6 + 4.5 = 9.1 kNm (added to γwMwind)

Ultimate loads and moments at foundation

Case 1. Gravity dead + live

Vfloor = 110.1 × 6.0/2 = 330.3 kN

FEd,floor = 100.7 × 6.0/2 = 302.1 kN

Vroof = 0.6 × 330.3 = 198.2 kN

FEd,roof = 0.6 × 302.1 = 181.2 kN

N = 330.3 + 198.2 + 1.4 × 14.0 = 548 kN

NEd = 302.1 + 181.3 + 1.25* × 14.6 = 502 kN

Refer to Figure 3.42b

Node moment M0 = Ve, where e = 150 + 80

G = 1.25 is for critical beam load Exp. 6.10b

= 230 mm from the centre of column

e = 230 mm

M0 = 330.3 × 0.23 = 76.0 kNm × 54% as

M0Ed = 302.1 × 0.23 × 54% × 50% c/o

distributed x 50% c/o = 20.5 kNm

= 18.8 kNm

{Eq. 35} Madd = N au = 330.3 × 0.0739K

{5.8.8.2(3)} M2 = NEd e2 = 302.1 × 0.071

+ 198.2 × 0.327K = 89.2K kNm

+ 181.3 × 0.299 = 75.7 kNm

M = 20.5 + 89.2 = 109.7 kNm

MEd = 18.8 + 75.7 + 9.1 = 103.6 kNm

{3.8.2.4} Mmin = 548 × 0.020 = 11.0 kNm

{6.1(4)} MEd,min = 502 × 0.020 = 10.0 kNm

N/fcu bh = 548 × 103/50 × 3002 = 0.12

NEd/fck bh = 502 × 103/40 × 3002 = 0.14

M/fcu bh2 = 109.7 × 106/50 × 3003 = 0.081

MEd/fck bh2 = 103.6 × 106/40 × 3003 = 0.096

Figure 3.43a gives K = 1

Figure 3.43b confirms K = 1

Asc = 0.12 × 50 × 300 × 300/500 = 1080 mm2

As = 0.15 × 40 × 300 × 300/500 = 1080 mm2

Case 2. Dead + live + wind

γf = all 1.2

Table 3.3. Exp. 6.10b is critical over 6.10a

γG = 1.25, γQ = 1.5, γW = 0.75

Vfloor = 1.2 × 74.32 × 6.0/2 = 267.5 kN

NEd as case 1. MEd as case 1 + wind MEd,w

Vroof = 0.6 × 267.5 = 160.5 kN

N = 267.5 + 160.5 + 1.2 × 14.0 = 445 kN

M0 = 267.5 × 0.23 × 54% × 50% = 16.6 kNm

Madd = 267.5 × 0.0739K + 160.5 × 0.327K = 72.2K kNm

Mwind = 1.2 × 31.5 = 37.8 kNm

MEd,w = 0.75 × 31.5 = 23.6 kNm

M = 16.6 + 72.2 + 37.8 = 126.6 kNm

MEd = 103.6 + 23.6 = 127.2 kNm

N/fcu bh = 0.10 and M/fcu bh2 = 0.094

NEd/fck bh = 0.14 and MEd/fck bh2 = 0.118

Asc = 0.19 × 50 × 300 × 300/500 = 1710 mm2

As = 0.21 × 40 × 300 × 300/500 = 1,512 mm2

γf = 1.0 and 1.4

γG = 1.25, γW = 1.5

Vfloor = 1.0 × 44.32 × 6.0/2 = 133.0 kN

FEd,floor = 44.5 × 6.0/2 = 133.5 kN

Vroof = 0.6 × 133.0 = 79.8 kN

FEd,roof = 0.6 × 133.5 = 80.1 kN

N = 133.0 + 79.8 + 14.0 = 226.8 kN

NEd = 133.5 + 80.1 + 14.6 = 228.2 kN

M0 = 133.0 × 0.23 × 54% × 50% = 8.3 kNm

M0Ed = 133.5 × 0.23 × 54% × 50% = 8.3 kNm

Madd = 133.0 × 0.0739K + 79.8 × 0.327K

M2 = 133.5 × 0.071 + 80.1 × 0.299 = 33.4 kNm

= 35.9K kNm

MEd,i = 133.5 × 0.015 + 80.1 × 0.025 = 4.0 kNm

Mwind = 1.4 × 31.5 = 44.1 kNm

MEd,w = 1.5 × 31.5 = 47.2 kNm

M = 8.3 + 35.9 + 44.1 = 88.3 kNm

MEd = 8.3 + 33.4 + 4.0 + 47.2 = 92.9 kNm

N/fcu bh = 0.05 and M/fcu bh2 = 0.065

NEd/fck bh = 0.063 and MEd/fck bh2 = 0.086

Asc = 0.13 × 50 × 300 × 300/500 = 1170 mm2

As = 0.19 × 40 × 300 × 300/500 = 1368 mm2

Maximum Asc = 1710 mm2 (Case 2)

Maximum As = 1,512 mm2 (Case 2)

Use 4 no. H25 bars (1963)

Use 4 no. H25 bars (1963)

{3.12.7.1} Links ≥ φ/4 use 8 mm

{9.5.3(1)} Links ≥ φ/4 use 8 mm

Spacing s = 12φ = 300 mm

{9.5.3(3)} s = min {20φ, b, h, 300} = 300 mm

{9.5.3(4)} s = 0.6s within h at top and bottom

{3.12.6.2} Asc,max = 0.08 × 3002 = 7,200 mm2

{9.5.2(3)} As,max = 0.04 × 3002 = 3,600 mm2

Asc,min = 0.004 × 3002 = 360 mm2

{9.5.2(2)} As,min = 0.002 × 3002 = 180 mm2 or 0.1 × 502 × 103/435 = 115 mm2

The Eurocodes require 12% less Asc, due mainly to reduced γf (1.25 & 1.5 versus 1.4 & 1.6) which reduces N by 9%. First- and second-order bending moments are roughly equal.

#### 3.6.3  Prestressed concrete slab

Calculate the service and ultimate moment of resistance, and the ultimate shear capacity of a 1200 mm wide × 200 mm deep prestressed concrete solid floor slab. The slab is pretensioned using 6 no. 12.5 mm plus 6 no. 9.3 mm diameter standard 7-wire helical strands at 30 mm bottom cover, plus 4 no. 5 mm diameter wires at 25 mm top cover. The tendons are Class 2 relaxation and stressed initially at 70%. The exposure is internal with relative humidity = 50%, fire resistance is 60 minutes, and design life is 50 years. Bearing length is 100 mm. The slab is designed as a Class 2 member for permissible tension to BS 8110. The floor carries office loading for characteristic dead loads of 1.5 kN/m2 finishes and 0.6 kN/m2 services/ceiling, a superimposed live load of 4.0 kN/m2 and demountable partitions of 1.0 kN/m2.

The section properties are cross-sectional area Ac = 232,040 mm2; second m.o.a Ix-x = 779 × 106 mm4; depth = 200 mm; centroid from bottom yb = 99.4 mm; breadth at top, bottom and at centroid = 1154, 1197 and 1134 mm, respectively. Height to the centroid of all tendons ys = 47.0 mm. Height to the centroid of tendons in tension zone yT = 35.7 mm.

Figure 3.43   (a) Reinforced concrete column design chart to BS 8110. (b) Reinforced concrete column design chart to BS EN 1992-1-1.

Use fck/fcu = 45/55, at transfer fck(t)/fci = 30/35, fpk/fpu = 1770 N/mm2, cement CEM I class 52.5R and a 10 mm coarse gravel aggregate. The self-weight of the precast concrete is determined by the manufacturer as 24.5 kN/m3 giving a self-weight of 4.90 kN/m2. In this exercise, ignore the reduced compression acting at the level of the strands due to self-weight and dead loads in the calculation of creep losses.

BS 8110 solution

Eurocodes solution

Durability. BS 8500-1. Table A.4 for 50 years

Durability. BS 8500-1. Table A.4 for 50 years

Exposure XC1. Cover c = 15 + Δc

Exposure XC1. Cnom = 15 + ΔCdev

{7.3} Δc = 5 mm

{4.4.1.3(3)}Cover controlled ΔCdev = 5 mm

Cover to tendons c ≥ 20 mm

Cover to tendons Cnom ≥ 20 mm

Fire. 1 h

Fire. R60. BS EN 1992-1-2

{Table 3.4} c = 20 mm

{Table 5.8} Depth h ≥ 80 mm < 200 mm

{5.2(5)}Axis a = 25 + 15 − Δa = 29 mm

{Exp. 5.3} Δa = 0.1 (500 – θcr) = 10.7 mm

{Fig. 5.1, curve 3}θcr = 390°C for

{Exp. 5.2} kpcr) = 805/1,770 = 0.455

σp,fi = 0.523 × 1,770/1.15 = 805 N/mm2

where Ed,fi/Ed = (7.0 + 0.3 × 5.0)/(1.25 × 7.0 + 1.5 × 5.0) = 0.523

{4.3}using ψfi = quasi-permanent ψ2 = 0.3

with Gk = 4.9 + 1.5 + 0.6 = 7.0 kN/m2 and Qk = 5.0 kN/m2

Centroid to steel tendons in fire zone, a = yT = 35.7 mm > 29 mm

Young's modulus

{Part 2, 7.2} Ec = 20 + 0.2 × 55 = 31 kN/mm2

{Table 3.1} Ecm = 22 (53/10)0.3 = 36.3 kN/mm2

{4.8.3.1} Eci = 20 + 0.2 × 35 = 27 kN/mm2

Ecm(t) = 22 (38/10)0.3 = 32.8 kN/mm2

{BS 5896, Table 6}Es = 195 kN/mm2 strand

{3.3.6(3)} Ep = 195,000 N/mm2 strand

Although Es for wire = 205 kN/mm2 use same as for strand

mi = 195/27 = 7.22. m = 195/31 = 6.29

m(t) = 195/32.8 = 5.94. m = 195/36.3 = 5.37

Section properties

Zb = 779 × 106/99.4 = 7.836 × 106 mm3

Zt = 779 × 106/101.6 = 7.744 × 106 mm3

e = 99.4 – 47.0 = 52.4 mm

Compound (subscript co) section properties using concrete + (m − 1) Aps

m − 1 = 5.29

m-1 = 4.37

yb,co = 98.3 mm; Ix-x,co = 799.6 × 106 mm4

yb,co = 98.5 mm; Ix-x,co = 796.0 × 106 mm4

Zb,co = 8.134 × 106 mm3; Zt,co = 7.862 × 106 mm3

Zb,co = 8.082 × 106 mm3; Zt,co = 7.842 × 106 mm3

Flexural capacity – service limit of stress

Aps = 6 × 52 + 6 × 93 + 4 × 19.6 = 948.5 mm2

Initial σi = 0.7 × 1770 = 1239 N/mm2

σpi = 0.7 × 1770 = 1,239 N/mm2

Pi = 1239 × 948.5 = 1,175,241 N

Ppi = 1239 × 948.5 = 1,175,241 N

{4.8.2.1} relaxation loss = 1.2 × 2% = 0.024

{3.3.2(7)} Relaxation at t = 20 h

{Exp. 3.29} for Class 2. µ = 0.7; ρ1000 = 2.5%

Loss Δσpr = 1239 × 0.66 × 2.5 × e(9.1 × 0.7)

(20/1000)0.75(1 – 0.7) = 4.95 N/mm2

fcc after relaxation loss = +8.99 N/mm2

σc after relaxation loss = +9.17 N/mm2

Shortening loss = 8.99 × 7.22/1,239 = 0.0524

{Exp. 3.29} Δσel = 9.17 × 5.94 = 54.46 N/mm2

σpm0 = 1,239.0 – 4.95 – 54.46 = 1,179.6 N/mm2

Transfer Rtr = 1 – 0.024 – 0.0524 = 0.924

Transfer Rtr = 1,179.6/1,239 = 0.952

Transfer Pt = 0.924 × 1,175,241 = 1,085,467 N

Ppm0 = 0.924 × 1,175,241 = 1,085,467 N

Prestress at transfer

{Table 3.1} fctm = 0.3 × 452/3 = 3.80 N/mm2

{4.3.5.2} fcti = 0.45√35 = 2.66 N/mm2

{Exp. 3.4} fctm(t) = (38/53) × 3.80 = 2.72 N/mm2

fbci = +11.94 N/mm2 < 0.5 × 35 = 17.5 OK

σb(t) = +12.31 N/mm2 < 0.6 × 30 = 18.0 OK

fbti = −2.67 N/mm2 > −2.66 say OK

σt(t) = −2.75 N/mm2 > −2.72 say OK

{4.8.5.2} Creep φ = 1.8

{Exp. B.2} φ(t,to) = 2.31 × 1.39 × 0.62 × 0.99

=1.98

where {Exp. B.4} β (fcm) = 2.31

{Exp. B.3b/B.8c} φRH = 1.39, with ho = 400 mm

{Exp. B.5}β (ti) = 0.62, where ti = 7.6 days at

50°C curing for 20 h

{Exp. B.7}βc (t,ti) = 0.99 for t = 20,833 days and {Exp. B.8b}βH (days) = 803

σc after initial loss = +8.77 N/mm2

{4.8.2.1} Loss = 1.8 × 0.924 × 0.0524 = 0.087

{Exp. 5.46} Δσp,c = 1.98 × 8.77 × 5.37/1.103 =

84.40 N/mm2

where denominator in code Exp. 5.46 = 1.103

{4.8.4} Shrinkage strain esh = 300 × 10–6

{Exp. 3.8} ∊cs =cd +ca = 404 +88= 492 × 10–6

where {Exp. 3.9} ∊cd = 0.985 × 0.73 × 566 × 10–6

= 404 × 10–6

βds (t,ts) = 0.985 with t as earlier, ts = 1 day

{Table 3.3} kn = 0.73 for ho = 400 mm

{Exp. B.11} ∊cd,o = 0.85 × (220 + 110 × 6) ×

e-(0.11 × 53/10) × 1.356 = 566 × 10–6

βRH = 1.55 × [1 − (50/100)3] = 1.356

{Exp. 3.12} ∊ca = 2.5 × (45 − 10) × 10–6 = 88 × 10–6

Loss = 300 × 10–6 × 195,000/1,239 = 0.0472

{Exp. 5.46} Δσp,sh = 492 × 10–6 × 195,000/1.103

= 87.0 N/mm2

{5.10.6(1b)} Long-term relaxation

{Exp. 3.29} µ = 1,179/1,770 = 0.666

{Exp. 3.29} σpr = 1,179 × 0.66 × 2.5 × e(9.1 × 0.666)

(500,000/1,000)0.75(1 − 0.666) = 39.65 N/mm2

{Exp. 5.46} Δσp,r = 0.8 × 39.65/1.103

= 28.7 N/mm2

σpo = 1,179.6 − 84.4 − 87.0 − 28.7 = 979.5 N/mm2

Rwk = 0.924 − 0.087 − 0.0472 = 0.789 (21.1%)

Rwk = 979.5/1,239 = 0.7905 (20.95%)

Pf = 0.789 × 1,175,241 = 927,621 N

Ppo = 0.791 × 1,175,241 = 929,076 N

Prestress in service

fbc = +10.20 N/mm2 < 0.33 × 55 = 18.15 OK

σb = +10.22 N/mm2 < 0.45 × 45 = 20.25 OK

fbt = −2.28 N/mm2 > 0.45 √55 = −3.34 OK

σt = −2.28 N/mm2 > −3.80 OK

Service moment of resistance Msr

Btm Msr = (10.20 + 3.34) × 8.134 = 110.1 kNm

MsR = (10.22 + 3.80) × 8.082 = 113.3 kNm

Top Msr = (2.28 + 18.15) × 7.862 = 160.6 kNm

MsR = (2.28 + 20.25) × 7.842 = 176.7 kNm*

{4.3.7.1} Ultimate moment of resistance Mur

{3.3.6(7)} Ultimate moment of resistance MRd

Refer to this book first ed. for full analysis

Refer to Section 4.3.4 for full analysis

fc = 0.45 fcu = 24.75 N/mm2 and λ = 0.9

fcd = 0.567 fck = 25.5 N/mm2 and λ = 0.8

Aps in tension zone = 870 mm2; d = 164.3 mm

Prestrain after losses ∊po = 0.005015

{5.10.9} ∊po = 0.004946

Refer to stress versus strain diagrams in the following

Strain ∊p = 0.012970; stress fp = 1539 N/mm2

p = 0.012700; stress fp = 1,442 N/mm2

{4.3.7.3} X = 50.1 mm; z = 141.8 mm

X = 51.2 mm; z = 143.8 mm

Mur = 870 × 1539 × 141.8 × 10–6 = 189.9 kNm

MRd = 870 × 1442 × 143.8 × 10–6 = 180.5 kNm

{4.3.8.1} Ultimate shear capacity Vco

{6.2.2(2)} Ultimate shear capacity VRd,c

{4.3.8.4} x = 100 + 99.4 = 199.4 mm

{6.2.2(2)} lx = 100 + 99.4 = 199.4 mm

Mean diameter of strands = 10.9 mm

{Exp. 8.16 and 8.18} lpt2 = 1.2 × 0.19 × 979.5 ×

lp = 10.9 × 240/√35 = 442 mm

10.9/4.06 = 721 mm

{Exp. 8.15} fpbt = 3.2 × 1.0 × 1.27 = 4.06 N/mm2

{8.10.2.2} fctd(t) = 0.7 × 2.72/1.5 = 1.27 N/mm2

x/lp = 199.4/442 = 0.451

αl = lx/lpt2 = 199.4/721 = 0.276

fcx = 927,621/232,040 = 4.00 N/mm2

σcp = 0.9 × 929,076/232,040 = 3.60 N/mm2

fcpx = 4.00 [0.451 × (2 − 0.451)] = 2.79 N/mm2

where {2.4.2.2(1)}γp,fav = 0.9

ft = 0.24√55 = 1.78 N/mm2

{3.1.6.2(P)} fctd =0.3 × 452/3 × 0.7/1.5 = 1.77 N/mm2

first m.o.a. Sx-x = 5.8326 × 106 mm3

{Eq. 54} Vco = 0.67 × 1134 × 200 ×

{Exp. 6.4} VRd,c = (779 × 1,134/5.8326) ×

√(1.782 + 0.8 × 2.79 × 1.78) = 406.1 kN

√(1.772 + 0.276 × 3.60 × 1.77) = 335.3 kN

1. MsR to the Eurocodes is greater than Class 2 BS 8110 because fctm is 13% greater than fct. However, there is no greater stress than fctm allowed in the Eurocodes, such as Class 3 (0.2) in BS 8110.
2. MRd < Mur because fp is not allowed to reach maximum design stress in the Eurocodes stress versus strain idealisation.
3. VRd,c < Vco because lpt2 is 63% greater than lp, and the build-up of prestress is linear in the Eurocodes rather than parabolic in BS 8110.

#### 3A Appendix A: Summary of Eurocode EC2: Design of Concrete Structures – General Rules and Rules for Buildings, BS EN 1992, Part 1-1

This code effectively replaces BS 8110, Parts 1 to 3, although the execution of work (tolerances, setting out, etc.) is found in BS EN 13670:2009, Execution of concrete structures. The division between commonplace and special design work separated in BS 8110 Parts 1 and 2 no longer exists, and there are no N-M interaction charts for column design. The last point reflects the fact that EC2 is a limit state code of principles rather than methods. The current amendment was published in February 2014. The UK Technical Committee B/525 (sub-committee 2) is currently engaged in a revision of the code.

Precast concrete is not treated as a separate design and construction method although, as with BS 8110, there are certain aspects of design, such as bearings, anchorage at supports, bursting, floor systems, compression/tension/shear joints, connections, pocket foundations, and corbels, collected in a separate section, in this case Section 10.

The format of BS EN 1992-1-1, as with all material based on the Eurocodes, is as follows:

 Section 1 Scope – references; assumptions; definitions; symbols. Note that symbols are often only defined here and not in the text Section 2 Basis of design – requirement related to BS EN 1990, Annex B; requirements related to BS EN 1991-1; material properties; PSFs γc and γs, load combinations and equilibrium Section 3 Materials – (concrete, rebar, tendons) strength, stress – strain models, deformation, shrinkage and creep; fatigue; anchorage; prestressing Section 4 Durability – environmental and exposure classes; cover to reinforcement Section 5 Structural analysis – load cases; imperfections, sway; structural models; linear elastic, plastic and non-linear analysis; redistribution; second-order effects with axial load (columns, walls); prestressing – stressing; forces; losses; service and ultimate; fatigue Section 6 ULS – bending, shear, torsion and punching shear; strut-and-tie models; anchorage and laps; partially loaded areas (localised bearings); fatigue Section 7 Serviceability limit state – crack control, spacing and crack width, deflections Section 8 Detailing in general – rebars – bar spacing, anchorage, laps, links details; prestressing tendons – anchorage, transmission length, development length Section 9 Detailing in particular – maximum and minimum areas; anchorage at supports; shear, torsion and surface reinforcement; solid and flat slabs, columns and walls, deep beams and stability ties Section 10 Precast concrete elements and structures – materials; losses of prestress; bearings; anchorage at supports; bursting; floor systems; compression/tension/shear joints; half joints; pocket foundations; corbels Section 12 Plain and lightly reinforced concrete – reduction factors for strength; precast walls and infill shear walls, construction joints, strip and pad footings Informative annexes – (A) improved PSFs; (B) creep and shrinkage strains in detail; (C) reinforcement properties; (D) prestressing tendons losses; (E) strength classes for durability; (F) tensile stresses in rebars in biaxial and shear stress fields; (G) soil-structure; (H) second-order effects; (I) flat slab and shear walls; (J) regions of discontinuity

The code is not prescriptive, and it is necessary to turn to calculation methodology given in documents published for example by The Concrete Centre, for example calculation of area of flexural and shear reinforcement in beams and N–M charts for r.c. columns.

The main issues relating to the design of precast concrete structures in NA to BS EN 1992-1-1 are

 2.4.2.2(1) Partial factor for prestress at ULS γP,fav = 0.9. 3.1.2(2)P Value of Cmax. Shear strength of concrete classes higher than C50/60 should be determined by tests, etc. 3.1.6(1)P Value of αcc = 0.85 for compression in flexure and axial loading and 1.0 for other phenomena, that is in bending fcd = 0.85 fck/1.5 = 0.567 fck but in shear fcd = 0.667 fck. 4.4.1.3(3) Δcdev under controlled conditions, such as steel mounts known as soldiers in front of hollow core slabs machines, may be reduced to 10 mm > Δcdev > 5 mm. 5.1.3(1)P Simplified load arrangements. Consider the two following arrangements for ‘all spans’ and alternate spans: (i) all spans carrying γGGk + γQQk + Pk; and (ii) alternate spans carrying γGGk + γQQk + Pk, other spans carrying only γGGk + Pk; the same value of γG should be used throughout the structure. For one-way spanning slabs, use the ‘all spans’ loaded if (i) area of each bay > 30 m2; (ii) Qk/Gk ≤ 1.25; and (iii) Qk < 5 kN/m2 excluding partitions. 5.5(4) Moment redistribution formula: values for steels with fyk ≤ 500 N/mm2, k1 = 0.4, k2 = 0.6 + 0.0014/∊cu2. Then if ∊cu2 = 0.0035, k2 = 1. Code Exp. 5.10a for fck ≤ 50 N/mm2, δ = k1 + k2 xu/d for zero moment redistribution δ = 1 = 0.4 + xu/d, then ≤ 0.6, that is for the balanced section, the limiting depth of the neutral axis is 0.6d. 5.10.9(1)P For pre-tensioning, rsup = 1.0 and rinf = 1.0, that is there are no modifications to the action of prestress. 6.2.3(3) Values of ν1 = ν unless the design stress of the shear reinforcement < 0.8 fyk, ν1 is modified. 7.2 The different limits of compressive stress in service depending on durability requirements and the avoidance of non-linear creep in prestressed sections in flexure. 7.3.1(5) Limitations of crack width wmax. Use Table National Annex NA4. This reduces wmax in r.c. sections to 0.3 mm and, in prestressed sections, the limiting permissible tension in service to zero for exposure class greater than XC1, although the value of the imposed live load may be reduced. 7.4.2(2) Values of basic span/depth ratios. Use Table NA5 which gives additional information and limits. 8.3(2) Minimum mandrel diameter Φm,min. Use in Table NA6a and Table NA6b, which contain additional information regarding scheduling reinforcement. 8.8(1) Additional rules for large diameter bars Φlarge > 40 mm. 9.5.2(1) Minimum diameter of longitudinal reinforcement in columns Φmin = 10 mm. 9.5.2(3) Maximum area of longitudinal reinforcement in columns. The designer should consider the practical upper limit taking into account the ability to place the concrete around the rebar, that is when casting columns horizontally mould, the maximum area is often around 8 to 10% of the area of concrete. This issue is considered further in the PD 6687-1:2010 (PD 6687-1 2010). 9.7(1) Minimum area of distribution reinforcement in deep beams = 0.2 % in each face. 9.10.2.2(2) Force to be resisted by peripheral tie. q1 = (20 + 4n0) where n0 is the number of storeys; q2 = 60 kN. 9.10.2.3(3) Minimum tensile force internal tie. Ftie,int = [(qk + gk)/7.5](lr/5)(Ft) ≥ Ft kN/m, with full definitions. Maximum spacing of internal ties = 1.5lr. 9.10.2.3(4) Internal ties on floors without screed. Ftie = [(gk + qk)/7.5](lr/5)Ft ≥ Ft kN/m. 9.10.2.4(2) Horizontal ties to external columns and/or walls at each floor level: Ftie,col = the greater of 2Ft ≤ ls/2.5Ft and 3% NEd at that level. 12.3.1(1) Values of αcc,pl = 0.6 and αct,pl = 0.6 (plain concrete). Annex J Design of corbels use PD 6687-1:2010.

#### Summary of Eurocode EC2: Design for fire BS EN 1992, Part 1-2

This code provides the following two alternatives for designing r.c. and prestressed concrete elements and structures for the actions of fire:

1. Performance-based design procedures in Section 2 to 4
2. Prescriptive rules, that is design aids such as tables and diagrams in Section 5.

The design procedure gives an analytical procedure taking into account the behaviour of the structural system at elevated temperatures, the potential heat exposure and the beneficial effects of active and passive fire protection systems, together with the consequences of of failure. The main text, together with informative annexes A to E, includes most of the principal concepts and rules necessary for structural fire design of concrete structures.

 Section 1 Scope – references; assumptions; definitions; symbols Section 2 Basis of design – requirements; actions; material properties; verification methods Section 3 Material properties – at elevated temperatures; concrete with siliceous and calcareous aggregates; thermal elongation of bars and tendons Section 4 Design procedures – simplified and advanced calculation methods; shear, torsion and anchorage; spalling; joints; protective layers Section 5 Tabulated data – columns, walls, beams and slabs; fire thickness and axis distance to bars Section 6 High-strength concrete – calculation models and tabulated data for columns, walls, beams and slabs Informative annexes (A) Temperature profiles; (B) Simplified calculation methods; (C) Buckling of columns under fire conditions; (D) Calculations for shear, torsion and anchorage; (E) Simplified calculations for beams and slabs

Section 1 Scope – references; assumptions; definitions; symbols.

The main issues relating to the design of precast concrete structures in NA to BS EN 1992-1-2 are

 3.2.3(5) Values for the parameters of the stress–strain relationship of reinforcing steel at elevated temperatures. Use Class N (Table 3.2a). 3.2.4(2) Ditto cold worked (wires and strands) prestressing steel at elevated temperatures. Use Class A. 5.6.1(1) Web thickness. Use dimensions for Class WA.

#### 2B Appendix B: Summary of Relevant Items in PD 6687-1:2010

This PD gives guidance on some specific items that were not published in the concrete Eurocodes or were in need of additional or noncontradictory additional information. Background research is cited in many cases. It is not to be regarded as a British standard. This PD gives noncontradictory complementary information for use with BS BS EN 1992 Parts 1-1 and 1-2 and their UK NAs.

The main items in the PD relating to the design of precast concrete structures are

#### References

Bhatt, P., MacGinley, T. J. and Choo, B. S. 2014. Reinforced Concrete Design to Eurocodes: Design Theory and Examples, 4th ed., CRC Press/Taylor & Francis, London, UK., 880pp.
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BS EN 1990. 2002. Eurocode 0: Basis of structural design, BSI, London, UK., March 2006.
NA to BS EN 1990. 2002. UK National Annex to Eurocode 0: Basis of structural design, BSI, London, UK., December 2004.
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NA to BS EN 1991-1-1. 2002. UK National Annex to Eurocode 1: Actions on Structures – Part 1-1: General Actions – Densities, self-weight, imposed loads for buildings, BSI, London, UK., December 2005.
BS EN 1992-1-1. 2004. Eurocode 2: Design of Concrete Structures – Part 1-1: General rules and rules for buildings, BSI, London, UK., February 2014.
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BS EN 1993-1-1. 2005. Eurocode 3. Design of steel structures. General rules and rules for buildings, BSI, London, UK., September 2006.
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Elliott, K. S., Davies, G., Ferreira, M., Mahdi, A. A. and Gorgun, H. 2003a. Can precast concrete structures be designed as semi-rigid frames? – Part 1, The experimental evidence, The Structural Engineer, London, 81(16), 14–27.
Elliott, K. S., Davies, G., Ferreira, M., Mahdi, A. A. and Gorgun, H. 2003b. Can precast concrete structures be designed as semi-rigid frames? – Part 2, Theoretical equations and applications for frame design, The Structural Engineer, London, 81(16), 28–37.
Elliott, K. S., Davies, G., Mahdi, A. A., Gorgun, H., Virdi. K. and Ragupathy, P. 1998. Precast concrete semi-rigid beam-to column connections in skeletal frames. COST C1 International Conference on Control of the Semi-rigid Behaviour of Civil Engineering Structural Connections, Liege, Belgium, pp. 45–54.
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Straman, J. P. 1990. Precast concrete cores and shear walls. Prefabrication of Concrete Structures, International Seminar, Delft, the Netherlands, 25–26 October 1990, pp. 41–54.

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